University of New England

Faculty of Arts and Sciences

School of Science and Technology

PMTH332

ABSTRACT ALGEBRA

Lecture Notes

Trimester 2, 2014

c�University of New EnglandCRICOS Provider No: 00003G

Contents

Introduction iii

Introduction to group theory iii

Chapter 0. Notation, Sets and Functions, Universal Properties 1

1. Logical Notation 1

2. Sets 1

3. Functions 4

4. Partitions and Equivalence Relations 13

5. Universal Properties 15

Chapter 1. Groups 21

Chapter 2. Basic Properties 27

Chapter 3. Comparing Groups 31

Chapter 4. Subgroups 35

Chapter 5. Generators of Groups and Subgroups 41

Chapter 6. Cosets, Quotient Groups and Lagrange’s Theorem 47

Chapter 7. Isomorphism Theorems 55

Chapter 8. Direct Products of Groups 59

Chapter 9. Cyclic Groups 61

Chapter 10. The Fundamental Theorem of Finitely Generated Abelian

Groups 67

Chapter 11. Groups Acting on Sets 75

Chapter 12. The Sylow Theorems 81

Chapter 13. Rings and Ring Homomorphisms 85

Chapter 14. Constructing Rings 93

1. Product Rings 93

iii

Lecture Notes

2. Quotient Rings 94

3. The Field of Fractions of an Integral Domain 96

Chapter 15. Polynomial Rings 101

Chapter 16. Maximal and Prime Ideals 107

Chapter 17. Principal Ideal Domains 111

Chapter 18. Unique Factorization Domains 117

Chapter 19. Constructions with Rule and Compass 125

1. Field Extensions 125

2. Constructible Numbers 129

3. Exercises 132

iv

Introduction

Group theory is vital not only to many branches of mathematics and statistics,

but also to theoretical physics, chemistry and computer science. For example sym-

metry groups and group actions play important rôles in both physics and geometry.

Algebraic topology studies spaces and maps between them via their homotopy, ho-

mology and cohomology groups.

You have already met many examples of groups and rings. The set of integers

together with addition is the prime example of a group and the set of integers

together with addition and multiplication is the prime example of a ring. There

are many other examples arising from our number systems and from geometry.

We investigate the following questions.

(1) What is a group and what is a ring?

(2) How do we compare di↵erent groups and how do we compare di↵erent

rings?

(3) How can we construct groups and rings from given ones?

(4) When are two groups or rings essentially the same?

We cover basic properties, subgroups and subrings, quotient structures, prod-

ucts of groups and rings, isomorphism theorems, cyclic groups, the Fundamental

Theorem of Finitely Generated Abelian Groups, the Sylow Theorems, fields of

fractions, polynomial rings and constructions with rule and compass.

Chapter 0 contains background material concerning sets and functions, parti-

tions and equivalence relations and universal properties. These concepts are used

throughout.

Chapters 1 to 12 are concerned with groups while Chapters 13 to 19 are con-

cerned with rings.

Introduction to group theory

Like vector spaces, groups are sets with additional structure, and we compare

di↵erent groups by means of functions which respect this structure. In the case

of groups the additional structure is given by a binary operation on the set in

question. This binary operation must satisfy certain conditions or axioms.

v

Lecture Notes

We say that the groups G and H are “essentially the same”, or isomorphic, if

and only if there are structure preserving functions ' : G ! H and : H ! Gsuch that ' = idG and ' = idH .

We obtain sets from given ones by considering subsets, quotient sets and Carte-

sian products of sets.

As with vector spaces, where not every subset of a given vector space is a

subspace, not every subset of a given group is a subgroup. We shall investigate

under what conditions subsets of a given group are subgroups.

In order to provide a group structure for the Cartesian product of the underlying

sets of two groups, it is enough to define the binary operation componentwise. The

projections onto the factors are then automatically struture preserving functions.

Given a subgroup H of the group G, we say that x 2 G is equivalent to y 2 GmoduloH if and only if xy�1 2 H. Equivalence moduloH is, in fact, an equivalencerelation on G. The quotient set G/⇠ carries a group structure such that the quotientmap ⌘ : G ! G/⇠, x 7! [x] is a homomorphism if and only if H is “normal” in G.

Given any two groups it can be extremely di�cult to decide whether they are

isomorphic. But we can answer this question for certain classes of groups. For

example, two cyclic groups are isomorphic if and only if they are of the same order.

The Structure Theorem for Finitely Generated Abelian Groups states that ev-

ery finitely generated abelian group is isomorphic to a product of cyclic groups.

This allows us to determine whether two finitely generated abelian groups are iso-

morphic.

