Gordon James, Martin Liebeck Representations and Characters of Groups, Second Edition 2001

Representations and Characters of Groups

Now in its second edition, this text provides a modern introduction to

the representation theory of ®nite groups. The authors have revised the

popular ®rst edition and added a considerable amount of new material.

The theory is developed in terms of modules, since this is appropriate

for more advanced work, but considerable emphasis is placed upon

constructing characters. The character tables of many groups are given,

including all groups of order less than 32, and all simple groups of

order less than 1000.

Among the applications covered are Burnside's paqb theorem, the

use of character theory in studying subgroup structure and permutation

groups, and a description of how to use representation theory to

investigate molecular vibration.

Each chapter is accompanied by a variety of exercises, and full

solutions to all the exercises are provided at the end of the book. This

will be ideal as a text for a course in representation theory, and in

view of the applications of the subject, will be of interest to mathema-

ticians, chemists and physicists alike.

REPRESENTATIONS AND

CHARACTERS OF GROUPS

GORDON JAMES and MARTIN LIEBECKDepartment of Mathematics,

Imperial College, London

Second Edition

PUBLISHED BY CAMBRIDGE UNIVERSITY PRESS (VIRTUAL PUBLISHING) FOR AND ON BEHALF OF THE PRESS SYNDICATE OF THE UNIVERSITY OF CAMBRIDGE The Pitt Building, Trumpington Street, Cambridge CB2 IRP 40 West 20th Street, New York, NY 10011-4211, USA 477 Williamstown Road, Port Melbourne, VIC 3207, Australia http://www.cambridge.org © Cambridge University Press 1993, 2001 This edition © Cambridge University Press (Virtual Publishing) 2003 First published in printed format 1993 Second edition 2001 A catalogue record for the original printed book is available from the British Library and from the Library of Congress Original ISBN 0 521 81205 4 hardback Original ISBN 0 521 00392 X paperback ISBN 0 511 01700 6 virtual (netLibrary Edition)

Contents

Preface page vii

1 Groups and homomorphisms 1

2 Vector spaces and linear transformations 14

3 Group representations 30

4 FG-modules 38

5 FG-submodules and reducibility 49

6 Group algebras 53

7 FG-homomorphisms 61

8 Maschke's Theorem 70

9 Schur's Lemma 78

10 Irreducible modules and the group algebra 89

11 More on the group algebra 95

12 Conjugacy classes 104

13 Characters 117

14 Inner products of characters 133

15 The number of irreducible characters 152

16 Character tables and orthogonality relations 159

17 Normal subgroups and lifted characters 168

18 Some elementary character tables 179

19 Tensor products 188

20 Restriction to a subgroup 210

21 Induced modules and characters 224

22 Algebraic integers 244

23 Real representations 263

24 Summary of properties of character tables 283

25 Characters of groups of order pq 288

26 Characters of some p-groups 298

27 Character table of the simple group of order 168 311

v

28 Character table of GL(2, q) 322

29 Permutations and characters 337

30 Applications to group theory 348

31 Burnside's Theorem 361

32 An application of representation theory to molecular vibration 367

Solutions to exercises 397

Bibliography 454

Index 455

vi Representations and characters of groups

Preface

We have attempted in this book to provide a leisurely introduction to

the representation theory of groups. But why should this subject

interest you?

Representation theory is concerned with the ways of writing a group

as a group of matrices. Not only is the theory beautiful in its own right,

but it also provides one of the keys to a proper understanding of ®nite

groups. For example, it is often vital to have a concrete description of a

particular group; this is achieved by ®nding a representation of the

group as a group of matrices. Moreover, by studying the different

representations of the group, it is possible to prove results which lie

outside the framework of representation theory. One simple example: all

groups of order p2 (where p is a prime number) are abelian; this can be

shown quickly using only group theory, but it is also a consequence of

basic results about representations. More generally, all groups of order

paqb ( p and q primes) are soluble; this again is a statement purely about

groups, but the best proof, due to Burnside, is an outstanding example

of the use of representation theory. In fact, the range of applications of

the theory extends far beyond the boundaries of pure mathematics, and

includes theoretical physics and chemistry ± we describe one such

application in the last chapter.

The book is suitable for students who have taken ®rst undergraduate

courses involving group theory and linear algebra. We have included two

preliminary chapters which cover the necessary background material.

The basic theory of representations is developed in Chapters 3±23, and

our methods concentrate upon the use of modules; although this accords

with the more modern style of algebra, in several instances our proofs

differ from those found in other textbooks. The main results are elegant

and surprising, but at ®rst sight they sometimes have an air of mystery

vii

about them; we have chosen the approach which we believe to be the

most transparent.

We also emphasize the practical aspects of the subject, and the text is

illustrated with a wealth of examples. A feature of the book is the wide

variety of groups which we investigate in detail. By the end of Chapter

28, we have presented the character tables of all groups of order less

than 32, of all p-groups of order at most p4, and of all the simple

groups of order less than 1000.

Every chapter is accompanied by a set of Exercises, and the solutions

to all of these are provided at the end of the book.

We would like to thank Dr Hans Liebeck for his careful reading of

our manuscript and the many helpful suggestions which he made.

Preface to Second EditionIn this second edition, we have included two new chapters; one

(Chapter 28) deals with the character tables of an in®nite series of groups,

and the other (Chapter 29) covers aspects of the representation theory of

permutation groups. We have also added a considerable amount of new

material to Chapters 20, 23 and 30, and made minor amendments else-

where.

viii Representations and characters of groups

1

1

Groups and homomorphisms

This book is devoted to the study of an aspect of group theory, so we

begin with a reÂsumeÂ of facts about groups, most of which you should

know already. In addition, we introduce several examples, such as

dihedral groups and symmetric groups, which we shall use extensively

to illustrate the later theory. An elementary course on abstract algebra

would normally cover all the material in the chapter, and any book on

basic group theory will supply you with further details. One or two

results which we shall use only infrequently are demoted to the

exercises at the end of the chapter ± you can refer to the solutions if

necessary.

Groups

A group consists of a set G, together with a rule for combining any

two elements g, h of G to form another element of G, written gh; this

rule must satisfy the following axioms:

(1) for all g, h, k in G,

(gh)k � g(hk);

(2) there exists an element e in G such that for all g in G,

eg � ge � g;

(3) for all g in G, there exists an element gÿ1 in G such that

ggÿ1 � gÿ1 g � e:

We refer to the rule for combining elements of G as the product

operation on G.

Axiom (1) states that the product operation is associative; the

element e in axiom (2) is an identity element of G; and gÿ1 is an

inverse of g in axiom (3).

It is elementary to see that G has just one identity element, and that

every g in G has just one inverse. Usually we write 1, rather than e,

for the identity element of G.

The product of an element g with itself, gg, is written g2; similarly

g3 � g2 g, gÿ2 � (gÿ1)2, and so on. Also, g0 � 1.

If the number of elements in G is ®nite, then we call G a ®nite

group; the number of elements in G is called the order of G, and is

written |G|.

1.1 Examples

(1) Let n be a positive integer, and denote by C the set of all complex

numbers. The set of nth roots of unity in C, with the usual multi-

plication of complex numbers, is a group of order n. It is written as

Cn and is called the cyclic group of order n. If a � e2ði=n, then

Cn � f1, a, a2, : : : , anÿ1g,and an � 1.

(2) The set Z of all integers, under addition, is a group.

(3) Let n be an integer with n > 3, and consider the rotation and

re¯ection symmetries of a regular n-sided polygon.

There are n rotation symmetries: these are r0, r1, . . . , rnÿ1 where rk

is the (clockwise) rotation about the centre O through an angle 2ðk=n.

There are also n re¯ection symmetries: these are re¯ections in the n

lines passing through O and a corner or the mid-point of a side of the

polygon.

These 2n rotations and re¯ections form a group under the product

operation of composition (that is, for two symmetries f and g, the

product fg means `®rst do f, then do g'). This group is called the

dihedral group of order 2n, and is written D2n.

Let A be a corner of the polygon. Write b for the re¯ection in the

2 Representations and characters of groups

line through O and A, and write a for the rotation r1. Then the n

rotations are

1, a, a2, : : : , anÿ1

(where 1 denotes the identity, which leaves the polygon ®xed); and the

n re¯ections are

b, ab, a2b, : : : , anÿ1b:

Thus all elements of D2n are products of powers of a and b ± that is,

D2n is generated by a and b.

Check that

an � 1, b2 � 1 and bÿ1ab � aÿ1:

These relations determine the product of any two elements of the

group. For example, we have ba j � aÿ jb (using the relation ba �aÿ1b), and hence

(aib)(a jb) � aiba jb � aiaÿ jbb � aiÿ j:

We summarize all this in the presentation

D2n � ha, b: an � 1, b2 � 1, bÿ1ab � aÿ1i:(4) For n a positive integer, the set of all permutations of

{1, 2, . . . , n}, under the product operation of composition, is a group.

It is called the symmetric group of degree n, and is written Sn. The

order of Sn is n!.

(5) Let F be either R (the set of real numbers) or C (the set of

complex numbers). The set of all invertible n 3 n matrices with entries

in F, under matrix multiplication, forms a group. This group is called

the general linear group of degree n over F, and is denoted by

GL(n, F). It is an in®nite group. The identity of GL(n, F) is of course

the identity matrix, which we denote by In or just I.

A group G is said to be abelian if gh � hg for all g and h in G.

While Cn and Z are abelian, most of the other examples given above

are non-abelian groups.

Subgroups

Let G be a group. A subset H of G is said to be a subgroup if H is

itself a group under the product operation inherited from G. We use

the notation H < G to indicate that H is a subgroup of G.

Groups and homomorphisms 3

It is easy to see that a subset H of a group G is a subgroup if and

only if the following two conditions hold:

(1) 1 2 H, and

(2) if h, k 2 H then hkÿ1 2 H.

1.2 Examples

(1) For every group G, both {1} and G are subgroups of G.

(2) Let G be a group and g 2 G. The subset

hgi � fgn: n 2 Zgis a subgroup of G, called the cyclic subgroup generated by g. If

gn � 1 for some n > 1, then kgl is ®nite. In this case, let r be the

least positive integer such that g r � 1; then r is equal to the number

of elements in kgl ± indeed,

hgi � f1, g, g 2, : : : , grÿ1g:We call r the order of the element g.

If G � kgl for some g 2 G then we call G a cyclic group. The

groups Cn and Z in Examples 1.1 are cyclic.

(3) Let G be a group and let a, b 2 G. De®ne H to be the subset of G

consisting of all elements which are products of powers of a and b ±

that is, all elements of the form

ai1 b j1 ai2 b j2 : : :ain b jn

for some n, where ik , jk 2 Z for 1 < k < n. Then H is a subgroup of

G; we call H the subgroup generated by a and b, and write

H � ha, bi:Given any ®nite set S of elements of G, we can similarly de®ne hSi,the subgroup of G generated by S.

This construction gives a powerful method of ®nding new groups as

subgroups of given groups, such as general linear or symmetric groups.

We illustrate the construction in the next example, and again in

Example 1.5 below.

(4) Let G � GL(2, C), the group of invertible 2 3 2 matrices with

entries in C, and let

A � i 0

0 ÿi

� �, B � 0 1

ÿ1 0

� �:


Put H � kA, Bl, the subgroup of G generated by A and B. Check that

A4 � I , A2 � B2, Bÿ1 AB � Aÿ1:

Using the third relation, we see that every element of H has the form

Ai Bj for some integers i, j; and using the ®rst two relations, we can

take 0 < i < 3 and 0 < j < 1. Hence H has at most eight elements.

Since the matrices

Ai Bj (0 < i < 3, 0 < j < 1)

are all distinct, in fact jH j � 8.

The group H is called the quaternion group of order 8, and is

written Q8. The above three relations determine the product of any two

elements of Q8, so we have the presentation

Q8 � hA, B: A4 � I , A2 � B2, Bÿ1 AB � Aÿ1i:(5) A transposition in the symmetric group Sn is a permutation which

interchanges two of the numbers 1, 2, . . . , n and ®xes the other n ÿ 2

numbers. Every permutation g in Sn can be expressed as a product of

transpositions. It can be shown that either all such expressions for g

have an even number of transpositions, or they all have an odd number

of transpositions; we call g an even or an odd permutation, accord-

ingly. The subset

An � fg 2 Sn: g is an even permutationgis a subgroup of Sn, called the alternating group of degree n.

Direct products

We describe a construction which produces a new group from given

ones.

Let G and H be groups, and consider

G 3 H � f(g, h): g 2 G and h 2 Hg:De®ne a product operation on G 3 H by

(g, h)(g9, h9) � (gg9, hh9)

for all g, g9 2 G and all h, h9 2 H. With this product operation, G 3 H

is a group, called the direct product of G and H.


More generally, if G1, . . . , Gr are groups, then the direct product

G1 3 : : : 3 Gr is

f(g1, : : : , gr): gi 2 Gi for 1 < i < rg,with product operation de®ned by

(g1, : : : , gr)(g91, : : : , g9r) � (g1 g91, : : : , grg9r):

If all the groups Gi are ®nite, then G1 3 . . . 3 Gr is also ®nite, of

order |G1| . . . |Gr |.

1.3 Example

The group C2 3 . . . 3 C2 (r factors) has order 2r and all its non-

identity elements have order 2.

Functions

A function from one set G to another set H is a rule which assigns a

unique element of H to each element of G. In this book, we generally

apply functions on the right ± that is, the image of g under a function

W is written as gW, not as Wg. We often indicate that W is a function

from G to H by the notation W: G! H. By an expression W: g! h,

where g 2 G and h 2 H, we mean that h � gW.

A function W: G! H is invertible if there is a function ö: H! G

such that for all g 2 G, h 2 H,

(gW)ö � g and (hö)W � h:

Then ö is called the inverse of W, and is written as Wÿ1. A function Wfrom G to H is invertible if and only if it is both injective (that is,

g1W � g2W for g1, g2 2 G implies that g1 � g2) and surjective (that is,

for every h 2 H there exists g 2 G such that gW � h). An invertible

function is also called a bijection.

Homomorphisms

Given groups G and H, those functions from G to H which `preserve

the group structure' ± the so-called homomorphisms ± are of particular

importance.

If G and H are groups, then a homomorphism from G to H is a

function W: G! H which satis®es

(g1 g2)W � (g1W)(g2W) for all g1, g2 2 G:


An invertible homomorphism is called an isomorphism. If there is an

isomorphism W from G to H, then G and H are said to be isomorphic,

and we write G � H; also, Wÿ1 is an isomorphism from H to G, so

H � G.

The following example displays a technique which can often be used

to prove that certain functions are homomorphisms.

1.4 Example

Let G � D2n � ka, b: an � b2 � 1, bÿ1ab � aÿ1l, and write the 2n

elements of G in the form aib j with 0 < i < n ÿ 1, 0 < j < 1. Let H

be any group, and suppose that H contains elements x and y which

satisfy

xn � y2 � 1, yÿ1xy � xÿ1:

We shall prove that the function W: G! H de®ned by

W: aib j ! xi y j (0 < i < nÿ 1, 0 < j < 1)

is a homomorphism.

Suppose that 0 < r < n ÿ 1, 0 < s < 1, 0 < t < n ÿ 1, 0 < u < 1.

Then

arbsatbu � aib j

for some i, j with 0 < i < n ÿ 1, 0 < j < 1. Moreover, i and j are

determined by repeatedly using the relations

an � b2 � 1, bÿ1ab � aÿ1:

Since we have xn � y2 � 1, yÿ1xy � xÿ1, we can also deduce that

xrysxtyu � xi y j:

Therefore,

(arbsatbu)W � (aib j)W � xi y j � xrysxtyu

� (arbs)W . (atbu)W,

and so W is a homomorphism.

We now demonstrate the technique of Example 1.4 in action.

1.5 Example

Let G � S5 and let x, y be the following permutations in G:

x � (1 2 3 4 5), y � (2 5)(3 4):


(Here we adopt the usual cycle notation ± thus, (1 2 3 4 5) denotes the

permutation 1! 2! 3! 4! 5! 1, and so on.) Check that

x5 � y2 � 1, yÿ1xy � xÿ1:

Let H be the subgroup kx, yl of G. Using the above relations, we see

that

H � fxi y j: 0 < i < 4, 0 < j < 1g,a group of order 10.

Now recall that

D10 � ha, b: a5 � b2 � 1, bÿ1ab � aÿ1i:By Example 1.4, the function W: D10 ! H de®ned by

W: aib j ! xi y j (0 < i < 4, 0 < j < 1)

is a homomorphism. Since W is invertible, it is an isomorphism. Thus,

H � kx, yl � D10.

Cosets

Let G be a group and let H be a subgroup of G. For x in G, the

subset

Hx � fhx: h 2 Hgof G is called a right coset of H in G. The distinct right cosets of H

in G form a partition of G (that is, every element of G is in precisely

one of the cosets).

Suppose now that G is ®nite, and let Hx1, . . . , Hxr be all the

distinct right cosets of H in G. For all i, the function

h! hxi (h 2 H)

is a bijection from H to Hxi, and so jHxij � jH j. Since

G � Hx1 [ : : : [ Hxr, and

Hxi \ Hxj is empty if i 6� j,

we deduce that

jGj � rjH j:In particular, we have


1.6 Lagrange's Theorem

If G is a ®nite group and H is a subgroup of G, then jH j divides |G|.

The number r of distinct right cosets of H in G is called the index

of H in G, and is written as jG: H j. Thus

jG: H j � jGj=jH jwhen G is ®nite.

Normal subgroups

A subgroup N of a group G is said to be a normal subgroup of G

if gÿ1 Ng � N for all g 2 G (where gÿ1 Ng � fgÿ1 ng: n 2 Ng); we

write N v G to indicate that N is a normal subgroup of G.

Suppose that N v G and let G=N be the set of right cosets of N in

G. The importance of the condition gÿ1 Ng � N (for all g 2 G) is that

it can be used to show that for all g, h 2 G, we have

fxy: x 2 Ng and y 2 Nhg � Ngh:

Hence we can de®ne a product operation on G=N by

(Ng)(Nh) � Ngh for all g, h 2 G:

This makes G=N into a group, called the factor group of G by N.

1.7 Examples

(1) For every group G, the sub-groups {1} and G are normal sub-

groups of G.

(2) For n > 1, we have An v Sn. If n > 2 then there are just two right

cosets of An in Sn, namely

An � fg 2 Sn: g eveng, and

An(1 2) � fg 2 Sn: g oddg:Thus |Sn:An| � 2, and so Sn=An � C2.

(3) Let G � D8 � ka, b: a4 � b2 � 1, bÿ1ab � aÿ1l and let N �ka2l � {1, a2}. Then N v G and

G=N � fN , Na, Nb, Nabg:Since (Na)2 � (Nb)2 � (Nab)2 � N, we see that G=N � C2 3 C2.

The subgroup kal is also normal in G, but the subgroup H � kbl is

not normal in G, since b 2 H while aÿ1ba � a2b =2 H.


Simple groups

A group G is said to be simple if G 6� {1} and the only normal

subgroups of G are {1} and G. For example, the cyclic group Cp, with

p a prime number, is simple. We shall give examples of non-abelian

simple groups in later chapters ± the smallest one is A5.

If G is a ®nite group which is not simple, then G has a normal

subgroup N such that both N and G/N have smaller order than G; and

in a sense, G is `built' out of these two smaller groups. Continuing

this process with the smaller groups, we eventually see that G is `built'

out of a collection of simple groups. (This is analogous to the fact that

every positive integer is built out of its prime factors.) Thus, simple

groups are fundamental to the study of ®nite groups.

Kernels and images

To conclude the chapter, we relate normal subgroups and factor groups

to homomorphisms. Let G and H be groups and suppose that

W: G! H is a homomorphism. We de®ne the kernel of W by

Ker W � fg 2 G: gW � 1g:(1:8)

Then Ker W is a normal subgroup of G. Also, the image of W is

Im W � fgW: g 2 Gg,(1:9)

and Im W is a subgroup of H.

The following result describes the way in which the kernel and

image of W are related.

1.10 Theorem

Suppose that G and H are groups and let W: G! H be a homomorph-

ism. Then

G=Ker W � Im W:

An isomorphism is given by the function

Kg ! gW (g 2 G)

where K � Ker W.


1.11 Example

The function W: Sn ! C2 given by

W: g ! 1, if g is an even permutation,

ÿ1, if g is an odd permutation,

�is a homomorphism. We have Ker W � An, and for n > 2, Im W � C2.

We know from Example 1.7(2) that Sn/An � C2, illustrating Theorem

1.10.

Summary of Chapter 1

1. Examples of groups are

Cn � ka: an � 1l,D2n � ka, b: an � b2 � 1, bÿ1ab � aÿ1l,Q8 � ka, b: a4 � 1, a2 � b2, bÿ1ab � aÿ1l,Sn � the symmetric group of degree n,

An � the alternating group of degree n,

GL(n, C) � the group of invertible n 3 n matrices over C,

G1 3 . . . 3 Gr, the direct product of the groups G1, . . . , Gr.

2. A normal subgroup N of G is a subgroup such that gÿ1 Ng � N for

all g in G. The factor group G=N consists of the right cosets Ng

(g 2 G), with multiplication

(Ng)(Nh) � Ngh:

3. A homomorphism W: G! H is a function such that

(g1 g2)W � (g1W)(g2W)

for all g1, g2 in G. The kernel, Ker W, is a normal subgroup of G,

and the image, Im W, is a subgroup of H. The factor group G=Ker Wis isomorphic to Im W.

Exercises for Chapter 1

1. Show that if G is an abelian group which is simple, then G is

cyclic of prime order.

2. Suppose that G and H are groups, with G simple, and that

W: G! H is a surjective homomorphism. Show that either W is an

isomorphism or H � {1}.


3. Suppose that G is a subgroup of Sn, and that G is not contained

in An. Prove that G \ An is a normal subgroup of G, and

G= (G \ An) � C2:

4. Let

G � D8 � ha, b: a4 � b2 � 1, bÿ1ab � aÿ1i, and

H � Q8 � hc, d: c4 � 1, c2 � d2, dÿ1cd � cÿ1i:

(a) Let x, y be the permutations in S4 which are given by

x � (1 2), y � (3 4),

and let K be the subgroup kx, yl of S4. Show that both the

functions ö: G! K and ø: H! K, de®ned by

ö: arbs ! xrys,

ø: crds ! xrys (0 < r < 3, 0 < s < 1),

are homomorphisms. Find Kerö and Kerø.

(b) Let X, Y be the 2 3 2 matrices which are given by

X � 0 i

i 0

� �, Y � 0 ÿ1

1 0

� �,

and let L be the subgroup hX , Y i of GL(2, C). Show that just

one of the functions ë: G! L and ì: H! L, de®ned by

ë: arbs ! X rY s,

ì: crds ! X rY s (0 < r < 3, 0 < s < 1),

is a homomorphism. Prove that this homomorphism is an

isomorphism.

5. Prove that D4m � D2m 3 C2 if m is odd.

6. (a) Show that every subgroup of a cyclic group is cyclic.

(b) Let G be a ®nite cyclic group, and let n be a positive integer

which divides |G|. Prove that

fg 2 G: gn � 1g

is a cylic subgroup of G of order n.

(c) If G is a ®nite cyclic group and x, y are elements of G with

the same order, show that x is a power of y.


7. Show that the set of non-zero complex numbers, under the usual

multiplication, is a group. Prove that every ®nite subgroup of this

group is cyclic.

8. Show that every group of even order contains an element of

order 2.

9. Find elements A and B of GL(2, C) such that A has order 8, B

has order 4, and

B2 � A4 and Bÿ1 AB � Aÿ1:

Show that the group kA, Bl has order 16.

(Hint: compare Q8 in Example 1.2(4).)

10. Suppose that H is a subgroup of G with |G:H| � 2. Prove that

H v G.


14

2

Vector spaces and linear transformations

An attractive feature of representation theory is that it combines two

strands of mainstream mathematics, namely group theory and linear

algebra. For reference purposes, we gather the results from linear

algebra concerning vector spaces, linear transformations and matrices

which we shall use later. Most of the material will be familiar to you

if you have taken a ®rst course on linear algebra, so we omit the

proofs. An exception occurs in the last section, where we deal with

projections; here, we explain in detail how the results work, in case

you have not come across projections before.

Vector spaces

Let F be either R (the set of real numbers) or C (the set of complex

numbers). A vector space over F is a set V, together with a rule for

adding any two elements u, v of V to form an element u � v of V, and

a rule for multiplying any element v of V by any element ë of F to

form an element ëv of V. (The latter rule is called scalar multi-

plication.) Moreover, these rules must satisfy:

(2.1) (a) V is an abelian group under addition;

(b) for all u, v in V and all ë, ì in F,

(1) ë(u � v) � ëu � ëv,

(2) (ë � ì)v � ëv � ìv,

(3) (ëì)v � ë(ìv),

(4) 1v � v.

The elements of V are called vectors, and those of F are called scalars.

We write 0 for the identity element of the abelian group V under

addition.

2.2 Examples

(1) Let R2 denote the set of all ordered pairs (x, y) where x and y are

real numbers. De®ne addition and scalar multiplication on R2 by

(x, y)� (x9, y9) � (x� x9, y� y9),

ë(x, y) � (ëx, ëy):

Then R2 is a vector space over R.

(2) More generally, for each positive integer n, we consider row

vectors

(x1, x2, : : : , xn)

where x1, x2, . . . , xn belong to F. We denote the set of all such row

vectors by F n, and de®ne addition and scalar multiplication on F n by

(x1, : : : , xn)� (x91, : : : , x9n) � (x1 � x91, : : : , xn � x9n),

ë(x1, : : : , xn) � (ëx1, : : : , ëxn):

Then F n is a vector space over F.

Bases of vector spaces

Let v1, . . . , vn be vectors in a vector space V over F. A vector v in V

is a linear combination of v1, . . . , vn if

v � ë1v1 � : : :� ënvn

for some ë1, . . . , ën in F. The vectors v1, . . . , vn are said to span V if

every vector in V is a linear combination of v1, . . . , vn.

We say that v1, . . . , vn are linearly dependent if

ë1v1 � : : :� ënvn � 0

for some ë1, . . . , ën in F, not all of which are zero; otherwise,

v1, . . . , vn are linearly independent.

The vectors v1, . . . , vn form a basis of V if they span V and are

linearly independent.

Throughout this book, we shall consider only vector spaces V which

are ®nite-dimensional ± this means that V has a basis consisting of

®nitely many vectors, as above. It turns out that any two bases of V

have the same number of vectors. The number of vectors in a basis of

V is called the dimension of V and is written as dim V. If V � {0} then

dim V � 0. The vector space V is n-dimensional if dim V � n.

Vector spaces and linear transformations 15

2.3 Example

Let V � F n. Then

(1, 0, 0, : : : , 0), (0, 1, 0, : : : , 0), : : : , (0, 0, 0, : : : , 1)

is a basis of V, so dim V � n. Another basis is

(1, 0, 0, : : : , 0), (1, 1, 0, : : : , 0), : : : , (1, 1, 1, : : : , 1):

Given a basis v1, . . . , vn of a vector space V, each vector v in V can

be written in a unique way as

v � ë1v1 � : : :� ënvn,

with ë1, . . . , ën in F. The vector v therefore determines the scalars

ë1, . . . , ën. Except in the case where V � {0}, there are many bases of

V. Indeed, the next result says that any linearly independent vectors can

be extended to a basis.

(2.4) If v1, . . . , vk are linearly independent vectors in V,

then there exist vk�1, . . . , vn in V such that

v1, . . . , vn form a basis of V.

Subspaces

A subspace of a vector space V over F is a subset of V which is itself

a vector space under the addition and scalar multiplication inherited

from V. For a subset U of V to be a subspace, it is necessary and

suf®cient that all the following conditions hold:

(2.5) (1) 0 2 U;

(2) if u, v 2 U then u � v 2 U;

(3) if ë 2 F and u 2 U then ëu 2 U.

2.6 Examples

(1) {0} and V are subspaces of V.

(2) Let u1, . . . , ur be vectors in V. We de®ne sp (u1, . . . , ur) to be the

set of all linear combinations of u1, . . . , ur; that is,

sp (u1, : : : , ur) � fë1u1 � : : :� ërur: ë1, : : : , ër 2 Fg:By (2.5), sp (u1, . . . , ur) is a subspace of V, and it is called the

subspace spanned by u1, . . . , ur.


Notice that the following fact is a consequence of (2.4).

(2.7) Suppose that U is a subspace of the vector space V.

Then dim U < dim V. Also, dim U � dim V if and

only if U � V.

Direct sums of subspaces

If U1, . . . , Ur are subspaces of a vector space V, then the sum

U1 � . . . � Ur is de®ned by

U1 � : : :� Ur � fu1 � : : :� ur: ui 2 Ui for 1 < i < rg:By (2.5), U1 � . . . � Ur is a subspace of V.

We say that the sum U1 � . . . � Ur is a direct sum if every element

of the sum can be written in a unique way as u1 � . . . � ur with

ui 2 Ui for 1 < i < r. If the sum is direct, then we write it as

U1 � : : :� Ur:

2.8 Examples

(1) Suppose that v1, . . . , vn is a basis of V, and for 1 < i < n, let Ui

be the subspace spanned by vi. Then

V � U1 � : : :� Un:

(2) Let U be a subspace of V and let v1, . . . , vk be a basis of U.

Extend v1, . . . , vk to a basis v1, . . . , vn of V (see (2.4)), and let W �sp (vk�1, . . . , vn). Then

V � U � W :

From this construction it follows that there are in®nitely many sub-

spaces W with V � U � W, unless U is {0} or V.

The next result is frequently useful when dealing with the direct sum

of two subspaces. You should consult the solutions to Exercises 2.3 and

2.4 if you have dif®culty with the proof.

(2.9) Suppose that V � U � W, that u1, . . . , ur is a basis of

U and that w1, . . . , ws is a basis of W. Then the

following three conditions are equivalent:

(1) V � U � W,

(2) u1, . . . , ur, w1, . . . , ws is a basis of V,

(3) U \ W � {0}.


Our next result, involving the direct sum of several subspaces, can

be deduced immediately from the de®nition of a direct sum.

(2.10) Suppose that U, W, U1, . . . , Ua, W1, . . . , Wb are

subspaces of a vector space V. If V � U � W and

also

U � U1 � : : :� Ua, and

W � W1 � : : :� Wb,

thenV � U1 � : : :� Ua � W1 � : : :� Wb:

We now introduce a construction for vector spaces which is analo-

gous to the construction of direct products for groups.

Let U1, . . . , Ur be vector spaces over F, and let

V � f(u1, : : : , ur): ui 2 Ui for 1 < i < rg:De®ne addition and scalar multiplication on V as follows: for all ui, u9iin Ui (1 < i < r) and all ë in F, let

(u1, : : : , ur)� (u91, : : : , u9r) � (u1 � u91, : : : , ur � u9r),

ë(u1, : : : , ur) � (ëu1, : : : , ëur):

With these de®nitions, V is a vector space over F. If, for 1 < i < r, we

putU 9i � f(0, : : : , ui, : : : , 0): ui 2 Uig

(where the ui is in the ith position), then it is immediate that

V � U 91 � : : :� U 9r:

We call V the external direct sum of U1, . . . , Ur, and, abusing notation

slightly, we write

V � U1 � : : :� Ur:

Linear transformations

Let V and W be vector spaces over F. A linear transformation from V

to W is a function W: V! W which satis®es

(u� v)W � uW� vW for all u, v 2 V , and

(ëv)W � ë(vW) for all ë 2 F and v 2 V :


Just as a group homomorphism preserves the group multiplication, so a

linear transformation preserves addition and scalar multiplication.

Notice that if W: V! W is a linear transformation and v1, . . . , vn is

a basis of V, then for ë1, . . . , ën in F we have

(ë1v1 � : : :� ënvn)W � ë1(v1W)� : : :� ën(vnW):

Thus, W is determined by its action on a basis. Furthermore, given any

basis v1, . . . , vn of V and any n vectors w1, . . . , wn in W, there is a

unique linear transformation ö: V! W such that viö� wi for all i; the

linear transformation ö is given by

(ë1v1 � : : :� ënvn)ö � ë1w1 � : : :� ënwn:

We sometimes construct a linear transformation ö: V! W in this way,

by specifying the values of ö on a basis of V, and then saying `extend

the action of ö to be linear'.

Kernels and images

Suppose that W: V! W is a linear transformation. The kernel of W(written Ker W) and the image of W (written Im W) are de®ned as

follows:

Ker W � fv 2 V : vW � 0g,(2:11)

Im W � fvW: v 2 Vg:Using (2.5), it is easy to check that Ker W is a subspace of V and Im Wis a subspace of W. Their dimensions are connected by the following

equation, which is known as the Rank±Nullity Theorem:

dim V � dim (Ker W)� dim (Im W):(2:12)

2.13 Examples

(1) If W: V! W is de®ned by vW � 0 for all v 2 V, then W is a linear

transformation, and

Ker W � V , Im W � f0g:(2) If W: V! V is de®ned by vW � 3v for all v 2 V, then W is a linear

transformation, and

Ker W � f0g, Im W � V :


(3) If W: R3 ! R2 is given by

(x, y, z)W � (x� 2y� z, ÿy� 3z)

for all x, y, z 2 R, then W is a linear transformation; we have

Ker W � sp ((7, ÿ3, ÿ1)), Im W � R2,

so dim (Ker W) � 1 and dim (Im W) � 2.

Invertible linear transformations

Again, let V and W be vector spaces over F. A linear transformation Wfrom V to W is injective if and only if Ker W � {0}, and hence W is

invertible precisely when W is surjective and Ker W � {0}. It turns out

that the inverse of an invertible linear transformation is also a linear

transformation (see Exercise 2.1).

If there exists an invertible linear transformation from V to W, then

V and W are said to be isomorphic vector spaces. By applying (2.12),

we see that isomorphic vector spaces have the same dimension. By

also taking (2.7) into account, we obtain the next result (see

Exercise 2.2).

(2.14) Let W be a linear transformation from V to itself. Then the

following three conditions are equivalent:

(1) W is invertible;

(2) Ker W � {0};

(3) Im W � V.

Endomorphisms

A linear transformation from a vector space V to itself is called an

endomorphism of V.

Suppose that W and ö are endomorphisms of V and ë 2 F. We de®ne

the functions W � ö, Wö and ëW from V to V by

v(W� ö) � vW� vö,(2:15)

v(Wö) � (vW)ö,

v(ëW) � ë(vW),

for all v 2 V. Then W � ö, Wö and ëW are endomorphisms of V. We

write W2 for WW.


2.16 Examples

(1) The identity function 1V de®ned by

1V : v! v for all v 2 V

is an endomorphism of V. If W is an endomorphism of V, then so is

W ÿ ë1V , for all ë 2 F. Note that

Ker (Wÿ ë1V ) � fv 2 V : vW � ëvg:(2) Let V � R2, and let W, ö be the functions from V to V de®ned by

(x, y)W � (x� y, xÿ 2y),

(x, y)ö � (xÿ 2y, ÿ2x� 4y):

Then W and ö are endomorphisms of V, and W � ö, Wö, 3W and W2 are

given by

(x, y)(W� ö) � (2xÿ y, ÿx� 2y),

(x, y)(Wö) � (ÿx� 5y, 2xÿ 10y),

(x, y)(3W) � (3x� 3y, 3xÿ 6y),

(x, y)W2 � (2xÿ y, ÿx� 5y):

Matrices

Let V be a vector space over F, and let W be an endomorphism of V.

Suppose that v1, . . . , vn is a basis of V and call it B . Then there are

scalars aij in F (1 < i < n, 1 < j < n) such that for all i,

viW � ai1v1 � : : :� ainvn:

2.17 De®nition

The n 3 n matrix (aij) is called the matrix of W relative to the basis

B , and is denoted by [W]B .

2.18 Examples

(1) If W � 1V (so that vW � v for all v 2 V), then [W]B � In for all

bases B of V, where In denotes the n 3 n identity matrix.

(2) Let V � R2 and let W be the endomorphism (x, y)! (x� y,

xÿ 2y) of V. If B is the basis (1, 0), (0, 1) of V and B 9 is the basis


(1, 0), (1, 1) of V, then

[W]B � 1 1

1 ÿ2

� �, [W]B 9 � 0 1

3 ÿ1

� �:

If we wish to indicate that the entries in a matrix A come from F,

then we describe A as a matrix over F.

Given two m 3 n matrices A � (aij) and B � (bij) over F, their sum

A � B is the m 3 n matrix over F whose ij-entry is aij � bij for all i,

j; and for ë 2 F, the matrix ëA is the m 3 n matrix over F obtained

from A by multiplying all the entries by ë.

As you know, the product of two matrices is de®ned in a less

transparent way. Given an m 3 n matrix A � (aij) and an n 3 p matrix

B � (bij), their product AB is the m 3 p matrix whose ij-entry isXn

k�1

aikbkj:

2.19 Example

Let

A � ÿ1 2

3 1

� �, B � 0 ÿ4

2 ÿ1

� �:

Then

A� B �ÿ1 ÿ2

5 0

!, 3A �

ÿ3 6

9 3

!, AB �

4 2

2 ÿ13

!,

BA �ÿ12 ÿ4

ÿ5 3

!:

The matrix of the sum or product of two endomorphisms (relative to

some basis) is related to the matrices of the individual endomorphisms

in the way you would expect:

(2.20) Suppose that B is a basis of the vector space V, and

that W and ö are endomorphisms of V. Then

[W� ö]B � [W]B � [ö]B , and

[Wö]B � [W]B [ö]B :


Also, for all scalars ë,

[ëW]B � ë[W]B :

We showed you in (2.17) how to get a matrix from an endomorph-

ism of a vector space V, given a basis of V. It is easy enough to

reverse this process and use a matrix to de®ne an endomorphism. We

concentrate on a particular way of doing this. Suppose that A is an

n 3 n matrix over F, and let V � Fn, the vector space of row vectors

(x1, . . . , xn) with each xi in F. Then for all v in V, the matrix product

vA also lies in V. The following remark is easily justi®ed.

(2.21) If A is an n 3 n matrix over F, then the function

v! vA (v 2 F n)

is an endomorphism of Fn.

2.22 Example

Let

A � 1 ÿ1

3 2

� �:

Then A gives us an endomorphism W of F2, where

(x, y)W � (x, y)1 ÿ1

3 2

� �� (x� 3y, ÿx� 2y):

Invertible matrices

An n 3 n matrix A is said to be invertible if there exists an n 3 n

matrix B with AB � BA � In. Such a matrix B, if it exists, is unique;

it is called the inverse of A and is written as Aÿ1. Write det A for the

determinant of A. Then a necessary and suf®cient condition for A to

be invertible is that det A 6� 0.

The connection between invertible endomorphisms and invertible

matrices is straightforward, and follows from (2.20): given a basis Bof V, an endomorphism W of V is invertible if and only if the matrix

[W]B is invertible.

Invertible matrices turn up when we relate two bases of a vector

space. An invertible matrix converts one basis into another, and this

same matrix is used to describe the way in which the matrix of an


endomorphism depends upon the basis. The precise meaning of these

remarks is revealed in the de®nition (2.23) and the result (2.24) below.

2.23 De®nition

Let v1, . . . , vn be a basis B of the vector space V, and let v91, . . . , v9nbe a basis B 9 of V. Then for 1 < i < n,

v9i � ti1v1 � : : :� tinvn

for certain scalars tij. The n 3 n matrix T � (tij) is invertible, and is

called the change of basis matrix from B to B 9.

The inverse of T is the change of basis matrix from B 9 to B .

(2.24) If B and B 9 are bases of V and W is an endomorphism of V,

then

[W]B � Tÿ1[W]B 9T ,

where T is the change of basis matrix from B to B 9.

2.25 Example

Suppose that V � R2. Let B be the basis (1, 0), (0, 1) and B 9 the

basis (1, 0), (1, 1) of V. Then

T � 1 0

1 1

� �, Tÿ1 � 1 0

ÿ1 1

� �:

If W is the endomorphism

W: (x, y)! (x� y, xÿ 2y)

of V, as in Example 2.18(2), then

[W]B � 1 1

1 ÿ2

� �� 1 0

ÿ1 1

� �0 1

3 ÿ1

� �1 0

1 1

� �� Tÿ1[W]B 9T :

Eigenvalues

Let V be an n-dimensional vector space over F, and suppose that W is

an endomorphism of V. The scalar ë is said to be an eigenvalue of Wif vW � ëv for some non-zero vector v in V. Such a vector v is called

an eigenvector of W.


Now ë is an eigenvalue of W if and only if Ker (W ÿ ë1V ) 6� {0},

which occurs if and only if W ÿ ë1V is not invertible. Therefore, if Bis a basis of V, then the eigenvalues of W are those scalars ë in F

which satisfy the equation

det ([W]B ÿ ëIn) � 0:

Solving this equation involves ®nding the roots of a polynomial of

degree n. Since every non-constant polynomial with coef®cients in C

has a root in C, we deduce the following result.

(2.26) Let V be a non-zero vector space over C, and let W be an

endomorphism of V. Then W has an eigenvalue.

2.27 Examples

(1) Let V � C2 and let W be the endomorphism of V which is given by

(x, y)W � (ÿy, x):

If B is the basis (1, 0), (0, 1) of V, then

[W]B � 0 1

ÿ1 0

� �:

We have det ([W]B ÿ ëI2) � ë2 � 1, so i and ÿi are the eigenvalues of

W. Corresponding eigenvectors are (1, ÿi) and (1, i). Note that if B 9 is

the basis (1, ÿi), (1, i) of V, then

[W]B 9 � i 0

0 ÿi

� �:

(2) Let V � R2 and let W again be the endomorphism which is given

by

(x, y)W � (ÿy, x):

This time, V is a vector space over R, and W has no eigenvalues in R.

Thus we depend upon F being C in result (2.26).

For an n 3 n matrix A over F, the element ë of F is said to be an

eigenvalue of A if vA � ëv for some non-zero row vector v in Fn. The

eigenvalues of A are those elements ë of F which satisfy

det (Aÿ ëIn) � 0:


2.28 Example

We say that an n 3 n matrix A � (aij) is diagonal if aij � 0 for all i

and j with i 6� j. We often display such a matrix in the form

A �ë1

. ..

ën

0B@1CA

which indicates, in addition, that aii � ëi for 1 < i < n. For this

diagonal matrix A, the eigenvalues are ë1, . . . , ën.

Projections

If a vector space V is a direct sum of two subspaces U and W, then

we can construct a special endomorphism of V which depends upon

the expression V � U � W:

2.29 Proposition

Suppose that V � U � W. De®ne ð: V! V by

(u� w)ð � u for all u 2 U , w 2 W :

Then ð is an endomorphism of V. Further,

Imð � U , Kerð � W and ð2 � ð:

Proof Since every vector in V has a unique expression in the form

u � w with u 2 U, w 2 W, it follows that ð is a function on V.

Let v and v9 belong to V. Then v � u � w and v9 � u9 � w9 for

some u, u9 in U and w, w9 in W. We have

(v� v9)ð � (u� u9� w� w9)ð � u� u9

� (u� w)ð� (u9� w9)ð

� vð� v9ð:

Also, for ë in F,

(ëv)ð � (ëu� ëw)ð � ëu � ë(vð):

Therefore, ð is an endomorphism of V.

Clearly Imð # U; and since uð � u for all u in U, we have Im � U.

Also,

0

0


(u� w)ð � 0, u � 0, u� w 2 W ,

and so Kerð � W.

Finally,

(u� w)ð2 � uð � u � (u� w)ð,

and so ð2 � ð. j

2.30 De®nition

An endomorphism ð of a vector space V which satis®es ð2 � ð is

called a projection of V.

2.31 Example

The endomorphism

(x, y)! (2x� 2y, ÿxÿ y)

of R2 is a projection.

We now show that every projection can be constructed using a direct

sum, as in Proposition 2.29.

2.32 Proposition

Suppose that ð is a projection of a vector space V. Then

V � Imð� Kerð:

Proof If v 2 V then v � vð � (v ÿ vð). Clearly the ®rst term vðbelongs to Imð; and the second term v ÿ vð lies in Kerð, since

(vÿ vð)ð � vðÿ vð2 � vðÿ vð � 0:

This establishes that V � Imð � Kerð.

Now suppose that v lies in Imð \ Kerð. As v 2 Imð, we have

v � uð for some u 2 V. Therefore

vð � uð2 � uð � v:

Since v 2 Kerð, it follows that v � vð � 0. Thus

Imð \ Kerð � f0g,and (2.9) now shows that V � Imð � Kerð. j


2.33 Example

If ð: (x, y)! (2x � 2y, ÿx ÿ y) is the projection of R2 which appears

in Example 2.31, then

Imð � f(2x, ÿx): x 2 Rg, Kerð � f(x, ÿx): x 2 Rg:


1. All our vector spaces are ®nite-dimensional over F, where F � C or

R. For example, F n is the set of row vectors (x1, . . . , xn) with each

xi in F, and dimF n � n.

2. V � U1 � . . . � Ur if each Ui is a subspace of V, and every element

v of V has a unique expression of the form v � u1 � . . . � ur

(ui 2 Ui).

Also, V � U � W if and only if V � U � W and U \ W � {0}.

3. A linear transformation W: V! W satis®es

(u� v)W � uW� vW and (ëv)W � ë(vW)

for all u, v in V and all ë in F. Ker W is a subspace of V and Im Wis a subspace of W, and

dim V � dim (Ker W)� dim (Im W):

4. A linear transformation W: V! W is invertible if and only if

Ker W � {0} and Im W � W.

5. Given a basis B of the n-dimensional vector space V, there is a

correspondence between the endomorphisms W of V and the n 3 n

matrices [W]B over F.

Given two bases B and B 9 of V, and an endomorphism W of V,

there exists an invertible matrix T such that

[W]B � Tÿ1[W]B 9T :

6. Eigenvalues ë of an endomorphism W satisfy vW � ëv for some non-

zero v in V.

7. A projection is an endomorphism ð of V which satis®es ð2 � ð.


1. Show that if V and W are vector spaces and W: V! W is an

invertible linear transformation then Wÿ1 is a linear transformation.


2. Suppose that W is an endomorphism of the vector space V. Show

that the following are equivalent:

(1) W is invertible;

(2) Ker W � {0};

(3) Im W � V.

3. Let U and W be subspaces of the vector space V. Prove that

V � U � W if and only if V � U � W and U \ W = {0}.

4. Let U and W be subspaces of the vector space V. Suppose that

u1, . . . , ur is a basis of U and w1, . . . , ws is a basis of W. Show

that V � U � W if and only if u1, . . . , ur, w1, . . . , ws is a basis

of V.

5. (a) Let U1, U2 and U3 be subspaces of a vector space V, with

V � U1 � U2 � U3. Show that

V � U1 � U2 � U3 ,U1 \ (U2 � U3) � U2 \ (U1 � U3) � U3 \ (U1 � U2) � f0g:

(b) Give an example of a vector space V with three subspaces U1,

U2 and U3 such that V � U1 � U2 � U3 and

U1 \ U2 � U1 \ U3 � U2 \ U3 � f0g,but V 6� U1 � U2 � U3.

6. Suppose that U1, . . . , Ur are subspaces of the vector space V, and

that V � U1 � . . . � Ur. Prove that

dim V � dim U1 � : : :� dim Ur:

7. Give an example of a vector space V with endomorphisms W and ösuch that V � Im W � Ker W, but V 6� Imö � Kerö.

8. Let V be a vector space and let W be an endomorphism of V. Show

that W is a projection if and only if there is a basis B of V such

that [W]B is diagonal, with all diagonal entries equal to 1 or 0.

9. Suppose that W is an endomorphism of the vector space V and

W2 � 1V . Show that V � U � W, where

U � fv 2 V : vW � vg, W � fv 2 V : vW � ÿvg:Deduce that V has a basis B such that [W]B is diagonal, with all

diagonal entries equal to �1 or ÿ1.


30

3

Group representations

A representation of a group G gives us a way of visualizing G as a

group of matrices. To be precise, a representation is a homomorphism

from G into a group of invertible matrices. We set out this idea in

more detail, and give some examples of representations. We also

introduce the concept of equivalence of representations, and consider

the kernel of a representation.

Representations

Let G be a group and let F be R or C. Recall from the ®rst chapter

that GL (n, F) denotes the group of invertible n 3 n matrices with

entries in F.

3.1 De®nition

A representation of G over F is a homomorphism r from G to

GL (n, F), for some n. The degree of r is the integer n.

Thus if r is a function from G to GL (n, F), then r is a representa-

tion if and only if

(gh)r � (gr)(hr) for all g, h 2 G:

Since a representation is a homomorphism, it follows that for every

representation r: G! GL (n, F), we have

1r � In, and

gÿ1r � (gr)ÿ1 for all g 2 G,

where In denotes the n 3 n identity matrix.

3.2 Examples

(1) Let G be the dihedral group D8 � ka, b: a4 � b2 � 1, bÿ1ab �aÿ1l. De®ne the matrices A and B by

A � 0 1

ÿ1 0

� �, B � 1 0

0 ÿ1

� �and check that

A4 � B2 � I , Bÿ1 AB � Aÿ1:

It follows (see Example 1.4) that the function r: G! GL (2, F) which

is given by

r: aib j ! Ai Bj (0 < i < 3, 0 < j < 1)

is a representation of D8 over F. The degree of r is 2.

The matrices gr for g in D8 are given in the following table:

g 1 a a2 a3

gr1 0

0 1

� �0 1

ÿ1 0

� � ÿ1 0

0 ÿ1

� �0 ÿ1

1 0

� �

g b ab a2b a3b

gr1 0

0 ÿ1

� �0 ÿ1

ÿ1 0

� � ÿ1 0

0 1

� �0 1

1 0

� �

(2) Let G be any group. De®ne r: G! GL (n, F) by

gr � In for all g 2 G,

where In is the n 3 n identity matrix, as usual. Then

(gh)r � In � InIn � (gr)(hr)

for all g, h 2 G, so r is a representation of G. This shows that every

group has representations of arbitrarily large degree.

Equivalent representations

We now look at a way of converting a given representation into another

one.

Group representations 31

Let r: G! GL (n, F) be a representation, and let T be an invertible

n 3 n matrix over F. Note that for all n 3 n matrices A and B, we

have

(Tÿ1 AT )(Tÿ1 BT ) � Tÿ1(AB)T :

We can use this observation to produce a new representation ó from r;

we simply de®ne

gó � Tÿ1(gr)T for all g 2 G:

Then for all g, h 2 G,

(gh)ó � Tÿ1((gh)r)T

� Tÿ1((gr)(hr))T

� Tÿ1(gr)T . Tÿ1(hr)T

� (gó )(hó ),

and so ó is, indeed, a representation.

3.3 De®nition

Let r: G! GL (m, F) and ó : G! GL (n, F) be representations of G

over F. We say that r is equivalent to ó if n � m and there exists an

invertible n 3 n matrix T such that for all g 2 G,

gó � Tÿ1(gr)T :

Note that for all representations r, ó and ô of G over F, we have

(see Exercise 3.4):

(1) r is equivalent to r;

(2) if r is equivalent to ó then ó is equivalent to r;

(3) if r is equivalent to ó and ó is equivalent to ô, then r is

equivalent to ô.

In other words, equivalence of representations is an equivalence rela-

tion.

3.4 Examples

(1) Let G � D8 � ka, b: a4 � b2 � 1, bÿ1ab � aÿ1l, and consider the

representation r of G which appears in Example 3.2(1). Thus ar � A


and br � B, where

A � 0 1

ÿ1 0

� �, B � 1 0

0 ÿ1

� �:

Assume that F � C, and de®ne

T � 1p2

1 1

i ÿi

� �:

Then

Tÿ1 � 1p2

1 ÿi

1 i

� �:

In fact, T has been constructed so that Tÿ1AT is diagonal; we have

Tÿ1 AT � i 0

0 ÿi

� �, Tÿ1 BT � 0 1

1 0

� �,

and so we obtain a representation ó of D8 for which

aó � i 0

0 ÿi

� �, bó � 0 1

1 0

� �:

The representations r and ó are equivalent.

(2) Let G � C2 � ka: a2 � 1l and let

A � ÿ5 12

ÿ2 5

� �:

Check that A2 � I. Hence r: 1! I, a! A is a representation of G. If

T � 2 ÿ3

1 ÿ1

� �,

then

Tÿ1 AT � 1 0

0 ÿ1

� �,

and so we obtain a representation ó of G for which

1ó � 1 0

0 1

� �, aó � 1 0

0 ÿ1

� �,

and ó is equivalent to r.


There are two easily recognized situations where the only represent-

ation which is equivalent to r is r itself; these are when the degree of

r is 1, and when gr � In for all g in G. However, there are usually

lots of representations which are equivalent to r.

Kernels of representations

We conclude the chapter with a discussion of the kernel of a

representation r: G! GL (n, F). In agreement with De®nition 1.8, this

consists of the group elements g in G for which gr is the identity

matrix. Thus

Ker r � fg 2 G: gr � Ing:Note that Ker r is a normal subgroup of G.

It can happen that the kernel of a representation is the whole of G,

as is shown by the following de®nition.

3.5 De®nition

The representation r: G! GL (1, F) which is de®ned by

gr � (1) for all g 2 G,

is called the trivial representation of G.

To put the de®nition another way, the trivial representation of G is

the representation where every group element is sent to the 1 3 1

identity matrix.

Of particular interest are those representations whose kernel is just

the identity subgroup.

3.6 De®nition

A representation r: G ! GL (n, F) is said to be faithful if

Ker r � {1}; that is, if the identity element of G is the only element g

for which gr � In.

3.7 Proposition

A representation r of a ®nite group G is faithful if and only if Im r is

isomorphic to G.


Proof We know that Ker r v G and by Theorem 1.10, the factor group

G/ Ker r is isomorphic to Im r. Therefore, if Ker r � {1} then

G � Im r. Conversely, if G � Im r, then these two groups have the

same (®nite) order, and so |Ker r| � 1; that is, r is faithful. j

3.8 Examples

(1) The representation r of D8 given by

(aib j)r � 0 1

ÿ1 0

� �i1 0

0 ÿ1

� � j

as in Example 3.2(1) is faithful, since the identity is the only element

g which satis®es gr � I. The group generated by the matrices

0 1

ÿ1 0

� �and

1 0

0 ÿ1

� �is therefore isomorphic to D8.

(2) Since Tÿ1AT � In if and only if A � In, it follows that all

representations which are equivalent to a faithful representation are

faithful.

(3) The trivial representation of a group G if faithful if and only if

G � {1}.

In Chapter 6 we shall show that every ®nite group has a faithful

representation.

The basic problem of representation theory is to discover and under-

stand representations of ®nite groups.


1. A representation of a group G is a homomorphism from G into

GL(n, F), for some n.

2. Representations r and ó of G are equivalent if and only if there

exists an invertible matrix T such that for all g 2 G,

gó � Tÿ1(gr)T :

3. A representation is faithful if it is injective.



1. Let G be the cyclic group of order m, say G � ka: am � 1l. Suppose

that A 2 GL (n, C), and de®ne r: G! GL (n, C) by

r: ar ! Ar (0 < r < mÿ 1):

Show that r is a representation of G over C if and only if Am � I.

2. Let

A � 1 0

0 1

� �, B � 1 0

0 e2ði=3

� �, C � 0 1

ÿ1 ÿ1

� �and let G � ka: a3 � 1l � C3. Show that each of the functions

r j: G! GL (2, C) (1 < j < 3), de®ned by

r1: ar ! Ar,

r2: ar ! Br,

r3: ar ! Cr (0 < r < 2),

is a representation of G over C. Which of these representations are

faithful?

3. Suppose that G � D2n � ka, b: an � b2 � 1, bÿ1ab � aÿ1l, and

F � R or C. Show that there is a representation r: G! GL (1, F)

such that ar � (1) and br � (ÿ1).

4. Suppose that r, ó and ô are representations of G over F. Prove:

(1) r is equivalent to r;

(2) if r is equivalent to ó, then ó is equivalent to r;

(3) if r is equivalent to ó, and ó is equivalent to ô, then r is

equivalent to ô.

5. Let G � D12 � ka, b: a6 � b2 � 1, bÿ1ab � aÿ1l. De®ne the matrices

A, B, C, D over C by

A � eið=3 0

0 eÿið=3

� �, B � 0 1

1 0

� �,

C � 1=2p

3=2

ÿp3=2 1=2

� �, D � 1 0

0 ÿ1

� �:

Prove that each of the functions rk : G! GL (2, C) (k � 1, 2, 3, 4),

given by


r1: arbs ! ArBs,

r2: arbs ! A3r(ÿB)s,

r3: arbs ! (ÿA)rBs,

r4: arbs ! C r Ds (0 < r < 5, 0 < s < 1),

is a representation of G. Which of these representations are faithful?

Which are equivalent?

6. Give an example of a faithful representation of D8 of degree 3.

7. Suppose that r is a representation of G of degree 1. Prove that

G=Ker r is abelian.

8. Let r be a representation of the group G. Suppose that g and h are

elements of G such that (gr)(hr) � (hr)(gr). Does it follow that

gh � hg?


38

4

FG-modules

We now introduce the concept of an FG-module, and show that there

is a close connection between FG-modules and representations of G

over F. Much of the material in the remainder of the book will be

presented in terms of FG-modules, as there are several advantages to

this approach to representation theory.

FG-modules

Let G be a group and let F be R or C.

Suppose that r: G! GL (n, F) is a representation of G. Write

V � F n, the vector space of all row vectors (ë1, . . . , ën) with ëi 2 F.

For all v 2 V and g 2 G, the matrix product

v(gr),

of the row vector v with the n 3 n matrix gr, is a row vector in V

(since the product of a 1 3 n matrix with an n 3 n matrix is again a

1 3 n matrix).

We now list some basic properties of the multiplication v(gr). First,

the fact that r is a homomorphism shows that

v((gh)r) � v(gr)(hr)

for all v 2 V and all g, h 2 G. Next, since 1r is the identity matrix, we

have

v(1r) � v

for all v 2 V. Finally, the properties of matrix multiplication give

(ëv)(gr) � ë(v(gr)),

(u� v)(gr) � u(gr)� v(gr)

for all u, v 2 V, ë 2 F and g 2 G.

4.1 Example

Let G � D8 � ka, b: a4 � b2 � 1, bÿ1ab � aÿ1l, and let r: G!GL (2, F) be the representation of G over F given in Example 3.2(1).

Thus

ar � 0 1

ÿ1 0

� �, br � 1 0

0 ÿ1

� �:

If v � (ë1, ë2) 2 F2 then, for example,

v(ar) � (ÿë2, ë1),

v(br) � (ë1, ÿë2),

v(a3r) � (ë2, ÿë1):

Motivated by the above observations on the product v(gr), we now

de®ne an FG-module.

4.2 De®nition

Let V be a vector space over F and let G be a group. Then V is an

FG-module if a multiplication vg (v 2 V, g 2 G) is de®ned, satisfying

the following conditions for all u, v 2 V, ë 2 F and g, h 2 G:

(1) vg 2 V;

(2) v(gh) � (vg)h;

(3) v1 � v;

(4) (ëv)g � ë(vg);

(5) (u � v)g � ug � vg.

We use the letters F and G in the name `FG-module' to indicate

that V is a vector space over F and that G is the group from which we

are taking the elements g to form the products vg (v 2 V).

Note that conditions (1), (4) and (5) in the de®nition ensure that for

all g 2 G, the function

v! vg (v 2 V )

is an endomorphism of V.

FG-modules 39

4.3 De®nition

Let V be an FG-module, and let B be a basis of V. For each g 2 G,

let

[g]B

denote the matrix of the endomorphism v! vg of V, relative to the

basis B .

The connection between FG-modules and representations of G over

F is revealed in the following basic result.

4.4 Theorem

(1) If r: G! GL(n, F) is a representation of G over F, and V � F n,

then V becomes an FG-module if we de®ne the multiplication vg by

vg � v(gr) (v 2 V , g 2 G):

Moreover, there is a basis B of V such that

gr � [g]B for all g 2 G:

(2) Assume that V is an FG-module and let B be a basis of V.

Then the function

g ! [g]B (g 2 G)

is a representation of G over F.

Proof (1) We have already observed that for all u, v 2 F n, ë 2 F and

g, h 2 G, we have

v(gr) 2 F n,

v((gh)r) � (v(gr))(hr),

v(1r) � v,

(ëv)(gr) � ë(v(gr)),

(u� v)(gr) � u(gr)� v(gr):

Therefore, F n becomes an FG-module if we de®ne

vg � v(gr) for all v 2 F n, g 2 G:

Moreover, if we let B be the basis

(1, 0, 0, : : : , 0), (0, 1, 0, : : : , 0), : : : , (0, 0, 0, : : : , 1)

of F n, then gr � [g]B for all g 2 G.


(2) Let V be an FG-module with basis B . Since v(gh) � (vg)h for

all g, h 2 G and all v in the basis B of V, it follows that

[gh]B � [g]B [h]B :

In particular,

[1]B � [g]B [gÿ1]B

for all g 2 G. Now v1 � v for all v 2 V, so [1]B is the identity matrix.

Therefore each matrix [g]B is invertible (with inverse [gÿ1]B ).

We have proved that the function g! [g]B is a homomorphism

from G to GL (n, F) (where n � dim V ), and hence is a representation

of G over F. j

Our next example illustrates part (1) of Theorem 4.4.

4.5 Examples

(1) Let G � D8 � ka, b: a4 � b2 � 1, bÿ1ab � aÿ1l and let r be the

representation of G over F given in Example 3.2(1), so

ar � 0 1

ÿ1 0

� �, br � 1 0

0 ÿ1

� �:

Write V � F2. By Theorem 4.4(1), V becomes an FG-module if we

de®ne

vg � v(gr) (v 2 V , g 2 G):

For instance,

(1, 0)a � (1, 0)0 1

ÿ1 0

� �� (0, 1):

If v1, v2 is the basis (1, 0), (0, 1) of V, then we have

v1a � v2, v1b � v1,

v2a � ÿv1, v2b � ÿv2:

If B denotes the basis v1, v2, then the representation

g ! [g]B (g 2 G)

is just the representation r (see Theorem 4.4(1) again).

(2) Let G � Q8 � ka, b: a4� 1, a2 � b2, bÿ1ab � aÿ1l. In Example

FG-modules 41

1.2(4) we de®ned Q8 to be the subgroup of GL (2, C) generated by

A � i 0

0 ÿi

� �and B � 0 1

ÿ1 0

� �,

so we already have a representation of G over C. To illustrate Theorem

4.4(1) we must this time take F � C. We then obtain a CG-module

with basis v1, v2 such that

v1a � iv1, v1b � v2,

v2a � ÿiv2, v2b � ÿv1:

Notice that in the above examples, the vectors v1a, v2a, v1b and v2b

determine vg for all v 2 V and g 2 G. For instance, in Example 4.5(1),

(v1 � 2v2)ab � v1ab� 2v2ab

� v2bÿ 2v1b

� ÿv2 ÿ 2v1:

A similar remark holds for all FG-modules V: if v1, . . . , vn is a basis

of V and g1, . . . , gr generate G, then the vectors vi g j (1 < i < n,

1 < j < r) determine vg for all v 2 V and g 2 G.

Shortly, we shall show you various ways of constructing FG-modules

directly, without using a representation. To do this, we turn a vector

space V over F into an FG-module by specifying the action of group

elements on a basis v1, . . . , vn of V and then extending the action to

be linear on the whole of V; that is, we ®rst de®ne vig for each i and

each g in G, and then de®ne

(ë1v1 � : : :� ënvn)g (ëi 2 F)

to be

ë1(v1 g)� : : :� ën(vng):

As you might expect, there are restrictions on how we may de®ne the

vectors vig. The next result will often be used to show that our chosen

multiplication turns V into an FG-module.

4.6 Proposition

Assume that v1, . . . , vn is a basis of a vector space V over F. Suppose

that we have a multiplication vg for all v in V and g in G which


satis®es the following conditions for all i with 1 < i < n, for all

g, h 2 G, and for all ë1, . . . , ën 2 F:

(1) vig 2 V;

(2) vi(gh) � (vi g)h;

(3) vi1 � vi;

(4) (ë1v1 � . . . � ënvn)g � ë1(v1 g) � . . . � ën(vn g).

Then V is an FG-module.

Proof It is clear from (3) and (4) that v1 � v for all v 2 V.

Conditions (1) and (4) ensure that for all g in G, the function

v! vg (v 2 V) is an endomorphism of V. That is,

vg 2 V ,

(ëv)g � ë(vg),

(u� v)g � ug � vg,

for all u, v 2 V, ë 2 F and g 2 G. Hence

(ë1u1 � : : :� ënun)h � ë1(u1 h)� : : :� ën(unh)(4:7)

for all ë1, . . . , ën 2 F, all u1, . . . , un 2 V and all h 2 G.

Now let v 2 V and g, h 2 G. Then v � ë1v1 � . . . � ënvn for some

ë1, . . . , ën 2 F, and

v(gh) � ë1(v1(gh))� : : :� ën(vn(gh)) by condition (4)

� ë1((v1 g)h)� : : :� ën((vng)h) by condition (2)

� (ë1(v1 g)� : : :� ën(vng))h by (4:7)

� (vg)h by condition (4):

We have now checked all the axioms which are required for V to be

an FG-module. j

Our next de®nitions translate the concepts of the trivial represent-

ation and a faithful representation into module terms.

4.8 De®nitions

(1) The trivial FG-module is the 1-dimensional vector space V over F

with

vg � v for all v 2 V , g 2 G:

FG-modules 43

(2) An FG-module V is faithful if the identity element of G is the

only element g for which

vg � v for all v 2 V :

For instance, the FD8-module which appears in Example 4.5(1) is

faithful.

Our next aim is to use Proposition 4.6 to construct faithful FG-

modules for all subgroups of symmetric groups.

Permutation modules

Let G be a subgroup of Sn, so that G is a group of permutations of

{1, . . . , n}. Let V be an n-dimensional vector space over F, with basis

v1, . . . , vn. For each i with 1 < i < n and each permutation g in G,

de®ne

vig � vig:

Then vig 2 V and vi1 � vi. Also, for g, h in G,

vi(gh) � vi( gh) � v(ig)h � (vig)h:

We now extend the action of each g linearly to the whole of V; that is,

for all ë1, . . . , ën in F and g in G, we de®ne

(ë1v1 � : : :� ënvn)g � ë1(v1 g)� : : :� ën(vng):

Then V is an FG-module, by Proposition 4.6.

4.9 Example

Let G � S4 and let B denote the basis v1, v2, v3, v4 of V. If g � (1 2),

then

v1 g � v2, v2 g � v1, v3 g � v3, v4 g � v4:

And if h � (1 3 4), then

v1 h � v3, v2 h � v2, v3 h � v4, v4 h � v1:

We have

[g]B �0 1 0 0

1 0 0 0

0 0 1 0

0 0 0 1

0BB@1CCA, [h]B �

0 0 1 0

0 1 0 0

0 0 0 1

1 0 0 0

0BB@1CCA:


4.10 De®nition

Let G be a subgroup of Sn. The FG-module V with basis v1, . . . , vn

such that

vig � vig for all i, and all g 2 G,

is called the permutation module for G over F. We call v1, . . . , vn the

natural basis of V.

Note that if we write B for the basis v1, . . . , vn of the permutation

module, then for all g in G, the matrix [g]B has precisely one non-

zero entry in each row and column, and this entry is 1. Such a matrix

is called a permutation matrix.

Since the only element of G which ®xes every vi is the identity, we see

that the permutation module is a faithful FG-module. If you are aware of

the fact that every group G of order n is isomorphic to a subgroup of Sn,

then you should be able to see that G has a faithful FG-module of

dimension n. We shall go into this in more detail in Chapter 6.

4.11 Example

Let G � C3 � ka: a3 � 1l. Then G is isomorphic to the cyclic subgroup

of S3 which is generated by the permutation (1 2 3). This alerts us to

the fact that if V is a 3-dimensional vector space over F, with basis v1,

v2, v3, then we may make V into an FG-module in which

v11 � v1, v21 � v2, v31 � v3,

v1a � v2, v2a � v3, v3a � v1,

v1a2 � v3, v2a2 � v1, v3a2 � v2:

Of course, we de®ne vg, for v an arbitrary vector in V and g � 1, a

or a2, by

(ë1v1 � ë2v2 � ë3v3)g � ë1(v1 g)� ë2(v2 g)� ë3(v3 g)

for all ë1, ë2, ë3 2 F. Proposition 4.6 can be used to verify that V is an

FG-module, but we have been motivated by the de®nition of permuta-

tion modules in our construction.

FG-modules and equivalent representations

We conclude the chapter with a discussion of the relationship between

FG-modules and equivalent representations of G over F. An FG-

FG-modules 45

module gives us many representations, all of the form

g ! [g]B (g 2 G)

for some basis B of V. The next result shows that all these representa-

tions are equivalent to each other (see De®nition 3.3); and moreover,

any two equivalent representations of G arise from some FG-module in

this way.

4.12 Theorem

Suppose that V is an FG-module with basis B , and let r be the

representation of G over F de®ned by

r: g ! [g]B (g 2 G):

(1) If B 9 is a basis of V, then the representation

ö: g ! [g]B 9 (g 2 G)

of G is equivalent to r.

(2) If ó is a representation of G which is equivalent to r, then there

is a basis B 0 of V such that

ó : g ! [g]B 0 (g 2 G):

Proof (1) Let T be the change of basis matrix from B to B 9 (see

De®nition 2.23). Then by (2.24), for all g 2 G, we have

[g]B � Tÿ1[g]B 9T :

Therefore ö is equivalent to r.

(2) Suppose that r and ó are equivalent representations of G. Then

for some invertible matrix T, we have

gr � Tÿ1(gó )T for all g 2 G:

Let B 0 be the basis of V such that the change of basis matrix from Bto B 0 is T. Then for all g 2 G,

[g]B � Tÿ1[g]B 0T ,

and so gó � [g]B 0. j

4.13 Example

Again let G � C3 � ka: a3 � 1l. There is a representation r of G which

is given by


1r � 1 0

0 1

� �, ar � 0 1

ÿ1 ÿ1

� �, a2r � ÿ1 ÿ1

1 0

� �:

(To see this, simply note that (ar)2 � a2r and (ar)3 � I; see Exercise

3.2.)

If V is a 2-dimensional vector space over C, with basis v1, v2 (which

we call B ), then we can turn V into a CG-module as in Theorem

4.4(1) by de®ning

v11 � v1, v1a � v2, v1a2 � ÿv1 ÿ v2,

v21 � v2, v2a � ÿv1 ÿ v2, v2a2 � v1:

We then have

[1]B � 1 0

0 1

� �, [a]B � 0 1

ÿ1 ÿ1

� �, [a2]B � ÿ1 ÿ1

1 0

� �:

Now let u1 � v1 and u2 � v1 � v2. Then u1, u2 is another basis of V,

which we call B 9. Since

u11 � u1, u1a � ÿu1 � u2, u1a2 � ÿu2,

u21 � u2, u2a � ÿu1, u2a2 � u1 ÿ u2,

we obtain the representation ö: g! [g]B 9 where

[1]B 9 � 1 0

0 1

� �, [a]B 9 � ÿ1 1

ÿ1 0

� �, [a2]B 9 � 0 ÿ1

1 ÿ1

� �:

Note that if

T � 1 0

1 1

� �then for all g in G, we have

[g]B � Tÿ1[g]B 9T ,

and so r and ö are equivalent, in agreement with Theorem 4.12(1).


1. An FG-module is a vector space over F, together with a multi-

plication by elements of G on the right. The multiplication satis®es

properties (1)±(5) of De®nition 4.2.

FG-modules 47

2. There is a correspondence between representations of G over F

and FG-modules, as follows.

(a) Suppose that r: G! GL (n, F) is a representation of G.

Then F n is an FG-module, if we de®ne

vg � v(gr) (v 2 F n, g 2 G):

(b) If V is an FG-module, with basis B , then r: g! [g]B is a

representation of G over F.

3. If G is a subgroup of Sn, then the permutation FG-module has

basis v1, . . . , vn, and vi g � vig for all i with 1 < i < n, and all g

in G.


1. Suppose that G � S3, and that V � sp (v1, v2, v3) is the permutation

module for G over C, as in De®nition 4.10. Let B 1 be the basis

v1, v2, v3 of V and let B 2 be the basis v1 � v2 � v3, v1 ÿ v2,

v1 ÿ v3. Calculate the 3 3 3 matrices [g]B 1and [g]B 2

for all g in

S3. What do you notice about the matices [g]B 2?

2. Let G � Sn and let V be a vector space over F. Show that V

becomes an FG-module if we de®ne, for all v in V,

vg � v, if g is an even permutation,

ÿv, if g is an odd permutation.

�3. Let Q8 � ka, b: a4 � 1, b2 � a2, bÿ1ab � aÿ1l, the quaternion

group of order 8. Show that there is an RQ8-module V of

dimension 4 with basis v1, v2, v3, v4 such that

v1a � v2, v2a � ÿv1, v3a � ÿv4, v4a � v3, and

v1b � v3, v2b � v4, v3b � ÿv1, v4b � ÿv2:

4. Let A be an n 3 n matrix and let B be a matrix obtained from A

by permuting the rows. Show that there is an n 3 n permutation

matrix P such that B � PA. Find a similar result for a matrix

obtained from A by permuting the columns.


49

5

FG-submodules and reducibility

We begin the study of FG-modules by introducing the basic building

blocks of the theory ± the irreducible FG-modules. First we require

the notion of an FG-submodule of an FG-module.

Throughout, G is a group and F is R or C.

FG-submodules

5.1 De®nition

Let V be an FG-module. A subset W of V is said to be an FG-

submodule of V if W is a subspace and wg 2 W for all w 2 W and all

g 2 G.

Thus an FG-submodule of V is a subspace which is also an FG-

module.

5.2 Examples

(1) For every FG-module V, the zero subspace {0}, and V itself, are

FG-submodules of V.

(2) Let G � C3 � ka: a3 � 1l, and let V be the 3-dimensional FG-

module de®ned in Example 4.11. Thus, V has basis v1, v2, v3, and

v11 � v1, v21 � v2, v31 � v3,

v1a � v2, v2a � v3, v3a � v1,

v1a2 � v3, v2a2 � v1, v3a2 � v2:

Put w � v1 � v2 � v3, and let W � sp (w), the 1-dimensional subspace

spanned by w. Since

w1 � wa � wa2 � w,

W is an FG-submodule of V. However, sp (v1 � v2) is not an FG-

submodule, since

(v1 � v2)a � v2 � v3 =2 sp (v1 � v2):

Irreducible FG-modules

5.3 De®nition

An FG-module V is said to be irreducible if it is non-zero and it has

no FG-submodules apart from {0} and V.

If V has an FG-submodule W with W not equal to {0} or V, then V

is reducible.

Similarly, a representation r: G! GL (n, F) is irreducible if the

corresponding FG-module F n given by

vg � v(gr) (v 2 F n, g 2 G)

(see Theorem 4.4(1)) is irreducible; and r is reducible if F n is

reducible.

Suppose that V is a reducible FG-module, so that there is an FG-

submodule W with 0 , dim W , dim V. Take a basis B 1 of W and

extend it to a basis B of V. Then for all g in G, the matrix [g]B has

the form

X g 0

Yg Zg

0@ 1A(5:4)

for some matrices Xg, Yg and Zg, where Xg is k 3 k (k � dim W).

A representation of degree n is reducible if and only if it is

equivalent to a representation of the form (5.4), where Xg is k 3 k and

0 , k , n.

Notice that in (5.4), the functions g! Xg and g! Zg are represen-

tations of G: to see this, let g, h 2 G and multiply the matrices [g]B

and [h]B given by (5.4). Notice also that if V is reducible then

dim V > 2.

5.5 Examples

(1) Let G � C3 � ka: a3 � 1l and let V be the 3-dimensional FG-

module with basis v1, v2, v3 such that


v1a � v2, v2a � v3, v3a � v1,

as in Example 4.11. We saw in Example 5.2(2) that V is a reducible

FG-module, and has an FG-submodule W � sp (v1 � v2 � v3). Let Bbe the basis v1 � v2 � v3, v1, v2 of V. Then

[1]B �1 0 0

0 1 0

0 0 1

0BB@1CCA, [a]B �

1 0 0

0 0 1

1 ÿ1 ÿ1

0BB@1CCA,

[a2]B �1 0 0

1 ÿ1 ÿ1

0 1 0

0BB@1CCA:

This reducible representation gives us two other representations: at the

`top left' we have the trivial representation and at the `bottom right'

we have the representation which is given by

1! 1 0

0 1

� �, a! 0 1

ÿ1 ÿ1

� �, a2 ! ÿ1 ÿ1

1 0

� �:

(2) Let G � D8 and let V � F2 be the 2-dimensional FG-module

described in Example 4.5(1). Thus G � ka, bl, and for all (ë, ì) 2 V we

have

(ë, ì)a � (ÿì, ë), (ë, ì)b � (ë, ÿì):

We claim that V is an irreducible FG-module. To see this, suppose

that there is an FG-submodule U which is not equal to V. Then

dim U < 1, so U � sp ((á, â)) for some á, â 2 F. As U is an FG-

module, (á, â)b is a scalar multiple of (á, â), and hence either á � 0

or â � 0. Since (á, â)a is also a scalar multiple of (á, â), this forces

á � â � 0, so U � {0}. Consequently V is irreducible, as claimed.


1. If V is an FG-module, and W is a subspace of V which is itself an

FG-module, then W is an FG-submodule of V.

2. The FG-module V is irreducible if it is non-zero and the only FG-

submodules are {0} and V.

FG-submodules and reducibility 51


1. Let G � C2 � ka: a2 � 1l, and let V � F2. For (á, â) 2 V, de®ne

(á, â)1 � (á, â) and (á, â)a � (â, á). Verify that V is an FG-module

and ®nd all the FG-submodules of V.

2. Let r and ó be equivalent representations of the group G over F.

Prove that if r is reducible then ó is reducible.

3. Which of the four representations of D12 de®ned in Exercise 3.5 are

irreducible?

4. De®ne the permutations a, b, c 2 S6 by

a � (1 2 3), b � (4 5 6), c � (2 3)(4 5),

and let G � ka, b, cl.(a) Check that

a3 � b3 � c2 � 1, ab � ba,

cÿ1ac � aÿ1 and cÿ1bc � bÿ1:

Deduce that G has order 18.

(b) Suppose that å and ç are complex cube roots of unity. Prove

that there is a representation r of G over C such that

ar � å 0

0 åÿ1

� �, br � ç 0

0 çÿ1

� �, cr � 0 1

1 0

� �:

(c) For which values of å, ç is r faithful?

(d) For which values of å, ç is r irreducible?

5. Let G � C13. Find a CG-module which is neither reducible nor

irreducible.


53

6

Group algebras

The group algebra of a ®nite group G is a vector space of dimension

|G| which also carries extra structure involving the product operation

on G. In a sense, group algebras are the source of all you need to

know about representation theory. In particular, the ultimate goal of

representation theory ± that of understanding all the representations of

®nite groups ± would be achieved if group algebras could be fully

analysed. Group algebras are therefore of great interest.

After de®ning the group algebra of G, we shall use it to construct

an important faithful representation, known as the regular representation

of G, which will be explored in greater detail later on.

The group algebra of G

Let G be a ®nite group whose elements are g1, . . . , gn, and let F be

R or C.

We de®ne a vector space over F with g1, . . . , gn as a basis, and

we call this vector space FG. Take as the elements of FG all

expressions of the form

ë1 g1 � : : :� ën gn (all ëi 2 F):

The rules for addition and scalar multiplication in FG are the natural

ones: namely, if

u �Xn

i�1

ëi gi and v �Xn

i�1

ìi gi

are elements of FG, and ë 2 F, then

u� v �Xn

i�1

(ëi � ìi)gi and ëu �Xn

i�1

(ëëi)gi:

With these rules, FG is a vector space over F of dimension n, with

basis g1, . . . , gn. The basis g1, . . . , gn is called the natural basis of

FG.

6.1 Example

Let G � C3 � ka: a3 � el. (To avoid confusion with the element 1 of

F, we write e for the identity element of G, in this example.) The

vector space CG contains

u � eÿ a� 2a2 and v � 12e� 5a:

We have

u� v � 32e� 4a� 2a2, 1

3u � 1

3eÿ 1

3a� 2

3a2:

Sometimes we write elements of FG in the formXg2G

ë g g (ë g 2 F):

Now, FG carries more structure than that of a vector space ± we

can use the product operation on G to de®ne multiplication in FG as

follows: Xg2G

ë g g

! Xh2G

ìh h

!�

Xg,h2G

ë gìh(gh)

�Xg2G

Xh2G

(ëhìhÿ1 g)g

where all ë g, ìh 2 F.

6.2 Example

If G � C3 and u, v are the elements of CG which appear in Example

6.1, then

uv � (eÿ a� 2a2)(12e� 5a)

� 12e� 5aÿ 1

2aÿ 5a2 � a2 � 10a3

� 212

e� 92aÿ 4a2:


6.3 De®nition

The vector space FG, with multiplication de®ned byXg2G

ë g g

! Xh2G

ìh h

!�X

g,h2G

ë gìh(gh)

(ë g, ìh 2 F), is called the group algebra of G over F.

The group algebra FG contains an identity for multiplication, namely

the element 1e (where 1 is the identity of F and e is the identity of

G). We write this element simply as 1.

6.4 Proposition

Multiplication in FG satis®es the following properties, for all

r, s, t 2 FG and ë 2 F:

(1) rs 2 FG;

(2) r(st) � (rs)t;

(3) r1 � 1r � r;

(4) (ër)s � ë(rs) � r(ës);

(5) (r � s)t � rt � st;

(6) r(s � t) � rs � rt;

(7) r0 � 0r � 0.

Proof (1) It follows immediately from the de®nition of rs that

rs 2 FG.

(2) Let

r �Xg2G

ë g g, s �Xg2G

ì g g, t �Xg2G

í g g,

(ë g, ì g, í g 2 F). Then

(rs)t �X

g,h,k2G

ë gìhík(gh)k

�X

g,h,k2G

ë gìhík g(hk)

� r(st):

We leave the proofs of the other equations as easy exercises. j

Group algebras 55

In fact, any vector space equipped with a multiplication satisfying

properties (1)±(7) of Proposition 6.4 is called an algebra. We shall be

concerned only with group algebras, but it is worth pointing out that

the axioms for an algebra mean that it is both a vector space and a

ring.

The regular FG-module

We now use the group algebra to de®ne an important FG-module.

Let V � FG, so that V is a vector space of dimension n over F,

where n � |G|. For all u, v 2 V, ë 2 F and g, h 2 G, we have

vg 2 V ,

v(gh) � (vg)h,

v1 � v,

(ëv)g � ë(vg),

(u� v)g � ug � vg,

by parts (1), (2), (3), (4) and (5) of Proposition 6.4, respectively.

Therefore V is an FG-module.

6.5 De®nition

Let G be a ®nite group and F be R or C. The vector space FG, with

the natural multiplication vg (v 2 FG, g 2 G), is called the regular

FG-module.

The representation g! [g]B obtained by taking B to be the natural

basis of FG is called the regular representation of G over F.

Note that the regular FG-module has dimension equal to |G|.

6.6 Proposition

The regular FG-module is faithful.

Proof Suppose that g 2 G and vg � v for all v 2 FG. Then 1g � 1, so

g � 1, and the result follows. j

6.7 Example

Let G � C3 � ka: a3 � el. The elements of FG have the form


ë1e� ë2a� ë3a2 (ëi 2 F):

We have

(ë1e� ë2a� ë3a2)e � ë1e� ë2a� ë3a2,

(ë1e� ë2a� ë3a2)a � ë3e� ë1a� ë2a2,

(ë1e� ë2a� ë3a2)a2 � ë2e� ë3a� ë1a2:

By taking matrices relative to the basis e, a, a2 of FG, we obtain the

regular representation of G:

e!1 0 0

0 1 0

0 0 1

0@ 1A, a!0 1 0

0 0 1

1 0 0

0@ 1A, a2 !0 0 1

1 0 0

0 1 0

0@ 1A:FG acts on an FG-module

You will remember that an FG-module is a vector space over F,

together with a multiplication vg for v 2 V and g 2 G (and the multi-

plication satis®es various axioms). Now, it is sometimes helpful to

extend the de®nition of the multiplication so that we have an element

vr of V for all elements r in the group algebra FG. This is done in

the following natural way.

6.8 De®nition

Suppose that V is an FG-module, and that v 2 V and r 2 FG; say

r �P g2G ì g g (ì g 2 F). De®ne vr by

vr �Xg2G

ì g(vg):

6.9 Examples

(1) Let V be the permutation module for S4, as described in Example

4.9. If

r � ë(1 2)� ì(1 3 4) (ë, ì 2 F)

then

v1 r � ëv1(1 2)� ìv1(1 3 4) � ëv2 � ìv3,

v2 r � ëv1 � ìv2,

(2v1 � v2)r � ëv1 � (2ë� ì)v2 � 2ìv3:

Group algebras 57

(2) If V is the regular FG-module, then for all v 2 V and r 2 FG, the

element vr is simply the product of v and r as elements of the group

algebra, given by De®nition 6.3.

Compare the next result with Proposition 6.4.

6.10 Proposition

Suppose that V is an FG-module. Then the following properties hold

for all u, v 2 V, all ë 2 F and all r, s 2 FG:

(1) vr 2 V;

(2) v(rs) � (vr)s;

(3) v1 � v;

(4) (ëv)r � ë(vr) � v(ër);

(5) (u � v)r � ur � vr;

(6) v(r � s) � vr � vs;

(7) v0 � 0r � 0.

Proof All parts except (2) are straightforward, and we leave them to

you. We shall give a proof of part (2), assuming the other parts.

Let v 2 V, and let r, s 2 FG with

r �Xg2G

ë g g, s �Xh2G

ìh h:

Then

v(rs) � vXg,h

ë gìh(gh)

!

�Xg,h

ë gìh(v(gh)) by (4) and (6)

�Xg,h

ë gìh((vg)h)

�X

g

ë g(vg)

! Xh

ìh h

!by (4), (5), (6)

� (vr)s: j



1. The group algebra FG of G over F consists of all linear combina-

tions of elements of G, and has a natural multiplication de®ned

on it.

2. The vector space FG, with the natural multiplication vg (v 2 FG,

g 2 G) is the regular FG-module.

3. The regular FG-module is faithful.


1. Suppose that G � D8 � ka, b: a4 � b2 � 1, bÿ1ab � aÿ1l.(a) Let x and y be the following elements of CG:

x � a� 2a2, y � b� abÿ a2:

Calculate xy, yx and x2.

(b) Let z � b � a2b. Show that zg � gz for all g in G. Deduce that

zr � rz for all r in CG.

2. Work out matrices for the regular representation of C2 3 C2 over F.

3. Let G � C2. For r and s in CG, does rs � 0 imply that r � 0 or

s � 0?

4. Assume that G is a ®nite group, say G � {g1, . . . , gn}, and write c

for the elementP

ni�1 gi of CG.

(a) Prove that ch � hc � c for all h in G.

(b) Deduce that c2 � |G|c.

(c) Let W: CG! CG be the linear transformation sending v to vc

for all v in CG. What is the matrix [W]B , where B is the basis

g1, . . . , gn of CG?

5. If V is an FG-module, prove from the de®nition that

0r � 0 for all r 2 FG, and

v0 � 0 for all v 2 V ,

where the symbol 0 is used for the zero elements of V and FG.

Show that for every ®nite group G, with |G| . 1, there exists an

FG-module V and elements v 2 V, r 2 FG such that vr � 0, but

neither v nor r is 0.

Group algebras 59

6. Suppose that G � D6 � ka, b: a3 � b2 � 1, bÿ1ab � aÿ1l, and let

ù � e2ði=3. Prove that the 2-dimensional subspace W of CG, de®ned

by

W � sp (1� ù2a� ùa2, b� ù2ab� ùa2b),

is an irreducible CG-submodule of the regular CG-module.


61

7

FG-homomorphisms

For groups and vector spaces, the `structure-preserving' functions are,

respectively, group homomorphisms and linear transformations. The

analogous functions for FG-modules are called FG-homomorphisms,

and we introduce these in this chapter.

FG-homomorphisms

7.1 De®nition

Let V and W be FG-modules. A function W: V! W is said to be an

FG-homomorphism if W is a linear transformation and

(vg)W � (vW)g for all v 2 V , g 2 G:

In other words, if W sends v to w then it sends vg to wg.

Note that if G is a ®nite group and W: V! W is an FG-homomorph-

ism, then for all v 2 V and r �P g2G ë g g 2 FG, we have

(vr)W � (vW)r

since

(vr)W �Xg2G

ë g(vg)W �Xg2G

ë g(vW)g � (vW)r:

The next result shows that FG-homomorphisms give rise to FG-

submodules in a natural way.

7.2 Proposition

Let V and W be FG-modules and let W: V! W be an FG-homomorph-

ism. Then Ker W is an FG-submodule of V, and Im W is an FG-

submodule of W.

Proof First note that Ker W is a subspace of V and Im W is a subspace

of W, since W is a linear transformation.

Let v 2 Ker W and g 2 G. Then

(vg)W � (vW)g � 0g � 0,

so vg 2 Ker W. Therefore Ker W is an FG-submodule of V.

Now let w 2 Im W, so that w � vW for some v 2 V. For all g 2 G,

wg � (vW)g � (vg)W 2 Im W,

and so Im W is an FG-submodule of W. j

7.3 Examples

(1) If W: V! W is de®ned by vW � 0 for all v 2 V, then W is an FG-

homomorphism, and Ker W � V, Im W � {0}.

(2) Let ë 2 F, and de®ne W: V! V by vW � ëv for all v 2 V. Then W is

an FG-homomorphism. Provided ë 6� 0, we have Ker W � {0}, Im W � V.

(3) Suppose that G is a subgroup of Sn. Let V � sp (v1, . . . , vn) be the

permutation module for G over F (see De®nition 4.10), and let

W � sp (w) be the trivial FG-module (see De®nition 4.8). We construct

an FG-homomorphism W from V to W. De®ne

W:Xn

i�1

ëivi !Xn

i�1

ëi

!w (ëi 2 F):

Thus viW � w for all i. Then W is a linear transformation, and for all

v �Pëivi 2 V and all g 2 G, we have

(vg)W �X

ëivig

!W �

Xëi

!w,

and

(vW)g �X

ëi

!wg �

Xëi

!w:

Therefore W is an FG-homomorphism. Here,

Ker W �Xn

i�1

ëivi:Xn

i�1

ëi � 0

( ),

Im W � W :


By Proposition 7.2, Ker W is an FG-submodule of the permutation

module V.

Isomorphic FG-modules

7.4 De®nition

Let V and W be FG-modules. We call a function W: V! W an FG-

isomorphism if W is an FG-homomorphism and W is invertible. If there

is such an FG-isomorphism, then we say that V and W are isomorphic

FG-modules and write V � W.

In the next result, we check that if V � W then W � V.

7.5 Proposition

If W: V! W is an FG-isomorphism, then the inverse Wÿ1: W! V is

also an FG-isomorphism.

Proof Certainly Wÿ1 is an invertible linear transformation, so we need

only show that Wÿ1 is an FG-homomorphism. For w 2 W and g 2 G,

((wWÿ1)g)W � ((wWÿ1)W)g as W is an FG-homomorphism

� wg

� ((wg)Wÿ1)W:

Hence (wWÿ1)g � (wg)Wÿ1, as required. j

Suppose that W: V! W is an FG-isomorphism. Then we may use Wand Wÿ1 to translate back and forth between the isomorphic FG-

modules V and W, and prove that V and W share the same structural

properties. We list some examples below:

(1) dim V � dim W (since v1, . . . , vn is a basis of V if and only if

v1W, . . . , vnW is a basis of W);

(2) V is irreducible if and only if W is irreducible (since X is an FG-

submodule of V if and only if XW is an FG-submodule of W);

(3) V contains a trivial FG-submodule if and only if W contains a

trivial FG-submodule (since X is a trivial FG-submodule of V if

and only if XW is a trivial FG-submodule of W).

FG-homomorphisms 63

Just as we often regard isomorphic groups as being identical, we

frequently disdain to distinguish between isomorphic FG-modules. For

the moment, though, we continue simply to emphasize the similarity

between isomorphic FG-modules. In the next result, we show that iso-

morphic FG-modules correspond to equivalent representations.

7.6 Theorem

Suppose that V is an FG-module with basis B , and W is an FG-

module with basis B 9. Then V and W are isomorphic if and only if the

representations

r: g ! [g]B and ó : g ! [g]B 9

are equivalent.

Proof We ®rst establish the following fact:

(7.7) The FG-modules V and W are isomorphic if and only

if there are a basis B 1 of V and a basis B 2 of W

such that

[g]B 1� [g]B 2

for all g 2 G:

To see this, suppose ®rst that W is an FG-isomorphism from V to W,

and let v1, . . . , vn be a basis B 1 of V; then v1W, . . . , vnW is a basis

B 2 of W. Let g 2 G. Since (vig)W � (viW)g for each i, it follows that

[g]B 1� [g]B 2

.

Conversely, suppose that v1, . . . , vn is a basis B 1 of V and w1,

. . . , wn is a basis B 2 of W such that [g]B 1� [g]B 2

for all g 2 G.

Let W be the invertible linear transformation from V to W for which

viW � wi for all i. Let g 2 G. Since [g]B 1� [g]B 2

, we deduce that

(vig)W � (viW)g for all i, and hence W is an FG-isomorphism. This

completes the proof of (7.7).

Now assume that V and W are isomorphic FG-modules. By (7.7),

there are a basis B 1 of V and a basis B 2 of W such that

[g]B 1� [g]B 2

for all g 2 G. De®ne a representation ö of G by

ö: g! [g]B 1. Then by Theorem 4.12(1), ö is equivalent to both r and

ó. Hence r and ó are equivalent.

Conversely, suppose that r and ó are equivalent. Then by Theorem

4.12(2), there is a basis B 0 of V such that gó � [g]B 0 for all g 2 G;


that is, [g]B 9 � [g]B 0 for all g 2 G. Therefore V and W are isomorphic

FG-modules, by (7.7). j

7.8 Example

Let G � ka: a3 � 1l, a cyclic group of order 3, and let W denote the

regular FG-module. Then 1, a, a2 is a basis of W; call it B 9. We have

[1]B 9 �1 0 0

0 1 0

0 0 1

0BB@1CCA, [a]B 9 �

0 1 0

0 0 1

1 0 0

0BB@1CCA,

[a2]B 9 �0 0 1

1 0 0

0 1 0

0BB@1CCA:

Compare the FG-module V de®ned in Example 4.11, with basis

v1, v2, v3 such that

v1a � v2, v2a � v3, v3a � v1:

Writing B for the basis v1, v2, v3 of V, we have

[g]B � [g]B 9 for all g 2 G:

According to (7.7), the FG-modules V and W are therefore isomorphic.

Indeed, the function

W: ë1v1 � ë2v2 � ë3v3 ! ë11� ë2a� ë3a2 (ëi 2 F)

is an FG-isomorphism from V to W.

7.9 Example

Let G � D8 � ka, b: a4 � b2 � 1, bÿ1ab � aÿ1l. In Example 3.4(1) we

encountered two equivalent representations r and ó of G, where

ar � 0 1

ÿ1 0

� �, br � 1 0

0 ÿ1

� �and

aó � i 0

0 ÿi

� �, bó � 0 1

1 0

� �:

FG-homomorphisms 65

Let V be the CG-module with basis v1, v2 for which

v1a � v2, v1b � v1,

v2a � ÿv1, v2b � ÿv2

(see Example 4.5(1)), and, in a similar way, let W be the CG-module

with basis w1, w2 for which

w1a � iw1, w1b � w2,

w2a � ÿiw2, w2b � w1

Thus, if we write B for the basis v1, v2 of V and B 9 for the basis w1,

w2 of W, then for all g 2 G we have

r: g ! [g]B and ó : g ! [g]B 9:

According to Theorem 7.6, the CG-modules V and W are isomorphic,

since r and ó are equivalent. To verify this directly, let W: V! W be

the invertible linear transformation such that

W: v1 ! w1 � w2,

v2 ! iw1 ÿ iw2:

Then (v ja)W � (v jW)a and (v jb)W � (v jW)b for j � 1, 2, and hence W is

a CG-isomorphism from V to W. (Compare Example 3.4(1).)

Direct sums

We conclude the chapter with a discussion of direct sums of FG-

modules, and we show that these give rise to FG-homomorphisms.

Let V be an FG-module, and suppose that

V � U � W ,

where U and W are FG-submodules of V. Let u1, . . . , um be a basis

B 1 of U, and w1, . . . , wn be a basis B 2 of W. Then by (2.9),

u1, . . . , um, w1, . . . , wn is a basis B of V, and for g 2 G,

[g]B �[g]B 1

0

0 [g]B 2

0@ 1A:More generally, if V � U1 � . . . � Ur, a direct sum of FG-submodules

Ui, and B i is a basis of Ui, then we can amalgamate B 1, . . . , B r to


obtain a basis B of V, and for g 2 G,

[g]B �[g]B 1

. ..

[g]B r

0B@1CA:(7:10)

The next result shows that direct sums give rise naturally to FG-

homomorphisms.

7.11 Proposition


V � U1 � : : :� Ur

where each Ui is an FG-submodule of V. For v 2 V, we have

v � u1 � . . . � ur for unique vectors ui 2 Ui, and we de®ne ði: V! V

(1 < i < r) by setting

vði � ui:

Then each ði is an FG-homomorphism, and is also a projection of V.

Proof Clearly ði is a linear transformation; and ði is an FG-homo-

morphism, since for v 2 V with v � u1 � . . . � ur (u j 2 Uj for all j),

and g 2 G, we have

(vg)ði � (u1 g � : : :� urg)ði � uig � (vði)g:

Also,

vð2i � uiði � ui � vði,

so ð2i � ði. Thus ði is a projection (see De®nition 2.30). j

We now present a technical result concerning sums of irreducible

FG-modules which will be used at a later stage.

7.12 Proposition


V � U1 � : : :� Ur,

where each Ui is an irreducible FG-submodule of V. Then V is a direct

sum of some of the FG-submodules Ui.

0

0

FG-homomorphisms 67

Proof The idea is to choose as many as we can of the FG-submodules

U1, . . . , Ur so that the sum of our chosen FG-submodules is direct.

To this end, choose a subset Y � {W1, . . . , Ws} of {U1, . . . , Ur} which

has the properties that

W1 � : : :� Ws is direct (i:e: equal to W1 � : : :� Ws), but

W1 � : : :� Ws � Ui is not direct, if Ui =2 Y :

Let

W � W1 � : : :� Ws:

We claim that Ui # W for all i. If Ui 2 Y this is clear, so assume that

Ui =2 Y. Then W � Ui is not a direct sum, so W \ Ui 6� {0}. But W \ Ui

is an FG-submodule of Ui, and Ui is irreducible; therefore

W \ Ui � Ui, and so Ui # W, as claimed.

Since Ui � W for all i with 1 < i < r, we have V � W �W1� . . . �Ws, as required. j

Finally, we remark that if V1, . . . , Vr are FG-modules, then we can

make the external direct sum V1� . . . �Vr (see Chapter 2) into an FG-

module by de®ning

(v1, : : : , vr)g � (v1 g, : : : , vrg)

for all vi 2 Vi (1 < i < r) and all g 2 G.


1. If V and W are FG-modules and W: V! W is a linear transformation

which satis®es

(vg)W � (vW)g

for all v 2 V, g 2 G, then W is an FG-homomorphism.

2. Kernels and images of FG-homomorphisms are FG-modules.

3. Isomorphic FG-modules correspond to equivalent representations.


1. Let U, V and W be FG-modules, and let W: U! V and ö: V! W

be FG-homomorphisms. Prove that Wö: U! W is an FG-homo-

morphism.


2. Let G be the subgroup of S5 which is generated by (1 2 3 4 5).

Prove that the permutation module for G over F is isomorphic to

the regular FG-module.

3. Assume that V is an FG-module. Prove that the subset

V0 � fv 2 V : vg � v for all g 2 Ggis an FG-submodule of V. Show that the function

W: v!Xg2G

vg (v 2 V )

is an FG-homomorphism from V to V0. Is it necessarily surjective?

4. Suppose that V and W are isomorphic FG-modules. De®ne the FG-

submodules V0 and W0 of V and W as in Exercise 3. Prove that V0

and W0 are isomorphic FG-modules.

5. Let G be the subgroup of S4 which is generated by (1 2) and (3 4).

Is the permutation module for G over F isomorphic to the regular

FG-module?

6. Let G � C2 � kx: x2 � 1l.(a) Show that the function

W: á1� âx! (áÿ â)(1ÿ x) (á, â 2 F)

is an FG-homomorphism from the regular FG-module to itself.

(b) Prove that W2 � 2W.

(c) Find a basis B of FG such that

[W]B � 2 0

0 0

� �:

FG-homomorphisms 69

70

8

Maschke's Theorem

We now come to our ®rst major result in representation theory, namely

Maschke's Theorem. A consequence of this theorem is that every FG-

module is a direct sum of irreducible FG-submodules, where as usual

F � R or C. (The assumption on F is important ± see Example 8.2(2)

below.) This essentially reduces representation theory to the study of

irreducible FG-modules.

Maschke's Theorem

8.1 Maschke's Theorem

Let G be a ®nite group, let F be R or C, and let V be an FG-module.

If U is an FG-submodule of V, then there is an FG-submodule W of V

such that

V � U � W :

Before proving Maschke's Theorem, we illustrate it with some

examples.

8.2 Examples

(1) Let G � S3 and let V � sp (v1, v2, v3) be the permutation module

for G over F (see De®nition 4.10). Put

u � v1 � v2 � v3 and U � sp (u):

Then U is an FG-submodule of V, since ug � u for all g 2 G.

There are many subspaces W of V such that V � U � W, for instance

sp (v2, v3) and sp (v1, v2 ÿ 2v3). But there is, in fact, only one FG-

submodule W of V with V � U � W. We shall ®nd this W in an

example after proving Maschke's Theorem (but you may like to look

for it yourself now).

(2) The conclusion of Maschke's Theorem can fail if F is not R or C.

For example, let p be a prime number, let G � C p � ka: a p � 1l, and

take F to be the ®eld of integers modulo p. Check that the function

a j ! 1 0

j 1

� �( j � 0, 1, : : : , pÿ 1)

is a representation from G to GL (2, F). The corresponding FG-module

is V � sp (v1, v2), where, for 0 < j < p ÿ 1,

v1a j � v1,

v2a j � jv1 � v2:

Clearly, U � sp (v1) is an FG-submodule of V. But there is no FG-

submodule W such that V � U � W, since U is the only 1-dimensional

FG-submodule of V, as can easily be seen.

Proof of Maschke's Theorem 8.1 We are given U, an FG-submodule of

the FG-module V. Choose any subspace W0 of V such that

V � U � W0:

(There are many choices for W0 ± simply take a basis v1, . . . , vm

of U, extend it to a basis v1, . . . , vn of V, and let W0 �sp (vm�1, . . . , vn).)

For v 2 V, we have v � u � w for unique vectors u 2 U and w 2 W0,

and we de®ne ö: V! V by setting vö � u. By Proposition 2.29, ö is

a projection of V with kernel W0 and image U.

We aim to modify the projection ö to create an FG-homomorphism

from V to V with image U. To this end, de®ne W: V! V by

vW � 1

jGjXg2G

vgögÿ1 (v 2 V ):(8:3)

It is clear that W is an endomorphism of V and Im W # U.

We show ®rst that W is an FG-homomorphism. For v 2 V and x 2 G,

(vx)W � 1

jGjXg2G

(vx)gögÿ1:

Maschke's Theorem 71

As g runs over the elements of G, so does h � xg. Hence

(vx)W � 1

jGjXh2G

vhöhÿ1x

� 1

jGjXh2G

vhöhÿ1

!x

� (vW)x:

Thus W is an FG-homomorphism.

Next, we prove that W2 � W. First note that for u 2 U, g 2 G, we

have ug 2 U, and so (ug)ö � ug. Using this,

uW � 1

jGjXg2G

ugögÿ1 � 1

jGjXg2G

(ug)gÿ1 � 1

jGjXg2G

u � u:(8:4)

Now let v 2 V. Then vW 2 U, so by (8.4) we have (vW)W � vW.

Consequently W2 � W, as claimed.

We have now established that W: V! V is a projection and an FG-

homomorphism. Moreover, (8.4) shows that Im W � U. Let W � Ker W.

Then W is an FG-submodule of V by Proposition 7.2, and V � U � W

by Proposition 2.32.

This completes the proof of Maschke's Theorem. j

8.5 Example

Let G � S3 and let V � sp (v1, v2, v3) be the permutation module, with

submodule U � sp (v1 � v2 � v3), as in Example 8.2(1). We use the

proof of Maschke's Theorem to ®nd an FG-submodule W of V such

that V � U � W.

First, let W0 � sp (v1, v2). Then V � U � W0 (but of course W0 is not

an FG-submodule). The projection ö onto U is given by

ö: v1 ! 0, v2 ! 0, v3 ! v1 � v2 � v3:

Check now that the FG-homomorphism W given by (8.3) is

W: vi ! 13(v1 � v2 � v3) (i � 1, 2, 3):

The required FG-submodule W is then Ker W, so

W � sp (v1 ÿ v2, v2 ÿ v3):

(In fact, W � Pëivi:

Pëi � 0

� , the FG-submodule constructed in

Example 7.3(3).)


Note that if B is the basis v1 � v2 � v3, v1, v2 of V, then for all

g 2 G, the matrix [g]B has the form

[g]B �j 0 0

j j j

j j j

0@ 1A:The zeros re¯ect the fact that U is an FG-submodule of V (see (5.4)).

If instead we use v1 � v2 � v3, v1 ÿ v2, v2 ÿ v3 as a basis B 9, then

we get

[g]B 9 �j 0 0

0 j j

0 j j

0@ 1A,

because sp (v1 ÿ v2, v2 ÿ v3) is also an FG-submodule of V.

This example illustrates the matrix version of Maschke's Theorem:

for an arbitrary ®nite group G, if we can choose a basis B of an FG-

module V such that [g]B has the form

� 0

� �

0@ 1Afor all g 2 G (see (5.4)), then we can ®nd a basis B 9 such that [g]B 9

has the form

� 0

0 �

0@ 1Afor all g 2 G.

To put this another way, suppose that r is a reducible representation

of a ®nite group G over F of degree n. Then we know that r is

equivalent to a representation of the form

g !Xg 0

Y g Z g

0@ 1A (g 2 G),

for some matrices Xg, Yg, Zg, where Xg is k 3 k with 0 , k , n.


Maschke's Theorem asserts further that r is equivalent to a represent-

ation of the form

g !Ag 0

0 Bg

0@ 1A,

where A g is also a k 3 k matrix.

Consequences of Maschke's Theorem

We now use Maschke's Theorem to show that every non-zero FG-

module is a direct sum of irreducible FG-submodules. (By an irreduci-

ble FG-submodule, we simply mean an FG-submodule which is an

irreducible FG-module.)

8.6 De®nition

An FG-module V is said to be completely reducible if V �U1� . . . �Ur, where each Ui is an irreducible FG-submodule of V.

8.7 Theorem

If G is a ®nite group and F � R or C, then every non-zero FG-module

is completely reducible.

Proof Let V be a non-zero FG-module. The proof goes by induction

on dim V. The result is true if dim V � 1, since V is irreducible in this

case.

If V is irreducible then the result holds, so suppose that V is

reducible. Then V has an FG-submodule U not equal to {0} or V.

By Maschke's Theorem, there is an FG-submodule W such that

V � U � W. Since dim U , dim V and dim W , dim V, we have, by

induction,

U � U1 � : : :� Ur, W � W1 � : : :� Ws,

where each Ui and Wj is an irreducible FG-module. Then by (2.10),

V � U1 � : : :� Ur � W1 � : : :� Ws,

a direct sum of irreducible FG-modules. j

Another useful consequence of Maschke's Theorem is the next

proposition.


8.8 Proposition

Let V be an FG-module, where F � R or C and G is a ®nite group.

Suppose that U is an FG-submodule of V. Then there exists a surjective

FG-homomorphism from V onto U.

Proof By Maschke's Theorem, there is an FG-submodule W of V such

that V � U � W. Then the function ð: V! U which is de®ned by

ð: u� w! u (u 2 U , w 2 W )

is an FG-homomorphism onto U, by Proposition 7.11. j

Theorem 8.7 tells us that every non-zero FG-module is a direct sum

of irreducible FG-modules. Thus, in order to understand FG-modules,

we may concentrate upon the irreducible FG-modules. We begin our

study of these in the next chapter.


Assume that G is a ®nite group and F � R or C.

1. Maschke's Theorem says that for every FG-submodule U of an FG-

module V, there is an FG-submodule W with

V � U � W :

2. Every non-zero FG-module V is a direct sum of irreducible FG-

modules:

V � U1 � : : :� Ur:


1. Let G � kx: x3 � 1l � C3, and let V be the 2-dimensional CG-

module with basis v1, v2, where

v1x � v2, v2x � ÿv1 ÿ v2:

(This is a CG-module, by Exercise 3.2.)

Express V as a direct sum of irreducible CG-submodules.

2. If G � C2 3 C2, express the group algebra RG as a direct sum of

1-dimensional RG-submodules.


3. Find a group G, a CG-module V and a CG-homomorphism

W: V! V such that V 6� Ker W � Im W.

4. Let G be a ®nite group and let r: G! GL (2, C) be a representation

of G. Suppose that there are elements g, h in G such that the

matrices gr and hr do not commute. Prove that r is irreducible.

(You may care to revisit Example 5.5(2) and Exercises 5.1, 5.3, 5.4,

6.6 in the light of this result.)

5. Suppose that G is the in®nite group

1 0

n 1

� �: n 2 Z

� �and let V be the CG-module C2, with the natural multiplication by

elements of G (so that for v 2 V, g 2 G, the vector vg is just the

product of the row vector v with the matrix g).

Show that V is not completely reducible.

(This shows that Maschke's Theorem fails for in®nite groups ±

compare Example 8.2(2).)

6. An alternative proof of Maschke's Theorem for CG-modules.

Let V be a CG-module with basis v1, . . . , vn and suppose that U

is a CG-submodule of V. De®ne a complex inner product ( , ) on

V as follows (see (14.2) for the de®nition of a complex inner

product): for ëi, ì j 2 C,Xn

i�1

ëivi,Xn

j�1

ì jv j

!�Xn

i�1

ëiìi:

De®ne another complex inner product [ , ] on V by

[u, v] �Xx2G

(ux, vx) (u, v 2 V ):

(1) Verify that [ , ] is a complex inner product, which satis®es

[ug, vg] � [u, v] for all u, v 2 V and g 2 G:

(2) Suppose that U is a CG-submodule of V, and de®ne

U? � fv 2 V : [u, v] � 0 for all u 2 Ug:Show that U? is a CG-submodule of V.


(3) Deduce Maschke's Theorem. (Hint: it is a standard property

of complex inner products that V � U � U? for all subspaces U

of V.)

7. Prove that for every ®nite simple group G, there exists a faithful

irreducible CG-module.


78

9

Schur's Lemma

Schur's Lemma is a basic result concerning irreducible modules.

Though simple in both statement and proof, Schur's Lemma is funda-

mental to representation theory, and we give an immediate application

by determining all the irreducible representations of ®nite abelian

groups.

Schur's Lemma concerns CG-modules rather than RG-modules, and

since much of the ensuing theory depends on it, we shall deal with

CG-modules for the remainder of the book (except in Chapter 23).

Throughout, G denotes a ®nite group.

Schur's Lemma

9.1 Schur's Lemma

Let V and W be irreducible CG-modules.

(1) If W: V! W is a CG-homomorphism, then either W is a CG-

isomorphism, or vW � 0 for all v 2 V.

(2) If W: V! V is a CG-isomorphism, then W is a scalar multiple of

the identity endomorphism 1V .

Proof (1) Suppose that vW 6� 0 for some v 2 V. Then Im W 6� {0}. As

Im W is a CG-submodule of W by Proposition 7.2, and W is irreducible,

we have Im W � W. Also by Proposition 7.2, Ker W is a CG-submodule

of V; as Ker W 6� V and V is irreducible, Ker W � {0}. Thus W is

invertible, and hence is a CG-isomorphism.

(2) By (2.26), the endomorphism W has an eigenvalue ë 2 C, and so

Ker (W ÿ ë1V ) 6� {0}. Thus Ker (W ÿ ë1V ) is a non-zero CG-submodule

of V. Since V is irreducible, Ker (W ÿ ë1V ) � V. Therefore

v(Wÿ ë1V ) � 0 for all v 2 V :

That is, W � ë1V , as required. j

Part (2) of Schur's Lemma has the following converse.

9.2 Proposition

Let V be a non-zero CG-module, and suppose that every CG-homo-

morphism from V to V is a scalar multiple of 1V . Then V is

irreducible.

Proof Suppose that V is reducible, so that V has a CG-submodule U

not equal to {0} or V. By Maschke's Theorem, there is a CG-

submodule W of V such that

V � U � W :

Then the projection ð: V! V de®ned by (u � w)ð � u for all u 2 U,

w 2 W is a CG-homomorphism (see Proposition 7.11), and is not a

scalar multiple of 1V , which is a contradiction. Hence V is

irreducible. j

We next interpret Schur's Lemma and its converse in terms of

representations.

9.3 Corollary

Let r: G! GL (n, C) be a representation of G. Then r is irreducible if

and only if every n 3 n matrix A which satis®es

(gr)A � A(gr) for all g 2 G

has the form A � ëIn with ë 2 C.

Proof As in Theorem 4.4(1), regard Cn as a CG-module by de®ning

vg � v(gr) for all v 2 Cn, g 2 G.

Let A be an n 3 n matrix over C. The endomorphism v! vA of

Cn is a CG-homomorphism if and only if

(vg)A � (vA)g for all v 2 Cn, g 2 G;

Schur's Lemma 79

that is, if and only if

(gr)A � A(gr) for all g 2 G:

The result now follows from Schur's Lemma 9.1 and Proposition 9.2.

j

9.4 Examples

(1) Let G � C3 � ka: a3 � 1l, and let r: G! GL (2, C) be the repre-

sentation for which

ar � 0 1

ÿ1 ÿ1

� �(see Exercise 3.2). Since the matrix

0 1

ÿ1 ÿ1

� �commutes with all gr (g 2 G), Corollary 9.3 implies that r is

reducible.

(2) Let G � D10 � ka, b: a5 � b2 � 1, bÿ1ab � aÿ1l, and let ù � e2ði=5.

Check that there is a representation r: G! GL (2, C) for which

ar � ù 0

0 ùÿ1

� �, br � 0 1

1 0

� �:

Assume that the matrix

A � á âã ä

� �commutes with both ar and br. The fact that (ar)A � A(ar) forces

â � ã � 0; and then (br)A � A(br) gives á � ä. Hence

A � á 0

0 á

� �� áI :

Consequently r is irreducible, by Corollary 9.3.


Representation theory of ®nite abelian groups

Let G be a ®nite abelian group, and let V be an irreducible CG-

module. Pick x 2 G. Since G is abelian,

vgx � vxg for all g 2 G,

and hence the endomorphism v! vx of V is a CG-homomorphism. By

Schur's Lemma 9.1(2), this endomorphism is a scalar multiple of the

identity 1V , say ëx1V . Thus

vx � ëxv for all v 2 V :

This implies that every subspace of V is a CG-submodule. As V is

irreducible, we deduce that dim V � 1. Thus we have proved

9.5 Proposition

If G is a ®nite abelian group, then every irreducible CG-module has

dimension 1.

The next result is a major structure theorem for ®nite abelian groups.

We shall not prove it here, but refer you to Chapter 9 of the book of

J. B. Fraleigh listed in the Bibliography.

9.6 Theorem

Every ®nite abelian group is isomorphic to a direct product of cyclic

groups.

We shall determine the irreducible representations of all direct

products

Cn13 Cn2

3 : : :3 Cnr

where n1, . . . , nr are positive integers. By Theorem 9.6, this covers

the irreducible representations of all ®nite abelian groups.

Let G � Cn13 . . . 3 Cn r

, and for 1 < i < r, let ci be a generator

for Cni. Write

gi � (1, : : : , ci, : : : , 1) (ci in ith position):

Then

G � hg1, : : : , g ri, with g ni

i � 1 and gigj � gjgi for all i, j:

Now let r: G! GL (n, C) be an irreducible representation of G

Schur's Lemma 81

over C. Then n � 1 by Proposition 9.5, so for 1 < i < r, there exists

ëi 2 C such that

gir � (ëi)

(where of course (ëi) is a 1 3 1 matrix). As gi has order ni, we have

ëni

i � 1; that is, ëi is an nith root of unity. Also, the values ë1, . . . , ër

determine r, since for g 2 G, we have g � gi11 . . . gir

r for some

integers i1, . . . , ir, and then

gr � (gi11 : : :g

irr )r � (ëi1

1 : : :ëi r

r ):(9:7)

For a representation r of G satisfying (9.7) for all i1, . . . , ir, write

r � rë1,:::,ë r:

Conversely, given any nith roots of unity ëi (1 < i < r), the function

gi11 : : :g

irr ! (ëi1

1 : : :ëi r

r )

is a representation of G. There are n1 n2 . . . nr such representations,

and no two of them are equivalent.

We have proved the following theorem.

9.8 Theorem

Let G be the abelian group Cn13 . . . 3 Cn r

. The representations

rë1,:::,ë rof G constructed above are irreducible and have degree 1.

There are |G| of these representations, and every irreducible represent-

ation of G over C is equivalent to precisely one of them.

9.9 Examples

(1) Let G � Cn � ka: an � 1l, and put ù � e2ði=n. The n irreducible

representations of G over C are rù j (0 < j < n ÿ 1), where

akrù j � (ù jk) (0 < k < nÿ 1):

(2) The four irreducible CG-modules for G � C2 3 C2 � kg1, g2l are

V1, V2, V3, V4, where Vi is a 1-dimensional space with basis vi

(i � 1, 2, 3, 4) and

v1 g1 � v1, v1 g2 � v2;

v2 g1 � v2, v2 g2 � ÿv2;

v3 g1 � ÿv3, v3 g2 � v3;

v4 g1 � ÿv4, v4 g2 � ÿv4:


Diagonalization

Let H � kgl be a cyclic group of order n, and let V be a non-zero

CH-module. By Theorem 8.7,

V � U1 � : : :� Ur,

a direct sum of irreducible CH-submodules Ui of V. Each Ui has

dimension 1, by Proposition 9.5; let ui be a vector spanning Ui. Put

ù � e2ði=n. Then for each i, there exists an integer mi such that

uig � ùmi ui:

Thus if B is the basis u1, . . . , ur of V, then

[g]B �ùm1

. ..

ùmr

0B@1CA:(9:10)

The following useful result is an immediate consequence of this.

9.11 Proposition

Let G be a ®nite group and V a CG-module. If g 2 G, then there is a

basis B of V such that the matrix [g]B is diagonal. If g has order n,

then the entries on the diagonal of [g]B are nth roots of unity.

Proof Let H � kgl. As V is also a CH-module, the result follows from

(9.10). j

Some further applications of Schur's Lemma

Our next application concerns an important subspace of the group

algebra CG.

9.12 De®nition

Let G be a ®nite group. The centre of the group algebra CG, written

Z(CG), is de®ned by

Z(CG) � fz 2 CG: zr � rz for all r 2 CGg:Using (2.5), it is easy to check that Z(CG) is a subspace of CG.

For abelian groups G, the centre Z(CG) is the whole group algebra.

For arbitrary groups G, we shall see that Z(CG) plays a crucial role in

0

0

Schur's Lemma 83

the study of representations of G (for example, its dimension is equal

to the number of irreducible representations of G ± see Chapter 15).

9.13 Example

The elements 1 andP

g2G g lie in Z(CG). Indeed, if H is any normal

subgroup of G, then Xh2H

h 2 Z(CG):

To see this, write z �Ph2H h. Then for all g 2 G,

gÿ1zg �Xh2H

gÿ1 hg �Xh2H

h � z,

and so zg � gz. Consequently zr � rz for all r 2 CG.

For example, if G � D6 � ka, b: a3 � b2 � 1, bÿ1ab � aÿ1l, then

{1}, kal and G are normal subgroups of G, so the elements

1, 1� a� a2 and 1� a� a2 � b� ab� a2b

lie in Z(CG). We shall see later that these elements in fact form a

basis of Z(CG).

We use Schur's Lemma to prove the following important property of

the elements of Z(CG).

9.14 Proposition

Let V be an irreducible CG-module, and let z 2 Z(CG). Then there

exists ë 2 C such that

vz � ëv for all v 2 V :

Proof For all r 2 CG and v 2 V, we have

vrz � vzr,

and hence the function v! vz is a CG-homomorphism from V to V.

By Schur's Lemma 9.1(2), this CG-homomorphism is equal to ë1V for

some ë 2 C, and the result follows. j

Some elements of the centre of CG are provided by the centre of G,

which we now de®ne.


9.15 De®nition

The centre of G, written Z(G), is de®ned by

Z(G) � fz 2 G: zg � gz for all g 2 Gg:Clearly Z(G) is a normal subgroup of G, and is a subset of Z(CG).

Although we have seen in Proposition 6.6 that for every ®nite group

G there is a faithful CG-module, it is not necessarily the case that

there is a faithful irreducible CG-module. Indeed, the following result

shows that the existence of a faithful irreducible CG-module imposes a

strong restriction on the structure of G.

9.16 Proposition

If there exists a faithful irreducible CG-module, then Z(G) is cyclic.

Proof Let V be a faithful irreducible CG-module. If z 2 Z(G) then z

lies in Z(CG), and hence by Proposition 9.14, there exists ëz 2 C such

that

vz � ëzv for all v 2 V :

Since V is faithful, the function

z! ëz (z 2 Z(G))

is an injective homomorphism from Z(G) into the multiplicative group

C� of non-zero complex numbers. Therefore Z(G) � {ëz: z 2 Z(G)},

which, being a ®nite subgroup of C�, is cyclic (see Exercise 1.7). j

We remark that the converse of Proposition 9.16 is false, since in

Exercise 25.6, we give an example of a group G such that Z(G) is

cyclic but there exists no faithful irreducible CG-module.

9.17 Example

If G is an abelian group, then G � Z(G), and so by Proposition 9.16,

there is no faithful irreducible CG-module unless G is cyclic. For

example, C2 3 C2 has no faithful irreducible representation (compare

Example 9.9(2)).

The irreducible representations of non-abelian groups are more

dif®cult to construct than those of abelian groups. In particular, they

Schur's Lemma 85

do not all have degree 1, as is shown by the following converse to

Proposition 9.5.

9.18 Proposition

Suppose that G is a ®nite group such that every irreducible CG-module

has dimension 1. Then G is abelian.

Proof By Theorem 8.7, we can write

CG � V1 � : : :� Vn,

where each Vi is an irreducible CG-submodule of the regular CG-

module CG. Then dim Vi � 1 for all i, since we are assuming that all

irreducible CG-modules have dimension 1. For 1 < i < n, let vi be a

vector spanning Vi. Then v1, . . . , vn is a basis of CG; call it B . For

all x, y 2 G, the matrices [x]B and [y]B are diagonal, and hence they

commute. Since the representation

g ! [g]B (g 2 G)

of G is faithful (see Proposition 6.6), we deduce that x and y

commute. Hence G is abelian, as required. j


1. Schur's Lemma states that every CG-homomorphism between irredu-

cible CG-modules is either zero or a CG-isomorphism. Also, the

only CG-homomorphisms from an irreducible CG-module to itself

are scalar multiples of the identity.

2. The centre Z(CG) of the group algebra CG consists of those

elements which commute with all elements of CG. The elements of

Z(CG) act as scalar multiples of the identity on all irreducible CG-

modules.

3. All irreducible CG-modules for a ®nite abelian group G have

dimension 1, and there are precisely |G| of them.


1. Write down the irreducible representations over C of the groups C2,

C3 and C2 3 C2.


2. Let G � C4 3 C4.

(a) Find a non-trivial irreducible representation r of G such that

g2r � (1) for all g 2 G.

(b) Prove that there is no irreducible representation ó of G such

that gó � (ÿ1) for all elements g of order 2 in G.

3. Let G be the ®nite abelian group Cn13 . . . 3 Cn r

. Prove that G has

a faithful representation of degree r. Can G have a faithful

representation of degree less than r?

4. Suppose that G � D8 � ka, b: a4 � b2 � 1, bÿ1ab � aÿ1l. Check that

there is a representation r of G over C such that

ar � ÿ7 10

ÿ5 7

� �, br � ÿ5 6

ÿ4 5

� �:

Find all 2 3 2 matrices M such that M(gr) � (gr)M for all g 2 G.

Hence determine whether or not r is irreducible.

Do the same for the representation ó of G, where

aó � 5 ÿ6

4 ÿ5

� �, bó � ÿ5 6

ÿ4 5

� �:

5. Show that if V is an irreducible CG-module, then there exists ë 2 C

such that

vXg2G

g

!� ëv for all v 2 V :

6. Let G � D6 � ka, b: a3 � b2 � 1, bÿ1ab � aÿ1l. Write ù � e2ði=3,

and let W be the irreducible CG-submodule of the regular CG-

module de®ned by

W � sp (1� ù2a� ùa2, b� ù2ab� ùa2b)

(see Exercise 6.6).

(a) Show that a � aÿ1 2 Z(CG).

(b) Find ë 2 C such that

w(a� aÿ1) � ëw

for all w 2 W. (Compare Proposition 9.14.)

Schur's Lemma 87

7. Which of the following groups have a faithful irreducible represent-

ation?

(a) Cn (n a positive integer);

(b) D8;

(c) C2 3 D8;

(d) C3 3 D8.


89

10

Irreducible modules and the group algebra

Let G be a ®nite group and CG be the group algebra of G over C.

Consider CG as the regular CG-module. By Theorem 8.7, we can

write

CG � U1 � : : :� Ur

where each Ui is an irreducible CG-module. We shall show in this

chapter that every irreducible CG-module is isomorphic to one of the

CG-modules U1, . . . , Ur. As a consequence, there are only ®nitely

many non-isomorphic irreducible CG-modules (a result which has

already been established for abelian groups in Theorem 9.8). Also, in

theory, to ®nd all irreducible CG-modules, it is suf®cient to decompose

CG as a direct sum of irreducible CG-submodules. However, this is

not really a practical way of ®nding the irreducible CG-modules, unless

G is a small group.

Irreducible submodules of CG

We begin with another consequence of Maschke's Theorem.

10.1 Proposition

Let V and W be CG-modules and let W: V! W be a CG-homomorph-

ism. Then there is a CG-submodule U of V such that V � Ker W � U

and U � Im W.

Proof Since Ker W is a CG-submodule of V by Proposition 7.2, there is

by Maschke's Theorem a CG-submodule U of V such that V �Ker W� U . De®ne a function W: U! Im W by

uW � uW (u 2 U ):

We show that W is a CG-isomorphism from U to Im W. Clearly W is a

CG-homomorphism, since W is a CG-homomorphism. If u 2 Ker W then

u 2 Ker W \ U � {0}; hence Ker W � {0}. Now let w 2 Im W; so w � vWfor some v 2 V. Write v � k � u with k 2 Ker W, u 2 U. Then

w � vW � kW� uW � uW � uW:

Therefore Im W � Im W. We have now established that W: U! Im W is

an invertible CG-homomorphism. Thus U � Im W, as required.

10.2 Proposition

Let V be a CG-module, and write

V � U1 � : : :� Us,

a direct sum of irreducible CG-submodules Ui. If U is any irreducible

CG-submodule of V, then U � Ui for some i.

Proof For u 2 U, we have u � u1 � . . . � us for unique vectors ui 2 Ui

(1 < i < s). De®ne ði: U! Ui by setting uði � ui. Choosing i such

that ui 6� 0 for some u 2 U, we have ði 6� 0.

Now ði is a CG-homomorphism (see Proposition 7.11). As U and

Ui are irreducible, and ði 6� 0, Schur's Lemma 9.1(1) implies that ði is

a CG-isomorphism. Therefore U � Ui. j

Of course it can happen that U is an irreducible CG-submodule of

U1 � . . . � Us (each Ui irreducible) without U being equal to any Ui,

as the following example shows.

10.3 Example

Let G be any group and let V be a 2-dimensional CG-module, with

basis v1, v2, such that vg � v for all v 2 V and g 2 G. Then

V � U1 � U2,

where U1 � sp (v1) and U2 � sp (v2) are irreducible CG-submodules.

However, U � sp (v1 � v2) is an irreducible CG-submodule which is

not equal to U1 or U2.

10.4 De®nitions

(1) If V is a CG-module and U is an irreducible CG-module, then we

say that U is a composition factor of V if V has a CG-submodule

which is isomorphic to U.


(2) Two CG-modules V and W are said to have a common composition

factor if there is an irreducible CG-module which is a composition

factor of both V and W.

We now come to the main result of the chapter, which shows that

every irreducible CG-module is a composition factor of the regular

CG-module.

10.5 Theorem

Regard CG as the regular CG-module, and write

CG � U1 � : : :� Ur,

a direct sum of irreducible CG-submodules. Then every irreducible

CG-module is isomorphic to one of the CG-modules Ui.

Proof Let W be an irreducible CG-module, and choose a non-zero

vector w 2 W. Observe that {wr: r 2 CG} is a CG-submodule of W;

since W is irreducible, it follows that

W � fwr: r 2 CGg:(10:6)

Now de®ne W: CG! W by

rW � wr (r 2 CG):

Clearly W is a linear transformation, and Im W � W by (10.6). Moreover,

W is a CG-homomorphism, since for r, s 2 CG,

(rs)W � w(rs) � (wr)s � (rW)s:

By Proposition 10.1, there is a CG-submodule U of CG such that

CG � U � Ker W and U � Im W � W :

As W is irreducible, so is U. By Proposition 10.2 we have U � Ui for

some i; then W � Ui, and the result is proved. j

Theorem 10.5 shows that there is a ®nite set of irreducible CG-

modules such that every irreducible CG-module is isomorphic to one

of them. We record this fact in the following corollary.

10.7 Corollary

If G is a ®nite group, then there are only ®nitely many non-isomorphic

irreducible CG-modules.

Irreducible modules and the group algebra 91

According to Theorem 10.5, to ®nd all the irreducible CG-modules

we need only decompose the regular CG-module as a direct sum of

irreducible CG-submodules. We now do this for a couple of examples;

however, this is not a practical method for studying CG-modules in

general.

10.8 Examples

(1) Let G � C3 � ka: a3 � 1l, and write ù � e2ði=3. De®ne v0, v1,

v2 2 CG by

v0 � 1� a� a2,

v1 � 1� ù2a� ùa2,

v2 � 1� ùa� ù2a2,

and let Ui � sp (vi) for i � 0, 1, 2. Then v1a � a � ù2a2 � ù1 � ùv1,

and similarly

via � ùivi for i � 0, 1, 2:

Hence Ui is a CG-submodule of CG for i � 0, 1, 2.

It is easy to check that v0, v1, v2 is a basis of CG, and hence

CG � U0 � U1 � U2,

a direct sum of irreducible CG-submodules Ui. By Theorem 10.5,

every irreducible CG-module is isomorphic to U0, U1 or U2. The

irreducible representation of G corresponding to Ui is the representation

rùi of Example 9.9(1).

(2) Let G � D6 � ka, b: a3 � b2 � 1, bÿ1ab � aÿ1l. We decompose CG

as a direct sum of irreducible CG-submodules. Let ù � e2ði=3 and

de®ne

v0 � 1� a� a2, w0 � bv0 (� b� ba� ba2),

v1 � 1� ù2a� ùa2, w1 � bv1,

v2 � 1� ùa� ù2a2, w2 � bv2:

As in (1) above, via � ùivi for i � 0, 1, 2, and so sp (vi) and sp (wi)

are Ckal-modules. Next, note that

v0b � w0, w0b � v0,

v1b � w2, w1b � v2,

v2b � w1, w2b � v1:


Therefore, sp (v0, w0), sp (v1, w2) and sp (v2, w1) are Ckbl-modules, and

hence are CG-submodules of CG. By the argument in Example 5.5(2),

the CG-submodules U3 � sp (v1, w2) and U4 � sp (v2, w1) are irreduci-

ble. However, sp(v0, w0) is reducible, as U1 � sp(v0 � w0) and U2 �sp(v0 ÿ w0) are CG-submodules.

Now v0, v1, v2, w0, w1, w2 is a basis of CG, and hence

CG � U1 � U2 � U3 � U4,

a direct sum of irreducible CG-submodules. Note that U1 is the trivial

CG-module, and U1 is not isomorphic to U2, the other 1-dimensional

Ui. But U3 � U4 (there is a CG-isomorphism sending v1 ! w1,

w2 ! v2).

We conclude from Theorem 10.5 that there are exactly three non-

isomorphic irreducible CG-modules, namely U1, U2 and U3. Corre-

spondingly, every irreducible representation of D6 over C is equivalent

to precisely one of the following:

r1: a! (1), b! (1);

r2: a! (1), b! (ÿ1);

r3: a! ù 0

0 ùÿ1

� �, b! 0 1

1 0

� �:


1. Every irreducible CG-module occurs as a composition factor of the

regular CG-module.

2. There are only ®nitely many non-isomorphic irreducible CG-

modules.


1. Let G be a ®nite group. Find a CG-submodule of CG which is

isomorphic to the trivial CG-module. Is there only one such CG-

submodule?

2. Let G � C4. Express CG as a direct sum of irreducible CG-

submodules. (Hint: copy the method of Example 10.8(1).)

Irreducible modules and the group algebra 93

3. Let G � D8 � ka, b: a4 � b2 � 1, bÿ1ab � aÿ1l. Find a 1-dimen-

sional CG-submodule, sp (u1) say, of CG such that

u1a � u1, u1b � ÿu1:

Find also 1-dimensional CG-submodules, sp (u2) and sp (u3), such

that

u2a � ÿu2, u2b � u2, and

u3a � ÿu3, u3b � ÿu3:

4. Use the method of Example 10.8(2) to ®nd all the irreducible

representations of D8 over C.

5. Suppose that V is a non-zero CG-module such that V � U1 � U2,

where U1 and U2 are isomorphic CG-modules. Show that there is a

CG-submodule U of V which is not equal to U1 or U2, but is

isomorphic to both of them.

6. Let G � Q8 � ka, b: a4 � 1, b2 � a2, bÿ1ab � aÿ1l, and let V be the

CG-module given in Example 4.5(2). Thus V has basis v1, v2 and

v1a � iv1, v1b � v2,

v2a � ÿiv2, v2b � ÿv1:

Show that V is irreducible, and ®nd a CG-submodule of CG which

is isomorphic to V.


95

11

More on the group algebra

We now go further into the structure of the group algebra CG of a

®nite group G. As in Chapter 10, we write

CG � U1 � : : :� Ur,

a direct sum of irreducible CG-modules Ui. In Theorem 10.5 we

proved that every irreducible CG-module U is isomorphic to one of

the Ui. The question arises: how many of the Ui are isomorphic to U?

There is an elegant and signi®cant answer to this question: the number

is precisely dim U (see Theorem 11.9).

Our proof of Theorem 11.9 is based on a study of the vector space

of CG-homomorphisms from one CG-module to another.

The space of CG-homomorphisms

11.1 De®nition

Let V and W be CG-modules. We write HomCG (V , W ) for the set of

all CG-homomorphisms from V to W.

De®ne addition and scalar multiplication on HomCG (V , W ) as

follows: for W, ö 2 HomCG (V , W ) and ë 2 C, de®ne W � ö and ëW by

v(W� ö) � vW� vö,

v(ëW) � ë(vW)

for all v 2 V . Then W� ö, ëW 2 HomCG (V , W ). With these de®nitions,

it is easily checked that HomCG (V, W ) is a vector space over C.

We begin our study of the vector space HomCG (V, W) with an easy

consequence of Schur's Lemma.

11.2 Proposition

Suppose that V and W are irreducible CG-modules. Then

dim (HomCG (V , W )) � 1, if V � W ,

0, if V 6� W :

�

Proof If V 6� W then this is immediate from Schur's Lemma 9.1(1).

Now suppose that V � W, and let W: V! W be a CG-isomorphism.

If ö 2 HomCG (V , W ), then öWÿ1 is a CG-isomorphism from V to V,

so by Schur's Lemma 9.1(2), there exists ë 2 C such that

öWÿ1 � ë1V :

Then ö � ëW, and so HomCG (V , W ) � fëW: ë 2 Cg, a 1-dimensional

space. j

For the next result, recall the de®nition of a composition factor of a

CG-module from 10.4.

11.3 Proposition

Let V and W be CG-modules, and suppose that HomCG(V , W ) 6� f0g.Then V and W have a common composition factor.

Proof Let W be a non-zero element of HomCG (V , W ). Then

V � Ker W� U for some non-zero CG-module U, by Maschke's Theo-

rem. Let X be an irreducible CG-submodule of U. Since XW 6� {0},

Schur's Lemma 9.1(1) implies that XW � X. Therefore X is a common

composition factor of V and W. j

The next few results show how to calculate the dimension of

HomCG (V , W ) in general. The key step is the following proposition.

11.4 Proposition

Let V , V1, V2 and W , W1, W2 be CG-modules. Then

(1) dim (HomCG (V, W1 � W2)) �dim (HomCG (V, W1)) � dim (HomCG(V, W2)),

(2) dim (HomCG (V1 � V2, W )) �dim (HomCG (V1, W)) � dim (HomCG(V2, W )).


Proof (1) De®ne the functions ð1: W1 � W2 ! W1 and ð2:

W1 � W2 ! W2 by

(w1 � w2)ð1 � w1, (w1 � w2)ð2 � w2

for all w1 2 W1, w2 2 W2. By Proposition 7.11, ð1 and ð2 are

CG-homomorphisms. If W 2 HomCG (V , W1 � W2), then Wð1 2HomCG(V , W1) and Wð2 2 HomCG (V, W2) (see Exercise 7.1).

We now de®ne a function f from HomCG (V, W1 � W2) to the

(external) direct sum of HomCG (V, W1) and HomCG (V, W2) by

f : W! (Wð1, Wð2) (W 2 HomCG (V , W1 � W2)):

Clearly f is a linear transformation. We show that f is invertible.

Given öi 2 HomCG (V, Wi) (i � 1, 2), the function

ö: v! vö1 � vö2 (v 2 V )

lies in HomCG (V, W1 � W2), and the image of ö under f is (ö1, ö2).

Hence f is surjective.

If W 2 Ker f, then vWð1 � 0 and vWð2 � 0 for all v 2 V, so vW �vW(ð1 � ð2) � 0. Therefore W � 0, so Ker f � {0} and f is injective.

We have established that f is an invertible linear transformation from

HomCG (V, W1 � W2) to HomCG (V, W1) � HomCG (V, W2). Consequently

these two vector spaces have equal dimensions, and (1) follows.

(2) For W 2 HomCG (V1 � V2, W ), de®ne WVi: Vi ! W (i � 1, 2) to

be the restriction of W to Vi; that is, WViis the function

viWVi� viW (vi 2 Vi):

Then WVi2 HomCG (Vi, W ) for i � 1, 2.

Now let h be the function from HomCG (V1 � V2, W ) to

HomCG (V1, W )� HomCG (V2, W ) which is given by

h: W! (WV1, WV2

) (W 2 HomCG (V1 � V2, W )):

Clearly h is an injective linear transformation. Given öi 2HomCG(Vi, W ) (i � 1, 2), the function

ö: v1 � v2 ! v1ö1 � v2ö2 (vi 2 Vi for i � 1, 2)

lies in HomCG (V1 � V2, W ) and has image (ö1, ö2) under h. Hence h

is surjective. We have shown that h is an invertible linear transforma-

tion, and (2) follows. j

More on the group algebra 97

Now suppose that we have CG-modules V, W, Vi, Wj (1 < i < r,

1 < j < s). By an obvious induction using Proposition 11.4, we have

(11.5) (1) dim (HomCG (V, W1 � . . . � Ws))

�Xs

j�1

dim (HomCG (V, Wj)),

(2) dim(HomCG(V1� . . . �Vr, W ))

�Xr

i�1

dim (HomCG (Vi, W)).

These in turn imply

(3) dim (HomCG (V1 � . . . � Vr, W1 � . . . � Ws))

�Xr

i�1

Xs

j�1

dim (HomCG (Vi, Wj)).

By applying (3) when all Vi and Wj are irreducible, and using

Proposition 11.2, we can ®nd dim (HomCG (V, W )) in general. In the

following corollary we single out the case where one of the CG-

modules is irreducible.

11.6 Corollary

Let V be a CG-module with

V � U1 � : : :� Us,

where each Ui is an irreducible CG-module. Let W be any irreducible

CG-module. Then the dimensions of HomCG (V , W ) and

HomCG (W , V ) are both equal to the number of CG-modules Ui such

that Ui � W.

Proof By (11.5),

dim (HomCG (V , W )) �Xs

i�1

dim (HomCG (Ui, W )), and

dim (HomCG (W , V )) �Xs

i�1

dim (HomCG (W , Ui)):


And by Proposition 11.2,

dim (HomCG (Ui, W )) � dim (HomCG (W , Ui)) � 1, if Ui � W ,

0, if Ui 6� W :

�The result follows. j

11.7 Example

For G � D6, we saw in Example 10.8(2) that

CG � U1 � U2 � U3 � U4,

a direct sum of irreducible CG-submodules, with U3 � U4 but U3 not

isomorphic to U1 or U2. Thus by Corollary 11.6, we have

dim (HomCG (CG, U3)) � dim (HomCG (U3, CG)) � 2:

You are asked in Exercise 11.5 to ®nd bases for these two vector

spaces of CG-homomorphisms.

The next proposition investigates the space of CG-homomorphisms

from the regular CG-module to any other CG-module. When combined

with Corollary 11.6, it will give the main result of this chapter.

11.8 Proposition

If U is a CG-module, then

dim (HomCG (CG, U )) � dim U :

Proof Let d � dim U. Choose a basis u1, . . . , ud of U. For 1 < i < d,

de®ne öi: CG! U by

röi � uir (r 2 CG):

Then öi 2 HomCG (CG, U) since for all r, s 2 CG,

(rs)öi � ui(rs) � (uir)s � (röi)s:

We shall prove that ö1, . . . , öd is a basis of HomCG (CG, U ).

Suppose that ö 2 HomCG(CG, U ). Then

1ö � ë1u1 � : : :� ëd ud

for some ëi 2 C. Since ö is a CG-homomorphism, for all r 2 CG we

have


rö � (1r)ö � (1ö)r

� ë1u1 r � : : :� ëd ud r

� r(ë1ö1 � : : :� ëdöd):

Hence ö � ë1ö1 � . . . � ëdöd . Therefore ö1, . . . , öd span

HomCG (CG, U ).

Now assume that

ë1ö1 � : : :� ëdöd � 0 (ëi 2 C):

Evaluating both sides at the identity 1, we have

0 � 1(ë1ö1 � : : :� ëdöd)

� ë1u1 � : : :� ëd ud ,

which forces ëi � 0 for all i. Hence ö1, . . . , öd is a basis of

HomCG (CG, U ), which therefore has dimension d. j

We now come to the main theorem of the chapter, which tells us

how often each irreducible CG-module occurs in the regular CG-

module.

11.9 Theorem

Suppose that

CG � U1 � : : :� Ur,

a direct sum of irreducible CG-submodules. If U is any irreducible

CG-module, then the number of CG-modules Ui with Ui � U is equal

to dim U.

Proof By Proposition 11.8,

dim U � dim (HomCG (CG, U )),

and by Corollary 11.6, this is equal to the number of Ui with Ui � U.

j

11.10 Example

Recall again from Example 10.8(2) that if G � D6 then

CG � U1 � U2 � U3 � U4,

where U1, U2 are non-isomorphic 1-dimensional CG-modules, and


U3, U4 are isomorphic irreducible 2-dimensional CG-modules. This

illustrates Theorem 11.9:

U1 occurs once, dim U1 � 1;

U2 occurs once, dim U2 � 1;

U3 occurs twice, dim U3 � 2:

We conclude the chapter with a signi®cant consequence of

Theorem 11.9 concerning the dimensions of all irreducible CG-modules.

11.11 De®nition

We say that the irreducible CG-modules V1, : : : , Vk form a com-

plete set of non-isomorphic irreducible CG-modules if every irreducible

CG-module is isomorphic to some Vi, and no two of V1, : : : , Vk

are isomorphic. (By Corollary 10.7, for any ®nite group G there exists

a complete set of non-isomorphic irreducible CG-modules.)

11.12 Theorem

Let V1, : : : , Vk form a complete set of non-isomorphic irreducible

CG-modules. Then Xk

i�1

(dim Vi)2 � jGj:

Proof Let CG � U1 � . . . � Ur, a direct sum of irreducible CG-sub-

modules. For 1 < i < k, write di � dim Vi. By Theorem 11.9, for each

i, the number of CG-modules Uj with Uj � Vi is equal to di. Therefore

dim CG � dim U1 � : : :� dim Ur

�Xk

i�1

di(dim Vi) �Xk

i�1

d2i :

As dim CG � |G|, the result follows. j

11.13 Example

Let G be a group of order 8, and let d1, . . . , dk be the dimensions of

all the irreducible CG-modules. By Theorem 11.12,Xk

i�1

d2i � 8:


Observe that the trivial CG-module is irreducible of dimension 1, and

so di � 1 for some i. Hence the possibilities for d1, . . . , dk are

1, 1, 1, 1, 1, 1, 1, 1 and

1, 1, 1, 1, 2:

Both these possibilities do occur: the ®rst holds when G is an abelian

group (see Proposition 9.5), and the second when G � D8 (see

Exercise 10.4).

We shall see later that dim Vi divides |G| for all i, and this fact,

combined with Theorem 11.12, is quite a powerful tool in ®nding the

dimensions of irreducible CG-modules.


1. dim (HomCG (V1 � : : :� Vr, W1 � : : :� Ws))

�Xr

i�1

Xs

j�1

dim (HomCG (Vi, Wj)):

2. dim (HomCG (CG, U )) � dim U .

3. Let CG � U1 � . . . � Ur, a direct sum of irreducible CG-modules,

and let U be any irreducible CG-module. Then the number of Ui

with Ui � U is equal to dim U.

4. If V1, : : : , Vk is a complete set of non-isomorphic irreducible CG-

modules, then Xk

i�1

(dim Vi)2 � jGj:


1. If G is a non-abelian group of order 6, ®nd the dimensions of all

the irreducible CG-modules.

2. If G is a group of order 12, what are the possible degrees of all the

irreducible representations of G?

Find the degrees of the irreducible representations of D12.

(Hint: use Exercise 5.3.)

3. Let G be a ®nite group. Find a basis for HomCG (CG, CG).


4. Suppose that G � Sn and V is the n-dimensional permutation

module for G over C, as de®ned in 4.10. If U is the trivial CG-

module, show that HomCG (V, U) has dimension 1.

5. Let G � D6 and let CG � U1 � U2 � U3 � U4, a direct sum of

irreducible CG-modules, as in Example 10.8(2). Find a basis for

HomCG (CG, U3) and a basis for HomCG (U3, CG).

6. Let V1, : : : , Vk be a complete set of non-isomorphic irreducible

CG-modules, and let V, W be arbitrary CG-modules. Assume that

for 1 < i < k,

di � dim (HomCG (V , Vi)) and ei � dim (HomCG (W , Vi)):

Show that dim (HomCG (V , W )) �Pki�1 diei.


104

12

Conjugacy classes

We take a break from representation theory to discuss some topics in

group theory which will be relevant in our further study of represent-

ations. After de®ning conjugacy classes, we develop enough theory to

determine the conjugacy classes of dihedral, symmetric and alternating

groups. At the end of the chapter we prove a result linking the

conjugacy classes of a group to the structure of its group algebra.

Throughout the chapter, G is a ®nite group.

Conjugacy classes

12.1 De®nition

Let x, y 2 G. We say that x is conjugate to y in G if

y � gÿ1xg for some g 2 G:

The set of all elements conjugate to x in G is

xG � fgÿ1xg: g 2 Gg,and is called the conjugacy class of x in G.

Our ®rst result shows that two distinct conjugacy classes have no

elements in common.

12.2 Proposition

If x, y 2 G, then either xG � yG or xG \ yG is empty.

Proof Suppose that xG \ yG is not empty, and pick z 2 xG \ yG. Then

there exist g, h 2 G such that

z � gÿ1xg � hÿ1 yh:

Hence x � ghÿ1 yhgÿ1 � kÿ1 yk, where k � hgÿ1. So

a 2 xG ) a � bÿ1xb for some b 2 G

) a � bÿ1 kÿ1 ykb

) a � cÿ1 yc where c � kb

) a 2 yG:

Therefore xG # yG. Similarly yG # xG (using y � kxkÿ1), and so

xG � yG. j

Since every element x of G lies in the conjugacy class xG (as

x � 1ÿ1x1 with 1 2 G), G is the union of its conjugacy classes and so

we deduce immediately

12.3 Corollary

Every group is a union of conjugacy classes, and distinct conjugacy

classes are disjoint.

Another way of seeing this Corollary 12.3 is to observe that

conjugacy is an equivalence relation, and that the conjugacy classes are

the equivalence classes.

12.4 De®nition

If G � xG1 [ . . . [ xG

l , where the conjugacy classes xG1 , . . . , xG

l are

distinct, then we call x1, . . . , xl representatives of the conjugacy classes

of G.

12.5 Examples

(1) For every group G, 1G � {1} is a conjugacy class of G.

(2) Let G � D6 � ka, b: a3 � b2 � 1, bÿ1ab � aÿ1l. The elements of G

are 1, a, a2, b, ab, a2b. Since gÿ1ag is a or a2 for every g 2 G, and

bÿ1ab � a2, we have

aG � fa, a2g:Also, aÿibai � aÿ2ib for all integers i, so

bG � fb, ab, a2bg:Thus the conjugacy classes of G are

f1g, fa, a2g, fb, ab, a2bg:

Conjugacy classes 105

(3) If G is abelian then gÿ1xg � x for all x, g 2 G, and so xG � {x}.

Hence every conjugacy class of G consists of just one element.

The next proposition is often useful when calculating conjugacy

classes.

12.6 Proposition

Let x, y 2 G. If x is conjugate to y in G, then xn is conjugate to yn in

G for every integer n, and x and y have the same order.

Proof Observe that for a, b 2 G, we have

gÿ1abg � (gÿ1ag)(gÿ1bg):

Hence gÿ1xng � (gÿ1xg)n. Suppose that x is conjugate to y in G, so

that y � gÿ1xg for some g 2 G. Then yn � gÿ1xng and therefore xn is

conjugate to yn in G. Let x have order m. Then ym � gÿ1xmg � 1, and

for 0 , r , m, yr � gÿ1xrg 6� 1, so y also has order m. j

Conjugacy class sizes

The next theorem determines the sizes of the conjugacy classes in G

in terms of certain subgroups which we now de®ne.

12.7 De®nition

Let x 2 G. The centralizer of x in G, written CG(x), is the set of

elements of G which commute with x; that is,

CG(x) � fg 2 G: xg � gxg:(So also CG(x) � {g 2 G: gÿ1xg � x}.)

It is easy to check that CG(x) is a subgroup of G (Exercise 12.1).

Observe that x 2 CG(x) and indeed, kxl # CG(x) for all x 2 G.

12.8 Theorem

Let x 2 G. Then the size of the conjugacy class xG is given by

jxGj � jG: CG(x)j � jGj=jCG(x)j:In particular, |xG | divides |G|.


Proof Observe ®rst that for g, h 2 G, we have

gÿ1xg � hÿ1xh, hgÿ1x � xhgÿ1

, hgÿ1 2 CG(x)

, CG(x)g � CG(x)h:

By dint of this, we may de®ne an injective function f from xG to the

set of right cosets of CG(x) in G by

f : gÿ1xg ! CG(x)g (g 2 G):

Clearly f is surjective. Hence f is a bijection, proving that |xG | �|G:CG(x)|. j

Before summarizing our results on conjugacy classes, we make the

observation that

jxGj � 1, gÿ1xg � x for all g 2 G(12:9)

, x 2 Z(G),

where Z(G) is the centre of G, as de®ned in 9.15.

We have now proved all parts of the following result.

12.10 The Class Equation

Let x1, . . . , xl be representatives of the conjugacy classes of G. Then

jGj � jZ(G)j �X

xi=2Z(G)

jxGi j,

where |xGi | � |G:CG(xi)|, and both |Z(G)| and |xG

i | divide |G|.

Conjugacy classes of dihedral groups

We illustrate the use of Theorem 12.8 by ®nding the conjugacy classes

of all dihedral groups.

Let G � D2n, the dihedral group of order 2n. Thus

G � ha, b: an � b2 � 1, bÿ1ab � aÿ1i:In ®nding the conjugacy classes of G, it is convenient to consider

separately the cases where n is odd and where n is even.

(1) n odd

First consider ai (1 < i < n ÿ 1). Since CG(ai) contains kal,jG: CG(ai)j < jG: haij � 2:


Also bÿ1aib � aÿi, so {ai, aÿi} # (ai)G. As n is odd, ai 6� aÿi, and so

|(ai)G | > 2. Using Theorem 12.8, we have

2 > jG: CG(ai)j � j(ai)Gj > 2:

Hence equality holds here, and

CG(ai) � hai, (ai)G � fai, aÿig:Next, CG(b) contains {1, b}; and as bÿ1aib � aÿi, no element ai or aib

(with 1 < i < n ÿ 1) commutes with b. Thus

CG(b) � f1, bg:

Therefore by Theorem 12.8, |bG | � n. Since all the elements ai have

been accounted for, bG must consist of the remaining n elements of G.

That is,

bG � fb, ab, : : : , anÿ1bg:

We have shown

(12.11) The dihedral group D2n (n odd) has precisely 12(n� 3) con-

jugacy classes:

{1}, {a, aÿ1}, . . . , {a(nÿ1)=2, aÿ(nÿ1)=2}, {b, ab, . . . , anÿ1b}.

(2) n even

Write n � 2m. As bÿ1amb � aÿm � am, the centralizer of am in G

contains both a and b, and hence CG(am) � G. Therefore the con-

jugacy class of am in G is just famg. As in case (1), (ai)G � {ai, aÿi}

for 1 < i < m ÿ 1.

For every integer j,

a jbaÿ j � a2 jb, a j(ab)aÿ j � a2 j�1b:

It follows that

bG � fa2 jb: 0 < j < mÿ 1g, (ab)G � fa2 j�1b: 0 < j < mÿ 1g:Hence

(12.12) The dihedral group D2n (n even, n � 2m) has precisely m � 3

conjugacy classes:

{1}, {am}, {a, aÿ1}, . . . , {amÿ1, aÿm�1},

{a2 jb: 0 < j < m ÿ 1}, {a2 j�1b: 0 < j < m ÿ 1}.


Conjugacy classes of Sn

We shall later need to know the conjugacy classes of the symmetric

group Sn. Our ®rst observation is simple but crucial.

12.13 Proposition

Let x be a k-cycle (i1 i2 . . . ik) in Sn, and let g 2 Sn. Then gÿ1xg is the

k-cycle (i1 g i2 g . . . ikg).

Proof Write A � {i1, . . . , ik}. For ir 2 A,

irg(gÿ1xg) � ir xg � ir�1 g (or i1 g if r � k):

Also, for 1 < i < n and i =2 A,

ig(gÿ1xg) � ixg � ig:

Hence gÿ1(i1 i2 . . . ik)g � (i1 g i2 g . . . ikg), as required. j

Now consider an arbitrary permutation x 2 Sn. Write

x � (a1 : : : ak 1)(b1 : : : bk 2

) : : : (c1 : : : cks),

a product of disjoint cycles, with k1 > k2 > . . . > ks. By Proposition

12.13, for g 2 Sn we have

(12.14)

gÿ1xg � gÿ1(a1 : : : ak1)ggÿ1(b1 : : : bk2

)g : : : gÿ1(c1 : : : cks)g

� (a1 g : : : ak1g)(b1 g : : : bk2

g) : : : (c1 g : : : cksg):

We call (k1, . . . , ks) the cycle-shape of x, and note that x and gÿ1xg

have the same cycle-shape. On the other hand, given any two permuta-

tions x, y of the same cycle-shape, say

x � (a1 : : : ak1) : : : (c1 : : : cks

),

y � (a91 : : : a9k1) : : : (c91 : : : c9ks

),

(products of disjoint cycles), there exists g 2 Sn sending

a1 ! a91, . . . , cks! c9ks

, and so by (12.14),

gÿ1xg � y:

We have proved the following result.


12.15 Theorem

For x 2 Sn, the conjugacy class xSn of x in Sn consists of all permuta-

tions in Sn which have the same cycle-shape as x.

12.16 Examples

(1) The conjugacy classes of S3 are

(2) The conjugacy class of (1 2)(3 4) in S4 consists of all the elements

of cycle-shape (2, 2) and is

f(1 2)(3 4), (1 3)(2 4), (1 4)(2 3)g:(3) There are precisely ®ve conjugacy classes of S4, with represent-

atives (see De®nition 12.4):

1, (1 2), (1 2 3), (1 2)(3 4), (1 2 3 4):

To calculate the sizes of the conjugacy classes, we simply count the

number of 2-cycles, 3-cycles, and so on. The number of 2-cycles is

equal to the number of pairs that can be chosen from {1, 2, 3, 4},

which is 42

ÿ � � 6. (The notation nr� � means the binomial coef®cient

n!=(r!(nÿ r)!).) The number of 3-cycles is 4 3 2 (4 for the choice of

®xed point and 2 because there are two 3-cycles ®xing a given point).

Similarly, there are three elements of cycle-shape (2, 2) and there are

six 4-cycles. Thus for G � S4, the conjugacy class representatives g,

the conjugacy class sizes |gG | and the centralizer orders |CG(g)|

(obtained using Theorem 12.8) are as follows:

Representative g 1 (1 2) (1 2 3) (1 2)(3 4) (1 2 3 4)Class size | gG | 1 6 8 3 6

|CG(g)| 24 4 3 8 4

We check our arithmetic by noting that

jS4j � 1� 6� 8� 3� 6:

Class Cycle-shape

{1} (1){(1 2), (1 3), (2 3)} (2)

{(1 2 3), (1 3 2)} (3)


(4) Similarly, the corresponding table for G � S5 is

Rep. g 1 (1 2) (1 2 3) (1 2)(3 4) (1 2 3 4) (1 2 3)(4 5) (1 2 3 4 5)

|gG | 1 10 20 15 30 20 24

|CG(g)| 120 12 6 8 4 6 5

Conjugacy classes of An

Given an even permutation x 2 An, we have seen in Theorem 12.15

that the conjugacy class xSn consists of all permutations in Sn which

have the same cycle-shape as x. The conjugacy class x An of x in An,

given by

x An � fgÿ1xg: g 2 Ang,is of course contained in xS n ; however, x An might not be equal to xSn .

For an easy example where equality does not hold, consider x �(1 2 3) 2 A3; here x A3 � fxg, while xS3 � {x, xÿ1}.

The next result determines precisely when xAn and xSn are equal, and

what happens when equality fails.

12.17 Proposition

Let x 2 An with n . 1.

(1) If x commutes with some odd permutation in Sn, then xSn � xAn .

(2) If x does not commute with any odd permutation in Sn then xSn

splits into two conjugacy classes in An of equal size, with represent-

atives x and (1 2)ÿ1x(1 2).

Proof (1) Assume that x commutes with an odd permutation g. Let

y 2 xSn , so that y � hÿ1xh for some h 2 Sn. If h is even then y 2 xAn ;

and if h is odd then gh 2 An and

y � hÿ1xh � hÿ1 gÿ1xgh � (gh)ÿ1x(gh),

so again y 2 xAn . Thus xSn # xAn , and so xSn � xAn .

(2) Assume that x does not commute with any odd permutation.

Then

CSn(x) � CAn

(x):


Hence by Theorem 12.8,

jxAn j � jAn: CAn(x)j � 1

2jSn: CAn

(x)j (as jAnj � 12jSnj)

� 12jSn: CSn

(x)j � 12jxSn j:

Next, we observe that

fhÿ1xh: h is oddg � ((1 2)ÿ1x(1 2))An

since every odd permutation has the form (1 2)a for some a 2 An. Now

xSn � fhÿ1xh: h is eveng [ fhÿ1xh: h is oddg� xAn [ ((1 2)ÿ1x(1 2))An :

Since |xAn | � 12|xSn |, the conjugacy classes xAn and ((1 2)ÿ1x(1 2))An must

be disjoint and of equal size, as we wished to show. j

12.18 Examples

(1) We ®nd the conjugacy classes of A4. The elements of A4 are the

identity, together with the permutations of cycle-shapes (2, 2) and (3).

Since (1 2)(3 4) commutes with the odd permutation (1 2), Proposition

12.17 implies that

(1 2)(3 4)A4 � (1 2)(3 4)S4 � f(1 2)(3 4), (1 3)(2 4), (1 4)(2 3)g:However, the 3-cycle (1 2 3) commutes with no odd permutation: for if

gÿ1(1 2 3)g � (1 2 3) then (1 2 3) � (1g 2g 3g) by Proposition 12.13,

so g is 1, (1 2 3) or (1 3 2), an even permutation. Hence by Proposition

12.17, (1 2 3)S4 splits into two conjugacy classes in A4 of size 4, with

representatives (1 2 3) and (1 2)ÿ1(1 2 3)(1 2) � (1 3 2).

Thus the conjugacy classes of A4 are

Representative 1 (1 2)(3 4) (1 2 3) (1 3 2)Class size 1 3 4 4Centralizer order 12 4 3 3

(2) We ®nd the conjugacy classes of A5. The non-identity even

permutations in S5 are those of cycle-shapes (3), (2, 2) and (5). The

elements (1 2 3) and (2 3)(4 5) commute with the odd permutation

(4 5); but (1 2 3 4 5) commutes with no odd permutation. (Check this

by using the argument in (1) above.) Hence by Proposition 12.17, the


conjugacy classes of A5 are represented by 1, (1 2 3), (1 2)(3 4),

(1 2 3 4 5) and (1 2)ÿ1(1 2 3 4 5)(1 2) � (1 3 4 5 2). Using Proposition

12.17(2), we see that the class sizes and centralizer orders are as

follows:

Representative 1 (1 2 3) (1 2)(3 4) (1 2 3 4 5) (1 3 4 5 2)Class size 1 20 15 12 12Centralizer order 60 3 4 5 5

Normal subgroups

Normal subgroups are related to conjugacy classes by the following

elementary result.

12.19 Proposition

Let H be a subgroup of G. Then H v G if and only if H is a union of

conjugacy classes of G.

Proof If H is a union of conjugacy classes, then

h 2 H , g 2 G) gÿ1 hg 2 H ,

so gÿ1 Hg # H. Thus H v G.

Conversely, if H v G then for all h 2 H, g 2 G, we have

gÿ1 hg 2 H, and so hG # H. Therefore

H �[h2H

hG,

and so H is a union of conjugacy classes of G. j

12.20 Example

We ®nd all the normal subgroups of S4. Let H v S4. Then by

Proposition 12.19, H is a union of conjugacy classes of S4. As we saw

in Example 12.16(3), these conjugacy classes have sizes 1, 6, 8, 3, 6.

Since jH j divides 24 by Lagrange's Theorem, and 1 2 H, there are just

four possibilities:

jH j � 1, 1� 3, 1� 8� 3 or 1� 6� 8� 3� 6:


In the ®rst case H � {1}, in the last case H � S4, and in the third

case H � A4. In the case where jH j � 1 � 3, we have

H � 1S4 [ (1 2)(3 4)S4 � f1, (1 2)(3 4), (1 3)(2 4), (1 4)(2 3)g:This is easily checked to be a subgroup of S4; we write it as V4

(V stands for `Viergruppe', meaning `four-group').

We have now shown that S4 has exactly four normal subgroups:

f1g, S4, A4 and V4 � f1, (1 2)(3 4), (1 3)(2 4), (1 4)(2 3)g:

The centre of a group algebra

In this ®nal section we link the conjugacy classes of the group G to

the centre of the group algebra CG. Recall from De®nition 9.12 that

the centre of CG is

Z(CG) � fz 2 CG: zr � rz for all r 2 CGg:We know that Z(CG) is a subspace of the vector space CG. There is a

convenient basis for this subspace which can be described in terms of

the conjugacy classes of G.

12.21 De®nition

Let C1, . . . , Cl be the distinct conjugacy classes of G. For 1 < i < l,

de®ne

Ci �Xg2Ci

g 2 CG:

The elements C1, . . . , Cl of CG are called class sums.

12.22 Proposition

The class sums C1, . . . , Cl form a basis of Z(CG).

Proof First we show that each Ci belongs to Z(CG). Let Ci consist of

the r distinct conjugates yÿ11 gy1, . . . , yÿ1

r gyr of an element g, so

Ci �Xr

j�1

yÿ1j gyj:

For all h 2 G,

hÿ1Ci h �Xr

j�1

hÿ1 yÿ1j gyjh:


As j runs from 1 to r, the elements hÿ1 yÿ1j gyjh run through Ci, since

hÿ1 yÿ1j gyjh � hÿ1 yÿ1

k gykh, yÿ1j gyj � yÿ1

k gyk :

Hence Xr

j�1

hÿ1 yÿ1j gyjh � Ci,

and so hÿ1Ci h � Ci. That is,

Ci h � hCi:

Therefore each Ci commutes with all h 2 G, hence with allPh2G ëh h 2 CG, and so Ci 2 Z(CG).

Next, observe that C1, . . . , Cl are linearly independent: for ifPli�1 ëiCi � 0 (ëi 2 C), then all ëi � 0 as the classes C1, . . . , Cl are

pairwise disjoint by Corollary 12.3.

It remains to show that C1, . . . , Cl span Z(CG). Let r �Pg2G ë gg 2 Z(CG). For h 2 G, we have rh � hr, so hÿ1 rh � r. That

is, Xg2G

ë g hÿ1 gh �Xg2G

ë g g:

So for every h 2 G, the coef®cient ë g of g is equal to the coef®cient

ëhÿ1 gh of hÿ1 gh. That is to say, the function g! ë g is constant on

conjugacy classes of G. It follows that r �P li�1 ëiCi where ëi is the

coef®cient ë gifor some gi 2 Ci. This completes the proof. j

12.23 Examples

(1) From Example 12.16(1), a basis for Z(CS3) is

1, (1 2)� (1 3)� (2 3), (1 2 3)� (1 3 2):

(2) From (12.12), a basis for Z(CD8) is

1, a2, a� a3, b� a2b, ab� a3b:


1. Every group is a union of conjugacy classes, and distinct conjugacy

classes are disjoint.

2. For an element x of a group G, the centralizer CG(x) is the set of


elements of G which commute with x. It is a subgroup of G, and

the number of elements in the conjugacy class xG is equal to

|G:CG(x)|.

3. The conjugacy classes of Sn correspond to the cycle-shapes of

permutations in Sn.

4. If x 2 An then xSn � xAn if and only if x commutes with some odd

permutation in Sn.

5. The class sums in CG form a basis for the centre of CG.


1. If G is a group and x 2 G, show that CG(x) is a subgroup of G

which contains Z(G).

2. Let G be a ®nite group and suppose that g 2 G and z 2 Z(G). Prove

that the conjugacy classes gG and (gz)G have the same size.

3. Let G � Sn.

(a) Prove that j(1 2)Gj � n2� � and ®nd CG((1 2)). Verify that your

solution satis®es Theorem 12.8.

(b) Show that j(1 2 3)Gj � 2 n3� � and j(1 2)(3 4)Gj � 3 n

4� �.(c) Now let n � 6. Show that

j(1 2 3)(4 5 6)Gj � 40 and j(1 2)(3 4)(5 6)Gj � 15,

and ®nd the sizes of the other conjugacy classes of S6. (There

are 11 conjugacy classes in all.)

4. What are the cycle-shapes of those permutations x 2 A6 for which

x A6 6� xS6 ?

5. Show that A5 is a simple group. (Hint: use the method of Example

12.20.)

6. Find the conjugacy classes of the quaternion group Q8. Give a basis

of the centre of the group algebra CQ8.

7. Let p be a prime number, and let n be a positive integer. Suppose

that G is a group of order pn.

(a) Use the Class Equation 12.10 to show that Z(G) 6� {1}.

(b) Suppose that n > 3 and that |Z(G)| � p. Prove that G has a

conjugacy class of size p.


117

13

Characters

Suppose that r: G! GL(n, C) is a representation of the ®nite group

G. With each n 3 n matrix gr (g 2 G) we associate the complex

number given by adding all the diagonal entries of the matrix, and call

this number ÷(g). The function ÷: G! C is called the character of the

representation r. Characters of representations have many remarkable

properties, and they are the fundamental tools for performing calcula-

tions in representation theory. For example, we shall show later that

two representations have the same character if and only if they are

equivalent. Moreover, basic problems, such as deciding whether or not

a given representation is irreducible, can be resolved by doing some

easy arithmetic with the character of the representation. These facts are

surprising, since from the de®nition of a representation r: G!GL(n, C), it appears that we must keep track of all the n2 entries in

each matrix gr, whereas the character records just one number for

each matrix.

The theory of characters will occupy a considerable portion of the

rest of the book. In this chapter we present some basic properties and

examples.

The trace of a matrix

13.1 De®nition

If A � (aij) is an n 3 n matrix, then the trace of A, written tr A, is

given by

tr A �Xn

i�1

aii:

That is, the trace of A is the sum of the diagonal entries of A.

13.2 Proposition

Let A � (aij) and B � (bij) be n 3 n matrices. Then

tr (A� B) � tr A� tr B, and

tr (AB) � tr (BA):

Moreover, if T is an invertible n 3 n matrix, then

tr (Tÿ1 AT ) � tr A:

Proof The ii-entry of A � B is aii � bii, and the ii-entry of AB isPnj�1 aijbji. Therefore

tr (A� B) �Xn

i�1

(aii � bii) �Xn

i�1

aii �Xn

i�1

bii � tr A� tr B,

and

tr (AB) �Xn

i�1

Xn

j�1

aijbji �Xn

j�1

Xn

i�1

bjiaij � tr (BA):

For the last part,

tr (Tÿ1 AT ) � tr ((Tÿ1 A)T )

� tr (T (Tÿ1 A)) (by the second part )

� tr A: j

Notice that, unlike the determinant function, the trace function is not

multiplicative; that is, tr (AB) need not equal (tr A)(tr B).

Characters

13.3 De®nition

Suppose that V is a CG-module with a basis B . Then the character of

V is the function ÷: G! C de®ned by

÷(g) � tr [g]B (g 2 G):

The character of V does not depend on the basis B , since if B and

B 9 are bases of V, then

[g]B 9 � Tÿ1[g]B T


for some invertible matrix T (see (2.24)), and so by Proposition 13.2,

tr [g]B 9 � tr [g]B for all g 2 G:

Naturally enough, we de®ne the character of a representation

r: G! GL(n, C) to be the character ÷ of the corresponding CG-

module Cn, namely

÷(g) � tr (gr) (g 2 G):

13.4 De®nition

We say that ÷ is a character of G if ÷ is the character of some CG-

module. Further, ÷ is an irreducible character of G if ÷ is the character

of an irreducible CG-module; and ÷ is reducible if it is the character

of a reducible CG-module.

You will have noticed that we are writing characters as functions on

the left. That is, we write ÷(g) and not g÷.

13.5 Proposition

(1) Isomorphic CG-modules have the same character.

(2) If x and y are conjugate elements of the group G, then

÷(x) � ÷(y)

for all characters ÷ of G.

Proof (1) Suppose that V and W are isomorphic CG-modules. Then by

(7.7), there are a basis B 1 of V and a basis B 2 of W such that

[g]B 1� [g]B 2

for all g 2 G:

Consequently tr [g]B 1� tr [g]B 2

for all g 2 G, and so V and W have

the same character.

(2) Assume that x and y are conjugate elements of G, so that

x � gÿ1 yg for some g 2 G. Let V be a CG-module, and let B be a

basis of V. Then

[x]B � [gÿ1 yg]B � [g]ÿ1B [y]B [g]B :

Hence by Proposition 13.2, we have tr [x]B � tr [y]B . Therefore ÷(x) �÷(y), where ÷ is the character of V. j

The result corresponding to Proposition 13.5(1) for representations is

that equivalent representations have the same character.

Characters 119

Later, we shall prove an astonishing converse of Proposition 13.5(1):

if two CG-modules have the same character, then they are isomorphic.

13.6 Examples

(1) Let G � D8 � ka, b: a4 � b2 � 1, bÿ1ab � aÿ1l, and let r: G!GL(2, C) be the representation for which

ar � 0 1

ÿ1 0

� �, br � 1 0

0 ÿ1

� �(see Example 3.2(1)). Let ÷ be the character of this representation. The

following table records g, gr and ÷(g) as g runs through G. (We

obtain ÷(g) by adding the two entries on the diagonal of gr.)

g 1 a a2 a3

gr1 0

0 1

� �0 1

ÿ1 0

� � ÿ1 0

0 ÿ1

� �0 ÿ1

1 0

� �÷(g) 2 0 ÿ2 0

g b ab a2b a3b

gr1 0

0 ÿ1

� �0 ÿ1

ÿ1 0

� � ÿ1 0

0 1

� �0 1

1 0

� �÷(g) 0 0 0 0

(2) Let G �S3, and take V to be the 3-dimensional permutation module

for G over C (see De®nition 4.10). Let B be the natural basis of V;

thus B is the basis v1, v2, v3, where vig � vig for 1 < i < 3 and all

g 2 G. The matrices [g]B (g 2 G) are given by Exercise 4.1. We

record these matrices, together with the character ÷ of V.

g 1 (1 2) (1 3)

[g]B

1 0 0

0 1 0

0 0 1

0@ 1A 0 1 0

1 0 0

0 0 1

0@ 1A 0 0 1

0 1 0

1 0 0

0@ 1A÷(g) 3 1 1


g (2 3) (1 2 3) (1 3 2)

[g]B

1 0 0

0 0 1

0 1 0

0@ 1A 0 1 0

0 0 1

1 0 0

0@ 1A 0 0 1

1 0 0

0 1 0

0@ 1A÷(g) 1 0 0

(3) Let G � C3 � ha: a3 � 1 i. By Theorem 9.8, G has just three

irreducible characters ÷1, ÷2, ÷3, with values

g 1 a a2

÷1(g) 1 1 1÷2(g) 1 ù ù2

÷3(g) 1 ù2 ù

where ù � e2ði=3.

(4) Let G � D6 � ka, b: a3 � b2 � 1, bÿ1ab � aÿ1l (so G � S3). In

Example 10.8(2), we found a complete set of non-isomorphic irreduci-

ble CG-modules U1, U2, U3. Thus if ÷i is the character of Ui for

1 < i < 3, then the irreducible characters of G are ÷1, ÷2 and ÷3. The

values of these characters on the elements of G can be calculated from

the corresponding representations r1, r2, r3 given in Example 10.8(2),

and they are as follows:

g 1 a a2 b ab a2b

÷1(g) 1 1 1 1 1 1÷2(g) 1 1 1 ÿ1 ÿ1 ÿ1÷3(g) 2 ÿ1 ÿ1 0 0 0

Notice that in all the above examples, the characters given take few

distinct values. This re¯ects the fact that by Proposition 13.5(2), every

character is constant on conjugacy classes of G. Moreover, it is much

quicker to write down the single complex number ÷(g) for the group

element g than to record the matrix which corresponds to g. Never-

theless, the character encapsulates a great deal of information about

the representation. This will become clear as the theory of characters

develops.

Characters 121

13.7 De®nition

If ÷ is the character of the CG-module V, then the dimension of V is

called the degree of ÷.

13.8 Examples

(1) In Example 13.6(1) we gave a character of D8 of degree 2; in

13.6(2) we gave a character of S3 of degree 3; and in 13.6(4) we saw

that the irreducible characters of D6 have degrees 1, 1 and 2.

(2) If V is any 1-dimensional CG-module, then for each g 2 G there is

a complex number ë g such that

vg � ë gv for all v 2 V :

The character ÷ of V is given by

÷(g) � ë g (g 2 G)

and ÷ has degree 1. Characters of degree 1 are called linear characters;

they are, of course, irreducible characters.

Observe that Theorem 9.8 gives all the irreducible characters of

®nite abelian groups; in particular, they are all linear characters.

Every linear character of G is a homomorphism from G to the

multiplicative group of non-zero complex numbers. In fact, these are

the only non-zero characters of G which are homomorphisms (see

Exercise 13.4).

(3) The character of the trivial CG-module (see De®nition 4.8(1)) is a

linear character, called the trivial character of G. We denote it by 1G.

Thus1G: g ! 1 for all g 2 G:

Given any group G, we therefore know at least one of the irreducible

characters of G, namely the trivial character. Finding all the irreducible

characters is usually dif®cult.

The values of a character

The next result gives information about the complex numbers ÷(g),

where ÷ is a character of G and g 2 G.

13.9 Proposition

Let ÷ be the character of a CG-module V. Suppose that g 2 G and g

has order m. Then


(1) ÷(1) � dim V;

(2) ÷(g) is a sum of mth roots of unity;

(3) ÷(gÿ1) � ÷(g);

(4) ÷(g) is a real number if g is conjugate to gÿ1.

Proof (1) Let n � dim V, and let B be a basis of V. Then the matrix

[1]B of the identity element 1 relative to B is equal to In, the n 3 n

identity matrix. Consequently

÷(1) � tr [1]B � tr In � n,

and so ÷(1) � dim V.

(2) By Proposition 9.11 there is a basis B of V such that

[g]B �ù1

. ..

ùn

0B@1CA

where each ùi is an mth root of unity. Therefore

÷(g) � ù1 � : : :� ùn,

a sum of mth roots of unity.

(3) We have

[gÿ1]B �ùÿ1

1

. ..

ùÿ1n

0B@1CA

and so ÷(gÿ1) � ùÿ11 � . . . � ùÿ1

n . Every complex mth root of unity ùsatis®es ùÿ1 � ù, since for all real W,

(eiW)ÿ1 � eÿiW,

which is the complex conjugate of eiW. Therefore

÷(gÿ1) � ù1 � : : :� ùn � ÷(g):

(4) If g is conjugate to gÿ1 then ÷(g) � ÷(gÿ1) by Proposition

13.5(2). Also ÷(gÿ1) � ÷(g) by (3), and so ÷(g) � ÷(g); that is, ÷(g) is

real. j

When the element g of G has order 2, we can be much more

speci®c about the possibilities for ÷(g):

0

0

0

0

Characters 123

13.10 Corollary

Let ÷ be a character of G, and let g be an element of order 2 in G.

Then ÷(g) is an integer, and

÷(g) � ÷(1) mod 2:

Proof By Proposition 13.9, we have

÷(g) � ù1 � : : :� ùn,

where n � ÷(1) and each ùi is a square root of unity. Then each ùi is

�1 or ÿ1. Suppose r of them are �1, and s are ÿ1, so that

÷(g) � r ÿ s, and ÷(1) � r � s:

Certainly then, ÷(g) 2 Z, and since r ÿ s � r � s ÿ 2s � r � s mod 2,

we have ÷(g) � ÷(1) mod 2. j

Our next result gives the ®rst inkling of the importance of characters,

showing that we can determine the kernel of a representation just from

knowledge of its character.

13.11 Theorem

Let r: G! GL(n, C) be a representation of G, and let ÷ be the

character of r.

(1) For g 2 G,

j÷(g)j � ÷(1), gr � ëIn for some ë 2 C:

(2) Ker r � {g 2 G: ÷(g) � ÷(1)}.

Proof (1) Let g 2 G, and suppose that g has order m. If gr � ëIn with

ë 2 C, then ë is an mth root of unity, and ÷(g) � në, so

|÷(g)| � n � ÷(1).

Conversely, suppose that |÷(g)| � ÷(1). By Proposition 9.11, there is

a basis B of Cn such that

[g]B �ù1

. ..

ùn

0B@1CA

where each ùi is an mth root of unity. Then

j÷(g)j � jù1 � : : :� ùnj � ÷(1) � n:(13:12)

0

0


Note now that for any complex numbers z1, . . . , zn, we have

jz1 � : : :� znj < jz1j � : : :� jznj,with equality if and only if the arguments of z1, . . . , zn are all equal.

(To see this, consider the picture

in the Argand diagram.) Since |ùi| � 1 for all i, we deduce from

(13.12) that ùi � ù j for all i, j. Thus

[g]B �ù1

. ..

ù1

0B@1CA � ù1 In:

Hence for all bases B 9 of Cn we have [g]B 9 � ù1 In, and so

gr � ù1 In. This completes the proof of (1).

(2) If g 2 Ker r then gr � In, and so ÷(g) � n � ÷(1).

Conversely, suppose that ÷(g) � ÷(1). Then by (1), we have gr � ëIn

for some ë 2 C. This implies that ÷(g) � ë÷(1), whence ë � 1. There-

fore gr � In, and so g 2 Ker r. Part (2) follows. j

Motivated by Theorem 13.11(2), we de®ne the kernel of a character

as follows.

13.13 De®nition

If ÷ is a character of G, then the kernel of ÷, written Ker ÷, is de®ned

by

Ker ÷ � fg 2 G: ÷(g) � ÷(1)g:By Theorem 13.11(2), if r is a representation of G with character ÷,

then Ker r � Ker ÷. In particular, Ker ÷ v G. We call ÷ a faithful

character if Ker ÷ � {1}.

13.14 Examples

(1) According to Example 13.6(4), the irreducible characters of the

group G � D6 � ka, b: a3 � b2 � 1, bÿ1ab � aÿ1l are ÷1, ÷2, ÷3, with

the following values:

0

0

Characters 125

g 1 a a2 b ab a2b

÷1(g) 1 1 1 1 1 1÷2(g) 1 1 1 ÿ1 ÿ1 ÿ1÷3(g) 2 ÿ1 ÿ1 0 0 0

Then Ker ÷1 � G, Ker ÷2 � kal and Ker ÷3 � {1}. In particular, ÷3 is a

faithful irreducible character of D6.