In order to understand the structure of a given group it is often useful to

study subgroups. The Sylow Theorems are concerned with certain subgroups called

Sylow p–subgroups, where p is a prime. Since isomorphisms preserve Sylow p–

subgroups, we may conclude that two groups cannot be isomorphic if, for a certain

prime p, the two groups have non–isomorphic Sylow p–subgroups. This is just one

example of a property which is preserved by isomorphism and therefore allows us to

distinguish non–isomorphic groups. If on the other hand two groups are isomorphic,

the Isomorphism Theorems often help construct isomorphisms between them.

vi

CHAPTER 0

Notation, Sets and Functions, Universal Properties

1. Logical Notation

It is sometimes convenient to use logical notation. We list the notation we use.

P =) Q for “if P , then Q”, or “Q whenever P”, or “P only if Q”;P () Q for “P if and only if Q”, that is to say P and Q are logicallyequivalent;

P :() Q for “P is defined to be equivalent to Q”;8 for “For every . . . ”;9 for “There is at least one . . . ”;9! for “There is a unique . . . ”, or “There is one and only one . . . ”.

2. Sets

The mathematics we study in this course can be expressed entirely in terms of

sets and functions between sets. We therefore begin with a summary of the set-

theoretical concepts and definitions used in the course. We also use the occasion

to summarise notational conventions we use throughout this course.

A set is almost any reasonable collection of things. We shall not even attempt

a more formal definition in this course. The things in the collection are called the

elements of the set in question. We write

x 2 A

to denote that x is an element of the set A and

x /2 A

to denote that x is not an element of the set A. Note that we do not exclude

the possibility that x be a set in its own right, except that x cannot be A: We

explicitly exclude A 2 A.Two sets are considered to be the same when they comprise precisely the same

elements, in other words, when every element of the first set is also an element of

the second and vice versa. Formally,

1

Lecture Notes

Definition 0.1. Given two sets A and B, we say that A is a subset of B if

and only if x 2 B whenever x 2 A. We write

A ✓ B

whenever this is the case. We say that A = B if and only if A ✓ B and B ✓ A.B is called a proper subset of A if B is a subset of A, but B 6= A. In such a

case we write

B ⇢ A.

One of the axioms of set theory is that any subset of any set is again a set.

Using our notational conventions, given two sets A and B,

A = B :() ((x 2 A) , (x 2 B)) .

When we wish to describe a set, we can do so by listing all of its elements.

Thus, if the set A has precisely a, b and c as its elements, then we write

A = { a, b, c }.

Another way of describing a set is by prescribing a number of conditions for

membership of the set. In this case we write

{ x | P (x), Q(x), . . . }

to denote that the set in question consists of all those x for which P (x), Q(x), . . .

all hold.

Notice that by the above the sets {a, b}, {a, b, b, b} and {a, a, a, a, a, a, b} are allthe same set.

There are several operations on sets.

Definition 0.2. The power set of the set A comprises all the subsets of A. It

is denoted by P(A).

One of the axioms of set theory is that the power set of any set is again a set.

B 2 P(A) if and only if B ✓ A

Definition 0.3. The union of two sets A and B consists of all those objects

which are in one, or other (or both) and is denoted by

A [ B.

Using the notation above,

A [ B := { x | x 2 A or x 2 B }.

Here := has been used to signify that the expression on the left hand side is defined

to be equal to the expression on the right hand side.

2

Lecture Notes

Definition 0.4. The intersection of the sets A and B consists of all those

objects which are elements of both and is denoted by

A \ B.

In other words,

A \ B := {x | x 2 A and x 2 B }.

Definition 0.5. Those elements of A that are not also elements of B form a

set in their own right, which we denote by

A \B,

so that

A \B := {x 2 A | x /2 B }.

Definition 0.6. Given two sets A and B, their Cartesian product A ⇥ Bconsists of all ordered pairs, with the first of each pair an element of A and the

second of B. Symbolically,

A⇥ B := { (x, y) | x 2 A, y 2 B }.

We can extend unions, intersections and Cartesian products to larger collections

of sets than merely pairs.

Definition 0.7. An indexed family of sets, with indexing set ⇤, consists of a

collection of sets, containing one set, say A�, for each element � of the indexing

set ⇤. We write {A� | � 2 ⇤}.

Definition 0.8. Given an indexed family of sets, say

{A� | � 2 ⇤ },

the union, intersection and (Cartesian) product are defined respectively by[

�2⇤

A� := {x | there is a � 2 ⇤ with x 2 A� }(1)

\

�2⇤

A� := {x | x 2 A� for every � 2 ⇤ }(2)

Y

�2⇤

A� := {(x�)�2⇤ | x� 2 A� for all � 2 ⇤}.(3)

Here (x�)�2⇤ denotes a generalised sequence, namely, an ordered choice of ele-

ments x�, one for each � 2 ⇤. Ordered pairs arise when ⇤ = {1, 2} and sequenceswhen ⇤ = N.