(2) Let G � D8 � ka, b: a4 � b2 � 1, bÿ1ab � aÿ1l, and let ÷ be the

character of G given in Example 13.6(1):

g 1 a a2 a3 b ab a2b a3b

÷(g) 2 0 ÿ2 0 0 0 0 0

Then Ker ÷ � {1}, so ÷ is a faithful character. And since |÷(a2)| �|ÿ2| � ÷(1), Theorem 13.11(1) implies that if r: G! GL(2, C) is a

representation with character ÷, then a2r � ÿI.

We next prove a result which is sometimes useful for constructing a

new character from a given one. For a character ÷ of G, de®ne

÷: G! C by

÷(g) � ÷(g) (g 2 G):

Thus the values of ÷ are the complex conjugates of the values of ÷.

13.15 Proposition

Let ÷ be a character of G. Then ÷ is a character of G. If ÷ is

irreducible, then so is ÷.

Proof Suppose that ÷ is the character of a representation r: G!GL(n, C). Thus

÷(g) � tr (gr) (g 2 G):

If A � (aij) is an n 3 n matrix over C, then we de®ne A to be the

n 3 n matrix (aij). Observe that if A � (aij) and B � (bij) are n 3 n

matrices over C, then

(AB) � A B,(13:16)


since the ij-entry of AB is Xn

k�1

aik bkj,

which is equal to the complex conjugate ofP

nk�1 aikbkj, the ij-entry of

AB.

It follows from (13.16) that the function r: G! GL(n, C) de®ned by

gr � (gr) (g 2 G)

is a representation of G. Since

tr (gr) � tr (gr) � tr (gr) � ÷(g) (g 2 G),

the character of the representation r is ÷.

It is clear that if r is reducible then r is reducible. Hence ÷ is

irreducible if and only if ÷ is irreducible. j

The regular character

13.17 De®nition

The regular character of G is the character of the regular CG-module.

We write the regular character as ÷reg.

In Theorem 13.19, we shall express the regular character in terms of

the irreducible characters of G. First we need a preliminary result.

13.18 Proposition

Let V be a CG-module, and suppose that

V � U1 � : : :� Ur,

a direct sum of irreducible CG-modules Ui. Then the character of V is

equal to the sum of the characters of the CG-modules U1, . . . , Ur.

Proof This is immediate from (7.10). j

13.19 Theorem

Let V1, : : : , Vk be a complete set of non-isomorphic irreducible

CG-modules (see De®nition 11.11), and for i � 1, . . . , k let ÷i be the

character of Vi and di � ÷i(1). Then

÷reg � d1÷1 � : : :� dk÷k :

Characters 127

Proof By Theorem 11.9,

CG � (V1 � : : :� V1)� (V2 � : : :� V2)� : : :� (Vk � : : :� Vk),

where for each i there are di factors Vi. Now the result follows from

Proposition 13.18. j

The values of ÷reg on the elements of G are easily described, and are

given in the next result.

13.20 Proposition

If ÷reg is the regular character of G, then

÷reg(1) � jGj, and

÷reg(g) � 0 if g 6� 1:

Proof Let g1, . . . , gn be the elements of G, and let B be the basis

g1, . . . , gn of CG. By Proposition 13.9(1), ÷reg(1) � dim CG � |G|.

Now let g 2 G with g 6� 1. Then for 1 < i < n, we have gig � gj

for some j with j 6� i. Therefore the ith row of the matrix [g]B has

zeros in every place except column j; in particular, the ii-entry is zero

for all i. It follows that

÷reg(g) � tr [g]B � 0: j

13.21 Example

We illustrate Theorem 13.19 and Proposition 13.20 for the group

G � D6. By Example 13.6(4), the irreducible characters of G are ÷1,

÷2, ÷3:

g 1 a a2 b ab a2b

÷1(g) 1 1 1 1 1 1÷2(g) 1 1 1 ÿ1 ÿ1 ÿ1÷3(g) 2 ÿ1 ÿ1 0 0 0

We calculate ÷1 � ÷2 � 2÷3:

(÷1 � ÷2 � 2÷3)(g) 6 0 0 0 0 0


This is the regular character of G, by Theorem 13.19; and it takes the

value |G| on 1, and the value 0 on all non-identity elements of G,

illustrating Proposition 13.20.

Permutation characters

In the case where G is a subgroup of the symmetric group Sn, there is

an easy construction using the permutation module which produces a

character of degree n, and we now describe this.

Suppose that G is a subgroup of Sn, so that G is a group of

permutations of {1, . . . , n}. The permutation module V for G over C

has basis v1, . . . , vn, where for all g 2 G,

vi g � vig (1 < i < n)

(see De®nition 4.10). Let B denote the basis v1, . . . , vn. Then the ii-

entry in the matrix [g]B is 0 if ig 6� i, and is 1 if ig � i. Therefore

the character ð of the permutation module V is given by

ð(g) � (the number of i such that ig � i):

For g 2 G, let

fix (g) � fi: 1 < i < n and ig � ig:Then

ð(g) � jfix(g)j (g 2 G):(13:22)

We call ð the permutation character of G.

13.23 Example

Let G � S4. Then by Example 12.16(3), G has ®ve conjugacy classes,

with representatives

1, (1 2), (1 2 3), (1 2)(3 4), (1 2 3 4):

The permutation character ð takes the values

gi 1 (1 2) (1 2 3) (1 2)(3 4) (1 2 3 4)

ð(gi) 4 2 1 0 0

Characters 129

13.24 Proposition

Let G be a subgroup of Sn. Then the function í: G! C de®ned by

í(g) � jfix (g)j ÿ 1 (g 2 G)

is a character of G.

Proof Let v1, . . . , vn be the basis of the permutation module V as

above, and let

u � v1 � : : :� vn, and U � sp (u):

Observe that ug � u for all g 2 G, so U is a CG-submodule of V.

Indeed, U is isomorphic to the trivial CG-module, so the character of

U is the trivial character 1G (see Example 13.8(3)). By Maschke's

Theorem 8.1, there is a CG-submodule W of V such that

V � U � W :

Let í be the character of W. Then

ð � 1G � í,

so |®x( g)| � 1 � í(g) for all g 2 G, and therefore

í(g) � jfix(g)j ÿ 1 (g 2 G): j

13.25 Example

Let G � A4, a subgroup of S4. By Example 12.18(1), the conjugacy

classes of G are represented by

1, (1 2)(3 4), (1 2 3), (1 3 2):

The values of the character í of G are

gi 1 (1 2)(3 4) (1 2 3) (1 3 2)

í(gi) 3 ÿ1 0 0


1. A character is obtained from a representation by taking the trace of

each matrix.

2. Characters are constant on conjugacy classes.


3. Isomorphic CG-modules have the same character.

4. For all characters ÷ of G, and all g 2 G, the complex number ÷(g)

is a sum of roots of unity, and ÷(gÿ1) � ÷(g).

5. The character of a representation determines the kernel of the

representation.

6. The regular character ÷reg of G takes the value |G| on the identity

and the value 0 on all other elements of G.

7. If G is a subgroup of Sn, then the function í which is given by

í(g) � jfix(g)j ÿ 1 (g 2 G)



1. Let G � D12 � ka, b: a6 � b2 � 1, bÿ1ab � aÿ1l, and let r1, r2 be

the representations of G for which

ar1 � ù 0

0 ùÿ1

� �, br1 � 0 1

1 0

� �(where ù � e2ði=3)

and

ar2 � ÿ1 0

0 1

� �, br2 � 1 0

0 ÿ1

� �:

Find the characters of r1 and r2. Find also Ker r1 and Ker r2;

check that your answers are consistent with Theorem 13.11.

2. Find all the irreducible characters of C4. Write the regular char-

acter of C4 as a linear combination of these.

3. Let ÷ be the character of the 7-dimensional permutation module

for S7. Find ÷(x) for x � (1 2) and for x � (1 6)(2 3 5).

4. Prove that the only non-zero characters of G which are homo-

morphisms are the linear characters.

5. Assume that ÷ is an irreducible character of G. Suppose that

z 2 Z(G) and that z has order m. Prove that there exists an mth

root of unity ë 2 C such that for all g 2 G,

÷(zg) � ë÷(g):

6. Prove that if ÷ is a faithful irreducible character of the group G,

then Z(G) � {g 2 G: |÷(g)| � ÷(1)}.

Characters 131

7. Let r be a representation of the group G over C.

(a) Show that ä: g! det (gr) (g 2 G) is a linear character of G.

(b) Prove that G/Ker ä is abelian.

(c) Assume that ä(g) � ÿ1 for some g 2 G. Show that G has a

normal subgroup of index 2.

8. Let g be a group of order 2k, where k is an odd integer. By

considering the regular representation of G, show that G has a

normal subgroup of index 2.

9. Let ÷ be a character of a group G, and let g be an element of

order 2 in G. Show that either

(1) ÷(g) � ÷(1) mod 4, or

(2) G has a normal subgroup of index 2.

(Compare Corollary 13.10. Hint: use Exercise 7.)

10. Prove that if x is a non-identity element of the group G, then

÷(x) 6� ÷(1) for some irreducible character ÷ of G.


133

14

Inner products of characters

We establish some signi®cant properties of characters in this chapter,

and in particular we prove the striking result (Theorem 14.21) that if

two CG-modules have the same character then they are isomorphic.

Also, we describe a method for decomposing a given CG-module as a

direct sum of CG-submodules, using characters.

The proofs rely on an inner product involving the characters of a

group, and we describe this ®rst.

Inner products

The characters of a ®nite group G are certain functions from G to C.

The set of all functions from G to C forms a vector space over C, if

we adopt the natural rules for adding functions and multiplying func-

tions by complex numbers. That is, if W, ö are functions from G to C,

and ë 2 C, then we de®ne W � ö: G! C by

(W� ö)(g) � W(g)� ö(g) (g 2 G)

and we de®ne ëW: G! C by

ëW(g) � ë(W(g)) (g 2 G):

(We write these functions on the left to agree with our notation for

characters.)

14.1 Example

Let G � C3 � ka: a3 � 1l, and suppose that W: G! C and ö: G! C

are given by

This means that W(1) � 2, W(a) � i, W(a2) � ÿ1 and ö(1) � ö(a) �ö(a2) � 1. Then W � ö and 3W are given by

We shall often think of functions from G to C as row vectors, as in

this example.

The vector space of all functions from G to C can be equipped with

an inner product in a way which we shall describe shortly. The

de®nition of an inner product on a vector space over C runs as

follows. With every ordered pair of vectors W, ö in the vector space,

there is associated a complex number kW, öl which satis®es the follow-

ing conditions:

(14.2) (a) kW, öl � hö, Wi for all W, ö;

(b) kë1W1 � ë2W2, öl � ë1kW1, öl � ë2kW2, öl for all ë1, ë2 2 C

and all vectors W1, W2, ö;

(c) kW, Wl . 0 if W 6� 0.

Notice that condition (a) implies that kW, Wl is always real, and that

conditions (a) and (b) give

hö, ë1è1 � ë2W2i � ë1hö, W1i � ë2hö, W2ifor all ë1, ë2 2 C and all vectors W1, W2, ö.

We now introduce an inner product on the vector space of all

functions from G to C. This will be of basic importance in our study

of characters.

14.3 De®nition

Suppose that W and ö are functions from G to C. De®ne

hW, öi � 1

jGjXg2G

W(g)ö(g):

1 a a2

W � ö 3 1 � i 03W 6 3i ÿ3

1 a a2

W 2 i ÿ1ö 1 1 1


It is transparent that the conditions of (14.2) hold, so k , l is an inner

product on the vector space of functions from G to C.

14.4 Example

As in Example 14.1, suppose that G � C3 � ka: a3 � 1l and that W and

ö are given by

Then

hè, öi � 13(2 . 1� i . 1ÿ 1 . 1) � 1

3(1� i),

hè, èi � 13(2 . 2� i . i� (ÿ1) . (ÿ1)) � 2,

hö, öi � 13(1 . 1� 1 . 1� 1 . 1) � 1:

Inner products of characters

We can exploit the fact that characters are constant on conjugacy

classes to simplify slightly the calculation of the inner product of two

characters.

14.5 Proposition

Assume that G has exactly l conjugacy classes, with representatives

g1, . . . , gl. Let ÷ and ø be characters of G.

(1) h÷, øi � hø, ÷i � 1

jGjXg2G

÷(g)ø(gÿ1), and this is a real number:

(2) h÷, øi �Xl

i�1

÷(gi)ø(gi)

jCG(gi)j :

Proof (1) We have ø(g) � ø(gÿ1) for all g 2 G, by Proposition

13.9(3). Therefore

h÷, øi � 1

jGjXg2G

÷(g)ø(gÿ1):

1 a a2

W 2 i ÿ1ö 1 1 1

Inner products of characters 135

Since {gÿ1: g 2 G} � G, we also have

h÷, øi � 1

jGjXg2G

÷(gÿ1)ø(g) � hø, ÷i:

Since kø, ÷l � h÷, øi, it follows that h÷, øi is real. (We shall prove

later that h÷, øi is, in fact, an integer.)

(2) Recall that gGi denotes the conjugacy class of G which contains

gi. Since characters are constant on conjugacy classes,Xg2 gG

i

÷(g)ø(g) � jgGi j÷(gi)ø(gi):

Now

G �[l

i�1

gGi and jgG

i j � jGj=jCG(gi)j,

by Corollary 12.3 and Theorem 12.8. Hence

h÷, øi � 1

jGjXg2G

÷(g)ø(g) � 1

jGjXl

i�1

Xg2 gG

i

÷(g)ø(g)

�Xl

i�1

jgGi jjGj ÷(gi)ø(gi)

�Xl

i�1

1

jCG(gi)j ÷(gi)ø(gi): j

14.6 Example

The alternating group A4 has four conjugacy classes, with represent-

atives

g1 � 1, g2 � (1 2)(3 4), g3 � (1 2 3), g4 � (1 3 2)

(see Example 12.18(1)). We shall see in Chapter 18 that there are

characters ÷ and ø of A4 which take the following values on the

representatives gi:

gi g1 g2 g3 g4

|CG(gi)| 12 4 3 3

÷ 1 1 ù ù2

ø 4 0 ù2 ù


(where ù � e2ði=3). Using part (2) of Proposition 14.5, we have

h÷, øi � 1 . 4

12� 1 . 0

4� ù . ù2

3� ù2 . ù

3� 0,

hø, øi � 4 . 4

12� 0 . 0

4� ù2 . ù2

3� ù . ù

3� 2:

We advise you to check also that k÷, ÷l � 1, and to ®nd the inner

products of ÷ and ø with the trivial character (which takes the value 1

on all elements of A4).

We are now going to pave the way to proving the key fact (Theorem

14.12) that the irreducible characters of G form an orthonormal set of

vectors in the vector space of functions from G to C; that is, for

distinct irreducible characters ÷ and ø of G, we have h÷, ÷i � 1 and

h÷, øi � 0.

Recall from Chapter 10 that the regular CG-module is a direct sum

of irreducible CG-submodules, say

CG � U1 � : : :� Ur,

and that every irreducible CG-module is isomorphic to one of the CG-

modules U1, . . . , Ur. There are several ways of choosing CG-sub-

modules W1 and W2 of CG such that CG � W1 � W2 and W1 and W2

have no common composition factor (see De®nition 10.4). For example,

we may take W1 to be the sum of those irreducible CG-submodules Ui

which are isomorphic to a given irreducible CG-module, and then let

W2 be the sum of the remaining CG-modules Ui. We shall investigate

some consequences of writing CG like this; therefore, we temporarily

adopt the following Hypothesis:

14.7 Hypothesis

Let CG � W1 � W2, where W1 and W2 are CG-submodules which have

no common composition factor. Write 1 � e1 � e2 where e1 2 W1 and

e2 2 W2.

Among other results, we shall derive a formula for e1 in terms of

the character of W1.

We ®rst look at the effect of applying the elements e1 and e2 of CG

to W1 and W2.


14.8 Proposition

For all w1 2 W1 and w2 2 W2, we have

w1e1 � w1, w2e1 � 0,

w1e2 � 0, w2e2 � w2:

Proof If w1 2 W1 then the function w2 ! w1w2 (w2 2 W2) is clearly a

CG-homomorphism from W2 to W1. But W2 and W1 have no common

composition factor, so every CG-homomorphism from W2 to W1 is

zero, by Proposition 11.3. Therefore w1w2 � 0 for all w1 2 W1,

w2 2 W2. Similarly w2w1 � 0. In particular, w1e2 � w2e1 � 0. Now

w1 � w11 � w1(e1 � e2) � w1e1, and

w2 � w21 � w2(e1 � e2) � w2e2,

and this completes the proof. j

14.9 Corollary

For the elements e1 and e2 of CG which appear in Hypothesis 14.7,

we have

e21 � e1, e2

2 � e2 and e1e2 � e2e1 � 0:

Proof In Proposition 14.8, take w1 � e1 and w2 � e2. j

Next, we evaluate e1.

14.10. Proposition

Let ÷ be the character of the CG-module W1 which appears in

Hypothesis 14.7. Then

e1 � 1

jGjXg2G

÷(gÿ1)g:

Proof Let x 2 G. The function

W: w! we1xÿ1 (w 2 CG)

is an endomorphism of CG. We shall calculate the trace of W in two

ways.


First, for w1 2 W1 and w2 2 W2 we have, in view of Proposition 14.8,

w1W � w1e1xÿ1 � w1xÿ1,

w2W � w2e1xÿ1 � 0:

Thus W acts on W1 by w1 ! w1xÿ1 and on W2 by w2 ! 0. By the

de®nition of the character ÷ of W1, the endomorphism w1 ! w1xÿ1 of

W1 has trace equal to ÷(xÿ1), and of course the endomorphism w2 ! 0

of W2 has trace 0. Therefore

tr W � ÷(xÿ1):

Secondly, e1 2 CG, so

e1 �Xg2G

ë g g

for some ë g 2 C. By Proposition 13.20, the endomorphism w! wgxÿ1

(w 2 CG) of CG has trace 0 if g 6� x and has trace |G| if g � x.

Hence, as W: w! wP

g2G ë g gxÿ1, we have

tr W � ëxjGj:Comparing our two expressions for tr W, we see that for all x 2 G,

ëx � ÷(xÿ1)=jGj:Therefore

e1 � 1

jGjXg2G

÷(gÿ1)g:j

14.11 Corollary

Let ÷ be the character of the CG-module W1 which appears in

Hypothesis 14.7. Then

h÷, ÷i � ÷(1):

Proof Using the de®nition 6.3 of the multiplication in CG, we deduce

from Proposition 14.10 that the coef®cient of 1 in e21 is

1

jGj2Xg2G

÷(gÿ1)÷(g) � 1

jGj h÷, ÷i:


On the other hand, we know from Corollary 14.9 that e21 � e1, and the

coef®cient of 1 in e1 is ÷(1)/|G|. Hence k÷, ÷l � ÷(1), as required.

j

We can now prove the main theorem concerning the inner product

k , l.

14.12 Theorem

Let U and V be non-isomorphic irreducible CG-modules, with char-

acters ÷ and ø, respectively. Then

h÷, ÷i � 1, and

h÷, øi � 0:

Proof Recall from Theorem 11.9 that CG is a direct sum of irreducible

CG-submodules, say

CG � U1 � : : :� Ur,

where the number of CG-submodules Ui which are isomorphic to U is

dim U. Let m � dim U, and de®ne W to be the sum of the m

irreducible CG-submodules Ui which are isomorphic to U; let X be the

sum of the remaining CG-submodules Ui. Then

CG � W � X :

Moreover, every composition factor of W is isomorphic to U, and no

composition factor of X is isomorphic to U. In particular, W and X

have no common composition factor. The character of W is m÷, since

W is the direct sum of m CG-submodules, each of which has

character ÷.

We now apply Corollary 14.11 to the character of W, and obtain

hm÷, m÷i � m÷(1):

As ÷(1) � dim U � m, this yields h÷, ÷i � 1.

Next, let Y be the sum of those CG-submodules Ui of CG which

are isomorphic to either U or V, and let Z be the sum of the remaining

CG-submodules Ui. Then

CG � Y � Z,


and Y and Z have no common composition factor. The character of Y

is m÷ � nø, where n � dim V. By Corollary 14.11,

m÷(1)� nø(1) � hm÷� nø, m÷� nøi� m2h÷, ÷i � n2hø, øi � mn(h÷, øi � hø, ÷i):

Now h÷, ÷i � hø, øi � 1, by the part of the theorem which we have

already proved; and ÷(1) � m, ø(1) � n. Therefore

h÷, øi � hø, ÷i � 0:

By Proposition 14.5(1), k÷, øl � kø, ÷l, and hence k÷, øl � 0. j

Applications of Theorem 14.12

Let G be a ®nite group, and let V1, : : : , Vk be a complete set of non-

isomorphic irreducible CG-modules (see De®nition 11.11). If ÷i is the

character of Vi (1 < i < k), then by Theorem 14.12, we have

h÷i, ÷ ji � äij for all i, j,(14:13)

where äij is the Kronecker delta function (that is, äij is 1 if i � j and

is 0 if i 6� j). In particular, this implies that the irreducible characters

÷1, . . . , ÷k are all distinct.

Now let V be a CG-module. By Theorem 8.7, V is equal to a direct

sum of irreducible CG-submodules. Each of these is isomorphic to

some Vi, so there are non-negative integers d1, . . . , dk such that

V � (V1 � : : :� V1)� (V2 � : : :� V2)� : : :(14:14)

� (Vk � : : :� Vk),

where for each i, there are di factors Vi:

Therefore the character ø of V is given by

ø � d1÷1 � : : :� dk÷k :(14:15)

Using (14.13), we obtain from this

hø, ÷ii � h÷i, øi � di for 1 < i < k, and(14:16)

hø, øi � d21 � : : :� d2

k :

Summarizing, we have


14.17 Theorem

Let ÷1, . . . , ÷k be the irreducible characters of G. If ø is any

character of G, thenø � d1÷1 � : : :� dk÷k

for some non-negative integers d1, . . . , dk . Moreover,

di � hø, ÷ii for 1 < i < k, and

hø, øi �Xk

i�1

d2i :

14.18 Example

Recall from Example 13.6(4) that the irreducible characters of S3 � D6

are ÷1, ÷2, ÷3, taking the following values on the conjugacy class

representatives 1, (1 2), (1 2 3):

Now let ø be the character of the 3-dimensional permutation module

for S3. By Example 13.6(2), we know that

ø(1) � 3, ø(1 2) � 1, ø(1 2 3) � 0:

Therefore, by Proposition 14.5(2),

hø, ÷1i � 3 . 1

6� 1 . 1

2� 0 � 1:

Similarly, kø, ÷2l � 0 and kø, ÷3l � 1. Thus by Theorem 14.17,

ø � ÷1 � ÷3:

(This can of course be checked immediately by comparing the values

of ø and ÷1 � ÷3 on each conjugacy class representative.)

A more substantial calculation along these lines is given in Example

15.7.

gi 1 (1 2) (1 2 3)|CS3

(gi)| 6 2 3

÷1 1 1 1÷2 1 ÿ1 1÷3 2 0 ÿ1


We shall see many more applications of the important Theorem

14.17.

14.19 De®nition

Suppose that ø is a character of G, and that ÷ is an irreducible

character of G. We say that ÷ is a constituent of ø if kø, ÷l 6� 0. Thus,

the constituents of ø are the irreducible characters ÷i of G for which

the integer di in the expression ø � d1÷1 � . . . � dk÷k is non-zero.

The next result is another signi®cant consequence of Theorem 14.12.

It gives us a quick and effective method of determining whether or not

a given CG-module is irreducible.

14.20 Theorem

Let V be a CG-module with character ø. Then V is irreducible if and

only if kø, øl � 1.

Proof If V is irreducible then kø, øl � 1 by Theorem 14.12.

Conversely, assume that kø, øl � 1. We have

ø � d1÷1 � : : :� dk÷k

for some non-negative integers di, and by (14.16),

1 � hø, øi � d21 � : : :� d2

k :

It follows that one of the integers di is 1 and the rest are zero. Then

by (14.14), V � Vi for some i, and so V is irreducible. j

We are now in a position to prove the remarkable result that `a CG-

module is determined by its character'. It is this fact which motivates

our study of characters in much of the rest of the book, for it means

that many questions about CG-modules can be answered using char-

acter theory.

14.21 Theorem

Suppose that V and W are CG-modules, with characters ÷ and ø,

respectively. Then V and W are isomorphic if and only if ÷ � ø.

Proof In Proposition 13.5 we proved the elementary fact that if V � W

then ÷ � ø. It is the converse which is the substantial part of this

theorem.


Thus, suppose that ÷ � ø. Again let V1, : : : , Vk be a complete set

of non-isomorphic irreducible CG-modules with characters ÷1, . . . , ÷k .

We know by (14.14) that there are non-negative integers ci, di

(1 < i < k) such that

V � (V1 � : : :� V1)� (V2 � : : :� V2)� : : :� (Vk � : : :� Vk)

with ci factors Vi for each i, and

W � (V1 � : : :� V1)� (V2 � : : :� V2)� : : :� (Vk � : : :� Vk)

with di factors Vi for each i. By (14.16),

ci � h÷, ÷ii, di � hø, ÷ii (1 < i < k):

Since ÷ � ø, it follows that ci � di for all i, and hence V � W. j

14.22 Example

Let G � C3 � ka: a3 � 1l, and let r1, r2, r3, r4 be the representations

of G over C for which

ar1 �ù 0

0 ù

!, ar2 �

ù 0

0 ùÿ1

!,

ar3 �0 1

ÿ1 ÿ1

!, ar4 �

1 ùÿ1

0 ù

!(ù � e2ði=3). The characters øi of the representations ri (i � 1, 2, 3, 4)

are

Hence by Theorem 14.21, the representations r2 and r3 are equivalent,

but there are no other equivalences among r1, r2, r3 and r4.

The next theorem is another consequence of Theorem 14.12.

1 a a2

ø1 2 2ù 2ù2

ø2 2 ÿ1 ÿ1ø3 2 ÿ1 ÿ1ø4 2 1 � ù 1 � ù2


14.23 Theorem

Let ÷1, . . . , ÷k be the irreducible characters of G. Then ÷1, . . . , ÷k

are linearly independent vectors in the vector space of all functions

from G to C.

Proof Assume that

ë1÷1 � : : :� ëk÷k � 0 (ëi 2 C):

Then for all i, using (14.13) we have

0 � hë1÷1 � : : :� ëk÷k , ÷ii � ëi:

Therefore ÷1, . . . , ÷k are linearly independent. j

We now relate inner products of characters to the spaces of CG-

homomorphisms which we constructed in Chapter 11.

14.24 Theorem

Let V and W be CG-modules with characters ÷ and ø, respectively.

Then

dim (HomCG (V , W )) � h÷, øi:

Proof We know from (14.14) that there are non-negative integers ci, di

(1 < i < k) such that

V � (V1 � : : :� V1)� (V2 � : : :� V2)� : : :� (Vk � : : :� Vk)

with ci factors Vi for each i, and

W � (V1 � : : :� V1)� (V2 � : : :� V2)� : : :� (Vk � : : :� Vk)

with di factors Vi for each i. By Proposition 11.2, for any i, j we have

dim (HomCG (Vi, Vj)) � äij:

Hence, using (11.5)(3) we see that

dim (HomCG (V , W )) �Xk

i�1

cidi:


On the other hand,

÷ �Xk

i�1

ci÷i and ø �Xk

i�1

di÷i

and so (14.13) implies that

h÷, øi �Xk

i�1

cidi:

The result follows. j

Decomposing CG-modules

It is sometimes of practical importance to be able to decompose a

given CG-module into a direct sum of CG-submodules, and we now

describe a process for doing this.

Once more we adopt Hypothesis 14.7:

CG � W1 � W2, where the CG-modules W1 and W2

have no common composition factor; and 1 � e1 � e2

with e1 2 W1, e2 2 W2.

Let V be any CG-module. We can write V � V1 � V2, where every

composition factor of V1 is a composition factor of W1 and every

composition factor of V2 is a composition factor of W2.

14.25 Proposition

With the above notation, for all v1 2 V1 and v2 2 V2 we have

v1e1 � v1, v2e1 � 0,

v1e2 � 0, v2e2 � v2:

Proof If v1 2 V1 then the function w2 ! v1w2 (w2 2 W2) is clearly a

CG-homomorphism from W2 to V1. Since W2 and V1 have no common

composition factor, we deduce the stated results just as in the proof of

Proposition 14.8. j

14.26 Proposition

If ÷ is an irreducible character of G, and V is any CG-module, then

VXg2G

÷(gÿ1)g

!


is equal to the sum of those CG-submodules of V which have character

÷ (where for r 2 CG, we de®ne Vr � fvr: v 2 Vg).

Proof Write

CG � U1 � : : :� Ur,

a direct sum of irreducible CG-submodules Ui. Let W1 be the sum of

those CG-submodules Ui which have character ÷, and let W2 be the

sum of the remaining CG-submodules Ui. Then the character of W1 is

m÷ where m � ÷(1), by Theorem 11.9. Also W1 and W2 satisfy

Hypothesis 14.7, and by Proposition 14.10, the element e1 of W1 is

given by

e1 � m

jGjXg2G

÷(gÿ1)g:

Let V1 be the sum of those CG-submodules of V which have

character ÷. Then Proposition 14.25 shows that Ve1 � V1. Clearly we

may omit the constant multiplier m/|G|, so

V1 � VXg2G

÷(gÿ1)g

!:

j

Once the irreducible characters of our group G are known, Proposi-

tion 14.26 provides a useful practical tool for ®nding CG-submodules

of a given CG-module V. The procedure is as follows:

(14.27) (1) Choose a basis v1, . . . , vn of V.

(2) For each irreducible character ÷ of G, calculate the vectors

vi(P

g2G÷(gÿ1)g) for 1 < i < n, and let V÷ be the sub-

space of V spanned by these vectors.

(3) Then V is the direct sum of the CG-modules V÷ as ÷ runs

over the irreducible characters of G. The character of V÷ is

a multiple of ÷.

We illustrate this method with a couple of simple examples. Some

more complicated uses of the method can be found in Chapter 32.

14.28 Examples

(1) Let G be any ®nite group and let V be any non-zero CG-module.

Taking ÷ to be the trivial character of G in Proposition 14.26, we see

that


VXg2G

g

!is the sum of all the trivial CG-submodules of V. For example, let

G � Sn and let V be the permutation module, with basis v1, . . . , vn

such that vig � vig for all i and all g 2 G. Then

VXg2G

g

!� sp (v1 � : : :� vn):

Hence V has a unique trivial CG-submodule.

(2) Let G be the subgroup of S4 which is generated by

a � (1 2 3 4) and b � (1 2)(3 4):

Then G � D8 (compare Example 1.5). Here is a list of the irreducible

characters ÷1, . . . , ÷5 of D8 (see Example 16.3(3)):

1 a a2 a3 b ab a2b a3b

÷1 1 1 1 1 1 1 1 1÷2 1 1 1 1 ÿ1 ÿ1 ÿ1 ÿ1÷3 1 ÿ1 1 ÿ1 1 ÿ1 1 ÿ1÷4 1 ÿ1 1 ÿ1 ÿ1 1 ÿ1 1÷5 2 0 ÿ2 0 0 0 0 0

Let V be the permutation module for G, with basis v1, v2, v3, v4

such that vig � vig for all i and all g 2 G.

For 1 < i < 5, let

ei � ÷i(1)

8

Xg2G

÷i(gÿ1)g:

For example, e5 � 12(1 ÿ a2). Then

Ve1 � sp (v1 � v2 � v3 � v4),

Ve2 � 0,

Ve3 � 0,

Ve4 � sp (v1 ÿ v2 � v3 ÿ v4),

Ve5 � sp (v1 ÿ v3, v2 ÿ v4):


We have

V � Ve1 � Ve4 � Ve5,

and so we have expressed V as a direct sum of irreducible CG-

submodules whose characters are ÷1, ÷4 and ÷5, respectively.

You might like to check that

e1 � : : :� e5 � 1,

e2i � ei for 1 < i < 5,

eiej � 0 for i 6� j:

Compare these results with Corollary 14.9.

Note that the procedure described in (14.27) does not in general

enable us to write a given CG-module as a direct sum of irreducible

CG-submodules (since V÷ is not in general irreducible).


1. The inner product of two functions W, ö from G to C is given by

hW, öi � 1

jGjXg2G

W(g)ö(g):

2. The irreducible characters ÷1, . . . , ÷k of G form an orthonormal

set; that is, h÷i, ÷ ji � äij for all i, j.

3. Every CG-module is determined by its character.

4. If ÷1, . . . , ÷k are the irreducible characters of G, and ø is any

character, then

ø � d1÷1 � : : :� dk÷k where di � hø, ÷ii:Each di is a non-negative integer. Also, ø is irreducible if and only

if kø, øl � 1.



1. Let G � S4. We shall see in Chapter 18 that G has characters ÷ and

ø which take the following values on the conjugacy classes:

Classrepresentative

1 (1 2) (1 2 3) (1 2)(3 4) (1 2 3 4)

|CG(gi)| 24 4 3 8 4

÷ 3 ÿ1 0 3 ÿ1ø 3 1 0 ÿ1 ÿ1

Calculate h÷, ÷i, h÷, øi and hø, øi. Which of ÷ and ø is irreducible?

2. Let G � Q8 � ka, b: a4 � 1, b2 � a2, bÿ1ab � aÿ1l, and let r1, r2,

r3 be the representations of G over C for which

ar1 �i 0

0 ÿi

!, br1 �

0 1

ÿ1 0

!,

ar2 �0 i

i 0

!, br2 �

0 ÿ1

1 0

!,

ar3 �ÿ1 0

0 1

!, br3 �

1 0

0 ÿ1

!:

Show that r1 and r2 are equivalent, but r3 is not equivalent to r1

or r2.

3. Suppose that r and ó are representations of G, and that for each g

in G there is an invertible matrix Tg such that

gó � Tÿ1g (gr)Tg:

Prove that there is an invertible matrix T such that for all g in G,

gó � Tÿ1(gr)T :

4. Suppose that ÷ is a non-zero, non-trivial character of G, and that

÷(g) is a non-negative real number for all g in G. Prove that ÷ is

reducible.

5. If ÷ is a character of G, show that

h÷reg, ÷i � ÷(1):


6. If ð is the permutation character of Sn, prove that

hð, 1Sni � 1:

(Hint: you may ®nd Exercise 11.4 relevant.)

7. Let ÷1, . . . , ÷k be the irreducible characters of the group G, and

suppose that

ø � d1÷1 � : : :dk÷k

is a character of G. What can you say about the integers di in the

cases kø, øl � 1, 2, 3 or 4?

8. Suppose that ÷ is a character of G and that for every g 2 G, ÷(g) is

an even integer. Does it follow that ÷ � 2ö for some character ö?


152

15

The number of irreducible characters

We devote this chapter to the theorem which states that the number of

irreducible characters of a ®nite group is equal to the number of

conjugacy classes of the group, and to some consequences of this

theorem. Together with the material from Chapter 14, the theorem

provides machinery for investigating characters which is used in the

remainder of the book.

Throughout, G is as usual a ®nite group.

Class functions

15.1 De®nition

A class function on G is a function ø: G! C such that ø(x) � ø(y)

whenever x and y are conjugate elements of G (that is, ø is constant

on conjugacy classes).

By Proposition 13.5(2), the characters of G are class functions on G.

The set C of all class functions on G is a subspace of the vector space

of all functions from G to C. A basis of C is given by those functions

which take the value 1 on precisely one conjugacy class and zero on

all other classes. Thus, if l is the number of conjugacy classes of G,

then

dim C � l:(15:2)

15.3 Theorem

The number of irreducible characters of G is equal to the number of

conjugacy classes of G.

Proof Let ÷1, . . . , ÷k be the irreducible characters of G, and let l be

the number of conjugacy classes of G. By Theorem 14.23, ÷1, . . . , ÷k

are linearly independent elements of C, so (15.2) implies that k < l.

In order to prove the reverse inequality l < k, we consider the

regular CG-module. If V1, : : : , Vk is a complete set of non-iso-

morphic irreducible CG-modules, we know from Theorem 8.7 that

CG � W1 � : : :� Wk ,

where for each i, Wi is isomorphic to a direct sum of copies of Vi.

Since CG contains the identity element 1, we can write

1 � f 1 � : : :� f k

with f i 2 Wi for 1 < i < k.

Now let z 2 Z(CG), the centre of CG. By Proposition 9.14, for each

i there exists ëi 2 C such that for all v 2 Vi,

vz � ëiv:

Hence wz � ëiw for all w 2 Wi, and in particular,

f iz � ëi f i (1 < i < k):

It follows that

z � 1z � ( f 1 � : : :� f k)z � f 1z� : : :� f kz

� ë1 f 1 � : : :� ëk f k :

This shows that Z(CG) is contained in the subspace of CG spanned by

f 1, : : : , f k . Since Z(CG) has dimension l by Proposition 12.22, we

deduce that l < k. This completes the proof that k � l. j

15.4 Corollary

The irreducible characters ÷1, : : : , ÷k of G form a basis of the vector

space of all class functions on G. Indeed, if ø is a class function, then

ø �Xk

i�1

ëi÷i

where ëi � kø, ÷il for 1 < i < k.

Proof Since ÷1, . . . , ÷k are linearly independent, they span a subspace

of C of dimension k. By (15.2), dim C � l, which is equal to k by

The number of irreducible characters 153

Theorem 15.3. Hence ÷1, . . . , ÷k span C, and so they form a basis of

C. The last part follows, using (14.13). j

Corollary 15.4 has the following useful consequence.

15.5 Proposition

Suppose that g, h 2 G. Then g is conjugate to h if and only if

÷(g) � ÷(h) for all characters ÷ of G.

Proof If g is conjugate to h then ÷(g) � ÷(h) for all characters ÷ of

G, by Proposition 13.5(2).

Conversely, suppose that ÷(g) � ÷(h) for all characters ÷. Then by

Corollary 15.4, ø(g) � ø(h) for all class functions ø on G. In

particular, this is true for the class function ø which takes the value 1

on the conjugacy class of g and takes the value 0 elsewhere. Then

ø(g) � ø(h) � 1, and so g is conjugate to h. j

15.6 Corollary

Suppose that g 2 G. Then g is conjugate to gÿ1 if and only if ÷(g) is

real for all characters ÷ of G.

Proof Since ÷(g) is real if and only if ÷(g) � ÷(gÿ1) (see Proposition

13.9(3)), the result follows immediately from Proposition 15.5. j

We conclude the chapter with an example illustrating some practical

methods of expressing characters and class functions of a group as

combinations of irreducible characters. As in previous examples, we

regard a character ÷ of G as a row vector, whose k entries are the

values of ÷ on the k conjugacy classes of G.

15.7 Example

We shall see in Section 18.4 that there is a certain group G of order

12 which has exactly six conjugacy classes with representatives

g1, . . . , g6 (where g1 � 1), and six irreducible characters ÷1, . . . , ÷6

given as follows:


gi g1 g2 g3 g4 g5 g6

|CG(gi)| 12 12 6 6 4 4

÷1 1 1 1 1 1 1÷2 1 ÿ1 ÿ1 1 i ÿi÷3 1 1 1 1 ÿ1 ÿ1÷4 1 ÿ1 ÿ1 1 ÿi i÷5 2 2 ÿ1 ÿ1 0 0÷6 2 ÿ2 1 ÿ1 0 0

Suppose we are given characters ÷ and ø of G as follows:

g1 g2 g3 g4 g5 g6

÷ 3 ÿ3 0 0 i ÿiø 4 0 0 4 0 0

Then it is easy to spot that

÷ � ÷2 � ÷6, ø � ÷1 � ÷2 � ÷3 � ÷4:

For example, the second entry in the row vector for ÷ is equal to

minus the ®rst entry. Inspecting the values of the irreducible characters

÷i, we see that ÷ must be a combination of ÷2, ÷4 and ÷6. The correct

answer now comes quickly to mind.

In fact, given any character ö of G whose degree is not large

compared with the degrees of the ÷i, it is not hard to use tactical

guesswork to express ö as a combination of the irreducible characters.

The reason for this is that the required coef®cients are known to be

non-negative integers, and the entries in the column corresponding to

g1 � 1 are positive integers (indeed, they are the degrees of the ÷i).

We suggest that you use the `guesswork method' to express the

following characters ë, ì of G as combinations of ÷1, . . . , ÷6:

g1 g2 g3 g4 g5 g6

ë 2 ÿ2 ÿ2 2 0 0ì 4 4 1 1 0 0


How do we cope with a class function or with a more dif®cult

character, like the following one?

g1 g2 g3 g4 g5 g6

ö 11 3 ÿ3 5 ÿ1 � 2i ÿ1 ÿ 2i

The answer is to use the inner product k , l. We know from Corollary

15.4 that the coef®cients ëi in the expression

ö � ë1÷1 � : : :� ë6÷6

are given by

ëi � hö, ÷ii (1 < i < 6):

Using Proposition 14.5(2), we calculate these inner products:

hö, ÷1i � 11 . 1

12� 3 . 1

12�ÿ3 . 1

6� 5 . 1

6� (ÿ1� 2i) . 1

4

� (ÿ1ÿ 2i) . 1

4� 1,

hö, ÷2i � 11 . 1

12� 3 . (ÿ1)

12� (ÿ3) . (ÿ1)

6� 5 . 1

6� (ÿ1� 2i) . (ÿi)

4

� (ÿ1ÿ 2i) . i

4� 3,

hö, ÷3i � 11 . 1

12� 3 . 1

12�ÿ3 . 1

6� 5 . 1

6� (ÿ1� 2i) . (ÿ1)

4

� (ÿ1ÿ 2i) . (ÿ1)

4� 2,

and similarly kö, ÷4l � 1, kö, ÷5l � 2 and kö, ÷6l � 0. Therefore

ö � ÷1 � 3÷2 � 2÷3 � ÷4 � 2÷5:


1. The number of irreducible characters of a group is equal to the

number of conjugacy classes of the group.


2. The irreducible characters ÷1, . . . , ÷k of G form a basis of the

vector space of all class functions on G. If ø is a class function,

then

ø �Xk

i�1

ëi÷i where ëi � hø, ÷ii:


1. The three irreducible characters of S3 are ÷1, ÷2, ÷3:

Let ÷ be the class function on S3 with the following values:

Express ÷ as a linear combination of ÷1, ÷2 and ÷3. Is ÷ a character

of S3?

2. Let ø1, ø2 and ø3 be the class functions on S3 taking the following

values:

Express ø1, ø2 and ø3 as linear combinations of the irreducible

characters ÷1, ÷2 and ÷3 of S3.

1 (1 2) (1 2 3)

÷1 1 1 1÷2 1 ÿ1 1÷3 2 0 ÿ1

1 (1 2) (1 2 3)

÷ 19 ÿ1 ÿ2

1 (1 2) (1 2 3)

ø1 1 0 0ø2 0 1 0ø3 0 0 1


3. Suppose that G is the group of order 12 in Example 15.7, with

conjugacy class representatives g1, . . . , g6 and irreducible charac-

ters ÷1, . . . , ÷6 as in that example. Let ø be the class function on

G taking the following values:

Express ø as a linear combination of ÷1, . . . , ÷6. Is ø a character

of G?

4. Let G be a group of order 12.

(a) Show that G cannot have exactly 9 conjugacy classes. (Hint:

show that Z(G) cannot have order 6.)

(b) Using the solution to Exercise 11.2, prove that G has 4, 6 or 12

conjugacy classes. Find groups G in which each of these

possibilities is realized.

g1 g2 g3 g4 g5 g6

ø 6 0 3 ÿ3 ÿ1 ÿ i ÿ1 � i


159

16

Character tables and orthogonality relations

The irreducible characters of a ®nite group G are class functions, and

the number of them is equal to the number of conjugacy classes of G.

It is therefore convenient to record all the values of all the irreducible

characters of G in a square matrix. This matrix is called the character

table of G. The entries in a character table are related to each other in

subtle ways, many of which are encapsulated in the orthogonality

relations (Theorem 16.4). Much of the later material in the book will

be devoted to understanding character tables. The motivation for this is

Theorem 14.21, which tells us that every CG-module is determined by

its character. Thus, many problems in representation theory can be

solved by considering characters.

Character tables

16.1 De®nition

Let ÷1, : : : , ÷k be the irreducible characters of G and let g1, : : : , gk

be representatives of the conjugacy classes of G. The k 3 k matrix

whose ij-entry is ÷i(g j) (for all i, j with 1 < i < k, 1 < j < k) is

called the character table of G.

It is usual to number the irreducible characters and conjugacy classes

of G so that ÷1 � 1G, the trivial character, and g1 � 1, the identity

element of G. Beyond this, the numbering is arbitrary. Note that in the

character table, the rows are indexed by the irreducible characters of G

and the columns are indexed by the conjugacy classes (or, in practice,

by conjugacy class representatives).

16.2 Proposition

The character table of G is an invertible matrix.

Proof This follows immediately from the fact that the irreducible

characters of G, and hence also the rows of the character table, are

linearly independent (Theorem 14.23). j

16.3 Examples

(1) Let G � D6 � ka, b: a3 � b2 � 1, bÿ1ab � aÿ1l. The irreducible

characters of G are given in Example 13.6(4). We take 1, a, b as

representatives of the conjugacy classes of G, and then the character

table of G is

(2) We can write down the character table of any ®nite abelian group

using Theorem 9.8. For example, the character table of C2 �ha: a2 � 1i is

and the character table of C3 � ka: a3 � 1l is

(3) Let G � D8 � ka, b: a4 � b2 � 1, bÿ1ab � aÿ1l. You found all the

irreducible representations of G in Exercise 10.4. The conjugacy classes

1 a b

÷1 1 1 1÷2 1 1 ÿ1÷3 2 ÿ1 0

1 a

÷1 1 1÷2 1 ÿ1

1 a a2

÷1 1 1 1÷2 1 ù ù2 (ù � e2ði=3)÷3 1 ù2 ù


of G are given by (12.12), and representatives are 1, a2, a, b, ab.

Hence the character table of G is

1 a2 a b ab

÷1 1 1 1 1 1÷2 1 1 1 ÿ1 ÿ1÷3 1 1 ÿ1 1 ÿ1÷4 1 1 ÿ1 ÿ1 1÷5 2 ÿ2 0 0 0

The character tables of all dihedral groups will be found in

Chapter 18.

The orthogonality relations

We have already seen many uses for the relations (14.13),

h÷r, ÷si � ärs,

among the irreducible characters ÷1, . . . , ÷k of G. These relations can

be expressed in terms of the rows of the character table, by writing

them as Xk

i�1

÷r(gi)÷s(gi)

jCG(gi)j � ärs

(see Proposition 14.5(2)). Similar relations exist between the columns

of the character table, and these are given by part (2) of our next

result.

16.4 Theorem

Let ÷1, . . . , ÷k be the irreducible characters of G, and let g1, : : : , gk

be representatives of the conjugacy classes of G. Then the following

relations hold for any r, s 2 {1, . . . , k}.

(1) The row orthogonality relations:Xk

i�1

÷r(gi)÷s(gi)

jCG(gi)j � ärs:

(2) The column orthogonality relations:Xk

i�1

÷i(gr)÷i(gs) � ärsjCG(gr)j:

Character tables and orthogonality relations 161

Proof The row orthogonality relations have already been proved. They

are recorded here merely for comparison with the column relations.

For 1 < s < k, let øs be the class function which satis®es

øs(gr) � ärs (1 < r < k):

By Corollary 15.4, øs is a linear combination of ÷1, . . . , ÷k , say

øs �Xk

i�1

ëi÷i (ëi 2 C):

We know that h÷i, ÷ ji � äij, so

ëi � høs, ÷ii � 1

jGjXg2G

øs(g)÷i(g):

Now øs(g) � 1 if g is conjugate to gs, and øs(g) � 0 otherwise; also

there are jGj=jCG(gs)j elements of G which are conjugate to gs, by

Theorem 12.8. Hence

ëi � 1

jGjXg2 gG

s

øs(g)÷i(g) � ÷i(gs)

jCG(gs)j :

Therefore

ärs � øs(gr) �Xk

i�1

ëi÷i(gr) �Xk

i�1

÷i(gr)÷i(gs)

jCG(gs)j ,

and the column orthogonality relations follow. j

16.5 Examples

We illustrate the column orthogonality relations.

(1) Let G � D6. We copy the character table of G from Example

16.3(1), and this time we record the order of the centralizer CG(gi)

next to each conjugacy class representative gi:

gi 1 a b|CG(gi)| 6 3 2

÷1 1 1 1÷2 1 1 ÿ1÷3 2 ÿ1 0


Consider the sumsP3

i�1 ÷i(g r)÷i(gs) for various cases:

r � 1, s � 2: 1 . 1� 1 . 1� 2 . (ÿ1) � 0;

r � 2, s � 2: 1 . 1� 1 . 1� (ÿ1) . (ÿ1) � 3;

r � 1, s � 3: 1 . 1� 1 . (ÿ1)� 2 . 0 � 0:

In each case, we read down columns r and s of the character table,

taking the products of the numbers which appear. The sum of the

products is 0 if r 6� s, and is the number at the top of the column (that

is, the order of the centralizer of gr) if r � s.

(2) Suppose we are given the following part of the character table of a

group G of order 12 which has exactly four conjugacy classes:

(where ù � e2ði=3). We shall use the column orthogonality relations to

determine the last row of the character table.

The entries in the ®rst column of the character table are the degrees

of the irreducible characters, so they are positive integers. By the

column orthogonality relations with r � s � 1, the sum of the squares

of these numbers is 12 (this also follows from Theorem 11.12). Hence

the last entry in the ®rst column is 3.

Let x denote the number at the foot of the second column. The

column orthogonality relation

X4

i�1

÷i(g1)÷i(g2) � 0

gives

1 . 1� 1 . 1� 1 . 1� 3x � 0:

Therefore x � ÿ1.

By considering the orthogonality relations between the ®rst column

and columns 3 and 4, we obtain the complete character table as

gi g1 g2 g3 g4

|CG( gi)| 12 4 3 3

÷1 1 1 1 1÷2 1 1 ù ù2

÷3 1 1 ù2 ù÷4


follows:

Notice that the orthogonality relations hold between all pairs of

columns, although our calculation has used only those relations which

involve the ®rst column. For example,

X4

i�1

÷i(g2)÷i(g2) � 1 . 1� 1 . 1� 1 . 1� (ÿ1) . (ÿ1) � 4,

X4

i�1

÷i(g3)÷i(g3) � 1 . 1� ù . ù� ù2 . ù2 � 0 . 0 � 3,

X4

i�1

÷i(g3)÷i(g4) � 1 . 1� ù . ù2 � ù2 . ù� 0 . 0 � 0:

We shall see later that the character table which we have constructed

here is that of A4.

Those column orthogonality relations which involve the ®rst column

of the character table were proved in Chapter 13, since Theorem 13.19

and Proposition 13.20 give

Xk

i�1

di÷i(g) �jGj, if g � 1,

0, if g 6� 1,

8<:where di � ÷i(1). By taking the complex conjugate of each side of this

equation, we get

Xk

i�1

÷i(1)÷i(g) �jGj, if g � 1,

0, if g 6� 1,

8<:

gi g1 g2 g3 g4

|CG(gi)| 12 4 3 3

÷1 1 1 1 1÷2 1 1 ù ù2

÷3 1 1 ù2 ù÷4 3 ÿ1 0 0


and these are just the column orthogonality relations which involve the

®rst column.

Rows versus columns

Notice that in Example 16.5(2), where we were given three of the four

irreducible characters of G, we calculated the values of the last

character one at a time using the column orthogonality relations. An

alternative approach would have been to use the row orthogonality

relations h÷i, ÷4i � äi4 to obtain four equations in the four unknown

values ÷4(g j) (1 < j < 4). Although the calculation with the column

orthogonality relations was easier to perform, it is a fact that the

column orthogonality relations contain precisely the same information

as the row orthogonality relations, as we shall now show.

The character table of G is a k 3 k matrix, and we adjust the entries

÷i(g j) in this matrix to obtain another k 3 k matrix M, by letting the

ij-entry of M be

÷i(gj)

jCG(gj)j1=2:

Let M t denote the transpose of the complex conjugate of M.

Now the rs-entry in M M t isXk

i�1

÷r(gi)÷s(gi)

jCG(gi)j � ärs,

by the row orthogonality relations, so M M t � I . Indeed, the equation

M M t � I is just another way of expressing the row orthogonality

relations. On the other hand, the rs-entry in M t M is

1

jCG(gr)j1=2jCG(gs)j1=2

Xk

i�1

÷i(gr)÷i(gs) � ärs,

by the column orthogonality relations, so M t M � I .

Since the properties M t M � I and M M t � I of a square matrix M

are equivalent to each other, we see that the row and column

orthogonality relations are equivalent.

We could have used the above argument to deduce the column

orthogonality relations from the row ones. More importantly, the row

and column orthogonality relations encapsulate the same information,


so when we are working with character tables, we can deduce exactly

the same results using either set of relations.


Let G be a ®nite group with irreducible characters ÷1, . . . , ÷k and

conjugacy class representatives g1, . . . , gk.

1. The character table of G is the k 3 k matrix with ij-entry ÷i(g j).

2. The row orthogonality relations state that for all r, s,Xk

i�1

÷r(gi)÷s(gi)

jCG(gi)j � ärs:

3. The column orthogonality relations state that for all r, s,Xk

i�1

÷i(gr)÷i(gs) � ärsjCG(gr)j:


1. Write down the character table of C2 3 C2.

2. A certain group G of order 8 is known to have a total of ®ve

conjugacy classes, with representatives g1, . . . , g5, and four linear

characters ÷1, . . . , ÷4 taking the following values:

gi g1 g2 g3 g4 g5

|CG(gi)| 8 8 4 4 4

÷1 1 1 1 1 1÷2 1 1 1 ÿ1 ÿ1÷3 1 1 ÿ1 1 ÿ1÷4 1 1 ÿ1 ÿ1 1

Find the complete character table of G.

3. There exists a group G of order 10 which has precisely four

conjugacy classes, with representatives g1, . . . , g4, and has irreduci-

ble characters ÷1, ÷2 as follows:


where á � (ÿ1�p5)=2 and â � (ÿ1ÿp5)=2.

Find the complete character table of G.

(Hint: ®rst ®nd the values of the remaining irreducible characters on

g1, then on g4 ± use Corollary 13.10.)

4. A certain group G has two columns of its character table as

follows:

where g1 � 1 and æ 2 C.

(a) Find æ.

(b) Find another column of the character table.

5. Let ÷1, . . . , ÷k be the irreducible characters of G. Show that

Z(G) � g 2 G:Xk

i�1

÷i(g)÷i(g) � jGj( )

:

6. Let G be a ®nite group with conjugacy class representatives

g1, : : : , gk and character table C. Show that det C is either real or

purely imaginary, and that

jdet Cj2 �Yk

i�1

jCG(gi)j:

Find �(det C) when G � C3.

gi g1 g2 g3 g4

|CG(gi)| 10 5 5 2

÷1 1 1 1 1÷2 2 á â 0

gi g1 g2

|CG(gi)| 21 7

÷1 1 1÷2 1 1÷3 1 1÷4 3 æ÷5 3 æ


168

17

Normal subgroups and lifted characters

If N is a normal subgroup of the ®nite group G, and N 6� {1}, then

the factor group G=N is smaller than G. The characters of G=N should

therefore be easier to ®nd than the characters of G. In fact, we can use

the characters of G=N to get some of the characters of G, by a process

which is known as lifting. Thus, normal subgroups help us to ®nd

characters of G. In the opposite direction, it is also true that the

character table of G enables us to ®nd the normal subgroups of G; in

particular, it is easy to tell from the character table whether or not G

is simple.

The linear characters of G (i.e. the characters of degree 1) are

obtained by lifting the irreducible characters of G=N in the case where

N is the derived subgroup of G. (The derived subgroup is de®ned

below in De®nition 17.7.) The linear characters, in turn, can be used to

get new irreducible characters from a given irreducible character, in a

way which we shall describe.

Lifted characters

We begin by constructing a character of G from a character of G=N .

17.1 Proposition

Assume that N v G, and let ~÷ be a character of G=N . De®ne

÷: G! C by

÷(g) � ~÷(Ng) (g 2 G):

Then ÷ is a character of G, and ÷ and ~÷ have the same degree.

Proof Let ~r: G=N ! GL (n, C) be a representation of G=N with

character ~÷. The function r: G! GL (n, C) which is given by the

composition

g ! Ng ! (Ng)~r (g 2 G)

is a homomorphism from G to GL (n, C). Thus r is a representation of

G. The character ÷ of r satis®es

÷(g) � tr (gr) � tr ((Ng)~r) � ~÷(Ng)

for all g 2 G. Moreover, ÷(1) � ~÷(N), so ÷ and ~÷ have the same

degree. j

17.2 De®nition

If N v G and ~÷ is a character of G=N , then the character ÷ of G

which is given by

÷(g) � ~÷(Ng) (g 2 G)

is called the lift of ~÷ to G.

17.3 Theorem

Assume that N v G. By associating each character of G=N with its

lift to G, we obtain a bijective correspondence between the set of

characters of G=N and the set of characters ÷ of G which satisfy

N < Ker ÷. Irreducible characters of G=N correspond to irreducible

characters of G which have N in their kernel.

Proof If ~÷ is a character of G=N , and ÷ is the lift of ~÷ to G, then

~÷(N) � ÷(1). Also, if k 2 N then

÷(k) � ~÷(Nk) � ~÷(N ) � ÷(1),

so N < Ker ÷.

Now let ÷ be a character of G with N < Ker ÷. Suppose that

r: G! GL (n, C) is a representation of G with character ÷. If g1,

g2 2 G and Ng1 � Ng2 then g1 gÿ12 2 N, so (g1 gÿ1

2 )r � I, and hence

g1r � g2r. We may therefore de®ne a function ~r: G=N ! GL (n, C)

by

(Ng)~r � gr (g 2 G):

Then for all g, h 2 G we have

((Ng)(Nh))~r � (Ngh)~r � (gh)r � (gr)(hr)

� ((Ng)~r)((Nh)~r),

Normal subgroups and lifted characters 169

so ~r is a representation of G=N . If ~÷ is the character of ~r then

~÷(Ng) � ÷(g) (g 2 G):

Thus ÷ is the lift of ~÷.

We have now established that the function which sends each char-

acter of G=N to its lift to G is a bijection between the set of

characters of G=N and the set of characters of G which have N in

their kernel. It remains to show that irreducible characters correspond

to irreducible characters. To see this, let U be a subspace of Cn, and

note that

u(gr) 2 U for all u 2 U , u(Ng)~r 2 U for all u 2 U :

Thus, U is a CG-submodule of Cn if and only if U is a C(G=N )-

submodule of Cn. The representation r is therefore irreducible if and

only if the representation ~r is irreducible. Hence ÷ is irreducible if and

only if ~÷ is irreducible. j

If we know the character table of G=N for some normal subgroup N

of G, then Theorem 17.3 enables us to write down as many irreducible

characters of G as there are irreducible characters of G=N .

17.4 Example

Let G � S4 and

N � V4 � f1, (1 2)(3 4), (1 3)(2 4), (1 4)(2 3)g,so that N v G (see Example 12.20). If we put a � N(1 2 3) and

b � N(1 2) then

G=N � ha, bi and a3 � b2 � N , bÿ1ab � aÿ1,

so G=N � D6. We know from Example 16.3(1) that the character table

of G=N is

N N (1 2) N(1 2 3)

~÷1 1 1 1~÷2 1 ÿ1 1~÷3 2 0 ÿ1


To calculate the lift ÷ of a character ~÷ of G=N , we note that

÷((1 2)(3 4)) � ~÷(N ) since (1 2)(3 4) 2 N ,

÷((1 2 3 4)) � ~÷(N (1 3)) since N (1 2 3 4) � N (1 3):

Hence the lifts of ~÷1, ~÷2, ~÷3 are ÷1, ÷2, ÷3, which are given by

1 (1 2) (1 2 3) (1 2)(3 4) (1 2 3 4)

÷1 1 1 1 1 1÷2 1 ÿ1 1 1 ÿ1÷3 2 0 ÿ1 2 0

Then ÷1, ÷2, ÷3 are irreducible characters of G, since ~÷1, ~÷2, ~÷3 are

irreducible characters of G=N .

Finding normal subgroups

The character table contains accessible information about the structure

of a group, as our next two propositions will demonstrate. First we

shall show how to ®nd all the normal subgroups of G, once the

character table of G is known. Recall that we can easily locate the

kernel of an irreducible character ÷ from the character table, since

Ker ÷ � fg 2 G: ÷(g) � ÷(1)g(see De®nition 13.13). Also Ker ÷ v G. Of course, any subgroup which

is the intersection of the kernels of irreducible characters is a normal

subgroup too. The following proposition shows that every normal

subgroup arises in this way.

17.5 Proposition

If N v G then there exist irreducible characters ÷1, : : : , ÷s of G

such that

N �\s

i�1

Ker ÷i:

Proof If g belongs to the kernel of each irreducible character of G,

then ÷(g) � ÷(1) for all characters ÷, so g � 1 by Proposition 15.5.

Hence the intersection of the kernels of all the irreducible characters of

G is {1}.


Now let ~÷1, . . . , ~÷s be the irreducible characters of G=N . By the

above observation, \s

i�1

Ker ~÷i � fNg:

For 1 < i < s, let ÷i be the lift to G of ~÷i. If g 2 Ker ÷i then

~÷i(N ) � ÷i(1) � ÷i(g) � ~÷i(Ng),

and so Ng 2 Ker ~÷i. Therefore if g 2T

Ker ÷i then Ng 2T

Ker ~÷i �{N}, and so g 2 N. Hence

N �\s

i�1

Ker ÷i: j

It is particularly easy to tell from the character table of G whether

or not G is simple:

17.6 Proposition

The group G is not simple if and only if

÷(g) � ÷(1)

for some non-trivial irreducible character ÷ of G, and some non-

identity element g of G.

Proof Suppose there is a non-trivial irreducible character ÷ such that

÷(g) � ÷(1) for some non-identity element g. Then g 2 Ker ÷, so

Ker ÷ 6� {1}. If r is a representation of G with character ÷, then

Ker ÷ � Ker r by Theorem 13.11(2). Since ÷ is non-trivial and irreduci-

ble, Ker r 6� G; hence Ker ÷ 6� G. Thus Ker ÷ is a normal subgroup of

G which is not equal to {1} of G, and so G is not simple.

Conversely, suppose that G is not simple, so that there is a normal

subgroup N of G with N 6� {1} and N 6� G. Then by Proposition 17.5,

there is an irreducible character ÷ of G such that Ker ÷ is not {1} or

G. As Ker ÷ 6� G, ÷ is non-trivial; and taking 1 6� g 2 Ker ÷, we have

÷(g) � ÷(1). j

Linear characters

Recall that a linear character of a group is a character of degree 1. We

shall show how to ®nd all linear characters of any group G, since the


®rst move in constructing the character table of G is often to write

down the linear characters. As a preliminary step, it is necessary to

determine the derived subgroup of G, which is de®ned in the following

way.

17.7 De®nition

For a group G, let G9 be the subgroup of G which is generated by all

elements of the form

gÿ1 hÿ1 gh (g, h 2 G):

Then G9 is called the derived subgroup of G.

We abbreviate gÿ1 hÿ1 gh as [g, h]. Thus

G9 � h[g, h]: g, h 2 Gi:

17.8 Examples

(1) If G is abelian then [g, h] � 1 for all g, h 2 G, so G9 � {1}.

(2) Let G � S3. Clearly [g, h] is always an even permutation, so

G9 < A3. If g � (1 2) and h � (2 3) then [g, h] � (1 2 3). Hence

G9 � h(1 2 3)i � A3.

We are going to show that G9 v G and that the linear characters of

G are the lifts to G of the irreducible characters of G=G9. One step is

provided by the following proposition.

17.9 Proposition

If ÷ is a linear character of G, then G9 < Ker ÷.

Proof Let ÷ be a linear character of G. Then ÷ is a homomorphism

from G to the multiplicative group of non-zero complex numbers.

Therefore, for all g, h 2 G,

÷(gÿ1 hÿ1 gh) � ÷(g)ÿ1÷(h)ÿ1÷(g)÷(h) � 1:

Hence G9 < Ker ÷. j

Next, we explore some group-theoretic properties of the derived

subgroup.


17.10 Proposition

Assume that N v G.

(1) G9 v G.

(2) G9 < N if and only if G=N is abelian. In particular, G=G9 is

abelian.

Proof (1) Note that for all a, b, x 2 G, we have

xÿ1(ab)x � (xÿ1ax)(xÿ1bx), and

xÿ1aÿ1x � (xÿ1ax)ÿ1:

Now G9 consists of products of elements of the form [g, h] and their

inverses. Therefore, to prove that G9 v G it is suf®cient by the ®rst

sentence to prove that xÿ1[g, h]x 2 G9 for all g, h, x 2 G. But

xÿ1[g, h]x � xÿ1 gÿ1 hÿ1 ghx

� (xÿ1 gx)ÿ1(xÿ1 hx)ÿ1(xÿ1 gx)(xÿ1 hx)

� [xÿ1 gx, xÿ1 hx]:

Therefore G9 v G.

(2) Let g, h 2 G. We have

ghgÿ1 hÿ1 2 N , Ngh � Nhg , (Ng)(Nh) � (Nh)(Ng):

Hence G9 < N if and only if G=N is abelian. Since we have proved

that G9 v G, we deduce that G=G9 is abelian. j

It follows from Proposition 17.10 that G9 is the smallest normal

subgroup of G with abelian factor group.

Given the derived subgroup G9, we can obtain the linear characters

of G by applying the next theorem.

17.11 Theorem

The linear characters of G are precisely the lifts to G of the irreducible

characters of G=G9. In particular, the number of distinct linear

characters of G is equal to jG=G9j, and so divides |G|.

Proof Let m � jG=G9j. Since G=G9 is abelian, Theorem 9.8 shows that

G=G9 has exactly m irreducible characters ~÷1, . . . , ~÷m, all of degree 1.

The lifts ÷1, . . . , ÷m of these characters to G also have degree 1, and

by Theorem 17.3 they are precisely the irreducible characters ÷ of G


such that G9 < Ker ÷. In view of Proposition 17.9, the characters

÷1, . . . , ÷m are therefore all the linear characters of G. j

17.12 Example

Let G � Sn. We shall show that G9 � An. If n � 1 or 2 then Sn is

abelian, so G9 � {1} � An. We proved that S93 � A3 in Example

17.8(2), so we assume that n > 4.

As Sn=An � C2, we have G9 < An by Proposition 17.10(2). If

g � (1 2), h � (2 3) and k � (1 2)(3 4), then

[g, h] � (1 2 3), [h, k] � (1 4)(2 3):

Since G9 v G, all the elements in (1 2 3)G and (1 4)(2 3)G belong to

G9. Therefore, by Theorem 12.15, G9 contains all 3-cycles and all

elements of cycle-shape (2, 2). But every product of two transpositions

is equal to the identity, a 3-cycle or an element of cycle-shape (2, 2);

and An consists of permutations, each of which is the product of an

even number of transpositions. Therefore An < G9. We have now

proved that G9 � An.

17.13 Example

We ®nd the linear characters of Sn (n > 2). From the last example, we

know that S9n � An. Since Sn=S9n � fAn, An(1 2)g � C2, the group

Sn=S9n has two linear characters ~÷1 and ~÷2, where

~÷1(An(1 2)) � 1,

~÷2(An(1 2)) � ÿ1:

Therefore by Theorem 17.11, Sn has exactly two linear characters ÷1,

÷2, which are given by

÷1 � 1Sn,

÷2(g) �1, if g 2 An,

ÿ1, if g =2 An:

(

Not only are the linear characters of G important in being irreducible

characters, but they can also be used to construct new irreducible

characters from old, as the next result shows.


17.14 Proposition

Suppose that ÷ is a character of G and ë is a linear character of G.

Then the product ÷ë, de®ned by

÷ë(g) � ÷(g)ë(g) (g 2 G)

is a character of G. Moreover, if ÷ is irreducible, then so is ÷ë.

Proof Let r: G! GL (n, C) be a representation with character ÷.

De®ne rë: G! GL (n, C) by

g(rë) � ë(g)(gr) (g 2 G):

Thus g(rë) is the matrix gr multiplied by the complex number ë(g).

Since r and ë are homomorphisms it follows easily that rë is a

homomorphism. The matrix g(rë) has trace ë(g) tr (gr), which is

ë(g)÷(g). Hence rë is a representation of G with character ÷ë.

Now for all g 2 G, the complex number ë(g) is a root of unity, so

ë(g)ë(g) � 1. Therefore

h÷ë, ÷ëi � 1

jGjXg2G

÷(g)ë(g)÷(g)ë(g)

� 1

jGjXg2G

÷(g)÷(g) � h÷, ÷i:

By Theorem 14.20, it follows that ÷ë is irreducible if and only if ÷ is

irreducible. j

The general case of a product of two characters will be discussed in

Chapter 19.


1. Characters of G=N correspond to characters ÷ of G for which

N < Ker ÷. The character of G which corresponds to the character ~÷of G=N is the lift of ~÷, and is given by ÷(g) � ~÷(Ng) (g 2 G).

2. The normal subgroups of G can be found from the character table

of G.

3. The linear characters of G are precisely the lifts to G of the

irreducible characters of G=G9.



1. Let G � Q8 � ka, b: a4 � 1, b2 � a2, bÿ1ab � aÿ1l.(a) Find the ®ve conjugacy classes of G.

(b) Find G9, and construct all the linear characters of G.

(c) Complete the character table of G.

Compare your table with the character table of D8 (Example

16.3(3)).

2. Let a and b be the following permutations in S7:

a � (1 2 3 4 5 6 7), b � (2 3 5)(4 7 6):

Let G � ka, bl. Check that

a7 � b3 � 1, bÿ1ab � a2:

(a) Show that G has order 21.

(b) Find the conjugacy classes of G.

(c) Find the character table of G.

3. Show that every group of order 12 has 3, 4 or 12 linear characters,

and hence cannot be simple.

4. A certain group G of order 12 has precisely six conjugacy classes,

with representatives g1, . . . , g6 (where g1 � 1), and has irreducible

characters ÷, ö with values as follows:

g1 g2 g3 g4 g5 g6

÷ 1 ÿi i 1 ÿ1 ÿ1ö 2 0 0 ÿ1 ÿ1 2

Use Proposition 17.14 to complete the character table of G. What

are the sizes of the conjugacy classes of G?

5. The character table of D8 is as shown (see Example 16.3(3)):

1 a2 a, a3 b, a2b ab, a3b

÷1 1 1 1 1 1÷2 1 1 1 ÿ1 ÿ1÷3 1 1 ÿ1 1 ÿ1÷4 1 1 ÿ1 ÿ1 1÷5 2 ÿ2 0 0 0


Express each normal subgroup of D8 as an intersection of kernels

of irreducible characters, as in Proposition 17.5.

6. You are given that the group

T4n � ha, b: a2n � 1, an � b2, bÿ1ab � aÿ1ihas order 4n. (It is known as a dicyclic group.)

(a) Show that if å is any (2n)th root of unity in C, then there is a

representation of T4n over C which sends

a! å 0

0 åÿ1

� �, b! 0 1

å n 0

� �:

(b) Find all the irreducible representations of T4n.

7. For n > 1, the group

U6n � ha, b: a2n � b3 � 1, aÿ1ba � bÿ1ihas order 6n.

(a) Let ù � e2ði=3. Show that if å is any (2n)th root of unity in C,

then there is a representation of U6n over C which sends

a! 0 åå 0

� �, b! ù 0

0 ù2

� �:

(b) Find all the irreducible representations of U6n.

8. Let n be an odd positive integer. The group

V8n � ha, b: a2n � b4 � 1, ba � aÿ1bÿ1, bÿ1a � aÿ1bihas order 8n.

(a) Show that if å is any nth root of unity in C, then there is a

representation of V8n over C which sends

a! å 0

0 ÿåÿ1

� �, b! 0 1

ÿ1 0

� �:

(b) Find all the irreducible representations of V8n.


179

18

Some elementary character tables

We now illustrate the techniques we have presented so far by construct-

ing the character tables of several groups, including the groups S4 and

A4, and all dihedral groups.

18.1 The group S4

In Example 17.4, we produced three irreducible characters ÷1, ÷2, ÷3 of

S4 by lifting characters of the factor group S4=V4. We shall now use

Proposition 17.14, which deals with the product of a character with a

linear character, to complete the character table of S4.

Let ÷4 be the character

÷4(g) � jfix (g)j ÿ 1 (g 2 S4)

which is given in Proposition 13.24. By Proposition 17.14, the product

÷4÷2 is also a character of S4. The values of ÷2, ÷4 and ÷4÷2 are as

follows:

gi 1 (1 2) (1 2 3) (1 2)(3 4) (1 2 3 4)|CG(gi)| 24 4 3 8 4

÷2 1 ÿ1 1 1 ÿ1÷4 3 1 0 ÿ1 ÿ1÷4÷2 3 ÿ1 0 ÿ1 1

Note that

h÷4, ÷4i � 9

24� 1

4� 1

8� 1

4� 1,

so ÷4 is irreducible. The character ÷4÷2 is also irreducible, either by

using the same calculation or by quoting the result of Proposition

17.14. Let ÷5 � ÷4÷2. Since S4 has ®ve conjugacy classes, and we have

produced ®ve irreducible characters, we have now found the complete

character table of S4, as shown.

18.2 The group A4

Let G � A4, the alternating group of degree 4. Then |G| � 12, and G

has four conjugacy classes, with representatives

1, (1 2)(3 4), (1 2 3), (1 3 2)

(see Example 12.18(1)).

Let í be the character of A4 given by Proposition 13.24, so that

í(g) � |®x (g)| ÿ 1 for all g 2 A4. The values of í are as follows:

gi 1 (1 2)(3 4) (1 2 3) (1 3 2)|CG(gi)| 12 4 3 3

í 3 ÿ1 0 0

Note that

hí, íi � 9

12� 1

4� 1,

so í is an irreducible character of G of degree 3.

Since G has four irreducible characters, and the sum of the squares

of their degrees is 12, there must be exactly three linear characters of

G. Thus jG=G9j � 3 by Theorem 17.11. It is not dif®cult to con®rm

this by showing that

Character table of S4

gi 1 (1 2) (1 2 3) (1 2)(3 4) (1 2 3 4)|CG(gi)| 24 4 3 8 4

÷1 1 1 1 1 1÷2 1 ÿ1 1 1 ÿ1÷3 2 0 ÿ1 2 0÷4 3 1 0 ÿ1 ÿ1÷5 3 ÿ1 0 ÿ1 1


G9 � V4 � f1, (1 2)(3 4), (1 3)(2 4), (1 4)(2 3)g:

Now G=G9 � fG9, G9(1 2 3), G9(1 3 2)g � C3, and the character table

of G=G9 is

(where ù � e2ði=3). The lifts of ~÷1, ~÷2, ~÷3 to G, together with the

character ÷4 � í, give the complete character table of A4:

18.3 The dihedral groups

Let G be the dihedral group D2n of order 2n, with n > 3, so that

G � ha, b: an � b2 � 1, bÿ1ab � aÿ1i:We shall derive the character table of G.

Write å � e2ði=n. For each integer j with 1 < j , n=2, de®ne

Aj � å j 0

0 åÿ j

� �, Bj � 0 1

1 0

� �:

Check that

Anj � B2

j � I , Bÿ1j AjBj � Aÿ1

j :

It follows that by de®ning r j: G! GL(2, C) by

(arbs)r j � (Aj)r(Bj)

s (r, s 2 Z),

we obtain a representation r j of G for each j with 1 < j , n=2.

G9 G9(1 2 3) G9(1 3 2)

~÷1 1 1 1~÷2 1 ù ù2

~÷3 1 ù2 ù

Character table of A4

gi 1 (1 2)(3 4) (1 2 3) (1 3 2)|CG(gi)| 12 4 3 3

÷1 1 1 1 1÷2 1 1 ù ù2

÷3 1 1 ù2 ù÷4 3 ÿ1 0 0

Some elementary character tables 181

Each r j is an irreducible representation, either by the proof of

Example 5.5(2) or by applying the result of Exercise 8.4.

If i and j are distinct integers with 1 < i , n=2 and 1 < j , n=2,

then å i 6� å j and å i 6� åÿ j, so ari and ar j have different eigenvalues.

Therefore there is no matrix T with ari � Tÿ1(ar j)T, and so ri and r j

are not equivalent.

Let ø j be the character of r j. We have now constructed distinct

irreducible characters ø j of G, one for each j which satis®es

1 < j , n=2.

At this point it is convenient to consider separately the cases where

n is odd and where n is even.

Case 1: n odd

By (12.11) the conjugacy classes of D2n (n odd) are

f1g, far, aÿrg(1 < r < (nÿ 1)=2), fasb: 0 < s < nÿ 1g:Thus there are (n� 3)=2 conjugacy classes.

The (nÿ 1)=2 irreducible characters

ø1, ø2, : : : , ø(nÿ1)=2

each have degree 2. As G has (n � 3)/2 irreducible characters in all,

there are two more to be found.

Since kal v G and G=hai � C2, we obtain two linear characters ÷1,

÷2 of G by lifting the irreducible characters of G=hai to G. These

characters ÷1 and ÷2 are given by ÷1 � 1G and

÷2(g) � 1 if g � ar for some r,

ÿ1 if g � arb for some r:

�We have now found all the irreducible characters of D2n (n odd).

(Incidentally, we have proved that D92n � kal for n odd, in view of

Theorem 17.11.)

The character table of D2n (n odd) is therefore as follows (where

å � e2ði=n):

gi 1 ar (1 < r < (n ÿ 1)/2) b|CG(gi)| 2n n 2

÷1 1 1 1÷2 1 1 ÿ1ø j 2 å jr � åÿ jr 0(1 < j < (n ÿ 1)/2)


Case 2: n even

If n is even, say n � 2m, then the conjugacy classes of D2n, as

supplied by (12.12), are

f1g, famg, far, aÿrg(1 < r < mÿ 1), fasb: s eveng, fasb: s oddg:

Hence G has m � 3 irreducible characters, of which m ÿ 1 are given

by

ø1, ø2, : : : , ømÿ1:

To ®nd the remaining four irreducible characters, we ®rst note that

ha2i � fa j: j eveng is a normal subgroup of G and

G=ha2i � fha2i, ha2ia, ha2ib, ha2iabg

� C2 3 C2:

Therefore G has four linear characters ÷1, ÷2, ÷3, ÷4 (and G9 � ka2l).Since these linear characters are the lifts of the irreducible characters

of G/ka2l, they are easy to calculate, and their values appear in the

following complete character table of D2n (n even, n � 2m, å � e2ði=n).

gi 1 am ar (1 < r < m ÿ 1) b ab|CG(gi)| 2n 2n n 4 4

÷1 1 1 1 1 1÷2 1 1 1 ÿ1 ÿ1÷3 1 (ÿ1)m (ÿ1)r 1 ÿ1÷4 1 (ÿ1)m (ÿ1)r ÿ1 1ø j 2 2(ÿ1) j å jr � åÿ jr 0 0(1 < j < m ÿ 1)

18.4 Another group of order 12

We shall now describe a non-abelian group G of order 12 which is not

isomorphic to either A4 or D12, and we shall construct the character

table of G. It is in fact known that every non-abelian group of order


12 is isomorphic to A4, D12 or G, but we shall not prove this result

here.

Let a and b be the following permutations in S12:

a � (1 2 3 4 5 6)(7 8 9 10 11 12),

b � (1 7 4 10)(2 12 5 9)(3 11 6 8),

and let G � ka, bl, a subgroup of S12. Since a has order 6 and b =2 kal,the group G has at least 12 elements, namely

ar, arb (0 < r < 5):

Check that a and b satisfy

a6 � 1, a3 � b2, bÿ1ab � aÿ1:

It follows from these relations that every element of G has the form

arbs with 0 < r < 5, 0 < s < 1 as given above, and so |G| � 12.

The relations further imply that

CG(a) � hai, CG(a3) � G, CG(b) � f1, a3, b, a3bg:These, and similar facts, help us to ®nd the conjugacy classes of G,

which are tabulated below:

Conjugacy class Representative gi |CG(gi)|

{1} 1 12

{a3} a3 12

{a, aÿ1} a 6

{a2, aÿ2} a2 6

{b, a2b, a4b} b 4

{ab, a3b, a5b} ab 4

Therefore G has six irreducible characters.

Observe that ka2l � {1, a2, a4} v G, and

G=ha2i � fha2i, ha2ia, ha2ib, ha2iabg:Since ka2la � ka2lb2, we have G=ha2i � C4. By lifting the irreducible

characters of C4 to G, we obtain the linear characters ÷1, ÷2, ÷3, ÷4 of

G given below:


gi 1 a3 a a2 b ab|CG(gi)| 12 12 6 6 4 4

÷1 1 1 1 1 1 1÷2 1 ÿ1 ÿ1 1 i ÿi÷3 1 1 1 1 ÿ1 ÿ1÷4 1 ÿ1 ÿ1 1 ÿi i÷5 á1 á2 á3 á4 á5 á6

÷6 â1 â2 â3 â4 â5 â6

It remains to ®nd the values ár, âr taken by the last two irreducible

characters ÷5, ÷6. For this, we shall use the column orthogonality

relations, Theorem 16.4(2).

Observe that á1, â1 are the degrees of ÷5, ÷6, so they are positive

integers; also a3 is an element of order 2, so á2 and â2 are integers by

Corollary 13.10. By the column orthogonality relations applied to

columns 1 and 2, we have

4� á21 � â2

1 � 12,

4� á22 � â2

2 � 12,

á1á2 � â1â2 � 0:

Since á1, â1 are positive integers, the ®rst equation gives á1 � â1 � 2.

The other two equations then imply that á2 � ÿâ2 � �2. Since we

have not yet distinguished between ÷5 and ÷6, we may take á2 � 2 and

â2 � ÿ2.

For r . 2, the column orthogonality relations

X6

i�1

÷i(gr)÷i(g1) � 0 andX6

i�1

÷i(gr)÷i(g2) � 0

now give us two equations involving 2ár � 2âr and 2ár ÿ 2âr, respec-

tively, so we can solve them for ár and âr. Explicitly:

r � 3: 2á3 � 2â3 � 0, 4� 2á3 ÿ 2â3 � 0;

r � 4: 4� 2á4 � 2â4 � 0, 2á4 ÿ 2â4 � 0;

r � 5: 2á5 � 2â5 � 0, 2á5 ÿ 2â5 � 0;

r � 6: 2á6 � 2â6 � 0, 2á6 ÿ 2â6 � 0:


Hence

á3 � ÿ1, â3 � 1,

á4 � ÿ1, â4 � ÿ1,

á5 � 0, â5 � 0,

á6 � 0, â6 � 0:

The complete character table of G is therefore as follows:

gi 1 a3 a a2 b abCG(gi)| 12 12 6 6 4 4

÷1 1 1 1 1 1 1÷2 1 ÿ1 ÿ1 1 i ÿi÷3 1 1 1 1 ÿ1 ÿ1÷4 1 ÿ1 ÿ1 1 ÿi i÷5 2 2 ÿ1 ÿ1 0 0÷6 2 ÿ2 1 ÿ1 0 0

We can deduce that G is not isomorphic to A4 or D12 from the fact

that the character table of G is different from those of A4 and D12.

It is instructive to note that we produced the last two irreducible

characters of G by simply using the orthogonality relations, without

constructing the corresponding CG-modules. This is typical of more

advanced calculations, and illustrates the fact that it is usually much

easier to construct an irreducible character of a group than to obtain an

irreducible representation. (In fact, it is not hard to construct the

representations of the above group G with characters ÷5 and ÷6 ± see

Exercise 17.6.)


In this chapter we gave the character tables of various groups, as

follows.

1. Section 18.1: the group S4.

2. Section 18.2: the group A4.

3. Section 18.3: the dihedral groups.



1. Regard D8 as a subgroup of S4 permuting the four corners of a

square, as in Example 1.1(3). Let ð be the corresponding permuta-

tion character of D8. Find the values of ð on the elements of D8,

and express ð as a sum of irreducible characters.

2. Write down explicitly the character table of D12, and show that all

its entries are integers.

Use the character table to ®nd seven distinct normal subgroups of

D12. (Hint: use Proposition 17.5.)

3. Let G � T4n � ha, b: a2n � 1, an � b2, bÿ1ab � aÿ1i, as in

Exercise 17.6. Find the character table of G.

(Hint: use the result of Exercise 17.6. It is a good idea to do the

cases n odd and n even separately.)

4. Let G � U6n � ka, b: a2n � b3 � 1, aÿ1ba � bÿ1l, as in Exercise

17.7. Find the character table of G.

5. Let G � V8n � ha, b: a2n � b4 � 1, ba � aÿ1bÿ1, bÿ1a � aÿ1bi,with n odd, as in Exercise 17.8. Find the character table of G.


188

19

Tensor products

The idea of multiplying a character of a group G by a linear character of

G was introduced at the end of Chapter 17, and it can be extended to

include the product of any pair of characters ÷ and ø. The value of the

product ÷ø on an element g of G is simply ÷(g)ø(g). It is therefore

straightforward to calculate the product, but a little ingenuity is required

in order to justify the conclusion that the product ÷ø is a character of G.

The plan is to take CG-modules V and W with characters ÷ and ørespectively, and to put them together to form a new CG-module, called

the tensor product of V and W, which has character ÷ø.

An important special case of the product ÷ø occurs when ÷ � ø, so

we consider the character ÷2, and more generally ÷3, ÷4, and so on. If

÷ is not linear, then the degrees of ÷, ÷2, . . . increase, and by taking

successive powers of ÷ we obtain arbitrarily many new characters.

Potentially, then, we have a chance of getting a large proportion of the

character table of G from just one non-linear character ÷ of G; and

indeed, products of characters provide a very good source of new

characters from given ones. We shall illustrate this by constructing the

character tables of S5 and S6.

At the end of the chapter, we apply tensor products in a different

way, to ®nd all the irreducible characters of a direct product G 3 H,

given those of G and H.

Tensor product spaces

Let V and W be vector spaces over C with bases v1, . . . , vm and

w1, : : : , wn, respectively. For each i, j with 1 < i < m, 1 < j < n, we

introduce a symbol vi wj. The tensor product space V W is de®ned

to be the mn-dimensional vector space over C with a basis given by

fvi wj: 1 < i < m, 1 < j < ng:Thus V W consists of all expressions of the formX

i, j

ëij(vi wj) (ëij 2 C):

For v 2 V and w 2 W with v �Pmi�1 ëivi and w �Pn

j�1 ì jwj

(ëi, ì j 2 C), we de®ne v w 2 V W by

v w �X

i, j

ëiì j(vi wj):

For example,

(2v1 ÿ v2) (w1 � w2)

� 2v1 w1 � 2v1 w2 ÿ v2 w1 ÿ v2 w2:

Do not be misled by the notation into believing that every element of

V W has the form v w, because this is not the case. For instance,

it is impossible to express

v1 w1 � v2 w2

in the form v w.

19.1 Proposition

(1) If v 2 V, w 2 W and ë 2 C, then

v (ëw) � (ëv) w � ë(v w):

(2) If x1, . . . , xa 2 V and y1, . . . , yb 2 W, thenXa

i�1

xi

!

Xb

j�1

yj

0@ 1A �Xi, j

xi yj:

Proof (1) Let v �Pmi�1 ëivi and w �Pn

j�1 ì jw j. Then

v (ëw) �X

i

ëivi

!

Xj

ëì jw j

!�X

i, j

ëëiì j(vi wj),

(ëv) w �X

i

ëëivi

!

Xj

ì jw j

!�X

i, j

ëëiì j(vi wj),

ë(v w) � ëX

i, j

ëiì j(vi wj) �X

i, j

ëëiì j(vi wj):

Tensor products 189

Therefore v (ëw) � (ëv) w � ë(v w).

The proof of part (2) is equally straightforward, and we leave it as

an exercise. j

Our construction of V W depended upon choosing a basis of V

and a basis of W at the beginning; the next proposition shows that

other bases of V and W work equally well.

19.2 Proposition

If e1, . . . , em is a basis of V and f1, . . . , fn is a basis of W, then the

elements in

fei f j: 1 < i < m, 1 < j < nggive a basis of V W.

Proof Write

vi �Xm

k�1

ëik ek , wj �Xn

l�1

ì jl f l (ëik , ì jl 2 C):

Then by Proposition 19.1, we have

vi wj �Xk, l

ëikì jl(ek f l):

Now the elements vi wj (1 < i < m, 1 < j < n) are a basis of

V W , and hence the mn elements ek fl (1 < k < m, 1 < l < n)

span V W. Since V W has dimension mn, it follows that the

elements ek fl are also a basis of V W.

Tensor product modules

We have introduced the tensor product of two vector spaces, so we are

now in a position to de®ne the tensor product of two CG-modules.

Let G be a ®nite group and let V and W be CG-modules with bases

v1, . . . , vm and w1, : : : , wn, respectively. We know that the elements

vi wj (1 < i < m, 1 < j < n)

give a basis of V W. The multiplication of vi wj by an element of


G is de®ned in the following simple way, which is then extended

linearly to a multiplication on the whole of V W.

19.3 De®nition

Let g 2 G. For all i, j, de®ne

(vi wj)g � vi g w j g

and, more generally, letXi, j

ëij(vi wj)

!g �

Xi, j

ëij(vig wjg)

for arbitrary complex numbers ëij.

19.4 Proposition

For all v 2 V, w 2 W and all g 2 G, we have

(v w)g � vg wg:

Proof Let v �Pmi�1 ëivi and w �Pn

j�1 ì jwj. Then

(v w)g �X

i, j

ëiì j(vi wj)

!g by Proposition 19:1

�X

i, j

ëiì j(vi g wjg)

�X

i

ëivi g

!

Xj

ì jwjg

!by Proposition 19:1

� vg wg: j

You should be warned that (v w)r 6� vr wr for most elements r in

CG. For example, consider what happens when r is a scalar multiple of g.

19.5 Proposition

The rule for multiplying an element of V W by an element of G,

given in De®nition 19.3, makes the vector space V W into a CG-

module.

Tensor products 191

Proof Let 1 < i < m, 1 < j < n, and g, h 2 G. Then

(vi wj)g � vi g wjg 2 V W ,

(vi wj)(gh) � vi(gh) wj(gh)

� (vi g)h (wjg)h

� (vi g wjg)h by Proposition 19:4

� ((vi wj)g)h,

(vi wj)1 � vi wj,

and Xi, j

ëij(vi wj)

!g �

Xi, j

ëij((vi wj)g):

Therefore all the conditions of Proposition 4.6 are ful®lled, and V W

is a CG-module. j

We now calculate the character of V W.

19.6 Proposition

Let V and W be CG-modules with characters ÷ and ø, respectively.

Then the character of the CG-module V W is the product character

÷ø, where

÷ø(g) � ÷(g)ø(g) for all g 2 G:

Proof Let g 2 G. By Proposition 9.11 we can choose a basis e1, . . . ,

em of V and a basis f1, . . . , fn of W such that

eig � ëiei (1 < i < m) and f jg � ì j f j (1 < j < n)

for some complex numbers ëi, ì j. Then

÷(g) �Xm

i�1

ëi, ø(g) �Xn

j�1

ì j:

Now for 1 < i < m and 1 < j < n,

(ei f j)g � eig f jg � ëiì j(ei f j),

and by Proposition 19.2, these vectors ei fj form a basis of V W.