The axioms of set theory assert that the union, intersection and Cartesian

product of any indexed set of sets is again a set.

3

Lecture Notes

A number of sets occur with such frequency that special notation has been in-

troduced for them. These include the sets N, Z, Q, R and C consisting respectivelyof all natural numbers, all integers, all rational numbers, all real numbers and all

complex numbers. Observe that

N ⇢ Z ⇢ Q ⇢ R ⇢ C.

Explicitly,

N : = { 0, 1, 2, 3, . . . }(4)Z : = { . . .� 3,�2� 1, 0, 1, 2, 3, . . . }(5)

Q : = { x 2 R | x = pq

for some p, q 2 Z, with q 6= 0 }(6)

We write ; for the empty set, which is the (unique!) set with no elements. Notethat it is a subset of every set, that is, if X is any set, then ; ✓ X.

3. Functions

To compare sets, we have the notion of a function or map or mapping.

Definition 0.9. A function, map, or mapping consists of three separate data,

namely

(i) a domain, that is, a set on which the function is defined,

(ii) a codomain, that is, a set in which the function takes its values, and

(iii) the assignment to each element of the domain of definition of a uniquely

determined element from the set in which the function takes it values.

This is conveniently depicted diagrammatically by

f : X �! Y,

or

Xf���! Y.

HereX is the domain of definition, Y is the set in which the function takes its values

and f is the name of the function. (Note that the function f need not be given in

terms of a mathematical formula.) We write X = dom(f) and Y = codom(f) to

indicate that X is the domain and Y the codomain of f .

Nevertheless, we do often denote the function by f alone, but only when there is

no danger of confusion. If we wish to express explicitly that the function, f : X �!Y, assigns the element y 2 Y to the element x 2 X, then we write f : x 7�! y

4

Lecture Notes

or, equivalently, y = f(x). (This latter form is certainly familiar to the reader.)

Sometimes the two parts are combined as

f : X �! Y, x 7�! y

or as

f : X �! Y

x 7�! y.

Definition 0.10. If f assigns y 2 Y to x 2 X, then we say that y is the imageof x under f or just the image of x.

Two functions f and g are equal, that is f = g, if and only if

(i) dom(f) = dom(g)

(ii) codom(f) = codom(g)

(iii) f(x) = g(x) for every x 2 dom(f).

In other words, to be the same, two functions must share both domain and

codomain as well as agreeing everywhere.

Note that a function (or map, or mapping) is not just a formula. In fact

there are distinct functions whose domains agree, which agree at every point (and

therefore have the same range), the only di↵erence between them is that they have

di↵erent codomains. Thus, they only di↵er in the values they do not take!

We shall meet an example soon. At this stage, it may seem peculiarly pedantic

to distinguish such functions, but there are important algebraic and geometric

examples, whose detailed study lies beyond the scope of these notes. We shall

adhere to the practice of specifying functions in a formally correct manner, so that

it can become ordinary matter of course for the reader do so as well.

Definition 0.11. The range or image of the function f : X �! Y is thesubset im(f) of Y defined by

im(f) := { y 2 Y | y = f(x) for some x 2 X } = { f(x) | x 2 X }.

Notice that im(f) ✓ codom(f) always holds, with equality holding only some-times. For example if f : R ! R is defined by f(x) := 1 for every x 2 R, thenim(f) = {1} 6= R = codom(f).

Definition 0.12. Given a function f : X ! Y and subsets A of X and B ofY , we define

f(A) := { y 2 Y | y = f(x) for some x 2 A }

f�1(B) := {x 2 X | f(x) 2 B }.

5

Lecture Notes

Then f(A) is called the image of A under f and f�1(B) is called the inverse image

of B under f , or the pre-image of B under f .

When the subsets in question are singletons, say A = {a} and B = {b}, wewrite f(a) and f�1(b) instead of f({a}) and f�1({b}).

For y 2 Y , the subset f�1(y) of X is called the fibre of f over y.

For A,B ✓ X and G,H ✓ Y there are important relationships between f(A [B), f(A\B), f�1(G[H) and f�1(G\H) on the one hand, and f(A)[f(B), f(A)\f(B), f�1(G)[f�1(H), and f�1(G)\f�1(H) on the other. These are investigatedin the exercises.