Hence, if ö is the character of V W then


ö(g) �X

i, j

ëiì j �X

i

ëi

! Xj

ì j

!� ÷(g)ø(g),

as required. j

19.7 Corollary

The product of two characters of G is again a character of G.

19.8 Example

The character table of S4 was given in Section 18.1. We reproduce it

here, and calculate ÷3÷4 and ÷4÷4.

gi 1 (1 2) (1 2 3) (1 2)(3 4) (1 2 3 4)|CG(gi)| 24 4 3 8 4

÷1 1 1 1 1 1÷2 1 ÿ1 1 1 ÿ1÷3 2 0 ÿ1 2 0÷4 3 1 0 ÿ1 ÿ1÷5 3 ÿ1 0 ÿ1 1

÷3÷4 6 0 0 ÿ2 0÷4÷4 9 1 0 1 1

We see that

÷3÷4 � ÷4 � ÷5, and

÷4÷4 � ÷1 � ÷3 � ÷4 � ÷5:

Powers of characters

Corollary 19.7 shows that if ÷ is a character of G then so is ÷2, where

÷2 � ÷÷, the product of ÷ with itself. More generally, for every non-

negative integer n, we de®ne ÷ n by

÷ n(g) � (÷(g))n for all g 2 G:

Thus ÷0 � 1G. An inductive proof using Corollary 19.7 shows that ÷ n

is a character of G. When ÷ is a faithful character (that is, Ker � {1}),

the powers of ÷ carry a lot of information about the whole character

table of G, as can be seen from Theorem 19.10 below.

Tensor products 193

In the course of the proof of Theorem 19.10, we shall need the

following result concerning the so-called `Vandermonde matrix'.

(19.9) If á1, . . . , ár are distinct complex numbers, then the matrix

A �

1 á1 á21 : : : árÿ1

1

1 á2 á22 : : : árÿ1

2

: : : : : : :

1 ár á2r : : : árÿ1

r

0BBB@1CCCA

is invertible.

We ®rst sketch a proof of this result.

Suppose that x1, . . . , xr are indeterminates, and consider

Ä � det

1 x1 x21 : : : xrÿ1

1

1 x2 x22 : : : xrÿ1

2

: : : : : : :

1 xr x2r : : : xrÿ1

r

0BBB@1CCCA:

If i 6� j and xi � xj then two rows of the given matrix are equal, so

Ä � 0. It follows that Ä is divisible byYi , j

(xi ÿ xj) � (x1 ÿ x2)(x1 ÿ x3) : : : (x1 ÿ xr)

3 (x2 ÿ x3) : : : (x2 ÿ xr)

..

.

3 (xrÿ1 ÿ xr):

Now the coef®cient of xrÿ11 xrÿ2

2 : : :xrÿ1 in this product is 1: for in the

way we have displayed the product, the term xrÿ11 must come from all

the factors in the ®rst row, xrÿ22 must come from all the factors in the

second row, and so on. On the other hand, to obtain xrÿ11 xrÿ2

2 : : : xrÿ1

in the expansion of the determinant Ä, we must take xrÿ11 from the

®rst row of the matrix, xrÿ22 from the second row, and so on. Hence

the coef®cient of xrÿ11 xrÿ2

2 . . . xrÿ1 in Ä is �1. It follows that

Ä � �Yi , j

(xi ÿ xj):


To obtain (19.9), we substitute ái for xi (1 < i < r), and deduce that

the matrix A is invertible since its determinant is non-zero.

19.10 Theorem

Let ÷ be a faithful character of G, and suppose that ÷(g) takes

precisely r different values as g varies over all the elements of G. Then

every irreducible character of G is a constituent of one of the powers

÷0, ÷1, . . . , ÷ rÿ1.

Proof Let the r values taken by ÷ be á1, . . . , ár, and for 1 < i < r,

de®neGi � fg 2 G: ÷(g) � áig:

Take á1 � ÷(1), so that G1 � Ker ÷. As ÷ is faithful, G1 � {1}.

Now let ø be an irreducible character of G. We must show that

h÷ j, øi 6� 0 for some j with 0 < j < r ÿ 1.

For 1 < i < r, let

âi �Xg2Gi

ø(g),

and note that â1 � ø(1) 6� 0. Then for all j > 0,

h÷ j, øi � 1

jGjXg2G

(÷(g)) j ø(g) � 1

jGjXr

i�1

(ái)jâi:

Let A be the r 3 r matrix with ij-entry (ái)jÿ1, and let b be the row

vector which is given byb � (â1, : : : , âr):

Now A is invertible by (19.9), and b 6� 0 since â1 6� 0; hence bA 6� 0.

But the ( j� 1)th entry in the row vector bA is equal to jGjh÷ j, øi,and thus h÷ j, øi 6� 0 for some j with 0 < j < r ÿ 1, as we wished to

prove. j

19.11 Examples

(1) If G 6� {1} and ÷ is the regular character of G, then ÷(g) takes just

two different values (see Proposition 13.20), so Theorem 19.10 says

that every irreducible character of G is a constituent of 1G or ÷; we

know this already, by Theorem 10.5.

(2) Let G � S4, and refer to Example 19.8. Let ÷ � ÷4. Then ÷(g)

takes four different values. We have seen that

÷2 � ÷1 � ÷3 � ÷4 � ÷5

Tensor products 195

and we ®nd that

h÷3, ÷2i � 1:

Thus ÷0, ÷1, ÷2, ÷3 (indeed, in this case, just ÷2, ÷3) have between

them as constituents all the irreducible characters ÷1, . . . , ÷5 of G,

illustrating Theorem 19.10.

Decomposing ÷2

In view of Theorem 19.10, it is of some importance to be able to

decompose powers of a character ÷ into sums of irreducible characters.

We are going to provide a method for decomposing ÷2, the square of

÷. This special case is particularly useful in ®nding irreducible char-

acters, as we shall see.

Let V be a CG-module with character ÷. By Proposition 19.6, the

module V V has character ÷2. Let v1, . . . , vn be a basis of V, and

de®ne a linear transformation T: V V! V V by

(vi v j)T � v j vi for all i, j

and extending linearly ± that is,Xi, j

ëij(vi v j)

!T �

Xi, j

ëij(v j vi):

Check that for all v, w 2 V, we have

(v w)T � w v:

Hence T is independent of the choice of basis.

Now de®ne subsets of V V as follows:

S(V V ) � fx 2 V V : xT � xg,A(V V ) � fx 2 V V : xT � ÿxg,

Since T is linear, it is easy to see that S(V V) and A(V V) are

subspaces of V V (indeed, they are eigenspaces of T). The subspace

S(V V) is called the symmetric part of V V, and the subspace

A(V V ) is known as the antisymmetric part of V V.

19.12 Proposition

The subspaces S(V V) and A(V V) are CG-submodules of V V.

Also,V V � S(V V )� A(V V ):


Proof For all ëij 2 C and g 2 G,Xi, j

ëij(vi v j)

!Tg �

Xi, j

ëij(v j g vi g)

�X

i, j

ëij(vi g v j g)T

�X

i, j

ëij(vi v j)

!gT :

Therefore T is a CG-homomorphism from V V to itself. Hence, for

x 2 S(V V), y 2 A(V V) and g 2 G, we have

(xg)T � (xT )g � xg, and

(yg)T � (yT )g � ÿyg,

so xg 2 S(V V) and yg 2 A(V V). Thus S(V V) and A(V V)

are CG-submodules of V V.

If x 2 S(V V) \ A(V V) then x � xT � ÿx, so x � 0. Further, for

all x 2 V we have

x � 12(x� xT )� 1

2(xÿ xT ):

Since T 2 is the identity, 12(x� xT ) 2 S(V V ) and 1

2(xÿ xT ) 2

A(V V ). Therefore,

V V � S(V V )� A(V V ): j

Note that the symmetric part of V V contains all vectors which

have the form v w � w v with v, w 2 V, while the antisymmetric

part of V V contains all vectors of the form v w ÿ w v. We now

present bases of the symmetric and antisymmetric parts of V V

which consist of elements like these.

19.13 Proposition

Let v1, . . . , vn be a basis of V.

(1) The vectors vi v j � v j vi (1 < i < j < n) form a basis of

S(V V ). The dimension of S(V V ) is n(n� 1)=2.

(2) The vectors vi v j ÿ v j vi (1 < i , j < n) form a basis of

A(V V). The dimension of A(V V ) is n(nÿ 1)=2.

Tensor products 197

Proof Clearly the vectors vi v j � v j vi (1 < i < j < n) are linearly

independent elements of S(V V ), and the vectors vi v j ÿ v j vi

(1 < i , j < n) are linearly independent elements of A(V V ). Hence

dim S(V V ) > n(n� 1)=2, dim A(V V ) > n(nÿ 1)=2:

By Proposition 19.12,

dim S(V V ) � dim A(V V ) � dim V V � n2:

Hence the above inequalities are equalities, and the result follows. j

De®ne ÷S to be the character of the CG-module S(V V ), and ÷A

to be the character of the CG-module A(V V ). By Proposition 19.12,

÷2 � ÷S � ÷A:

The next result gives the values of the characters ÷S and ÷A.

19.14 Proposition

For g 2 G, we have

÷S(g) � 12(÷2(g)� ÷(g2)), and

÷A(g) � 12(÷2(g)ÿ ÷(g2)):

Proof By Proposition 9.11 we can choose a basis e1, . . . , en of V such

that ei g � ëiei (1 < i < n) for some complex numbers ëi. Then

(ei ej ÿ ej ei)g � ëië j(ei ej ÿ ej ei),

and hence from Proposition 19.13(2),

÷A(g) �Xi , j

ëië j:

Now ei g2 � ë2i ei, so ÷(g) �Pi ëi and ÷(g2) �Pi ë2

i . Therefore

÷2(g) � (÷(g))2 �X

i

ë2i � 2

Xi , j

ëië j � ÷(g2)� 2÷A(g):

Hence

÷A(g) � 12(÷2(g)ÿ ÷(g2)):

Also, ÷2 � ÷S � ÷A, which implies that

÷S(g) � ÷2(g)ÿ ÷A(g) � 12(÷2(g)� ÷(g2)): j


19.15 Example

Let G � S4. The character table of G is given in Example 19.8. Let

÷ � ÷4. The values of ÷, and the values of ÷S and ÷A, given by

Proposition 19.14, appear below.

gi 1 (1 2) (1 2 3) (1 2)(3 4) (1 2 3 4)|CG(gi)| 24 4 3 8 4

÷ 3 1 0 ÿ1 ÿ1÷S 6 2 0 2 0÷A 3 ÿ1 0 ÿ1 1

We ®nd that ÷S � ÷1 � ÷3 � ÷4 and ÷A � ÷5.

The techniques which we have developed so far give a useful method

for ®nding new irreducible characters of a group, given one or two

irreducible characters to start with. The strategy is simple:

(1) Given a character ÷, form ÷S and ÷A, and use inner products to

analyse ÷S and ÷A for new irreducible characters.

(2) If ø is a new character found in (1), then form øS and øA and

repeat.

We illustrate this strategy with two examples.

19.16 Example The character table of S5

Let G � S5, the symmetric group of degree 5. By Example 12.16(4), G

has conjugacy class representatives gi, conjugacy class sizes and

centralizer orders |CG(gi)| as follows:

gi 1 (1 2) (1 2 3) (1 2)(3 4) (1 2 3 4) (1 2 3)(4 5) (1 2 3 4 5)Class size 1 10 20 15 30 20 24|CG(gi)| 120 12 6 8 4 6 5

Thus G has exactly seven irreducible characters.

(a) Linear characters By Example 17.13, G9 � A5 and G has exactly

two linear characters ÷1, ÷2, obtained by lifting the irreducible char-

acters of G=G9. We have

Tensor products 199

÷1 � 1G, and

÷2(g) � 1, if g is an even permutation,

ÿ1, if g is an odd permutation:

�(b) Permutation character Proposition 13.24 gives us a character ÷3

of G with values

÷3(g) � jfix (g)j ÿ 1 (g 2 G):

Observe that

h÷3, ÷3i � 42

120� 22

12� 12

6� (ÿ1)2

6� (ÿ1)2

5� 1:

Hence ÷3 is irreducible, by Theorem 14.20. Next, Proposition 17.14

shows that ÷4 � ÷3÷2 is also an irreducible character.

At this point we have the following portion of the character table of

G:

gi 1 (1 2) (1 2 3) (1 2)(3 4) (1 2 3 4) (1 2 3)(4 5) (1 2 3 4 5)|CG(gi)| 120 12 6 8 4 6 5

÷1 1 1 1 1 1 1 1÷2 1 ÿ1 1 1 ÿ1 ÿ1 1÷3 4 2 1 0 0 ÿ1 ÿ1÷4 4 ÿ2 1 0 0 1 ÿ1

(c) Tensor products We now use tensor products to construct the last

three irreducible characters of G.

Write ÷ � ÷3. By Proposition 19.14 the values of the characters ÷S

and ÷A are

gi 1 (1 2) (1 2 3) (1 2)(3 4) (1 2 3 4) (1 2 3)(4 5) (1 2 3 4 5)|CG(gi)| 120 12 6 8 4 6 5

÷S 10 4 1 2 0 1 0÷A 6 0 0 ÿ2 0 0 1

Thus,


h÷A, ÷Ai � 36

120� 4

8� 1

5� 1,

and so ÷A is a new irreducible character, which we call ÷5.

Next,

h÷S , ÷1i � 10

120� 4

12� 1

6� 2

8� 1

6� 1,

h÷S , ÷3i � 40

120� 8

12� 1

6ÿ 1

6� 1, and

h÷S , ÷Si � 100

120� 16

12� 1

6� 4

8� 1

6� 3,

Therefore,

÷S � ÷1 � ÷3 � ø,

where ø is an irreducible character of degree 5. Let ÷6 � ø, so that

÷6 � ÷S ÿ ÷1 ÿ ÷3.

Finally, ÷7 � ÷6÷2 is a different irreducible character of degree 5.

We have now found all seven irreducible characters of S5. The

character table of S5 is as shown.


gi 1 (1 2) (1 2 3) (1 2)(3 4) (1 2 3 4) (1 2 3)(4 5) (1 2 3 4 5)|CG(gi)| 120 12 6 8 4 6 5

÷1 1 1 1 1 1 1 1÷2 1 ÿ1 1 1 ÿ1 ÿ1 1÷3 4 2 1 0 0 ÿ1 ÿ1÷4 4 ÿ2 1 0 0 1 ÿ1÷5 6 0 0 ÿ2 0 0 1÷6 5 1 ÿ1 1 ÿ1 1 0÷7 5 ÿ1 ÿ1 1 1 ÿ1 0

19.17 Example The character table of S6

In this example, we use techniques similar to those of the previous

example to ®nd 8 of the 11 irreducible characters of the symmetric

group S6; we then ®nd the last three irreducible characters by using the

orthogonality relations.

Let G � S6, of order 720. For ease of printing, it is convenient to

label each conjugacy class by the cycle-shape of its elements. Using

Tensor products 201

this notation, the conjugacy class sizes and centralizer orders are as

follows (see Exercise 12.3):

Cycle-shape (1) (2) (3) (2,2) (4) (3,2) (5) (2,2,2) (3,3) (4,2) (6)Class size 1 15 40 45 90 120 144 15 40 90 120|CG(gi)| 720 48 18 16 8 6 5 48 18 8 6

Since G has 11 conjugacy classes, it has 11 irreducible characters.

(a) Linear characters As with all symmetric groups Sn (n > 2), the

derived subgroup is An, and we get exactly two linear characters ÷1

and ÷2, where

÷1 � 1G,

÷2(g) � 1, if g is even,

ÿ1, if g is odd

�(see Example 17.13).

(b) Permutation character and tensor products The function ÷3 which

is given by

÷3(g) � jfix (g)j ÿ 1 (g 2 G)

is a character of G, by Proposition 13.24. Let ÷ � ÷3. The values of ÷,

÷S and ÷A are as follows:

Class (1) (2) (3) (2,2) (4) (3,2) (5) (2,2,2) (3,3) (4,2) (6)|CG(gi)| 720 48 18 16 8 6 5 48 18 8 6

÷ � ÷3 5 3 2 1 1 0 0 ÿ1 ÿ1 ÿ1 ÿ1÷S 15 7 3 3 1 1 0 3 0 1 0÷A 10 2 1 ÿ2 0 ÿ1 0 ÿ2 1 0 1

We calculate that

h÷3, ÷3i � 1, h÷A, ÷Ai � 1,

h÷S , ÷1i � 1, h÷S , ÷3i � 1,

h÷S , ÷Si � 3:


Therefore ÷3 is irreducible; so is ÷4 � ÷3÷2. Also, ÷5 � ÷A is irreduci-

ble, as is ÷6 � ÷5÷2. Further,

÷S � ÷1 � ÷3 � ÷7,

where ÷7 is another irreducible character, of degree 9. Finally,

÷8 � ÷7÷2 is also irreducible.

The irreducible characters ÷1, . . . , ÷8 are recorded in the following

portion of the character table of G.

Class (1) (2) (3) (2,2) (4) (3,2) (5) (2,2,2) (3,3) (4,2) (6)|CG(gi)| 720 48 18 16 8 6 5 48 18 8 6

÷1 1 1 1 1 1 1 1 1 1 1 1÷2 1 ÿ1 1 1 ÿ1 ÿ1 1 ÿ1 1 1 ÿ1÷3 5 3 2 1 1 0 0 ÿ1 ÿ1 ÿ1 ÿ1÷4 5 ÿ3 2 1 ÿ1 0 0 1 ÿ1 ÿ1 1÷5 10 2 1 ÿ2 0 ÿ1 0 ÿ2 1 0 1÷6 10 ÿ2 1 ÿ2 0 1 0 2 1 0 ÿ1÷7 9 3 0 1 ÿ1 0 ÿ1 3 0 1 0÷8 9 ÿ3 0 1 1 0 ÿ1 ÿ3 0 1 0

(c) Orthogonality relations We now use the column orthogonality

relations to complete the character table of G. It will be shown later

(Corollary 22.16) that all the entries in the character tables of all

symmetric groups are integers, but for the moment we know for certain

only that ÷(g) is an integer if g2 � 1 (see Corollary 13.10). It is

therefore convenient ®rst to concentrate on elements of order 2, so that

we can guarantee that the solutions to the equations which we deal

with are integers.

Let s denote the permutation (1 2) and t denote the permutation

(1 2)(3 4). Thus the conjugacy classes of s and t correspond to the

second and fourth columns of the character table, respectively, in the

ordering which we have adopted. From Corollary 13.10 we know that

÷(s) and ÷(t) are integers for all characters ÷ of G.

Ingeniously, we call the three irreducible characters of G which have

yet to be found ÷9, ÷10 and ÷11.

The column orthogonality relations give

X11

i�1

÷i(s)2 � 48:

Tensor products 203

Hence

÷9(s)2 � ÷10(s)2 � ÷11(s)2 � 2:

We can assume, without loss of generality, that

÷9(s)2 � ÷10(s)2 � 1, ÷11(s)2 � 0:

Now ÷9÷2 is an irreducible character, and is not equal to any of

÷1, . . . , ÷8. Moreover, since ÷9÷2(s) � ÿ÷9(s), we see that ÷9÷2 is not

÷9 or ÷11. Therefore,

÷9÷2 � ÷10:

Once more, we lose no generality in assuming that

÷9(s) � 1, ÷10(s) � ÿ1:

The plan is now to ®nd ÷i(1) and ÷i(t) for i � 9, 10, 11. That is, we

aim to evaluate the integers a, b, c, d, e, f in the following portion of

the character table:

The column orthogonality relations giveX11

i�1

÷i(1)÷i(s) � 0,X11

i�1

÷i(s)÷i(t) � 0,

X11

i�1

÷i(t)÷i(t) � 16,X11

i�1

÷i(1)÷i(t) � 0,

whence

aÿ b � 0, d ÿ e � 0,

d2 � e2 � f 2 � 2, ad � be� cf � 10:

The only solution to these equations in integers with a . 0 and b . 0

is

d � e � 1, f � 0, a � b � 5:

Element 1 s tClass (1) (2) (2,2)

÷9 a 1 d÷10 b ÿ1 e÷11 c 0 f


Finally, we ®nd that c � 16 by using the relation

X11

i�1

÷i(1)2 � 720:

The above portion of the character table is therefore

We can now determine the three unknown entries in each further

column, since the column orthogonality relations will give three

independent equations in these unknowns (as the above 3 3 3 matrix is

invertible). Having done these calculations, we ®nd that the complete

character table of S6 is as shown.

Direct products

We conclude the chapter by showing that tensor products can be used

to determine the character table of a direct product G 3 H, given the

character tables of G and H.

Let V be a CG-module, with basis v1, . . . , vm, and let W be a

Element 1 s tClass (1) (2) (2,2)

÷9 5 1 1÷10 5 ÿ1 1÷11 16 0 0


Class (1) (2) (3) (2,2) (4) (3,2) (5) (2,2,2) (3,3) (4,2) (6)|CG(gi)| 720 48 18 16 8 6 5 48 18 8 6

÷1 1 1 1 1 1 1 1 1 1 1 1÷2 1 ÿ1 1 1 ÿ1 ÿ1 1 ÿ1 1 1 ÿ1÷3 5 3 2 1 1 0 0 ÿ1 ÿ1 ÿ1 ÿ1÷4 5 ÿ3 2 1 ÿ1 0 0 1 ÿ1 ÿ1 1÷5 10 2 1 ÿ2 0 ÿ1 0 ÿ2 1 0 1÷6 10 ÿ2 1 ÿ2 0 1 0 2 1 0 ÿ1÷7 9 3 0 1 ÿ1 0 ÿ1 3 0 1 0÷8 9 ÿ3 0 1 1 0 ÿ1 ÿ3 0 1 0÷9 5 1 ÿ1 1 ÿ1 1 0 ÿ3 2 ÿ1 0÷10 5 ÿ1 ÿ1 1 1 ÿ1 0 3 2 ÿ1 0÷11 16 0 ÿ2 0 0 0 1 0 ÿ2 0 0

Tensor products 205

CH-module, with basis w1, . . . , wn. For all i, j with 1 < i < m and

1 < j < n, and all g 2 G, h 2 H, de®ne

(vi wj)(g, h) � vi g wjh

and extend this de®nition linearly to the whole of V W, that is, for

ëij 2 C, Xi, j

ëij(vi wj)

!(g, h) �

Xi, j

ëij(vi g wjh):

As in Proposition 19.4, we ®nd that

(v w)(g, h) � vg wh,

for all v 2 V, w 2 W. Then a proof similar to that of Proposition 19.5

shows that V W is a C(G 3 H)-module.

Let ÷ be the character of V and ø be the character of W. By the

proof of Proposition 19.6, the character of V W is ÷ 3 ø, where

(÷ 3 ø)(g, h) � ÷(g)ø(h) (g 2 G, h 2 H):

19.18 Theorem

Let ÷1, . . . , ÷a be the distinct irreducible characters of G and let

ø1, . . . , øb be the distinct irreducible characters of H. Then G 3 H

has precisely ab distinct irreducible characters, and these are

÷i 3 ø j (1 < i < a, 1 < j < b):

Proof For all i, j, k, l,

h÷i 3 ø j, ÷k 3 ÷ liG3 H � 1

jG 3 H jXg2Gh2H

÷i(g)ø j(h)÷k(g)ø l(h)

� 1

jGjXg2G

÷i(g)÷k(g)

!1

jH jXh2H

ø j(h)ø l(h)

!� h÷i, ÷kiGhø j, ø liH � äikä jl:

(Here the subscripts G 3 H, G and H indicate inner products of

characters of G 3 H, G and H, respectively.) Thus the ab characters

÷i 3 ø j are distinct and irreducible.

Next, note that for all g, x 2 G and h, y 2 H, we have

(x, y)ÿ1(g, h)(x, y) � (xÿ1 gx, yÿ1 hy):


Hence elements (g, h) and (g9, h9) of G 3 H are conjugate if and only

if the elements g and g9 are conjugate in G and the elements h and

h9 are conjugate in H. Consequently, if g1, . . . , ga are representatives

of the conjugacy classes of G and h1, . . . , hb are representatives of the

conjugacy classes of H, then the elements

(gi, hj) (1 < i < a, 1 < j < b)

are representatives of the conjugacy classes of G 3 H. In particular,

G 3 H has precisely ab conjugacy classes.

By Theorem 15.3, G 3 H has exactly ab irreducible characters, so

the irreducible characters ÷i 3 ø j which we have found must be all the

irreducible characters of G 3 H. j

19.19 Example The character table of S3 3 C2

The character table of S3 (� D6) is given in Example 16.3(1). We

reproduce it here, alongside the character table of C2.

The conjugacy classes of S3 3 C2 are represented by

(1, 1), ((1 2), 1), ((1 2 3), 1), (1, ÿ1), ((1 2), ÿ1), ((1 2 3), ÿ1),

and by Theorem 19.18, the character table of S3 3 C2 is as shown.


gi 1 (1 2) (1 2 3)|CG(gi)| 6 2 3

÷1 1 1 1÷2 1 ÿ1 1÷3 2 0 ÿ1

Character table of C2

hi 1 ÿ1|CH (hi)| 2 2

ø1 1 1ø2 1 ÿ1

Character table of S3 3 C2

(gi, hj) (1, 1) ((1 2), 1) ((1 2 3), 1) (1, ÿ1) ((1 2), ÿ1) ((1 2 3), ÿ1)|CG3 H (gi, hj)| 12 4 6 12 4 6

÷1 3 ø1 1 1 1 1 1 1÷2 3 ø1 1 ÿ1 1 1 ÿ1 1÷3 3 ø1 2 0 ÿ1 2 0 ÿ1÷1 3 ø2 1 1 1 ÿ1 ÿ1 ÿ1÷2 3 ø2 1 ÿ1 1 ÿ1 1 ÿ1÷3 3 ø2 2 0 ÿ1 ÿ2 0 1

Tensor products 207

Compare the solution to Exercise 18.2, where we give the character

table of D12 (Exercise 1.5 shows that D12 � S3 3 C2).


1. The product of any two characters of G is a character of G.

2. If ÷ is a character of G, then so are ÷S and ÷A, where

÷S(g) � 12(÷2(g)� ÷(g2)),

÷A(g) � 12(÷2(g)ÿ ÷(g2))

for all g 2 G.

3. The irreducible characters of G 3 H are those characters ÷ 3 ø,

where ÷ is an irreducible character of G and ø is an irreducible

character of H. The values of ÷ 3 ø are given by

(÷ 3 ø)(g, h) � ÷(g)ø(h)

for all g 2 G, h 2 H.


1. Let ÷, ø and ö be characters of the group G. Show that

h÷ø, öi � h÷, øöi � hø, ÷öi:2. Suppose that ÷ and ø are irreducible characters of G. Prove that

h÷ø, 1Gi � 1, if ÷ � ø,

0, if ÷ 6� ø:

�3. Let ÷ be a character of G which is not faithful. Show that there is

some irreducible character ø of G such that k÷ n, øl � 0 for all

integers n with n > 0.

(This shows that the hypothesis that ÷ is faithful cannot be dropped

from Theorem 19.10.)

4. In Example 20.13 of the next chapter we shall show that there exist

irreducible characters ÷ and ö of A5 which take the following

values:


1 (1 2 3) (1 2)(3 4) (1 2 3 4 5) (1 3 4 5 2)

÷ 5 ÿ1 1 0 0ö 3 0 ÿ1 (1�p5)=2 (1ÿp5)=2

Calculate the values of ÷S , ÷A, öS and öA. Express these characters

as linear combinations of the irreducible characters ø1, . . . , ø5 of

A5 which are given in Example 20.13.

5. A certain group G or order 24 has precisely seven conjugacy classes

with representatives g1, . . . , g7; further, G has a character ÷ with

values as follows:

gi g1 g2 g3 g4 g5 g6 g7

|CG(gi)| 24 24 4 6 6 6 6

÷ 2 ÿ2 0 ÿù2 ÿù ù ù2

where ù � e2ði=3. Moreover, g21, g2

2, g23, g2

4, g25, g2

6, g27 are con-

jugate to g1, g1, g2, g5, g4, g4, g5, respectively.

Find ÷S and ÷A, and show that both are irreducible.

By forming products of the irreducible characters found so far,

®nd the character table of G.

6. Write down the character table of D6 3 D6.

Tensor products 209

210

20

Restriction to a subgroup

In this chapter and the next, we are going to look at ways of relating the

representations of a group to the representations of its subgroups. Here, we

introduce the elementary idea of restricting a CG-module to a subgroup H

of G, and illustrate its use. The case where H is a normal subgroup of G is

of particular interest, and Clifford's Theorem 20.8 gives important informa-

tion in this case. We apply this result in the situation where H is of index 2

in G, which occurs, for example, when G � Sn and H � An.

Restriction

Let H be a subgroup of the ®nite group G. Then CH is a subset of CG. If

V is a CG-module, then V is also a CH-module, since properties (1)±(5) of

De®nition 4.2 certainly hold for all g, h 2 H if they hold for all g, h 2 G.

This simple way of converting a CG-module into a CH-module is known

as restricting from G to H. If V is a CG-module then we write the

corresponding CH-module as V # H , and call it the restriction of V to H.

The character of V # H is obtained from the character ÷ of V by

evaluating ÷ on the elements of H only. We write this character of H

as ÷ # H , and refer to it as the restriction of ÷ to H. More generally,

if f: G! C is any function, then f # H denotes the restriction of f to

H (so that ( f # H)(h) � f (h) for all h 2 H).

20.1 Example

Let G � D8 � ka, b: a4 � b2 � 1, bÿ1ab � aÿ1l. As in Example 4.5(1),

let V be the CG-module with basis v1, v2 for which

v1a � v2, v1b � v1,

v2a � ÿv1, v2b � ÿv2:

If H is the subgroup {1, a2, b, a2b} of G, then V # H is the CH-

module with basis v1, v2 for which

v1a2 � ÿv1, v1b � v1,

v2a2 � ÿv2, v2b � ÿv2:

The character ÷ of V is given by

g 1 a a2 a3 b ab a2b a3b

÷(g) 2 0 ÿ2 0 0 0 0 0

and the character ÷ # H of V # H is given by

If V is a CG-module and H is a subgroup of G, then

dim V � dim (V # H). However, it might be the case that V is an

irreducible CG-module while V # H is not an irreducible CH-module;

Example 20.1 illustrates this fact. On the other hand, if V # H is an

irreducible CH-module then V is an irreducible CG-module; for if U

is a CG-submodule of V, then U # H is a CH-submodule of V # H .

20.2 Example

Let G � S5 and let H be the subgroup A4 of G consisting of all even

permutations of {1, 2, 3, 4} ®xing 5. By 18.2, the character table of H

is

h 1 a2 b a2b

÷(h) 2 ÿ2 0 0

gi 1 (1 2)(3 4) (1 2 3) (1 3 2)jC H (gi)j 12 4 3 3

ø1 1 1 1 1ø2 1 1 ù ù2

ø3 1 1 ù2 ùø4 3 ÿ1 0 0

Restriction to a subgroup 211

(where ù � e2ði=3). The character table of G is given in Example

19.16, with irreducible characters labelled ÷1, . . . , ÷7.

For each i with 1 < i < 7, we calculate the character ÷i # H as a

sum of irreducible characters ø j. From Example 19.16 we see that

÷1 # H � ÷2 # H , ÷3 # H � ÷4 # H and ÷6 # H � ÷7 # H :

Therefore we need only consider ÷1 # H, ÷3 # H , ÷5 # H and ÷6 # H .

These have values

It is easy to spot that

÷1 # H � ø1,

÷3 # H � ø1 � ø4,

÷5 # H � 2ø4,

÷6 # H � ø2 � ø3 � ø4:

Constituents of a restricted character

To help us discuss the way in which a restricted character ÷ # H can

be expressed in terms of the irreducible characters of H, we introduce

the following notation.

20.3 De®nitions

The inner product k , lG is the inner product on the vector space of

functions from G to C which we have de®ned earlier, and k , l H is the

inner product on the vector space of functions from H to C, de®ned

similarly. Thus, if W1 and W2 are functions from G to C, then

hW1, W2iG � 1

jGjXg2G

W1(g)W2(g),

1 (1 2)(3 4) (1 2 3) (1 3 2)

÷1 # H 1 1 1 1÷3 # H 4 0 1 1÷5 # H 6 ÿ2 0 0÷6 # H 5 1 ÿ1 ÿ1


and if ö1 and ö2 are functions from H to C, then

hö1, ö2iH � 1

jH jXh2H

ö1(h)ö2(h):

If ÷ is a character of G and ø1, . . . , ør are the irreducible

characters of the subgroup H of G, then by Theorem 14.17,

÷ # H � d1ø1 � : : :� drør

for some non-negative integers d1, : : : , d r which are given by

di � h÷ # H , øiiH :

We say that øi is a constituent of ÷ # H if the coef®cient di in the

above expression is non-zero.

The next proposition shows that every irreducible character of H is a

constituent of the restriction of some irreducible character of G.

20.4 Proposition

Let H be a subgroup of G and let ø be a non-zero character of H.

Then there exists an irreducible character ÷ of G such that

h÷ # H , øiH 6� 0:

Proof Let ÷1, . . . , ÷k be the irreducible characters of G. Recall from

Theorem 13.19 and Proposition 13.20 that the regular character ÷reg of

G satis®es

÷reg(g) � jGj if g � 1,

0 if g 6� 1,

�and ÷reg �

Xk

i�1

÷i(1)÷i:

Now

0 6� jGjjH jø(1) � h÷reg # H , øiH �Xk

i�1

÷i(1)h÷i # H , øiH :

Therefore h÷i # H , øiH 6� 0 for some i. j

Suppose that we know the character table of G. In the light of

Proposition 20.4, we could hope to ®nd the character table of the

subgroup H by restricting the irreducible characters ÷ of G to H.

Unfortunately, it may be very dif®cult in practice to write down the

restrictions ÷ # H in terms of irreducible characters of H. The best


chance of doing this occurs when the index jG: H j(� jGj=jH j) is

small, since the restrictions ÷ # H then do not have many constituents,

as the following result shows.

20.5 Proposition

Let H be a subgroup of G, let ÷ be an irreducible character of G, and

let ø1, : : : , ør be the irreducible characters of H. Then ÷ # H �d1ø1 � : : : � d rør, where the non-negative integers d1, . . . , d r

satisfy Xr

i�1

d2i < jG: H j:(20:6)

Moreover, we have equality in (20.6) if and only if ÷(g) � 0 for all

elements g of G which lie outside H.

Proof By Theorem 14.17, we haveXr

i�1

d2i � h÷ # H , ÷ # HiH � 1

jH jXh2H

÷(h)÷(h):

Also, since ÷ is irreducible,

1 � h÷, ÷iG � 1

jGjXg2G

÷(g)÷(g)

� 1

jGjXh2H

÷(h)÷(h)� K

� jH jjGjXr

i�1

d2i � K,

where K � (1=jGj)P g=2H÷(g)÷(g): Now K > 0, and K � 0 if and only

if ÷(g) � 0 for all g with g =2 H. The conclusions of the proposition

follow at once. j

We can say more about the constituents of ÷ # H in the case where

H is a normal subgroup of G. For example, we will see that all the

constituents of ÷ # H have the same degree.

The way to exploit the fact that H v G us revealed in the following

proposition.


20.7 Proposition

Suppose that H v G. Let V be an irreducible CG-module and U be an

irreducible CH-submodule of V # H. For every g 2 G let Ug �fug: u 2 Ug. Then

(1) The set Ug is an irreducible CH-submodule of V # H and

dim Ug � dim U .

(2) As a CH-module, V is a direct sum of some of the CH-

modules Ug.

(3) If g1, g2, g 2 G and Ug1 and Ug2 are isomorphic CH-modules,

then Ug1 g and Ug2 g are isomorphic CH-modules.

Proof (1) Clearly, Ug is a subspace of V; and since H v G we have

ghgÿ1 2 H for all h 2 H, so

(ug)h � u(ghgÿ1)g 2 Ug (u 2 U),

proving that Ug is a CH-submodule of V # H. Further, if W is a CH-

submodule of Ug, then Wgÿ1 is a CH-submodule of U; since U is

irreducible, Wgÿ1 � {0} or U, whence W � {0} or Ug. Therefore, Ug

is an irreducible CH-module, as claimed. Moreover, u! ug (u 2 U) is

an invertible linear transformation from U to Ug, so dim U � dim Ug.

(2) The sum of all the subspaces Ug with g 2 G is a CG-sub-

module of V. Therefore, since V is irreducible, we have

V �Xg2G

Ug:

Then by Proposition 7.12, V is a direct sum of some of the CH-

modules Ug.

(3) Now let ö be a CH-isomorphism from Ug1 to Ug2. De®ne

è : Ug1 g! Ug2 g by

è : wg ! (wö)g (w 2 Ug1):

Then è is clearly an isomorphism of vector spaces. Suppose that

h 2 H . Then gh � h9g for some h9 2 H, and

(wgh)è � (wh9g)è � (wh9ö)g � (wö)h9g

� (wö)gh � (wgè)h:

Therefore, è is a CH-isomorphism, and the proof of the proposition is

complete. j


We now come the fundamental theorem on the restriction of a

character to a normal subgroup.

20.8 Clifford's Theorem

Suppose that H v G and that ÷ is an irreducible character of G. Then

(1) all the constituents of ÷ # H have the same degree; and

(2) if ø1, . . . , øm are the constituents of ÷ # H, then

÷ # H � e(ø1 � : : : � øm)

for some positive integer e.

Proof Let V be a CG-module with character ÷. Then it follows from

Proposition 20.7, parts (1) and (2), that all the constituents of ÷ # H

have the same degree.

Let e � h÷ # H , ø1i. Then V contains a CH-module X1 whose

character is eø1, and which is therefore a direct sum of e isomorphic

CH-modules, each having character ø1; say

X 1 � U1 � : : : � Ue:

By Proposition 20.7(3), if g 2 G then X1 g is a direct sum of

isomorphic CH-modules. On the other hand, V is a sum of CH-

modules of the form X1 g, by Proposition 20.7(2). Hence V has the

form

V � X 1 � : : : � X m

where each Xi is a direct sum of e isomorphic CH-modules, and

Xi 6� Xj if i 6� j. Therefore,

÷ # H � e(ø1 � : : : � øm): j

Our main applications of Clifford's Theorem will concern the case

where jG: H j � 2, but you might like to look at Corollary 22.14 in

Chapter 22 to see an advanced use of the theorem.

Normal subgroups of index 2

We are shortly going to give more precise information about the

constituents of ÷ # H when H is a normal subgroup of G of index 2

(that is, jG: H j � 2). Examples where this happens are G � Sn,


H � An, or G � D2n � ha, b: an � b2 � 1, bÿ1ab � aÿ1i, H � hai. In

fact, if H is a subgroup of index 2 in G, then H must be normal in G

(see Exercise 1.10).

When H is a normal subgroup of index 2 in G, the character tables

of G and H are closely related. We describe this relationship in

(20.13) below, and we then illustrate the results by ®nding the character

table of A5 from that of S5 (which we have already obtained in

Example 19.16).

20.9 Proposition

Suppose that H is a normal subgroup of index 2 in G, and let ÷ be an

irreducible character of G. Then either

(1) ÷ # H is irreducible, or

(2) ÷ # H is the sum of two distinct irreducible characters of H of the

same degree.

Proof If ø1, . . . , ør are the irreducible characters of H, then by

Proposition 20.5,

÷ # H � d1ø1 � : : :� drør,

wherePr

i�1d2i < 2. Since d1, : : : , d r are non-negative integers, we

deduce that either ÷ # H � øi for some i, or ÷ # H � øi � ø j for some

i, j with i 6� j. In the latter case, øi and ø j have the same degree, by

Clifford's Theorem 20.8 j

For practical purposes, it is often desirable to have more details

about the two cases in Proposition 20.9, and we shall supply these

next.

Since G=H � C2, we may lift the non-trivial linear character of

G=H to obtain a linear character ë of G which satis®es

ë(g) �1 if g 2 H ,

ÿ1 if g =2 H :

(Note that for all irreducible characters ÷ of G, ÷ and ÷ë are irreducible

characters of the same degree (see Proposition 17.14). Also,

÷ # H � ÷ë # H , since ë(h) � 1 for all h 2 H.

20.10 Proposition

Suppose that H is a normal subgroup of index 2 in G, and that ÷ is


an irreducible character of G. Then the following three conditions are

equivalent:

(1) ÷ # H is irreducible;

(2) ÷(g) 6� 0 for some g 2 G with g =2 H;

(3) the characters ÷ and ÷ë of G are not equal.

Proof We use Proposition 20.5; since jG: H j � 2, ÷ # H is irreducible

if and only if the inequality in (20.6) is strict, and this happens if and

only if ÷(g) 6� 0 for some g 2 G with g =2 H. Thus (1) is equivalent to

(2).

To see that (2) is equivalent to (3), observe that

÷ë(g) �÷(g) if g 2 H ,

ÿ÷(g) if g =2 H ,

(

so ÷(g) 6� 0 for some g with g =2 H if and only if ÷ë 6� ÷. j

According to Proposition 20.9, if H is a normal subgroup of G of

index 2, and ÷ is an irreducible character of G, then ÷ # H is the sum

of one or two irreducible characters of H. In the next proposition we

consider the ®rst possibility.

20.11 Proposition


an irreducible character of G for which ÷ # H is irreducible. If ö is an

irreducible character of G which satis®es

ö # H � ÷ # H ,

then either ö � ÷ or ö � ÷ë.

Proof We have

(÷� ÷ë)(g) �2÷(g) if g 2 H ,

0 if g =2 H :

(

Hence


h÷� ÷ë, öiG � 1

jGjXg2H

2÷(g)ö(g)

� 1

jH jXg2H

÷(g)ö(g)

� h÷ # H , ö # HiH :

Now k÷ # H, ö # Hl H � 1, since ÷ # H is irreducible and

ö # H � ÷ # H. Therefore k÷ � ÷ë, ölG � 1, and so either ö � ÷ or

ö � ÷ë. j

Finally, we look at the case where ÷ # H is reducible.

20.12 Proposition


an irreducible character of G for which ÷ # H is the sum of two

irreducible characters of H, say ÷ # H � ø1 � ø2. If ö is an irreduci-

ble character of G such that ö # H has ø1 or ø2 as a constituent,

then ö � ÷.

Proof In view of Proposition 20.10, ÷(g) � 0 for all g with g =2 H.

Therefore,

hö, ÷iG � 1

jGjXg2G

ö(g)÷(g) � 1

jGjXg2H

ö(g)÷(g)

� 12hö # H , ÷ # HiH :

If ö # H has ø1 or ø2 as a constituent, then hö # H , ÷ # HiH 6� 0, so

kö, ÷lG 6� 0, and hence ö � ÷. j

We summarize our results on subgroups of index 2, by explaining

how to list the irreducible characters of H on the assumption that we

know the character table of G.

(20.13) Let H be a normal subgroup of index 2 in the group G.

(1) Each irreducible character ÷ of G which is non-

zero somewhere outside H restricts to be an

irreducible character of H. Such characters of G

occur in pairs (÷ and ÷ë) which have the same

restriction to H (Propositions 20.10, 20.11).


(2) If ÷ is an irreducible character of G which is zero

everywhere outside H, then ÷ restricts to be the

sum of two distinct irreducible characters of H of

the same degree. The two characters of H which

we get from ÷ in this way come from no other

irreducible character of G (Propositions 20.9,

20.10, 20.12).

(3) Every irreducible character of H appears among

those obtained by restricting irreducible characters

of G, as in parts (1) and (2) (Proposition 20.4).

In case (2) of (20.13), extra work is needed to calculate the values

taken by the two constituents of ÷ # H . Fortunately, in practice case (1)

occurs more frequently than case (2).

20.14 Example The character table of A5

Write H � A5, and note that H is a normal subgroup of index 2 in

the group S5. The conjugacy classes of H and their sizes are given in

Example 12.18(2), and the irreducible characters ÷1, . . . , ÷7 of S5 can

be found in Example 19.16.

Observe that ÷1, ÷3 and ÷6 are non-zero somewhere outside H, so by

(20.13)(1), ÷1 # H , ÷3 # H and ÷6 # H are irreducible characters of H.

Call them ø1, ø2 and ø3, respectively. Also, ÷5(g) � 0 for all g =2 H,

so by (20.13)(2), ÷5 # H � ø4 � ø5 where ø4 and ø5 are distinct

irreducible characters of H of degree 3. Note that ÷2 # H � ÷1 # H ,

÷4 # H � ÷3 # H and ÷7 # H � ÷6 # H, and hence ø1, . . . , ø5 are the

distinct irreducible characters of H by (20.13)(3).

The results we have obtained so far have been deduced from our

summary (20.13) of facts about characters of subgroups of index 2.

They can also be veri®ed by calculating inner products. We have

established that the character table of H is

gi 1 (1 2 3) (1 2)(3 4) (1 2 3 4 5) (1 3 4 5 2)|CG(gi)| 60 3 4 5 5

ø1 1 1 1 1 1ø2 4 1 0 ÿ1 ÿ1ø3 5 ÿ1 1 0 0ø4 3 á2 á3 á4 á5

ø5 3 â2 â3 â4 â5


We use the column orthogonality relations to calculate the unknowns

ái and âi. The values of ái � âi for 2 < i < 5 are given by noting that

ø4 � ø5 � ÷5 # H (or by using the column orthogonality relations for

column 1 and column i). We get

á2 � â2 � 0, á3 � â3 � ÿ2, á4 � â4 � á5 � â5 � 1:

Using Proposition 12.13, we see that each element of A5 is conjugate

to its inverse. Hence by Proposition 13.9(4), all the numbers in the

character table are real. By the orthogonality relation for column i with

itself (2 < i < 5), we obtain

3 � 3� á22 � â2

2,

4 � 2� á23 � â2

3,

5 � 2� á24 � â2

4 � 2� á25 � â2

5:

Hence

á2 � â2 � 0, á3 � â3 � ÿ1,

and we ®nd that á4 and â4 are the solutions of the quadratic

x2 ÿ xÿ 1 � 0. Since we have not yet distinguished between ø4 and

ø5, we may take

á4 � 12(1�p5), â4 � 1

2(1ÿp5):

Similarly, á5 and â5 are 12(1 � p5). Since ø4 6� ø5, we have

á5 � 12(1ÿp5), â5 � 1

2(1�p5):

Thus the character table of A5 is as shown.

where á � 12(1 � p5), â � 1

2(1 ÿ p5).


gi 1 (1 2 3) (1 2)(3 4) (1 2 3 4 5) (1 3 4 5 2)|CG(gi)| 60 3 4 5 5

ø1 1 1 1 1 1ø2 4 1 0 ÿ1 ÿ1ø3 5 ÿ1 1 0 0ø4 3 0 ÿ1 á âø5 3 0 ÿ1 â á


Proposition 17.6 allows us to deduce from the character table that

A5 is a simple group.


Assume throughout that H is a subgroup of G.

1. If ÷ is a character of G, then ÷ # H is a character of H. The values

of ÷ # H are given by

(÷ # H)(h) � ÷(h)

for all h 2 H . In particular, ÷ and ÷ # H have the same degree.

2. The number of irreducible constituents of ÷ # H is bounded above

by jG : H j. Indeed, if ø1, : : : , ør are the irreducible characters of

H, and ÷ # H � d1ø1 � : : :� d rør, thenXr

i�1

d2i < jG : H j:

3. If H v G and ÷ is an irreducible character of G, then all the

constituents of ÷ # H have the same degree.


1. Let G � S4 and let H be the subgroup k(1 2 3 4), (1 3)l of G.

(a) Show that H � D8.

(b) For each irreducible character ÷ of G (given in Section 18.1),

express ÷ # H as a sum of irreducible characters of H.

2. Use the restrictions of the irreducible characters of S6, given in

Example 19.17, to ®nd the character table of A6.

(The seven conjugacy classes of A6 can be found by consulting the

solutions to Exercises 12.3 and 12.4.)

3. Let G be a group with an abelian subgroup H of index n. Prove

that ÷(1) < n for every irreducible character ÷ of G.

4. Suppose that G is a group with a subgroup H of index 3, and let ÷be an irreducible character of G. Prove that

h÷ # H , ÷ # HiH � 1, 2 or 3:

Give examples to show that each possibility can occur.


5. It is known that the complete list of degrees of the irreducible

characters of S7 is

1, 1, 6, 6, 14, 14, 14, 14, 15, 15, 20, 21, 21, 35, 35:

Also, A7 has nine conjugacy classes. Find the complete list of

degrees of the irreducible characters of A7.


224

21

Induced modules and characters

Throughout this chapter we assume that H is a subgroup of the ®nite

group G. We saw in the last chapter that restriction gives a simple way

of converting a CG-module into a CH-module. Much more subtle than

this is the process of induction, which constructs a CG-module from a

given CH-module, and induction is the main concern of this chapter.

As H is smaller than G, it is usually the case that CH-modules are

easier to understand and construct than CG-modules, so induction can

often give us an important handle on the representations of a group if

we know some representations of its subgroups. We shall see many

applications of this method in later chapters.

Before describing the process of induction, we require some results

which connect CH-homomorphisms with CG-homomorphisms.

CH-homomorphisms and CG-homomorphisms

Let U be a CH-submodule of the regular CH-module CH. If r 2 CG,

then

W: u! ru (u 2 U )

de®nes a CH-homomorphism from U to CG, since for all s 2 CH,

(us)W � rus � (uW)s. We now prove the striking fact that every CH-

homomorphism from U to CG has this simple form.

21.1 Proposition

Assume that H < G, and let U be a CH-submodule of CH. If W is a

CH-homomorphism from U to CG, then there exists r 2 CG such that

uW � ru for all u 2 U :

Proof By Maschke's Theorem 8.1, there is a CH-submodule W of CH

such that CH � U � W. De®ne ö: CH! CG by

ö: u� w! uW (u 2 U , w 2 W ):

Then ö is easily seen to be a CH-homomorphism. Let r � 1ö. For

u 2 U,uW � uö � (1u)ö � (1ö)u � ru,

and so W is of the required form. j

We give two corollaries of Proposition 21.1, the ®rst of which is just

the case H � G of the proposition.

21.2 Corollary

Let U be a CG-submodule of CG. Then every CG-homomorphism from

U to CG has the form

u! ru (u 2 U )

for some r 2 CG.

21.3 Corollary

Let U and V be CG-submodules of CG. Then the following two

statements are equivalent:

(1) U \ V � {0};

(2) there exists r 2 CG such that for all u 2 U, v 2 V,

ru � u and rv � 0:

Proof Assume that U \ V � {0}. Then the sum U � V is a direct sum,

so

u� v! u (u 2 U , v 2 V )

is a function; moreover, it is a CG-homomorphism from U � V to CG

(see Proposition 7.11). Therefore by Corollary 21.2, there exists

r 2 CG such that for all u 2 U, v 2 V,

r(u� v) � u:

Then ru � u if u 2 U, and rv � 0 if v 2 V.

Conversely, assume that for some r 2 CG we have ru � u and

rv � 0 for all u 2 U, v 2 V. If x 2 U \ V then rx � x and rx � 0, and

so x � 0. Consequently U \ V � {0}. j

Induced modules and characters 225

Induction from H to G

For any subset X of CG, we write X(CG) for the subspace of CG

which is spanned by all the elements xg with x 2 X, g 2 G. That is,

X (CG) � sp fxg: x 2 X , g 2 Gg:Clearly, X(CG) is then a CG-submodule of CG.

Remember that H is a subgroup of G, so CH is a subset of CG.

21.4 De®nition

Assume that H is a subgroup of G. Let U be a CH-submodule of

CH, and let U " G denote the CG-module U(CG). Then U " G is

called the CG-module induced from U.

21.5 Example

Let G � D6 � ka, b: a3 � b2 � 1, bÿ1ab � aÿ1l, and let H � kal, a cyc-

lic subgroup of G of order 3. Let ù � e2ði=3, and de®ne

W0 � sp (1� a� a2),

W1 � sp (1� ù2a� ùa2),

W2 � sp (1� ùa� ù2a2):

These are CH-submodules of CH (see Example 10.8(1)). Clearly,

W0 " G � sp (1� a� a2, b� ab� a2b),

W1 " G � sp (1� ù2a� ùa2, b� ù2ab� ùa2b),

W2 " G � sp (1� ùa� ù2a2, b� ùab� ù2a2b):

Recall from Example 10.8(2) that

CG � U1 � U2 � U3 � U4,

a direct sum of irreducible CG-modules Ui, where

U1 � sp (1� a� a2 � b� ab� a2b),

U2 � sp (1� a� a2 ÿ bÿ abÿ a2b),

U3 � sp (1� ù2a� ùa2, b� ù2ab� ùa2b),

U4 � sp (1� ùa� ù2a2, b� ùab� ù2a2b):

Thus

W0 " G � U1 � U2, W1 " G � U3, W2 " G � U4:


In particular, W0 " G is reducible, while W1 " G and W2 " G are

irreducible.

We now show that isomorphic CH-modules give isomorphic induced

CG-modules.

21.6 Proposition

Assume that H < G. Suppose that U and V are CH-submodules of CH

and that U is CH-isomorphic to V. Then U " G is CG-isomorphic to

V " G.

Proof Let W: U! V be a CH-isomorphism. By Proposition 21.1, there

exists r 2 CG such that uW � ru for all u 2 U, and also there exists

s 2 CG such that vWÿ1 � sv for all v 2 V. Consequently

sru � u and rsv � v for all u 2 U , v 2 V :

If a 2 U " G then a is a linear combination of elements ug

(u 2 U, g 2 G), so ra is a linear combination of elements rug, and

hence ra 2 V " G. Therefore

ö: a! ra (a 2 U " G)

is a function from U " G to V " G. Moreover, ö is a CG-homomorph-

ism, as (aö)g � rag � (ag)ö (a 2 U " G, g 2 G). Since sru � u and

rsv � v for all u 2 U, v 2 V, we have

sra � a, rsb � b for all a 2 U " G, b 2 V " G:

Hence the function

b! sb (b 2 V " G)

is the inverse of ö. Therefore ö is a CG-isomorphism, proving that

U " G is CG-isomorphic to V " G. j

The next proposition and its corollary enable us to de®ne the

induced module U " G for an arbitrary CH-module U.

21.7 Proposition

Assume that U and V are CH-submodules of CH with U \ V � {0}.

Then (U " G) \ (V " G) � f0g.


Proof By Corollary 21.3, there exists r 2 CG such that ru � u and

rv � 0 for all u 2 U, v 2 V. Then for all u 2 U, v 2 V and all g 2 G,

rug � ug and rvg � 0:

Since U " G is spanned by elements of the form ug (u 2 U, g 2 G),

this implies that

ru9 � u9 for all u9 2 U " G,

and similarly,

rv9 � 0 for all v9 2 V " G:

Therefore (U " G) \ (V " G) � f0g by Corollary 21.3. j

21.8 Corollary

Let U be a CH-submodule of CH, and suppose that

U � U1 � : : :� Um,

a direct sum of CH-submodules Ui. Then

U " G � (U1 " G)� : : :� (Um " G):

Proof We prove this by induction on m. It is trivial for m � 1. Now

U � U1 � V, where V � U2 � : : : � Um. The de®nition of U " G

implies that

U " G � (U1 " G)� (V " G):

Therefore by Proposition 21.7,

U " G � (U1 " G)� (V " G):

By induction, V " G � (U2 " G)� : : : � (Um " G), and hence, using

(2.10), we obtain

U " G � (U1 " G)� : : :� (Um " G),

as required. j

We can now de®ne the induced module U " G for an arbitrary CH-

module U (where U is not necessarily a CH-submodule of CH).

21.9 De®nition

Let U be a CH-module. Then (by Theorems 8.7 and 10.5),

U � U1 � : : :� Um


for certain irreducible CH-submodules Ui of CH. De®ne U " G to be

the following (external) direct sum:

U " G � (U1 " G)� : : :� (Um " G):

Proposition 21.6 and Corollary 21.8 ensure that this de®nition is

consistent with De®nition 21.4.

We emphasize that the de®nition of the induced module U " G in

the case where U is a CH-submodule of CH is a natural one:

U " G � U (CG):

We shall always prove results for general induced modules U " G by

®rst dealing with the special case where U is a CH-submodule of CH,

and then applying the fact (which is immediate from De®nition 21.9)

that

(U1 � : : :� Um) " G � (U1 " G)� : : :� (Um " G):(21:10)

Our ®rst major result on general induced modules is known as `the

transitivity of induction'.

21.11 Theorem

Suppose that H and K are subgroups of G such that H < K < G. If U

is a CH-module, then

(U " K) " G � U " G:

Proof Assume ®rst that U is a CH-submodule of CH. Then U(CK) is

spanned by elements of the form

uk (u 2 U , k 2 K):

Therefore, (U(CK))(CG) is spanned by elements of the form

ukg (u 2 U , k 2 K, g 2 G):

Since K < G, it follows that (U(CK))(CG) � U(CG). That is,

(U " K) " G � U " G:(21:12)

Now let U be an arbitrary CH-module. Then

U � U1 � : : :� Um

for certain irreducible CH-submodules Ui of CH. By (21.10),

U " K � (U1 " K)� : : :� (Um " K):


Therefore

(U " K) " G � (U1 " K) " G � : : :� (Um " K) " G

by (21:10)

� (U1 " G)� : : :� (Um " G) by (21:12)

� U " G by Definition 21.9. j

Induced characters

21.13 De®nition

If ø is the character of a CH-module U, then the character of the

induced CG-module U " G is denoted by ø " G, and is called the

character induced from ø.

The next example illustrates an important relationship between

induced characters and restrictions of characters.

21.14 Example

Let G � S5 and let H be the subgroup A4 of G, as in Example 20.2.

We showed in that example that if ÷1, . . . , ÷7 are the irreducible

characters of G (given in Example 19.16), and ø1, . . . , ø4 are the

irreducible characters of H (given in 18.2) then

÷1 # H � ø1,

÷2 # H � ø1,

÷3 # H � ø1 � ø4,

÷4 # H � ø1 � ø4,

÷5 # H � 2ø4,

÷6 # H � ø2 � ø3 � ø4,

÷7 # H � ø2 � ø3 � ø4:

By Theorem 14.17, the coef®cients which appear here are the values of

h÷i # H , ø jiH for appropriate i, j. We record these coef®cients in a


matrix whose ij-entry is h÷i # H , ø jiH :

1CCCCCCCCA

0BBBBBBBB@

ø1 ø2 ø3 ø4

÷1 1 0 0 0

÷2 1 0 0 0

÷3 1 0 0 1

÷4 1 0 0 1

÷5 0 0 0 2

÷6 0 1 1 1

÷7 0 1 1 1

The rows of this matrix tell us how the irreducible characters of G

restrict to H. For example, row 3 gives

÷3 # H � 1 . ø1 � 0 . ø2 � 0 . ø3 � 1 . ø4:

Remarkably, it turns out that the columns of the matrix tell us how

the irreducible characters of H induce to G. To be precise, the seven

integers in column 1 give

ø1 " G � 1 . ÷1 � 1 . ÷2 � 1 . ÷3 � 1 . ÷4 � 0 . ÷5 � 0 . ÷6 � 0 . ÷7:

Similarly,ø2 " G � ø3 " G � ÷6 � ÷7, and

ø4 " G � ÷3 � ÷4 � 2÷5 � ÷6 � ÷7:

Thus the ij-entry in our matrix, which we already know to be equal to

h÷i # H , ø jiH , is also equal to h÷i, ø j " GiG.

In fact, it is true that

h÷, ø " GiG � h÷ # H , øiH

for all characters ÷ of G and ø of H, and we devote the next section

to a proof of this result, which is known as the Frobenius Reciprocity

Theorem.

The Frobenius Reciprocity Theorem

Before proving the theorem, we need the following preliminary result.

21.15 Proposition

Assume that H < G. Let U be a CH-submodule of CH, and let V be a

CG-submodule of CG. Then the vector spaces

HomCG (U " G, V ) and HomC H (U , V # H)

have equal dimensions.


Proof Suppose that W 2 HomCG (U " G, V ). Then by Corollary 21.2,

there is an element r 2 CG such that

sW � rs for all s 2 U " G:

De®ne W: U ! CG to be the restriction of W to U; that is,

uW � ru for all u 2 U :

Then W 2 HomC H (U , V # H). Thus the function

W! W

is a linear transformation from HomCG (U " G, V ) to

HomC H (U , V # H). We shall show that this linear transformation is

invertible.

Let ö 2 HomC H (U , V # H). Then by Proposition 21.1, there exists

r 2 CG such that uö � ru for all u 2 U. De®ne W from U " G to CG

by

sW � rs (s 2 U " G):

Then W 2 HomCG (U " G, V ). Moreover, ö � W. Therefore the function

W! W is surjective.

Finally, note that if r1, r2 2 CG and r1u � r2u for all u 2 U, then

r1s � r2s for all s 2 U " G, since s is a linear combination of

elements ug with u 2 U, g 2 G. Hence the function W! W is injective,

and so it is an invertible linear transformation from HomCG(U " G, V )

to HomC H (U , V # H). These two vector spaces therefore have the

same dimension, as required. j

21.16 The Frobenius Reciprocity Theorem

Assume that H < G. Let ÷ be a character of G and let ø be a

character of H. Then

hø " G, ÷iG � hø, ÷ # HiH :

Proof First assume that the characters ÷ and ø are irreducible. Then

there is a CH-submodule U of CH which has character ø, and there

is a CG-submodule V of CG which has character ÷. By Theorem

14.24, we have

hø " G, ÷iG � dim (HomCG (U " G, V )), and

hø, ÷ # HiH � dim (HomC H (U , V # H)):


From Proposition 21.15, we therefore deduce that

hø " G, ÷iG � hø, ÷ # HiH(21:17)

in the special case we are considering, namely when ÷ and ø are

irreducible.

For the general case, let ÷1, . . . , ÷k be the irreducible characters of

G and let ø1, . . . , øm be the irreducible characters of H. Then for

some integers di, ej we have

÷ �Xk

i�1

di÷i and ø �Xm

j�1

ejø j:

Therefore

hø " G, ÷iG �Xm

j�1

ejø j " G,Xk

i�1

di÷i

* +G

�Xm

j�1

Xk

i�1

ejdihø j " G, ÷iiG

�Xm

j�1

Xk

i�1

ejdihø j, ÷i # HiH by (21:17)

�Xm

j�1

ejø j,Xk

i�1

di÷i # H

* +H

� hø, ÷ # HiH :

This completes the proof of the Frobenius Reciprocity Theorem. j

21.18 Corollary

If f is a class function on G, and ø is a character of H, then

hø " G, f iG � hø, f # HiH :

Proof This follows at once from the Frobenius Reciprocity Theorem,

since by Corollary 15.4, the characters of G span the vector space of

class functions on G. j

The values of induced characters

We now show how to evaluate induced characters. Let ø be a character

of the subgroup H of G, and for convenience of notation, de®ne the


function _ø: G! C by

_ø(g) �ø(g) if g 2 H ,

0 if g =2 H :

(

21.19 Proposition

The values of the induced character ø " G are given by

(ø " G)(g) � 1

jH jXy2G

_ø(yÿ1 gy)

for all g 2 G.

Proof Let f: G! C be the function given by

f (g) � 1

jH jXy2G

_ø(yÿ1 gy) (g 2 G):

We aim to prove that f � ø " G.

If w 2 G then

f (wÿ1 gw) � 1

jH jXy2G

_ø(yÿ1wÿ1 gwy) � f (g)

since wy runs through G as y runs through G. Therefore f is a class

function, and so by Corollary 15.4, it is suf®cient to show that

h f , ÷iG � hø " G, ÷iG for all irreducible characters ÷ of G.

Let ÷ be an irreducible character of G. Then

h f , ÷iG � 1

jGjXg2G

f (g)÷(g)

� 1

jGj1

jH jXg2G

Xy2G

_ø(yÿ1 gy)÷(g):

Put x � yÿ1 gy. Then

h f , ÷iG � 1

jGj1

jH jXx2G

Xy2G

_ø(x)÷(yxyÿ1):

� 1

jH jXx2H

ø(x)÷(x)


since _ø(x) � 0 if x =2 H, and ÷(yxyÿ1) � ÷(x) for all y 2 G. Therefore

h f , ÷iG � hø, ÷ # HiH

and so by the Frobenius Reciprocity Theorem,

h f , ÷iG � hø " G, ÷iG:

This was the equation required to show that f � ø " G, so the proof is

complete. j

21.20 Corollary

If ø is a character of the subgroup H of G, then the degree of ø " G

is given by

(ø " G)(1) � jGjjH jø(1):

Proof This follows immediately by evaluating (ø " G)(1) using Prop-

osition 21.19. Alternatively, the stated degree of ø " G can be found

using just the de®nition of induced modules (see Exercise 21.3). j

For practical purposes, a formula for the values of induced characters

different from that given in Proposition 21.19 is more useful, and we

shall derive this next (it is given in Proposition 21.23 below).

For x 2 G, de®ne the class function f Gx on G by

f Gx (y) � 1 if y 2 xG

0 if y =2 xG

�(y 2 G):

Thus f is the characteristic function of the conjugacy class xG.

21.21 Proposition

If ÷ is a character of G and x 2 G, then

h÷, f Gx iG �

÷(x)

jCG(x)j :


Proof We have

h÷, f Gx iG �

1

jGjXg2G

÷(g) f Gx (g)

� 1

jGjXg2xG

÷(g) � jxGjjGj ÷(x)

� ÷(x)

jCG(x)j by Theorem 12:8: j

Note that a result similar to Proposition 21.21 was used in the proof

of Theorem 16.4.

If H < G and h 2 H then h H � hG; but if g 2 G then gG may

contain 0, 1, 2 or more conjugacy classes of H. To put this another

way, we have:

(21.22) Suppose that x 2 G.

(1) If no element of xG lies in H, then f Gx # H � 0.

(2) If some element of xG lies in H, then there are elements

x1, . . . , xm 2 H such that

f Gx # H � f H

x1� : : :� f H

xm:

Statement (2) just says that H \ xG breaks up into m conjugacy

classes of H, with representatives x1, : : : , xm.

21.23 Proposition

Let ø be a character of the subgroup H of G, and suppose that x 2 G.

(1) If no element of xG lies in H, then (ø " G)(x) � 0.

(2) If some element of xG lies in H, then

(ø " G)(x) � jCG(x)j ø(x1)

jCH (x1)j � : : :�ø(xm)

jCH (xm)j� �

,

where x1, . . . , xm 2 H and f Gx # H � f H

x1� . . . � f H

xm(as in (21.22)).

Proof By Proposition 21.21 and Corollary 21.18, we have

(ø " G)(x)

jCG(x)j � hø " G, f Gx iG � hø, f G

x # HiH :


If no element of xG lies in H, then f Gx # H � 0, and hence

(ø " G)(x) � 0. And if some element of xG lies in H, and

f Gx # H � f H

x1� : : : � f H

xmas in (21.22)(2), then

(ø " G)(x)

jCG(x)j � hø, f Hx1� : : :� f H

xmiH

� hø, f Hx1iH � : : :� hø, f H

xmiH

� ø(x1)

jCH (x1)j � : : :�ø(xm)

jCH (xm)j by Proposition 21:21:

The result follows. j

21.24 Example

Let G � S4 and let H � ka, bl, where

a � (1 2 3 4), b � (1 3):

Then H � D8, since a4 � b2 � 1 and bÿ1ab � aÿ1. By (12.12), the

conjugacy classes of H are

f1g,

fa2 � (1 3)(2 4)g,

fa � (1 2 3 4), a3 � (1 4 3 2)g,

fb � (1 3), a2b � (2 4)g,

We have

f G1 # H � f H

1 , f G(1 3) # H � f H

(1 3), f G(1 2 3) # H � 0,

f G(1 2)(3 4) # H � f H

(1 3)(2 4) � f H(1 2)(3 4), and

f G(1 2 3 4) # H � f H

(1 2 3 4):

For example, the statement

f G(1 2)(3 4) # H � f H

(1 3)(2 4) � f H(1 2)(3 4)

records the fact that the G-conjugacy class (1 2)(3 4)G contains exactly

two H-conjugacy classes, with representatives (1 3)(2 4) and (1 2)(3 4).


The orders of the centralizers of the elements of H are as follows:

h 1 (1 3)(2 4) (1 2 3 4) (1 3) (1 2)(3 4)|CG(h)| 24 8 4 4 8|CH (h)| 8 8 4 4 4

Suppose that ø is a character of H. Then according to Proposition

21.23, we have

(ø " G)(1) � 24ø(1)

8,

(ø " G)((1 3)) � 4ø((1 3))

4,

(ø " G)((1 2 3)) � 0,

(ø " G)((1 2)(3 4)) � 8ø((1 3)(2 4))

8� ø((1 2)(3 4))

4

� �,

(ø " G)((1 2 3 4)) � 4ø((1 2 3 4))

4:

Referring to Example 16.3(3) for the irreducible characters ÷1, . . . , ÷5

of H � D8, we therefore have

1 (1 2) (1 2 3) (1 2)(3 4) (1 2 3 4)

÷1 " G 3 1 0 3 1÷2 " G 3 ÿ1 0 ÿ1 1÷3 " G 3 1 0 ÿ1 ÿ1÷4 " G 3 ÿ1 0 3 ÿ1÷5 " G 6 0 0 ÿ2 0

In the next example, we use induced characters to ®nd the character

table of a group of order 21.

21.25 Example (cf. Exercise 17.2)

De®ne permutations a, b in S7 by

a � (1 2 3 4 5 6 7), b � (2 3 5)(4 7 6)


and let G be the subgroup ka, bl of S7. Check that

a7 � b3 � 1, bÿ1ab � a2:

It follows from these relations that the elements of G are all of the

form aibj with 0 < i < 6, 0 < j < 2. Also, G has order 21.

We aim to ®nd the character table of G. First we ®nd the conjugacy

classes. Since hai < CG(a), 7 divides jCG(a)j; and since b =2 CG(a),

jCG(a)j, 21. Hence jCG(a)j � 7, and similarly jCG(b)j � 3. Using this,

we see that the conjugacy classes of G are

f1g,fa, a2, a4g,fa3, a5, a6g,faib: 0 < i < 6g,faib2: 0 < i < 6g:

We take 1, a, a3, b and b2 to be representatives of the conjugacy

classes. Notice that G has exactly ®ve irreducible characters.

Since kal v G and G=hai � C3, we obtain three linear characters ÷1,

÷2, ÷3 of G as the lifts of the linear characters of G=hai. Their values

are shown below:

g 1 a a3 b b2

|CG(g)| 21 7 7 3 3

÷1 1 1 1 1 1÷2 1 1 1 ù ù2

÷3 1 1 1 ù2 ù

where ù � e2ði=3.

Let H � kal. We shall obtain the last two irreducible characters of G

by inducing linear characters of H. Let ç � e2ði=7. For 1 < k < 6, there

is a character øk of H given by

øk(a j) � ç jk (0 < j < 6):

To use the formula in Proposition 21.23 for calculating øk " G, note

that

f Ga # H � f H

a � f Ha2 � f H

a4


since no two of the elements a, a2, a4 are conjugate in H. Hence by

Proposition 21.23,

(ø1 " G)(a) � ç� ç2 � ç4

and similarly

(ø1 " G)(a3) � ç3 � ç5 � ç6, (ø1 " G)(1) � 3:

And since no G-conjugate of b or b2 lies in H,

(ø1 " G)(b) � (ø1 " G)(b2) � 0:

Similarly

(ø3 " G)(1) � 3, (ø3 " G)(a) � ç3 � ç5 � ç6,

(ø3 " G)(a3) � ç� ç2 � ç4, and (ø3 " G)(b) � (ø3 " G)(b2) � 0:

Thus if we write ÷4 � ø1 " G and ÷5 � ø3 " G, then the values of

÷4 and ÷5 are

1 a a3 b b2

÷4 3 ç � ç2 � ç4 ç3 � ç5 � ç6 0 0÷5 3 ç3 � ç5 � ç6 ç � ç2 � ç4 0 0

Now ÷4 # H � ø1 � ø2 � ø4 and ÷5 # H � ø3 � ø5 � ø6. Therefore

÷4 6� ÷5, since ø1, . . . , ø6 are linearly independent.

We now calculate that

h÷4, ÷4iG � 9

21� 2

7� 2

7� 0

3� 0

3� 1,

and similarly h÷5, ÷5iG � 1. Thus ÷4 and ÷5 are our last two irreducible

characters, and the character table of G is as shown.

Character table of ha, b: a7 � b3 � 1, bÿ1ab � a2i

g 1 a a3 b b2

|CG(gi)| 21 7 7 3 3

÷1 1 1 1 1 1÷2 1 1 1 ù ù2

÷3 1 1 1 ù2 ù÷4 3 ç � ç2 � ç4 ç3 � ç5 � ç6 0 0÷5 3 ç3 � ç5 � ç6 ç � ç2 � ç4 0 0



Assume that H is a subgroup of G.

1. For each CH-module U, an induced CG-module U " G can be

de®ned. If U is a CH-module of CH, then U " G is simply U(CG).

2. If ø is a character of H then the induced character ø " G is given

by

(ø " G)(g) � 1

jH jXy2G

_ø(yÿ1 gy):

In particular, the degree of ø:G is |G:H|ø(1).

3. If no element of gG lies in H, then

(ø " G)(g) � 0:

If some element of gG lies in H, then

(ø " G)(g) � jCG(g)j ø(x1)

jCH (x1)j � : : :�ø(xm)

jCH (xm)j� �

where f Gg # H � f H

x1� : : : � f H

xm.

4. The Frobenius Reciprocity Theorem states that

hø " G, ÷iG � hø, ÷ # HiH ,

where ø is a character of H and ÷ is a character of G.


1. Let G � D8 � ka, b: a4 � b2 � 1, bÿ1ab � aÿ1l and let H be the

subgroup ka2, bl. De®ne U to be the 1-dimensional subspace of CH

spanned by

1ÿ a2 � bÿ a2b:

(a) Check that U is a CH-submodule of CH.

(b) Find a basis of the induced CG-module U " G.

(c) Write down the character of the CH-module U and the char-

acter of the CG-module U " G. Is U " G irreducible?

2. Let G � S4 and let H be the subgroup k(1 2 3)l � C3.

(a) If ÷1, . . . , ÷5 are the irreducible characters of G, as given in


Section 18.1, work out the restrictions ÷i # H (1 < i < 5) as

sums of the irreducible characters ø1, ø2, ø3 of C3.

(b) Calculate the induced characters ø j " G (1 < j < 3) as sums of

the irreducible characters ÷i of G.

3. Show direct from the de®nition that if H < G and ø is a character

of H, then

(ø " G)(1) � jGjjH jø(1):

4. Let H be a subgroup of G, let ø be a character of H, and let ÷ be

a character of G. Prove that

(ø(÷ # H)) " G � (ø " G)÷:

(Hint: consider the inner product of each side with an arbitrary

irreducible character of G, and use the Frobenius Reciprocity

Theorem.)

5. Let G � S7 and let H � ka, bl, where

a � (1 2 3 4 5 6 7), b � (2 3 5)(4 7 6),

as in Example 21.25. Let ö and ø be the irreducible characters of

H which are given by

gi 1 a a3 b b2

|CH (gi)| 21 7 7 3 3

ö 1 1 1 1 1ø 3 ç � ç2 � ç4 ç3 � ç5 � ç6 0 0

where ç � e2ði=7 (see Example 21.25).

You are given that jCG(a)j � 7 and jCG(b)j � 18. Calculate the

values of the induced characters ö " G and ø " G.

6. Suppose that H is a subgroup of G, and let ÷1, . . . , ÷k be the

irreducible characters of G. Let ø be an irreducible character of H.

Show that the integers d1, : : : , d k , which are given by ø " G

� d1÷1 � : : : dk÷k , satisfyXk

i�1

d2i < jG : H j:

(Compare Proposition 20.5.)


7. Suppose that H is a normal subgroup of index 2 in G, and let ø be

an irreducible character of H. Discover and prove results for ø " G

which are similar to those presented in Chapter 20 for the restriction

of irreducible characters of G to H.


244

22

Algebraic integers

Among the properties of characters which may be regarded as funda-

mental, perhaps the most opaque is that which states that the degree of

an irreducible character of a ®nite group G must divide the order of

G. This is one of several results which we shall prove in this chapter,

using algebraic integers. Most of the results concern arithmetic proper-

ties of character values. We discuss properties of a group element

g 2 G which ensure that ÷(g) is an integer for all characters ÷ of G.

And we prove some useful congruence properties; for example, if p is

a prime number and g 2 G is an element of order pr for some r, then

÷(g) � ÷(1) mod p for any character ÷ of G for which ÷(g) is an

integer.

Algebraic integers

22.1 De®nition

A complex number ë is an algebraic integer if and only if ë is an

eigenvalue of some matrix, all of whose entries are integers.

Thus, for ë to be an algebraic integer, we require that

det (Aÿ ëI) � 0

for some square matrix A with integer entries. Equivalently, for the

same matrix A, we have

uA � ëu

for some non-zero row vector u.

We remark that ë is an algebraic integer if and only if ë is a root of

a polynomial of the form

x n � anÿ1x nÿ1 � : : :� a1x� a0

where a0, : : : , anÿ1 are integers (see Exercise 22.7). In fact, alge-

braic integers are usually de®ned in this way.

22.2 Examples

(1) Every integer n is an algebraic integer, since n is an eigenvalue of

the 1 3 1 matrix (n).

(2)p

2 is an algebraic integer, since it is an eigenvalue of the

matrix0 1

2 0

� �:

(3) If ë is an algebraic integer, then so are ÿë and the complex

conjugate ë of ë. To see this, note that if A is an integer matrix and u

is a row vector with uA � ëu, then

u(ÿA) � (ÿë)u and uA � ëu,

where u is the row vector obtained from u by replacing each entry by

its complex conjugate.

(4) Let A be the n 3 n matrix given by

A �

0 0 0 : : : 0 0 1

1 0 0 : : : 0 0 0

0 1 0 : : : 0 0 0

: : :0 0 0 : : : 1 0 0

0 0 0 : : : 0 1 0

0BBBBBB@

1CCCCCCA:Suppose that ù is an nth root of unity, and let u be the row vector

(1, ù, ù2, . . . , ùnÿ1). Then

uA � (ù, ù2, : : : , ùnÿ1, 1) � ùu:

This shows that every nth root of unity is an algebraic integer.

22.3 Theorem

If ë and ì are algebraic integers, then ëì and ë � ì are also

algebraic integers.

Proof There exist square matrices A and B, all of whose entries are

integers, and non-zero row vectors u and v, such that

uA � ëu, vB � ìv:

Suppose that A is an m 3 m matrix and B is an n 3 n matrix.

Algebraic integers 245

Let e1, . . . , em be a basis of Cm and f 1, : : : , f n be a basis of Cn.

Then the vectors ei fj (1 < i < m, 1 < j < n) form a basis of the

tensor product space V � Cm Cn. De®ne an endomorphism A B of

V by

(ei f j)(A B) � eiA f jB (1 < i < m, 1 < j < n),

extending linearly (that is, (P

ëij(ei f j))(A B) �Pëij(eiA f jB)).

It is easy to check as in the proof of Proposition 19.4 that for all

vectors x 2 Cm, y 2 Cn, we have

(x y)(A B) � xA yB:

Hence(u v)(A B) � uA vB � ëu ìv � ëì(u v):

Therefore ëì is an eigenvalue of A B. Since the matrix of A B

relative to the basis ei fj (1 < i < m, 1 < j < n) has integer entries,

it follows that ëì is an algebraic integer.

Let Im and In denote the identity m 3 m and n 3 n matrices,

respectively. Then

(u v)(A In � Im B) � uA vIn � uIm vB

� ëu v� u ìv

� (ë� ì)(u v),

and we deduce as above that ë � ì is an algebraic integer. j

Theorem 22.3 shows that the set of all algebraic integers forms a

subring of C. The next result provides a link between algebraic integers

and characters.

22.4 Corollary

If ÷ is a character of G and g 2 G, then ÷(g) is an algebraic integer.

Proof By Proposition 13.9, ÷(g) is a sum of roots of unity. Each root

of unity is an algebraic integer, by Example 22.2(4), so any sum of

roots of unity is an algebraic integer by Theorem 22.3. Hence ÷(g) is

an algebraic integer. j

22.5 Proposition

If ë is both a rational number and an algebraic integer, then ë is an

integer.


Proof Suppose that ë is a rational number which is not an integer. We

shall show that ë is not an algebraic integer, which is enough to

establish the proposition.

Write ë � r=s, where r and s are coprime integers and s 6� �1. Let

p be a prime number which divides s. For every n 3 n matrix A of

integers, the entries of sA ÿ rI which are not on the diagonal are

divisible by s, and hence also by p. Therefore

det (sAÿ rI) � (ÿr)n � mp

for some integer m. As p does not divide r (since r and s are

coprime), we deduce that det (sA ÿ rI) 6� 0. Thus

det (Aÿ ëI) � 1

s

� �n

det (sAÿ rI) 6� 0,

and hence ë is not an algebraic integer. j

The next result is an immediate consequence of Corollary 22.4 and

Proposition 22.5.

22.6 Corollary

Let ÷ be a character of G and let g 2 G. If ÷(g) is a rational number,

then ÷(g) is an integer.

In passing, note that we have, as a special case of Proposition 22.5,

the well known result thatp

2 is irrational. (Example 22.2(2) shows

thatp

2 is an algebraic integer.)

The degree of every irreducible character divides |G|

To prepare for the proof that |G| is divisible by the degree of each

irreducible character of G, we establish two preliminary lemmas. Recall

from De®nition 12.21 that if C is a conjugacy class of G, then

C �Xx2C

x 2 CG:

22.7 Lemma

Suppose that g 2 G and that C is the conjugacy class of G which

contains g. Let U be an irreducible CG-module, with character ÷. Then


uC � ëu for all u 2 U ,

where

ë � jGjjCG(g)j

÷(g)

÷(1):

Proof Since C lies in the centre of CG (see Proposition 12.22), we

know by Proposition 9.14 that there exists ë 2 C such that uC � ëu for

all u 2 U; that is,

uXx2C

x � ëu for all u 2 U :

Consequently if B is a basis of U, thenXx2C

[x]B � ëI :

Taking the traces of both sides of this equation, we obtainXx2C

÷(x) � ë÷(1),

and since ÷ is constant on the conjugacy class C, this yields

jCj÷(g) � ë÷(1):

Thus ë � jCj÷(g)=÷(1). As |C| � |G:CG(g)| by Theorem 12.8, the result

follows. j

22.8 Lemma

Let r �P g2G á gg 2 CG, where each á g is an integer. Suppose that u

is a non-zero element of CG such that

ur � ëu,

where ë 2 C. Then ë is an algebraic integer.

Proof Let g1, . . . , gn be the elements of G. Then for 1 < i < n, we

have

gir �Xn

j�1

aijgj


for certain integers aij. (In fact, aij � á g where g � gÿ1i gj.) The

statement that ur � ëu (with u 6� 0) says that ë is an eigenvalue of the

integer matrix A � (aij). Therefore ë is an algebraic integer. j

22.9 Example

Let G � Cn � kx: xn � 1l, and de®ne

u � 1� ùxÿ1 � ù2xÿ2 � : : :� ùnÿ1x 2 CG,

where ù is an nth root of unity. Then

ux � ùu

and so Lemma 22.8 con®rms that ù is an algebraic integer.

Notice that this example is just a reworking of Example 22.2(4).

22.10 Corollary

If ÷ is an irreducible character of G and g 2 G, then

ë � jGjjCG(g)j

÷(g)

÷(1)

is an algebraic integer.

Proof Let U be an irreducible CG-submodule of CG with character ÷,

and let C be the sum of the elements in the conjugacy class of G

which contains g. Then uC � ëu for all u 2 U, by Lemma 22.7.

Therefore ë is an algebraic integer, by Lemma 22.8. j

22.11 Theorem

If ÷ is an irreducible character of G, then ÷(1) divides |G|.

Proof Let g1, . . . , gk be representatives of the conjugacy classes of G.

Then for all i, both

jGjjCG(gi)j

÷(gi)

÷(1)and ÷(gi)

are algebraic integers, by Corollaries 22.10 and 22.4. Hence by

Theorem 22.3, Xk

i�1

jGjjCG(gi)j

÷(gi)÷(gi)

÷(1)


is an algebraic integer. This algebraic integer is equal to jGj=÷(1), by

the row orthogonality relations, Theorem 16.4(1). As jGj=÷(1) is a

rational number. Proposition 22.5 now implies that jGj=÷(1) is an

integer. That is, ÷(1) divides |G|. j

22.12 Examples

(1) If p is a prime number and G is a group of order pn for some n,

then ÷(1) is a power of p for all irreducible characters ÷ of G.

In particular, if jGj � p2 then ÷(1) � 1 for all irreducible characters

÷. (Note that ÷(1) , p, as the sum of the squares of the degrees of the

irreducible characters is equal to |G|.) Hence, using Proposition 9.18,

we recover the well known result that groups of order p2 are abelian.

(2) Let G be a group of order 2 p, where p is prime. By Theorem

22.11, the degree of each irreducible character of G is 1 or 2 (it

cannot be p for the reason noted in (1) above). By Theorem 17.11, the

number of linear characters of G divides |G|. Hence, either the degrees

of the irreducible characters of G are all 1, or they are

1, 1, 2, : : : , 2 (with ( pÿ 1)=2 degrees 2):

(3) If G � Sn then every prime p which divides the degree of an

irreducible character of G also divides n!, and hence satis®es p < n.

Theorem 22.11 also has the following interesting consequences

concerning irreducible characters of simple groups. (Recall that a group

G is simple if it has no normal subgroups apart from {1} and G.)

22.13 Corollary

No simple group has an irreducible character of degree 2.

Proof Suppose that G is a simple group which has an irreducible

character ÷ of degree 2. Let r: G! GL(2, C) be a representation of

G with character ÷. Since Ker r v G and G is simple, we have

Ker r � f1g, and so r is injective.

First, observe that G is non-abelian, by Proposition 9.5. Hence

G9 6� 1, and so G9 � G as G is simple. Therefore by Theorem 17.11,

G has no non-trivial linear characters. But g! det (gr) is a linear

character of G (see Exercise 13.7(a)), and this implies that

det (gr) � 1 for all g 2 G:


Now G has even order, by Theorem 22.11. Therefore G contains an

element x of order 2 (see Exercise 1.8).

Consider the 2 3 2 matrix xr. As r is injective, xr has order 2; and

by Proposition 9.11, there is a 2 3 2 matrix T such that Tÿ1(xr)T is a

diagonal matrix with diagonal entries �1. Since det (xr) � 1, we

conclude that

Tÿ1(xr)T � ÿ1 0

0 ÿ1

� �:

Thus

xr � T (ÿI)Tÿ1 � ÿI :

Consequently (xr)(gr) � (gr)(xr) for all g 2 G. As r is injective, this

means that xg � gx for all g 2 G, and hence

hxi v G:

This is a contradiction, as G is simple. j

Our next result again shows that information about character degrees

can sometimes be used to learn about the structure of a group. This

time, we assume that every irreducible character of G has degree a

power of a prime p, and we deduce that G has an abelian normal p-

complement N; that is, N is an abelian normal subgroup of G, and jN jis coprime to p, while jG: N j is a power of p.

22.14 Corollary

Suppose that p is a prime and the degree of every irreducible character

of G is a power of p. Then G has an abelian normal p-complement. In

particular, G is not simple unless G has prime order.

Proof The result is correct if G is abelian (see Theorem 9.6), so we

assume that G is non-abelian.

Theorems 11.12 and 17.11 give us the equation

jGj � jG=G9j �X

÷(1)2,

the sum being over the irreducible characters ÷ of G for which

÷(1) . 1.

Since G is non-abelian, ÷(1) . 1 for some irreducible character ÷ of

G. Then ÷(1) is divisible by p, by our hypothesis, so p divides |G| by


Theorem 22.11; and we deduce from the equation above that p divides

the order of the abelian group G=G9. Since every ®nite abelian group

is isomorphic to a direct product of cyclic groups, it follows that G=G9

has a subgroup of index p. Hence G has a normal subgroup H of

index p.

Let ø be an irreducible character of H. Then h÷ # H , øi 6� 0 for

some irreducible character ø of H, by Proposition 20.4. Next, Clif-

ford's Theorem 20.8 shows that ø(1) divides ÷(1), so ø(1) is a power

of p. We may now apply induction on |G| to deduce that H has an

abelian normal p-complement N.

We have that |N| is coprime to p and jG : N j is a power of p, so it

remains to prove that N v G.

Suppose that g 2 G and the order of g is coprime to p. Then

g 2 H , since otherwise p divides the order of gH which in turn

divides the order of g; a similar argument shows that g 2 N. Hence N

consists of those elements of G whose order is coprime to p, and from

this fact it follows easily that N v G.

Finally, assume that G is simple, so either N � {1} or N � G. If

N � f1g then G is a p-group so Z(G) 6� {1} (see Exercise 12.7);

because G is simple, we have Z(G) � G, so G is abelian. On the other

hand, if N � G then G is again abelian. But an abelian simple group

has prime order, by Exercise 1.1. Therefore, G has prime order. j

A condition which ensures that ÷(g) is an integer

In Theorem 22.16 below we give a group-theoretic condition on an

element g of G which implies that ÷(g) is an integer for every

character ÷ of G. This result implies, for example, that for all n, every

entry in the character table of Sn is an integer (see Corollary 22.17).

Bearing in mind the dif®culties we encountered in constructing the

character tables of Sn for small values of n (we reached n � 6 in

Example 19.17), Theorem 22.16 is evidently a useful result.

Before proving Theorem 22.16, we require a preliminary lemma

concerning roots of unity. If a and b are positive integers, then we

denote their highest common factor by (a, b). Also, for integers d and

n, we write d|n to denote the fact that d divides n.

22.15 Lemma

If ù is an nth root of unity, then


X1<i<n,(i,n)�1

ùi

is an integer.

Proof We prove the result by induction on n. It is trivial for n � 1.

Also, if ù � 1 then the result is immediate. So suppose that ù is an

nth root of unity and ù 6� 1. Then ù is a root of the polynomial

(xn ÿ 1)=(xÿ 1) � x nÿ1 � : : :� x� 1: ThereforeP

ni�1 ù

i � 0.

Now we partition the sumP

ni�1 ù

i according to the highest common

factor d of i and n:

0 �Xn

i�1

ùi �Xdjn

X1<i<n(i,n)�d

ùi �Xdjn

X1< j<n=d,( j,n=d)�1

ùdj:

If djn then ùd is an (n=d)th root of unity, and if in addition d . 1,

then by our induction hypothesis,X1< j<n=d,( j,n=d)�1

ùdj 2 Z:

It follows that X1<i<n,(i,n)�1

ùi �Xn

i�1

ùi ÿXdjn,

d . 1

X1< j<n=d,( j,n=d)�1

ùdj 2 Z,

as required. j

22.16 Theorem

Let g be an element of order n in G. Suppose that g is conjugate to gi

for all i with 1 < i < n and (i, n) � 1. Then ÷(g) is an integer for all

characters ÷ of G.