Definition 0.13. The identity function, on the set X, denoted idX , is the

function

idX : X �! X, x 7�! x,

which we may also write as idX (x) := x.

Notice that both the domain and codomain must be precisely X for this defi-

nition to specify the identity function.

Definition 0.14. If X is a subset of Y , then the inclusion (of X in Y ) is the

map

iY

X: X �! Y, x 7�! x.

It is often denoted simply by i when the context makes the domain and codomain

clear.

We can characterise subsets in terms of functions.

Theorem 0.15. The set X is a subset of the set Y if and only if

X �! Y, x 7�! x

defines a function.

Proof. Exercise. ⇤

If we use the equational notation, we again have i(x) = x. But here, the x on

the left of the equality sign is viewed as an element of the set X, whereas on the

right hand side it is viewed as an element of the set Y .

We have just characterised subsets of a fixed set X in terms of functions with

codomain X, assigning a specific function to each subset of X. We can also char-

acterise subsets of X in terms of functions whose domain is X.

6

Lecture Notes

Definition 0.16. Given a function f : X ! Y and a subset A of X we canuse f to define a function f |A : A ! Y , called the restriction of f to A by meansof

f |A (x) = f(x) for every x 2 A.

Note that unless of course A = X, this is not the same function as f , even

though the two functions agree everywhere they are both defined.

Functions can sometimes be composed.

Definition 0.17. Given functions f : X ! Y and g : Y ! Z their composi-tion, denoted by g � f , is the function defined by

g � f : X �! Z, x 7�! g (f(x)) ,

as long as codom(f) = dom(g) = Y .

In such a case, it is easy to see that

dom(g � f) = dom(f)(7)

codom(g � f) = codom(g)(8)

im(g � f) ✓ im(g).(9)

Equality need not hold in 9. To see this consider the functions f : R ! R, x 7! 1and g : R ! R, y 7! y. Clearly im(g � f) = {1} 6= R = im(g).

Given A ✓ X and a function f : X ! Y , the restriction f |A : A ! Y is, infact, the composition of f and the inclusion of A into X:

f |A= f � iX

A.

The composition of functions is associative, that is to say given h : W ! X, g :X ! Y and f : Y ! Z, the two functions (f�g)�h : W ! Z and f�(g�h) : W ! Zare, in fact, one and the same function. To see this, note that the domains and

codomains agree and take w 2 W . Then

((f � g) � h) (w) := (f � g) (h(w))

:= f (g (h(w)))

=: f ((g � h)(w))

=: (f � (g � h)) (w).

7

Lecture Notes

It follows immediately from the definitions that if f : X ! Y is any function,then

idY � f = f

f � idX = f.

We say that the identity functions act as neutral elements with respect to compo-

sition.

Definition 0.18. The function f : X ! Y is said to be invertible if there is afunction g : Y ! X such that f � g = idY and g � f = idX , that is, if f (g(y)) = yfor every y 2 Y and g (f(x)) = x for every x 2 X. In this case g is said to be theinverse of f , and is denoted by f�1.

This notation (as well as the use of the definite article!) is justified by the fact

that a function cannot have more than one inverse, as we prove below.

Note that in our definition of invertibility of the function f : X ! Y , thefunction g : Y ! X needed to satisfy two conditions. We consider these separately,and introduce terminology for this purpose.

Definition 0.19. The function g : Y ! X is a left inverse of f : X ! Y ifand only if g �f = idX and the function h : Y ! X is a right inverse of f : X ! Yif and only if f � h = idY .

Theorem 0.20. If f : X ! Y has both a left and a right inverse, then thesemust be the same, and hence f is invertible with a uniquely determined inverse.

Proof. Suppose that f : X ! Y has a left inverse g : Y ! X and a rightinverse h : Y ! X. Then

g = g � idY = g � (f � h) = (g � f) � h = idX � h = h.

⇤

To decide whether the function f : X ! Y has an inverse does not seem tobe an easy task at first glance. If we blindly follow our definition, we would try

all possible functions from Y to X and see which, if any, satisfy the conditions in

the definition. This is not completely satisfactory. It would be preferable to be

able to determine from intrinsic properties of f — that is, properties of f alone,

without reference to other functions — whether it admits an inverse. Such an

intrinsic criterion is available, as we show. To do so, we introduce some important

properties of functions.

Definition 0.21. The function f : X ! Y is said to be8

Lecture Notes

(i) 1–1 or injective or mono if and only if it follows from f(x) = f(x0) that

x = x0;

(ii) onto or surjective or epi if and only if given any y 2 Y there is an x 2 Xwith f(x) = y — in other words im(f) = codom(f);

(iii) 1–1 and onto or bijective or iso if and only if it is both 1–1 and onto.

Thus a function is injective if and only if it distinguishes di↵erent elements of

its domain: di↵erent elements of its domain are mapped to di↵erent elements of its

codomain.

Similarly, a function is surjective if and only if its range coincides with its

codomain.

Example 0.22. We write R+0 for {x 2 R | x � 0 }.

(i) f : R ! R, x 7! x2 is neither injective nor surjective, as f(1) = f(�1)and there is no x with f(x) = �4.

(ii) g : R ! R+0 , x 7! x2 is not injective, but it is surjective, as f(1) = f(�1)and every non-negative real number can be written as the square of a real

number.

(iii) h : R+0 ! R, x 7! x2 is injective, but it not surjective, as f(x) = f(x0)if and only if x2 = x02 if and only if x0 = ±x if and only if x0 = x as, bydefinition, x, x0 � 0. On the other hand, there is no x with f(x) = �4.

(ii) g : R+0 ! R+0 , x 7! x2 is both injective, and surjective, as should be clearfrom parts (ii) and (iii).

Theorem 0.23. Take a non-empty set X and a function f : X ! Y . Then fhas

(i) a left inverse if and only if it is injective (or 1–1),

(ii) a right inverse if and only if it is surjective (or onto) and

(iii) an inverse if and only if it is bijective.

Proof. (i) We first suppose that f : X ! Y has a left inverse, say, g : Y ! X.9

Lecture Notes

To see that f must then be injective (that is 1–1), suppose that f(x) = f(x0).

Then

x = idX (x) = (g � f)(x) = g (f(x)) = g (f(x0)) = (g � f)(x0) = idX (x0) = x0,

as required.

For the converse, suppose that f : X ! Y is injective. Define g : Y ! X by

g(y) :=

8

<

:

x if y = f(x)

x0 otherwise, where x0 is some fixed element of X.

We must first show that g so defined is, in fact a function. The only possible

di�culty is that g might assign more than one element of X, say x and x0, to some

element y of Y . From the definition of g, this could only happen when y 2 im(f).This, in turn, is only possible when y = f(x) = f(x0). But then x = x0, since f is

injective.

That g � f = idX follows from the definition of g.(ii) We first suppose that f : X ! Y has a right inverse, say, g : Y ! X.

To see that f must be surjective, take y 2 Y and put x := g(y). Then

f(x) = f (g(y)) = (f � g)(y) = idY (y) = y,

showing that f is, indeed surjective.

For the converse, suppose that f : X ! Y is surjective. Define g : Y ! X bychoosing for each y 2 Y an element, say xy of X with f(xy) = y. This is alwayspossible is because f is surjective.

This g is obviously a function.

Now take y 2 Y . Then (f � g)(y) = f (g(y)) = f(xy) = y, by the definition ofg. Thus f � g = idY .(iii) This follows from Theorem (0.20) and parts (i) and (ii) here. ⇤

We can express whether f is injective or surjective in terms of its fibres.

Lemma 0.24. The function f : X �! Y is(i) surjective if and only if each fibre f�1(y) (y 2 Y ) is non-empty;

(ii) injective if and only if each fibre f�1(y) (y 2 Y ) contains at most oneelement.

Proof. Obvious. ⇤

Example 0.25. Theorem 0.23 illustrates one of the ways in which two functions

can have the same domain and agree everywhere without being the same function.

10

Lecture Notes

Let X be a non-empty proper subset of the set Y , so that X ⇢ Y . Then thetwo functions

idX : X �! X, x 7�! x(10)

iY

X: X �! Y, x 7�! x(11)

share a common domain and agree at every point, so that they have the same range,

namely, X. But they cannot be the same function. For whereas idX is invertible

— it is its own inverse — Theorem 0.23 tells us that iY

Xcannot be invertible, since

it fails to be surjective and so cannot have even a right inverse.

Observation 0.26. While only some functions are surjective or injective, every

function can be written naturally as a surjective function followed by an injective

one. Given the function f : X ! Y , if we put R := im(f), then, by definition,R ✓ Y and so we have

iY

R: R �! Y.

It is now easy to verify that if we define

f̂ : X �! R, x 7�! f(x),

then

(i) iY

Ris injective,

(ii) f̂ is surjective and

(iii) f = iY

R� f̂ .

[This is called the mono-epi factorisation of f .]

This factorisation justifies the common practice of being sloppy with the com-

position of functions, by allowing functions f and g to be composed as long as

im(f) ⇢ dom(g) even when codom(f) 6= dom(g). Our (more formal) treatmentaccounts for this, for if im(f) ✓ dom(g), we can compose g with idom(g)

R� f̂ , to form

g � idom(g)R

� f̂ .