Proof Let V be a CG-module with character ÷ of degree m. By

Proposition 9.11, there is a basis B of V such that

[g]B �ù1

. ..

ùm

0B@1CA0

0


where ù1, . . . , ùm are nth roots of unity. For 1 < i < n, the matrix

[gi]B has diagonal entries ùi1, . . . , ùi

m, and so

÷(gi) � ùi1 � : : :� ùi

m:

Therefore by Lemma 22.15, X1<i<n,(i,n)�1

÷(gi) 2 Z:

As g is conjugate to gi if 1 < i < n and (i, n) � 1, we have

÷(gi) � ÷(g) for such i, and hence

s÷(g) 2 Z,

where s is the number of integers i with 1 < i < n and (i, n) � 1.

Consequently ÷(g) is a rational number, and so ÷(g) is an integer by

Corollary 22.6. j

We remark that using Galois theory it is possible to prove the

converse of Theorem 22.16, namely that if ÷(gi) 2 Z for all characters

÷ of G, then g is conjugate to gi whenever i is coprime to n.

22.17 Corollary

All the character values of symmetric groups are integers.

Proof If g 2 Sn and i is coprime to the order of g, then the permuta-

tions g and gi have the same cycle-shape, and hence are conjugate by

Theorem 12.15. The result now follows from Theorem 22.16. j

The p9-part of a group element

The rest of the chapter is devoted to some important congruence

properties of character values. For example, one particularly useful

consequence of our results is that if p is a prime number, g is an

element of G of order pr for some r, and ÷ is a character of G such

that ÷(g) 2 Z, then ÷(g) � ÷(1) mod p.

Before going into the character theory, we need to de®ne the p9-part

of a group element. The de®nition will emerge from the following

lemma.


22.18 Lemma

Let p be a prime number and let g 2 G. Then there exist x, y 2 G such

that

(1) g � xy � yx,

(2) the order of x is a power of p, and

(3) the order of y is coprime to p.

Moreover, the elements x and y of G which satisfy conditions (1)±(3)

are unique.

Proof Let the order of g be upv, where u, v 2 Z and (u, p) � 1. Then

there exist integers a, b such that

au� bpv � 1:

Put x � gau and y � gbpv. Then

xy � yx � gau�bpv � g,

xpv � gaupv � 1,

yu � gbupv � 1:

Hence the order of x is a power of p and the order of y divides u, so

is coprime to p. Therefore x and y satisfy conditions (1)±(3).

Now suppose that x9, y9 2 G also satisfy (1)±(3); that is, g �x9y9 � y9x9, the order of x9 is a power of p and the order of y9 is

coprime to p. We must show that x � x9 and y � y9.

We have

x9g � x9y9x9 � gx9,

so x9 commutes with g, hence also with gau � x. Since both x and x9

have order a power of p, it follows that xÿ1x9 has order a power of p.

Similarly, y9 commutes with y and y(y9)ÿ1 has order coprime to p.

Finally, xy � g � x9y9, so

xÿ1x9 � y(y9)ÿ1:

If z � xÿ1x9 � y(y9)ÿ1, then we have shown that the order of z is both

a power of p and coprime to p. Therefore z � 1, and so x � x9 and

y � y9, as required. j


22.19 De®nition

We call the element y which appears in Lemma 22.18 the p9-part of

g.

We extract the following statement from the proof of Lemma 22.18.

(22.20) Let the order of g be upv, where u, v 2 Z and

(u, p) � 1, and choose integers a, b with au � bpv � 1.

Then the p9- part of g is gbpv.

For example, if p � 2 and g has order 6, then the p9-part of g is

gÿ2; the expression g � xy in Lemma 22.18 has x � g3, y � gÿ2.

A little ring theory

To prepare for our main result on congruence properties of character

values, we need a few basic facts about a certain subring of C in

which all our character values will lie.

Let n be a positive integer and let æ � e2ði=n. De®ne Z[æ] to be the

subring of C generated by Z and æ; that is,

Z[æ] � f f (æ): f (x) 2 Z[x]g:Clearly, every element of Z[æ] is an integer combination of the powers

1, æ, æ2, . . . , ænÿ1, so in fact

Z[æ] � f f (æ): f (x) 2 Z[x] of degree < nÿ 1g:Now let p be a prime number and let

pZ[æ] � fpr: r 2 Z[æ]g,a principal ideal of Z[æ].

22.21 Proposition

There are only ®nitely many ideals I of Z[æ] which contain pZ[æ].

Proof Consider the factor ring Z[æ]= pZ[æ]. By de®nition, this has as

its elements all the cosets pZ[æ] � r with r 2 Z[æ]. Every such coset

contains an element of the form

a0 � a1æ� : : :� anÿ1ænÿ1, with ai 2 Z, 0 < ai < pÿ 1 for all i:

As there are only ®nitely many such elements, we conclude that

Z[æ]=pZ[æ] is ®nite. The ideals of Z[æ] which contain pZ[æ] are in


bijective correspondence with the ideals of Z[æ]=pZ[æ] (the corre-

spondence being I ! I= pZ[æ]). Therefore there are only ®nitely many

such ideals, and the proof is complete. j

We deduce from Proposition 22.21 that there is a maximal ideal of

P of Z[æ] which contains pZ[æ]; that is, P is a proper ideal which is

contained in no larger proper ideal of Z[æ]. (A proper ideal of Z[æ] is

an ideal which is not equal to Z[æ].)

We now prove two easy results about the maximal ideal P.

22.22 Proposition

If r, s 2 Z[æ] and rs 2 P, then either r 2 P or s 2 P. In particular, if

rn 2 P for some positive integer n, then r 2 P.

Proof Assume that rs 2 P and r =2 P. We must show that s 2 P.

Since r =2 P, the ideal rZ[æ] � P of Z[æ] strictly contains P. As P is

maximal, we therefore have

rZ[æ]� P � Z[æ]:

Consequently, there exist a 2 Z[æ], b 2 P such that

1 � ra� b:

Then

s � rsa� sb:

As rs 2 P and b 2 P, it follows that s 2 P, as required.

For the last statement of the proposition, assume that r n 2 P. Since

r n � rr nÿ1, this implies that either r 2 P or r nÿ1 2 P. Repeating this

argument, we conclude that r 2 P. j

22.23 Proposition

We have P \ Z � pZ.

Proof Let m 2 P \ Z. If p Bj m then there are integers a, b with

am� bp � 1; but this implies that 1 2 P, which is false, since

P 6� Z[æ]. Thus pjm, which establishes that P \ Z � pZ. Since p 2 P,

we also have pZ � P \ Z. j


Congruences

At last we are ready to prove our results on congruences of character

values. Let G be a group of order n and let æ � e2ði=n. The ring Z[æ]

is of interest because all the character values of G lie in Z[æ] (see

Proposition 9.11). As in the previous section, let p be a prime number

and let P be a maximal ideal of Z[æ] containing pZ[æ].

22.24 Theorem

Let g 2 G and let y be the p9- part of g. If ÷ is any character of G,

then

÷(g)ÿ ÷(y) 2 P:

Proof Suppose that g has order m � upv, where u, v 2 Z and

(u, p) � 1. Choose integers a, b with au � bpv � 1. Then y � gbpv(see

(22.20)).

The orders of g and of y divide n � |G|, so ÷(g) and ÷(y) are both

sums of nth roots of unity, and hence lie in Z[æ].

Now let ù be an mth root of unity (so ù 2 Z[æ] as mjn). Then

ù � ùau�bpv, and so

ù pv � ùaupv. ùbp2v � ùbp2v

:

Consider the number (ù ÿ ùbpv) pv

. By the Binomial Theorem,

(ùÿ ùbpv

) pv � ù pv ÿ pvù pvÿ1

ùbpv � : : :� pv

r

� �ù pvÿ r

ùrbpv

� : : :� (ÿ1) pv

ùbp2v

:

For 0 , r , pv, the binomial coef®cient ( pv

r ) is divisible by p. Hence

(ùÿ ùbpv

) pv � ù pv � (ÿ1) pv

ùbp2v � pá,

where á 2 Z[æ]. Moreover, since ù pv � ùbp2v, we have

ù pv � (ÿ1) pv

ùbp2v �0, if p 6� 2,

2ù pv, if p � 2,

(so it follows that

(ùÿ ùbpv

) pv 2 pZ[æ]:

Thus (ù ÿ ùbpv) pv

lies in the maximal ideal P. Application of Proposi-

tion 22.22 now forces


(ùÿ ùbpv

) 2 P:(22:25)

By Proposition 9.11, there are mth roots of unity ù1, . . . , ùd such

that

÷(g) � ù1 � : : :� ùd and ÷(y) � ùbpv

1 � : : :� ùbpv

d :

Then

÷(g)ÿ ÷(y) � (ù1 ÿ ùbpv

1 )� : : :� (ùd ÿ ùbpv

d ),

which, by (22.25), lies in P. j

22.26 Corollary

Let p be a prime number. Suppose that g 2 G and that y is the p9- part

of g. If ÷ is a character of G such that ÷(g) and ÷(y) are both

integers, then

÷(g) � ÷(y) mod p:

Proof As ÷(g) and ÷(y) are both integers, Theorem 22.24 and Proposi-

tion 22.23 give

÷(g)ÿ ÷(y) 2 P \ Z � pZ:

Therefore ÷(g) � ÷(y) mod p. j

22.27 Corollary

Let p be a prime number. Suppose that g 2 G and the order of g is a

power of p. If ÷ is a character of G such that ÷(g) 2 Z, then

÷(g) � ÷(1) mod p:

Proof As the order of g is a power of p, the p9-part of g is 1, so the

result is immediate from Corollary 22.26. j

Notice that Corollary 13.10 is the special case of Corollary 22.27 in

which g has order 2.

We shall use the congruence results 22.24±22.27 extensively in our

character calculations in Chapters 25±7. For the moment, we just

illustrate the results with reference to some character tables which we

already know.


22.28 Example

Recall from Example 20.14 that the character table of A5 is as shown.

where á � (1�p5)=2, â � (1ÿp5)=2.

If g � (1 2 3) then Corollary 22.26 implies that ÷(g) � ÷(1) mod 3

whenever ÷(g) 2 Z. Thus the corresponding entries in columns 1 and 2

of the character table are congruent modulo 3, as can be seen by

inspecting the table. Similarly the entries in columns 1 and 3 are

congruent modulo 2. Also

÷i((1 2 3 4 5)) � ÷i(1) mod 5 for i � 1, 2, 3:

However, ÷4((1 2 3 4 5)) � á =2 Z. We illustrate Theorem 22.24 for this

value. If we take p � 5 and g � (1 2 3 4 5), then the p9-part of g is

1, and

÷4(g)ÿ ÷4(1) � áÿ 3 � 12(1�p5ÿ 6)

� p5 . 12(1ÿp5) � â

p5:

Put æ � e2ði=60, and let P be a maximal ideal of Z[æ] containing 5Z[æ].

Then (p

5)2 2 P, sop

5 2 P by Proposition 22.22. Since â 2 Z[æ] (see

Proposition 9.11), we have âp

5 2 P. That is,

÷4(g)ÿ ÷4(1) 2 P:

This illustrates Theorem 22.24.


1. Character values are algebraic integers.

2. The degree of every irreducible character of G divides |G|.


1 (1 2 3) (1 2)(3 4) (1 2 3 4 5) (1 3 4 5 2)

÷1 1 1 1 1 1÷2 4 1 0 ÿ1 ÿ1÷3 5 ÿ1 1 0 0÷4 3 0 ÿ1 á â÷5 3 0 ÿ1 â á


3. If g is conjugate to gi for all integers i which are coprime to the

order of g, then ÷(g) is an integer, for all characters ÷.

4. Let p be a prime number. If g 2 G and y is the p9-part of g, then

÷(g) � ÷(y) mod p for all characters ÷ of G such that ÷(g) and ÷(y)

are integers.


1. Let G be a group of order 15. Use Theorems 11.12, 17.11 and

22.11 to show that every irreducible character of G has degree 1.

Deduce that G is abelian.

2. Prove that the number of conjugacy classes in a group of order 16

is 7, 10 or 16.

3. Let p and q be prime numbers with p . q, and let G be a non-

abelian group of order pq.

(a) Find the degrees of all the irreducible characters of G.

(b) Show that |G9| � p.

(c) Show that q divides p ÿ 1 and that G has q� (( pÿ 1)=q)

conjugacy classes.

4. Let G be a group and let ö be a character of G such that

ö(g) � ö(h) for all non-identity elements g and h of G.

(a) Show that ö � a1G � b÷reg for some a, b 2 C.

(b) Show that a � b and a � b|G| are integers.

(c) Show that if ÷ is a non-trivial irreducible character of G, then

b÷(1) is an integer.

(d) Deduce that both a and b are integers.

5. Suppose that G is a group of odd order. This exercise shows that

the only irreducible character ÷ of G such that ÷ � ÷ is the trivial

character.

(a) Prove that if g 2 G and g � gÿ1, then g � 1.

(b) Now let ÷ be an irreducible character of G with ÷ � ÷. Prove

that

h÷, 1Gi � 1

jGj (÷(1)� 2á),

where á is an algebraic integer.

(c) Deduce that ÷ � 1G.


6. It is often possible to calculate the character table from limited

arithmetic information about the group G. This exercise illustrates

this point with the group G � S5.

A certain group G of order 120 has exactly seven conjugacy

classes, and contains an element g of order 5 such that jCG(g)j � 5.

Moreover, g, g2, g3 and g4 are conjugate in G.

(a) Show that for every irreducible character ÷ of G, ÷(g) is 0, 1

or ÿ1.

(b) Use Corollary 22.27 to deduce that G has two irreducible

characters of degree 5.

(c) Find ÷(1) and ÷(g) for all irreducible characters ÷ of G.

(d) You are given that all entries in the character table of G are

integers, and that the conjugacy classes of G have represent-

atives g1, . . . , g7 with orders and centralizer orders as follows:

Using Corollary 22.26 and the column orthogonality relations, ®nd

the character table of G.

7. Prove that a complex number ë is an algebraic integer if and only if

ë is a root of a polynomial of the form

x n � anÿ1x nÿ1 � : : :� a1x� a0,

where each ar (0 < r < n ÿ 1) is an integer.

gi g1 g2 g3 g4 g5 g6 g7

Order of gi 1 2 2 3 4 6 5|CG(gi)| 120 12 8 6 4 6 5


263

23

Real representations

Since Chapter 9, we have always taken our representations to be over

the ®eld C of complex numbers. However, some results in representa-

tion theory work equally well for the ®eld R of real numbers. There is

a subtle interplay between representations over C and representations

over R, which we shall explore in this chapter.

Often, characters of CG-modules are real-valued, and the ®rst main

result of the chapter describes the number of real-valued irreducible

characters of G.

Let r be a representation of G. If all the matrices gr (g 2 G) have

real entries, then of course the character of r is real-valued. However,

the converse is not true: it can happen that the character of r is real-

valued, while there is no representation ó equivalent to r with all the

matrices gó having real entries. Various criteria for whether or not a

character corresponds to a representation over R lead us to the

remarkable Frobenius±Schur Count of Involutions. This is used in the

last section to prove a famous result of Brauer and Fowler concerning

centralizers of involutions in ®nite simple groups.

The material in this chapter is perhaps at a slightly more advanced

level than that in the rest of the book, and is not used in the ensuing

chapters, which consist largely of the calculation of character tables

and applications of character theory. Nevertheless, the subject of real

representations not only is elegant and interesting, but also gives

delicate information about characters which often comes into play in

more dif®cult calculations.

Real characters

An element g of the ®nite group G is said to be real if g is conjugate

to gÿ1; and if g is real, then the conjugacy class gG is also said to be

real. Notice that if a conjugacy class is real, then it contains the

inverse of each of its elements, since (gÿ1)G � fxÿ1: x 2 gGg.On the other hand, a character ÷ of G is real if ÷(g) is real for all

g 2 G. Thus for example, the conjugacy class {1} of G is real, and the

trivial character of G is real.

23.1 Theorem

The number of real irreducible characters of G is equal to the number

of real conjugacy classes of G.

Proof Let X denote the character table of G, and let X denote the

complex conjugate of the matrix X.

For every irreducible character ÷ of G, the complex conjugate ÷ is

also an irreducible character (see Proposition 13.15), so X can be

obtained from X by permuting the rows. Hence there is a permutation

matrix P such that

PX � X

(see Exercise 4.4).

For every conjugacy class gG of G, the entries in the column of X

which corresponds to gG are the complex conjugates of the entries in

the column of X which corresponds to (gÿ1)G. Therefore X can be

obtained from X be permuting the columns, and so there is a permuta-

tion matrix Q such that

XQ � X

By Proposition 16.2, X is invertible. Therefore

Q � Xÿ1 X � Xÿ1 PX :

Consequently P and Q have the same trace, by Proposition 13.2. Since

the trace of a permutation matrix is equal to the number of points

®xed by the corresponding permutation, we have

the number of real irreducible characters of G is tr (P),

andthe number of real conjugacy classes of G is tr (Q):

As these numbers are equal, the result is proved. j

Part of the following corollary was obtained by a different method in

Exercise 22.5.


23.2 Corollary

The group G has a non-trivial real irreducible character if and only if

the order of G is even.

Proof If G has odd order, then no non-identity element of G is real

(see the solution to Exercise 23.1). Therefore by Theorem 23.1, the

only real character of G is the trivial character.

If G has even order, then by Exercise 1.8, G has an element g of

order 2. Hence G has at least two real conjugacy classes, {1} and gG,

and so G has at least two real irreducible characters by Theorem

23.1. j

Characters which can be realized over R

Let ÷ be a character of the group G. We say that ÷ can be realized

over R if there is a representation r: G! GL (n, C) with character ÷,

such that all the entries in each matrix gr (g 2 G) are real. This is the

same as saying that there is some CG-module V with character ÷, and

there is a basis v1, . . . , vn of V, such that for all g 2 G and 1 < i < n,

vi g is a linear combination of v1, . . . , vn with real coef®cients.

23.3 Examples

(1) Let G � D8 � ka, b: a4 � b2 � 1, bÿ1ab � aÿ1l, and let ÷ be the

irreducible character of G of degree 2 (see Example 16.3(3)). Then ÷can be realized over R, since

ar � 0 1

ÿ1 0

� �, br � ÿ1 0

0 1

� �provides a representation r of G with character ÷ such that all the

matrices gr (g 2 G) have real entries.

(2) Let G � Q8 � ka, b: a4 � 1, b2 � a2, bÿ1ab � aÿ1l, and let ÷ be the

irreducible character of G of degree 2 (see Exercise 17.1). The values

of ÷ are as follows:

1 a2 a b ab

÷ 2 ÿ2 0 0 0

Real representations 265

Thus ÷ is real. In fact, ÷ cannot be realized over R, but it is at the

moment unclear how to prove this. (We shall eventually establish this

in Example 23.18(3) below.)

Although every character which can be realized over R is perforce a

real character, Example 23.3(2) tells us that the converse is false.

RG-modules

Recall that in Chapter 4 we de®ned an FG-module, where F is R or

C. Thus an RG-module is a vector space over R, with a multiplication

by elements of G satisfying the conditions of De®nition 4.2. In this

section we shall study the relationship between RG-modules and CG-

modules.

23.4 Examples

Let V be a 2-dimensional vector space over R, with basis v1, v2.

(1) V becomes an RD8-module if we de®ne

v1a � v2, v1b � ÿv1,

v2a � ÿv1, v2b � v2

(compare Example 23.3(1)).

(2) V becomes an RC3-module, where C3 � kx: x3 � 1l, if we de®ne

v1x � v2,

v2x � ÿv1 ÿ v2:

(This gives the representation x! 0 1

ÿ1 ÿ1

� �of Exercise 3.2.)

Every RG-module can easily be converted into a CG-module.

Simply take a basis v1, . . . , vn of the RG-module, and consider the

vector space over C with basis v1, . . . , vn. This new vector space is

clearly a CG-module (with vi g de®ned as before). The construction is

even easier to understand in terms of representations: if r:

G! GL (n, R) is a representation then for each g 2 G, the matrix grhas its entries in R, and hence also in C. Therefore we obtain a

representation r: G! GL (n, C). Notice that a character ÷ of G can be


realized over R if and only if there exists an RG-module with

character ÷.

Rather more subtle than this is the construction of an RG-module from

a given CG-module. Let V be a CG-module with basis v1, . . . , vn,

and let g 2 G. There exist complex numbers z jk such that

v j g �Xn

k�1

zjkvk (1 < j < n):

Now let VR be the vector space over R with basis

v1, : : : , vn, iv1, : : : , ivn:

Write z jk � xjk � iyjk with xjk , yjk 2 R. We de®ne a multiplication of

VR by g by putting

v j g �Xn

k�1

(xjkvk � yjk(ivk)), and(23:5)

(iv j)g �Xn

k�1

(ÿyjkvk � xjk(ivk)) (1 < j < n),

and extending linearly to de®ne vg for all v 2 VR. In this way we

de®ne vg for all v 2 VR and all g 2 G. Regarding v j as an element of

the CG-module V, we have

(v j g)h � v j(gh) for all g, h 2 G, 1 < j < n:

It follows easily that, regarding v j and iv j as elements of VR, we have

(v j g)h � v j(gh) and ((iv j)g)h � (iv j)(gh):

Hence using Proposition 4.6, we see that (23.5) makes VR into an RG-

module.

If ÷ is the character of V, then

÷(g) �Xn

k�1

zkk :

The character of VR, evaluated at g, is

2Xn

k�1

xkk � ÷(g)� ÷(g):

Hence the character of VR is ÷ � ÷.

We summarize the basic properties of VR in the next proposition.


23.6 Proposition

Let V be a CG-module with character ÷.

(1) The RG-module VR has character ÷ � ÷; in particular, dim VR

� 2 dim V.

(2) If V is an irreducible CG-module and VR is a reducible RG-

module, then ÷ can be realized over R.

Proof We have already proved part (1).

For part (2), suppose that V is an irreducible CG-module and VR is

a reducible RG-module. Then by part (1), VR � U � W where U is an

RG-module with character ÷ and W is an RG-module with character ÷.

Thus there is an RG-module, namely U, with character ÷, and so ÷ can

be realized over R. j

23.7 Examples

(1) Let G � C3 � kx: x3 � 1l, and let V be the 1-dimensional CG-

module with basis v1 such that

v1x � 12(ÿ1� i

p3)v1

(note that 12(ÿ1� i

p3) � e2ði=3). Then VR has basis v1, iv1, and with

respect to this basis, x is represented by the matrix

ÿ1=2p

3=2

ÿp3=2 ÿ1=2

� �:

(2) Let G � D8 � ka, b: a4 � b2 � 1, bÿ1ab � aÿ1l, and let V be the 2-

dimensional CG-module with basis v1, v2 such that

v1a � iv1, v1b � v2,

v2a � ÿiv2, v2b � v1:

Then VR has basis v1, v2, v3, v4, where v3 � iv1, v4 � iv2. With

respect to this basis, we obtain the representation r, where

ar �0 0 1 0

0 0 0 ÿ1

ÿ1 0 0 0

0 1 0 0

0BB@1CCA, br �

0 1 0 0

1 0 0 0

0 0 0 1

0 0 1 0

0BB@1CCA:

The subspace of VR which is spanned by v1 � v4 and v2 � v3 is an

RG-submodule. Therefore the character of V can be realized over R,


by Proposition 23.6(2). In fact, we already know this from Example

23.3(1).

Bilinear forms

The question of whether or not a given character can be realized over

R turns out to be related to the existence of a certain bilinear form on

the corresponding CG-module.

Let V be a vector space over F, where F is R or C. A bilinear form

â on V is a function which associates with each ordered pair (u, v) of

vectors in V an element â(u, v) of F, and which has the following

properties:

â(ë1u1 � ë2u2, v) � ë1â(u1, v)� ë2â(u2, v),

â(u, ë1v1 � ë2v2) � ë1â(u, v1)� ë2â(u, v2),

for all u, v, u1, u2, v1, v2 2 V and ë1, ë2 2 F. (Thus for ®xed u, v, the

functions x! â(x, v) and y! â(u, y) are both linear ± hence the term

bilinear.)

The bilinear form â is symmetric if

â(u, v) � â(v, u) for all u, v 2 V :

And the bilinear form â is skew-symmetric if

â(u, v) � ÿâ(v, u) for all u, v 2 V :

If V is an FG-module, then a bilinear form â on V is said to be G-

invariant if

â(ug, vg) � â(u, v) for all u, v 2 V and g 2 G:

Our next result shows that every RG-module has a G-invariant

symmetric bilinear form with a strong positivity property. A similar

result for CG-modules was given in Exercise 8.6.

23.8 Theorem

If V is an RG-module, then there exists a G-invariant symmetric

bilinear form â on V such that

â(v, v) . 0 for all non-zero v 2 V :


Proof Let v1, . . . , vn be a basis of V. For u �Pnj�1ë jv j, v �Pn

j�1ì jv j 2 V with ë j, ì j 2 R, de®ne

ã(u, v) �Xn

j�1

ë jì j:

Then ã is a symmetric bilinear form on V. Moreover, for non-zero

v 2 V,

ã(v, v) �Xn

j�1

ì2j . 0:

Now let

â(u, v) �Xx2G

ã(ux, vx) (u, v 2 V ):

Again, â is a symmetric bilinear form on V, and â(v, v) . 0 for all

non-zero v 2 V.

If g 2 G, then gx runs through G as x runs through G, and hence

â(ug, vg) �Xx2G

ã(ugx, vgx) � â(u, v):

Therefore â is G-invariant and the theorem is proved. j

23.9 Proposition

Let V be an RG-module and let â be a G-invariant bilinear form on V.

If U is an RG-submodule of V, then so is

W � fw 2 V : â(u, w) � 0 for all u 2 Ug:

Proof It is easy to see that W is a subspace of V. Now let w 2 W and

g 2 G. For all u 2 U, we have ugÿ1 2 U, so

â(u, wg) � â(ugÿ1, wggÿ1) � â(ugÿ1, w) � 0:

Thus wg 2 W, so W is an RG-submodule of V. j

23.10 Proposition

Suppose that â is a G-invariant symmetric bilinear form on the RG-

module V, and that there exist u, v 2 V with â(u, u) . 0 and

â(v, v) , 0. Then V is a reducible RG-module.


Proof Theorem 23.8 supplies us with a G-invariant symmetric bilinear

form â1 on V such that

â1(w, w) . 0 for all non-zero w 2 V :

By a general result on bilinear forms (see Exercise 23.7), there is a

basis v1, . . . , vn of V such that

â1(vi, v j) � â(vi, v j) � 0 if i 6� j,

andâ1(vi, vi) � 1 for all i,

â(v1, v1) . 0,

â(v2, v2) , 0:

Let â(v1, v1) � x, and de®ne ã by

ã(u, v) � â1(u, v)ÿ 1

xâ(u, v) (u, v 2 V ):

Since â and â1 are G-invariant symmetric bilinear forms on V, so is ã.

But for all v �Pni�1ëivi 2 V (ëi 2 R), we have

ã(v, v1) � ë1ã(v1, v1) � 0:

Therefore, if we de®ne

W � fw 2 V : ã(v, w) � 0 for all v 2 Vg,then W is non-zero, and is an RG-submodule of V by Proposition 23.9.

Moreover,

ã(v2, v2) � 1ÿ 1

xâ(v2, v2) . 0,

so W 6� V. Therefore V is a reducible RG-module. j

We can now relate bilinear forms to the question of whether or not

a given character of G can be realized over R.

23.11 Theorem

Let ÷ be an irreducible character of G. The following two conditions

are equivalent:

(1) ÷ can be realized over R;

(2) there exists a CG-module V with character ÷, and a non-zero G-

invariant symmetric bilinear form on V.


Proof We ®rst show that (2) implies (1). Let V be a CG-module with

character ÷, and suppose that â is a non-zero G-invariant symmetric

bilinear form on V. There exist u, v 2 V with â(u, v) � â(v, u) 6� 0.

Sinceâ(u� v, u� v) � â(u, u)� â(v, v)� 2â(u, v),

there exists w 2 V with â(w, w) 6� 0. Let â(w, w) � z and v1 � zÿ1=2w.

Thenâ(v1, v1) � 1:

Extend v1 to a basis v1, . . . , vn of V. Then v1, . . . , vn, iv1, . . . , ivn

is a basis of the RG-module VR.

De®ne a function W from VR to V by

W:Xn

j�1

ë jv j �Xn

j�1

ì j(iv j)!Xn

j�1

(ë j � iì j)v j (ë j, ì j 2 R):

Then W is a bijection, and for all w1, w2, v 2 VR, all ë 2 R and all

g 2 G, we have

(w1 � w2)W � w1W� w2W,(23:12)

(ëv)W � ë(vW),

(vg)W � (vW)g:

Now de®ne a function ~â on ordered pairs of elements of VR by

~â(u, v) � the real part of â(uW, vW) (u, v 2 VR):

You can readily check, using the properties (23.12), that ~â is a G-

invariant symmetric bilinear form on VR. Notice that

~â(v1, v1) � 1 and ~â(iv1, iv1) � ÿ1:

Therefore VR is a reducible RG-module, by Proposition 23.10. It now

follows from Proposition 23.6(2) that ÷ can be realized over R. This

establishes that (2) implies (1) in the theorem.

Conversely, suppose that ÷ can be realized over R, and let U be an

RG-module with character ÷. By Theorem 23.8, there is a non-zero G-

invariant symmetric bilinear form ã on U. Let u1, : : : , un be a basis

of U, and let V be the vector space over C with basis u1, : : : , un. As

explained earlier, V is a CG-module (with ui g de®ned as for U).

De®ne ã̂ on V by

ã̂Xn

j�1

ë ju j,Xn

k�1

ìk uk

!�Xn

j�1

Xn

k�1

ë jìkã(u j, uk)


(where ë j, ìk 2 C). Then ã̂ is a non-zero G-invariant symmetric

bilinear form on the CG-module V, and V has character ÷. Thus (1)

implies (2), and the proof of the theorem is complete. j

The indicator function

We now associate with each irreducible character ÷ of G a certain

number, called the indicator of ÷, which is always 0, 1 or ÿ1. We shall

see later that this number tells us whether or not ÷ can be realized

over R.

Observe that

h÷2, 1Gi � 1

jGjXg2G

÷(g)÷(g) � h÷, ÷i:

Therefore, for irreducible characters ÷, we have

h÷2, 1Gi �0, if ÷ is not real,

1, if ÷ is real:

(Let V be a CG-module with character ÷. Recall from Chapter 19 that

÷2 is the character of the CG-module V V, and

÷2 � ÷S � ÷A,

where ÷S is the character of the symmetric part of V V, and ÷A is

the character of the antisymmetric part of V V. Hence if

h÷2, 1Gi � 1, then precisely one of ÷S and ÷A has 1G as a constituent.

23.13 De®nition

If ÷ is an irreducible character of G, then we de®ne the indicator é÷ of

÷ by

é÷ �0, if ÷ is not a constituent of ÷S or ÷A,

1, if 1G is a constituent of ÷S ,

ÿ1, if 1G is a constituent of ÷A:

8>><>>:We call é the indicator function on the set of irreducible characters of

G. Note that é÷ 6� 0 if and only if ÷ is real.

The next result gives a signi®cant property of the indicator function,

relating it to the internal structure of the group G.


23.14 Theorem

For all x 2 G, X÷

(é÷)÷(x) � jfy 2 G: y2 � xgj,

where the sum is over all irreducible characters ÷ of G.

Proof De®ne a function W: G! C by

W(x) � jfy 2 G: y2 � xgj (x 2 G):

Note that W is a class function on G, since for g 2 G we have

y2 � x, (gÿ1 yg)2 � gÿ1xg:

Therefore by Corollary 15.4, W is a linear combination of the irreduci-

ble characters of G.

The de®nition of é÷ gives

é÷ � h÷S ÿ ÷A, 1Gi

� 1

jGjXg2G

÷(g2) by Proposition 19:14

� 1

jGjXx2G

Xg2G: g2�x

÷(g2)

� 1

jGjXx2G

W(x)÷(x)

� hW, ÷i:Therefore, W �P(é÷)÷, and the result follows. j

23.15 Example

Let G � S3. The character table of G is

1 (1 2) (1 2 3)

÷1 1 1 1÷2 1 ÿ1 1÷3 2 0 ÿ1


Using Proposition 19.14 we calculate that é÷ � 1 for each irreducible

character ÷ of G, soP

(é÷)÷ � ÷1 � ÷2 � ÷3, which takes the following

values:

Sure enough, in accordance with Theorem 23.14, four elements of G

square to be 1, namely 1, (1 2), (1 3) and (2 3); no elements square to

be (1 2); and one element, (1 3 2), squares to be (1 2 3).

Back to reality

We now relate the indicator function to the previous material on

bilinear forms. Using this, we show that the indicator of an irreducible

character determines whether or not it can be realized over R, and

deduce the Frobenius±Schur Count of Involutions.

23.16 Theorem

Let V be an irreducible CG-module with character ÷.

(1) There exists a non-zero G-invariant bilinear form on V if and

only if é÷ 6� 0.

(2) There exists a non-zero G-invariant symmetric bilinear form on

V if and only if é÷ � 1.

(3) There exists a non-zero G-invariant skew-symmetric bilinear form

on V if and only if é÷ � ÿ1.

Proof In this proof we regard C as a 1-dimensional vector space over

C, and de®ne a multiplication of C by elements of G by

ëg � ë (ë 2 C, g 2 G):

In this way, C becomes a trivial CG-module.

(1) Suppose that é÷ 6� 0. Then 1G is a constituent of ÷2, and hence

the CG-module V V has a trivial CG-submodule. By Proposition 8.8,

there is a non-zero CG-homomorphism from V V onto this trivial

CG-submodule, and hence there is a non-zero CG-homomorphism Wfrom V V onto the trivial CG-module C.

1 (1 2) (1 2 3)

÷1 � ÷2 � ÷3 4 0 1


Now de®ne â by

â(u, v) � (u v)W (u, v 2 V ):

Then â is a non-zero bilinear form on V, and for u, v 2 V and g 2 G,

we have

â(ug, vg) � (ug vg)W � ((u v)g)W

� ((u v)W)g � (u v)W � â(u, v):

Thus â is G-invariant.

Conversely, suppose that there is a non-zero G-invariant bilinear form

â on V. Let v1, : : : , vn be a basis of V, so that vi v j

(1 < i < n, 1 < j < n) form a basis of V V. De®ne W: V V! C by

putting

(vi v j)W � â(vi, v j) (1 < i < n, 1 < j < n)

and extending linearly to the whole of V V. For g 2 G, we have

((vi v j)g)W � (vi g v j g)W � â(vi g, v j g)

� â(vi, v j) as â is G-invariant

� (vi v j)W:

Hence W is a non-zero CG-homomorphism from V V onto the trivial

CG-module C. Thus, by Proposition 10.1, V V has a trivial CG-

submodule. If follows that ÷2 has the trivial character 1G as a constitu-

ent, and therefore é÷ 6� 0.

(2) Suppose that é÷ � 1. Then 1G is a constituent of ÷S , which is the

character of the CG-module S(V V), the symmetric part of V V. As

in (1), it follows by Proposition 8.8 that there is a non-zero CG-

homomorphism W from S(V V) onto the trivial CG-module C. De®ne

â(u, v) � (u v� v u)W (u, v 2 V ):

Then â is a non-zero G-invariant symmetric bilinear form on V.

Conversely, suppose that there exists a non-zero G-invariant sym-

metric bilinear form â on V. Let v1, . . . , vn be a basis of V, and de®ne

W: S(V V)! C by putting

(vi v j � v j vi)W � â(vi, v j) (1 < i, j < n)

and extending linearly. Since â is symmetric, W is well-de®ned; and Wis a non-zero CG-homomorphism from S(V V) onto the trivial CG-


module C. Hence ÷S has the trivial character 1G as a constituent, and

so é÷ � 1.

(3) The proof of (3) is very similar to that of (2), and is omitted. j

We can now relate real representations of G to involutions in G,

where by an involution we mean an element of order 2.

23.17 Corollary (The Frobenius±Schur Count of Involutions)

For each irreducible character ÷ of G, we have

é÷ �0, if ÷ is not real,

1, if ÷ can be realized over R,

ÿ1, if ÷ is real, but ÷ cannot be realized over R:

8>><>>:Moreover, for all x 2 G,X

÷

(é÷)÷(x) � jfy 2 G: y2 � xgj,

where the sum is over all irreducible characters ÷ of G. In particular,X÷

(é÷)÷(1) � 1� t,

where t is equal to the number of involutions in G.

Proof When we de®ned the indicator function, we showed that é÷ 6� 0

if and only if ÷ is real. And Theorems 23.11 and 23.16(2) show that ÷can be realized over R if and only if é÷ � 1. This proves that é÷ is

determined as in the statement of the corollary.

The expression forP

÷(é÷)÷(x) was obtained in Theorem 23.14.

Putting x � 1, we see thatP

÷(é÷)÷(1) is equal to the number of

elements y of G satisfying y2 � 1. These elements are just the

involutions in G, together with the identity, so the number of them is

precisely 1 � t. j

We conclude with some examples illustrating the use of Corollary

23.17.

23.18 Examples

(1) Let ÷ be a linear character. Then é÷ � 1 if ÷ is real, and é÷ � 0 if

÷ is non-real.


For an abelian group, the Frobenius±Schur Count of Involutions

shows that the number of real irreducible (linear) characters is equal to

the number of real conjugacy classes, since in this case g is conjugate

to gÿ1 if and only if g2 � 1. This special case of Theorem 23.1 can

be proved directly without much dif®culty (see Exercise 23.2).

(2) We know that all the irreducible characters of D8 �ka, b: a4 � b2 � 1, bÿ1ab � aÿ1l can be realized over R (see Example

23.3(1), and note that all four linear characters are real). Thus é÷ � 1

for all irreducible characters ÷ of D8, and soX÷

(é÷)÷(1) � 1� 1� 1� 1� 2 � 6:

The ®ve involutions in D8 which are predicted by the Frobenius±

Schur Count of Involutions are a2, b, ab, a2b and a3b.

(3) In the group Q8 � ka, b: a4 � 1, a2 � b2, bÿ1ab � aÿ1l, there is just

one involution, namely a2. Now é÷ � 1 for each of the four linear

characters, and X÷

(é÷)÷(1) � 2

by the Frobenius±Schur Count of Involutions. Therefore, if ø is the

irreducible character of degree 2, then éø � ÿ1. In particular ø cannot

be realized over R.

(4) The symmetric group S4 has ten elements whose square is 1, namely

the identity, the six elements which are conjugate to (1 2), and the three

elements which are conjugate to (1 2)(3 4). Since the degrees 1, 1, 2, 3,

3 of the irreducible characters of S4 sum to be 10 (see Section 18.1), we

see that all the characters of S4 can be realized over R.

The Brauer±Fowler Theorem

We now apply Corollary 23.17 to give a proof of a famous theorem of

Brauer and Fowler.

23.19 Brauer±Fowler Theorem

Let n be a positive integer. Then there exist only ®nitely many non-

isomorphic ®nite simple groups containing an involution with central-

izer of order n.


Despite its fairly elementary proof, this result is of great historical

importance in ®nite group theory. It led Brauer to propose the

programme of determining, for each ®nite group C, all the simple

groups G possessing an involution u with CG(u) � C. This programme

was the start of the modern attempt to classify all ®nite simple groups,

which was ®nally completed in the early 1980s. For further information

about this, see the book by D. Gorenstein listed in the Bibliography.

In Exercise 10 at the end of the chapter you are asked to carry out

Brauer's programme in the case where C � C2. This should not trouble

you too much. In Chapter 30, Theorem 30.8, you will ®nd a much

more sophisticated case, in which C � D8.

For proof of Theorem 23.19 we require two preliminary lemmas.

23.20 Lemma

If a1, . . . , an are real numbers, thenP

a2i >

Pai� �2=n.

Proof This follows from the Cauchy±Schwarz inequality

kvk kwk > jv:wj, taking v � (a1, . . . , an) and w � (1, . . . , 1). j

23.21 Lemma

Let G be a group of even order m, and let t be the number of

involutions in G (so t . 0 by Exercise 8 of Chapter 1). Write

a � (mÿ 1)=t. Then G contains a non-identity element x such that

jG : CG(x)j < a2.

Proof By Corollary 23.17, we have

t <X÷

÷(1)

where the sum is over all non-trivial irreducible characters ÷ of G.

Writing k for the number of irreducible characters of G, we deduce

using Lemma 23.20 and Theorem 11.12 that

t2 < (X÷

÷(1))2 < (k ÿ 1)X

÷(1)2 � (k ÿ 1)(mÿ 1),

and hence mÿ 1 < (k ÿ 1)(mÿ 1)2=t2 � (k ÿ 1)a2. Now k ÿ 1 is

the number of non-identity conjugacy classes of G. If every non-

identity conjugacy class has size more then a2, then (k ÿ 1)a2 .


jGj ÿ 1 � mÿ 1, a contradiction. Therefore some non-identity class xG

has size at most a2. Then jG : CG(x)j < a2, giving the result. j

Proof of Theorem 23.19 Suppose G is simple and contains an involu-

tion u such that jCG(u)j � n. Let jGj � m, and let t be the number of

involutions in G. By Proposition 12.6, every element of the conjugacy

class uG is an involution, and hence

t > juGj � jG : CG(u)j � m=n:

Therefore (mÿ 1)=t , n, and so by Lemma 23.21, there is a non-

identity element x 2 G such that jG : CG(x)j, n2.

Let H � CG(x). If H � G then x lies in Z(G), the centre of G,

which is a normal subgroup of G. Since G is simple it follows that

G � Z(G), so G is abelian and therefore G � C2.

Now suppose that H 6� G. Write r � jG : H j, so r , n2. By Exer-

cise 9 at the end of the chapter, there is a non-trivial homomorphism èfrom G to the symmetric group Sr. As G is simple, the normal

subgroup Ker è � f1g. Thus G is isomorphic to a subgroup of Sr,

hence of Sn2 . In particular, given n, there are only ®nitely many

possibilities for G. j


1. The number of real irreducible characters of G is equal to the

number of real conjugacy classes of G.

Let é be the indicator function, and let ÷ be an irreducible

character of G.

2: é÷ �

0, if ÷ is non-real,

1, if there exists an RG-module U with character ÷,

with character ÷

ÿ1, if ÷ is real, but there does not exist an RG-module:

8>>>><>>>>:3: é÷ �

0, if 1G is not a constituent of ÷S or ÷A,

1, if 1 G is a constituent of ÷S ,

ÿ1, if 1G is a constituent of ÷A:

8><>:4.P

÷(é÷)÷(1) � |{g 2 G: g2 � 1}|.



1. Prove that if G is a group of odd order then no non-identity

element of G is real.

2. Let G be a ®nite abelian group. Use the description of the

irreducible characters of G, given in Theorem 9.8, to prove directly

that the number of real irreducible characters of G is equal to the

number of elements g in G for which g2 � 1.

3. Let G � D2n and consult Section 18.3 for the character table of G.

How many elements g of G satisfy g2 � 1? Deduce that é÷ � 1 for

all irreducible characters ÷ of G.

4. Let r be an irreducible representation of degree 2 of a group G,

and let ÷ be the character of r. Prove that ÷A(g) � det (gr) for all

g 2 G. Deduce that é÷ � ÿ1 if and only if det (gr) � 1 for all

g 2 G.

5. Let G � T4n � ha, b: a2n � 1, an � b2, bÿ1ab � aÿ1i, as in Exer-

cise 17.6. Let V be a 2-dimensional vector space over C with basis

v1, v2 and let å be a (2n)th root of unity in C with å 6� �1.

Exercise 17.6 shows that V becomes an irreducible CG-module if

we de®ne

v1a � åv1, v1b � v2,

v2a � åÿ1v2, v2b � å nv1:

Let ÷ be the character of this CG-module V.

(a) Note that å n � �1. Use Exercise 4 to show that é÷ � 1 if

å n � 1 and é÷ � ÿ1 if å n � ÿ1.

(b) Let â be the bilinear form on V for which

â(v1, v1) � â(v2, v2) � 0,

â(v1, v2) � 1, â(v2, v1) � å n:

Prove that the bilinear form â is G-invariant, and use Theorem

23.16 to provide a second proof that é÷ � 1 if å n � 1 and

é÷ � ÿ1 if å n � ÿ1.

(c) Prove that an is the only element of order 2 in T4n.

(d) Use the character table of G, which appears in the solution to

Exercise 18.3, to ®nd é÷ for each irreducible character ÷ of G.

Check that


X÷

(é÷)÷(1) � 2,

in agreement with the Frobenius±Schur Count of Involutions.

6. Prove that if ÷ is an irreducible character of a group G, and

é÷ � ÿ1, then ÷(1) is even.

(Hint: the solution uses a well known result about skew-symmetric

bilinear forms.)

7. Suppose that V is a vector space over R and that â1 and â are

symmetric bilinear forms on V. Assume that â1(w, w) . 0 for all

non-zero w in V. Prove that there is a basis e1, : : : , en of V such

that

â1(ei, ei) � 1 for all i, and

â1(ei, ej) � â(ei, ej) � 0 for all i 6� j:

8. Schur's Lemma is crucial for the development of the theory of

CG-modules. This exercise indicates the extent to which results

like Schur's Lemma hold for RG-modules.

Let V and W be irreducible RG-modules.

(a) Prove that if W: V! W is an RG-homomorphism then either Wis an RG-isomorphism or vW � 0 for all v 2 V.

(b) Prove that if W: V! V is an RG-isomorphism and V remains

irreducible as a CG-module, then W � ë1V for some real num-

ber ë.

. (c) Give an example of a group G, an irreducible RG-module V

and an RG-homomorphism W: V! V which is not a multiple

of 1V .

9. Let G be a group with a subgroup H of index n. Let Ù be the set

of n right cosets Hx of H in G. For g 2 G, de®ne a function

r g :Ù! Ù by (Hx)r g � Hxg for all x 2 G.

Prove that r g is a permutation of Ù, and that the function r : g! r g

is a homomorphism from G to the symmetric group on Ù.

Show that the kernel of r isT

x2Gxÿ1Hx.

Deduce that if a group G has a subgroup H of index n, then

there is a homomorphism G! Sn with kernel contained in H.

10. Suppose that G is a ®nite group containing and involution t with

CG(t) � C2. Prove that |G : G9| � 2. Deduce that if G is simple,

then G � C2.


283

24

Summary of properties of character tables

In this short chapter we present no new results, but instead we gather

together from previous chapters various properties which are helpful

when we try to ®nd the character table of a particular group. In the

next four chapters we shall calculate several character tables in detail.

Usually we begin by working out the conjugacy classes and centrali-

zer orders of our given ®nite group G. The size of the character table

is determined by the number k of conjugacy classes of G; the character

table is then a k 3 k matrix, with columns indexed by the conjugacy

classes of G (the ®rst column corresponding to the conjugacy class

{1}), and with rows indexed by the irreducible characters of G.

When doing calculations, we commonly come across a new character

÷, which may or may not be irreducible. We can then calculate h÷, ÷i,which is given by

h÷, ÷i � 1

jGjXg2G

÷(g)÷(g):

The character ÷ is irreducible if and only if h÷, ÷i � 1 (see Theorem

14.20). If ÷ is reducible then we calculate h÷, ÷ii for each of the

irreducible characters ÷i which we already know, and then

÷ÿX

i

h÷, ÷ii÷i

will also be a character. We can thus determine whether ÷ is a linear

combination of the irreducible characters we already know; and if it is

not, then we can obtain from ÷ a linear combination of irreducible

characters, all of which are new.

We have developed a number of methods for producing characters ÷

on which to perform such calculations. For example, every subgroup of

Sn has a permutation character (see (13.22)); the product of two

characters is again a character (Proposition 19.6); given a character øwe can form the symmetric and antisymmetric parts of its square, øS

and øA (see Proposition 19.14); and if H is a subgroup of G then we

can restrict characters of G to H, and induce characters of H to G.

These and other properties of characters are summarized in the follow-

ing list.

Properties of characters

Assume that ÷1, . . . , ÷k are the irreducible characters of G.

(1) (Example 13.8(3)) There is a (trivial) character ÷ of G which is

given by

÷(g) � 1 for all g 2 G:

(2) (Theorem 17.11) The group G has precisely jG=G9j linear char-

acters. These are the characters ÷ given by

÷(g) � ø(gG9) (g 2 G)

as ø varies over the irreducible (linear) characters of G=G9.

(3) (Theorem 17.3) As a generalization of (2), if N v G and ø is an

irreducible character of G=N , then we get an irreducible character

÷ of G which is given by

÷(g) � ø(gN ) (g 2 G)

(÷ is the lift of ø). This method gives precisely those irreducible

characters of G which have N contained in their kernel.

(4) (Theorem 19.18) If G � G1 3 G2 then all the irreducible charac-

ters ÷ of G are given by

÷(g1, g2) � ö1(g1)ö2(g2) (g1 2 G1, g2 2 G2),

as öi varies over the irreducible characters of Gi (i � 1, 2).

(5) (Proposition 13.24) If G is a subgroup of Sn, then the function

í: G! C de®ned by

í(g) � jfix (g)j ÿ 1 (g 2 G)



(6) (Theorems 11.12 and 22.11) The entries ÷i(1) (1 < i < k) in the

®rst column of the character table of G are positive integers, and

satisfy

Xk

i�1

÷i(1)2 � jGj:

Moreover, each integer ÷i(1) divides |G|.

(7) (Row orthogonality relations, Theorem 16.4(1)) For all i, j, we

have h÷i, ÷ ji � äij.

(8) (Column orthogonality relations, Theorem 16.4(2)) For all g,

h 2 G, we have

Xk

i�1

÷i(g)÷i(h) �jCG(g)j, if g and h are conjugate,

0, otherwise:

(

(9) (Exercise 13.5) If ÷ is an irreducible character of G and z 2 Z(G),

then there exists a root of unity å such that for all g 2 G,

÷(zg) � å÷(g):

(10) (Proposition 13.9(2)) If g is an element of order n in G, and ÷ is

a character of G, then ÷(g) is a sum of nth roots of unity.

Moreover, |÷(g)| < ÷(1).

(11) (Proposition 13.9(3, 4)) If g 2 G and ÷ is a character of G, then

÷(gÿ1) � ÷(g):

In particular, if g is conjugate to gÿ1 then ÷(g) is real for all

characters ÷ of G.

(12) (Corollary 15.6) If g 2 G and g is not conjugate to gÿ1, then

÷(g) is non-real for some character ÷ of G.

(13) (Theorem 22.16) Let g 2 G. If g is conjugate to gi for all positive

integers i which are coprime to the order of g, then ÷(g) is an

integer for all characters ÷ of G.

(14) (Corollary 22.26) Suppose that p is a prime number and that y is

the p9-part of the element g of G. If ÷ is a character of G such

that ÷(g) and ÷(y) are both integers, then

÷(g) � ÷(y) mod p:

Summary of properties of character tables 285

In particular, if the order of g is a power of p, then

÷(g) � ÷(1) mod p:

(15) (Proposition 13.15) If ÷ is an irreducible character of G, then so

is ÷, where

÷(g) � ÷(g) (g 2 G):

(16) (Theorem 23.1) The number of real irreducible characters of G is

equal to the number of real conjugacy classes of G.

(17) (Proposition 17.14) If ÷ is an irreducible character of G and ë is

a linear character of G, then ÷ë is an irreducible character of G,

where

÷ë(g) � ÷(g)ë(g) (g 2 G):

(18) (Proposition 19.6) If ÷ and ø are characters of G, then so is the

product ÷ø, where

÷ø(g) � ÷(g)ø(g) (g 2 G):

(19) (Proposition 19.14) If ÷ is a character of G, then so are ÷S and

÷A, where for all g 2 G,

÷S(g) � 12(÷2(g)� ÷(g2)),

÷A(g) � 12(÷2(g)ÿ ÷(g2)):

(20) (De®nition 21.13, Proposition 21.23) If H is a subgroup of G and

ø is a character of H, then ø " G is a character of G, with

values given by Proposition 21.23.

(21) (Chapter 20) If H is a subgroup of G and ø is a character of G,

then ø # H is a character of H, where

(ø # H)(h) � ø(h) (h 2 H):

We have seen that the character table of a group G gives group-

theoretic information about G. For example, the ®rst column determines

|G| and jG=G9j (by (6) and (2)). We can see from the character table

whether or not G is simple (Proposition 17.6); indeed, we can ®nd all

the normal subgroups of G (Proposition 17.5). Two important normal

subgroups are G9 and Z(G); these can be determined in the following

ways. The derived subgroup G9 consists of those elements g in G

which satisfy ÷(g) � 1 for all linear characters ÷ of G. The centre

Z(G) can be found by noting which elements g of G satisfy


P÷(g)÷(g) � |G|, the sum being over all irreducible characters ÷ of G.

In Chapter 30 we shall see some more impressive results about

subgroups of G, which can be deduced from the character table.

As a ®nal remark, it is of course true that isomorphic groups have

the same character table; however, the converse is false: in Exercise

17.1 we gave examples of non-isomorphic groups, D8 and Q8, with the

same character table.

Summary of properties of character tables 287

288

25

Characters of groups of order pq

By the end of the next chapter, we shall have determined the character

tables of all groups of order less than 32. A number of these groups

are so-called Frobenius groups, and in this chapter we shall describe a

class of Frobenius groups and ®nd the character tables of the groups in

this class. In particular, this will give the character tables of all groups

whose order is the product of two prime numbers.

Throughout the chapter, p will denote a prime number.

Primitive roots modulo p

Recall that the set

Z p � f0, 1, : : : , pÿ 1g,with addition and multiplication modulo p, is a ®eld; that is, Z p is an

abelian group under addition, and Z�p � Z p ÿ f0g is an abelian group

under multiplication. Clearly Z p is a cyclic group under addition,

generated by 1. It is also true, but not at all obvious, that Z�p is cyclic:

25.1 Theorem

The multiplicative group Z�p is cyclic; that is, there exists an integer n

such that

n pÿ1 � 1 mod p, and

nr 6� 1 mod p for 0 , r , pÿ 1:

An integer n of order p ÿ 1 in Z�p is called a primitive root modulo

p. We shall not provide a proof of Theorem 25.1, but for a good

account, we refer you to Theorem 45.3 of the book by J. B. Fraleigh

listed in the Bibliography.

25.2 Example

The number 2 is a primitive root modulo 3, 5, 11 and 13, but not

modulo 7; and 3 is a primitive root modulo 7.

As an immediate consequence of Theorem 25.1 we have

25.3 Proposition

If q| p ÿ 1 then there is an integer u such that u has order q modulo p

± that is, such that

uq � 1 mod p, and

ur 6� 1 mod p for 0 , r , q:

Frobenius groups of order pq, where q| p 2 1

25.4 Example

De®ne

G � 1 y

0 x

� �: x 2 Z�p, y 2 Z p

� �:

Under matrix multiplication, G is a group of order p( p ÿ 1) (see

Exercise 25.1).

Now let q| p ÿ 1, and let u be an element of order q in the

multiplicative group Z�p. De®ne

A � 1 1

0 1

� �, B � 1 0

0 u

� �,

and let F � hA, Bi, the subgroup of G generated by A and B. Then

Bÿ1 AB � 1 u

0 1

� �� Au,

and so we have the relations

Ap � Bq � I , Bÿ1 AB � Au:(25:5)

Using these relations, we see that every element of F is of the form

Ai Bj with 0 < i < p ÿ 1, 0 < j < q ÿ 1. These pq elements are dis-

Characters of groups of order pq 289

tinct, so jFj � pq. Moreover the relations (25.5) determine all products

in F, so we have the presentation

F � hA, B: Ap � Bq � I , Bÿ1 AB � Aui:

25.6 De®nition

If p is a prime and q| p ÿ 1, then we write F p,q for the group of order

pq with presentation

F p,q � ha, b: a p � bq � 1, bÿ1ab � aui,where u is an element of order q in Z�p.

It is not hard to show that, up to isomorphism, F p,q does not depend

on which integer u of order q we choose (see Exercise 25.3).

The groups F p,q belong to a wider class of groups known as

Frobenius groups. We shall not give the general de®nition of these

here, as we shall only be dealing with F p,q; the interested reader can

®nd more information in the book by D. S. Passman listed in the

Bibliography.

The next result classi®es all groups whose order is the product of

two distinct prime numbers.

25.7 Proposition

Suppose that G is a group of order pq, where p and q are prime

numbers with p . q. Then either G is abelian, or q divides p ÿ 1 and

G � F p,q.

Proof Assume that G is non-abelian. It follows from Exercise 22.3

that q divides p ÿ 1 and G has a normal subgroup H of order p.

(Alternatively, these facts follow readily from Sylow's Theorems (see

Chapter 18 of the book by J. B. Fraleigh listed in the Bibliography).)

Both H and G=H are cyclic, since they have prime order. Suppose

that H � kal and G=H � hHbi; then G is generated by a and b. Since

bq 2 H but b does not have order pq (as G is non-abelian), it follows

that b has order q.

Now H v G, so bÿ1ab � au for some integer u. Further,

a � bÿqabq � au q

and so uq � 1 mod p. Thus the order of u in the group Z�p divides q.


If the order of u were 1 then we would have bÿ1ab � a, and G would

be abelian. Therefore the order of u is q. We have now established that

a p � bq � 1, bÿ1ab � au, order of u in Z�p is q:

Hence G � F p,q. j

25.8 Example

By Proposition 25.7, every group of order 15 is abelian (indeed,

isomorphic to C3 3 C5); and the groups of order 21 are C3 3 C7 and

F7,3.

The character table of Fp,q

We have, in fact, already found the character tables of certain of the

groups F p,q: the dihedral group of order 2 p is the case where q � 2,

and in Example 21.25 we dealt with F7,3. We now construct the

character table of F p,q in general. Thus let

G � Fp,q � ha, b: ap � bq � 1, bÿ1ab � auiwhere p is prime, q| p ÿ 1 (q not necessarily prime), and u has order q

modulo p.

Let S be the subgroup of Z�p consisting of the powers of u. Thus

jSj � q. Write r � ( pÿ 1)=q, and choose coset representatives

v1, : : : , vr for S in Z�p.

25.9 Proposition

The conjugacy classes of G � F p,q are

f1g,(av i )G � favi s: s 2 Sg (1 < i < r),

(bn)G � fambn: 0 < m < pÿ 1g (1 < n < qÿ 1):

Proof The equation

bÿ javb j � avu j

shows that av is conjugate to avs for all s 2 S. Therefore the conjugacy

class of avi has size at least q; also the size of this conjugacy class is

equal to |G: CG(avi )|, and since kal < CG(avi ), this size is at most q.

Hence (avi )G has size q, and has the form stated in the proposition.


Since CG(bn) contains kbl, and kbl has index p in G, it follows that

for n 6� 0 mod q, we have |CG(bn)| � q, and so the conjugacy class of

bn has size p. On the other hand, as G=hai is abelian, every conjugate

of bn has the form ambn for some m. Hence

(bn)G � fambn: 0 < m < pÿ 1gand the proof is complete. j

By Proposition 25.9, G has q � r conjugacy classes, so we seek

q � r irreducible characters.

First, observe that the derived subgroup G9 � kal, so G=G9 has order

q and therefore by Theorem 17.11, G has precisely q linear characters.

These are given by ÷n (0 < n < q ÿ 1), where

÷n(axb y) � e2ðiny=q (0 < x < pÿ 1, 0 < y < qÿ 1):

We shall show that G has r irreducible characters of degree q.

Let å � e2ði= p. For v 2 Z�p, denote by øv the character of kal which

is given by

øv(ax) � åvx (0 < x < pÿ 1):

We calculate the values of the induced character øv " G, using Proposi-

tion 21.23. We obtain

(øv " G)(axby) � 0 if 1 < y < qÿ 1, and

(øv " G)(ax) �Xs2S

åvsx (0 < x < pÿ 1):

Note that øv " G has degree q, and

øv " G � øvs " G if s 2 S:

For each coset representative v j (1 < j < r) of S in Z�p, let

ö j � øv j" G:

We now prove that each ö j is irreducible. By the Frobenius Recipro-

city Theorem 21.16, for all s 2 S,

hö j # hai, øv j sihai � hö j, øv j s " GiG � hö j, ö jiG:Hence

ö j # hai � hö j, ö jiGXs2S

øv j s � ÷,


where ÷ is either 0 or a character of kal. Taking degrees, it follows

that

ö j(1) > jSjhö j, ö jiG:Since ö j(1) � q � jSj, we deduce that kö j, ö jlG � 1. This proves that

ö j is irreducible, and also that

ö j # hai � hö j, ö jiGXs2S

øv j s:

By Theorem 14.23, the characters øv (v 2 Z�p) are linearly indepen-

dent, and hence ö1 # kal, . . . , ör # kal are distinct. Consequently the

irreducible characters ö1, . . . , ör are distinct.

We have now found q � r distinct irreducible characters ÷n, ö j of G

(0 < n < q ÿ 1, 1 < j < r), so we have the complete character table of

G. We summarize in the following theorem.

25.10 Theorem

Let p be a prime number, q| p ÿ 1 and r � ( pÿ 1)=q. Then the group

F p,q � ha, b: a p � bq � 1, bÿ1ab � aui� faxby: 0 < x < pÿ 1, 0 < y < qÿ 1g

has q � r irreducible characters. Of these, q have degree 1 and are

given by

÷n(axby) � e2ðiny=q (0 < n < qÿ 1)

and r have degree q and are given by

ö j(axby) � 0 if 1 < y < qÿ 1,

ö j(ax) �

Xs2S

e2ðiv j sx= p,

for 1 < j < r, where v1S, : : : , vrS are the cosets in Z�p of the

subgroup S generated by u.

We conclude by illustrating Theorem 25.10 in some examples.

25.11 Example

Let

G � F p, pÿ1 � ha, b: a p � b pÿ1 � 1, bÿ1ab � aui


where u is a primitive root modulo p. Then G has p ÿ 1 linear

characters, and one irreducible character ö of degree p ÿ 1, with

values given by

ö(axby) � 0 if 1 < y < pÿ 2,

ö(ax) � ÿ1 if 1 < x < pÿ 1:

25.12 Example

Let a, b 2 S5 be the permutations

a � (1 2 3 4 5), b � (2 3 5 4):

Check that

a5 � b4 � 1, bÿ1ab � a2:

Hence if G � ka, bl, then G � F5,4, and so by the previous example

the character table of G is as shown.

25.13 Example

We consider the case p � 13, q � 4. Here

F13,4 � ha, b: a13 � b4 � 1, bÿ1ab � a5i:Write å � e2ði=13, and let

á � å� å5 � å8 � å12,

â � å2 � å3 � å10 � å11,

ã � å4 � å6 � å7 � å9:

By Theorem 25.10, the character table of F13,4 is as shown opposite.

In Example 21.25 we found the character table of F7,3. You may like

Character table of F5,4

gi 1 a b b2 b3

|CG(gi)| 20 5 4 4 4

÷0 1 1 1 1 1÷1 1 1 i ÿ1 ÿi÷2 1 1 ÿ1 1 ÿ1÷3 1 1 ÿi ÿ1 iö 4 ÿ1 0 0 0


to check that this agrees with the description of the character table

provided by Theorem 25.10.


1. Suppose that p is prime and q divides p ÿ 1. Let u be an element

of order q in Z�p. Then

Fp,q � ha, b: ap � bq � 1, bÿ1ab � aui:

The irreducible characters of Fp,q are described in Theorem 25.10.

2. Let p and q be prime numbers with p . q. If G has order pq, then

either G is abelian or G � F p,q.


1. Let p be a prime number. Prove that

1 y

0 x

� �: x 2 Z�p, y 2 Z p

� �,

under matrix multiplication, is a group of order p( p ÿ 1).

2. Write down the character table of the non-abelian group F11,5 of

order 55.

3. Let p and q be positive integers, with p prime and q| p ÿ 1. Let u

and v be two integers which are of order q modulo p, and de®ne

Character table of F13,4

gi 1 a a2 a4 b b2 b3

|CG(gi)| 52 13 13 13 4 4 4

÷0 1 1 1 1 1 1 1÷1 1 1 1 1 i ÿ1 ÿi÷2 1 1 1 1 ÿ1 1 ÿ1÷3 1 1 1 1 ÿi ÿ1 iö1 4 á â ã 0 0 0ö2 4 â ã á 0 0 0ö3 4 ã á â 0 0 0


G1 � ha, b: ap � bq � 1, bÿ1ab � aui,G2 � ha, b: ap � bq � 1, bÿ1ab � avi:

Prove that G1 � G2.

(This justi®es the comment which follows the de®nition of Fp,q in

25.6.)

4. Suppose that p is a prime number, with p 6� 2. Let q � ( p ÿ 1)=2

and let

G � Fp,q � ha, b: ap � bq � 1, bÿ1ab � aui,where u is an element of order q modulo p.

(a) Show that there exists an integer m such that um � ÿ1 mod p if

and only if p � 1 mod 4.

(b) Deduce that a is conjugate to aÿ1 if and only if p � 1 mod 4.

(c) Using the orthogonality relations, show that the two irreducible

characters ö1, ö2 of G of degree q have values

12(ÿ1�p(äp))

on the element a, where ä � 1 if p � 1 mod 4, and ä � ÿ1 if

p � ÿ1 mod 4.

(d) Deduce that if å � e2ði= p thenXs2Q

ås � (ÿ1�p(äp)),

where Q is the set of quadratic residues modulo p (that is,

Q � f12, 22, : : : , (( pÿ 1)=2)2g).5. Let E be the group of order 18 which is given by

E � ha, b, c: a3 � b3 � c2 � 1, ab � ba, cÿ1ac � aÿ1,

cÿ1bc � bÿ1i,as in Exercise 5.4. Note that ka, bl is a normal subgroup of E

which is isomorphic to C3 3 C3. By inducing linear characters of

this subgroup, obtain the character table of E.

6. Show that the group E of Exercise 5 has the properties that Z(E) is

cyclic, but E has no faithful irreducible representation. (Thus, E

provides a counterexample to the converse of Proposition 9.16.)

7. (a) Find a group whose irreducible character degrees are


1, 1, 1, 3, 3, 3, 3:

(b) Find a group whose irreducible character degrees are

1, 1, 1, 1, 1, 1, 3, 3, 3, 3, 3, 3, 3, 3:

(c) Find a group whose irreducible character degrees are

1, 1, 1, 1, 1, 1, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 6, 6, 6, 6:

8. Let G be the group of order 54 which is given by

G � ha, b: a9 � b6 � 1, bÿ1ab � a2i:Find the character table of G.


298

26

Characters of some p-groups

Throughout this chapter, p will be a prime number. We shall show

how to obtain the character tables of all groups of order pn for n < 4.

The method consists of examining the characters of those p-groups

which contain an abelian subgroup of index p, and before explaining

the method, we show that all groups of order pn with 1 < n < 4 do,

indeed, have an abelian subgroup of index p. We later give explicitly

the irreducible characters of all groups of order p3 and of all groups

of order 16. At the end of the chapter we point out, with references,

that we have found the character tables of all groups of order less

than 32.

Elementary properties of p-groups

A p-group is a group whose order is a power of the prime number p.

In the ®rst lemma we collect several well known properties of p-

groups. Recall that Z(G) denotes the centre of G (see De®nition 9.15).

26.1 Lemma

Let G be a group of order pn with n > 1.

(1) If {1} 6� H v G then H \ Z(G) 6� {1}. In particular, Z(G) 6� {1}.

(2) If K < Z(G) and G=K is cyclic, then G is abelian.

(3) If n < 2 then G is abelian.

Proof (1) Since H v G, H is a union of conjugacy classes of G, all

of which have size a power of p; and H \ Z(G) consists of those

conjugacy classes in H which have size 1. Therefore

jH j � jH \ Z(G)j � (a multiple of p):

As |H| is a multiple of p and |H \ Z(G)| 6� 0, we deduce that

H \ Z(G) 6� {1}.

(2) Suppose that G=K is cyclic, generated by gK. Let x1, x2 2 G.

Then

x1 � gik1, x2 � gjk2

for some integers i, j and some k1, k2 2 K. Since k1, k2 2 Z(G), it

follows that x1x2 � x2x1. Therefore G is abelian.

(3) By (1), jG=Z(G)j < pnÿ1. Hence if n < 2 then G=Z(G) is

cyclic and so G is abelian by (2). j

26.2 Lemma

Let G be a group of order pn with 1 < n < 4. Then G contains an

abelian subgroup of index p.

Proof The result is immediate if n � 1, so suppose that 2 < n < 4.

Assume that Z(G) contains a subgroup K of order pnÿ2. Then we

can ®nd a subgroup H of G such that K < H and jH j � pnÿ1. As

K < Z(H) and, by Lemma 26.1(2), H=Z(H) is not of order p, we

deduce that Z(H) � H . Therefore H is an abelian subgroup of index

p in G.

Now assume that Z(G) has no subgroup of order pnÿ2. Since

Z(G) 6� f1g by Lemma 26.1(1), the only possibility is that |G| � p4

and |Z(G)| � p. Then by Exercise 12.7, G has an element x whose

conjugacy class xG is of size p. Let H � CG(x). Then by Theorem

12.8, jH j � jGj=jxGj � p3. Moreover, Z(G) and kxl are distinct non-

identity subgroups of Z(H), and so Z(H) > p2. Hence again Z(H) � H

by Lemma 26.1(2), and H is an abelian subgroup of index p. j

For our ®nal result on the structure of p-groups, recall that the

derived subgroup of G is denoted by G9 (see De®nition 17.7).

26.3 Lemma

Let G be a non-abelian p-group which contains an abelian subgroup H

of index p. Then there exists a normal subgroup K of G such that

K < H \ G9 \ Z(G) and jKj � p:

Characters of some p-groups 299

Proof Since G is non-abelian, we have {1} 6� G9 v G, and hence

G9 \ Z(G) 6� {1} by Lemma 26.1(1). Let K be a subgroup of order p

in G9 \ Z(G). Now K < Z(G) implies that K v G and that KH is an

abelian subgroup of G (where KH � {kh: k 2 K, h 2 H}). Since G is

non-abelian and |G:H| � p, we have KH � H, and therefore K < H. j

Characters of p-groups with an abelian subgroup of index p

In view of Lemma 26.2, the next theorem provides us with all the

irreducible characters of non-abelian groups of order p3 or p4.

26.4 Theorem

Assume that G is a non-abelian p-group which contains an abelian

subgroup H of index p. Let K be a normal subgroup of G as in Lemma

26.3. Then every irreducible character of G is given by either

(1) the lift of an irreducible character of G=K, or

(2) ø " G, for some linear character ø of H which satis®es

K 6< Kerø.

Proof Let |G| � pn. By Theorem 17.3, the irreducible characters of

G=K lift to give precisely those irreducible characters of G which have

K in their kernel. The sum of the squares of the degrees of the

irreducible characters obtained in this way is jG=Kj � pnÿ1, by Theo-

rem 11.12.

We shall construct pnÿ2 ÿ pnÿ3 further irreducible characters of G,

each of degree p. Since

pnÿ1 � ( pnÿ2 ÿ pnÿ3) p2 � pn � jGj,(�)we shall then have obtained all the irreducible characters of G, again

by Theorem 11.12.

First note that if ÷ is a character of G of degree p, then either ÷ is

irreducible or ÷ is a sum of linear characters (since by Theorem 22.11,

the degree of every irreducible character of G is a power of p). In the

latter case, we have G9 < Ker ÷, as G9 is in the kernel of every linear

character, and hence K < G9 < Ker ÷. This establishes

if ÷(1) � p and K 6< Ker ÷, then ÷ is irreducible:(26:5)

We know by Theorem 9.8 that all the pnÿ1 irreducible characters of

the abelian group H are linear. Let Ö denote the set of linear


characters of H which do not have K in their kernel. Since the linear

characters of H which do have K in their kernel are precisely the lifts

of linear characters of H=K, we have

jÖj � pnÿ1 ÿ pnÿ2:

Let ø 2 Ö. By Proposition 21.23, since K < Z(G),

(ø " G)(k) � pø(k) for all k 2 K:

Thus ø " G has degree p and does not have K in its kernel. Therefore

by (26.5), ø " G is an irreducible character of G.

Suppose now that ø1 is a linear character of H such that

ø " G � ø1 " G. Then by the Frobenius Reciprocity Theorem 21.16,

1 � hø " G, ø1 " GiG � h(ø " G) # H , ø1iH :

Since (ø " G) # H has degree p, this implies that there are at most p

elements ø1 of Ö such that ø1 " G � ø " G. It follows that

fø " G: ø 2 Öggives at least jÖj= p � ( pnÿ1 ÿ pnÿ2)=p distinct irreducible characters

of G of degree p which do not have K in their kernel. As we saw

in (�), G has at most pnÿ2 ÿ pnÿ3 such characters. Therefore

fø " G: ø 2 Ög consists precisely of the pnÿ2 ÿ pnÿ3 irreducible

characters we seek, and the proof is complete. j

We now use Theorem 26.4 to give an explicit construction of the

irreducible characters of the non-abelian groups of order p3. We shall

then illustrate Theorem 26.4 further by constructing the character tables

of all the non-abelian groups of order 16.

Groups of order p3

By Theorem 9.6, the abelian groups of order p3 are

C p3 , C p2 3 Cp and Cp 3 Cp 3 Cp:

The character tables of these groups are given by Theorem 9.8.

Now let G be a non-abelian group of order p3. Write Z � Z(G). By

Lemma 26.1, Z 6� {1} and G=Z is non-cyclic. Hence G=Z � Cp 3 Cp

and Z � kzl � Cp. Choose aZ, bZ such that G=Z � haZ, bZi. Then


G=Z � farbsZ: 0 < r < pÿ 1, 0 < s < pÿ 1gand in particular, every element of G is of the form

arbszt

for some r, s, t with 0 < r, s, t < p ÿ 1.

26.6 Theorem

Let G � {arbszt: 0 < r, s, t < p ÿ 1} be a non-abelian group of order

p3, as above. Write å � e2ði= p. Then the irreducible characters of G

are

÷u,v (0 < u < pÿ 1, 0 < v < pÿ 1), and

öu (1 < u < pÿ 1),

where for all r, s, t,

÷u,v(arbszt) � å ru�sv,

öu(arbszt) �påut, if r � s � 0,

0, otherwise:

(

Proof By Theorem 9.8, the irreducible characters of G=Z are øu,v

(0 < u, v < p ÿ 1), where

øu,v(arbsZ) � å ru�sv:

The lift to G of øu,v is the linear character ÷u,v which appears in the

statement of the theorem.

Let H � ka, zl, so that H is an abelian subgroup of order p2. For

1 < u < p ÿ 1, choose a character øu of H which satis®es

øu(zt) � åut (0 < t < pÿ 1):

We shall calculate øu " G.

Let r be an integer with 1 < r < p ÿ 1. If ar is conjugate to an

element g of G, then arZ is conjugate to gZ in the abelian group

G=Z, so arZ � gZ, and therefore g � arzt for some t. Since ar =2 Z,

the conjugacy class (ar)G does not have size 1, and hence

(ar)G � farzt: 0 < t < pÿ 1g:


Then by Proposition 21.23,

(øu " G)(arzt) � øu(ar)� øu(arz)� : : :� øu(arz pÿ1)

� øu(ar)Xpÿ1

s�0

øu(zs)

� øu(ar)Xpÿ1

s�0

åus

� 0:

Also,(øu " G)(zt) � pøu(zt) � påut, and

(øu " G)(g) � 0 if g =2 H :

We have now established that if öu � øu " G, then öu takes the values

stated in the theorem.

We ®nd that

höu, öuiG � 1

p3

Xg2G

öu(g)öu(g)

� 1

p3

Xg2Z

öu(g)öu(g)

� 1

p3

Xg2Z

p2

� 1:

Therefore öu is irreducible.

Clearly the irreducible characters ÷u,v (0 < u, v < p ÿ 1) and

öu (1 < u < p ÿ 1) are all distinct, and the sum of the squares of their

degrees is

p2 . 12 � ( pÿ 1) . p2 � jGj:Hence we have found all the irreducible characters of G. j

Notice that the calculation in the proof of Theorem 26.6 is a special

case of the proof of Theorem 26.4 (with K � Z(G)).

In fact, up to isomorphism, there are precisely two non-abelian

groups of order p3. If p � 2, they are D8 and Q8. And if p is odd,

they are


H1 � ha, b: a p2 � b p � 1, bÿ1ab � a p�1i, and(26:7)

H2 � ha, b, z: a p � b p � z p � 1, az � za, bz � zb,

bÿ1ab � azi:We have Z(H1) � ka pl, Z(H2) � kzl. The elements a, b in H1 and H2

will serve for the elements a, b chosen in the statement of Theorem

26.6.

26.8 The groups of order 16

It is known that, up to isomorphism, there are precisely fourteen groups

of order 16 (see p. 134 of the book by Coxeter and Moser listed in the

Bibliography). We shall describe all these groups and their character

tables.

By Theorem 9.6, the abelian groups of order 16 are

C16, C8 3 C2, C4 3 C4, C4 3 C2 3 C2 and C2 3 C2 3 C2 3 C2,

and their character tables are given by Theorem 9.8.

For each of the nine non-abelian groups G of order 16 it is the case

that |G9 \ Z(G)| � 2 (see Exercise 26.7), so the subgroup K described

in Lemma 26.3 is given by

K � G9 \ Z(G):

Now G=K is a group of order 8. It is not C8 by Lemma 26.1(2), and

it is not Q8 by Exercise 26.8. Hence

G=K � D8, C4 3 C2 or C2 3 C2 3 C2:

We shall divide our descriptions into three parts, according to these

three possibilities for G=K. Our descriptions will be in terms of

presentations; it is possible to see, using Exercise 26.5, that all the nine

groups G1 , . . . , G9 given below do indeed have order 16.

(A) There are three non-abelian groups G of order 16 with

G=K � D8. These are

G1 � ha, b: a8 � b2 � 1, bÿ1ab � aÿ1i � D16,

G2 � ha, b: a8 � 1, b2 � a4, bÿ1ab � aÿ1i,G3 � ha, b: a8 � b2 � 1, bÿ1ab � a3i:


In each case K � ka4l. Each of the groups has seven conjugacy classes

C1, . . . , C7, and these are given in the following table.

C1 C2 C3 C4 C5 C6 C7

G1, G2 1 a4 a2, a6 a, a7 a3, a5 aib (i even) aib (i odd)G3 1 a4 a2, a6 a, a3 a5, a7 aib (i even) aib (i odd)

Using Theorem 26.4, we obtain the character tables of G1, G2 and

G3:

Class C1 C2 C3 C4 C5 C6 C7

Centralizerorder 16 16 8 8 8 4 4

1 1 1 1 1 1 11 1 1 1 1 ÿ1 ÿ11 1 1 ÿ1 ÿ1 1 ÿ11 1 1 ÿ1 ÿ1 ÿ1 12 2 ÿ2 0 0 0 02 ÿ2 0 á â 0 02 ÿ2 0 â á 0 0

where

á � p2 � ÿâ for G1, G2,

á � ip

2 � ÿâ for G3:

The ®rst ®ve characters are the lifts of the irreducible characters of

G=K � D8. The last two characters can be obtained as in Theorem

26.4(2) as induced characters ø " G, where ø is a linear character of

the abelian subgroup kal of index 2; alternatively, they can be found

by using the column orthogonality relations. Note that a is conjugate

to aÿ1 in G1 and G2, but not in G3; hence the values in the columns

C4 and C5 are all real for G1 and G2, but not for G3 (see Corollary

15.6).

(B) There are three non-abelian groups G of order 16 with G=K �C4 3 C2 (where, as before, K � G9 \ Z(G), of order 2). These are

G4 � ha, b, z: a4 � z, b2 � z2 � 1, bÿ1ab � azi,G5 � ha, b, z: a4 � 1, b2 � z, z2 � 1, bÿ1ab � azi,G6 � ha, b, z: a4 � 1, b2 � z2 � 1, bÿ1ab � az, az � za, bz � zbi:


The given presentations are somewhat cumbersome (for example, since

a4 � z in G4, z is redundant), but they are in a form which allows us

to describe the conjugacy classes C1, . . . , C10 of all three groups G4,

G5, G6 simultaneously:

C1 C2 C3 C4 C5 C6 C7 C8 C9 C10

1 z a2 a2z a, az a3, a3z b, bz a2b, a2bz ab, abz a3b, a3bz

In each case, K � kzl. The character tables of G4, G5 and G6 can again

be found using Theorem 26.4:

Class C1 C2 C3 C4 C5 C6 C7 C8 C9 C10

Centralizer order 16 16 16 16 8 8 8 8 8 8

1 1 1 1 1 1 1 1 1 11 1 ÿ1 ÿ1 i ÿi 1 ÿ1 i ÿi1 1 1 1 ÿ1 ÿ1 1 1 ÿ1 ÿ11 1 ÿ1 ÿ1 ÿi i 1 ÿ1 ÿi i1 1 1 1 1 1 ÿ1 ÿ1 ÿ1 ÿ11 1 ÿ1 ÿ1 i ÿi ÿ1 1 ÿi i1 1 1 1 ÿ1 ÿ1 ÿ1 ÿ1 1 11 1 ÿ1 ÿ1 ÿi i ÿ1 1 i ÿi2 ÿ2 á â 0 0 0 0 0 02 ÿ2 â á 0 0 0 0 0 0

where

á � 2i � ÿâ for G4,

á � 2 � ÿâ for G5, G6:

(C) Finally, there are three non-abelian groups G of order 16 with

G=K � C2 3 C2 3 C2 (where K � G9 \ Z(G), of order 2). These are

G7 � ha, b, z: a4 � b2 � z2 � 1, bÿ1ab � aÿ1, az � za, bz � zbi� D8 3 C2,

G8 � ha, b, z: a4 � z2 � 1, a2 � b2, bÿ1ab � aÿ1, az � za, bz � zbi� Q8 3 C2,

G9 � ha, b, z: a2 � b2 � z4 � 1, bÿ1ab � az2, az � za, bz � zbi:


Each of these groups has ten conjugacy classes, which are given by

C1 C2 C3 C4 C5 C6 C7 C8 C9 C10

G7, G8 1 a2 z a2z a, a3 az, a3z b, a2b bz, a2bz ab, a3b abz, a3bzG9 1 z2 z z3 a, az2 az, az3 b, bz2 bz, bz3 ab, abz2 abz, abz3

We have

K � ha2i for G7, G8,

hz2i for G9,

(and the character tables of G7, G8 and G9, given by Theorem 26.4, are

as follows.

Class C1 C2 C3 C4 C5 C6 C7 C8 C9 C10

Centralizer order 16 16 16 16 8 8 8 8 8 8

1 1 1 1 1 1 1 1 1 11 1 1 1 1 1 ÿ1 ÿ1 ÿ1 ÿ11 1 1 1 ÿ1 ÿ1 1 1 ÿ1 ÿ11 1 ÿ1 ÿ1 1 ÿ1 1 ÿ1 1 ÿ11 1 1 1 ÿ1 ÿ1 ÿ1 ÿ1 1 11 1 ÿ1 ÿ1 1 ÿ1 ÿ1 1 ÿ1 11 1 ÿ1 ÿ1 ÿ1 1 1 ÿ1 ÿ1 11 1 ÿ1 ÿ1 ÿ1 1 ÿ1 1 1 ÿ12 ÿ2 á â 0 0 0 0 0 02 ÿ2 â á 0 0 0 0 0 0

where

á � 2 � ÿâ for G7, G8,

á � 2i � ÿâ for G9:

26.9 The groups of order less than 32

At this point, we have in fact found the character tables of all groups

of order 31 or less. Apart from abelian groups and dihedral groups,

whose character tables are given by Theorem 9.8 and Section 18.3, the

groups, with references for their character tables, are as follows:


|G| G Reference forcharacter table

8 Q8 Exercise 17.112 A4 Section 18.2

T12 Exercise 18.316 G1, . . . , G9 Section 26.818 D6 3 C3 Theorem 19.18

E Exercise 25.520 T20 Exercise 18.3

F5,4 Theorem 25.1021 F7,3 Theorem 25.1024 D12 3 C2, A4 3 C2, T12 3 C2 Theorem 19.18

D8 3 C3, Q8 3 C3, D6 3 C4 Theorem 19.18S4 Section 18.1

SL (2, 3) Exercise 27.2T24 Exercise 18.3U24 Exercise 18.4V24 Exercise 18.5

27 H1, H2 Theorem 26.628 T28 Exercise 18.330 D6 3 C5, D10 3 C3 Theorem 19.18


In this chapter, we gave the irreducible characters of various non-

abelian p-groups G, as follows.

1. Theorem 26.4: p-groups which contain an abelian subgroup of

index p.

2. Theorem 26.6: groups of order p3.

3. Section 26.8: groups of order 16.


1. Suppose that G is a group of order pn ( p prime, n > 2), with an

abelian subgroup H of index p. Show that for some integer m > 2,

G has pm linear characters and pnÿ2 ÿ pmÿ2 irreducible characters

of degree p.


2. Let H be the group of order 27 which is given by

H � ha, b, z: a3 � b3 � z3 � 1, az � za, bz � zb, bÿ1ab � azi(see (26.7)).

Find the conjugacy classes of H, and use Theorem 26.6 to write

down the character table of H.

3. Let G be the group of order 32 which is given by

G � ha, b: a16 � 1, b2 � a8, bÿ1ab � aÿ1i:Using Theorem 26.4, or otherwise, ®nd the character table of G.

4. Let A, B, C, D be the following 4 3 4 matrices:

A �

ÿ1 0 0 0

0 1 0 0

0 0 1 0

0 0 0 ÿ1

0BBBBB@

1CCCCCA, B �

0 0 i 0

0 0 0 ÿi

i 0 0 0

0 ÿi 0 0

0BBBBB@

1CCCCCA,

C �

0 i 0 0

i 0 0 0

0 0 0 ÿi

0 0 ÿi 0

0BBBBB@

1CCCCCA, D �

0 0 0 1

0 0 1 0

0 1 0 0

1 0 0 0

0BBBBB@

1CCCCCA,

and let G � kA, B, C, Dl. Write Z � ÿI.

(a) Prove that all pairs of generators commute modulo hZi, and

deduce that G9 � hZi.(b) Show that for all g in G, g2 2 hZi, and deduce that G is a 2-

group of order at most 32.

(c) Prove that the given representation of G of degree 4 is

irreducible. (Hint: use Corollary 9.3.)

(d) Show that |G| � 32, and ®nd all the irreducible representations

of G.

5. Let G1, . . . , G9 be the non-abelian groups of order 16 with pre-

sentations as given in the text.

(a) Find faithful irreducible representations of degree 2 for G1, G2,

G3, G4 and G9.

(b) Why do the remaining groups G5, G6, G7 and G8 have no

faithful irreducible representations?


(c) Check that the following give faithful representations of G5

and G6:

G5: a!0 1 0

1 0 0

0 0 i

0@ 1A, b!i 0 0

0 ÿi 0

0 0 1

0@ 1A,

z!ÿ1 0 0

0 ÿ1 0

0 0 1

0@ 1A;

G6: a!i 0 0

0 ÿi 0

0 0 i

0@ 1A, b!0 1 0

1 0 0

0 0 1

0@ 1A,

z!ÿ1 0 0

0 ÿ1 0

0 0 1

0@ 1A:(d) Find faithful representations of degree 3 for G7 � D8 3 C2 and

G8 � Q8 3 C2.

(Note: This exercise can be used to con®rm that the presentations

of G1, . . . , G9 given in the text do indeed give groups of order 16.)

6. Prove that no two of the groups G1, . . . , G9 are isomorphic.

7. Let G be a non-abelian group of order p4.

(a) Prove that |Z(G)| � p or p2, and that if |Z(G)| � p2 then G

has p3 � p2 ÿ p conjugacy classes.

(b) Prove that |G9| � p or p2, and that if |G9| � p2 then G has

2 p2 ÿ 1 conjugacy classes.

(c) Deduce that |G9 \ Z(G)| � p.

8. (a) Prove that if G is any group, then G=Z(G) 6� Q8.

(Hint: assume that G=Z � haZ, bZ: a4 2 Z, a2 � b2 mod Z,

bÿ1ab � aÿ1 mod Zi. Prove that a2 commutes with b, and hence

that a2 2 Z.)

(b) Deduce from the result of Exercise 7 that if G is a group of

order 16, then G=(G9 \ Z(G)) 6� Q8.


311

27

Character table of the simple groupof order 168

Recall that a simple group is a non-trivial group G such that the only

normal subgroups of G are f1g and G itself. We discussed brie¯y in

Chapter 1 the signi®cance of simple groups in the theory of ®nite

groups. Examples of simple groups which we have met so far are

cyclic groups of prime order, A5 and A6. In fact the group A5, of

order 60, is the smallest non-abelian simple group. The next smallest

is a certain group of order 168, and in this chapter we shall describe

this group and ®nd its character table. The group belongs to a whole

family of simple groups, and we begin with a description of this

family.

Special linear groups

Let p be a prime number, and recall that Z p is the ®eld which consists

of the numbers 0, . . . , p ÿ 1, with addition and multiplication modulo

p. Denote by SL (2, p) the set of all 2 3 2 matrices M with entries in

Z p such that det M � 1. Then SL (2, p) is a group under matrix

multiplication, and is called the 2-dimensional special linear group

over Z p.

To calculate the order of the group SL (2, p), we count the

matrices

a b

c d

� �(a, b, c, d 2 Z p, ad ÿ bc � 1):

If c � 0, then there are p( p ÿ 1) choices for a, b, d which make

ad ÿ bc � 1 (since a, b are arbitrary, except that a 6� 0; and d is

determined by a). And there are p2( p ÿ 1) choices for a, b, c, d

with c 6� 0, such that ad ÿ bc � 1 (since a, d may be chosen

arbitrarily, c is any non-zero element of Z p; and then b is deter-

mined). Therefore

jSL (2, p)j � p( pÿ 1)� p2( pÿ 1)

� p( p2 ÿ 1):

If p � 2 then SL (2, p) has order 6, and it is easy to see that this

group is isomorphic to S3; so assume that p is an odd prime. By

Exercise 27.1, the centre of SL (2, p) is

Z � fI , ÿIg(where I is the 2 3 2 identity matrix). The factor group SL (2, p)=Z is

called the 2-dimensional projective special linear group, and is written

as PSL (2, p). Thus

PSL (2, p) � SL (2, p)=f�Ig:Since |SL (2, p)| � p( p2 ÿ 1), we have

jPSL(2, p)j � p( p2 ÿ 1)=2:

It is known that PSL (2, 3) � A4, PSL (2, 5) � A5, and that for p > 5,

the group PSL (2, p) is simple (see Theorem 8.19 of the book by J. J.

Rotman listed in the Bibliography).

The simple group G � PSL (2, 7) has order 168, and we shall

construct the character table of this group. After ®nding the conjugacy

classes of G, we shall ®nd the character table using only numerical

calculations, notably the orthogonality relations (Theorem 16.4) and

congruence properties (Corollary 22.26). The power of these techniques

is therefore well illustrated. In the exercises, we indicate other ways of

obtaining characters of G, using information about subgroups.