It is this which is commonly abbreviated to g � f , just as we often write f |A forf � iX

Awhen A ✓ X.

11

Lecture Notes

We often depict functions using diagrams. We say that the diagram

Af//

h ��

B

g✏✏C

commutes when h = g � f , in other words, when the composition g � f coincideswith h. Similarly, the diagram

Af���! B

j

?

?

y

?

?

y

g

C ���!k

D

commutes when k�j = g �f , in other words, when the compositions g �f and k�jcoincide. We can express the fact that the composition of functions is associative

— that is to say, if f : W ! X, g : X ! Y and h : Y ! Z are functions then(h � g) � f = h � (g � h) — by means of commutative diagrams, namely by statingthat

Wf���! X

g�f?

?

y

?

?

y

h�g

Y ���!h

Z

commutes, or, equivalently, that

Wf//

g�f

X

g✏✏

h�g

Y

h// Z

commutes.

Addition and multiplication of integers, rational numbers or real numbers are

special cases of binary operations.

Definition 0.27. A binary operation on the set X is a function

X ⇥X �! X.

Both groups and rings are sets with additional structure and in both cases the

additional structure is given by binary operations satisfying certain conditions.

12

Lecture Notes

4. Partitions and Equivalence Relations

For many purposes, two distinct objects can be indistinguishable, or the di↵er-

ences irrelevant: we treat them as equivalent. We next define this notion formally.

A relation R between the elements of the set X and those of the set Y can be

represented by the subset of X ⇥ Y comprising those pairs (x, y) (x 2 X, y 2 Y )for which x stands in the relation R to y. We usually write xRy to denote this.

An example is provided by the telephone book. Here we regard X as the set of

all subscribers, and Y as all telephone numbers.

If Y happens to coincide with X, we speak of a binary relation on X.

Definition 0.28. An equivalence relation on X is a binary relation ⇠ on X,which is reflexive, symmetric and transitive. That it to say, for all x, y, z 2 X, wehave

Reflexiveness: x ⇠ x.

Symmetry: x ⇠ y if and only if y ⇠ x.

Transitivity: If x ⇠ y and y ⇠ z, then x ⇠ z.Given an equivalence relation ⇠ on X, and x 2 X, we define

[x] := {t 2 X | x ⇠ t},

and call it the equivalence class of x. We call any z 2 [x] a representative of [x].Finally, define the quotient set, denoted by X/ ⇠, to be the set of all such

equivalence classes, so that

X/⇠ := {[x] | x 2 X}.

We then have a function, called the natural map, the quotient function or the

quotient map

⌘ : X �! X/⇠, x 7�! [x].

Example 0.29. We obtain the positive rational numbers from the positive

integers by introducing solutions to equations of the form

(12) ax = b

where a, b 2 Z+. We denote the uniquely determined solution of equation (12) byba (uniquely determined as ax

0 = b = ax implies x0 = x since a 6= 0). We mustidentify all expressions arising from equations with the same solutions. So assume

(13) cx = d

13

Lecture Notes

has the same solution x0 as (12). Then

ax0 = b

cx0 = d} =) { acx� = bc

acx� = ad} =) ad = bc

Conversely, we can show that ad = bc implies that (12) and (13) have the same

solution. (This is an easy exercise.) Thus we define

b

a⇠ d

c:() bc = ad.

Thenb

a⇠ b

asince ba = ab. Furthermore

b

a⇠ d

cimplies

d

c⇠ b

asince bc = ad ()

ad = bc. Finally, assume thatb

a⇠ d

cand

d

c⇠ f

e. Then bc = ad and de = cf and

hence bce = ade = acf . Thus cbe = caf so that be = af which impliesb

a⇠ f

e.

This shows that ⇠ is an equivalence relation. The set of positive rational numbersis equal to the set of equivalence classes.

Example 0.30. Given a natural number n we say that the integer k is congruent

modulo n to the integer l if and only if

k � l = ns for some s 2 Z.

We then write k ⌘ l mod n. It is an exercise to show that congruence modulon is an equivalence relation. The equivalence classes are also called residue classes

modulo n.

Another important notion is that of a partition of a set.

Definition 0.31. A partition of the set X is a collection of {X� | � 2 ⇤}of disjoint non-empty subsets of X, whose union is X. Thus {X� | � 2 ⇤} is apartition of X if and only if

(1) ; ⇢ X� ✓ X for each � 2 ⇤;

(2) X� \Xµ = ; whenever � 6= µ;

(3) X =[

�2⇤

X�.

These two notions, however di↵erent they may appear, are intimately related.

In fact they are two sides of the same coin, as the next theorem shows.

Theorem 0.32. Every equivalence relation on the set X determines a unique

partition of X, and conversely.

14

Lecture Notes

Proof. We outline a proof, leaving the details as an exercise for the reader.

Given the equivalence relation ⇠ on X, the equivalence classes form a partitionof X, that is every x 2 X belongs to some equivalence class, and if [x] \ [x0] 6= ;,then [x] = [x0].

If {X� | � 2 ⇤} is a partition of X, then

x ⇠ x0 if and only if x, x0 2 X� for some � 2 ⇤

defines an equivalence relation on X

Now show that if we start with an equivalence relation, construct the associated

partition, then the associated equivalence relation is the original one.

Finally, show that if we start with a partition, define the associated equivalence

relation, then the associated partition is the original one. ⇤

5. Universal Properties

We shall be using universal properties in this course. A universal property is one

which holds for all members of a class. These are extremely useful in mathematics,

since they may be used in quite general settings and anything enjoying a universal

property in a suitable context is uniquely defined by it.

We provide two examples to clarify the terminology and show the uniqueness

of the Cartesian product of two sets and the construction of the quotient set from

an equivalence relation on a given set.

5.1. The Cartesian Product of Two Sets. Recall that given two sets X

and Y , their Cartesian product, X ⇥ Y , is defined by

X ⇥ Y := { (x, y) | x 2 X, y 2 Y }.

In other words, X ⇥ Y is the set of all ordered pairs, the first member of which isan element of X and the second an element of Y .

One consequence of this definition is that for any sets X and Y , their Cartesian

product comes equipped with two natural maps called the canonical projections

onto the factors, namely,

prX : X ⇥ Y �! X, (x, y) 7�! x;(14)

prY : X ⇥ Y �! Y, (x, y) 7�! y.(15)

The Cartesian product together with the canonical projections enjoys the uni-

versal property that given any set and any pair of functions from that set, one

into the first factor of the Cartesian product, the other into the second, there is

a uniquely determined function from the given set to the Cartesian product such

15

Lecture Notes

that the given functions agree with the new function composed with the respective

canonical projections. More formally,

Theorem 0.33. Given any set W whatsoever and any pair of functions f :

W �! X and g : W �! Y , there is a unique function h : W �! X ⇥Y such that

f = prX � h and g = prY � h.

This is expressed diagrammatically by

Y

W

g;;

9!h //

f ##

X ⇥ Y.

prY

OO

prX✏✏X

We prove the theorem by choosing the obvious candidate for h and then veri-

fying that it satisfies all the requirements. Since this is likely to be the first time

the reader has met universal properties, the proof will be extremely detailed.

Proof. Any function from W to X ⇥ Y assigns to each element w of W anelement X ⇥ Y . This must be of the form (xw, yw) for some uniquely determinedxw 2 X and uniquely determined yw 2 Y . Applying the canonical projections inturn yields xw and yw respectively.

So, if the function h : W �! X ⇥ Y is to satisfy the requirements that f =prX � h and g = prY � h, we must have h(w) = (f(w), g(w)). This is enough toensure uniqueness.

Hence we define h : W �! X⇥Y by h(w) = (f(w), g(w)), and all that remainsis to verify that h, so defined, is, in fact, a function! We must show that to each

(and every!) element w of W h assigns a uniquely determined element of X ⇥ Y .So take any w 2 W .

Since f is a function, it assigns to w the uniquely determined element f(w) of

X. Since g is also a function, it assigns to the same w the uniquely determined

element g(w) of Y . By the definition of X ⇥ Y , these in turn uniquely determinethe element (f(w), g(w)) of X ⇥ Y . Thus h assigns to w the uniquely determinedelement (f(w), g(w) of X ⇥ Y , as required. ⇤

Given are X and Y . It is asserted that X ⇥ Y and prX : X ⇥ Y ! X andprY : X ⇥ Y ! Y can be found with the prescribed property. In other words,the property is enjoyed not by X ⇥ Y alone, but by X ⇥ Y together withprX : X ⇥ Y ! X and prY : X ⇥ Y ! Y .

16

Lecture Notes

Observation 0.34. The property is universal in the sense that it applies to

each and every set W , each and every function f : W ! X and each andevery function g : W ! Y .

We next show that the Cartesian product of two sets and the canonical projec-

tions are uniquely determined up to a unique bijection by the universal property

(0.33).

To see this, let W, qX : W ! X, qY : W ! Y also exhibit the same universalproperty. The following commutative diagram su�ces as proof.

Y Y Y Y

qY

x

?

?

prY

x

?