The conjugacy classes of PSL (2, 7)

27.1 Lemma

The group PSL (2, 7) has exactly six conjugacy classes. The following

table records representatives gi (1 < i < 6) for the conjugacy classes,

together with the order of gi, the order of CG(gi), and the size of the

conjugacy class containing gi.


Order of gi |CG(gi)| |gGi |

g1 � 1 0

0 1

� �Z 1 168 1

g2 � 0 1

ÿ1 0

� �Z 2 8 21

g3 � 2 ÿ2

2 2

� �Z 4 4 42

g4 � 2 0

0 4

� �Z 3 3 56

g5 � 1 1

0 1

� �Z 7 7 24

g6 � 1 ÿ1

0 1

� �Z 7 7 24

Proof For each i, we verify that gi has the stated order, and then use

direct calculation to ®nd all the elements of G which commute with

gi. Consider, for example, g4. Suppose that

a b

c d

� �Z

commutes with g4. Then

a b

c d

� �2 0

0 4

� �� 2 0

0 4

� �a b

c d

� �,

and hence b � c � 0. Consequently

CG(g4) � 1 0

0 1

� �Z,

2 0

0 4

� �Z,

4 0

0 2

� �Z

� �:

Similarly

CG(g2) � MZ: M � 1 0

0 1

� �,

0 ÿ1

1 0

� �,

3 2

2 4

� �,

3 ÿ2

ÿ2 4

� �,

�2 3

3 ÿ2

� �,

2 4

4 ÿ2

� �,

2 2

ÿ2 2

� �,

2 ÿ2

2 2

� ��:

Also, CG(gi) � kgil for i � 3, 5, 6.

Among g1, . . . , g6, the only elements with the same order are g5

Character table of the simple group of order 168 313

and g6; so no two of these six elements are conjugate, except possibly

g5 and g6. Suppose that gÿ1 g6 g � g5 with

g � a b

c d

� �Z 2 G:

Then gg5 � g6 g, and so

a a� b

c c� d

� �� aÿ c bÿ d

c d

� �with ad ÿ bc � 1:

It follows that c � 0, a 6� 0, d � aÿ1 and

a a� b

0 aÿ1

� �� a bÿ aÿ1

0 aÿ1

� �:

Therefore a2 � ÿ1, which is impossible for a 2 Z7. Thus g5 is not

conjugate to g6, and we have established that no two of the elements

g1, . . . , g6 are conjugate.

The size of the conjugacy class gGi is obtained by dividing 168 by

|CG(gi)| (Theorem 12.8). Since the sum of the sizes of the six

conjugacy classes gGi (1 < i < 6) is 168, these exhaust the conjugacy

classes of G. j

Notice that using Lemma 27.1, it is easy to check that G is indeed

simple, since any normal subgroup is a union of conjugacy classes (see

Proposition 12.19).

27.2 Corollary

(1) If 1 < i < 4 and ÷ is a character of G, then ÷(gi) is an integer.

(2) For some character ÷ of G, ÷(g5) is non-real.

Proof (1) By Lemma 27.1, for 1 < i < 4, gi is conjugate to (gi)k

whenever gi and (gi)k have the same order. Hence the conclusion

follows from Theorem 22.16.

(2) Notice that g6 � gÿ15 , so g5 is not conjugate to its inverse.

Therefore (2) follows from Corollary 15.6. j

The character table of G PSL (2, 7)

Since G has six conjugacy classes, it also has six irreducible charac-

ters. Let ÷1, . . . , ÷6 be the irreducible characters of G, where ÷1 is the


trivial character (so that ÷1(g) � 1 for all g 2 G). Recall that the

character table is the 6 3 6 matrix with ij-entry ÷i(gj).

We shall repeatedly exploit the column orthogonality relations,

Theorem 16.4(2), and the congruence properties given by Corollaries

22.26 and 22.27 for the elements g2, g3, g4, for which the character

values are known to be integers, by Corollary 27.2.

The entries in the column of g4 are integers, and the sum of the

squares of these integers is equal to |CG(g4)| � 3. The entries must

therefore be 1, �1, �1, 0, 0, 0 in some order. (We know that the entry

in the ®rst row is ÷1(g4) � 1.)

Similarly the entries in the column of g3 are 1, �1, �1, �1, 0, 0 in

some order, and the entries in column g2 are 1, �1, �1, �1, �2, 0 in

some order. Now for all characters ÷ of G, we have by Corollary

22.27,

÷(g2) � ÷(1) mod 2, and

÷(g3) � ÷(1) mod 2,

and so

÷(g2) � ÷(g3) mod 2:

Since we also know that X6

i�1

÷i(g3)÷i(g4) � 0,

we see that, with a suitable ordering of ÷2, . . . , ÷6, part of the

character table of G is as follows:

We shall determine the signs later. For the moment we concentrate on

the entries in the ®rst column of the character table (i.e. the degrees

÷i(1)). Let di � ÷i(1), so di is the entry on row i of column 1. By

Class representative g2 g3 g4

Centralizer order 8 4 3

÷1 1 1 1÷2 �1 �1 �1÷3 0 0 �1÷4 �1 �1 0÷5 �1 �1 0÷6 �2 0 0


Corollary 22.27, Theorem 22.11 and the fact thatP6

i�1 d2i � 168, we

have

d4 � 0 mod 3,

d4 � 1 mod 2,

d4 divides jGj � 168, and

d24 < 168:

The only positive integer d4 which satis®es these conditions is d4 � 3.

In the same way, d5 � 3.

Next,

d6 � 0 mod 3,

d6 � 0 mod 2,

d6 divides 168, and

d26 < 168:

Therefore d6 is 6 or 12. But

0 �X6

i�1

÷i(g2)di � 1� d2 � 3� 3� 2d6,

so as d22 < 168, we have d6 6� 12, and hence d6 � 6.

Now

1� d22 � d2

3 � 32 � 32 � 62 � 168,

so d22 � d2

3 � 113. The only solutions to this equation with d2, d3

positive integers have d2, d3 equal to 7, 8 in some order. Since

d2 � 1 mod 2, we have d2 � 7 and d3 � 8.

We have now found the ®rst column of the character table, and have

the following portion:

Class representative g1 g2 g3 g4

Centralizer order 168 8 4 3

÷1 1 1 1 1÷2 7 �1 �1 �1÷3 8 0 0 �1÷4 3 �1 �1 0÷5 3 �1 �1 0÷6 6 �2 0 0


The equations

X6

i�1

÷i(g1)÷i(gj) � 0 for j � 2, 3, 4

now enable us to determine the signs in the columns for g2, g3, g4.

We obtain

Class representative g1 g2 g3 g4

Centralizer order 168 8 4 3

÷1 1 1 1 1÷2 7 ÿ1 ÿ1 1÷3 8 0 0 ÿ1÷4 3 ÿ1 1 0÷5 3 ÿ1 1 0÷6 6 2 0 0

Next, the equation

1 � h÷2, ÷2i �X6

i�1

÷2(gi)÷2(gi)

jCG(gi)j

� 7:7

168� 1

8� 1

4� 1

3� ÷2(g5)÷2(g5)

7� ÷2(g6)÷2(g6)

7

gives ÷2(g5) � ÷2(g6) � 0. (Note that ÷2(g5) � ÷2(1) mod 7, but we

could not use this fact as we were not sure that ÷2(g5) was an integer.)

Also, for j � 5, 6,

0 �X6

i�1

÷i(g4)÷i(gj) � 1ÿ ÷3(gj)

and so ÷3(g5) � ÷3(g6) � 1.

By Corollary 27.2, there is an irreducible character ÷ of G such that

÷(g5) is non-real. For this character ÷, the complex conjugate ÷ will be

a different character of the same degree. Hence ÷4 and ÷5 (being the

only two irreducible characters with the same degree) must be complex

conjugates of each other.

Let ÷4(g5) � ÷5(g5) � z, and let ÷6(g5) � t. Thus the column for g5

is


Now

0 �X6

i�1

÷i(g2)÷i(g5) � 1ÿ zÿ z� 2t,

0 �X6

i�1

÷i(g3)÷i(g5) � 1� z� z,

7 �X6

i�1

÷i(g5)÷i(g5) � 2� 2zz� t t:

Solving these equations, we obtain

t � ÿ1, z � (ÿ1� ip

7)=2:

Since g6 � gÿ15 , we have ÷(g6) � ÷(g5) for all characters ÷ of G.

We have now completely determined the character table of

G � PSL (2, 7), as shown.

It is known that there are precisely ®ve non-abelian simple groups of

order less than 1000. We give you the character tables of all of these,

Class representative g5

Centralizer order 7

÷1 1÷2 0÷3 1÷4 z÷5 z÷6 t

Character table of PSL (2, 7)

Class representative g1 g2 g3 g4 g5 g6

Centralizer order 168 8 4 3 7 7

÷1 1 1 1 1 1 1÷2 7 ÿ1 ÿ1 1 0 0÷3 8 0 0 ÿ1 1 1÷4 3 ÿ1 1 0 á á÷5 3 ÿ1 1 0 á á÷6 6 2 0 0 ÿ1 ÿ1

where á � (ÿ1� ip

7)=2.


as follows:

G Order of G Reference for character table

A5 60 Example 20.13PSL (2, 7) 168 This chapter

A6 360 Exercise 20.2PSL (2, 8) 504 Exercise 28.3PSL (2, 11) 660 Exercise 27.6


1. SL (2, p) � a b

c d

� �: a, b, c, d 2 Z p, ad ÿ bc � 1

� �.

jSL (2, p)j � p( p2 ÿ 1).

2. PSL (2, p) � SL (2, p)=f�Ig.jPSL (2, p)j � p( p2 ÿ 1)=2 ( p odd).

3. We constructed the character table of PSL (2, 7), the simple group

of order 168.


1. Prove that Z(SL (2, p)) � f�Ig.2. Find the character table of SL (2, 3).

3. Deduce directly from the character table of PSL (2, 7) that this

group is simple.

4. In this exercise we present an alternative construction of the

character table of G � PSL (2, 7), given the conjugacy classes of G,

as in Lemma 27.1.

(a) De®ne the subgroup T of G, of order 21, as follows:

T � a b

0 aÿ1

� �Z : a 2 Z�7, b 2 Z7

� �(where Z � {�I}). Calculate the values of the induced char-

acter (1T ) " G, and show that

(1T ) " G � 1G � ÷,

where ÷ is an irreducible character of G.


(b) Let ë be a non-trivial linear character of T. Calculate the values

of ë " G and prove that this is an irreducible character

of G.

(c) By considering ÷S (see Proposition 19.14), obtain an irreducible

character of G of degree 6.

(d) From (a), (b), (c), we now have irreducible characters of G of

degrees 1, 7, 8 and 6. Use orthogonality relations to complete

the character table of G.

5. The character table of SL (2, 7).

Let G � SL (2, 7), the group of all 2 3 2 matrices of determinant

1, with entries in the ®eld Z7.

(a) Show that G has 11 conjugacy classes with representatives gi as

follows:

gi Order of gi |CG(gi)| | gGi |

g1 � 1 0

0 1

� �1 336 1

g2 � ÿ1 0

0 ÿ1

� �2 336 1

g3 � 0 1

ÿ1 0

� �4 8 42

g4 � 2 ÿ2

2 2

� �8 8 42

g5 � ÿ2 2

ÿ2 ÿ2

� �8 8 42

g6 � 2 0

0 4

� �3 6 56

g7 � ÿ2 0

0 ÿ4

� �6 6 56

g8 � 1 1

0 1

� �7 14 24

g9 � ÿ1 ÿ1

0 ÿ1

� �14 14 24

g10 � 1 ÿ1

0 1

� �7 14 24

g11 � ÿ1 1

0 ÿ1

� �14 14 24


(b) Use the character table of PSL (2,7) to write down the six

irreducible characters of G with kernel containing Z � {�I}.

(c) Let ÷7, ÷8, ÷9, ÷10, ÷11 be the remaining irreducible characters

of G. Show that for any j with 7 < j < 11 and any g 2 G, we

have ÷ j(g) � ÿ÷ j(ÿg).

(d) Prove that ÷ j(g3) � 0 for 7 < j < 11, and deduce that ÷ j(1) is

even.

(e) By considering the column of g6 in the character table, and

congruences modulo 3, show that the degrees of ÷7, . . . , ÷11 are

4, 4, 6, 6, 8, and ®nd ÷ j(g6) for 7 < j < 11.

(f) Let ø be one of the irreducible characters of degree 4. By

considering the values of øA on g1, g2, g3 and g6 (see

Proposition 19.14), prove that øA is equal to the irreducible

character of G of degree 6 whose kernel contains Z. Deduce the

values of the irreducible characters of degree 4 on all gi.

(g) Complete the character table of G.

6. The character table of PSL (2, 11).

Let G � PSL (2, 11). This group has eight conjugacy classes with

representatives g1, . . . , g8 having orders and centralizer orders as

follows:

Also, gÿ15 , gÿ1

6 , gÿ17 , gÿ1

8 are conjugate to g5, g6, g8, g7, respec-

tively.

Find the character table of G.

gi g1 g2 g3 g4 g5 g6 g7 g8

Order of gi 1 2 3 6 5 5 11 11|CG(gi)| 660 12 6 6 5 5 11 11


322

28

Character table of GL(2, q)

We are now going to calculate the character tables of an important

in®nite series of groups, and one of the exercises will show you how

to use the results of this chapter to determine the character tables of

in®nitely many simple groups. In the last chapter and its exercises, we

found the character tables of certain groups of 2 3 2 matrices with

entries in Z7 and Z11. We shall determine the character tables of some

matrix groups with entries from an arbitrary ®nite ®eld. At ®rst sight,

this is a daunting task, since the number of irreducible characters

increases with the size of the ®eld. However, we shall see that the

conjugacy classes of our groups fall into four families, as do the

irreducible characters. Consequently, we can display the character

values in a 4 3 4 matrix.

The ®elds Fq and Fq2

We consider ®nite ®elds, and we shall tell you which properties of

these ®elds we will use; if you are unfamiliar with ®nite ®elds then

you might like to consult the book by J. B. Fraleigh listed in the

Bibliography.

Recall that a ®eld (F, �, 3) is a set F with two binary operations

� and 3 such that the following properties hold. First, (F, �) is an

abelian group, with identity element 0. Secondly, if we write

F� � Fnf0g, then (F�, 3) is an abelian group, with identity element

1. Finally, the distributive law holds; that is (a� b)c � ac� bc for all

a, b, c 2 F. For example, R, C and Z p ( p prime) are ®elds, with the

usual de®nitions of � and 3.

The basic properties of ®nite ®elds which we will use without proof

are these:

(28.1) Let p be a prime and n be a positive integer, and write

q � pn. Then there exists a ®eld Fq of order q and

every ®eld of order q is isomorphic to Fq.

For every s 2 Fq the sum of s with itself p times is

zero; in short, ps � 0.

The group (F�q , 3) is cyclic.

Notice that the binomial coef®cients (pi ) with 1 < i < pÿ 1 are all

divisible by p; it follows that (s� t) p � s p � t p for all s, t 2 Fq, and

hence (s� t) p k � s p k � t p k

for all positive integers k. We use this

remark in the proof of the next proposition.

28.2 Proposition

Let F � Fq2 and S � fs 2 F : sq � sg.(1) The set S is a sub®eld of F of order q, and hence S � Fq.

(2) If r 2 F then r � rq, r1�q 2 S.

Proof (1) Suppose that s, t 2 S. Then (s� t)q � sq � tq � s� t, so

s� t 2 S. It is now easy to check that (S, �) and (Snf0g, 3) are

abelian groups, so S is a ®eld.

(2) Since F�q2 is a group of order q2 ÿ 1, we see that rq2 � r for all

r 2 F. This implies that (r � rq)q � rq � rq2 � r � rq and (r1�q)q �r1�q, so r � rq, r1�q 2 S. j

Hereafter, we shall identify the sub®eld S of Fq2 in Proposition 28.2

with the ®eld Fq.

We introduce the following useful notation.

(28.3) Let å be a generator of the cyclic group F�q2 and let

ù � e(2ði=(q2ÿ1)). Suppose that r 2 F�q2 . We may write

r � åm for some m and we let r � ùm.

Then r! r is an irreducible character of F�q2 . More-

over, every irreducible character of F�q2 has the form

r ! r j for some integer j.

You are now in a position to appreciate the statement of the main

result of this chapter, namely Theorem 28.5.

Character table of GL(2, q) 323

The conjugacy classes of GL(2, q)

The general linear group GL(2, q) is de®ned to be the group of

invertible 2 3 2 matrices with entries in Fq. The subgroup consisting of

all matrices of determinant 1 is the special linear group SL(2, q), and

we talked about some special linear groups in the last chapter. Here,

we are going to calculate the character table of GL(2, q).

Let G � GL(2, q), and remember that the matrix

a b

c d

� �belongs to G if and only if its rows are linearly independent. The

number of such matrices is found by noting that (a, b) can be any

non-zero row vector, giving us q2 ÿ 1 choices; and once (a, b) has

been chosen, (c, d) can be any row vector which is not a multiple of

(a, b), giving us q2 ÿ q choices. Therefore,

jGj � (q2 ÿ 1)(q2 ÿ q) � q(qÿ 1)2(q� 1):

There are four families of conjugacy classes of G, of which three

are easy to describe.

First,

a b

0 c

� �can be conjugate to

a9 b90 c9

� �only if fa, cg � fa9, c9g, since conjugate matrices have the same

eigenvalues. Keep this in mind during the following discussion.

The matrices

sI � s 0

0 s

� �(s 2 F�q )

belong to the centre of G. They give us qÿ 1 conjugacy classes of

size 1.

Next, consider the matrices

us � s 1

0 s

� �(s 2 F�q ):

Let


g � a b

c d

� �2 G:

Then

gus � as a� bs

cs c� ds

� �and usg � as d � bs

cs ds

� �so g belongs to the centralizer of us if and only if c � 0 and a � d.

Thus, the matrices us (s 2 F�q ) give us qÿ 1 conjugacy classes; the

centralizer order is (qÿ 1)q, so, by Theorem 12.8, each conjugacy

class contains q2 ÿ 1 elements.

Now, let

ds, t � s 0

0 t

� �2 G (s, t 2 F�q )

and note that

0 1

1 0

� �ÿ1

ds, t0 1

1 0

� �� d t,s:

On the other hand, if s 6� t, then we have that gds, t � ds, t g if and only

if b � c � 0. Thus, the matrices ds, t (s, t 2 F�q , s 6� t) give us

(qÿ 1)(qÿ 2)=2 conjugacy classes; the centralizer order is (qÿ 1)2, so

each conjugacy class contains q(q� 1) elements.

Finally, consider

vr � 0 1

ÿr1�q r � rq

� �(r 2 Fq2nFq):

By Proposition 28.2, vr 2 G. The characteristic polynomial of vr is

det(xI ÿ vr) � x(xÿ (r � rq))� r1�q � (xÿ r)(xÿ rq),

so vr has eigenvalues r and rq. Since r =2 Fq we see that vr lies in

none of the conjugacy classes we have constructed so far.

Now,

gvr �ÿbr1�q a� b(r � rq)

ÿdr1�q c� d(r � rq)

!and

vr g �c d

ÿar1�q � c(r � rq) ÿbr1�q � d(r � rq)

!:


Hence gvr � vr g only if c � ÿbr1�q and d � a� b(r � rq). If these

conditions hold, then

ad ÿ bc � a2 � ab(r � rq)� b2 r1�q � (a� br)(a� brq):

Since (a, b) 6� (0, 0) and r, rq =2 Fq, we see that a� br and a� brq

are non-zero. Therefore, g 2 CG(vr) if and only if

g � a b

ÿbr1�q a� b(r � rq)

� �:

Thus, jCG(vr)j � q2 ÿ 1, and the conjugacy class containing vr has

size q2 ÿ q.

The matrix v t has eigenvalues t and tq, so it is not conjugate to vr

unless t � r or t � rq. We therefore partition Fq2nFq into subsets

fr, rqg; each subset gives us a conjugacy class representative vr and

different subsets give us representatives of different conjugacy classes.

We have now found all the conjugacy classes of G.

28.4 Proposition

There are q2 ÿ 1 conjugacy classes in GL(2, q), described as follows.

Class rep. g sI us ds, t vr

|CG(g)| (q2 ÿ 1)(q2 ÿ q) (qÿ 1)q (qÿ 1)2 q2 ÿ 1

No. of classes qÿ 1 qÿ 1 (qÿ 1)(qÿ 2)=2 (q2 ÿ q)=2

The families of conjugacy class representatives sI and us are indexed

by elements s of F�q .

The family of conjugacy class representatives ds, t is indexed by

unordered pairs fs, tg of distinct elements of F�q .

The family of conjugacy class representatives vr is indexed by

unordered pairs fr, rqg of elements of F�q2nF�q .

Proof The conjugacy classes we have found account for

(qÿ 1)� (qÿ 1)(q2ÿ 1)� (qÿ 1)(qÿ 2)q(q� 1)=2� (q2ÿ q)(q2ÿ q)=2

elements altogether. But this sum is equal to the order of GL(2, q), so

we have found all the conjugacy classes. j


The characters of GL(2,q)

We are now in a position to describe the character table of G.

28.5 Theorem

Label the conjugacy classes of GL(2, q) as in Proposition 28.4, and let

r ! r be the function from F�q2 to C described in (28.3). Then the

irreducible characters of GL(2, q) are given by ëi, øi, øi, j, ÷i as

follows.

sI us ds, t vr

ëi s2i s2i (st)i r i(1�q)

øi qs2i 0 (st)i ÿr i(1�q)

øi, j (q� 1)si� j si� j si t j � s j t i 0÷i (qÿ 1)si ÿsi 0 ÿ(r i � r iq)

Here, we have the following restrictions on the subscripts.

(a) For ëi we have 0 < i < qÿ 2. Thus, there are qÿ 1 characters

ëi, each of degree 1.

(b) For øi we have 0 < i < qÿ 2. Thus, there are qÿ 1 characters

øi, each of degree q.

(c) For øi, j we have 0 < i , j < qÿ 2. Thus, there are

(qÿ 1)(qÿ 2)=2 characters øi, j, each of degree q� 1.

(d) For ÷i, we ®rst consider the set of integers j with

0 < j < q2 ÿ 1 and (q� 1) 6 j j; if j1 and j2 belong to this set

and j1 � j2q mod (q2 ÿ 1) then we choose precisely one of j1

and j2 to belong to the indexing set for the characters ÷i.

Hence, there are (q2 ÿ q)=2 characters ÷i, each of degree qÿ 1.

Before we embark upon the task of calculating the irreducible

characters of G, we present a proposition which will be useful later.

Recall that å is our chosen generator for F�q2 and

vå � 0, 1

ÿå1�q å� åq

� �:

28.6 Proposition

Let K � hvåi. Then jKj � q2 ÿ 1. The group K contains the qÿ 1

scalar matrices sI in G, and of the remaining q2 ÿ q elements of K,


precisely two belong to each of the conjugacy classes represented by

vr with r 2 F�q2nF�q .

Proof The eigenvalues of vå are å and åq, so vå has order q2 ÿ 1.

The eigenvalues of viå and of viq

å are å i and å iq. If å i 6� å iq then

å i =2 Fq and viå and viq

å must be conjugate to vå i . Hence two elements

of K belong to the conjugacy class of vå i .

If å i � å iq then viå � å iI , since vi

å has eigenvalues in Fq and viå is

diagonalizable, and this case accounts for the qÿ 1 scalar matrices. j

We shall construct, in turn, the irreducible characters ëi, øi, øi, j and

÷i which appear in Theorem 28.5.

28.7 Proposition

There are qÿ 1 linear characters ëi of G, and they are given in

Theorem 28.5.

Proof The map det : g ! det g is a homomorphism from G onto F�q .

As i varies between 0 and qÿ 2 inclusive, the functions

ëi : g ! (det g)i (g 2 G)

give qÿ 1 distinct linear characters of G, whose values appear in

Theorem 28.5. j

We will see later that the linear characters ëi (0 < i < qÿ 2) which

appear in Proposition 28.7 are all the linear characters of G.

28.8 Proposition

For all integers i, j there is a character øi, j of G whose values on the

conjugacy class representatives, as described in Proposition 28.4, are

as follows.

sI us ds, t vr

øi, j (q� 1)si� j si� j si t j � s j t i 0


Proof Let

B � a b

0 c

� �2 G

� �:

Then B is a subgroup of G with jBj � (qÿ 1)2q. De®ne ëi, j : B! C

by

ëi, j :s r

0 t

� �! si t j:

Then ëi, j is a character of B. We let øi, j � ëi, j " G.

We use Proposition 21.23 to calculate øi, j(g) for each conjugacy

class representative g, as follows.

g � sI : øi, j(g) � jCG(g)jjCB(g)j ëi, j(g)

g � us : øi, j(g) � jCG(g)jjCB(g)j ëi, j(g)

g � ds, t : øi, j(g) � jCG(g)j ëi, j(g)

jCB(g)j �ëi, j(g9)

jCB(g9)j� �

, where g9 � d t,s

g � vr : øi, j(g) � 0:

Hence, the values of øi, j are as stated in the proposition. j

28.9 Proposition

For each integer i, there is an irreducible character øi of G whose

values are given in Theorem 28.5. The characters øi for 0 < i

< qÿ 2 are all different.

Proof We shall demonstrate that the character øi,i which appears in

Proposition 28.8 gives us øi,i � ëi � øi. To this end, we calculate

høi,i, øi,ii and høi,i, ëii. Remember that the complex conjugate of si is

sÿi. We have

høi,i, øi,ii � (q� 1)2

(q2 ÿ 1)(q2 ÿ q)(qÿ 1)� 1

(qÿ 1)q(qÿ 1)

� 4

(qÿ 1)2

(qÿ 1)(qÿ 2)

2

� 2:

Here, the ®rst term corresponds to the conjugacy classes of elements


sI , and the calculation of this ®rst term involves the following three

observations.

(1) øi,i(sI)øi,i(sI) � (q� 1)2.

(2) jCG(sI)j � (q2 ÿ 1)(q2 ÿ q):

(3) There are qÿ 1 conjugacy classes with representatives of the

forms sI .

The remaining terms in høi,i, øi,ii are calculated in a similar

fashion.

Next,

høi,i, ëii � (q� 1)

(q2 ÿ 1)(q2 ÿ q)(qÿ 1)� 1

(qÿ 1)q(qÿ 1)

� 2

(qÿ 1)2

(qÿ 1)(qÿ 2)

2

� 1:

The facts that høi,i, ëii � 1 and høi,i, øi,ii � 2 imply that øi,i �ëi � øi for some irreducible character øi. Subtract ëi from øi,i to get

the values of øi as given in Theorem 28.5.

Let s be an element of F�q of order qÿ 1. Then øi : ds,1 ! si.

Hence the characters øi for 0 < i < qÿ 2 are all different. j

28.10 Proposition

Suppose that 0 < i , j < qÿ 2. Then the character øi, j which appears

in Proposition 28.8 is irreducible.

Proof We shall show that høi, j, øi, ji � 1. Using the values of øi, j

which are given in Proposition 28.8, we obtain høi, j, øi, ji �A� B� C, where

A � (q� 1)2

(q2 ÿ 1)(q2 ÿ q)(qÿ 1), B � 1

(qÿ 1)q(qÿ 1) and

C � 1

(qÿ 1)2

1

2

Xs 6� t

(si t j � s j t i)(sÿi tÿ j � sÿ j tÿi):

The coef®cent 12

appears in C because we have just one conjugacy

class for each unordered pair fs, tg of distinct elements of F�q .

To evaluate C, note that fds, t : s, t 2 F�q g is an abelian group of

order (qÿ 1)2, and if ó : ds, t ! si t j � s j t i then ó is a sum of two


inequivalent irreducible characters of this group. Thus, hó , ó i � 2:

That is,

1

(qÿ 1)24(qÿ 1)�

Xs 6� t

(si t j � s j t i)(sÿi tÿ j � sÿ j tÿi)

!� 2:

Hence,

C � qÿ 3

qÿ 1:

And now we ®nd that A� B� C � 1. Therefore, høi, j, øi, ji � 1, and

øi, j is irreducible. j

28.11 Corollary

The characters øi, j for 0 < i , j < qÿ 2 are distinct irreducible

characters of G.

Proof Suppose that 0 < i , j < qÿ 2 and 0 < i9 , j9 < qÿ 2, and

(i, j) 6� (i9, j9). We must prove that øi, j 6� øi9, j9.

Consider the group B and its linear characters ëi, j which were used

in the proof of Proposition 28.8. We have

ëi, j � ë j,i :s b

0 t

� �! si t j � s j t i:

Since ëi, j � ë j,i 6� ëi9, j9 � ë j9,i9, there exists complex (qÿ 1)th roots of

unity s and t such that

either s 6� t and si t j � s j t i 6� si9 t j9 � s j9 t i9

or s � t and si� j 6� si9� j9.

In either case, we see that øi, j differs from øi9, j9 on a conjugacy class

of G. Therefore, øi, j 6� øi9, j9. j

28.12 Proposition

For each integer i, there exists a character öi of G which takes the

following values.

sI us ds, t vr

öi q(qÿ 1)si 0 0 r i � r iq


Proof Let K � hvåi, as in Proposition 28.6, and consider the linear

character ái of K which sends the generator vå of K to å i. Suppose

that g 2 K and g is conjugate in G to vr. Then g has eigenvalues r

and rq. Hence ái(g) � r i or r iq and

ái(g)� ái(gq) � r i � r iq:

Let öi � ái " G.

In order to calculate öi, ®rst recall that ái " G is zero on all

elements which are not conjugate to an element of K. Thus, by

Proposition 28.6, öi is zero on the elements of the form us and

ds, t (s 6� t).

If g � sI with s 2 F�q then g 2 K and

öi(g) � jCG(g)jjCK(g)jái(g) � q(qÿ 1)si:

Suppose that r 2 Fq2nFq. Then, by Proposition 28.6, vr is conjugate to

an element of g of K. Also,

öi(g) � jCG(g)j ái(g)

jCK(g)j �ái(gq)

jCK (g)j� �

� ái(g)� ái(gq)

� r i � r iq:

Thus, öi has the values stated in the proposition. j

To be able to work out certain inner products involving our

characters öi, we shall the use the following lemma.

28.13 Lemma

Assume that i is an integer and (q� 1) 6 j i. ThenXr2F

q2nFq

(r i � r iq)(rÿi � rÿiq) � 2(qÿ 1)2:

Proof Note that

G1 � r 0

0 rq

� �: r 2 F�q2

� �and G2 � r 0

0 rq

� �: r 2 F�q

� �are abelian groups of orders q2 ÿ 1 and qÿ 1, respectively. Now,


r 0

0 rq

� �! r i � r iq

gives a character ÷ of degree 2 for each group. For G1, the character ÷is a sum of two inequivalent irreducible characters, since (q� 1) 6 j i

implies that å i 6� å iq; and for G2, the character ÷ is twice an irreducible

character, since rq � r for r 2 F�q . Taking the inner product of the

character ÷ of G1 with itself, we get

1

q2 ÿ 1

Xr2F�

q2

(r i � r iq)(rÿi � rÿiq) � 2

and doing the same for the character ÷ of G2, we get

1

qÿ 1

Xr2F�q

(r i � r iq)(rÿi � rÿiq) � 4:

Hence Xr2F

q2nFq

(r i � r iq)(rÿi � rÿiq) � 2(q2 ÿ 1)ÿ 4(qÿ 1)

� 2(qÿ 1)2: j

28.14 Proposition

For each integer i, let ÷i be the class function on G with the following

values.

sI us ds, t vr

÷i (qÿ 1)si ÿsi 0 ÿ(r i � r iq)

If (q� 1) 6 j i then ÷i is an irreducible character of G.

Proof We can justify the manoeuvre which we now perform only by

saying that it gives the correct answer.

Recall the characters øi, j, øi and öi given in Propositions 28.8, 28.9

and 28.12. Now, ÷i is the class function on G which is given by

÷i � ø0,ÿiøi ÿ ø0,i ÿ öi:


The table below allows us to verify this.

sI us ds, t vr

ø0,ÿi (q� 1)sÿi sÿi sÿi � tÿi 0øi qs2i 0 (st)i ÿr i(1�q)

ø0,ÿiøi q(q� 1)si 0 si � t i 0ø0,i (q� 1)si si si � t i 0öi q(qÿ 1)si 0 0 r i � r iq

÷i (qÿ 1)si ÿsÿi 0 ÿ(r i � r iq)

Next, assume that (q� 1) 6 j i. We work out h÷i, ÷ii using Lemma

28.13.

h÷i, ÷ii � (qÿ 1)2

(q2 ÿ 1)(q2 ÿ q)(qÿ 1)� 1

q2 ÿ q(qÿ 1)� (qÿ 1)2

q2 ÿ 1� 1:

Since ÷i is a linear combination of irreducible characters of G, with

integer coef®cients, and h÷i, ÷ii � 1 and ÷i(1) . 0, it follows that ÷i is

an irreducible character of G. j

28.15 Proposition

Suppose that i and j are integers with (q� 1) 6 j i and (q� 1) 6 j j and

j 6� i, iq mod(q2 ÿ 1). Then the characters ÷i and ÷ j of G are different.

Proof Let K � hvåi, as in Proposition 28.6, and consider the linear

character ái of K which sends the generator vå of K to å i. Suppose

that g 2 K.

If g � sI where s 2 F�q then (ái � áiq)(g) � 2si.

If g is conjugate to vr where r 2 Fq2nFq then (ái � áiq)(g) �r i � r iq.

Since j 6� i, iq mod(q2 ÿ 1), the characters ái � áiq and á j � á jq of

K are different, so either si 6� s j for some s 2 F�q or

r i � r iq 6� r j � r jq for some r 2 Fq2nFq. Therefore, ÷i 6� ÷ j, as we

wished to show. j

We have now completed the proof of Theorem 28.5, since we have

shown that the class functions given in the theorem are inequivalent

irreducible characters; and the number of them is q2 ÿ 1, which is the

same as the number of conjugacy classes of G.

It is possible to use the character table of GL(2, q) to ®nd the


character table of SL(2, q), since most of the irreducible characters

remain irreducible when restricted. We do not go fully into this, since

the answers are quite complicated, and they depend upon whether q is

a power of 2 or q � 1 mod 4 or q � 3 mod 4. In Exercise 28.2, though,

you are asked to consider the easiest case, namely that where q is a

power of 2. Since SL(2, q) � PSL(2, q) when q is a power of 2, this

gives the character tables of an in®nite series of simple groups

PSL(2, q).

Among the characters of SL(2, q), those with kernel containing the

centre of SL(2, q) provide the characters of the groups PSL(2, q) ±

compare Chapter 27 ± and so the character table of PSL(2, q) is rather

easier to ®nd than that of SL(2, q). A challenging exercise is to

determine the character table of PSL(2, q) when q � 1 mod 4 or

q � 3 mod 4 from the character table of GL(2, q).

Although the character table of GL(2, q) was ®rst given in 1907, it

was not until the 1950's that the character table of GL(3, q) was found.

Then, in 1955, J. A. Green determined the character table of GL(n, q)

for all positive integers n.


The character table of GL(2, q) has the following properties.

(a) There are qÿ 1 conjugacy classes with representatives of the form

sI � s 0

0 s

� �, and there are qÿ 1 irreducible characters of

degree 1.

(b) There are qÿ 1 conjugacy classes with representatives of the form

us � s 1

0 s

� �, and there are qÿ 1 irreducible characters of degree

q.

(c) There are (qÿ 1)(qÿ 2)=2 conjugacy classes with representatives

of the form ds, t � s 0

0 t

� �(s 6� t), and there are (qÿ 1)(qÿ 2)=2

irreducible characters of degree q� 1.

(d) There are (q2 ÿ q)=2 conjugacy classes with representatives of the


form vr � 0 1

ÿr1�q r � rq

� �, and there are (q2 ÿ q)=2 irreducible

characters of degree qÿ 1.


1. Use Theorem 28.5 to write down explicitly the character table of

GL(2, 3).

2. Suppose that q is a power of 2. Let Z � fsI : s 2 F�q g. Prove that

GL(2, q) � Z 3 SL(2, q):

Deduce the character table of SL(2, q) from that of GL(2, q). Prove

that if q 6� 2 then SL(2, q) is simple.

3. Use your solution to Exercise 28.2 to write down explicitly the

character table of PSL(2, 8).


337

29

Permutations and characters

We have already seen in Chapter 13 that if G is a permutation group,

i.e. a subgroup of Sn for some n, then G has a permutation character

ð de®ned by ð(g) � jfix(g)j for g 2 G, a fact which proved useful in

many of our subsequent character table calculations. In this chapter we

take the theory of permutation groups and characters somewhat further,

and develop some useful results, particularly about irreducible charac-

ters of symmetric groups (see Theorem 29.12 below).

Group actions

We begin with a more general notion than that of a permutation group.

If Ù is a set, denote by Sym(Ù) the group of all permutations of Ù.

In particular, if Ù � f1, 2, . . . , ng then Sym(Ù) � Sn.

De®nition

Let G be a group and Ù a set. An action of G on Ù is a

homomorphism ö: G! Sym(Ù). We also say that G acts on Ù(via ö).

29.1 Examples

(1) If G < Sn then the identity map is an action of G on f1, . . . , ng.(2) Let G � Sn and let Ù be the set consisting of all pairs fi, jg of

elements of f1, . . . , ng. De®ne ö: G! Sym(Ù) by setting

fi, jg(gö) � fig, jggfor all g 2 Sn and 1 < i , j < n. (So for example, (1 2)ö sends

f1, 3g ! f2, 3g.) Check that ö is an action of Sn; it is called the

action of Sn on pairs.

(3) Let G � GL(2, q), the group of invertible 2 3 2 matrices over the

®nite ®eld Fq, as de®ned in Chapter 28. Let V be the 2-dimensional

vector space over Fq consisting of all row vectors (a, b) with

a, b 2 Fq, and let Ù be the set of all 1-dimensional subspaces hvi of

V. De®ne ö: G! Sym(Ù) by setting

hvi(gö) � hvgifor all hvi 2 Ù and g 2 G. For example, if

g � 1 1

0 1

� �then gö sends

h(a, b)i ! h(a, a� b)i:Then ö is an action of G on Ù.

(4) Let G be a group with a subgroup H of index n, and let Ù be the

set of all right cosets Hx of H in G (so jÙj � n). De®ne ö:

G! Sym(Ù) by

(Hx)(gö) � Hxg

for all x, g 2 G. Then by Exercise 9 of Chapter 23, ö is an action of

G, and Kerö � T x2G xÿ1 Hx < H .

To simplify notation, if ö: G! Sym(Ù) is an action, for ù 2 Ù and

g 2 G we usually just write ùg instead of ù(gö). With this notation,

the fact that ö is a homomorphism simply says that ù(gh) � (ùg)h

for all ù 2 Ù and g, h 2 G.

Adopting this notation, de®ne a relation � on Ù as follows: for

á, â 2 Ù, we have á � â if and only if there exists g 2 G such that

ág � â. It is easy to see that � is an equivalence relation on Ù. The

equivalence classes are called the orbits of G on Ù. Thus Ù is the

disjoint union of the orbits of G. Write

orb(G, Ù)

for the number of orbits of G on Ù. The group G is said to be

transitive on Ù if orb(G, Ù) � 1; in other words, G is transitive if,

given any á, â 2 Ù, there exists g 2 G such that ág � â.

29.2 Examples

(1) Let G � C4, generated by x, say, and let ö: G! S8 be the action


de®ned by xö � (1 2 3 4)(5 6)(7 8) (and of course xkö �((1 2 3 4)(5 6)(7 8))k for any k). Then G has three orbits on

Ù � f1, . . . , 8g, namely f1, 2, 3, 4g, f5, 6g and f7, 8g.(2) The group G is transitive on the set Ù in each of Examples 29.1(2,

3, 4). This is clear in Example (2); to verify it for Example (3) you

need to convince yourself that for any two non-zero row vectors

v, w 2 V there is an invertible 2 3 2 matrix A 2 GL(2, q) such that

vA � w; and in Example (4), simply observe that, given two right

cosets Hx, Hy 2 Ù, the element g � xÿ1 y 2 G has the property that

(Hx)g � Hy.

Let G be a group acting on a set Ù. For ù 2 Ù, write ùG for the

orbit of G which contains ù, so ùG � fùg : g 2 Gg; and de®ne

Gù � fg 2 G : ùg � ùg:We call Gù the stabilizer of ù in G.

29.3 Proposition

The stabilizer Gù is a subgroup of G. Moreover, the size of the orbit

ùG is equal to the index of Gù in G; that is,

jùGj � jG : Gùj:

Proof If g, h 2 Gù then ù(gh) � (ùg)h � ùh � ù, hence gh 2 Gù.

Also gÿ1 2 Gù, and Gù contains the identity, so Gù is a subgroup.

Now let Ä be the set of right cosets Gùx of Gù in G. Observe that

for x, y 2 G,

Gùx � Gù y, xyÿ1 2 Gù , ùxyÿ1 � ù, ùx � ùy:

Hence we can de®ne an injective function ã : Ä! ùG by

ã(Gùx) � ùx

for all x 2 G. Clearly ã is also surjective, and hence jÄj � jùGj, as

required. j

Permutation characters

Let G be a group acting on a ®nite set Ù. Denote by CÙ the vector

space over C for which Ù is a basis. In other words, CÙ consists of

all expressions of the form

Permutations and characters 339

Xù2Ù

ëùù (ëù 2 C)

with the obvious addition and scalar multiplication. As in Chapter 13,

we can make CÙ into a CG-module, called the permutation module,

by de®ning Xëùù

!g �

Xëù(ùg)

for all g 2 G. We see just as in Chapter 13 (p. 129) that if ð is the

character of this permutation module, then for g 2 G,

ð(g) � jfixÙ(g)j,where fixÙ(g) � fù 2 Ù : ùg � ùg. We call ð the permutation char-

acter of G on Ù.

The next result, though elementary, is rather famous, and provides a

basic link between the permutation character and the action of G. It is

often referred to as `̀ Burnside's Lemma'', but is in fact due to Cauchy

and Frobenius.

29.4 Proposition

Let G be a group acting on a ®nite set Ù, and let ð be the

permutation character. Then

hð, 1Gi � 1

jGjXg2G

jfixÙ(g)j � orb(G, Ù):

Proof First note that

hð, 1Gi � 1

jGjXg2G

ð(g) � 1

jGjXg2G

jfixÙ(g)j:

Let Ä1, . . . , Ä t be the orbits of G on Ù, and for each i, pick ùi 2 Äi.

By Proposition 29.3, for 1 < i < t we have

jÄij � jùGi j � jG : Gùi

j:Hence jÄij jGùi

j � jGj. Now de®ne Ö � f(ù, g) : ù 2 Ù, g 2 G,

ùg � ùg. We calculate jÖj in two different ways. First, for each g,

the number of ù 2 Ù such that ùg � ù is equal to jfixÙ(g)j, hence


jÖj �Xg2G

jfixÙ(g)j:

Secondly, for each ù, the number of g 2 G such that ùg � ù is equal

to jGùj, hence

jÖj �Xù2ÙjGùj �

Xt

i�1

jÄij jGùij �

Xt

1

jGj � tjGj:

ThereforeP

g2G jfixÙ(g)j � tjGj, and the conclusion follows. j

29.5 Corollary

G is transitive on Ù if and only if hð, 1Gi � 1.

Now let G be a group, and suppose that G acts on two sets Ù1 and

Ù2, with corresponding permutation characters ð1 and ð2 respectively.

Then we can de®ne an action of G on the Cartesian product Ù1 3 Ù2

by setting

(ù1, ù2)g � (ù1 g, ù2 g)

for all ùi 2 Ùi, g 2 G. It is clear that fixÙ13Ù2(g) �

fixÙ1(g) 3 fixÙ2

(g) for any g 2 G. Hence if ð is the permutation

character of G on Ù1 3 Ù2, then

ð(g) � ð1(g)ð2(g)

for all g 2 G.

29.6 Proposition

Let G act on Ù1 and Ù2, with permutation characters ð1 and ð2

respectively. Then

hð1, ð2i � orb(G, Ù1 3 Ù2):

Proof We have

hð1, ð2i � 1

jGjXg2G

jfixÙ1(g)kfixÙ2

(g)j � 1

jGjXg2G

jfixÙ13Ù2(g)j,

which is equal to orb(G, Ù1 3 Ù2) by Proposition 29.4. j

In the rest of the chapter we apply Proposition 29.6 in a number of

situations, the ®rst being the case where Ù1 � Ù2.


Suppose G acts on Ù. Then G also acts on Ù 3 Ù in the way

de®ned above, namely (ù1, ù2)g � (ù1 g, ù2 g) for all ù1, ù2 2Ù, g 2 G.

29.7 De®nition

The number of orbits of G on Ù 3 Ù is called the rank of G on Ù,

written r(G, Ù). Thus

r(G, Ù) � orb(G, Ù 3 Ù):

The next result is immediate from Proposition 29.6.

29.8 Proposition

Let G act on Ù, with permutation character ð. Then

r(G, Ù) � hð, ði:

Now suppose G is transitive on Ù and jÙj. 1. Then

Ä � f(ù, ù) : ù 2 Ùgis an orbit of G on Ù 3 Ù, and hence certainly r(G, Ù) > 2. The case

where equality holds is of particular interest.

29.9 De®nition

Let G be transitive on Ù. Then G is said to be 2-transitive on Ù if

r(G, Ù) � 2.

In other words, G is 2-transitive if, for any ordered pairs (á1, á2)

and (â1, â2) in Ù 3 Ù, with á1 6� á2, â1 6� â2, there exists g 2 G such

that á1 g � â1 and á2 g � â2.

29.10 Corollary

If G is 2-transitive on Ù, with permutation character ð, then

ð � 1G � ÷,

where ÷ is an irreducible character of G.

Proof We have hð, 1Gi � 1 by Corollary 29.5, and hð, ði � 2 by

Proposition 29.8. The result follows, using Theorem 14.17. j


29.11 Examples

(1) The symmetric group Sn is 2-transitive on f1, . . . , ng. Also An is

2-transitive, provided n > 4. Hence these groups have an irreducible

character ÷ given by

÷(g) � jfix(g)j ÿ 1:

We have seen this irreducible character in a number of previous

examples (see 18.1, 19.16, 19.17).

(2) Consider the action of G � GL(2, q) given in Example 29.1(3).

Here Ù is the set of all 1-dimensional subspaces of the 2-dimensional

vector space V. We claim that G is 2-transitive on Ù. To see this, let

(hv1i, hv2i) and (hw1i, hw2i) be two pairs of distinct 1-spaces in Ù.

Then v1, v2 and w1, w2 are both bases of V. The linear transformation

from V to V which sends v1 ! w1, v2 ! w2 is therefore invertible,

giving an element of GL(2, q) which sends hv1i ! hw1i, hv2i ! hw2i.Hence G is 2-transitive on Ù, as claimed. Since jÙj � q� 1 (see

Exercise 1 at the end of the chapter), the irreducible character ÷ of G

given by Corollary 29.10 is the character ø0 of degree q in Theorem

28.5.

(3) Consider the action of Sn on pairs de®ned in Example 29.1(2), with

n > 4. This action is not 2-transitive, since, for example, there is no

element of Sn which sends (f1, 2g, f3, 4g) to (f1, 2g, f2, 3g). In fact

it is easy to see that the orbits of G � Sn on Ù 3 Ù are Ä, Ä1 and

Ä2, where Ä is as above, and

Ä1 � f(fi, jg, fk, lg) : jfi, jg \ fk, lgj � 1g,Ä2 � f(fi, jg, fk, lg) : jfi, jg \ fk, lgj � 0g:

Thus hð, ði � r(G, Ù) � 3, and so ð � 1G � ÷� ø, where ÷ and øare irreducible characters of Sn.

Some irreducible characters of Sn

By Theorem 12.15 we know that the conjugacy classes of Sn are in

bijective correspondence with the set of all possible cycle-shapes of

permutations. Each cycle-shape (including 1-cycles) is a sequence

ë � (ë1, . . . , ës) of positive integers ëi such that ë1 > ë2 > . . . > ës

and ë1 � . . . � ës � n, and we call such a sequence a partition of n.

By Theorem 15.3, the irreducible characters of Sn are also in

bijective correspondence with the partitions ë of n. A key aim is

therefore to construct, for each partition ë, an irreducible character ÷ë


of Sn, in a natural way. We shall apply the material of this chapter to

carry out this aim in the case where ë � (nÿ k, k), a 2-part partition

(see Theorem 29.13 below). The ideas can be developed to carry out

the aim in general, but we do not do this; if you want to see this, and

much more, on the character theory of Sn, we refer you to the book by

G. James listed in the Bibliography.

Let G � Sn and I � f1, 2, . . . , ng. For an integer k < n=2, de®ne

Ik to be the set consisting of all subsets of I of size k. Just as in

Example 29.1(2) we can de®ne an action of G on Ik as follows: for

any subset A � fi1, . . . , ikg 2 Ik and any g 2 G, let

Ag � fi1 g, . . . , ikgg:Let ðk be the permutation character of G in its action on Ik . Observe

that

ðk(1) � jIk j � n

k

� �:

29.12 Proposition

If l < k < n=2, then hðk , ð li � l � 1.

Proof By Proposition 29.6, hðk , ð li � orb(G, Ik 3 Il). The orbits of

G � Sn on Ik 3 Il are easily seen to be J0, J1, . . . Jl, where for

0 < s < l,

Js � f(A, B) 2 Ik 3 Il : jA \ Bj � sg:Hence orb(G, Ik 3 Il) � l � 1, giving the conclusion. j

29.13 Theorem

Let m � n=2 if n is even, and m � (nÿ 1)=2 if n is odd. Then Sn has

distinct irreducible characters ÷(n) � 1G, ÷(nÿ1,1), ÷(nÿ2,2), . . . , ÷(nÿm,m)

such that for all k < m,

ðk � ÷(n) � ÷(nÿ1,1) � . . . � ÷(nÿk,k):

In particular, ÷(nÿk,k) � ðk ÿ ðkÿ1.

Proof We prove the existence of irreducible characters ÷(n),

÷(nÿ1,1), . . . , ÷(nÿk,k) such that ðk � ÷(n) � ÷(nÿ1,1) � . . . � ÷(nÿk,k), by

induction on k. This holds for k � 1 by Corollary 29.10.

Now assume the statement holds for all values less than k. Then


there exist irreducible characters ÷(n), ÷(nÿ1,1), . . . , ÷(nÿk�1,kÿ1) such

that

ði � ÷(n) � ÷(nÿ1,1) � . . . � ÷(nÿi,i)

for all i , k. Now by Proposition 29.12,

hðk , 1Gi � 1, hðk , ð1i � 2, . . . , hðk , ðkÿ1i � k, hðk , ðki � k � 1:

It follows that ðk � ðkÿ1 � ÷ for some irreducible character ÷. Writing

÷ � ÷(nÿk,k), we have ðk � ÷(n) � ÷(nÿ1,1) � . . . � ÷(nÿk,k), as re-

quired. j

29.14 Examples

(1) The formula ÷(nÿk,k) � ðk ÿ ðkÿ1 makes it easy to calculate the

values of the characters ÷(nÿk,k). For example, the degree is

÷(nÿk,k)(1) � ðk(1)ÿ ðkÿ1(1) � n

k

� �ÿ n

k ÿ 1

� �:

As another example, suppose n � 7 and let us calculate the value of

the irreducible character ÷(5,2) on a 3-cycle:

÷(5,2)(123) � ð2(123)ÿ ð1(123) � jfix I2(123)j ÿ jfix I1

(123)j � 6ÿ 4 � 2:

(2) In the character table of S6 given in Example 19.17, the irreducible

characters ÷1, ÷3, ÷7, ÷9 are equal to ÷(6), ÷(5,1), ÷(4,2), ÷(3,3), respectively.


1. An action of G on Ù is a homomorphism G! Sym(Ù). The orbits

are the equivalence classes in Ù of the relation de®ned by

á � â, ág � â for some g 2 G. The size of the orbit ùG contain-

ing ù is jùGj � jG : Gùj.2. If G acts on Ù then CÙ is the permutation module, and the

corresponding character of G is ð, where ð(g) � jfixÙ(g)j. The

number of orbits is equal to hð, 1Gi.3. The rank r(G, Ù) is the number of orbits of G on Ù 3 Ù, and

r(G, Ù) � hð, ði. If G is 2-transitive then r(G, Ù) � 2 and ð �1G � ÷ with ÷ irreducible.

4. The irreducible characters of Sn are in bijective correspondence with

partitions of n. The irreducible characters ÷(nÿk,k) corresponding to

2-part partitions have values given by Theorem 29.13.



1. Let G be a ®nite group, and de®ne a function ö : G 3 G !Sym(G) by x((g, h)ö) � gÿ1xh for all x, g, h 2 G.

(a) Show that ö is an action of G 3 G on G, which is transitive.

(b) Find the stabilizer (G 3 G)1 of the identity 1 2 G, and ®nd the

kernel of ö.

(c) Show that the rank of this action r(G 3 G, G) is equal to the

number of conjugacy classes of G, and the permutation char-

acter ð is

ð �X÷

÷ 3 ÷,

where the sum is over all irreducible characters ÷ of G, and

÷ 3 ÷ is the irreducible character of G 3 G given by Theorem

19.18.

2. Show that if Ù is the set of all 1-dimensional subspaces of a 2-

dimensional vector space over Fq (as in Example 29.1(3)), then

jÙj � q� 1.

3. Let G � GL(2, q) and let V � F2q as in Example 29.1(2). Let

V� � V ÿ f0g, and de®ne an action ö : G! Sym(V�) by

v(gö) � vg for v 2 V�, g 2 G. Let ð be the permutation character

of G in this action.

Decompose ð as a sum of irreducible characters of GL(2, q) (the

latter are given by Theorem 28.5).

(Hint: one way to do this is to write down the values of ð on the

conjugacy classes of G, and take inner products with the irreducible

characters of G given in 28.5.)

4. Let G be a ®nite group, and let H1, H2 be subgroups of G. For

i � 1, 2 de®ne Ùi to be the set of right cosets of Hi in G, so that

G acts on Ùi as in Example 29.1(4); let ði be the permutation

character of G in the action on Ùi. Suppose that ð1 � ð2.

Prove that if G is abelian, then H1 � H2:

Give an example to show that this need not be the case in

general.

5. Let G be a ®nite group acting transitively on a set Ù of size greater

than 1. Prove that G contains an element g such that jfixÙ(g)j � 0.

(Such an element is called a ®xed-point-free element of G.)


6. Let n be a positive integer, and let Ù be the set of all ordered pairs

(i, j) with i, j 2 f1, . . . , ng and i 6� j. Let Sn act on Ù in the

obvious way (namely, (i, j)g � (ig, jg) for g 2 Sn), and let the

permutation character of Sn in this action be ð(nÿ2,1,1).

By considering inner products as in the proof of Theorem 29.13,

prove that

ð(nÿ2,1,1) � 1� 2÷(nÿ1,1) � ÷(nÿ2,2) � ÷,

where ÷ is an irreducible character.

Writing ÷ � ÷(nÿ2,1,1), calculate the degree of ÷(nÿ2,1,1), and calcu-

late its value on the elements (12) and (123) of Sn.

In the character table of S6 given in Example 19.17, which

irreducible character is equal to ÷(4,1,1)?


348

30

Applications to group theory

There are several ways of using the character theory of a group to

determine information about the structure of the group. The examples

which we have come across so far ± ®nding the centre of the group,

seeing whether or not the group is simple, and so on ± require little

calculation. In this chapter we present some rather deeper applications.

The ®rst involves doing arithmetic with character values to determine

certain numbers, known as the class algebra constants. These constants

carry information about the multiplication in G, and they can be used

to investigate the subgroup structure of G, as we shall demonstrate.

The second application takes this much further: the Brauer±Fowler

Theorem 23.19 motivates the study of simple groups containing an

involution with centralizer isomorphic to a given group C. Using a

little group theory and a lot of character theory we shall carry out such

a study in the case where C � D8, the dihedral group of order 8.

Class algebra constants

Let G be a ®nite group and let C1, . . . , Cl be the distinct conjugacy

classes of G. Recall from Proposition 12.22 that the class sums

C1, . . . , Cl form a basis for the centre of the group algebra CG

(where Ci �P

g2Cig).

30.1 Proposition

There exist non-negative integers aijk such that for 1 < i < l and

1 < j < l,

CiC j �Xl

k�1

aijk Ck :

Proof For g 2 Ck the coef®cient of g in the product CiC j is equal to

the number of pairs (a, b) with a 2 Ci, b 2 Cj and ab � g. This

number is a non-negative integer, and is independent of the chosen

element g of Ck . The result follows. j

Another way of looking at Proposition 30.1 is to note that CiC j

belongs to Z(CG), so it must be a linear combination of C1, . . . , Cl.

30.2 De®nition

The integers aijk in the formula

CiC j �Xl

k�1

aijk Ck

are the class algebra constants of G.

From their very de®nition, the numbers aijk carry information about

the multiplication in G:

(30:3) For all g 2 Ck and all i, j we have

aijk � the number of pairs (a, b) with a 2 Ci, b 2 Cj and ab � g:

Also, the constants aijk determine the product of any two elements in

the centre Z(CG) of the group algebra, since C1, . . . , Cl is a basis of

Z(CG). As the centre of the group algebra plays an important role in

representation theory, you might suspect that the class algebra constants

are determined by the character table of G. Our next theorem shows

that this is indeed the case.

30.4 Theorem

Let gi 2 Ci for 1 < i < l. Then for all i, j, k, we have

aijk � jGjjCG(gi)j jCG(gj)j

X÷

÷(gi)÷(gj)÷(gk)

÷(1)

where the sum is over all the irreducible characters ÷ of G.

Applications to group theory 349

Proof Let ÷ be an irreducible character of G, and let U be a CG-

module with character ÷. Then by Lemma 22.7, for all u 2 U we have

uCi � jGj÷(gi)

jCG(gi)j÷(1)u:

Therefore

uCiC j � jGj2jCG(gi)j jCG(gj)j

÷(gi)÷(gj)

(÷(1))2u

and

Xl

m�1

aijmuCm �Xl

m�1

aijm

jGj÷(gm)

jCG(gm)j÷(1)u:

Since CiC j �P

maijmCm, we deduce that

Xl

m�1

aijm

÷(gm)

jCG(gm)j �jGj

jCG(gi)j jCG(gj)j÷(gi)÷(gj)

÷(1):(30:5)

Pick k with 1 < k < l. Multiply both sides of equation (30.5) by ÷(gk)

and sum over all irreducible characters ÷ of G, to obtain

Xl

m�1

aijm

X÷

÷(gm)÷(gk)

jCG(gm)j �jGj

jCG(gi)j jCG(gj)jX÷

÷(gi)÷(gj)÷(gk)

÷(1):

By the column orthogonality relations, Theorem 16.4(2), this yields


X÷

÷(gi)÷(gj)÷(gk)

÷(1):

j

Examples

30.6 Example

In this example we shall use the class algebra constants to prove some

facts about the elements and subgroups of the symmetric group S4;

these results can readily be proved directly, but they serve as a useful

illustration of the method.

Let G � S4. By Section 18.1, the character table of G is as shown:


(1) We use Theorem 30.4 to calculate the class algebra constant

a555:

a555 � 24

4:4

1

1�ÿ1

1� 0

2�ÿ1

3� 1

3

� �� 0:

Hence, by (30.3), S4 does not possess elements a, b of order 4 such

that the product ab also has order 4. We deduce from this that S4 does

not have a subgroup which is isomorphic to the quaternion group Q8:

for Q8 does have two elements of order 4 with product of order 4.

(2) By Theorem 30.4,

a245 � 24

4:81� 1� 1

3� 1

3

� �� 2:

Hence S4 has elements a, b of order 2 such that ab has order 4.

Writing x � ab, we have

x4 � 1, aÿ1xa � ba � (ab)ÿ1 � xÿ1,

so ka, bl � D8. We deduce the fact (which we already know from

Exercise 18.1) that S4 has a subgroup which is isomorphic to D8.

(3) Finally,

a235 � 24

4:3(1� 1) � 4,

so S4 has elements a of order 2 and b of order 3 with ab of order 4.

In fact, it can be shown that S4 has a presentation as follows:

S4 � ha, b: a2 � b3 � (ab)4 � 1i:In other words, S4 is generated by a and b, and all products of

elements of S4 are determined by the given relations. We supply a


Class Ci C1 C2 C3 C4 C5

gi 1 (1 2) (1 2 3) (1 2)(3 4) (1 2 3 4)|CG(gi)| 24 4 3 8 4

÷1 1 1 1 1 1÷2 1 ÿ1 1 1 ÿ1÷3 2 0 ÿ1 2 0÷4 3 1 0 ÿ1 ÿ1÷5 3 ÿ1 0 ÿ1 1


proof in the solution to Exercise 30.6 ± in the meantime, you may

wish to puzzle out the relevance of the ®gure above.

30.7 Example

We use Theorem 30.4 to ®nd a subgroup H of the simple group

PSL (2, 7) with H isomorphic to S4. That such a subgroup exists is not

obvious, and it is quite tricky to construct directly.

We found in Chapter 27 that the character table of G � PSL (2, 7) is

as follows.


Class rep. gi g1 g2 g3 g4 g5 g6

Order of gi 1 2 4 3 7 7|CG(gi)| 168 8 4 3 7 7

÷1 1 1 1 1 1 1÷2 7 ÿ1 ÿ1 1 0 0÷3 8 0 0 ÿ1 1 1÷4 3 ÿ1 1 0 á á÷5 3 ÿ1 1 0 á á÷6 6 2 0 0 ÿ1 ÿ1


7)=2.


We calculate the class algebra constant a243. By Theorem 30.4,

a243 � 168

8:31� 1

7� 0� 0� 0� 0

� �� 8:

Hence, by (30.3), G contains elements x and y such that x has order

2, y has order 3 and xy has order 4. Let H be the subgroup kx, yl of

G. From Example 30.6, we know that

S4 � ha, b: a2 � b3 � (ab)4 � 1i:Hence there is a homomorphism ö from S4 onto H (ö sends a to x

and b to y). By Theorem 1.10, S4=Kerö � H . Now Kerö, being a

normal subgroup of S4, is {1}, V4, A4 or S4 (see Example 12.20), so

H is isomorphic to S4, S3, C2 or {1}. Since H has an element of order

4, namely xy, we conclude that

H � S4:

Thus we have shown that PSL (2, 7) has a subgroup which is iso-

morphic to S4.

The Brauer programme

The Brauer±Fowler Theorem 23.19 states that there are only ®nitely

many non-isomorphic ®nite simple groups containing an involution with

a given centralizer. This fact led Brauer to initiate a programme to

®nd, given a ®nite group C, all ®nite simple groups G possessing an

involution t such that CG(t) � C. This programme formed an important

part of the effort of many mathematicians to classify all the ®nite

simple groups, an effort which was ®nally completed in the early

1980s (see the book by D. Gorenstein listed in the Bibliography).

The next result carries out part of Brauer's programme in the case

where C � D8, a dihedral group of order 8. It determines the possible

orders of simple groups G having an involution t such that

CG(t) � D8. We have chosen to present this result because it provides

a wonderful illustration of the use of character theory in the service of

group theory.

30.8 Theorem

Let G be a ®nite non-abelian simple group which has an involution t

such that CG(t) � D8. Then G has order 168 or 360.


Observe that PSL(2, 7) is a simple group of order 168 having an

involution with centralizer D8 (see Lemma 27.1); and A6 is a simple

group of order 360 with this property (see Exercise 7 at the end of the

chapter). Using some rather more sophisticated group theory than that

covered in this book, one can show that PSL(2, 7) and A6 are the only

simple groups of order 168 or 360.

Before embarking upon the proof of Theorem 30.8, we require a

couple of preliminary results. The ®rst is Sylow's Theorem, a basic

result in ®nite group theory. We shall not prove this, but refer you to

Theorems 18.3 and 18.4 of the book by J. Fraleigh listed in the

Bibliography.

30.9 Sylow's Theorem

Let p be a prime number, and let G be a ®nite group of order pab,

where a, b are positive integers and p 6 j b. Then

(1) G contains a subgroup of order pa; such a subgroup is call a

Sylow p-subgroup of G;

(2) all Sylow p-subgroups are conjugate in G (i.e. if P, Q are Sylow p-

subgroups, then there exists g 2 G such that Q � gÿ1 Pg);

(3) if R is a subgroup of G with jRj � pc for some c, then there is a

Sylow p-subgroup of G containing R.

30.10 Lemma

Let G be a ®nite non-abelian simple group, and let P be a Sylow 2-

subgroup of G. Suppose Q is a subgroup of P with jP : Qj � 2. If u is

an involution in G, then u is conjugate to an element of Q.

Proof Suppose u is not conjugate to an element of Q. Let Ù be the

set of right cosets Qx of Q in G, and de®ne an action of G on Ù by

(Qx)g � Qxg for x, g 2 G (see Example 29.2(4)). Observe that

jÙj � 2jG : Pj � 2m, where m is odd since P is a Sylow 2-subgroup.

Now consider fixÙ(u) � fù 2 Ù : ùu � ùg If Qx 2 fixÙ(u), then

Qxu � Qx and hence xuxÿ1 2 Q, contrary to assumption. Hence

fixÙ(u) � Æ. This means that in its action on Ù, the involution u is a

product of m disjoint 2-cycles, hence is an odd permutation. Hence the

subgroup

fg 2 G : g acts as an even permutation on Ùg


is a normal subgroup of index 2 in G. This is impossible since G is

non-abelian and simple. This contradiction completes the proof. j

We also need to introduce the idea of a generalized character of a

group H . This is simply a class function of the form

ø �X

x

n÷÷

where the sum is over all the irreducible characters of H , and each

n÷ 2 Z. If n÷ > 0 for all ÷ then of course ø is a character, but this

need not be the case for a generalized character. In particular, the

degree ø(1) can be 0 or negative for a generalized character ø. Notice

also that the orthogonality relations give the usual inner products

hø, ÷i � n÷, hø, øi �X

n2÷

for a generalized character ø as above.

The generalized character ø can be expressed as a difference áÿ â,

where á and â are characters of H : take

á �Xn÷>0

n÷÷, â � ÿXn÷,0

n÷÷:

Finally, if H is a subgroup of a group G, we de®ne the induced

generalized character ø " G by

ø " G � (á " G)ÿ (â " G)

where ø � áÿ â as above. It is clear from this de®nition that the

formulae for the values of ø " G given in Proposition 21.19 and

Corollary 21.20 hold for generalized characters ø.

Proof of Theorem 30.8

Let G be a ®nite non-abelian simple group with an involution t such

that CG(t) � D � D8. Certainly t commutes with itself, so t 2 D; and

as t commutes with all elements of D, we have t 2 Z(D), the centre

of D. The centre of D8 is a cyclic group of order 2 (see (12.12)), and

hence Z(D) � hti.By Theorem 30.9(3), there is a Sylow 2-subgroup P of G such that

D < P. Then Z(P) < CG(t) � D, so Z(P) < Z(D) � hti. By Lemma

26.1(1) we have Z(P) 6� 1, and hence Z(P) � hti. Therefore P <


CG(t) � D, and so P � D. In other words, D is a Sylow 2-subgroup

of G.

Write D � ha, bi where a4 � b2 � 1 and bÿ1ab � aÿ1. Then t � a2.

Let C � hai be the cyclic subgroup of index 2 in D. By Lemma

30.10, every involution of G is conjugate to an involution in C. As

t � a2 is the only such involution, we conclude that tG is the unique

conjugacy class of involutions in G.

Next, let g 2 G and suppose that gÿ1cg 2 C for some non-identity

element c 2 C. Since t � c or c2, we must have gÿ1 tg � t, hence

g 2 CG(t) � D and so gÿ1Cg � C.

We summarise what we have proved so far:

(30.11) D is a Sylow 2-subgroup of G; tG is the unique

conjugacy class of involutions in G; for any g 2 G we

have C \ gÿ1Cg � f1g or C; and if C \ gÿ1Cg � C

then g 2 D.

This is all the group theory we will need for the proof. The rest is

character theory.

Let ë be the linear character of C such that ë(a) � i, and de®ne

è � (1C " D)ÿ (ë " D),

a generalized character of D. Then è takes the value 2 on a, aÿ1, the

value 4 on t, and 0 elsewhere. Referring to the character table of D8

in Example 16.3(3), we have è � ÷1 � ÷2 ÿ ÷5. (In particular,

è(1) � 0.) Hence hè, èi � 3.

We next establish

hè " G, è " Gi � 3:(30:12)

To see this, observe ®rst that by Frobenius reciprocity, hè " G,

è " Gi � h(è " G) # D, èi Now for 1 6� c 2 C, Proposition 21.19 gives

(è " G)(c) � 1

8

Xy2G

_è(yÿ1cy):

By (30.11), if yÿ1cy 2 C then y 2 D, whence yÿ1cy � c�1 and

è(yÿ1cy) � è(c). And if yÿ1cy 2 Dÿ C then è(yÿ1cy) � 0. It follows

that (è " G)(c) � è(c). Since è vanishes on Dÿ C, we therefore have

h(è " G) # D, èi � hè, èi � 3, giving (30.12).

Now hè " G, 1Gi � h1C ÿ ë, 1Ci � 1. Also (è " G)(1) � 0 (see Corol-

lary 21.20), and so it follows from (30.12) that


è " G � 1G � áÿ â,

where á, â are irreducible characters of G. Since we have shown that

(è " G)(t) � è(t) � 4, we have now proved the following.

(30.13) We have è " G � 1G � áÿ â, where á, â are irreduci-

ble, 1� á(1)ÿ â(1) � 0 and 1� á(t)ÿ â(t) � 4.

Note that by Corollary 13.10, á(t) and â(t) are integers.

We now introduce another class function of G into the picture. For

g 2 G, de®ne ã(g) to be number of ordered pairs (x, y) 2 tG 3 tG

such that g � xy. If we write tG � Ci and g lies in the conjugacy

class Ck of G, then ã(g) � aiik in the notation of (30.3). Hence

Theorem 30.4 yields the following.

(30.14) We have

ã � jGjjDj2X÷

÷(t)2

÷(1)÷,

where the sum is over all irreducible characters ÷ of G.

We shall calculate the inner product of ã and è " G in two ways.

First, from (30.13) and (30.14) we have

hè " G, ãi � jGj64

1� á(t)2

á(1)ÿ â(t)2

â(1)

!:(30:15)

On the other hand, by Frobenius Reciprocity, hè " G, ãi �h1C ÿ ë, ã # Ci. Consider ã(c) for 1 6� c 2 C. If c � xy with x, y 2 tG,

then xÿ1cx � yx � cÿ1, and hence x 2 D by (30.11); similarly y 2 D.

Now calculation in D8 shows that ã(c) � 4. Therefore

h1C ÿ ë, ã # Ci � 1

jCj :4:((1ÿ i)� 2� (1� i)) � 4:

Hence from (30.15) we deduce

jGj 1� á(t)2

á(1)ÿ â(t)2

â(1)

!� 28:(30:16)

This equation gives us enough number-theoretic information about

jGj to ®nish the proof fairly quickly. Write d � á(1) and e � á(t) 2 Z.

By (30.13) we have


â(1) � d � 1, â(t) � eÿ 3:

From the column orthogonality relations 16.4(2), we have

8 � jCG(t)j > 1� á(t)2 � â(t)2 � 1� e2 � (eÿ 3)2,

from which it follows that e � 1 or 2.

Suppose now that e � 1. Then (30.16) gives

jGj 1� 1

dÿ 4

d � 1

� �� 28,

whence

jGj � 28 d(d � 1)

(d ÿ 1)2:

Now the highest common factor hcf(d ÿ 1, d � 1) is 1 or 2, and

hcf(d ÿ 1, d) � 1. Hence (d ÿ 1)2 must divide 210, and so d ÿ 1 � 2r

with r < 5. Moreover, a Sylow 2-subgroup of G has order 8. It follows

that r � 3 and d � 9, giving jGj � 360, one of the possibilities in the

conclusion of Theorem 30.8.

Finally, suppose that e � 2. Then (30.16) yields

jGj � 28 d(d � 1)

(d � 2)2:

Reasoning as above, we deduce that d � 2 � 23, giving d � 6 and

jGj � 168. This completes the proof of Theorem 30.8. j


1. The class algebra constants aijk are given by

CiC j �X

k

aijk Ck :

They can be calculated from the character table, by using the

formula


X÷

÷(gi)÷(gj)÷(gk)

÷(1):

2. Given groups G and H, the class algebra constants of G can

sometimes be used to determine whether or not G has a subgroup

which isomorphic to H.


3. Using Sylow's Theorem, together with lots of ingenious character

theory, it can be shown that any simple group possessing an

involution with centralizer isomorphic to D8 must have order 168 or

360.


1. Use the character table of PSL (2, 7), given at the end of Chapter

27, to prove that PSL (2, 7) contains elements a and b such that a

has order 2, b has order 3 and ab has order 7.

2. Does PSL (2, 7) contain a subgroup isomorphic to D14?

(Hint: D14 � ka, b: a2 � b2 � 1, (ab)7 � 1l.)

For the next three exercises, you may assume that A5 is a simple

group, and that A5 has the following presentation:

A5 � ha, b: a2 � b3 � (ab)5 � 1i:3. The character table of PSL (2, 11) is given in the solution to

Exercise 27.6. Does PSL (2, 11) contain a subgroup which is iso-

morphic to A5?

4. Prove that A5 is characterized by its character table ± that is, if G

is a group with the same character table as A5 (see Example 20.13),

then G � A5.

5. Suppose that G is a group, and that G has the character table

shown.

(a) Show that G is a simple group of order 360.

(b) Use the Frobenius±Schur Count of Involutions to obtain an

g1 g2 g3 g4 g5 g6 g7

÷1 1 1 1 1 1 1 1÷2 5 1 ÿ1 2 ÿ1 0 0÷3 5 1 ÿ1 ÿ1 2 0 0÷4 8 0 0 ÿ1 ÿ1 á â÷5 8 0 0 ÿ1 ÿ1 â á÷6 9 1 1 0 0 ÿ1 ÿ1÷7 10 ÿ2 0 1 1 0 0

where á � (1�p5)=2, â � (1ÿp5)=2.


upper bound for the number of involutions in G, and deduce

that g2 has order 2 and g3 has order 4.

(c) Prove that G has a subgroup H which is isomorphic to A5.

(d) Using Exercise 23.9, show that G � A6.

6. Use the ®gure which appears in Example 30.6(3) to show that every

group G which is generated by two elements a and b which satisfy

a2 � b3 � (ab)4 � 1

has order at most 24.

7. Prove that PSL(2, 7) and A6 are simple groups of order 168, 360

respectively, both of which contain an involution with centralizer

isomorphic to D8.

8. Find a simple group G having an involution t such that

CG(t) � D16.

(Hint: look for a suitable simple group PSL(2, p).)


361

31

Burnside's Theorem

One of the most famous applications of representation theory is

Burnside's Theorem, which states that if p and q are prime numbers

and a and b are positive integers, then no group of order paqb is

simple. In the ®rst edition of his book Theory of groups of ®nite order

(1897), Burnside presented group-theoretic arguments which proved the

theorem for many special choices of the integers a, b, but it was only

after studying Frobenius's new theory of group representations that he

was able to prove the theorem in general. Indeed, many later attempts

to ®nd a proof which does not use representation theory were unsuc-

cessful, until H. Bender found one in 1972.

A preliminary lemma

We prepare for the proof of Burnside's Theorem with a lemma (31.2)

which is concerned with character values. In order to establish this

lemma we require some basic facts about algebraic integers and

algebraic numbers, which we now describe. We omit proofs of these ±

for a good account, see for instance the book by Pollard and Diamond

listed in the Bibliography.

An algebraic number is a complex number which is a root of some

non-zero polynomial over Q. We call a polynomial in x monic if the

coef®cient of the highest power of x in it is 1.

Let á be an algebraic number; and let p(x) be a monic polynomial

over Q of smallest possible degree having á as a root. Then p(x) is

unique and irreducible; it is called the minimal polynomial of á. The

roots of p(x) are called the conjugates of á.

For example, if ù is an nth root of unity then the minimal poly-

nomial of ù divides xn ÿ 1, and so every conjugate of ù is also an

nth root of unity.

If á is an algebraic integer, then á is a root of a monic polynomial

with integer coef®cients (see Chapter 22), and it turns out that the

minimal polynomial of á also has integer coef®cients.

We shall require the following fact about conjugates:

(31.1) Let á and â be algebraic numbers. Then every conju-

gate of á � â is of the form á9 � â9, where á9 is a

conjugate of á and â9 is a conjugate of â. Moreover, if

r 2 Q then every conjugate of rá is of the form rá9,

where á9 is a conjugate of á.

For an elementary proof of this, see Pollard and Diamond, Chapter

V, Section 3. Alternatively, (31.1) can be proved easily using some

Galois theory.

31.2 Lemma

Let ÷ be a character of a ®nite group G, and let g 2 G. Then

j÷(g)=÷(1)j < 1, and if

0 , j÷(g)=÷(1)j, 1

then ÷(g)=÷(1) is not an algebraic integer.

Proof Let ÷(1) � d. By Proposition 13.9 we have ÷(g) � ù1 �. . . � ùd , where each ùi is a root of unity, so

÷(g)=÷(1) � (ù1 � : : : � ùd)=d:

Since |÷(g)| � |ù1 � . . . � ùd | < |ù1| � . . . � |ùd | � d, it follows that

j÷(g)=÷(1)j < 1.

Now suppose that ÷(g)=÷(1) is an algebraic integer and

j÷(g)=÷(1)j, 1. We prove that ÷(g) � 0.

Write ã � ÷(g)=÷(1), and let p(x) be the minimal polynomial of ã,

so that

p(x) � xn � anÿ1x nÿ1 � : : : � a1x� a0

where ai 2 Z for all i. By (31.1), each conjugate of ã is of the form

(ù91 � : : :� ù9d)=d

where ù91, . . . , ù9d are roots of unity. Hence each conjugate of ã has


modulus at most 1. It follows that if ë is the product of all the

conjugates of ã (including ã), then

jëj, 1:

But the conjugates of ã are, by de®nition, the roots of the polynomial

p(x), and the product of all these roots is equal to �a0. Thus

ë � �a0:

Since a0 2 Z and |ë| , 1, it follows that a0 � 0. As p(x) is irreducible,

this implies that

p(x) � x,

which in turn forces ã � 0. Thus ÷(g) � 0, and the proof is complete.

j

Burnside's paqb Theorem

We deduce the main result, Theorem 31.4, from another interesting

theorem of Burnside.

31.3 Theorem

Let p be a prime number and let r be an integer with r > 1. Suppose

that G is a ®nite group with a conjugacy class of size pr. Then G is

not simple.

Proof Let g 2 G with |gG | � pr. Since pr . 1, G is not abelian and

g 6� 1. As usual, denote the irreducible characters of G by ÷1, . . . , ÷k ,

and take ÷1 to be the trivial character.

The column orthogonality relations, Theorem 16.4(2), applied to the

columns corresponding to 1 and g in the character table of G, give

1�Xk

i�2

÷i(g)÷i(1) � 0:

Therefore Xk

i�2

÷i(g) .÷i(1)

p� ÿ 1

p:

Now ÿ1=p is not an algebraic integer, by Proposition 22.5. Therefore,

for some i > 2, ÷i(g)÷i(1)= p is not an algebraic integer (see Theorem

22.3). Since ÷i(g) is an algebraic integer (Corollary 22.4), it follows

Burnside's Theorem 363

that ÷i(1)= p is not an algebraic integer; in other words, p does not

divide ÷i(1). Thus

÷i(g) 6� 0 and p 6 j ÷i(1):

As |gG | � pr, this means that ÷i(1) and |gG | are coprime integers, and

so there are integers a and b such that

ajG:CG(g)j � b÷i(1) � 1:

Hence

ajGj÷i(g)

jCG(g)j÷i(1)� b÷i(g) � ÷i(g)

÷i(1):

By Corollaries 22.10 and 22.4, the left-hand side of this equation is an

algebraic integer; and since ÷i(g) 6� 0, it is non-zero. Now Lemma 31.2

implies that

j÷i(g)=÷i(1)j � 1:

Let r be a representation of G with character ÷i. By Theorem

13.11(1), there exists ë 2 C such that

gr � ëI :

Let K � Ker r, so that K is a normal subgroup of G. Since ÷i is not

the trivial character, K 6� G. If K 6� {1} then G is not simple, as

required; so assume that K � {1}, that is, r is a faithful representation.

Since gr is a scalar multiple of the identity, gr commutes with hrfor all h 2 G. As r is faithful, it follows that g commutes with all

h 2 G; in other words,

g 2 Z(G):

Therefore Z(G) 6� {1}. As Z(G) is a normal subgroup of G and

Z(G) 6� G, we conclude that G is not simple. j

We now come to the main result of the chapter, Burnside's Theorem.

31.4 Burnside's paqb Theorem

Let p and q be prime numbers, and let a and b be non-negative

integers with a � b > 2. If G is a group of order paqb, then G is not

simple.

Proof First suppose that either a � 0 or b � 0. Then the order of G is

a power of a prime, so by Lemma 26.1(1) we have Z(G) 6� {1}.


Choose g 2 Z(G) of prime order. Then kgl v G and kgl is not equal to

{1} or G. Hence G is not simple.

Now assume that a . 0 and b . 0. By Sylow's Theorem 30.9, G has

a subgroup Q of order qb. We have Z(Q) 6� {1} by Lemma 26.1(1).

Let g 2 Z(Q) with g 6� 1. Then Q < CG(g), so

jgGj � jG:CG(g)j � pr

for some r. If pr � 1 then g 2 Z(G), so Z(G) 6� {1} and G is not

simple as before. And if pr . 1 then G is not simple, by Theorem

31.3. j

In fact Burnside's paqb Theorem leads to a somewhat more informa-

tive result about groups of order paqb:

(31.5) Every group of order paqb is soluble.

Here, by a soluble group we mean a group G which has subgroups

G0, G1, . . . , Gr with

1 � G0 , G1 , : : : , Gr � G

such that for 1 < i < r, Giÿ1 v Gi and the factor group Gi=Giÿ1 is

cyclic of prime order.

We sketch a proof of (31.5), using induction on a � b. The result is

clear if a � b < 1, so assume that a � b > 2 and let G be a group of

order paqb. By Burnside's Theorem 31.4, G has a normal subgroup H

such that H is not {1} or G. Both H and the factor group G=H have

order equal to a product of powers of p and q, and these orders are

less than paqb. Hence by induction, H and G=H are both soluble.

Therefore there are subgroups

1 � G0 v G1 v : : : v Gs � H ,

1 � Gs=H v Gs�1=H v : : : v Gr=H � G=H

with all factor groups Gi=Giÿ1 of prime order. Then the series

1 � G0 v G1 v : : : v Gr � G

shows that G is soluble.


1. If G has a conjugacy class of size pr ( p prime, r > 1), then G is

not simple.

Burnside's Theorem 365

2. If |G| � paqb ( p, q primes, a� b > 2), then G is not simple.


1. Show that a non-abelian simple group cannot have an abelian

subgroup of prime power index.

2. Prove that if G is a non-abelian simple group of order less than 80,

then |G| � 60.

(Hint: use Exercise 13.8.)


367

32

An application of representation theoryto molecular vibration

Representation theory is used extensively in many of the physical

sciences. Such applications come about because every physical system

has a symmetry group G, and certain vector spaces associated with the

system turn out to be RG-modules. For example, the vibration of a

molecule is governed by various differential equations, and the symme-

try group of the molecule acts on the space of solutions of these

equations. It is on this application ± the theory of molecular vibrations

± that we concentrate in this ®nal chapter. In order to keep our

treatment elementary, we stay within the framework of classical mech-

anics throughout. (Quantum mechanical effects can be incorporated

subsequently, but we shall not go into this ± for an account, consult

the book by D. S. Schonland listed in the Bibliography.)

Symmetry groups

Let V be R2 or R3, and for v, w 2 V, let d(v, w) denote the distance

between v and w ± in other words, if v � (x1, x2, . . .) and

w � (y1, y2, . . .), then

d(v, w) �r X

(xi ÿ yi)2

!:

An isometry of V is an invertible endomorphism W of V such that

d(vW, wW) � d(v, w) for all v, w 2 V :

The set of all isometries of V forms a group under composition, called

the orthogonal group of V, and denoted by O(V).

Any rotation of R3 about an axis through the origin is an example

of an isometry; so is any re¯ection in a plane through the origin. The

endomorphism ÿ1R3 which sends every vector v to ÿv is another

example of an element in the orthogonal group O(R3). It turns out that

the composition of two rotations is again a rotation, and that for every

isometry g in O(R3), either g or ÿg is a rotation (see Exercise 32.1).

The orthogonal group O(R3) therefore contains a subgroup of index 2

which consists of the rotations. The same is true of the group O(R2).

If Ä is a subset of V , where V � R2 or R3, then we de®ne G(Ä) to

be the set of isometries which leave Ä invariant ± that is,

G(Ä) � fg 2 O(V ): Äg � Äg(where Äg � {vg: v 2 Ä}). Then G(Ä) is a subgroup of O(V ), and is

called the symmetry group of Ä. The subgroup of G(Ä) consisting of

the rotations in G(Ä) is called the rotation group of Ä. The index of

the rotation group of Ä in the symmetry group G(Ä) is 1 or 2.

32.1 Example

Let V � R2, and let Ä be a regular n-sided polygon, with n > 3,

centred at the origin. The symmetry group of Ä is easily seen to be

the dihedral group D2n, which was de®ned as a group of n rotations

and n re¯ections preserving Ä (see Example 1.1(3)).

Now let V � R3, and again let Ä be a regular n-sided polygon

(n > 3) centred at the origin. This time, G(Ä) � D2n 3 C2; the extra

elements arise because there is an isometry which ®xes all points of Ä,

namely the re¯ection in the plane of Ä.

32.2 Example

Let Ä be a regular tetrahedron in R3 centred at the origin:


Label the corners of the tetrahedron 1, 2, 3, 4. We claim that each

permutation of the numbers 1, 2, 3, 4 corresponds to an isometry of Ä.

For example, the 2-cycle (1 2) corresponds to a re¯ection in the plane

which contains the origin and the edge 34; similarly each 2-cycle

corresponds to a re¯ection. Since S4 is generated by the 2-cycles, each

of the 24 permutations of 1, 2, 3, 4 corresponds to an isometry, as

claimed.

No non-identity endomorphism of R3 ®xes all the corners of Ä,

since Ä contains three linearly independent vectors. Therefore we have

found all the isometries, and G(Ä) � S4.

Notice that the rotation group of Ä is isomorphic to A4; for

example, (1 2)(3 4) corresponds to a rotation through ð about the axis

through the mid-points of the edges 12 and 34, and (1 2 3) corresponds

to a rotation through 2ð=3 about the axis through the origin and the

corner 4.

Finally, observe that the group G(Ä) is unchanged if we take Ä to

consist of just the four corners of the tetrahedron.

32.3 Example

In this example we describe the symmetry groups of the molecules

H2O (water), CH3Cl (methyl chloride) and CH4 (methane). The symme-

try group of a molecule is de®ned to be the group of isometries which

not only preserve the position of the molecule in space, but also send

each atom to an atom of the same kind.

The shapes of the three molecules are as follows.

An application of representation theory to molecular vibration 369

We always assume that the centroid of our molecule lies at the origin

in R3.

The CH4 molecule has four hydrogen atoms at the corners of a

regular tetrahedron, and a carbon atom at the centre of the tetrahedron.

So the symmetry group of the molecule CH4 is equal to the symmetry

group of the tetrahedron, as given in Example 32.2. This group is

isomorphic to S4, permuting the four hydrogen atoms among them-

selves and ®xing the carbon atom.

As for the CH3Cl molecule, this possesses a rotation symmetry a of

order 3 about the vertical axis, and three re¯ection symmetries in the

planes containing the C, Cl and one of the H atoms. If b is one of

these re¯ections, then the symmetry group is

f1, a, a2, b, ab, a2bgand is isomorphic to S3, permuting the three H atoms and ®xing the C

and Cl atoms.

Finally, the H2O molecule possesses two re¯ection symmetries, one

in the plane of the molecule, and one in a plane perpendicular to this

one passing through the O atom; and it has a rotation symmetry of

order 2. Hence the symmetry group is isomorphic to C2 3 C2.

Vibration of a physical system

We prepare for a description of the general problem with an example.

32.4 Example

Suppose we have a spring stretched between two points P and Q on a

smooth horizontal table, with equal masses m attached at the points of

trisection of the spring:

The masses are displaced slightly, and released. What can we say about

the subsequent motion of the system?

To investigate this problem, we let x1 and x2 be the displacements of

the two masses at time t. We measure x1 from left to right, and x2

from right to left, as indicated in the ®gure above. Let k be the


stiffness of the spring ± in other words, if the extension in the spring

is x, then the restoring force is kx.

The spring pulls the left-hand mass towards P with force kx1 and

towards Q with force ÿk(x1 � x2). By dealing with the right-hand mass

similarly, we obtain the following equations of motion of the system:

m�x1 � ÿkx1 ÿ k(x1 � x2) � ÿ2kx1 ÿ kx2,

m�x2 � ÿkx2 ÿ k(x1 � x2) � ÿkx1 ÿ 2kx2,

where �xi denotes the second derivative of xi with respect to t.

These are second order linear differential equations in two unknowns

x1 and x2, so the general solution involves four arbitrary constants. We

shall ®nd the general solution, using a method which can be applied in

a much wider context.

Write x � (x1, x2), �x � (�x1, �x2) and q � k=m. Then the equations of

motion are equivalent to the matrix equation

�x � xA, where A � ÿ2q ÿq

ÿq ÿ2q

� �:(32:5)

Notice that A is symmetric. Hence the eigenvalues of A are real, and

A has two linearly independent eigenvectors. It is this property which

we wish to emphasize and exploit in the present example.

Before we explicitly ®nd the eigenvectors of A, let us explain why

they allow us to solve the equation of motion (32.5).

Suppose that u is an eigenvector of A, with eigenvalue ÿù2. For an

arbitrary constant â, let

x � sin (ùt � â) u:

Then

�x � ÿù2 sin (ùt � â) u

� sin (ùt � â) uA (since uA � ÿù2u)

� xA:

Thus x is a solution of the equation of motion. If u1 and u2 are

linearly independent eigenvectors of A, with eigenvalues ÿù21 and

ÿù22, respectively, then

á1 sin (ù1 t � â1) u1 � á2 sin (ù2 t � â2) u2


is a solution of the equation of motion which involves four arbitrary

constants á1, á2, â1, â2, so it is the general solution.

We now adopt this line of attack in the problem to hand. For the

matrix given in (32.5), the eigenvalues are ÿ3q and ÿq, with corre-

sponding eigenvectors (1, 1) and (1, ÿ1). Therefore the general solution

of the equation of motion (32.5) is

á1 sin (p

(3q) t � â1) (1, 1)� á2 sin (p

q . t � â2) (1, ÿ1):

The solutions which involve just one eigenvector of A are called the

normal modes of vibration. They are as follows.

sin (p

(3q) t � â1) (1, 1)Mode 1:

Here, x1 � x2 � sin (p

(3q) t � â1) and the vibration is

sin (p

q . t � â2) (1, ÿ1)Mode 2:

Here, x1 � ÿx2 � sin (p

q . t � â2) and the vibration is

The general molecular vibration problem

Suppose we have a molecule consisting of n atoms which vibrate under

internal forces. At the equilibrium position of each atom, we assign

three coordinate axes, which we use to measure the displacement of

the atom. Thus, the state of the molecule at a given time is described

by a vector in the 3n-dimensional vector space R3n. We assume that

the internal forces are linear functions of the displacements. It follows

that when we apply Newton's Second Law of Motion, we obtain

equations which may be expressed in the form

�x � xA:(32:6)

(Compare (32.5).) Here x is the row vector in R3n which measures the

displacements of all the atoms, and A is a 3n 3 3n matrix with real

entries which are determined by the internal forces.

Assume, for the moment, that at each atom the three coordinate axes


which we have chosen are at right angles to each other. It can be shown,

from physical considerations, that in this special case the matrix A is

symmetric. In particular, A has real eigenvalues, and A has 3n linearly

independent eigenvectors. Now, the effect of changing coordinate axes is

merely to replace A by a matrix which is conjugate to A. Therefore we

have the following proposition, for the general case, where our chosen

coordinate axes are not necessarily at right angles to each other.

32.7 Proposition

All the eigenvalues of A are real, and A has 3n linearly independent

eigenvectors.

To solve the equation of motion (32.6), we look for normal modes

of the system, which we de®ne next.

32.8 De®nition

A normal mode of vibration for our molecule is a vector in R3n of one

of the following forms:

sin (ùt � â) u (â constant)(1)

where ÿù2 is a non-zero eigenvalue of A and u is a corresponding

eigenvector;

(t � â) u (â constant)(2)

where u is an eigenvector of A corresponding to the eigenvalue 0.

32.9 Proposition

Each normal mode of vibration is a solution of the equation of motion

(32.6), and for this solution all the atoms vibrate with the same

frequency. The general solution of the equation of motion is a linear

combination of the normal modes of vibration.

Proof If uA � ÿù2 A and x � sin (ùt � â) u, then

�x � ÿù2 sin (ùt � â) u � sin (ùt � â) uA � xA:

If uA � 0 and x � (t � â)u, then

�x � 0 � (t � â)uA � xA:

This proves that the normal modes of vibration are solutions of the


equation of motion (32.6). By construction, all the atoms vibrate with

the same frequency (namely, ù or 0) in a normal mode.

Note that A can have no strictly positive eigenvalue; for if ë were

such an eigenvalue, with eigenvector u, then x � epë tu would be a

solution to the equation of motion, which is nonsense. Therefore there

exist 3n linearly independent normal modes, by Proposition 32.7. Since

each normal mode involves an arbitrary constant, the general linear

combination of normal modes involves 6n arbitrary constants, so it is

the general solution to the equation of motion (32.6) (as (32.6) consists

of second order differential equations in 3n unknowns). j

Proposition 32.9 reduces the problem of solving the equations of

motion to that of ®nding all the eigenvalues and eigenvectors of the

3n 3 3n matrix A. However, this can be a huge and unwieldy task if

it is attempted directly ± even writing down the matrix A for a given

molecule can be a painful operation!

The symmetry group of the molecule and its representation theory

can often be used to simplify greatly the calculation of the eigenvectors

of A, and we shall describe a method for doing this.

Use of the symmetry group

We continue the discussion of the previous section. Let G be the

symmetry group of the molecule in question. Since G permutes the

atoms among themselves, each element of G acts as an endomorphism

of the space R3n of displacement vectors. Thus, R3n is an RG-module.

32.10 Example

Let g be the rotation of order 2 of the H2O molecule:

Assign coordinate axes at the initial positions of each atom as shown,

and for 1 < i < 9, let vi denote a unit vector along coordinate axis i.

Then g ®xes v1, negates v2 and v3, interchanges v4 and v7, and


interchanges v5 and v6 with the negatives of v8 and v9. Therefore g

acts on R9 as follows:

(x1, x2, x3, x4, x5, x6, x7, x8, x9)g

� (x1, ÿx2, ÿx3, x7, ÿx8, ÿx9, x4, ÿx5, ÿx6):

We return to the general set-up. The equations of motion are �x � xA,

and we are trying to ®nd the eigenvectors of A. The eigenspace for the

eigenvalue ë of A is, by de®nition,

fx 2 R3n: xA � ëxg:We can now present the crucial proposition which allows us to

exploit the symmetry group G of our molecule. In effect, it tells us

that the function x! xA (x 2 R3n) is an RG-homomorphism from R3n

to itself.

32.11 Proposition

For all g 2 G and x 2 R3n,

(xg)A � (xA)g,

and the eigenspaces of A are RG-submodules of R3n.

Proof Let ÿù2 be a non-zero eigenvalue of A, and let v be a

corresponding eigenvector. Then v speci®es the directions and relative

magnitudes of the displacements of the atoms from the equilibrium

position when the molecule is vibrating in a normal mode of frequency

ù. For all g in G, vg must also specify the directions and relative

magnitudes of displacements in a normal mode of frequency ù, since

the relative con®guration of the atoms is unaltered by applying g.

Therefore, vg is an eigenvector of A, with eigenvalue ÿù2. This shows

that the eigenspace for ÿù2 is an RG-submodule of R3n. A similar

argument applies to the eigenspace for the eigenvalue 0.

Choose a basis of R3n which consists of eigenvectors of A (see

Proposition 32.7), and let g 2 G. For all vectors v in the basis,

vA � ëv for some ë 2 R, and

(vg)A � ë(vg) � (ëv)g � (vA)g:

Hence (xg)A � (xA)g for all x 2 R3n. j

The idea now is to use representation theory to express the RG-

module R3n as a direct sum of irreducible RG-submodules, and hence


to determine the eigenspaces of A, and the normal modes of the

molecule.

We can use character theory to see which irreducible RG-modules

are contained in R3n; and if ÷ is the character of an irreducible RG-

module which occurs, then the elementXg2G

÷(gÿ1)g

sends R3n onto the sum of those irreducible RG-submodules of R3n

which have character ÷ (see (14.27)). (This procedure sometimes needs

to be modi®ed, since the character of the RG-module R3n might

contain an irreducible character which cannot be realized over R ± but

in practice, problems like this are uncommon.)

32.12 De®nition

Suppose that ÷ is the character of an irreducible RG-module. Let V÷

denote the sum of those irreducible RG-submodules of R3n which have

character ÷. We call V÷ a homogeneous component of R3n.

The problem of ®nding the eigenspaces of A is considerably simpli-

®ed as a consequence of our next proposition.

32.13 Proposition

Each homogeneous component V÷ of R3n is A-invariant ± that is,

xA 2 V÷ for all x 2 V÷:

Proof By Maschke's Theorem we may write R3n � V÷ � W for some

RG-module W, and no RG-submodule of W has character ÷. The

function

å: v� w! w (v 2 V÷, w 2 W )

is an RG-homomorphism. Therefore, by Proposition 32.11, the function

x! (xA)å (x 2 V÷)

is an RG-homomorphism from V÷ into W. By Proposition 11.3, this

function is zero, so xA 2 V÷ for all x 2 V÷. (Although Proposition 11.3

is stated in terms of CG-modules, its proof works equally well for

RG-modules ± compare Exercise 23.8.) j


32.14 Corollary

If V÷ is an irreducible RG-module, then all the non-zero vectors in V÷

are eigenvectors of A.

Proof (Compare the proof of Schur's Lemma.) Since V÷ is A-invariant,

we may choose v 2 V÷ such that v is an eigenvector of A, with

eigenvalue ë, say. Then the intersection of V÷ with the eigenspace for ëis a non-zero RG-submodule of V÷, so it must equal V÷. j

We now summarize the steps in the procedure for ®nding the normal

modes of vibration of a given molecule.

32.15 Summary

(1) Assign three coordinate axes at each of the n atoms of the

molecule, to obtain R3n.

(2) Calculate the symmetry group G of the molecule. Then R3n is an

RG-module.

(3) Calculate the character ÷ of the RG-module R3n and express ÷ as a

linear combination of the irreducible characters of G.

(4) Express R3n as a direct sum of homogeneous components. This can

be done by applying the elementP

g2G÷i(gÿ1)g to R3n for each

irreducible character ÷i of G which appears in ÷, or by some other

method.

(5) Consider, in turn, each homogeneous component V÷iof R3n, and

®nd the eigenvectors of A in V÷i(see Proposition 32.13). This involves

no extra work if V÷iis irreducible (see Corollary 32.14). If V÷i

is

reducible, then see Remark 32.19 below, or Exercise 32.7, to make

further progress.

(6) If v is an eigenvector of A, with eigenvalue ÿù2, then

sin (ùt � â) v (or (t � â)v if ù � 0)

is a normal mode, where â is an arbitrary constant. It is usually

necessary to know the equations of motion in order to determine the

frequency ù.

This programme can often be successfully completed, as we shall

illustrate in the examples which make up the rest of this chapter.


32.16 Example

We ®rst return to Example 32.4, with the spring and two vibrating

masses:

The symmetry group of this system is G � hg: g2 � 1i, where g is the

re¯ection in the mid-point of PQ. The displacement vectors (x1, x2)

form an RG-module R2.

Since (x1, x2)g � (x2, x1), the RG-submodules of R2 are sp (u1) and

sp (u2), where

u1 � (1, 1), u2 � (1, ÿ1):

It follows that the normal modes of the system are given by

sin (ù1 t � â1)(1, 1), sin (ù2 t � â2)(1, ÿ1),

where â1, â2 are constants and ù1, ù2 are the frequencies. This agrees

with the conclusion of Example 32.4.

Notice that we have determined the normal modes of vibration (but

not their frequencies) using the symmetry group alone.

32.17 Example

Consider a hypothetical triatomic molecule, where the three identical

atoms are at the corners of an equilateral triangle. For simplicity, we

consider only vibrations of the molecule in the plane, so we assign two

displacement coordinates to each atom. We choose to take our axes

along the edges of the triangle, as shown, as this eases the calcula-

tions:

Thus the position of the molecule is given by a vector (x1, . . . , x6) in

R6, where xi is the displacement along axis i (1 < i < 6).

The symmetry group (in two dimensions) of the molecule is the

dihedral group D6, generated by a rotation a of order 3 and a re¯ection


b (see Example 32.1). It is easy to work out the action of each element

of D6 on R6. For example, if b is the re¯ection which ®xes the top

atom, then

(x1, x2, x3, x4, x5, x6)b � (x2, x1, x6, x5, x4, x3):

We want to express the RD6-module R6 as a direct sum of

irreducible RD6-modules. To do this, we ®rst calculate the character ÷of the module. Since the rotation a does not ®x any of the atoms,

÷(a) � 0. And from the action of b given above, we see that ÷(b) � 0.

Thus the values of ÷ are

By Section 18.3, the character table of D6 is

Hence ÷ � ÷1 � ÷2 � 2÷3.

Thus, we seek to express R6 as a direct sum of RD6-submodules

with characters ÷1, ÷2, ÷3 and ÷3.

As a matter of notation, if v1, v2, v3 are 2-dimensional displacement

vectors for the three atoms, then we represent the displacement vector

(v1, v2, v3) 2 R6 pictorially by the diagram

We ®rst calculate the normal modes of the form (t � â)v, corre-

sponding to the eigenspace of A for eigenvalue 0. These include the

rotation and translation modes, which occur for every molecule.

1 a b

÷ 6 0 0

1 a b

÷1 1 1 1÷2 1 1 ÿ1÷3 2 ÿ1 0


Rotation mode In this mode, the molecule rotates with constant angular

velocity about the centre. The mode is given by (t � â)v, where

v � (1, ÿ1, 1, ÿ1, 1, ÿ1); pictorially,

We call sp (v) the rotation submodule of R6. If ÷R is the character of

sp (v), then

÷R(1) � 1, ÷R(a) � 1, ÷R(b) � ÿ1,

and so ÷R � ÷2. Indeed, sp (v) � R6å2, where

å2 �Xg2D6

÷2(gÿ1)g � 1� a� a2 ÿ bÿ abÿ a2b

(compare (14.27)).

Translation modes These are modes in which all atoms move in the

same constant direction with the same constant speed. The modes are

of the form (t � â)v, where v is a vector in the span of v1, v2 and v3,

these vectors being given pictorially by

(thus v1 � (ÿ1, 1, 0, ÿ1, 1, 0), v2 � (1, 0, ÿ1, 1, 0, ÿ1), v3 � (0, ÿ1,

1, 0, ÿ1, 1)).

Since v1 � v2 � v3 � 0, the subspace sp (v1, v2, v3) has dimension 2.

It is clearly an RD6-submodule of R6, and is called the translation

submodule; it does not contain the rotation submodule. Thus the

character of the translation submodule is part of ÷ ÿ ÷2 � ÷1 � 2÷3, so

the character must be ÷3.


Vibratory modes The remaining normal modes correspond to eigen-

spaces of the matrix A with non-zero eigenvalues, and are called

vibratory modes. The sum of these eigenspaces forms an RD6-submo-

dule R6vib of R6 (by Proposition 32.11), with character ÷vib, where

÷vib � ÷ÿ (÷2 � ÷3) � ÷1 � ÷3:

In particular, R6vib has dimension 3. Since no mode in R6

vib can have

any translation component, if w 2 R6vib then the total component of w

in each direction is zero; moreover, R6vib does not contain the rotation

submodule, so that total moment of each vector in R6vib about the

centre is zero. These constraints imply three independent linear equa-

tions in the coordinates of the vectors in R6vib, and since R6

vib has

dimension 3, every vector in R6 which satis®es these equations lies in

R6vib. Therefore a basis of R6

vib is u1, u2, u3, where

Clearly sp (u1 � u2 � u3) is an RD6-submodule of R6vib with charac-

ter ÷1. The vibratory mode given by u1 � u2 � u3 is sometimes called

the expansion±contraction mode (you will see the reason for this name

in the picture in (32.18(3)) below).

Finally, since D6 permutes the vectors u1, u2, u3 among themselves,

it is easy to see that sp (u1 ÿ u2, u1 ÿ u3) is an RD6-submodule of

R6vib. Its character is ÷3 and it gives us the last eigenspace for the

matrix A.

Our calculation of the normal modes is now complete, and we

summarize our ®ndings below.


(32.18) (1) Rotation mode:

(2) Translation modes: linear combinations of

(3) Vibratory mode: expansion±contraction mode

(4) Vibratory modes: linear combinations of

(We have chosen 2u2 ÿ u1 ÿ u3, 2u1 ÿ u2 ÿ u3 as the basis for the

vibratory modes in (4) merely because these modes look simpler than

u1 ÿ u2, u1 ÿ u3 pictorially.)

We emphasize that we have found the normal modes of vibration

without explicit knowledge of the equations of motion. In order to


check our results, we now calculate the equations of motion.

Let m be the mass of each atom, and assume that the magnitude of

the force between two atoms is k times the decrease in distance

between them.

For a general displacement (x1, x2, x3, x4, x5, x6), denote the new

positions of the atoms by P9, Q9, R9. From the diagram, the difference

in length between QR and Q9R9 is

(x4 � x5)� 12(x3 � x6):

(We always assume that x1, : : : , x6 are small compared with the

distance between the atoms, so that we may ignore second order

terms.)

Similarly,

PRÿ P9R9 � (x1 � x6)� 12(x2 � x5),

PQÿ P9Q9 � (x2 � x3)� 12(x1 � x4):

Hence the force on the molecule at P in the direction of the ®rst

coordinate axis is

ÿk(PRÿ P9R9) � ÿk(x1 � x6)ÿ 12k(x2 � x5):

Therefore,m

k�x1 � ÿ(x1 � x6)ÿ 1

2(x2 � x5):

In the same way,

m

k�x2 � ÿ(x2 � x3)ÿ 1

2(x1 � x4),


and we obtain similar equations for �x3, . . . , �x6. The matrix A for

which �x � xA is therefore given by

A � ÿk

m

1 1=2 1=2 0 0 1

1=2 1 1 0 0 1=2

0 1 1 1=2 1=2 0

0 1=2 1=2 1 1 0

1=2 0 0 1 1 1=2

1 0 0 1=2 1=2 1

0BBBBBB@

1CCCCCCA:

You should check that the vectors which we gave in (32.18) are indeed

eigenvectors of A.

32.19 Remark

In Example 32.17, the character ÷ of the RG-module R6 was given by

÷ � ÷1 � ÷2 � 2÷3:

All the non-zero vectors in the homogeneous components for ÷1 and ÷2

gave normal modes, since these homogenous components were irreduci-

ble (see Corollary 32.14). The homogeneous component V÷3for ÷3 was

reducible, but we were able to write it as a sum of two subspaces of

eigenvectors (those appearing in (32.18)(2) and (4)) because V÷3\ R6

vib

was an A-invariant RG-submodule of V÷3different from {0} and V÷3

.

This illustrates a method which sometimes helps to deal with reducible

homogeneous components. In our next example, the situation is more

complicated.

32.20 Example

We analyse the normal modes for the methane molecule CH4.

We determined the symmetry group G of this molecule in Example

32.2, where we found that G is isomorphic to S4. Label the corners of


the tetrahedron 1, 2, 3, 4, as shown below, and identify G with S4;

thus, for example, the rotations about the vertical axis through 1 are

written as1, (2 3 4), (2 4 3):

In order fully to exploit the symmetry of the methane molecule, at

each hydrogen atom we choose displacement axes along the edges of

the tetrahedron. Let v12, v13, v14 be unit vectors at corner 1 in the

directions of the edges 12, 13, 14, respectively; similarly, let v21, v23,

v24 be unit vectors at corner 2 in the directions of the edges 21, 23,

24, and so on, giving twelve vectors vij, in all.

We now introduce a new idea, by taking four unit vectors w1, w2,

w3 and w4 at the carbon atom, with wi pointing towards corner i

(1 < i < 4). Since w1 � w2 � w3 � w4 � 0, these four vectors span a

3-dimensional space.

Let V be the vector space over R with basis

v12, v13, v14, v21, v23, v24, v31, v32, v34, v41, v42, v43,

and let W be the vector space over R spanned by w1, w2, w3, w4. Then

V � R12, W � R3 and V and W are RG-modules. Our main task is to

®nd RG-submodules of R15 � V � W.

The action of G on V is easy to describe; for g in G, we have

vijg � vig,jg for all i, j:

Thus, G permutes our twelve basis vectors of V, and we can quickly

calculate the character ÷ of V:

1 (1 2) (1 2 3) (1 2)(3 4) (1 2 3 4)

÷ 12 2 0 0 0


For example, (1 2) ®xes the basis vectors v34 and v43 only; all the basis

vectors are moved by (1 2 3); and so on.

The group G acts on W as follows; for g in G, we have

wig � wig (1 < i < 4):

After recalling that w1 � . . . � w4 � 0, it is easy to calculate the

character ö of W ± it is

1 (1 2) (1 2 3) (1 2)(3 4) (1 2 3 4)

ö 3 1 0 ÿ1 ÿ1

By Section 18.1, the character table of S4 is as shown at the top of

p. 387. We ®nd that

÷ � ÷1 � ÷3 � 2÷4 � ÷5,

ö � ÷4:

By applying the elementsXg2G

÷i(gÿ1)g (i � 1, 3, 5, 4)

to R15, we can ®nd RG-submodules with characters ÷1, ÷3, ÷5 and 3÷4

(see (14.27)).

The RG-submodule W1 with character ÷1 is spanned byXi, j

vij

and this gives the expansion±contraction normal mode:

We next describe the RG-submodule W5 with character ÷5. Let

p1 � (v23 ÿ v32)� (v34 ÿ v43)� (v42 ÿ v24),

p2 � (v31 ÿ v13)� (v14 ÿ v41)� (v43 ÿ v34),


p3 � (v12 ÿ v21)� (v41 ÿ v14)� (v24 ÿ v42),

p4 � (v21 ÿ v12)� (v13 ÿ v31)� (v32 ÿ v23):

The vector pi gives a rotation about the axis through the corner i and

the centroid of the tetrahedron.

It should be clear from the pictures that for all i with 1 < i < 4 and


1 (1 2) (1 2 3) (1 2)(3 4) (1 2 3 4)

÷1 1 1 1 1 1÷2 1 ÿ1 1 1 ÿ1÷3 2 0 ÿ1 2 0÷4 3 1 0 ÿ1 ÿ1÷5 3 ÿ1 0 ÿ1 1


all g in G, we have pig � � pj for some j. Therefore, if we let

W5 � sp ( p1, p2, p3, p4), then W5 is an RG-submodule of V. Now

p1 � p2 � p3 � p4 � 0, so dim W5 � 3. Check that the character of W5

is ÷5. The RG-module W5 is the rotation submodule. (Compare, for

example, the picture for p4 with the picture for the rotation vector v in

Example 32.17.)

Now we construct the RG-submodule W3 of V with character ÷3. Let

q1 � (v12 � v21)� (v34 � v43)ÿ (v13 � v31)ÿ (v24 � v42),

q2 � (v13 � v31)� (v24 � v42)ÿ (v14 � v41)ÿ (v23 � v32),

q3 � (v14 � v41)� (v23 � v32)ÿ (v12 � v21)ÿ (v34 � v43):

(Each qi is associated with an `opposite pair of edges'.)

For all i with 1 < i < 4 and g in G, we have qi g � �q j for some j.

Let W3 � sp (q1, q2, q3). Then W3 is an RG-submodule of V. Since

q1 � q2 � q3 � 0, the dimension of W3 is 2; its character is ÷3.


In the RG-submodules W1, W5 and W3 which we have found so far,

all the non-zero vectors are eigenvectors of A, by Corollary 32.14.

We now come to the homogeneous component (V � W)÷4of R15.

De®ne the vectors r1, r2, r3, r4 by

r1 � (v12 � v21)� (v13 � v31)� (v14 � v41)

ÿ (v23 � v32)ÿ (v24 � v42)ÿ (v34 � v43),

r2 � (v12 � v21)� (v23 � v32)� (v24 � v42)

ÿ (v13 � v31)ÿ (v14 � v41)ÿ (v34 � v43),

r3 � (v13 � v31)� (v23 � v32)� (v34 � v43)

ÿ (v12 � v21)ÿ (v14 � v41)ÿ (v24 � v42),

r4 � (v14 � v41)� (v24 � v42)� (v34 � v43)

ÿ (v13 � v31)ÿ (v12 � v21)ÿ (v23 � v32):

(The vector ri is associated with corner i.)


For each g in G and i with 1 < i < 4, we have ri g � rig. Thus G

permutes the vectors r1, r2, r3, r4 among themselves. Note that

r1 � r2 � r3 � r4 � 0, so r1, r2, r3, r4 span a 3-dimensional RG-

submodule W4 of V. The character of W4 is ÷4 (see Proposition 13.24).

Next, de®ne the vectors s1, s2, s3, s4 by

s1 � (v12 � v13 � v14)ÿ (v21 � v31 � v41),

s2 � (v21 � v23 � v24)ÿ (v12 � v32 � v42),

s3 � (v31 � v32 � v34)ÿ (v13 � v23 � v43),

s4 � (v41 � v42 � v43)ÿ (v14 � v24 � v34):

We have

sig � sig (g 2 G, 1 < i < 4),

s1 � s2 � s3 � s4 � 0,


and s1, s2, s3, s4 span a 3-dimensional RG-submodule W94 of V with

character ÷4.

Now recall that w1, w2, w3, w4 span W; we have

wig � wig (g 2 G, 1 < i < 4),

w1 � w2 � w3 � w4 � 0,

and the character of W is ÷4. The sum of W4, W 94 and W is direct, so

(V � W )÷4� W4 � W 94 � W :

We now break off temporarily from studying the methane molecule,

in order to deal with the easier case of a molecule with identical atoms

at the corners of the tetrahedron, and no central atom. In this case, the

space W does not enter our calculations, and we can decompose

V÷4� W4 � W 94 in the following way.

(32.21) (1) The vectors r1 ÿ 2s1, r2 ÿ 2s2, r3 ÿ 2s3, r4 ÿ 2s4

span the 3-dimensional space of translation modes.

(2) The vectors r1, r2, r3, r4 span the subspace

V÷4\ R12

vib of V÷4, and so they give the ®nal 3-dimen-

sional space of eigenvectors (see Remark 32.19). The

normal mode (sinùt)r1 is sometimes called an `um-

brella mode'. To see why, look at the picture of the

vector r1!

We return now to the methane molecule.

The task in front of us is to ®nd the eigenvectors of A which lie in

(V � W )÷4� W4 � W 94 � W :

The solution of this problem depends, in fact, upon the constants which

appear in the equations of motion, so we cannot complete the work

using only representation theory. Since dim (V � W )÷4� 9, it is very

dif®cult to calculate the eigenvectors of A in (V � W )÷4directly. We

shall therefore press on and explain how to reduce the dif®culty to that

of calculating the eigenvectors of a 3 3 3 matrix.

Let H be the subgroup of S4 consisting of those permutations which

®x the number 1, and let

U1 � fv 2 (V � W )÷4: vh � v for all h 2 Hg:

Since (vh)A � (vA)h for all v 2 V÷4and all h 2 H, it follows that U1

is A-invariant.


We ®nd that

h3÷4 # H , 1 HiH � 3,

and so dim U1 � 3. But for all h 2 H, r1 h � r1, s1 h � s1 and

w1 h � w1. Therefore

U1 � sp (r1, s1, w1):

Once the equations of motion, and hence the matrix A, have been

calculated, it is possible to calculate the 3 3 3 matrix B of the action

of A on r1, s1, w1 (see Exercise 32.5). The eigenvectors of B then give

three eigenvectors of A.

Better still,

r1(1 2) � r2, s1(1 2) � s2, w1(1 2) � w2,

and since A commutes with the action of G, the space U2, de®ned by

U2 � sp (r2, s2, w2)

is A-invariant, and the matrix of A acting on r2, s2, w2 is again B. A

similar remark applies to U3, where

U3 � sp (r3, s3, w3):

Therefore, the process of calculating the eigenvectors of the 3 3 3

matrix B gives nine eigenvectors of A which form a basis of

(V � W )÷4.

One eigenvector of A acting on r1, s1, w1 is easy to ®nd, namely the

translation vector

r1 ÿ 2s1 � 3 cos Ww1,

where W is the angle between an edge of the tetrahedron and the line

joining a corner on the edge to the centroid. Thus we obtain the

translation submodule sp (r1 ÿ 2s1 � 3 cos Ww1, r2 ÿ 2s2 � 3 cos Ww2,

r3 ÿ 2s3 � 3 cos Ww3).

By means of representation theory, we have therefore reduced the

initial problem of ®nding the eigenvectors of a 15 3 15 matrix A to

that of calculating two of the eigenvectors of a 3 3 3 matrix. It is hard

to imagine a more spectacular application of representation theory with

which to conclude our text.



1. The symmetry group G of a molecule with n atoms consists of

those distance-preserving endomorphisms of R3 which send each

atom to an atom of the same kind.

2. The equations of motion of the molecule have the form

�x � xA,

where x 2 R3n and the 3n 3 3n matrix A has 3n linearly indepen-

dent eigenvectors.

3. If u is an eigenvector of A, with eigenvalue ÿù2, then x �sin (ùt � â)u (or x � (t � â)u if ù � 0) is a solution of the equa-

tions of motion, and is called a normal mode. All solutions are

linear combinations of normal modes.

4. The space R3n of displacement vectors is an RG-module. The action

of any g 2 G on R3n commutes with A.

5. To determine the eigenvectors of A (and hence all solutions of the

equations of motion), it is suf®cient to ®nd the eigenvectors of A

restricted to each homogeneous component V÷iof the RG-module

R3n. If V÷ iis irreducible, then all non-zero vectors in V÷i

are

eigenvectors of A.


1. Suppose that b 2 O(R3), and let e1 � (1, 0, 0), e2 � (0, 1, 0),

e3 � (0, 0, 1).

(a) Show that the matrix B of b with respect to the basis e1, e2, e3

of R3 satis®es BBt � I (where Bt denotes the transpose of B).

Deduce that det B � �1.

(b) Let C � (det B)B. Prove successively that

(i) C has a real eigenvalue,

(ii) C has a positive eigenvalue,

(iii) 1 is an eigenvalue of C.

(c) Deduce from (a) and (b) that b is a rotation if det B � 1, and

ÿb is a rotation otherwise.

(d) Show that if b is a rotation through an angle ö about some

axis, then tr B � 1 � 2 cosö.

2. Suppose that G is the symmetry group of some molecule in R3.


Show that the sum of the character ÷T of the translation submodule

and the character ÷R of the rotation submodule has value at g 2 G

given by

(÷T � ÷R)(g) �2(1� 2 cosö), if g is a rotation

through angle öabout some axis,

0, if g is not a rotation:

8>><>>:3. Consider the triatomic molecule studied in Example 32.17. Take the

axes for the displacement coordinates as shown below:

Calculate the equations of motion �x � xA with respect to these axes,

and verify that A is symmetric. (See the paragraph before Proposi-

tion 32.7.)

4. Consider the space spanned by the vectors r1, r2, r3, r4 given in

Example 32.20. Find a basis for this space which is simpler than

the one which we have used. What property of r1, r2, r3, r4

prompted us to use these vectors?

5. The purpose of this exercise is to derive the equations of motion of

the methane molecule, and so ®nd explicitly the 3 3 3 matrix B

which appears at the end of Example 32.20.

Label the corners of the tetrahedron 1, 2, 3, 4 and let 0 denote

the centroid of the tetrahedron. Work with 15 unit displacement

vectors

v12, v13, : : : , v43, w1, w2, w3

as described in Example 32.20, and let the position vector of the

molecule be Xi6� j

xijvij �X3

i�1

yiwi:

(a) Prove that cos (/ 012) � p(2=3) and cos (/ 102) � ÿ1=3.


(b) Show that the decrease in the length of the edge 12 from its

original length is

x12 � x21 � 12(x13 � x14 � x23 � x24),

with similar expressions for the edges 13, 14, 23, 24, 34.

Also, show that the length of the edge 01 has decreased by

p(2=3)(x12 � x13 � x14)� y1 ÿ 1

3(y2 � y3),

with similar expressions for the edges 02, 03.

Finally, show that the length of the edge 04 has decreased byp

(2=3)(x41 � x42 � x43)ÿ 13(y1 � y2 � y3):

(c) Let m1 denote the mass of a hydrogen atom, and m2 the mass

of a carbon atom. Assume that the magnitude of the force

between hydrogen atoms is k1 times the decrease in distance

between them, and the magnitude of the force between a

hydrogen atom and a carbon atom is k2 times the decrease in

distance between them.

Prove that

m1�x12 �ÿ k1[x12 � x21 � 12(x13 � x14 � x23 � x24)]

ÿ 13k2[x12 � x13 � x14 �p(3=2)(y1 ÿ 1

3y2 ÿ 1

3y3)],

with similar expressions for �x13, �x14, �x21, �x23, �x24, �x31, �x32, �x34.

Also, show

m1�x41 �ÿ k1[x14 � x41 � 12(x42 � x43 � x12 � x13)]

ÿ 13k2[x41 � x42 � x43 ÿ 1

3

p(3=2)(y1 � y2 � y3)],

with similar expressions for �x42 and �x43.

Finally, show

m2 �y1 � ÿk2[p

(2=3)(x12 � x13 � x14 ÿ x41

ÿ x42 ÿ x43)� 43y1],

with similar expressions for �y2 and �y3.

(d) The equations in part (c) determine the 15 3 15 matrix A in the

equations of motion �x � xA. Verify that the vectorsXi, j

vij, p1, p2, p3, q1, q2,


which appear in Example 32.20, are eigenvectors of A.

(e) Find the entries bij in the 3 3 3 matrix B which are given by

r1 A � b11 r1 � b12s1 � b13w1,

s1 A � b21 r1 � b22s1 � b23w1,

w1 A � b31 r1 � b32s1 � b33w1,

where the vectors r1, s1, w1 are as in Example 32.20.

(f) Verify that

(1, ÿ2,p

6)

is an eigenvector of B.

6. Consider a hypothetical molecule in which there are four identical

atoms at the corners of a square. Assume that the only internal

forces are along the sides of the square.

(a) Find the normal modes of the molecule.

(b) Calculate the equations of motion, �x � xA, and check that the

vectors you found in part (a) are, indeed, eigenvectors of A.

7. In this exercise, we derive a method for simplifying the problem of

®nding the eigenvectors of A when the homogeneous component V÷i

is reducible. (See 32.15(5).) We assume that ÷i is the character of

an irreducible RG-module which remains irreducible as a CG-

module.

Suppose that V÷ i� U1 � . . . � Um, a sum of m isomorphic irredu-

cible RG-modules. We reduce our problem to that of ®nding the

eigenvectors of an m 3 m matrix.

For 1 < i < m, let Wi be an RG-isomorphism from U1 to Ui.

(a) Prove that for all non-zero u in U1,

sp (uW1, : : : , uWm)

is an A-invariant vector space of dimension m.

(Hint: compose the function w! wA with a projection, and use

Exercise 23.8.)

(b) Let Au denote the matrix of the endomorphism w! wA of

sp (uW1, . . . , uWm) with respect to the basis uW1, . . . , uWm.

Prove that if u and v are non-zero elements of U1, then

Au � Av.

(c) Assume that the eigenvectors of the m 3 m matrix Au are

known. Show how to ®nd the eigenvectors of A in V÷i.


397

Solutions to exercises

Chapter 1

1. Note that all subgroups of G are normal, since G is abelian; and G 6� {1}since G is simple. Let g be a non-identity element of G. Then kgl is anormal subgroup of G, so kgl � G. If G were in®nite, then hg2i would be anormal subgroup different from G and {1}; hence G is ®nite. Let p be aprime number which divides |G|. Then hg pi is a normal subgroup of Gwhich is not equal to G. Therefore g p � 1, and so G is cyclic of primeorder.

2. Since G is simple and Ker W v G, either Ker W � f1g or Ker W � G. IfKer W � f1g then W is an isomorphism; and if Ker W � G then H � f1g.

3. First, G \ An � fg 2 G: g is even}, so G \ An v G. Since G \ An 6� G, wemay choose h 2 G with h =2 An. For all odd g in G, we have g �(ghÿ1)h 2 (G \ An)h. Therefore G \ An and (G \ An)h are the only rightcosets of G \ An in G, and G=(G \ An) � C2.

4. (a) Using the method of Example 1.4, it is routine to verify that ö and øare homomorphisms. Kerö � {1, a2} and Kerø � {1, c2}.

(b) Since b2ë � I but (bë)2 � Y 2 � ÿI , it follows that ë is not ahomomorphism. Check using the method of Example 1.4 again that ì isa homomorphism. Also Ker ì � {1} and Im ì � L, so ì is anisomorphism.

5. Let

D4m � ha, b: a2m � b2 � 1, bÿ1ab � aÿ1i, and

D2m � hc, d: cm � d2 � 1, dÿ1cd � cÿ1i,where m is odd. The elements of D2m 3 C2 are

(cid j, (ÿ1)k)

for 0 < i < m ÿ 1, 0 < j < 1, 0 < k < 1. Let x � (c(m�1)=2, ÿ1) and y �(d, 1). Check that

x2m � y2 � 1, yÿ1xy � xÿ1:

By Example 1.4, the function W: D4m ! D2m 3 C2 de®ned by

W: aib j ! xi y j (0 < i < 2mÿ 1, 0 < j < 1)

is a homomorphism. Since Im W � kx, yl, it contains

x2 � (c, 1) and xm � (1, ÿ1)

and hence Im W � D2m 3 C2. As |D4m| � |D2m 3 C2|, we conclude that W isan isomorphism.

6. (a) Let G � kal and suppose that 1 6� H < G. First observe that there existsi . 0 such that ai 2 H. Choose k as small as possible such that k . 0and ak 2 H . If 1 6� a j 2 H then j � qk � r for some integers q, r with0 < r , k. Hence ar � a jaÿqk � a j(ak)ÿq 2 H . Since r , k, we haver � 0. Therefore a j � akq and so H � kakl; thus H is cyclic.

(b) Assume that G � hai and jGj � dn. If g 2 G and gn � 1, then g � a j

for some integer j and dnj jn, so dj j; hence g 2 kadl. It follows that

fg 2 G: gn � 1g � hadi,which is a cyclic group of order n.

(c) If x and y are elements of order n in the ®nite cyclic group G, then x,y 2 H, where H � {g 2 G: gn � 1}. Now kxl and kyl have order n; alsoH has order n, by part (b). We deduce that

hxi � H � hyi:Thus x 2 hyi, and so x is a power of y.

7. Let G be the set of non-zero complex numbers. If g, h 2 G then gh 6� 0,so gh 2 G. If g, h, k 2 G then (gh)k � g(hk); also 1 2 G and 1g � g1 � gfor all g 2 G. Finally, if g 2 G then gÿ1 � 1=g 2 G, and gÿ1 g � ggÿ1 � 1.Thus G is a group under multiplication.

If H is a subgroup of G of order n, then hn � 1 for all h 2 H (since theorder of h divides n, by Lagrange's Theorem). Therefore

H < fg 2 G: gn � 1g � he2ði=ni:Since jH j � n � jhe2ði=nij, it follows that H � ke2ði=nl.

8. Partition G into subsets fg, gÿ1g (g 2 G). Each such subset has size 1 or2, and the identity element is in a subset of size 1. Hence, if jGj is eventhen there exists g 2 G such that g 6� 1 and the subset fg, gÿ1g has size 1;so g � gÿ1 and g has order 2.

9. De®ne matrices A, B as follows:

A � eið=4 0

0 eÿið=4

� �, B � 0 1

ÿ1 0

� �:

Check that A8 � I, B2 � A4 and Bÿ1AB � Aÿ1. These relations show thatevery element of the group kA, Bl has the form A j Bk with 0 < j < 7,0 < k < 1. Moreover,

A j � ei jð=4 0

0 eÿi jð=4

� �, A j B � 0 ei jð=4

ÿeÿi jð=4 0

� �:

Since these matrices, with 0 < j < 7, are all distinct, kA, Bl has order 16.


10. Suppose jG: H j � 2 and let g 2 G. If g 2 H then gÿ1 Hg � H. And ifg =2 H then H, Hg are the two right cosets of H in G, while H, gH arethe two left cosets. Therefore Hg � gH, and so gÿ1 Hg � H again. HenceH v G.

Chapter 2

1. Let u, w 2 W and ë 2 F. Since W is a linear transformation, we have

(uWÿ1 � wWÿ1)W � (uWÿ1)W� (wWÿ1)W � u� w,

(ë(wWÿ1))W � ë(wWÿ1)W � ëw:

Hence (u � w)Wÿ1 � uWÿ1 � wWÿ1 and (ëw)Wÿ1 � ë(wWÿ1), so Wÿ1 is alinear transformation.

2. (1)) (2): If W is invertible then W is injective, so Ker W � {0}.(2)) (3): If Ker W � {0} then dim (Im W) � dim V (by (2.12)), so Im W � V(by (2.7)).(3)) (1): Assume that Im W � V, so W is surjective. By (2.12), Ker W � {0}.If u, v 2 V and uW � vW then (u ÿ v)W � 0, so u ÿ v 2 Ker W � {0}, and sou � v. Thus W is injective. As W is surjective and injective, W is invertible.

3. First suppose that V � U � W. Then V � U � W. Let v 2 U \ W. Thenv � v � 0 � 0 � v and this gives us two expressions for v as the sum of anelement in U and an element in W. Since such expressions are unique,v � 0. Thus U \ W � {0}.

Now suppose that V � U � W and U \ W � {0}. If u1 � w1 � u2 � w2

with u1, u2 2 U and w1, w2 2 W, then u1 ÿ u2 � w2 ÿ w1 2 U \ W � {0};hence u1 � u2 and w1 � w2. This shows that V � U � W.

4. Assume ®rst that V � U � W. If v 2 V then v � u � w for some u 2 U andw 2 W; since u is a linear combination of u1, . . . , ur and w is a linearcombination of w1, . . . , ws, it follows that v is a linear combination ofu1, . . . , ur, w1, . . . , ws. Therefore u1, . . . , ur, w1, . . . , ws span V. Supposethat

ë1u1 � : : :� ërur � ì1w1 � : : :� ìsws � 0

with all ëi, ì j in F. Since V � U � W, the expression 0 � 0 � 0 is theunique expression for 0 as the sum of vectors in U and W, and so

ë1u1 � : : :� ër ur � ì1w1 � : : :� ìsws � 0:

As u1, . . . , ur are linearly independent, this forces ëi � 0 for all i;similarly ìi � 0 for all i. Therefore u1, . . . , ur, w1, . . . , ws are linearlyindependent; hence they form a basis of V.

Conversely, suppose that u1, . . . , ur, w1, . . . , ws is a basis of V. Ifv 2 U \ W then v � ë1u1 � . . . � ërur � ì1w1 � . . . � ìsws for some ëi,ì j 2 F; this gives ëi � ì j � 0 for all i, j, since u1, . . . , ur, w1, . . . , ws arelinearly independent. Thus v � 0 and so U \ W � {0}. It is easy to see thatV � U � W, so by Exercise 3, V � U � W.

5. (a) Assume ®rst that V � U1 � U2 � U3. Let u 2 U1 \ (U2 � U3). Thenu � u1 � u2 � u3 for some ui 2 Ui (1 < i < 3). Since u1 � 0 � 0 �0 � u2 � u3 and the sum U1 � U2 � U3 is direct, we have u1 �

Chapter 2 399

u2 � u3 � 0. Therefore U1 \ (U2 � U3) � {0}. Similarly,U2 \ (U1 � U3) � U3 \ (U1 � U2) � f0g.Now suppose that U1 \ (U2 � U3) � U2 \ (U1 � U3) �U3 \ (U1 � U2) � {0}. Assume that ui, u9i 2 Ui (1 < i < 3) andu1 � u2 � u3 � u91 � u92 � u93. Thenu1 ÿ u91 � (u92 ÿ u2) � (u93 ÿ u3) 2 U1 \ (U2 � U3) � {0}, so u1 � u91.Similarly, u2 � u92 and u3 � u93. Therefore V � U1 � U2 � U3.

(b) Let V � R2, and U1 � sp ((1, 0)), U2 � sp ((0, 1)), U3 � sp ((1, 1)).

6. By Exercise 4, if V � U � W then dim V � dim U � dim W. More generally,if V � U1 � . . . � Ur then V � U1 � (U2 � . . . � Ur) (see (2.10)); byinduction on r, dim (U2 � . . . � Ur) � dim U2 � . . . � dim Ur, sodim V � dim U1 � . . . � dim Ur.

7. Let V � R2. De®ne W, ö: V! V by

W: (x, y)! (x, 0) and ö: (x, y)! (y, 0):

Then Im W � sp ((1, 0)), Ker W � sp ((0, 1)), so V � Im W � Ker W; and Imö �Kerö � sp ((1, 0)), so V 6� Imö � Kerö.

8. Suppose ®rst that W is a projection. Then V � Im W � Ker W by Proposition2.32. Take a basis u1, . . . , ur for Im W and a basis w1, . . . , ws for Ker W.Then u1, . . . , ur, w1, . . . , ws is a basis, say B, of V, by Exercise 4. SinceuiW � ui for all i and wjW � 0 for all j, the matrix [W]B is diagonal, thediagonal entries being r 1's followed by s 0's.

Conversely, if [W]B has the given form, then clearly W2 � W, so W is aprojection.

9. Let v 2 V. Then

v � 12(v� vW)� 1

2(vÿ vW):

Observe that 12(v � vW)W � 1

2(vW � v), so 1

2(v � vW) 2 U. Similarly,

12(v ÿ vW) 2 W. Thus V � U � W. If v 2 U \ W then v � vW � ÿv, so v � 0.

Therefore V � U � W, by Exercise 3. The construction of the basis B issimilar to that in Solution 8.

Chapter 3

1. First, suppose that r is a representation of G. Then

I � 1r � (am)r � (ar)m � Am:

Conversely, assume that Am � I. Then (ai)r � Ai for all integers i(including i . m ÿ 1 and i , 0). Therefore for all integers i, j,

(aia j)r � (ai� j)r � Ai� j � Ai A j � (air)(a jr),

and so r is a representation.

2. Check that A3 � B3 � C3 � I . Hence by Exercise 1, each r j is arepresentation. The representations r2 and r3 are faithful, but r1 is not.


3. De®ne r by

(aib j)r � (ÿ1) j (0 < i < nÿ 1, 0 < j < 1):

It is easy to check that r is a representation of G.

4. (1) For all g 2 G, Iÿ1(gr)I � gr; hence r is equivalentto r.(2) If r is equivalent to ó then there is an invertible matrix T such thatgó � Tÿ1(gr)T for all g 2 G; then gr � (Tÿ1)ÿ1(gó)Tÿ1, so ó is equivalentto r.(3) If r is equivalent to ó and ó is equivalent to ô, then there are invertiblematrices S and T such that gó � Sÿ1(gr)S and gô � Tÿ1(gó)T for allg 2 G; then gô � (ST)ÿ1(gr)(ST), so r is equivalent to ô.

5. Check that in each of the cases (1) S � A, T � B, (2) S � A3, T � ÿB, (3)S � ÿA, T � B, (4) S � C, T � D, we have

S6 � T 2 � I , Tÿ1ST � Sÿ1:

It follows that each rk is a representation (see Example 1.4).The matrices Ar Bs (0 < r < 5, 0 < s < 1) are all different, so r1 is

faithful. Similarly r4 is faithful. But r2 and r3 are not faithful, sincea2r2 � I and a3r3 � I.

The representations r1 and r4 are equivalent: to see this, let

T � 1 1

ÿi i

� �:

Then Tÿ1(gr4)T � gr1 for all g 2 G. (To ®nd T, ®rst work out theeigenvectors of C.)

If j 6� 2, then a2r j 6� I; hence r2 is not equivalent to any of the others.And if j 6� 3, then a3r j 6� I; hence r3 is not equivalent to any of the others.

6. De®ne the matrices A and B by

A �0 1 0

ÿ1 0 0

0 0 1

0@ 1A, B �1 0 0

0 ÿ1 0

0 0 1

0@ 1A:Then the function r: ar bs ! Ar Bs (0 < r < 3, 0 < s < 1) is a faithfulrepresentation of D8 � ka, b: a4 � b2 � 1, bÿ1ab � aÿ1l. Compare Example3.2(1).

7. By Theorem 1.10, G=Ker r � Imr. But Imr < GL (1, F) and GL (1, F) isabelian. Therefore G=Ker r is abelian.

8. No: let G be any non-abelian group and let r be the trivial representation.

Chapter 3 401

Chapter 4

1.

g 1 (1 2) (1 3)

[g]B 1

1 0 0

0 1 0

0 0 1

0@ 1A 0 1 0

1 0 0

0 0 1

0@ 1A 0 0 1

0 1 0

1 0 0

0@ 1A[g]B 2

1 0 0

0 1 0

0 0 1

0@ 1A 1 0 0

0 ÿ1 0

0 ÿ1 1

0@ 1A 1 0 0

0 1 ÿ1

0 0 ÿ1

0@ 1A

g (2 3) (1 2 3) (1 3 2)

[g]B 1

1 0 0

0 0 1

0 1 0

0@ 1A 0 1 0

0 0 1

1 0 0

0@ 1A 0 0 1

1 0 0

0 1 0

0@ 1A[g]B 2

1 0 0

0 0 1

0 1 0

0@ 1A 1 0 0

0 ÿ1 1

0 ÿ1 0

0@ 1A 1 0 0

0 0 ÿ1

0 1 ÿ1

0@ 1AAll the matrices [g]B 2

have the form

1 0 0

0 j j

0 j j

0@ 1A:2. Let g 2 Sn. For all u, v in V and ë in F, we have

vg 2 V , v1 � v, (ëv)g � ë(vg), (u� v)g � ug � vg:

It remains to check (2) of De®nition 4.2. Let v 2 V and g, h 2 Sn. Assume®rst that gh 2 An. Then v(gh) � v; and (vg)h � v, since either vg � v � vh(if g, h 2 An) or vg � ÿv � vh (if g, h =2 An). Next, assume that gh =2 An.Then v(gh) � ÿv; and (vg)h � ÿv, since one of g, h is in An and theother is not. We have now checked all the conditions in De®nition 4.2, so Vis an FG-module.

3. Let A �0 1 0 0

ÿ1 0 0 0

0 0 0 ÿ1

0 0 1 0

0BB@1CCA and B �

0 0 1 0

0 0 0 1

ÿ1 0 0 0

0 ÿ1 0 0

0BB@1CCA.

Check that

A4 � I , B2 � A2, Bÿ1 AB � Aÿ1:

Hence r: aib j ! Ai Bj (0 < i < 3, 0 < j < 1) is a representation of Q8 overR. Let V � R4. By Theorem 4.4(1), V becomes an RQ8-module if we de®nevg � v(gr) for all v 2 V, g 2 Q8. If we put


v1 � (1, 0, 0, 0), v2 � (0, 1, 0, 0), v3 � (0, 0, 1, 0), v4 � (0, 0, 0, 1),

then for all i, via and vib are as required in the question.

4. You may ®nd it helpful ®rst to check that if

M �

0 1

1 0

1

. ..

1

0BBBBB@

1CCCCCAthen MA is obtained from A by swapping the ®rst two rows, and AM isobtained from A by swapping the ®rst two columns. To solve the exercise,let g be the permutation in Sn which has the property that for all i,

row i of B � row ig of A:

Let P be the n 3 n matrix ( pij) de®ned by

pij �1, if j � ig,

0, if j 6� ig:

8<:Then P is a permutation matrix, and the ij-entry of PA isXn

k�1

pikakj � aig, j:

Hence PA � B.If C is a matrix obtained from A by permuting the columns, then

C � AQ for some permutation matrix Q; the proof is similar to that for therows.

Chapter 5

1. It is easy to verify that V is an FG-module. Let U be a non-zero FG-submodule of V, and let (á, â) 2 U with (á, â) 6� (0, 0). Then(á, â) � (á, â)a � (á � â, á � â) 2 U, and (á, â) ÿ (á, â)a �(á ÿ â, â ÿ á) 2 U. Since at least one of á � â and á ÿ â is non-zero, wededuce that (1, 1) or (1, ÿ1) belongs to U. Hence the FG-submodules of Vare {0}, sp ((1, 1)), sp ((1, ÿ1)) and V.

2. Suppose that r has degree n and r is reducible. Then r is equivalent to arepresentation ô of the form

ô: g !X g 0

Y g Z g

0@ 1A (g 2 G)

where Xg is a k 3 k matrix and 0 , k , n. Then ó is equivalent to ô, sinceó is equivalent to r. Therefore ó is reducible.

3. Let G � D12 and let r1, . . . , r4 be the representations of G de®ned inExercise 3.5. First consider the FG-module V � F2, where vg � v(gr1) for

0

0

Chapter 5 403

v 2 V, g 2 G. Suppose that U is a non-zero FG-submodule of V. Then U isan FH-module, where H is the subgroup {1, b}. By the solution to Exercise1, either (1, 1) or (1, ÿ1) lies in U. Since (1, 1) and (1, 1)a are linearlyindependent, and also (1, ÿ1) and (1, ÿ1)a are linearly independent, itfollows that dim U > 2. Consequently U � V and so V is irreducible. (SeeExample 5.5(2) for an alternative argument.) Therefore r1 is irreducible;also r4 is irreducible, since r1 and r4 are equivalent.

Now let V � F2 with vg � v(gr2) for v 2 V, g 2 G. Then (1, 1)a �ÿ(1, 1) � (1, 1)b. Hence sp ((1, 1)) is an FG-submodule of V and r2 isreducible.

Finally, r3 is irreducible, by an argument similar to that for r1.

4. (a) It is easy to check the given relations. Using the relations, we may writeevery element of G in the form

aib jck (0 < i < 2, 0 < j < 2, 0 < k < 1):

Thus jGj < 18. However, it is clear that |ka, bl| � 9 and ka, bl 6� G.Hence, by Lagrange's Theorem, jGj is a multiple of 9 and jGj. 9.Therefore jGj � 18.

(b) Let

A � å 0

0 åÿ1

� �, B � ç 0

0 çÿ1

� �, C � 0 1

1 0

� �:

Check that A3 � B3 � C2 � I , AB � BA, Cÿ1 AC � Aÿ1 andCÿ1 BC � Bÿ1. Hence r is a representation (compare Example 1.4).

(c) For every element g of ka, bl, there exists a cube root î of unity suchthat

gr � î 0

0 îÿ1

� �:

But there are only three distinct cube roots of unity, so there existdistinct g1, g2 2 ka, bl with g1r � g2r. Therefore r is never faithful.

(d) Let V � C2 be the CG-module obtained by de®ning vg � v(gr) for allv 2 C2, g 2 G. If U is a non-zero CG-submodule of V, then U is a CH-submodule, where H is the subgroup {1, c}. Hence either (1, 1) or(1, ÿ1) lies in U, by the solution to Exercise 1; accordingly, let u be(1, 1) or (1, ÿ1) (so that u 2 U). Now u and ua are linearly independentunless å � 1, and u and ub are linearly independent unless ç � 1. Hence,if either å 6� 1 or ç 6� 1 then dim U � 2 and so r is irreducible. On theother hand, if å � ç � 1 then sp ((1, 1)) is a CG-submodule of V, so r isreducible.

5. Let V � {0} and let 0g � 0 for all g 2 G. Then V is neither reducible norirreducible.

Chapter 6

1: (a) xy � ÿ2:1ÿ a3 � ab� 3a2b� 2a3b,

yx � ÿ2:1ÿ a3 � b� 2a2b� 3a3b,

x2 � 4:1� a2 � 4a3:


(b) az � ab � a3b � a2ba � ba � za, and bz � 1 � a2 � zb. Henceaib jz � zaib j for all i, j and so gz � zg for all g 2 G. If r 2 CG thenr � P g ë g g with ë g 2 C, so rz �Pë g gz �Pë gzg � zr.

2. Let C2 3 C2 � ka, b: a2 � b2 � 1, ab � bal. Relative to the basis 1, a, b,ab of F(C2 3 C2), the regular representation r is given by

1r �1 0 0 0

0 1 0 0

0 0 1 0

0 0 0 1

0BBB@1CCCA, ar �

0 1 0 0

1 0 0 0

0 0 0 1

0 0 1 0

0BBB@1CCCA,

br �0 0 1 0

0 0 0 1

1 0 0 0

0 1 0 0

0BBB@1CCCA, (ab)r �

0 0 0 1

0 0 1 0

0 1 0 0

1 0 0 0

0BBB@1CCCA:

3. No: let G � ka: a2 � 1l, and take r � 1 � a, s � 1 ÿ a.

4. (a) As g runs through G, so do gh and hg. Hence ch � hc � c.(b) c2 � c

Ph2G h �Ph2G ch � jGjc.

(c) All the entries in [W]B are 1 (compare the solution to Exercise 2). Thereason is that for all i, j, there exists a unique h in G such thatgih � gj.

5. Note ®rst that if u is an element of a vector space, and u � u � u, thenu � 0. Now 0r � (0 � 0)r � 0r � 0r, and v0 � v(0 � 0) � v0 � v0; hence0r � v0 � 0.

Let V be the trivial FG-module and let 0 6� v 2 V and 1 6� g 2 G. If r �1 ÿ g, then vr � 0 and neither v nor r is 0.

6. Let v1 � 1 � ù2a � ùa2 and v2 � b � ù2ab � ùa2b. Check that v1a � ùv1,v2a � ù2v2, v1b � v2 and v2b � v1. Hence W is a CG-submodule of CG.Use the argument of either Example 5.5(2) or the solution to Exercise 5.3to prove that W is irreducible.

Chapter 7

1. For all u1, u2 2 U, ë 2 F and g 2 G, we have

(u1 � u2)Wö � (u1W� u2W)ö � u1(Wö)� u2(Wö),

(ëu1)Wö � (ë(u1W))ö � ë(u1(Wö)),

(u1 g)Wö � ((u1W)g)ö � ((u1W)ö)g � (u1(Wö))g:

2. Let a � (1 2 3 4 5) and let v1, . . . , v5 be the natural basis for thepermutation module for G over F. Then

W: ë1v1 � : : :� ë5v5 ! ë11� ë2a� ë3a2 � ë4a3 � ë5a4

is the required FG-isomorphism. (Note that viW � ai, so (via)W � vi�1W �ai�1 � (viW)a; hence W is an FG-homomorphism.)

3. It is easy to show that V0 is an FG-submodule of V. Let x 2 G. Then

Chapter 7 405

Xg2G

vxg �Xg2G

vg �Xg2G

vgx:

Hence (vx)W � vW � (vW)x; noting that VW � V0, we see that W is an FG-homomorphism from V to V0.

If v 2 V0 then (v=jGj)W � v; hence W is surjective.

4. Suppose that ö: V! W is an FG-isomorphism. Let g 2 G. For all v 2 V0,(vö)g � (vg)ö � vö, and so V0ö � W0. For all w 2 W0, (wöÿ1)g �(wg)öÿ1 � wöÿ1, so W0öÿ1 � V0. Hence the function ö, restricted to V0,is an FG-isomorphism from V0 to W0.

5. No: let v1, . . . , v4 be the natural basis of the permutation module V for Gover F. In the notation of Exercise 3,

V0 � sp (v1 � v2, v3 � v4) and (FG)0 � spXg2G

g

!:

Since V0 and (FG)0 have different dimensions, it follows from Exercise 4that V and FG are not isomorphic FG-modules.

6. (a) W is easily seen to be a linear transformation. Also

(á1� âx)xW � (â1� áx)W � (âÿ á)(1ÿ x) � (áÿ â)(1ÿ x)x

� (á1� âx)Wx:

Hence W is an FG-homomorphism.(b) (á ÿ â)(1 ÿ x)W � ((á ÿ â) ÿ (â ÿ á))(1 ÿ x) � 2(á ÿ â)(1 ÿ x). Hence

W2 � 2W.(c) Let B be the basis 1 ÿ x, 1 � x.

Chapter 8

1. V � sp (ÿùv1 � v2) � sp (ÿù2v1 � v2), where ù � e2ði=3. (Find eigenvectorsfor x.)

2. Let G � {1, a, b, ab} � C2 3 C2 (so a2 � b2 � 1, ab � ba). Then

RG � sp (1� a� b� ab)� sp (1� aÿ bÿ ab)� sp (1ÿ a� bÿ ab)

� sp (1ÿ aÿ b� ab):

3. Let G be any group, and let V be a 2-dimensional vector space over C withbasis v1, v2. De®ne vg � v for all v 2 V, g 2 G; this makes V into a CG-module. If we let

W: ëv1 � ìv2 ! ëv2 (ë, ì 2 C)

then W is a CG-homomorphism from V to V, and Ker W � Im W � sp (v2).

4. Suppose r is reducible. Then by Maschke's Theorem, r is equivalent to arepresentation ó of the form

gó � ë g 0

0 ì g

� �(ë g, ì g 2 C):

Then (gó)(hó) � (hó)(gó) for all g, h 2 G, since all diagonal matrices


commute with each other; hence also (gr)(hr) � (hr)(gr) for all g, h 2 G.This is a contradiction. Therefore r is irreducible.

5. U � sp ((1, 0)) is the only 1-dimensional CG-submodule of V, so there is noCG-submodule W of V with V � U � W.

6. (1) It is straightforward to verify for [ , ] the axioms of a complex innerproduct. For example, if u 6� 0 then (ux, ux) . 0 for all x 2 G, so[u, u] . 0. Also

[ug, vg] �Xx2G

(ugx, vgx) �Xx2G

(ux, vx) � [u, v]:

(2) It is easy to prove that U? is a subspace of V. Let v 2 U? and g 2 G.Then for all u 2 U,

[u, vg] � [ugÿ1, vggÿ1] by part (1)

� [ugÿ1, v] � 0 since ugÿ1 2 U :

Therefore vg 2 U?, and so U? is a CG-submodule of V.

(3) Let W � U?. Then V � U � W, and W is a CG-submodule of V bypart (2).

7. We know that the regular CG-module CG is faithful (Proposition 6.6). LetCG � U1 � . . . � Ur, where U1, . . . , Ur are irreducible CG-submodules ofCG. Then there exist i 2 {1, . . . , r} and g 2 G such that ug 6� u for someu 2 Ui (otherwise vg � v for all v 2 CG). De®ne

K � fx 2 G: vx � v for all v 2 Uig:Check that K is a normal subgroup of G; also K 6� G since g =2 K. Since Gis simple, we must therefore have K � {1}. This means that Ui is a faithfulirreducible CG-module.

Chapter 9

1. Let C2 � ka: a2 � 1l. Irreducible representations r1, r2:

1r1 � ar1 � (1); 1r2 � (1), ar2 � (ÿ1):

Let C3 � kb: b3 � 1l and let ù � e2ði=3. Irreducible representations r1, r2,r3:

1r1 � br1 � b2r1 � (1);

bir2 � (ùi);

bir3 � (ù2i):

Let C2 3 C2 � {(1, 1), (x, 1), (1, y), (x, y)}, where x2 � y2 � 1. Irreduciblerepresentations r1, r2, r3, r4:

Chapter 9 407

gr1 � (1) for all g 2 C2 3 C2;

(xi, y j)r2 � (ÿ1) j;

(xi, y j)r3 � (ÿ1)i;

(xi, y j)r4 � (ÿ1)i� j:

2. Let C4 3 C4 � k(x, 1), (1, y): x4 � y4 � 1l.(a) r: (xi, yj)! (ÿ1)i.(b) If g1 � (x2, 1) and g2 � (1, y2) then g1, g2 and g1 g2 all have order 2.

Since (g1 g2)ó � (g1ó)(g2ó) for all representations ó, we cannot have(g1 g2)ó � g1ó � g2ó � (ÿ1).

3. For 1 < j < r, let gj generate Cn j, and let å j � e2ði=n j . Then

r: (gi11 , : : : , gir

r )!åi1

1

. ..

åi rr

0BB@1CCA

is a faithful representation of Cn13 . . . 3 Cnr

of degree r.Yes: if r � 2, n1 � 2, n2 � 3, then

ó : (gi11 , gi2

2 )! (åi11 å

i22 )

is a faithful representation of degree 1 , r.

4. Check that A4 � B2 � I, Bÿ1AB � Aÿ1 when A � ar and B � br. Hence rgives a representation; similarly for ó.

If M(gr) � (gr)M for g � a and for g � b, then M � ëI for some ë 2 C.Hence r is irreducible (Corollary 9.3).

Notice that the matrix5 ÿ6

4 ÿ5

� �commutes with gó for all g 2 G; hence ó is reducible (Corollary 9.3).

5. Let z �Pg2G g. Then xz � z � zx for all x 2 G. Hence z 2 Z(CG), and theresult follows from Proposition 9.14.

6. (a) Clearly a commutes with a � aÿ1. Also bÿ1(a � aÿ1)b � aÿ1 � a, so bcommutes with a � aÿ1.

(b) Check that w(a � aÿ1) � ÿw for all w 2 W.

7. (a) Let Cn � kx: xn � 1l. Then r: xj ! (e2ði j=n) is a faithful irreduciblerepresentation of Cn.

(b) r: a ! 0 1

ÿ1 0

� �, b ! 1 0

0 ÿ1

� �gives a faithful irreducible representation of D8 (see Example 5.5(2)).

(c) The centre of C2 3 D8 is isomorphic to C2 3 C2, so is not cyclic.Therefore Proposition 9.16 shows that C2 3 D8 has no faithfulirreducible representation.

(d) Let C3 � kx: x3 � 1l and let ù � e2ði=3. Check that

r: (x, a)! 0 ùÿù 0

� �, (x, b)! ù 0

0 ÿù� �

0

0


gives a representation of C3 3 D8. It is irreducible (see for exampleExercise 8.4) and faithful.

Chapter 10

1. Let V � sp (P

g2G g). Then V is a trivial CG-submodule of CG. Nowsuppose that U is an arbitrary trivial CG-submodule of CG, so U � sp (u)for some u. Then ug � u for all g 2 G, so |G|u � u(

Pg2G g) �

(P

g2G g)u 2 V. Thus U � V, and so CG has exactly one trivial CG-submodule, namely V.

2. Let G � kx: x4 � 1l. Then

CG � sp (1� x� x2 � x3)� sp (1� ixÿ x2 ÿ ix3)

� sp (1ÿ x� x2 ÿ x3)� sp (1ÿ ixÿ x2 � ix3):

3. Let

u1 � 1� a� a2 � a3 ÿ bÿ abÿ a2bÿ a3b,

u2 � 1ÿ a� a2 ÿ a3 � bÿ ab� a2bÿ a3b,

u3 � 1ÿ a� a2 ÿ a3 ÿ b� abÿ a2b� a3b:

4. We decompose CG as a direct sum of irreducible CG-submodules. Let

v0 � 1� a� a2 � a3, v1 � 1� iaÿ a2 ÿ ia3,

v2 � 1ÿ a� a2 ÿ a3, v3 � 1ÿ iaÿ a2 � ia3

(compare the solution to Exercise 2). For 0 < j < 3, let wj � bv j. Then, asin Example 10.8(2), the subspaces sp (v0, w0), sp (v1, w3), sp (v2, w2) andsp (v3, w1) are CG-submodules of CG. We have

sp (v0, w0) � U0 � U1, sp (v2, w2) � U2 � U3,

where Ui � sp (ui) (0 < i < 3) and u1, u2, u3 are as in the solution toExercise 3, while u0 �

Pg2G g.

Let U4 � sp (v1, w3), U5 � sp (v3, w1). As in Example 5.5(2) (or seeExercise 8.4), U4 and U5 are irreducible CG-modules. Moreover U4 � U5,since there is a CG-isomorphism sending v1 ! w1, w3 ! v3.

Theorem 10.5 now shows that there are exactly ®ve non-isomorphicirreducible CG-modules, namely U0, U1, U2, U3 and U4. Therefore everyirreducible representation of D8 over C is equivalent to precisely one of thefollowing:

r0: a! (1), b! (1)

r1: a! (1), b! (ÿ1)

r2: a! (ÿ1), b! (1)

r3: a! (ÿ1), b! (ÿ1)

r4: a! i 0

0 ÿi

!, b! 0 1

1 0

!:

Chapter 10 409

5. Let W: U1 ! U2 be a CG-isomorphism. For ë 2 C, de®ne the functionöë: U1 ! V by

öë: u! u� ëuW (u 2 U1):

Then öë is easily seen to be a CG-homomorphism; moreover,

u 2 Keröë , u� ëuW � 0, u � 0,

since the sum U1 � U2 is direct. Thus U1 � Imöë. It is easy to check thatif ë 6� ì then Imöë 6� Imöì. Therefore we have constructed in®nitely manyCG-submodules Imöë of the required form.

6. V is irreducible, either by the method of Example 5.5(2) or by Exercise 8.4.Let u1 � 1 ÿ ia ÿ a2 � ia3, u2 � b ÿ iab ÿ a2b � ia3b. Then sp (u1, u2) is

a CG-submodule of CG which is isomorphic to V. A CG-isomorphism isgiven by v1 ! u1, v2 ! u2.

Chapter 11

1. Since G is non-abelian, not all the dimensions are 1 (see Proposition 9.18).Hence, by Theorem 11.12, the dimensions are 1, 1, 2.

2. Compare Example 11.13 to see that the possible answers are 112, 182, 1422

and 133 (where 112 means twelve 1s, 182 means eight 1s and one 2, and soon). It will be shown later (Exercises 15.4, 17.3) that 182 cannot occur.

By Exercise 5.3, D12 has at least two inequivalent irreduciblerepresentations of degree 2. Hence the answer for D12 is 1422.

3. For each g 2 G, de®ne ö g: CG! CG by rö g � gr (r 2 CG). Then{ö g: g 2 G} gives a basis of HomCG(CG, CG) (compare the proof ofProposition 11.8).

4. Let v1, . . . , vn be the natural basis of V. Then sp (v1 � . . . � vn) is theunique trivial CG-submodule of V (compare Exercise 10.1). Hence byCorollary 11.6, dim (HomCG(V, U)) � 1.

5. Let v1, w2 be the basis of U3 described in Example 10.8(2). De®ne W1 andW2 by rW1 � v1 r, rW2 � w2 r (r 2 CG). Then W1, W2 is a basis ofHomCG (CG, U3), by the proof of Proposition 11.8. Also, de®ne ö1, ö2 byuö1 � u, uö2 � bu (u 2 U3). Then ö1, ö2 is a basis ofHomCG (U3, CG).

6. Let V � X1 � . . . � Xr and W � Y1 � . . . � Ys, where each Xa and each Yb

is an irreducible CG-module. Then by (11.5)(3) and Proposition 11.2,dim (HomCG (V, W)) is equal to the number of ordered pairs (a, b) such thatX a � Yb. This, in turn, equalsXk

i�1

jf(a, b): Xa � Yb � Vigj:

Now the number of integers a with X a � Vi is dim (HomCG (V , Vi)) � di,by Corollary 11.6, and similarly the number of integers b with Yb � Vi isei. Therefore, dim (HomCG (V, W)) �Pk

i�1diei.


Chapter 12

1. Assume that g, h 2 CG(x). Then gx � xg and hx � xh, so hÿ1x � xhÿ1 andghÿ1x � gxhÿ1 � xghÿ1; thus ghÿ1 2 CG(x). Also 1x � x1, so 1 2 CG(x).Therefore CG(x) is a subgroup of G.

If z 2 Z(G) then zg � gz for all g 2 G, so zx � xz and z 2 CG(x).

2. Note that x 2 CG(g), x 2 CG(gz). Now the required result follows fromTheorem 12.8.

3. (a) (1 2)G � {(i j): 1 < i , j < n} and this set has size (n2 ). The centralizer

CG((1 2)) consists of all elements x and (1 2)x, where x is a permutation®xing 1 and 2. Thus |CG((1 2))| � 2´(n ÿ 2)!, in agreement with Theorem12.8 (since (n

2 ) � n!=(2:(nÿ 2)!)).(b) (1 2 3)G consists of all 3-cycles (i j k). There are (n

3 ) choices for thenumbers i, j, k (unordered). Each choice gives exactly two 3-cycles,namely (i j k) and (i k j).

(1 2)(3 4)G consists of all permutations of the form (i j)(k l ) with i, j,k, l distinct. There are (n

4 ) choices for the numbers i, j, k, l (unordered),and three permutations for each choice, namely (i j)(k l ), (i k)( j l ) and(i l )( j k).

(c) Every element of (1 2 3)(4 5 6)G has the form (1 i j)(k l m), with i, j, k, l,m distinct. There are ®ve choices for i; then four choices for j; then wecan make two different 3-cycles (k l m) and (k m l ) from the remainingnumbers. This gives 5 . 4 . 2 � 40 elements in all.

The elements of (1 2)(3 4)(5 6)G have the form (1 i)( j k)(l m). Thereare ®ve choices for i; then we can make three permutations ( j k)(l m),( j l )(k m) and ( j m)(k l ) of cycle shape (2, 2) from the remainingnumbers. Hence |(1 2)(3 4)(5 6)G | � 5 . 3 � 15.

The sizes of the conjugacy classes of S6 are given in the followingtable:

Cycle-shape (1) (2) (3) (22) (4) (3, 2) (5) (23) (32) (4, 2) (6)Class size 1 15 40 45 90 120 144 15 40 90 120

4. An element x of cycle-shape (5) has CS6(x) � kxl (note that |xS6 | � 144 and

use Theorem 12.8). Hence by Proposition 12.17, x A6 6� xS6 . For elements gof other cycle-shapes, gA6 � gS6 .

5. By Example 12.18(2), the conjugacy classes of A5 have sizes 1, 12, 12, 15,20. If H is a normal subgroup of A5 then jH j divides 60, and 1 2 H, andH is a union of conjugacy classes of A5. Hence jH j � 1 or 60; therefore A5

is simple.

6. We have Q8 � ka, b: a4 � 1, b2 � a2, bÿ1ab � aÿ1l. The conjugacy classesof Q8 are

f1g, fa2g, fa, a3g, fb, a2bg, fab, a3bg,and a basis of Z(CQ8) is

1, a2, a� a3, b� a2b, ab� a3b:

7. The class equation gives

Chapter 12 411

jGj � jZ(G)j �X

xi=2Z(G)

jxGi j:

(a) For xi =2 Z(G), |xGi | divides pn and |xG

i | 6� 1 by Theorem 12.8 and (12.9).Therefore p divides |xG

i |. Hence p divides |Z(G)|, so Z(G) 6� f1g.(b) If no conjugacy class of G has size p, then p2 divides |xG

i | for allxi =2 Z(G). If, in addition, |G| > p3, then by the class equation, p2 divides|Z(G)|. This is a contradiction.

Chapter 13

1. The characters ÷i of ri (i � 1, 2) are as follows:

Also Ker r1 � {1, a3} and Ker r2 � {1, a2, a4}.

2. Let C4 � kx: x4 � 1l. The irreducible characters ÷1, . . . , ÷4 of C4 are asfollows:

We have ÷reg � ÷1 � ÷2 � ÷3 � ÷4.

3. Since ÷(g) � |®x (g)|, we have ÷((1 2)) � 5 and ÷((1 6)(2 3 5)) � 2.

4. If ÷ is a non-zero character which is a homomorphism, then÷(1) � ÷(12) � (÷(1))2, so ÷(1) � 1.

5. Let r be a representation with character ÷. Then zr � ëI for some ë 2 C,by Proposition 9.14. Thus, for all g in G, (zg)r � (zr)(gr) � ë(gr), andhence ÷(zg) � ë÷(g). Moreover, I � 1r � zmr � (zr)m � ëmI, so ëm � 1.

6. Let r be a representation with character ÷. If g 2 Z(G) then gr � ëI forsome ë 2 C, by Proposition 9.14. Conversely, if gr � ëI for some ë 2 C,then (gr)(hr) � (hr)(gr) for all h 2 G, and hence g 2 Z(G) since r isfaithful.

We have now proved that gr � ëI for some ë 2 C if and only ifg 2 Z(G). The required result now follows from Theorem 13.11(1).

7. (a) For all g, h 2G, det ((gh)r) � det ((gr)(hr)) � det (gr) det (hr). Henceg! (det (gr)) is a representation of G over C of degree 1, and so ä isa linear character of G.

(b) G=Ker ä � Imä by Theorem 1.10, and Im ä is a subgroup of the

1 a3 a, a5 a2, a4 b, a2b, a4b ab, a3b, a5b

÷1 2 2 ÿ1 ÿ1 0 0÷2 2 0 0 2 0 ÿ2

1 x x2 x3

÷1 1 1 1 1÷2 1 i ÿ1 ÿi÷3 1 ÿ1 1 ÿ1÷4 1 ÿi ÿ1 i


multiplicative group C� of non-zero complex numbers, which isabelian. Therefore G=Ker ä is abelian.

(c) Im ä is a ®nite subgroup of C�, hence is cyclic, by Exercise 1.7. Alsoÿ1 2 Im ä, so Im ä has even order. Hence Im ä contains a subgroup Hof index 2. It is easy to check that {g 2 G: ä(g) 2 H} is a normalsubgroup of G of index 2.

8. Let r be the regular representation of G, and de®ne the character ä as inExercise 7. By Exercise 1.8, G has an element x of order 2. Order thenatural basis g1, . . . , g2k of CG so as to obtain a basis B in which gand gx are adjacent for all g 2 G. Then

[x]B �

0 1

1 0

0 1

1 0

. ..

0BBBBB@

1CCCCCA:There are k blocks (0

110), and since k is odd, det ([x]B ) � (ÿ1)k � ÿ1. Thus

ä(x) � ÿ1. The required result now follows from Exercise 7.

9. Let V be a CG-module with character ÷. We may choose a basis B of Vso that [g]B is diagonal with all diagonal entries �1 (see (9.10)). Saythere are r entries 1 and s entries ÿ1. De®ne ä as in Exercise 7. If s isodd then ä(g) � ÿ1, and G has a normal subgroup of index 2 by Exercise7. And if s is even then ÿs � s mod 4, so÷(g) � r ÿ s � r � s � ÷(1) mod 4.

10. As x 6� 1, we have ÷reg(x) 6� ÷reg(1) (see Proposition 13.20), so ÷i(x) 6� ÷i(1)for some irreducible character ÷i of G, by Theorem 13.19.

Chapter 14

1. Using Proposition 14.5(2), we obtain

h÷, ÷i � 3 . 3

24� (ÿ1)(ÿ1)

4� 0� 3 . 3

8� (ÿ1)(ÿ1)

4� 2,

h÷, øi � 3 . 3

24� (ÿ1) . 1

4� 0� 3 . (ÿ1)

8� (ÿ1)(ÿ1)

4� 0,

hø, øi � 3 . 3

24� 1 . 1

4� 0� (ÿ1)(ÿ1)

8� (ÿ1)(ÿ1)

4� 1:

Hence ø is irreducible by Theorem 14.20 (but ÷ is not).

2. Let ÷i be the character of ri (i � 1, 2, 3). The values of these charactersare as follows:

0

0

Conjugacy class 1 a2 a, a3 b, a2b ab, a3b

÷1 2 ÿ2 0 0 0÷2 2 ÿ2 0 0 0÷3 2 2 0 0 ÿ2

Chapter 14 413

By Theorem 14.21, r1 and r2 are equivalent, but r3 is not equivalent to r1

or r2.

3. The representations r and ó have the same character, by Proposition 13.2;hence r and ó are equivalent, by Theorem 14.21, and this gives therequired matrix T.

4. Let ÷1 be the trivial character of G. Then

h÷, ÷1i � 1

jGjXg2G

÷(g):

Since ÷(1) . 0 and by hypothesis ÷(g) > 0 for all g 2 G, we havek÷, ÷1l 6� 0. As ÷ 6� ÷1, Theorem 14.17 shows that ÷ is reducible.

5. We have

h÷reg, ÷i � 1

jGjXg2G

÷reg(g)÷(g):

But ÷reg(g) is |G| if g � 1 and is 0 if g 6� 1. Hence k÷reg, ÷l � ÷(1).

6. This follows at once from Exercise 11.4 and Theorem 14.24.

7. Recall that hø, øi �Pki�1d2

i . Hence if kø, øl � a where a � 1, 2 or 3, thenexactly a of the integers di are 1 and the rest are 0. If kø, øl � 4, theneither exactly four of the di are 1, or exactly one of the di is 2; the rest are0.

8. No: let G � C2 and ÷ � ÷reg, the regular character of C2.

Chapter 15

h÷, ÷1i � 16(19 . 1� 3 . (ÿ1) . 1� 2 . (ÿ2) . 1) � 2,1:

h÷, ÷2i � 16(19 . 1� 3 . (ÿ1)(ÿ1)� 2 . (ÿ2) . 1) � 3,

h÷, ÷3i � 16(19 . 2� 0� 2 . (ÿ2)(ÿ1)) � 7:

Hence ÷ � 2÷1 � 3÷2 � 7÷3. Since all the coef®cients here are non-negativeintegers, it follows that ÷ is a character of S3.

2. By the method used in the solution to Exercise 1, we obtain

ø1 � 16÷1 � 1

6÷2 � 1

3÷3,

ø2 � 12÷1 ÿ 1

2÷2,

ø3 � 13÷1 � 1

3÷2 ÿ 1

3÷3:

3. We ®nd that ø � ÿ÷2 � ÷3 � ÷5 � 2÷6. Since the coef®cient of ÷2 is anegative integer, ø is not a character of G.

4. (a) For all groups G, if x 2 G then the subgroup generated by x and theelements of Z(G) is abelian (since the elements of Z(G) commute withpowers of x). Hence, if G � Z(G) [ Z(G)x then G � Z(G). Therefore thecentre of a group G never has index 2 in G.Every abelian group of order 12 has 12 conjugacy classes. If |G| � 12


and G is non-abelian, then |Z(G)| divides 12 and |Z(G)| 6� 6 or 12, so|Z(G)| < 4. Therefore, at most 4 conjugacy classes of G have size 1 (see(12.9)); the remaining conjugacy classes have size at least 2, so therecannot be as many as 9 conjugacy classes in total.

(b) Since the number of irreducible representations is equal to the numberof conjugacy classes, it follows from the solution to Exercise 11.2 andpart (a) that G has 4, 6 or 12 conjugacy classes. If G is abelian (e.g.G � C4 3 C3) then G has 12 conjugacy classes; if G � D12 then G has6 conjugacy classes (see (12.12)); and if G � A4 then G has 4 conjugacyclasses (see Example 12.18(1)).

Chapter 16

1. Let C2 3 C2 � {(1, 1), (x, 1), (1, y), (x, y): x2 � y2 � 1}. The charactertable of C2 3 C2 is (cf. Exercise 9.1)

2. The last row of the character table is (cf. Example 16.5(2))

3. The complete character table of G is

The two unknown degrees ÷3(1), ÷4(1) are 1, 2 sinceP4

i�1(÷i(1))2 � 10.

Because g4 has order 2, Corollary 13.10, together with the relationP4i�1÷i(1)÷i(g4) � 0, gives the values on g4. Then

P4i�1÷i(g2)÷i(g4) � 0

gives ÷3(g2) � 1; similarly ÷3(g3) � 1. Finally,P4

i�1÷i(1)÷i(g2) � 0 gives

÷4(g2) � (ÿ1ÿp5)=2; similarly ÷4(g3) � (ÿ1�p5)=2.

(1, 1) (x, 1) (1, y) (x, y)

÷1 1 1 1 1÷2 1 1 ÿ1 ÿ1÷3 1 ÿ1 1 ÿ1÷4 1 ÿ1 ÿ1 1

g1 g2 g3 g4 g5

÷5 2 ÿ2 0 0 0

gi g1 g2 g3 g4

|CG(gi)| 10 5 5 2

÷1 1 1 1 1÷2 2 (ÿ1�p5)=2 (ÿ1ÿp5)=2 0÷3 1 1 1 ÿ1÷4 2 (ÿ1ÿp5)=2 (ÿ1�p5)=2 0

Chapter 16 415

4. (a)P5

i�1÷i(g1)÷i(g2) � 0 gives 3� 3æ� 3æ � 0, andP5

i�1÷i(g2)÷i(g2) � 7gives 3� 2ææ � 7.Hence æ � (ÿ1� i

p7)=2.

(b) The column which corresponds to the conjugacy class containing gÿ12

has values which are the complex conjugates of those in the column ofg2 (see Proposition 13.9(3)); since æ is non-real, this is a differentcolumn of the character table of G.

5. Let g 2 G. By the column orthogonality relations applied to the columncorresponding to g, we have

Pki�1÷i(g)÷i(g) � jCG(g)j. This number is

equal to |G| if and only if CG(g) � G, which occurs if and only ifg 2 Z(G).

6. The matrix C is obtained from C by rearranging the columns (seeProposition 13.9(3)). Therefore det C � �det C; if det C � det C then det Cis real, and if det C � ÿdet C then det C is purely imaginary.

By the column orthogonality relations, CtC is the k 3 k diagonalmatrix whose diagonal entries are |CG(gi)| (1 < i < k). Hencejdet Cj2 � Q jCG(gi)j.

If G � C3 then det C � �i3p

3. (The sign depends upon the ordering ofrows and columns.)

Chapter 17

1. (a) The conjugacy classes of Q8 are f1g, fa2g, fa, a3g, fb, a2bg andfab, a3bg.

(b) G9 � f1, a2g and G=G9 � fG9, G9a, G9b, G9abg � C2 3 C2. Thecharacter table of C2 3 C2 is given in the solution to Exercise 16.1.Hence the linear characters of G are

(c) Using the column orthogonality relations, the last irreducible character ofG is

The character table of Q8 is the same as that of D8.

2. It is easy to see that a7 � b3 � 1. Use Proposition 12.13 to see quickly thatbÿ1ab � a2.(a) Using the relations, every element of G has the form ambn with

gi 1 a2 a b ab|CG(gi)| 8 8 4 4 4

÷1 1 1 1 1 1÷2 1 1 1 ÿ1 ÿ1÷3 1 1 ÿ1 1 ÿ1÷4 1 1 ÿ1 ÿ1 1

1 a2 a b ab

÷5 2 ÿ2 0 0 0


0 < m < 6, 0 < n < 2; hence jGj < 21. But a has order 7 and b hasorder 3, so 21 divides jGj by Lagrange's Theorem. Therefore jGj � 21.

(b) The conjugacy classes of G are f1g, fa, a2, a4g, fa3, a5, a6g,famb: 0 < m < 6g and famb2: 0 < m < 6g.

(c) First, G9 � kal, so we get three linear characters of G:

where ù � e2ði=3. To ®nd the two remaining irreducible characters ÷4

and ÷5, we note that a is not conjugate to aÿ1; therefore for someirreducible character ÷, we have ÷(a) 6� ÷(a) (see Corollary 15.6). Hence÷4 and ÷5 must be complex conjugates of each other. Applying thecolumn orthogonality relations, we obtain


7)=2.

3. The number of linear characters of G divides jGj by Theorem 17.11. Nowconsult the solution to Exercise 11.2 to see that there are 3, 4 or 12 linearcharacters. If there are 12, then G is abelian (see Proposition 9.18), so G iscertainly not simple. If G has 3 or 4 linear characters then jG=G9j � 3 or 4and again G is not simple as G9 v G.

4. In the character table below, we have ÷1 � 1G, ÷2 � ÷, ÷3 � ÷2, ÷4 � ÷2÷3,÷5 � ö, ÷6 � ö÷; all of these are irreducible characters by Proposition17.14. The centralizer orders are obtained by using the orthogonalityrelations, and the class sizes jgG

i j come from the equationsjGj � jCG(gi)jjgG

i j (Theorem 12.8).

gi 1 a a3 b b2

|CG(gi)| 21 7 7 3 3

÷1 1 1 1 1 1÷2 1 1 1 ù ù2

÷3 1 1 1 ù2 ù

1 a a3 b b2

÷4 3 á á 0 0÷5 3 á á 0 0


|CG(gi)| 12 4 4 6 6 12|gG

i | 1 3 3 2 2 1

÷1 1 1 1 1 1 1÷2 1 ÿi i 1 ÿ1 ÿ1÷3 1 ÿ1 ÿ1 1 1 1÷4 1 i ÿi 1 ÿ1 ÿ1÷5 2 0 0 ÿ1 ÿ1 2÷6 2 0 0 ÿ1 1 ÿ2

Chapter 17 417

5. The normal subgroups of D8 are

D8 � Ker ÷1, hai � Ker ÷2,

ha2, bi � Ker ÷3, ha2, abi � Ker ÷4,

ha2i � Ker ÷2 \ Ker ÷3, f1g � Ker ÷5:

6. (a) Check that the given matrices satisfy the relevant relations:

å 0

0 åÿ1

!2n

� 1 0

0 1

!,

å 0

0 åÿ1

!n

� 0 1

å n 0

!2

,

0 1

å n 0

!ÿ1å 0

0 åÿ1

!0 1

å n 0

!� å 0

0 åÿ1

!ÿ1

:

Hence we have representations of T4n (cf. Example 1.4).(b) The representations in part (a) are irreducible unless å � �1, by Exercise

8.4. For å � e2ðir=2n, with r � 1, 2, . . . , n ÿ 1, we get n ÿ 1 irreduciblerepresentations, no two of which are equivalent, since they have distinctcharacters. Moreover G9 � ka2l, so jG=G9j � 4 and there are fourrepresentations of degree 1 (see Theorem 17.11). Now the sum of thesquares of the degrees of the irreducible representations we have foundso far is

(nÿ 1) . 22 � 4 . 12 � 4n:

Hence we have found all the irreducible representations, by Theorem11.12. (For further details on the representations of degree 1, see thesolution to Exercise 18.3; note that the structure of G=G9 depends uponwhether n is even or odd.)

7. (a) Check that the given matrices satisfy the relevant relations.(b) The given representations, for å � e2ðik=2n with 0 < k < n ÿ 1, are

irreducible (by Exercise 8.4), and inequivalent (consider the charactervalues on a2). Also G9 � kbl, so jG=G9j � 2n and there are 2nrepresentations of degree 1. The sum of the squares of the degrees ofthe irreducible representations we have found is

n . 22 � 2n . 12 � 6n,

so we have obtained all the irreducible representations.

8. (a) Check that the given matrices satisfy the relevant relations.(b) The given representations, for å � e2ði j=n with 0 < j < n ÿ 1, are

irreducible (by Exercise 8.4) and inequivalent (their characters aredistinct). Note that b2 does not belong to the kernel of any of theserepresentations.We get further representations by

a! ç 0

0 çÿ1

� �, b! 0 1

1 0

� �,

where ç is any (2n)th root of unity in C. For ç � e2ði j=2n with1 < j < n ÿ 1, these representations are irreducible and inequivalent.Moreover, they are not equivalent to any representation found earlier,since b2 is in the kernel of each of these representations.


Finally, G9 � ka2, b2l and G=G9 � C2 3 C2, so we get fourrepresentations of degree 1.

We have now found all the irreducible representations, since the sumof the squares of the irreducible representations given above is

n . 22 � (nÿ 1) . 22 � 4 . 12 � 8n:

Chapter 18

1. The character table of D8 is as shown.

(See Example 16.3(3) or Section 18.3.) Regarding D8 as the symmetrygroup of a square, take b to be a re¯ection in a diagonal of the square.Then ð takes the following values:

Hence ð � ÷1 � ÷3 � ÷5. (Compare Example 14.28(2), where we took b tobe a different re¯ection.)

2. Let ù � e2ði=6. Then ù � ùÿ1 � 1, ù2 � ùÿ2 � ù4 � ùÿ4 � ÿ1. Hence,using Section 18.3, the character table of D12 is as shown.

Character table of D8

gi 1 a2 a b ab|CG(gi)| 8 8 4 4 4

÷1 1 1 1 1 1÷2 1 1 1 ÿ1 ÿ1÷3 1 1 ÿ1 1 ÿ1÷4 1 1 ÿ1 ÿ1 1÷5 2 ÿ2 0 0 0

1 a2 a b ab

ð 4 0 0 2 0

Character table of D12

gi 1 a3 a a2 b ab|CG(gi)| 12 12 6 6 4 4

÷1 1 1 1 1 1 1÷2 1 1 1 1 ÿ1 ÿ1÷3 1 ÿ1 ÿ1 1 1 ÿ1÷4 1 ÿ1 ÿ1 1 ÿ1 1÷5 2 ÿ2 1 ÿ1 0 0÷6 2 2 ÿ1 ÿ1 0 0

Chapter 18 419

Seven normal subgroups of D12 are G � Ker ÷1, kal � Ker ÷2,ka2, bl � Ker ÷3, ka2, abl � Ker ÷4, ka2l � Ker ÷3 \ Ker ÷4, ka3l � Ker ÷6 and{1} � Ker ÷5.

3. The n � 3 conjugacy classes of G are

f1g, fang, far, aÿrg(1 < r < nÿ 1), fa2 jb: 0 < j < nÿ 1g,fa2 j�1b: 0 < j < nÿ 1g:

We get n ÿ 1 irreducible characters ø j (1 < j < n ÿ 1) of G fromExercise 17.6 as follows:

where ù � e2ði=2n.The remaining four irreducible characters of G are linear. If n is odd,

then G=G9 � hG9bi � C4 and the linear characters are

If n is even, then G=G9 � C2 3 C2 and the linear characters are

Note that T4 � C4, T8 � Q8 and T12 is the Example in Section 18.4.

4. The 3n conjugacy classes of G are, for 0 < r < n ÿ 1,

fa2rg, fa2r b, a2rb2g, fa2r�1, a2r�1b, a2r�1b2g:We have G9 � hbi and G=G9 � hG9ai � C2n. Hence we get 2n linearcharacters ÷ j (0 < j < 2n ÿ 1), as shown. Exercise 17.7 gives us nirreducible characters øk (0 < k < n ÿ 1).

gi 1 an ar (1 < r < n ÿ 1) b ab|CG(gi)| 4n 4n 2n 4 4

ø j 2 2(ÿ1) j ùrj � ùÿrj 0 0

gi 1 an ar (1 < r < n ÿ 1) b ab

÷1 1 1 1 1 1÷2 1 ÿ1 (ÿ1)r i ÿi÷3 1 1 1 ÿ1 ÿ1÷4 1 ÿ1 (ÿ1)r ÿi i

gi 1 an ar (1 < r < n ÿ 1) b ab

÷1 1 1 1 1 1÷2 1 1 1 ÿ1 ÿ1÷3 1 1 (ÿ1)r 1 ÿ1÷4 1 1 (ÿ1)r ÿ1 1


Observe that U6 � D6, U12 � T12 and U18 � D6 3 C3.

5. The 2n � 3 conjugacy classes of G are

f1g, fb2g, fa2r�1, aÿ2rÿ1b2g(0 < r < nÿ 1),

fa2s, aÿ2sg, fa2sb2, aÿ2sb2g(1 < s < (nÿ 1)=2),

fajbk : j even, k � 1 or 3g, and

fajbk : j odd, k � 1 or 3g:

Using Exercise 17.8, we get four linear characters ÷1, . . . , ÷4, n charactersø j (0 < j < n ÿ 1) of degree 2, and a further n ÿ 1 characters ö j

(1 < j < n ÿ 1) of degree 2, as shown below. For example, the charactertable of V24 is given at the top of p. 422.

Character table of U6n

gi a2r a2rb a2r�1

|CG(gi)| 6n 3n 2n

÷ j ù2 jr ù2 jr ù j(2r�1)

(0 < j < 2n ÿ 1)øk 2ù2kr ÿù2kr 0(0 < k < n ÿ 1)

Note: ù � e2ði=2n.

Character table of V8n

gi 1 b2 a2r�1 a2s a2sb2 b ab(0 < r < n ÿ 1) (1 < s < (nÿ 1)=2)

|CG(gi)| 8n 8n 4n 4n 4n 4 4

÷1 1 1 1 1 1 1 1÷2 1 1 1 1 1 ÿ1 ÿ1÷3 1 1 ÿ1 1 1 1 ÿ1÷4 1 1 ÿ1 1 1 ÿ1 1ø j 2 ÿ2 ù2 j(2r�1) ù4 js ÿù4 js 0 0(0 < j < n ÿ 1) ÿùÿ2 j(2r�1) �ùÿ4 js ÿùÿ4 js

ö j 2 2 ù j(2r�1) ù2 js ù2 js 0 0(1 < j < n ÿ 1) �ùÿ j(2r�1) �ùÿ2 js �ùÿ2 js

Note: ù � e2ði=2n

Chapter 18 421

Chapter 19

h÷ø, öi � 1

jGjXg2G

÷(g)ø(g)ö(g) � 1

jGjXg2G

÷(g)ø(g)ö(g) � h÷, øöi:1:

Similarly, k÷ø, öl � kø, ÷öl.

2. k÷ø, 1Gl � k÷, øl, by Exercise 1. The result now follows from Proposition13.15 and (14.13).

3. Let V be a CG-module with character ÷. Since ÷ is not faithful, there exists1 6� g 2 G with vg � v for all v 2 V. By Proposition 15.5 there is anirreducible character ø of G such that ø(g) 6� ø(1). Let n be an integerwith n > 0. Then wg � w for all w 2 V . . . V (n factors). Henceö(g) � ö(1) for all irreducible characters ö for which k÷ n, öl 6� 0.Therefore k÷ n, øl � 0.

4. Note that (1 2 3 4 5)2 is conjugate to (1 3 4 5 2) in A5. Using Proposition19.14 we obtain

Then

÷S � ø1 � ø2 � 2ø3,

÷A � ø2 � ø4 � ø5,

öS � ø1 � ø3,

öA � ø4:

Character table of V24

gi 1 b2 a a3 a5 a2 a2b2 b ab|CG(gi)| 24 24 12 12 12 12 12 4 4

÷1 1 1 1 1 1 1 1 1 1÷2 1 1 1 1 1 1 1 ÿ1 ÿ1÷3 1 1 ÿ1 ÿ1 ÿ1 1 1 1 ÿ1÷4 1 1 ÿ1 ÿ1 ÿ1 1 1 ÿ1 1ø0 2 ÿ2 0 0 0 2 ÿ2 0 0ø1 2 ÿ2 i

p3 0 ÿi

p3 ÿ1 1 0 0

ø2 2 ÿ2 ÿip

3 0 ip

3 ÿ1 1 0 0ö1 2 2 1 ÿ2 1 ÿ1 ÿ1 0 0ö2 2 2 ÿ1 2 ÿ1 ÿ1 ÿ1 0 0

1 (1 2 3) (1 2)(3 4) (1 2 3 4 5) (1 3 4 5 2)

÷S 15 0 3 0 0÷A 10 1 ÿ2 0 0öS 6 0 2 1 1öA 3 0 ÿ1 (1�p5)=2 (1ÿp5)=2


5. We have recorded ÷ as ÷5, ÷S as ÷4 and ÷A as ÷2, below. Since k÷i, ÷il � 1for i � 2, 4, 5, these characters are irreducible. The table also records thetrivial character ÷1, ÷3 � ÷2, ÷6 � ÷5 and ÷7 � ÷2÷5; these are irreducible byPropositions 13.15 and 17.14. Since G has seven conjugacy classes, thecharacter table is complete.

6. Taking D6 � ka, b: a3 � b2 � 1, bÿ1ab � aÿ1l, the character table ofD6 3 D6 is as shown.

Chapter 20

1. (a) Regard D8 as the subgroup of S4 which permutes the four corners of asquare, as in Example 1.1(3). Take b to be the re¯ection in the axisshown:

Character table of G (cf. Exercise 27.2)

gi g1 g2 g3 g4 g5 g6 g7

|CG(gi)| 24 24 4 6 6 6 6

÷1 1 1 1 1 1 1 1÷2 1 1 1 ù ù2 ù2 ù÷3 1 1 1 ù2 ù ù ù2

÷4 3 3 ÿ1 0 0 0 0÷5 2 ÿ2 0 ÿù2 ÿù ù ù2

÷6 2 ÿ2 0 ÿù ÿù2 ù2 ù÷7 2 ÿ2 0 ÿ1 ÿ1 1 1

Note: ù � e2ði=3

Character table of D6 3 D6

(gi, hj) (1, 1) (a, 1) (b, 1) (1, a) (a, a) (b, a) (1, b) (a, b) (b, b)|CG(gi, hj)| 36 18 12 18 9 6 12 6 4

÷1 3 ÷1 1 1 1 1 1 1 1 1 1÷2 3 ÷1 1 1 ÿ1 1 1 ÿ1 1 1 ÿ1÷3 3 ÷1 2 ÿ1 0 2 ÿ1 0 2 ÿ1 0÷1 3 ÷2 1 1 1 1 1 1 ÿ1 ÿ1 ÿ1÷2 3 ÷2 1 1 ÿ1 1 1 ÿ1 ÿ1 ÿ1 1÷3 3 ÷2 2 ÿ1 0 2 ÿ1 0 ÿ2 1 0÷1 3 ÷3 2 2 2 ÿ1 ÿ1 ÿ1 0 0 0÷2 3 ÷3 2 2 ÿ2 ÿ1 ÿ1 1 0 0 0÷3 3 ÷3 4 ÿ2 0 ÿ2 1 0 0 0 0

Chapter 20 423

Then a! (1 2 3 4), b! (1 3) gives the required isomorphism.(b) Take the irreducible characters ÷1, . . . , ÷5 of S4 as in Section 18.1, and

take the character table of H to be

(see Example 16.3(3) or Section 18.3). We obtain

÷1 # H � ø1, ÷2 # H � ø4, ÷3 # H � ø1 � ø4, ÷4 # H � ø3 � ø5,

÷5 # H � ø2 � ø5:

2. Let ÷1, . . . , ÷11 be the irreducible characters of S6, as in Example 19.17.Either by direct calculation, or using (20.13), we ®nd that the characters÷i # A6 (i � 1, 3, 5, 7, 9) are distinct irreducible characters of A6; thesegive the characters ø1, . . . , ø5 in our character table below. Also,k÷11 # A6, ÷11 # A6l � 2. Arguing as in Example 20.14, we obtain from÷11 # A6 the two irreducible characters which we have called ø6 and ø7.

Note: á � (1�p5)=2, â � (1ÿp5)=2

3. Let ø1, . . . , ør be the irreducible characters of H. Then ÷ # H � d1ø1 �. . . � drør for some non-negative integers di. Since each øi has degree 1,the inequality (20.6) gives

÷(1) � d1 � : : :� dr < d21 � : : :� d2

r < n:

4. The inequality k÷ # H, ÷ # Hl H < 3 follows at once from Proposition 20.5.Write d � k÷ # H, ÷ # Hl H . For examples with d � 1 or 2, take G � S3

and H a subgroup of order 2, and ÷ an irreducible character of G of degreed. For an example with d � 3, take G � A4, H � V4 and ÷ an irreduciblecharacter of G of degree 3 (see Section 18.2).

gi 1 (1 3)(2 4) (1 2 3 4) (1 3) (1 2)(3 4)|CH (gi)| 8 8 4 4 4

ø1 1 1 1 1 1ø2 1 1 1 ÿ1 ÿ1ø3 1 1 ÿ1 1 ÿ1ø4 1 1 ÿ1 ÿ1 1ø5 2 ÿ2 0 0 0


gi 1 (1 2 3) (1 2)(3 4) (1 2 3 4 5) (1 3 4 5 2) (1 2 3)(4 5 6) (1 2 3 4)(5 6)|CA6

(gi)| 360 9 8 5 5 9 4

ø1 1 1 1 1 1 1 1ø2 5 2 1 0 0 ÿ1 ÿ1ø3 10 1 ÿ2 0 0 1 0ø4 9 0 1 ÿ1 ÿ1 0 1ø5 5 ÿ1 1 0 0 2 ÿ1ø6 8 ÿ1 0 á â ÿ1 0ø7 8 ÿ1 0 â á ÿ1 0


5. See (20.13). Since 20 occurs only once in the list of degrees for S7, therestriction of the irreducible character of degree 20 to A7 must be the sumof two different irreducible characters of degree 10. From the remainingfourteen irreducible characters of S7, upon restriction to A7 we get at leastseven irreducible characters of A7; and we get precisely seven if and only ifthe restriction of each of the fourteen characters is irreducible. We are toldthat A7 has exactly nine conjugacy classes. Hence the irreducible charactersof A7 have degrees

1, 6, 14, 14, 15, 10, 10, 21, 35:

Chapter 21

1. (a) Let u � 1 ÿ a2 � b ÿ a2b. Then ua2 � ÿu and ub � u. Hence sp (u) is aCH-submodule of CH.

(b) Since every element of G belongs to H or to Ha, the elements u and uaform a basis of U " G.

(c) The character ø of U and the character ø " G of U " G are given by

Since hø " G, ø " Gi � 1, the induced module U " G is irreducible.

2. Label the characters of H as follows:

where ù � e2ði=3.(a) ÷1 # H � ÷2 # H � ø1, ÷3 # H � ø2 � ø3,

÷4 # H � ÷5 # H � ø1 � ø2 � ø3.(b) Using the Frobenius Reciprocity Theorem, we obtain

ø1 " G � ÷1 � ÷2 � ÷4 � ÷5, ø2 " G � ø3 " G � ÷3 � ÷4 � ÷5:

3. It is suf®cient to prove that if U is a CH-submodule of CH thendim (U " G) � jG : H jdim U . Let Hgj (1 < j < m) be the distinct right

1 a2 b a2b

ø 1 ÿ1 1 ÿ1

1 a2 a b ab

ø " G 2 ÿ2 0 0 0

1 (1 2 3) (1 3 2)

ø1 1 1 1ø2 1 ù ù2

ø3 1 ù2 ù

Chapter 21 425

cosets of H in G. Then U(CG) � Ug1 � . . . � Ugm, where Ugj � {ugj:u 2 U}. The sum Ug1 � . . . � Ugm is direct (since the elements in Ugj arelinear combinations of elements in the right coset Hgj). Also,dim (Ugj) � dim U (since u! ugj (u 2 U) is a vector space isomorphism).Hence dim(U " G) � dim(U (CG)) � m dim U .

4. Let ö be an irreducible character of G. By using the Frobenius ReciprocityTheorem twice, together with the result of Exercise 19.1 (also twice), weobtain

h(ø(÷ # H)) " G, öiG � hø(÷ # H), ö # HiH � hø, (÷ö) # HiH

� hø " G, ÷öiG � h(ø " G)÷, öiG:Since this holds for all irreducible characters ö of G, we deduce fromTheorem 14.17 that (ø(÷ # H)) " G � (ø " G)÷.

5. The values of ö " G and ø " G are given by Proposition 21.23. On elementsof cycle-shapes (1), (7) and (3, 3), the values are as follows, and on allother elements the values are zero.

6. We have |G: H|ø(1) � d1÷1(1) � . . . � dk÷k(1), where

di � hø " G, ÷iiG � hø, ÷i # HiH ,

by the Frobenius Reciprocity Theorem. Hence, since ø is irreducible,÷i # H � diø � â where either â is a character of H or â � 0. Thus÷i(1) > diø(1), and therefore

jG: H jø(1) > (d21 � : : :� d2

k)ø(1):

The required result follows.

7. By applying the result of Exercise 6, we deduce, as in the proof ofProposition 20.9, that either(1)ø " G is irreducible, or(2)ø " G is the sum of two different irreducible characters of the same

degree.Suppose ®rst that ø " G is irreducible, say ø " G � ÷. Then ÷(1) � 2ø(1)

and k÷ # H, øl H � 1, by the Frobenius Reciprocity Theorem; hence ÷ # H isreducible, say ÷ # H � ø � ö. Now suppose that ø9 is an irreduciblecharacter of H. We have

hø9 " G, ÷iG 6� 0, hø9, ÷ # HiH 6� 0, ø9 � ø or ö:

ThusIf ø " G is irreducible, then for precisely one other irreducible character

ö of H we have ø " G � ö " G. (Compare Proposition 20.11.)

Cycle-shape (1) (7) (3, 3)

ö " G 240 2 12ø " G 720 ÿ1 0


Suppose next that ø " G is reducible, say ø " G � ÷1 � ÷2. Then÷1(1) � ø(1) and k÷1 # H, øl H � 1; hence ÷1 # H � ø. Now suppose that ø9is an irreducible character of H. We have

hø9 " G, ÷1iG 6� 0, hø9, ÷1 # HiH 6� 0, ø9 � ø:

ThusIf ø " G � ÷1 � ÷2 where ÷1 and ÷2 are irreducible characters of G, and

ø9 is an irreducible character of H such that ø9 " G has ÷1 or ÷2 as aconstituent, then ø9 � ø. (Compare Proposition 20.12.)

Chapter 22

1. The number of linear characters divides 15 (Theorem 17.11), and the sumof the squares of the degrees of the irreducible characters is 15 (Theorem11.12); moreover, each degree divides 15 (Theorem 22.11). Hence everyirreducible character has degree 1, and so G is abelian by Proposition 9.18.

2. Use Theorems 11.12, 17.11 and 22.11 again. The degree of each irreduciblecharacter is 1 or 2, and if there are r characters of degree 1 and s ofdegree 2, then

r divides 16, and r . 12 � s . 22 � 16:

Hence r � 4 or 8 or 16, and r � s � 7 or 10 or 16.

3. (a) Since G is non-abelian, not every irreducible character has degree 1(Proposition 9.18). This time, Theorems 11.12, 17.11 and 22.11 showthat there are r irreducible characters of degree 1 and s irreduciblecharacters of degree q, where

r divides pq, 1 < s and r � sq2 � pq:

Hence r � q and s � ( p ÿ 1)/q.(b) jG9j � p by Theorem 17.11.(c) The number of conjugacy classes of G is r � s.(For more information on groups of order pq, see Chapter 25.)

4. (a) By hypothesis, there exist a, b 2 C such that ö(g) � a for all g 6� 1 andö(1) � a� bjGj. Then ö � a1G � b÷reg, since both sides of this equationtake the same values on all elements of G.

(b) We have

h1G, öi � 1

jGj (a� bjGj � (jGj ÿ 1)a) � a� b, and

h÷reg, öi � 1

jGj jGj(a� bjGj) � a� bjGj:

Since ö is a character, both k1G, öl and k÷reg, öl are integers.(c) If ÷ is a non-trivial irreducible character of G, then kö, ÷l 2 Z and

k1G, ÷l � 0. Hence kö ÿ a1G, ÷l 2 Z. But kö ÿ a1G, ÷l � kb÷reg, ÷l �bjGj÷(1)=jGj � b÷(1).

Chapter 22 427

(d) Since ÷(1) divides jGj, part (c) implies that b|G| is an integer. Therefore,by part (b), a, and hence also b, is an integer.

5. (a) If g 2 G then g has odd order, by Lagrange's Theorem. Therefore, ifg2 � 1 then g � 1.

(b) For all g 2 G, we have ÷(g)� ÷(gÿ1) � ÷(g)� ÷(g) � 2÷(g), and ÷(g)is an algebraic integer. Partition Gnf1g into subsets by putting eachelement with its inverse. Each such subset has size 2, by part (a). HenceX

g2G

÷(g) � ÷(1)� 2á

for some algebraic integer á. The stated result follows, since

h÷, 1Gi � 1

jGjXg2G

÷(g):

(c) If ÷ 6� 1G in part (b), then k÷, 1Gl � 0, and hence á � ÿ÷(1)=2. Butÿ÷(1)=2 is a rational number which is not an integer (since ÷(1) dividesjGj, hence is odd). This contradicts Proposition 22.5. Thus ÷ � 1G.

(Further information about the number of characters ÷ such that ÷ � ÷appears in Theorem 23.1 and Corollary 23.2.)

6. (a) By Theorem 22.16, ÷(g) is an integer for all characters ÷. Let ÷1, . . . ,÷7 be the irreducible characters of G, with ÷1 � 1G. By the columnorthogonality relations, we have

(I) 1�X7

i�2

(÷i(g))2 � 5, and (II) 1�X7

i�2

÷i(1)÷i(g) � 0:

From equation (I) we deduce that either ÷i(g) � 0, �1 for all i, or÷i(g) � �2 for exactly one i and ÷i(g) � 0 for all other i . 1; thesecond possibility is ruled out by equation (II).

(b) We deduce from part (a) that ÷i(g) � 0 for two values of i, say i � 2, 3,and ÷i(g) � �1 for 4 < i < 7. By Corollary 22.27,÷2(1) � ÷3(1) � 0 mod 5. Also

(III)X7

i�1

(÷i(1))2 � 120:

Since 52 � 102 . 120, we deduce that ÷2(1) � ÷3(1) � 5.(c) By Corollary 22.27, ÷i(1) � ÷i(g) mod 5 for all i, and from equation (III)

we have

(IV)X7

i�4

(÷i(1))2 � 69:

Hence the possibilities for the pairs of integers (÷i(1), ÷i(g)) with4 < i < 7 are (1, 1), (4, ÿ1), (6, 1). The only possibility which isconsistent with equation (IV) is that the values of ÷i(1) for 4 < i < 7 are1, 4, 4, 6 in some order.

(d) We now have the following part of the character table of G:


gi g1 g2 g3 g4 g5 g6 g7

Order of gi 1 2 2 3 4 6 5|CG(gi)| 120 12 8 6 4 6 5

÷1 1 1 1 1 1 1 1÷2 5 0÷3 5 0÷4 1 1÷5 4 ÿ1÷6 4 ÿ1÷7 6 1

We successively calculate the ®ve remaining columns of the charactertable. We are told that ÷i(gj) is an integer for all i, j.

(1) First, ÷i(g4) � ÷i(1) mod 3 andP

7i�1(÷i(g4))2 � 6. Hence the values of

÷i(g4) for 1 < i < 7 are 1, ÿ1, ÿ1, 1, 1, 1, 0, respectively.(2) Next, ÷i(g5) � ÷i(1) mod 2 and

P7i�1(÷i(g5))2 � 4. Hence ÷i(g5) � �1

for 1 < i < 4 and ÷i(g5) � 0 for 5 < i < 7. SinceP

7i�1÷i(1)÷i(g5) � 0,

we deduce that ÷4(g5) � ÿ1 and (without loss of generality)÷2(g5) � ÿ÷3(g5) � 1.

(3) Since ÷i(g3) � ÷i(1) mod 2 andP

7i�1(÷i(g3))2 � 8, we deduce that

÷i(g3) � �1 for 1 < i < 4 and that the values of ÷i(g3) for 5 < i < 7 are0, 0, �2 in some order. Also

P7i�1÷i(g3)÷i(gr) � 0 for r � 4, 7, from

which we see that ÷i(g3) � 1 for 1 < i < 4. From the relationP7i�1÷i(1)÷i(g3) � 0 we now deduce that the entries in column 3 are 1,

1, 1, 1, 0, 0, ÿ2 in order from the top.(4) We have ÷i(g6) � ÷i(g4) mod 2 and

P7i�1(÷i(g6))2 � 6. Therefore

÷i(g6) � �1 for 1 < i < 6 and ÷7(g6) � 0. By applying the columnorthogonality relations involving column 6 and columns 3, 4, 5 and 7 weobtain ÷2(g6) � ÿ÷3(g6) � ÷4(g6) � ÿ1 and (without loss of generality)÷5(g6) � ÿ÷6(g6) � 1.

(5) Only the entries in column 2 remain to be calculated. These can beobtained from the column orthogonality relations.

The character table of G is as shown.

gi g1 g2 g3 g4 g5 g6 g7

Order of gi 1 2 2 3 4 6 5|CG(gi)| 120 12 8 6 4 6 5

÷1 1 1 1 1 1 1 1÷2 5 ÿ1 1 ÿ1 1 ÿ1 0÷3 5 1 1 ÿ1 ÿ1 1 0÷4 1 ÿ1 1 1 ÿ1 ÿ1 1÷5 4 ÿ2 0 1 0 1 ÿ1÷6 4 2 0 1 0 ÿ1 ÿ1÷7 6 0 ÿ2 0 0 0 1

Chapter 22 429

7. Suppose ®rst that ë is an eigenvalue of an n 3 n matrix A, all of whoseentries are integers. Then det (A ÿ ëIn) � 0, and hence ë is a root of thepolynomial det (xIn ÿ A), which is of the form

xn � anÿ1xnÿ1 � : : :� a1x� a0 (ar 2 Z):

Conversely, assume that ë is a root of the polynomial p(x) � a0 � a1x �. . . � anÿ1x nÿ1 � x n (ar 2 Z). Let

A �

0 1 0 : : : 0

0 0 1 0

0 0 0 0

..

. ...

0 0 0 1

ÿa0 ÿa1 ÿa2 : : : ÿanÿ1

0BBBBBB@

1CCCCCCA:

Check that det (xIn ÿ A) � p(x). As p(ë) � 0, it follows that ë is aneigenvalue of A. Since A has integer entries, ë is therefore an algebraicinteger.

Chapter 23

1. Assume that x 2 G and x is real. Then gÿ1xg � xÿ1 for some g 2 G. Hencegÿ2xg2 � x, so g2 2 CG(x). Let m be the order of g. Since jGj is odd,m � 2n � 1 for some integer n, by Lagrange's Theorem. Then g �g2(n�1) 2 CG(x). Therefore xÿ1 � gÿ1xg � x. Since x2 � 1 and x has oddorder, it follows that x � 1.

2. Adopt the notation of Theorem 9.8. The character ÷ of G � Cn13 . . .

3 Cnrwhich is given by

÷(gi11 : : : gir

r ) � ëi11 : : : ëi r

r

is real if and only if ëi � �1 for all i with 1 < i < r. Now the nith root ofunity ëi can be ÿ1 if and only if ni is even. Hence the number of realirreducible characters is 2m, where m is the number of the integers n1, . . . ,nr which are even. However, the elements g of G which satisfy g2 � 1 areprecisely those elements gi1

1 . . . gi rr where for each j, either i j � 0 or n j is

even and i j � n j=2. The number of such elements is also 2m.

3. The elements g of D2n for which g2 � 1 are

1, aib (0 < i < nÿ 1) (and also an=2 if n is even):

This gives n � 1 elements if n is odd, and n � 2 elements if n is even.These numbers coincide with

P÷(1), summing over all the irreducible

characters ÷. Since é÷ < 1 for all ÷, it follows from the Frobenius±SchurCount of Involutions that we must have é÷ � 1 for all ÷.

4. Let ë1 and ë2 be the eigenvalues of gr. Then ÷A(g) �12((ë1 � ë2)2 ÿ (ë2

1 � ë22)) � ë1ë2 � det (gr) (see Proposition 19.14). Since

÷(1) � 2 we have ÷A(1) � 1. It now follows from the De®nition 23.13 of é÷that é÷ � ÿ1 if and only if ÷A � 1G. The result follows.


5. (a) First, it is easy to check that ÷(g) is real for all g 2 G, so é÷ � �1. Letr be the representation obtained by using the basis v1, v2 of V. Thendet (ar) � 1 and det (br) � ÿå n; hence det (gr) � 1 for all g 2 G if andonly if å n � ÿ1. The result now follows from Exercise 4.

(b) It is easy to check that if g � a or b and i, j 2 {1, 2} then

â(vig, v j) � â(vi, v jgÿ1):

For example, â(v1b, v1) � â(v2, v1) � å n � â(v1, å nv2) � â(v1, v1bÿ1).Hence â is G-invariant. The de®nition of â shows that â is symmetric ifå n � 1 and â is skew-symmetric if å n � ÿ1. The result now followsfrom Theorem 23.16.

(c) The elements of T4n are ai and aib (0 < i < 2n ÿ 1); a has order 2n andaib has order 4. Hence an is the only element of order 2.

(d) Refer the Exercise 18.3 for the characters ø j (1 < j < n ÿ 1) and ÷ j

(1 < j < 4) of T4n. Clearly é÷1 � é÷3 � 1, and é÷2 � é÷4 � 0 or 1,according to whether n is odd or even, respectively. By part (a) (or part(b)) and the construction of the characters ø j in Exercise 17.6, we getéø j � ÿ1 or 1, according to whether j is odd or even, respectively.Therefore

Pnÿ1j�1 (éø j)ø j(1) � 0 or ÿ2, according to whether n is odd or

even, respectively. HenceP

÷ (é÷)÷(1) � 2.

6. Let V be a CG-module with character ÷. Since é÷ � ÿ1, there exists a non-zero G-invariant skew-symmetric bilinear form â on V. As â is G-invariant,the subspace {u 2 V: â(u, v) � 0 for all v 2 V} is a CG-submodule of V; since V is irreducible it follows that

fu 2 V : â(u, v) � 0 for all v 2 Vg � f0g: (�)Pick a basis v1, . . . , vn of V and let A be the n 3 n matrix with ij-entryâ(vi, v j). Since â is skew-symmetric, we have At � ÿA. Thereforedet (At) � (ÿ1)ndet A, so det A � (ÿ1)n det A. Also A is invertible by (�), sodet A 6� 0. It follows that n is even; as n � ÷(1) the result is proved.

7. Choose a basis f1, . . . , fn of V and de®ne the symmetric n 3 n matricesA � (aij) and B � (bij) by

aij � â1( f i, f j), bij � â( f i, f j):

By applying the Gram±Schmidt orthogonalization process, we may constructa basis f 91, : : : , f 9n of V such that â1( f 9i, f 9j) � äij for all i, j. LetP � ( pij) be the n 3 n matrix which is given by

f 9i �X

j

pij f j:

Then PAPt � I n and PBPt is symmetric. By a well known property ofsymmetric matrices, there is an orthogonal matrix Q (i.e. QQt � I) suchthat Q(PBPt)Qÿ1 is diagonal. Write Q � (qij), and de®ne the basis e1, . . . ,en of V by

ei �X

j

qij f 9j:

Chapter 23 431

Then

â1(ei, ej) � äij, since QPAPtQt � In, and

â(ei, ej) � 0 if i 6� j, since QPBPtQt is diagonal:

8. (a) The proof is similar to that of part (1) of Schur's Lemma 9.1.(b) Let v1, . . . , vn be a basis of the RG-module V, and consider the vector

space V9 over C with basis v1, . . . , vn. Then V9 is a CG-module, whichwe are assuming to be an irreducible CG-module. By Schur's Lemmathere exists ë 2 C such that vW � ëv for all v 2 V9. But v1W � ëv1 2 V, soë 2 R.

(c) Let G � C3 � ka: a3 � 1l, and let V be the RG-submodule of the regularRG-module which is spanned by 1 ÿ a and 1 ÿ a2. Then V is anirreducible RG-module. De®ne W: V! V by vW � av (v 2 V).

9. r g is a permutation, as Hxg � Hyg) Hx � Hy, and r is a homomorphismas (Hx)(r gh) � Hxgh � (Hx)(r g)(rh). We have

g 2 ker r, Hxg � Hx, 8x 2 G,xgxÿ1 2 H , 8x 2 G, g 2 \x2G xÿ1Hx:

Finally, r is a homomorphism G! Sym(Ù) � Sn with kernelT

x2Gxÿ1Hx,which is contained in H.

10. Let c1, c2 be the columns of the character table of G corresponding to theclasses {1} and tG. By the orthogonality relation for c2 we haveP

÷i(t)2 � |CG(t)| � 2 (the sum over all irreducible characters ÷i, with÷1 � 1G), so we may take ÷1(t) � 1, ÷2(t) � �1 and ÷i(t) � 0 for i > 3.Now the orthogonality of c1 and c2 gives ÷1(1) � ÷2(1) � 1 and÷2(t) � ÿ1. Further, ÷1 and ÷2 are the only linear characters, since a linearcharacter must take the value �1 on t. Hence |G : G9| � 2 by Theorem17.11. For the last part, if G is simple then since G9 v G, we haveG9 � 1, i.e. G is abelian. Hence G � C2.

Chapter 25

1. It is clear that the given set of matrices has size p( p ÿ 1). Call it G. Forclosure, note that

1 y

0 x

� �1 y90 x9

� �� 1 y9� yx

0 xx9

� �;

associativity is a property of matrix multiplication, identity is

1 0

0 1

� �; inverse of

1 y

0 x

� �is

1 ÿyxÿ1

0 xÿ1

� �.

Therefore G is a group.

2. Let ç � e2ði=5 and å � e2ði=11, and write

á � å� å3 � å4 � å5 � å9, â � å2 � å6 � å7 � å8 � å10:


3. Recall that Z�p is cyclic, so by Exercise 1.6(c), there exists an integer msuch that um � v mod p. Also, m is coprime to q, since both u and v haveorder q modulo p. Hence bm has order q. Also, bÿmabm � aum � av. Letb9 � bm. Then

G1 � ha, b9: ap � b9q � 1, b9ÿ1ab9 � avi:Hence G1 � G2.

4. (a) Note that ÿ1 is the only element of order 2 in Z�p. Hence

um � ÿ1 mod p for some m

, the element u of Z�p has even order

, q is even

, p � 1 mod 4:

(b) By Proposition 25.9, aG � {aum

: m 2 Z}. Therefore

aÿ1 2 aG , um � ÿ1 mod p for some m, p � 1 mod 4:

(c) ÷i(a) � 1 for all the q linear characters ÷i of G. Hence

0 �X÷ irred

÷(1)÷(a) � q� qö1(a)� qö2(a):

Therefore ö1(a) � ö2(a) � ÿ1. Also, |CG(a)| � p, so

p �X÷

÷(a)÷(a) � q� ö1(a)ö1(a)� ö2(a)ö2(a):

Hence ö1(a)ö1(a)� ö2(a)ö2(a) � ( p� 1)=2.If p � 1 mod 4, then a is conjugate to aÿ1 and so ÷(a) is real for all

characters ÷, so (ö1(a))2 � (ö2(a))2 � ( p� 1)=2. Hence ö1(a) and ö2(a)are (ÿ1�p p)=2.

If p � ÿ1 mod 4, then a is not conjugate to aÿ1, and it follows fromCorollary 15.6 that ö1(a) and ö2(a) are not both real. Henceö2(a) � ö1(a). This time, 2ö1(a)ö1(a) � ( p� 1)=2, and we ®nd thatö1(a) and ö2(a) are (ÿ1� i

pp)=2.

Character table of F11, 5

gi 1 a a2 b b2 b3 b4

|CG(gi)| 55 11 11 5 5 5 5

÷1 1 1 1 1 1 1 1÷2 1 1 1 ç ç2 ç3 ç4

÷3 1 1 1 ç2 ç4 ç ç3

÷4 1 1 1 ç3 ç ç4 ç2

÷5 1 1 1 ç4 ç3 ç2 ç÷6 5 á â 0 0 0 0÷7 5 â á 0 0 0 0

Chapter 25 433

(d) By Theorem 25.10, ö1(a) �P( pÿ1)=2m�1 åu m

. Since Z�p is cyclic of orderp ÿ 1 and u has order ( pÿ 1)=2, it follows that {u, u2, . . . , u( pÿ1)=2} isprecisely the set of quadratic residues modulo p. The result now followsfrom part (c).

5. Let H � ka, bl. Then for all h 2 H, the conjugacy class hE consists of hand hÿ1. All the elements outside H form a single conjugacy class of E.Also, E9 � H, so E has exactly two linear characters, say ÷1 and ÷2.A typical non-trivial linear character of H is

where ù � e2ði=3. Then ÷ " E is the irreducible character ÷3 given in thetable which follows. The characters ÷4, ÷5 and ÷6 are obtained similarly.

6. Z(E) � {1}, and for all i with 1 < i < 6, there exist gi 2 E such that gi 6� 1but ÷i(gi) � ÷i(1) (so gi 2 Ker ÷i).

7. (a) F13,3 (see Theorem 25.10).(b) C2 3 F13,3 (see Theorem 19.18).(c) D6 3 F13,3 (see Theorem 19.18).

8. The conjugacy classes of G are

f1g, fa3, a6g, far: 3 B rg, farb2: 3 B rg, farb4: 3 B rg,farb2: r � 0, 3, 6g, farb4: r � 0, 3, 6g,farb: 0 < r < 8g, farb3: 0 < r < 8g, farb5: 0 < r < 8g:

Let H1 � kal. Then H1 v G and G=H1 � C6. Hence we get six linearcharacters ÷1, . . . , ÷6 of G, as shown.

Let H2 � ka3, b2l. Then H2 v G and G=H2 � hH2a, H2bi � D6. Lift theirreducible character of D6 of degree 2 to obtain ÷7 in the table below.Then ÷8 � ÷7÷2 and ÷9 � ÷7÷3 are also irreducible. The ®nal irreduciblecharacter ÷10 can be found by using the column orthogonality relations.

1 a a2 b b2 ab ab2 a2b a2b2

÷ 1 1 1 ù ù2 ù ù2 ù ù2

Character table of E

gi 1 a b ab a2b c|CG(gi)| 18 9 9 9 9 2

÷1 1 1 1 1 1 1÷2 1 1 1 1 1 ÿ1÷3 2 2 ÿ1 ÿ1 ÿ1 0÷4 2 ÿ1 2 ÿ1 ÿ1 0÷5 2 ÿ1 ÿ1 2 ÿ1 0÷6 2 ÿ1 ÿ1 ÿ1 2 0



Chapter 26

1. Let ÷ be an irreducible character of G. Then k÷ # H, øl H 6� 0 for someirreducible character ø of H. Therefore h÷, ø " GiG 6� 0 by the FrobeniusReciprocity Theorem. But ø(1) � 1, since H is abelian; and (ø " G)(1) � p,by Corollary 21.20. Hence ÷(1) < p, and so ÷(1) � 1 or p by Theorem22.11. Assume that G has r linear characters and s irreducible characters ofdegree p. Then

r � pm for some m, by Theorem 17:11, and

r � sp2 � pn, by Theorem 11:12:

Since s � pnÿ2 ÿ pmÿ2 and s is an integer, m is at least 2.

2. {1}, {z} and {z2} are conjugacy classes of H. For all other elements h ofH, the conjugacy class hH � {h, hz, hz2}.

Character table of G � ka, b: a9 � b6 � 1, bÿ1ab � a2l

gi 1 a3 a ab2 ab4 b b2 b3 b4 b5

|CG(gi)| 54 27 9 9 9 6 18 6 18 6

÷1 1 1 1 1 1 1 1 1 1 1÷2 1 1 1 ù2 ù ÿù ù2 ÿ1 ù ÿù2

÷3 1 1 1 ù ù2 ù2 ù 1 ù2 ù÷4 1 1 1 1 1 ÿ1 1 ÿ1 1 ÿ1÷5 1 1 1 ù2 ù ù ù2 1 ù ù2

÷6 1 1 1 ù ù2 ÿù2 ù ÿ1 ù2 ÿù÷7 2 2 ÿ1 ÿ1 ÿ1 0 2 0 2 0÷8 2 2 ÿ1 ÿù2 ÿù 0 2ù2 0 2ù 0÷9 2 2 ÿ1 ÿù ÿù2 0 2ù 0 2ù2 0÷10 6 ÿ3 0 0 0 0 0 0 0 0

Character table of H (a non-abelian group of order 27)

gi 1 z z2 a a2 b ab a2b b2 ab2 a2b2

|CH (gi)| 27 27 27 9 9 9 9 9 9 9 9

÷00 1 1 1 1 1 1 1 1 1 1 1÷01 1 1 1 1 1 ù ù ù ù2 ù2 ù2

÷02 1 1 1 1 1 ù2 ù2 ù2 ù ù ù÷10 1 1 1 ù ù2 1 ù ù2 1 ù ù2

÷11 1 1 1 ù ù2 ù ù2 1 ù2 1 ù÷12 1 1 1 ù ù2 ù2 1 ù ù ù2 1÷20 1 1 1 ù2 ù 1 ù2 ù 1 ù2 ù÷21 1 1 1 ù2 ù ù 1 ù2 ù2 ù 1÷22 1 1 1 ù2 ù ù2 ù 1 ù 1 ù2

ö1 3 3ù 3ù2 0 0 0 0 0 0 0 0ö2 3 3ù2 3ù 0 0 0 0 0 0 0 0


Chapter 26 435

3. G has 11 conjugacy classes: {1}, {a8}, {ar, aÿr} (1 < r < 7),{arb: r even}, {arb: r odd}. Here, the group K which appears in Theorem26.4 is {1, a8}, and G=K � D16. The character table of D16 is given inSection 26.8 (D16 � G1) and in Section 18.3. By lifting the irreduciblecharacters of D16, we obtain the characters ÷1, . . . , ÷7 of G as shown inthe table at the top of this page. Then the four characters ø j ( j � 1, 3, 5,7) come from inducing to G those linear characters ÷ of kal for which÷(a8) � ÿ1.

4. (a) Check that AB � ÿBA, AC � ÿCA, AD � DA, BC � CB, BD � ÿDB,CD � ÿDC. Hence Z 2 G, and G=hZi is abelian while G is non-abelian.Therefore G9 � hZi (see Proposition 17.10).

(b) A2 � ÿB2 � ÿC2 � D2 � I. Since G=hZi is abelian, it follows thatg2 2 hZi for all g 2 G. Hence every element of G has the formAi BjC k Dl Z m for some i, j, k, l, m 2 f0, 1g, so jGj < 32; also G is a2-group, since g4 � 1 for all g 2 G.

(c) A routine calculation shows that every matrix which commutes with eachof A, B, C and D has the form ëI for some ë 2 C. Hence by Corollary9.3, the given representation is irreducible.

(d) Since G has irreducible representations of degrees 1 and 4,jGj > 12 � 42 � 17. Combined with part (b), this shows that jGj � 32.Since G9 � hZi, G has precisely 16 representations of degree 1. Theseare as follows: for each (r, s, t, u) with r, s, t, u 2 {0, 1}, we get arepresentation

Ai BjC k Dl Z m ! (ÿ1)ir� js�kt� lu:

Together with the irreducible representation of degree 4, these are all theirreducible representations of G, by Theorem 11.12.

5. (a) Let å � e2ði=8. We obtain representations as follows:

Character table of G � ka, b: a16 � 1, b2 � a8, bÿ1ab � aÿ1l

gi 1 a8 a a2 a3 a4 a5 a6 a7 b ab|CG(gi)| 32 32 16 16 16 16 16 16 16 4 4

÷1 1 1 1 1 1 1 1 1 1 1 1÷2 1 1 1 1 1 1 1 1 1 ÿ1 ÿ1÷3 1 1 ÿ1 1 ÿ1 1 ÿ1 1 ÿ1 1 ÿ1÷4 1 1 ÿ1 1 ÿ1 1 ÿ1 1 ÿ1 ÿ1 1÷5 2 2 0 ÿ2 0 2 0 ÿ2 0 0 0÷6 2 2

p2 0 ÿp2 ÿ2 ÿp2 0

p2 0 0

÷7 2 2 ÿp2 0p

2 ÿ2p

2 0 ÿp2 0 0ø j 2 ÿ2 cj c2 j c3 j c4 j c5 j c6 j c7 j 0 0( j � 1, 3, 5, 7)

Note: cm � e2ðim=16 � eÿ2ðim=16 � 2 cos (mð/8)


G1: a! å 0

0 åÿ1

!, b! 0 1

1 0

!;

G2: a! å 0

0 åÿ1

!, b! 0 1

ÿ1 0

!;

G3: a! å 0

0 å3

!, b! 0 1

1 0

!;

G4: a! å 0

0 å5

!, b! 0 1

1 0

!;

G5: a! 1 0

0 ÿ1

!, b! 0 1

1 0

!, z! i 0

0 i

!:

(b) If j � 5, 6, 7 or 8 then Z(Gj) � {1, a2, z, a2z} � C2 3 C2. Since Z(Gj)is not cyclic, Gj has no faithful irreducible representation, by Proposition9.16.

(c) Check that the matrices satisfy the required relations, so giverepresentations. It is easy to see that the matrices generate groups withmore than eight elements, so the representations are faithful.

(d) The following give faithful representations:

G7: a!i 0 0

0 ÿi 0

0 0 1

0B@1CA, b!

0 1 0

1 0 0

0 0 1

0B@1CA, z!

1 0 0

0 1 0

0 0 ÿ1

0B@1CA;

G8: a!i 0 0

0 ÿi 0

0 0 1

0B@1CA, b!

0 1 0

ÿ1 0 0

0 0 1

0B@1CA, z!

1 0 0

0 1 0

0 0 ÿ1

0B@1CA:

6. The following table records the numbers of elements of orders 1, 2, 4 and 8in G1, . . . , G9:

Therefore no two of G1, . . . , G9 are isomorphic, except possibly G5 andG8. But G5=G59 � C2 3 C4, while G8=G89 � C2 3 C2 3 C2, so G5 6� G8.

7. (a) By Lemma 26.1(1) we have {1} 6� Z(G) 6� G. Also jG=Z(G)j 6� p byLemma 26.1(2). Therefore jZ(G)j � p or p2. Assume that jZ(G)j � p2.If g =2 Z(G) then Z(G) < CG(g) 6� G, and g 2 CG(g). Hence

G1 G2 G3 G4 G5 G6 G7 G8 G9

Order 1 1 1 1 1 1 1 1 1 1Order 2 9 1 5 3 3 7 11 3 5Order 4 2 10 6 4 12 8 4 12 10Order 8 4 4 4 8 0 0 0 0 0

Chapter 26 437

jCG(g)j � p3 and jgGj � p. Therefore, G has p2 � ( p4 ÿ p2)=pconjugacy classes.

(b) Assume that G has r irreducible characters of degree 1 and s irreduciblecharacters of degree p. Since

P÷(1)2 � p4 (Theorem 11.12), there are

no irreducible characters of degree greater than p, so r � sp2 � p4.Therefore jG=G9j � r � p2 or p3, and if r � p2 then r � s � 2 p2 ÿ 1.Part (b) follows, as r � s is equal to the number of conjugacy classes ofG.

(c) Note that G9 \ Z(G) 6� {1} by Lemma 26.1(1). By parts (a) and (b), if|Z(G)| � p2 then |G9| � p; and if |G9| � p2 then |Z(G)| � p. Hence|G9 \ Z(G)| � p.

8. (a) Let Z � Z(G), and assume that

G=Z � haZ, bZi, with a4 2 Z, a2 Z � b2 Z, bÿ1abZ � aÿ1 Z:

Then a2 � b2z for some z 2 Z, and hence ba2 � b3z � b2zb � a2b. Sincea2 commutes with a, b and all elements in Z, we have a2 2 Z.Therefore G=Z 6� Q8.

(b) If G is a non-abelian group of order 16, then by Exercise 7, eitherG9 \ Z(G) � G9, in which case G=(G9 \ Z(G)) is abelian, orG9 \ Z(G) � Z(G), in which case G=(G9 \ Z(G)) 6� Q8 by part (a).

Chapter 27

1. Assume that

z � a b

c d

� �2 Z(SL (2, p)):

Then

z1 1

0 1

� �� 1 1

0 1

� �z) c � 0, and

z0 1

ÿ1 0

� �� 0 1

ÿ1 0

� �z) c � ÿb, a � d:

Therefore z � aI; and since z 2 SL (2, p), we have a2 � 1, so a � �1.

2. Check that

1 1

0 1

� �and

1 ÿ1

0 1

� �are elements of G � SL (2, 3) of order 3 which are not conjugate to eachother. The element

ÿ1 0

0 ÿ1

� �lies in Z(G), and


0 1

ÿ1 0

� �

has order 4. Hence the following are conjugacy class representatives:

g1 g2 g3 g4

1 0

0 1

� � ÿ1 0

0 ÿ1

� �0 1

ÿ1 0

� �1 1

0 1

� �

Order of gi 1 2 4 3|CG(gi)| 24 24 4 6

g5 g6 g7

1 ÿ1

0 1

� � ÿ1 1

0 ÿ1

� � ÿ1 ÿ1

0 ÿ1

� �

Order of gi 3 6 6|CG(gi)| 6 6 6

We now describe how to construct the character table of G, which is givenbelow.

First observe that the vector space (Z3)2 has exactly four 1-dimensionalsubspaces, namely the spans of the vectors (0, 1), (1, 1), (2, 1) and (1, 0).The group G permutes these subspaces among themselves, so we obtain ahomomorphism ö: G! S4. Check that Kerö � {�I}. HenceG=f�Ig � Imö, a subgroup of S4 of order 12; therefore G=f�Ig � A4.The characters ÷1, ÷2, ÷3, ÷4 of G are obtained by lifting to G theirreducible characters of A4 (which are given in Section 18.2).

The values of ÷5, ÷6, ÷7 on the elements g1, g2, g3 can be deduced fromthe column orthogonality relations. Note that G has three real conjugacyclasses, so by Theorem 23.1, one of ÷5, ÷6 and ÷7 must be real. Assume,without loss of generality, that ÷5 is real. Then ÷5(g4) � á, where á is real.Also á 6� 0, by Corollary 22.27. Since ÷5÷2 and ÷5÷3 are irreduciblecharacters of G of degree 2, whose values on g4 are áù and áù2, theymust be ÷6 and ÷7 in some order, say ÷5÷2 � ÷6 and ÷5÷3 � ÷7. Theequation

Pj÷ j(g4)÷ j(g4) � 6 gives áá � 1. Since á is real, á � �1. Then

á � ÿ1 since ÷5(g4) � ÷5(1) mod 3. Now note that for j � 5, 6, 7, Exercise13.5 implies that ÷ j(g7) � ÿ÷ j(g4). Finally, ÷(g5) � ÷(g4) and ÷(g6) � ÷(g7)for all ÷.

Chapter 27 439

3. Apply Proposition 17.6.

4. (a) For the character table of T, notice that T is isomorphic to the group oforder 21 whose character table is found in Exercise 17.2 and Example21.25. Representatives of the conjugacy classes of T are h1, . . . , h5,where

h1 �1 0

0 1

!Z, h2 �

2 0

0 4

!Z, h3 �

4 0

0 2

!Z,

h4 �1 1

0 1

!Z, h5 �

1 ÿ1

0 1

!Z:

Two of the linear characters of T are

where ù � e2ði=3. The values of 1T : G and ë : G are as follows (seeProposition 21.23):

We ®nd that k1T : G, 1T : Gl � 2 and k1T : G, 1Gl � 1. Hence1T : G � 1G � ÷, where ÷ is an irreducible character of G. Also, kë : G,ë : Gl � 1, so ë : G is irreducible; write ö � ë : G.

Character table of SL (2, 3)

gi g1 g2 g3 g4 g5 g6 g7

|CG(gi)| 24 24 4 6 6 6 6

÷1 1 1 1 1 1 1 1÷2 1 1 1 ù ù2 ù2 ù÷3 1 1 1 ù2 ù ù ù2

÷4 3 3 ÿ1 0 0 0 0÷5 2 ÿ2 0 ÿ1 ÿ1 1 1÷6 2 ÿ2 0 ÿù ÿù2 ù2 ù÷7 2 ÿ2 0 ÿù2 ÿù ù ù2


hi h1 h2 h3 h4 h5

|CT (hi)| 21 3 3 7 7

1T 1 1 1 1 1ë 1 ù ù2 1 1


|CG(gi)| 168 8 4 3 7 7

1T " G 8 0 0 2 1 1ë " G 8 0 0 ÿ1 1 1


(c) The values of ÷ and ÷S are as shown below (see Proposition 19.14):

We ®nd that k÷S , 1Gl � k÷S , öl � k÷S , ÷l � 1. Hence there is a characteræ of G such that

÷S � 1G � ö� ÷� æ:

The values of æ are as shown above. We calculate that kæ, æl � 4, soeither æ � 2ø for some irreducible character ø, or æ is the sum of fourdistinct irreducible characters (cf. Exercise 14.7).Now 1G, ö and ÷ are three of the six irreducible characters of G, andnone is a constituent of æ. Since there are only six irreducible charactersin all, æ cannot therefore be the sum of four irreducible characters, andso æ � 2ø with ø irreducible. The values of ø are as shown above.

(d) The characters 1G, ö, ÷ and ø are the characters ÷1, ÷3, ÷2 and ÷6,respectively, in the character table of G given at the end of Chapter 27.The remaining irreducible characters ÷4, ÷5 can readily be calculatedusing the column orthogonality relations (noting that gi is real if andonly if 1 < i < 4).

5. (a) Compare the proof of Lemma 27.1. Note that because g2 lies in Z(G),gi and gig2 have the same centralizer for all i.

(b) By lifting, we obtain the characters ÷1, . . . , ÷6 in the character tableshown below.

(c) Use Exercise 13.5 (noting that ÷ j(ÿI) 6� ÷ j(I) for 7 < j < 11, since ÿI isnot in the kernel of these characters).

(d) SinceP

11j�1 ÷ j(g3)÷ j(g3) � 8, we deduce that ÷ j(g3) � 0 for 7 < j < 11.

(Alternatively, apply part (c).) Also, by Corollary 22.27, ÷ j(1) is even.(e) By Theorem 22.16, ÷(g6) 2 Z for all characters ÷. SinceP

11j�1 (÷ j(g6))2 � 6, the values of ÷ j(g6) for 7 < j < 11 must be �1,

�1, �1, 0, 0, in some order. By Corollary 22.27 again, two of ÷7, . . . ,÷11, say ÷9 and ÷10, have degrees divisible by 6. Further,P

11j�1 (÷ j(1))2 � 168, and 122 � 62 . 168, so ÷9(1) � ÷10(1) � 6.Next, ÷7(1)2 � ÷8(1)2 � ÷11(1)2 � 96. The only possibility is that two

of ÷7(1), ÷8(1), ÷11(1), say ÷7(1) and ÷8(1), are equal to 4, and÷11(1) � 8. The congruences ÷(1) � ÷(g6) mod 3 now give the remainingvalues on g6. Use part (c) to ®ll in the values on g2 and g7.

(f ) By Proposition 19.14, øA has the following values on g1, g2, g3 and g6:


|CG(gi)| 168 8 4 3 7 7

÷ 7 ÿ1 ÿ1 1 0 0÷S 28 4 0 1 0 0æ 12 4 0 0 ÿ2 ÿ2ø 6 2 0 0 ÿ1 ÿ1

g1 g2 g3 g6

øA 6 6 2 0

Chapter 27 441

The values of øA on g1 and g2 show that øA is a linear combination of÷1, ÷4, ÷5 and ÷6. Then the value of øA on g6 shows that ÷1 is not aconstituent of øA; ®nally, the value on g3 forces øA � ÷6.Now g2

4 is conjugate to g3. Hence(ø(g4)2 ÿ ø(g3))=2 � øA(g4) � ÷6(g4) � 0, and therefore, ø(g4) � 0.Similarly, ø(g5) � 0.Let x � ø(g8). Since g2

8 is conjugate to g8, we get(x2 ÿ x)=2 � øA(g8) � ÿ1. Therefore x � (1� i

p7)=2. Say

÷7(g8) � (1ÿ ip

7)=2. Then ÷8 � ÷7. For all ÷, ÷(g10) � ÷(g8); using thisfact and part (c), we ®ll in the values of ÷7 and ÷8.

(g) For i 6� 6, we haveP11

j�1 ÷ j(gi)÷ j(g6) � 0. This allows us to ®ll in thevalues of ÷11. Since g4 is conjugate to gÿ1

4 , ÷(g4) is real for all ÷. ThenP11j�1÷ j(g1)÷ j(g4) � 0 and

P11j�1÷ j(g4)2 � 8 imply that

÷9(g4) � ÿ÷10(g4) � �p2. Say ÷9(g4) � p2. The column orthogonalityrelations now let us ®nd the remaining values of ÷9 and ÷10, therebycompleting the character table of G.

6. Let Z � {�I} and de®ne the subgroup T of G, of order 55, by

T � a b

0 aÿ1

� �Z: a 2 Z�11, b 2 Z11

� �:

Then T is generated by

x � 1 1

0 1

� �Z and y � 2 0

0 6

� �Z,

and T has ®ve linear characters æ j (0 < j < 4), where

æ j: xuyv ! e2ði jv=5:

Character table of SL (2, 7)

gi g1 g2 g3 g4 g5 g6 g7 g8 g9 g10 g11

Order of gi 1 2 4 8 8 3 6 7 14 7 14|CG(gi)| 336 336 8 8 8 6 6 14 14 14 14

÷1 1 1 1 1 1 1 1 1 1 1 1÷2 7 7 ÿ1 ÿ1 ÿ1 1 1 0 0 0 0÷3 8 8 0 0 0 ÿ1 ÿ1 1 1 1 1÷4 3 3 ÿ1 1 1 0 0 á á á á÷5 3 3 ÿ1 1 1 0 0 á á á á÷6 6 6 2 0 0 0 0 ÿ1 ÿ1 ÿ1 ÿ1÷7 4 ÿ4 0 0 0 1 ÿ1 ÿá á ÿá á÷8 4 ÿ4 0 0 0 1 ÿ1 ÿá á ÿá á÷9 6 ÿ6 0

p2 ÿp2 0 0 ÿ1 1 ÿ1 1

÷10 6 ÿ6 0 ÿp2p

2 0 0 ÿ1 1 ÿ1 1÷11 8 ÿ8 0 0 0 ÿ1 1 1 ÿ1 1 ÿ1

Note: á � (ÿ1� ip

7)=2


The characters æ1 " G and æ2 " G are irreducible; they are ÷3 and ÷4 in thetable. (In calculating ÷3(g5), note that e2ði=5 � eÿ2ði=5 � (ÿ1�p5)=2:) Let÷1 � 1G. We have hæ0 " G, ÷1i � 1 and hæ0 " G, æ0 " Gi � 2. Henceæ0 " G � ÷1 � ÷2 for an irreducible character ÷2 of G.

We have now found four of the eight irreducible characters of G, namely÷1, ÷2, ÷3, ÷4.

SinceP

8j�1 ÷ j(g5)÷ j(g5) � 5, we deduce that the remaining irreducible

characters ÷5, ÷6, ÷7, ÷8 take the value 0 on g5. By Corollary 22.27,÷ j(1) � 0 mod 5 for 5 < j < 8. But

P8j�5 (÷ j(1))2 � 250; hence, without loss

of generality, ÷5(1), ÷6(1), ÷7(1), ÷8(1) are 10, 10, 5, 5, respectively.Note that ÷(gj) is an integer for 1 < j < 4 and all characters ÷, by

Theorem 22.16.Since ÷(1) � ÷(g3) mod 3 for all characters ÷, and

P8j�1 ÷ j(g3)2 � 6, the

values of ÷ j(g3) (5 < j < 8) are as shown.Now ÷(g4) � ÷(g3) mod 2 for all ÷, and

P8j�1 ÷ j(g4)2 � 6, so ÷ j(g4) � �1

for 5 < j < 8. Next, ÷(g2) � ÷(g4) mod 3 for all ÷, andP

8j�1 ÷ j(g2)2 � 12;

hence |÷(g2)| , 3 for all irreducible ÷. We may now conclude from the factsthat ÷(g2) � ÷(g1) mod 2 and

P8j�1 ÷ j(g2)2 � 12, that ÷ j(g2) � �2 for j � 5,

6 and ÷ j(g2) � �1 for j � 7, 8. By consideringP

8j�1 ÷ j(1)÷ j(g2) � 0, we

see that ÷7(g2) � ÷8(g2) � 1, and ÷5(g2), ÷6(g2) have opposite signs;without loss of generality, ÷5(g2) � 2 � ÿ÷6(g2).

We have now completed columns 1, 2, 3 and 5 of the character table.Since ÷(g4) � ÷(g2) mod 3 for all characters ÷, and

P8j�1 ÷ j(g4)2 � 6, we

can complete column 4.The column orthogonality relations now enable us to ®nish the character

table.


gi g1 g2 g3 g4 g5 g6 g7 g8

Order of gi 1 2 3 6 5 5 11 11|CG(gi)| 660 12 6 6 5 5 11 11

÷1 1 1 1 1 1 1 1 1÷2 11 ÿ1 ÿ1 ÿ1 1 1 0 0÷3 12 0 0 0 á â 1 1÷4 12 0 0 0 â á 1 1÷5 10 2 1 ÿ1 0 0 ÿ1 ÿ1÷6 10 ÿ2 1 1 0 0 ÿ1 ÿ1÷7 5 1 ÿ1 1 0 0 ã ã÷8 5 1 ÿ1 1 0 0 ã ã

Note: á � (ÿ1�p5)=2, â � (ÿ1ÿp5)=2 and ã � (ÿ1� ip

11)=2

Chapter 27 443

Chapter 28

1. We take g1, : : : , g8 as representatives of the conjugacy classes, where

g1 �1 0

0 1

!g2 �

2 0

0 2

!g3 �

1 1

0 1

!g4 �

2 1

0 2

!

g5 �1 0

0 2

!g6 �

0 1

2 0

!g7 �

0 1

1 2

!g8 �

0 1

1 1

!:

The character table of GL(2, 3) is then as follows.

gi g1 g2 g3 g4 g5 g6 g7 g8

|CG(gi)| 48 48 6 6 4 8 8 8

ë0 1 1 1 1 1 1 1 1ë1 1 1 1 1 ÿ1 1 ÿ1 ÿ1ø0 3 3 0 0 1 ÿ1 ÿ1 ÿ1ø1 3 3 0 0 ÿ1 ÿ1 1 1ø0,1 4 ÿ4 1 ÿ1 0 0 0 0÷1 2 ÿ2 ÿ1 1 0 0 i

��2p ÿi

��2p

÷2 2 2 ÿ1 ÿ1 0 2 0 0÷4 2 ÿ2 ÿ1 1 0 0 ÿi

��2p

i��2p

2. Every element r of Fq can be expressed as a square, since r � rq and q iseven.

Suppose thata b

c d

� �2 GL(2, q). We may write ad ÿ bc as s2 for some

s in F�q . Then

a b

c d

� �� s 0

0 s

� �a=s b=s

c=s d=s

� �:

The ®rst matrix in the product is sI and the second belongs to SL(2, q). Itnow follows easily that GL(2, q) � Z 3 SL(2, q) where Z � fsI : s 2 F�q g.

You should have no dif®culty in proving that the conjugacy classes ofSL(2, q) have representatives as follows.(a) The identity I has centralizer of order q3 ÿ q.

(b) The matrix u1 � 1 1

0 1

� �has centralizer of order q.

(c) There are (qÿ 2)=2 conjugacy classes with representatives

ds,sÿ1 � s 0

0 sÿ1

� �, indexed by unordered pairs fs, sÿ1g of elements

from FqnF2. Each such element has centralizer of order qÿ 1.


(d) There are q=2 conjugacy classes with representatives

vr � 0 1

1 r � rÿ1

� �, indexed by unordered pairs fr, rÿ1g of elements

from Fq2nFq such that r1�q � 1. Each such element has centralizer oforder q� 1.

By restricting characters from GL(2, q) to SL(2, q) you will quicklybe able to prove that the character table of SL(2, q) is as follows.

Here, we have used the function r! r de®ned in (28.3). The subscriptsfor ø0,i satisfy 1 < i < (qÿ 2)=2, and the subscripts for ÷i satisfy1 < i < q=2.

If q 6� 2 then the kernel of every non-trivial character is the identitysubgroup, and therefore SL(2, q) is simple.

3. Note ®rst that PSL(2, 8) � SL(2, 8).The polynomial x3 � x� 1 is irreducible over F2. Hence we may write

F8 � fa� bç� cç2 : a, b, c 2 F2 and ç3 � 1� çg:The pairs fs, sÿ1g of elements from F8nF2 are

fç, 1� ç2g, fç2, 1� ç� ç2g, f1� ç, ç� ç2g:These give us the conjugacy class representatives g3, g4, g5 below.

The irreducible monic quadratics over F8 with constant term 1 are

x2 � x� 1, x2 � çx� 1, x2 � ç2x� 1, x2 � (ç� ç2)x� 1:

The companion matrices for these quadratics give use the conjugacy classrepresentatives g6, g7, g8, g9 below

We can now list representatives g1, : : : , g9 of the conjugacy classes ofSL(2, 8), as follows.

g1 �1 0

0 1

!g2 �

1 1

0 1

!

g3 �ç 0

0 1� ç2

!g4 �

ç2 0

0 1� ç� ç2

!g5 �

1� ç 0

0 ç� ç2

!

g6 �0 1

1 1

!g7 �

0 1

1 ç

!g8 �

0 1

1 ç2

!g9 �

0 1

1 ç� ç2

!:

We may choose a generator å of F�64 so that å7 � åÿ7 � ç. Then

I u1 ds,sÿ1 vr

ë0 1 1 1 1ø0 q 0 1 ÿ1ø0,i q� 1 1 si � sÿi 0÷i qÿ 1 ÿ1 0 ÿ(r i � rÿi)

Chapter 28 445

å14 � åÿ14 � ç2, å21 � åÿ21 � 1 and å28 � åÿ28 � ç4 � ç� ç2. Thecharacter table of SL(2, 8) is then as follows.

Here, the 3 3 3 submatrices A and B are given by

A �2 cos(2ð=7) 2 cos(4ð=7) 2 cos(6ð=7)

2 cos(4ð=7) 2 cos(6ð=7) 2 cos(2ð=7)

2 cos(6ð=7) 2 cos(2ð=7) 2 cos(4ð=7)

0B@1CA

B �ÿ2 cos(2ð=9) ÿ2 cos(4ð=9) ÿ2 cos(8ð=9)

ÿ2 cos(4ð=9) ÿ2 cos(8ð=9) ÿ2 cos(2ð=9)

ÿ2 cos(8ð=9) ÿ2 cos(2ð=9) ÿ2 cos(4ð=9)

0B@1CA:

Chapter 29

1. (a) It is straightforward to check that ö is a homomorphism. For x, y 2 G,the element (x, y) 2 G 3 G sends x to y, so that action is transitive.

(b) (G 3 G)1 � f(g, g) : g 2 Gg, and kerö � f(z, z) : z 2 Z(G)g.(c) Every orbit of G 3 G on G 3 G contains an ordered pair of the form

(1, x), and if (g, h) sends (1, x) to (1, y) then g � h and y � gÿ1xg.Hence if C1, : : : , Ck are the conjugacy classes of G, and xi 2 Ci, then(1, xi) (1 < i < k) are orbit representatives for the action of G 3 G onG 3 G, and so the rank is equal to k.

Since x((g, h)ö) � x if and only if xhxÿ1 � g, we see that ð(g, h) �jfixG(g, h)j is equal to 0 if g is not conjugate to h, and is equal tojCG(x)j if g is conjugate to h (since in the latter case, if xhxÿ1 � g thenan arbitrary element y 2 G such that yhyÿ1 � g is of the form y � xcwith c 2 CG(x)). Hence using the column orthogonality relations we seethat ð �P÷ 3 ÷.

2. There are q2 ÿ 1 non-zero vectors in V, and each 1-dimensional subspace

gi g1 g2 g3 g4 g5 g6 g7 g8 g9

|CG(gi)| 504 8 7 7 7 9 9 9 9

ë0 1 1 1 1 1 1 1 1 1ø0 8 0 1 1 1 ÿ1 ÿ1 ÿ1 ÿ1ø0,1 9 1 0 0 0 0ø0,2 9 1 A 0 0 0 0ø0,3 9 1 0 0 0 0÷3 7 ÿ1 0 0 0 ÿ2 1 1 1÷1 7 ÿ1 0 0 0 1÷2 7 ÿ1 0 0 0 1 B÷4 7 ÿ1 0 0 0 1


contains qÿ 1 of them; also two 1-dimensional subspaces have no non-zerovectors in common. Hence jÙj � (q2 ÿ 1)=(qÿ 1) � q� 1.

3. Use the notation for the conjugacy classes and irreducible characters ofGL(2, q) given in Proposition 28.4 and Theorem 28.5. It is easy to checkthat ð takes the values q2 ÿ 1, qÿ 1, qÿ 1 on the classes withrepresentatives I, u1, d1, t respectively, and takes the value 0 on all otherclasses. Taking inner products we ®nd

hð, 1Gi � hð, ø0i � hð, ø0, ji � 1 (1 < j < qÿ 2):

As 1G � ø0 �Pqÿ2

1 ø0, j has degree q2 ÿ 1 � ð(1), we conclude that

ð � 1G � ø0 �Pqÿ2

1 ø0, j.

4. Observe that the coset H1x is ®xed by g if and only if xgxÿ1 2 H1. If G isabelian this amounts to g 2 H1, and hence we see that ð1(g) � 0 if g =2 H1

and ð1(g) � jG : H1j if g 2 H1. Thus H1 � fg 2 G : ð1(g) 6� 0g. Likewisefor H2; since ð1 � ð2 we deduce that H1 � H2.

As a counterexample for G non-abelian, takeG � D8 � ha, b : a4 � b2 � 1, bÿ1ab � aÿ1i with H1 � hbi, H2 � ha2bi.Then ð1 � ð2 but H1 6� H2.

5. By Proposition 29.4 we have 1 � 1jGjP

g2GjfixÙ(g)j, hencePjfixÙ(g)j �

jGj. Since jfixÙ(g)j is a non-negative integer for each g, andjfixÙ(1)j � jÙj. 1, we must have jfixÙ(g)j � 0 for some g.

6. Write ð � ð(nÿ2,1,1). Calculating inner products using Proposition 29.6, as inthe proof of Theorem 29.13, we ®nd

hð, ði � 7, hð, 1i � 1, hð, ð(nÿ1,1)i � 3, hð, ð(nÿ2,2)i � 4:

Using Theorem 29.13 it follows that ð(nÿ2,1,1) � 1� 2÷(nÿ1,1) � ÷(nÿ2,2) � ÷with ÷ irreducible. Hence ÷ � 1� ðÿ ð(nÿ1,1) ÿ ð(nÿ2,2), from which it iseasy to calculate that ÷(1) � 1

2(nÿ 1)(nÿ 2),

÷(12) � 12(nÿ 2)(nÿ 5), ÷(123) � 1

2(nÿ 4)(nÿ 5). For n � 6, ÷(4,1,1) is the

character ÷5 in Example 19.17.

Chapter 30

1. By Theorem 30.4, a245 � 168/(8´3) � 7. Hence, by (30.3), PSL (2, 7)contains elements a and b such that a has order 2, b has order 3 and abhas order 7.

2. No: a225 � (1� (ÿ1� ip

7)=6� (ÿ1ÿ ip

7)=6ÿ 4=6)168=(8:8) � 0, andsimilarly a226 � 0. Hence PSL (2,7) does not contain two involutions whoseproduct has order 7.

3. Yes. Number the conjugacy classes of PSL (2, 11) as in the solution toExercise 27.6. Then

a235 � 660

12:61� 1

11

� �� 10:

Therefore PSL (2, 11) contains elements x and y such that x, y and xy haveorders 2, 3 and 5, respectively. Let H be the subgroup kx, yl of PSL (2, 11).

Chapter 29 447

There is a homomorphism W from A5 onto H (W sends a! x, b! y).Since Ker W 3 A5 and A5 is simple, we deduce that H � A5.

4. Suppose that G is a group whose character table iswhere á � (1�p5)=2, â � (1ÿp5)=2.

For 1 < i < 5 we have jCG(gi)j �P5

j�1÷ j(gi)÷ j(gi). Therefore thecentralizers of g1, g2, g3, g4, g5 have orders 60, 3, 4, 5, 5, respectively.Hence the orders of g2, g4 and g5 are 3, 5 and 5; also the order of g3

must be 2, since for no other i (except i � 1) is |CG(gi)| even.Now a324 � 60=(4 . 3). Therefore G contains elements x and y such that x

has order 2, y has order 3 and xy has order 5. As in the solution toExercise 3, G has a subgroup H with H � A5. Since jGj � 60, we haveG � A5.

5. (a) Using the fact thatP

7j�1÷ j(gi)÷ j(gi) � |CG(gi)|, we ®nd that the

centralizer orders and class sizes are as follows:

Hence jGj � 360. Also G is simple, by Proposition 17.6.(b) By the Frobenius±Schur Count of Involutions (Corollary 23.17), the

number of involutions t in G is bounded by

1� t <X7

j�1

÷ j(1) � 46:

By considering jCG(gi)j, we see that gi has even order only for i � 2and 3. Since t < 45, only g2 can be an involution. Hence g3 has order4. (This information about the orders of g2 and g3 can also be deducedusing Sylow's Theorem.)

(c) Clearly g6 and g7 have order 5, and at least one of g4 and g5 has order3. If j � 4 or 5 and k � 6 or 7 then

g1 g2 g3 g4 g5

÷1 1 1 1 1 1÷2 4 1 0 ÿ1 ÿ1÷3 5 ÿ1 1 0 0÷4 3 0 ÿ1 á â÷5 3 0 ÿ1 â á

g1 g2 g3 g4 g5 g6 g7

|CG(gi)| 360 8 4 9 9 5 5|gG

i | 1 45 90 40 40 72 72


a2 jk � jGjjCG(g2)j jCG(gj)j

X÷

÷(g2)÷(gj)÷ (gk)

÷(1)

� 360

8 . 9� 5:

Therefore G contains elements x and y such that x has order 2, y hasorder 3 and xy has order 5. As in the solution to Exercise 3, thesubgroup H � kx, yl of G is isomorphic to A5.

(d) If g, h 2 G then

(gh)r: Hx! Hxgh, and

(gr)(hr): Hx! Hxg ! Hxgh:

Hence r is a homomorphism.(e) Since G is simple, Ker r � {1}. Hence G is isomorphic to a subgroup K

of S6. Since jS6: Kj � 6!=360 � 2, K must be A6.

6. Consider the ®gure in Example 30.6(3). We shall explain how to label thevertices by elements of G.Choose a vertex and label it 1. Label the vertices according to thefollowing inductive rule. Assume that v is a vertex and an adjacent vertex uis labelled by g. Then label v by

ga if the edge uv has no arrow,

gb if the edge uv has an arrow from u to v,

gbÿ1 if the edge uv has an arrow from v to u:

For example, if you decided to label the bottom left-hand vertex by 1, thenpart of the labelling would be

The relation a2 � 1 ensures that the labelling is consistent along unmarkededges; since b3 � 1, the labelling is consistent around triangles; and therelation abababab � 1 deals with the octagons.

Every element in G has the form given by the label of one of the 24vertices, so jGj < 24.

7. The conjugacy classes of PSL(2, 7) are given in Lemma 27.1 The elementg2 is an involution with centralizer of order 8 given in the proof of

Lemma 27.1. Letting a � 2 2

ÿ2 2

� �, b � 2 4

4 ÿ2

� �, we see that the

centralizer is generated by a and b, and a4 � b2 � 1, bÿ1ab � aÿ1ÿ, hencethe centralizer is isomorphic to D8.

As in Exercises 12.3 and 12.4, we see that CA6((12)(34)) has order 8 and

is generated by (1324) and (13)(24), hence as above is isomorphic to D8.

Chapter 30 449

8. Let G be the simple group PSL(2, 17). In the ®eld Z17 the element 4 is a

fourth root of unity, so t � 4 0

0 ÿ4

� �Z is an involution. Calculate that

CG(t) is generated by the group of diagonal matrices together with

b � 0 1

ÿ1 0

� �Z, hence is generated by b and a � 3 0

0 6

� �Z. As

a8 � b2 � 1 and bÿ1ab � aÿ1, we have CG(t) � D16.

Chapter 31

1. Assume that G has an abelian subgroup H of index pr ( p prime), and that|G| . p. If H � {1} then |G| � pr and so G is not simple (see Lemma26.1(1)). So assume that H 6� {1}; pick 1 6� h 2 H. Then H < CG(h) as His abelian, so |G:CG(h)| is a power of p. If |G:CG(h)| � 1 then khl v G andG is not simple. And if |G:CG(h)| . 1, then G is not simple by Theorem31.3.

2. By Burnside's Theorem, jGj is divisible by at least three distinct primes.Since 3 . 5 . 7 . 80, jGj is even. Then by Exercise 13.8, |G| is divisible by4. Since 4 . 3 . 7 . 80, the only possibility is that jGj � 4 . 3 . 5 � 60.

Chapter 32

1. (a) The fact that BBt � I follows from the observation that for all i, j,

d(eib, ejb) � d(ei, ej) � äij:

Since 1 � det I � (det B)(det Bt) � (det B)2, we have det B � �1.(b) (i) The eigenvalues of C are the roots of det (C ÿ xI), which is a cubic

polynomial over R. Therefore, C has one or three real eigenvalues. (ii)Moreover, the product of the eigenvalues of C is det C � 1. If C hasthree real eigenvalues, then they cannot all be negative; and if C has onereal eigenvalue ë and a pair of conjugate non-real eigenvalues ì, ì, thenëìì � 1, and hence ë . 0. Therefore C has a real positive eigenvalue,say ë. (iii) Let v be an eigenvector for ë. Then

d(v, v) � d(vC, vC) � d(ëv, ëv) � ë2d(v, v),

and so ë � 1.(c) Let c be the isometry v! vC. By (b), c ®xes a vector v; it is now easy

to convince yourself that c must be a rotation about the axis through v.The required result for b follows from the de®nition of c.

(d) Take three orthogonal axes, one of which is the axis of the rotation b.With respect to these axes, the matrix of b is

1 0 0

0 cosö sinö0 ÿsinö cosö

0@ 1A:Hence tr B � 1 � 2 cosö.


2. We regard G as a subgroup of O(R3). It is easy to see that the translationsubmodule T (which consists of all the translation modes) is isomorphic tothe RG-module given by the natural action of G on R3. Hence by part (d)of Exercise 1,

÷T (g) �

1� 2 cosö, if g is a rotation through ö, about some

axis,

ÿ(1� 2 cosö), if the element ÿ g of O(R3) is a rotation

through ö:

8>>>>>><>>>>>>:Now consider the rotation submodule R, which consists of all the rotationmodes. A rotation mode is speci®ed by a 3-dimensional vector öv, where vis a unit vector along the axis of the rotation and ö denotes the angle ofrotation, taken positive in the right-hand screw sense. Let g 2 G, andconsider g acting on öv. It sends v to vg, and if g is a rotation, itpreserves the sense of the rotation; however, if g is a re¯ection then ittransforms a right-hand screw to a left-hand screw, and hence sends öv to(ÿö)(vg). Therefore

÷R(g) �÷T (g), if g is a rotation,

ÿ÷T (g), if g is not a rotation,

(and so ÷T � ÷R has the required form.

3. The matrix A is

k

m

ÿ3=2 0 3=4 ÿp3=4 3=4p

3=4

0 ÿ1=2 ÿp3=4 1=4p

3=4 1=4

3=4 ÿp3=4 ÿ3=4p

3=4 0 0

ÿp3=4 1=4p

3=4 ÿ5=4 0 1

3=4p

3=4 0 0 ÿ3=4 ÿp3=4p3=4 1=4 0 1 ÿp3=4 ÿ5=4

0BBBBBB@

1CCCCCCA:4. A simpler basis is given by

12(r1 � r2) � (v12 � v21)ÿ (v34 � v43),

12(r1 � r3) � (v13 � v31)ÿ (v24 � v42),

12(r1 � r4) � (v14 � v41)ÿ (v23 � v32):

We chose r1, r2, r3, r4 to be the images of w1, w2, w3, w4 under an RG-isomorphism. (Compare the construction of the matrix B, towards the endof Example 32.20.)

5. (a) This is a routine geometrical exercise.(b) Let the displaced positions of the atoms be 09, 19, 29, 39, 49. The

distance of 19 from the plane through 1 perpendicular to 12 isx12 � 1

2(x13 � x14). Similarly, the distance of 29 from the plane through 2

perpendicular to 12 is x21 � 12(x23 � x24). Therefore 12 has decreased by

x12 � x21� 12(x13 � x14 � x23 � x24). The other calculations can be done in

the same way.(c) We express the force at each atom as a vector, and then write this vector

Chapter 32 451

as a linear combination of our three chosen unit vectors at the initialposition of the atom. Let dij denote the decrease in the length ij, ascalculated in part (b). Then, for example, at vertex 1, the contributionsto the component of the force vector in the direction 12 are as follows:ÿk1d12 from the force between atoms 1 and 2; zero from the forcebetween atoms 1 and 3 and from the force between atoms 1 and 4; andÿk2d10 sec (/012) from the force between atoms 1 and 0. Hence

m1�x12 � ÿk1d12 ÿ 13

p(3=2)k2d10:

Upon substituting for d12 and d10 from part (b), we obtain the givenexpression for �x12.

The other accelerations are calculated in the same way.(d) The entries in the 15 3 15 matrix A are the coef®cients which appear in

the equations of motion �x � xA. If you write down the matrix A, thenyou will easily verify that the given vectors are eigenvectors of A (witheigenvalues

ÿ(4k1 � k2)=m1, 0, 0, 0, ÿk1=m1, ÿk1=m1,

respectively).(e) The matrix B is

ÿ(6k1 � k2)=3m1 ÿ2k2=3m1 ÿ4k2

p2=(m2

p3)

ÿ(3k1 � k2)=3m1 ÿ2k2=3m1 ÿ4k2

p2=(m2

p3)

ÿk2

p3=(9m1

p2) ÿ2k2

p3=(9m1

p2) ÿ4k2=3m2

0@ 1A:(f ) You will ®nd that (1, ÿ2,

p6) is an eigenvector of B, with eigenvalue 0.

This agrees with the statement in Example 32.20 that the translationvector

r1 ÿ 2s1 � 3 cos Ww1

is an eigenvector of A.

6. (a) Assign coordinate axes along the edges of the square, as shown below.

The symmetry group is G � D8. Let a denote the rotation sendingP! Q! R! S! P, and let b denote the re¯ection in the axis PR.The character ÷ of the RG-module R8 is

1 a2 a b ab

÷ 8 0 0 0 0


We refer to the character table of D8 which is given in Example 16.3(3),and see that

÷ � ÷1 � ÷2 � ÷3 � ÷4 � 2÷5:

The rotation mode is (t � â)v, where

v � (1, ÿ1, 1, ÿ1, 1, ÿ1, 1, ÿ1) 2 V÷2:

The translation modes are (t � â)v, where v is in the span of v1, v2

and

v1 � (1, 0, 0, 1, ÿ1, 0, 0, ÿ1), v2 � (0, 1, ÿ1, 0, 0, ÿ1, 1, 0):

The homogeneous components V÷1, V÷3

and V÷4are spanned by the

vectors (1, 1, 1, 1, 1, 1, 1, 1), (1, 1, ÿ1, ÿ1, 1, 1, ÿ1, ÿ1) and (1, ÿ1,ÿ1, 1, 1, ÿ1, ÿ1, 1), respectively. The ®nal set of eigen-vectors is givenby V÷5

\ R8vib which is spanned by

(1, 0, 0, ÿ1, ÿ1, 0, 0, 1) and (0, 1, 1, 0, 0, ÿ1, ÿ1, 0):

(b) The matrix A is

ÿ k

m

1 0 0 0 0 0 0 1

0 1 1 0 0 0 0 0

0 1 1 0 0 0 0 0

0 0 0 1 1 0 0 0

0 0 0 1 1 0 0 0

0 0 0 0 0 1 1 0

0 0 0 0 0 1 1 0

1 0 0 0 0 0 0 1

0BBBBBBBBBB@

1CCCCCCCCCCA:

7. (a, b) Let åi (1 < i < m) be the projection which is given by

åi: u1 � : : :� um ! ui

(where uk 2 Uk for all k). Then

w! wAå jWÿ1j Wi (w 2 Ui)

gives an RG-homomorphism from Ui to Ui. By Exercise 23.8, there existëij 2 R such that for all w 2 Ui,

wAå j � ëijwWÿ1i W j:

SinceP

mj�1å j is the identity endomorphism of U1 � . . . � Um, we have

wA �Xm

j�1

ëijwWÿ1i W j for all w 2 Ui:

Now take in turn w � uWi and w � vWi to obtain the results of parts (a)and (b) of the question.

(c) Take a basis u1, . . . , un of U1. Assume that the eigenvectors of Au areknown. For all k with 1 < k < n, the eigenvectors of A insp (ukW1, . . . , ukWm) are given by the eigenvectors of Au (see part (b)).Hence we know all the eigenvectors of A in U1 � . . . � Um.

Chapter 32 453

454

Bibliography

Books mentioned in the text

H. S. M. Coxeter and W. J. O. Moser, Generators and Relations for DiscreteGroups (Fourth Edition), Springer-Verlag, 1980.

J. B. Fraleigh, A First Course in Abstract Algebra (Third Edition), Addison-Wesley, 1982.

D. Gorenstein, Finite Simple Groups: An Introduction to their Classi®cation,Plenum Press, New York, 1982.

G. D. James, The Representation Theory of the Symmetric Groups, Lecture Notesin Mathematics No. 682, Spring-Verlag, 1978.

D. S. Passman, Permutation Groups, Benjamin, 1968.H. Pollard and H. G. Diamond, The Theory of Algebraic Numbers (Second

Edition), Carus Mathematical Monographs No. 9, Mathematical Associationof America, 1975.

J. J. Rotman, An Introduction to the Theory of Groups (Third Edition), Allyn andBacon, 1984.

D. S. Schonland, Molecular Symmetry ± an Introduction to Group Theory and itsuses in Chemistry, Van Nostrand, 1965.

Suggestions for further reading

M. J. Collins, Representations and Characters of Finite Groups, CambridgeUniversity Press, 1990.

C. W. Curtis and I. Reiner, Methods of Representation Theory with Applicationsto Finite Groups and Orders, Volume I, Wiley-Interscience, 1981.

W. Feit, Characters of Finite Groups, Benjamin, 1967.I. M. Isaacs, Character Theory of Finite Groups, Academic Press, 1976.W. Ledermann, Introduction to Group Characters (Second Edition), Cambridge

University Press, 1987.J. P. Serre, Linear Representations of Finite Groups, Springer-Verlag, 1977.

455

Index

A4, 112, 130, 136, 181, 308A5, 10, 112, 116, 220, 312, 359A6, 116, 222, 354, 360A7, 223An, 5, 9, 11, 111, 343abelian group, 3, 11, 81, 82action, 337algebra, 55, 56algebraic integer, 244, 362algebraic number, 361alternating group, 5, 9, 11, 111antisymmetric part, 196, 273associative, 2

basis, 15natural, 45, 54

bijection, 6bilinear form, 269

symmetric, 269skew-symmetric, 269

Brauer±Fowler Theroem, 278Burnside's Lemma, 340Burnside's Theorem, 363, 364

C, 2Cn, 2, 82, 88centralizer, 106centre

of group, 85, 107, 116, 298of group algebra, 83, 114, 153

change of basis, 24character, 118

degree, 122, 247faithful, 125, 195generalized, 355induced, 230, 234, 236integer-valued, 253irreducible, 119kernel of, 125linear, 122, 172, 174

permutation, 129product, 176, 192real, 263realized over R, 265reducible, 119regular, 127, 128, 150trivial, 122

character table, 159A4, 181A5, 221, 359A6, 359, 424C2, 160C3, 160C4, 412C2 3 C2, 415Cn, 82D6, 160D8, 161D10, 415D12 � S3 3 C2, 207, 419D2n (n odd), 182D2n (n even), 183D6 3 D6, 423E, of order 18, 434F7,3, 240, 417F11,5, 433Fp,q, 291GL(2,q), 327PSL(2,7), 318PSL(2,8), 445PSL(2,11), 443Q8, 416S4, 180S5, 201, 262S6, 205SL(2,3), 440SL(2,7), 442SL(2,q), 445T12, 186T4n, 420

U6n, 421V24, 422V8n, 421direct product, 206order 16, 305, 306, 307order 27, 435order , 32, 308order p3, 301order pq, 291p-group, 300

class algebra constants, 349class equation, 107class function, 152class sum, 114Clifford's Theorem, 216complete set, 101completely reducible, 74composition, 2composition factor, 90

common, 91, 96congruences, 259conjugacy class, 104conjugate, 104, 361constituent, 143, 213coset, 8cycle notation, 8cycle-shape, 109cyclic group, 2, 4, 12, 82, 88

D2n, 2, 12, 107, 181degree, 30, 122, 249derived subgroup, 173diagonalization, 83dicyclic group T4n, 178, 187, 281, 420dihedral group, 2, 12, 107, 181dimension, 15direct product, 5, 206direct sum, 17, 66

external, 18

eigenvalue, 24eigenvector, 24endomorphism, 20equivalent, 32, 46even permutation, 5expansion±contraction mode, 381external direct sum, 18

F � R or C, 3Fn, 15Fp,q, 290FG, 53factor group, 9faithful character, 125, 195faithful module, 44, 56, 85faithful representation, 34FG-module, see module

FG-submodule, 49FG-homomorphism, 61FG-isomorphism, 63Frobenius group, 290Frobenius Reciprocity Theorem, 232Frobenius±Schur Count of Involutions, 277function, 6

bijective, 6injective, 6invertible, 6surjective, 6

GL(n,F), 3GL(2,q), 324, 343general linear group, 3group, 1

abelian, 3, 11, 81, 82alternating, 5, 9, 11, 111cyclic, 2, 4, 12, 82, 88dicyclic, 178dihedral, 2, 12, 107, 181factor, 9®nite, 2general linear, 3order, 2order p3, 302, 304orthogonal, 367projective special linear, 312quaternion, 5rotation, 368simple, 10, 250, 278, 311, 318, 353,

363, 364soluble, 365special linear, 311symmetric, 3, 109, 116, 175, 254symmetry, 368

group algebra, 55

H < G, 3H v G, 9HomCG(V, W), 95, 96homogeneous component, 376homomorphism, 6, 10, 61

ideal, 256maximal, 257proper, 257

index of subgroup, 9indicator function, 273induced character, 230, 234, 236induced module, 226, 228inner product, 134involution, 277, 353irreducible character, 119irreducible module, 50, 74, 79, 91irreducible representation, 50, 79isomorphism, 7, 20, 63


kernel, 10, 19, 34, 124, 125

Lagrange's Theorem, 9lift, 169linear character, 122, 173, 174linear transformation, 18linearly dependent, 15linearly independent, 15

Maschke's Theorem, 70, 76matrix, 21

change of basis, 24diagonal, 26identity, 3, 21invertible, 23permutation, 45

methane, 384minimal polynomial, 361module, 39

completely reducible, 74faithful, 44, 56, 85irreducible, 50, 79, 85permutation, 45, 62reducible, 50regular, 56trivial, 43

natural basis, 45, 54normal modes of vibration, 372, 373normal p-complement, 251normal subgroup, 9, 113, 171, 215, 216,

217

odd permutation, 5orbit, 338order of G, 2order of g, 4orthogonal group, 367orthogonality relations, 161

PSL(2,7), 312, 318, 319, 354, 359, 360PSL(2,11), 321, 359PSL(2, p), 312p-group, 298p9-part, 256, 258permutation, 3, 5

even, 5odd, 5

permutation module, 45, 62, 340permutation character, 129, 340permutation matrix, 45powers of characters, 193presentation, 3primitive root, 284product of characters, 176, 192

projection, 27, 67projective special linear group, 312

Q8, 5, 116, 177, 278, 416quaternion group, 5, 116, 177, 278, 416

R, 3rank, 342Rank±Nullity Theorem, 19real character, 263real conjugacy class, 263real element, 263reducible character, 119reducible module, 50reducible representation, 50regular character, 127, 128, 150regular module, 56regular representation, 56representation, 30

degree, 30, 249equivalent, 32, 46faithful, 34irreducible, 50, 79kernel of, 34, 124reducible, 50regular, 56trivial, 34

representatives, 105restriction, 210rotation group, 368rotation mode, 379rotation submodule, 380, 394

S4, 44, 110, 113, 180, 275S5, 111, 201, 262S6, 116, 205S7, 223Sn, 3, 109, 116, 175, 254, 343, 344SL(2,3), 319, 440SL(2,7), 320, 442SL(2, p), 311SL(2,q), 336, 445Schur's Lemma, 78simple group, 10, 250, 278, 311, 318,

363, 364skew-symmetric bilinear form, 269special linear group, 311stabilizer, 339subgroup, 3, 4

cyclic, 4derived, 173generated, 4normal, 9, 113, 171, 215, 216, 217

submodule, 49irreducible, 74

Sylow's Theorem, 354, 365symmetric bilinear form, 269

Index 457

symmetric group, 3, 109, 116, 175, 254symmetric part, 196, 273symmetry group, 368

T4n, 178, 187, 281, 420tensor product module, 190tensor product space, 188trace, 117transitive, 338, 341transitivity of induction, 229translation mode, 379translation submodule, 380, 394transposition, 5

trivial character, 122module, 43representation, 34

U6n, 178, 187, 421

V8n, 178, 187, 421Vandermonde matrix, 194vibratory modes, 381

water, 369, 374

Z, 2


Gordon James, Martin Liebeck Representations and Characters of Groups, Second Edition 2001

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Transcript of Gordon James, Martin Liebeck Representations and Characters of Groups, Second Edition 2001