?

qY

x

?

?

prY

x

?

?

W9!'���! X ⇥ Y 9! ���! W 9!'���! X ⇥ Y

qX

?

?

y

prY

?

?

y

qY

?

?

y

prX

?

?

y

X X X X

where we have written = for the appropriate identity maps.

To see how this constitutes a proof, we do some “diagram-chasing”, starting

at the left-hand copy of W . Let f : W ! X be the composition idX � qX andg : W ! Y be the composition idY �qY . Thus we have two functions defined on W ,one taking values in X, the other in Y . By the universal property of X ⇥ Y, prX :X ⇥ Y ! X, prY : X ⇥ Y ! Y, there is a unique function ' : W ! X ⇥ Y suchthat prX � ' = qX and prY � ' = qY . Similarly, starting from the left-hand copy ofX ⇥Y we find a unique : X ⇥Y ! W such that qX � = prX and qY � = prY .

We now show that these two functions, ' and — each of which is uniquely

determined — are in fact bijections. To do so, we show that � ' = idW and' � = idX⇥Y . (Actually, we only establish the first of these equalities, since thesecond follows mutatis mutandis.)

We again start from the left-hand copy of W . This time we let f : W ! X bethe composition idX � idX � qX and g : W ! Y the composition idY � idY � qY . Bythe universal property of W, qX : W ! X, qY : W ! Y , there is a unique function# : W ! W such that

qX � # = f = idX � idX � qX = qX

and

qY � # = g = idY � idY � qY = qY .

Now clearly idW shares this property with # and so # = idW . On the other hand,

qX � � ' = prX � ' = qX17

Lecture Notes

and

qY � � ' = prY � ' = qYby the calculations in the previous paragraph. Hence � ' satisfies the sameconditions as #, and so by the uniqueness property, � ' = # = idW . The sameargument, starting from the left-hand copy of X ⇥ Y shows that ' � = idX⇥Y .

This completes the proof.

We have executed this last piece of “diagram chasing” in detail because this is

a common method of proof in homological algebra, algebraic topology and their

applications. Observe that the elements of the sets involved play no rôle whatsoever

in the proof. This fact has important consequences, which are dealt with in category

theory, but which we do not pursue in this course.

Observation 0.35. This universal property has been described purely in terms

of functions between sets, without any mention of the elements of the sets. Hence

we can ask whether there are objects of interest displaying the analogous universal

property in any situation which can be represented by diagrams similar to the

one above. For example, if we take X and Y to be real vector spaces and insist

that the arrows in our diagram represent linear transformations (rather than just

functions), then the analogous situation consists of being given two real vector

spaces X and Y and seeking a real vector space X ⇥ Y and linear transformationsprX : X ⇥ Y ! X, prY : X ⇥ Y ! Y such that given any real vector space andany linear transformations f : W �! X and g : W �! Y there is a unique lineartransformation h : W �! X ⇥ Y with f = prX � h and g = prY � h.

5.2. Quotient Sets. Given an equivalence relation ⇠ on the set X, the quo-tient setX/ ⇠ and the quotient map ⌘ : X �! X/⇠ also enjoy a universal property:

Theorem 0.36. Let ⇠ be an equivalence relation on the set X. Given any setY and any function f : X ! Y with the property that

f(x) = f(x0) whenever x ⇠ x0,

there is a uniquely determined function f̃ : X/ ⇠! Y such that

f = f̃ � ⌘,

that is, f̃([x]) = f(x) for all [x] 2 X/⇠. This is expressed in the diagram

Xf//

⌘✏✏

Y.

X/⇠9!f̃

==

18

Lecture Notes

Proof. Given f : X ! Y with f(x) = f(x0) whenever x ⇠ x0, define

f̃ : X/⇠ �! Y, [x] 7�! f(x),

so that

f̃([x]) = f̃(⌘(x)) = (f̃ � ⌘)(x) = f(x).

This establishes the uniqueness of f̃ as well as the fact that condition (0.36) is

satisfied. It only remains to show that f̃ so defined is actually a function.

Since the definition of f̃([x]) depends on the choice of the representative x 2 X,the only way in which f̃ could fail to be a function is that for some x, x0 2 X, [x] =[x0], but f(x) 6= f(x0). However [x] = [x0] if and only if x ⇠ x0, in which casef(x) = f(x0) by hypothesis.

⇤

Exercises:

0.1 Prove that the set X is a subset of the set Y if and only if

X �! Y, x 7�! x

defines a function.

0.2 Given the function f : X ! Y and subsets A of X and B of Y , prove thefollowing statements.

(i) A ✓ f�1 (f(A)).

(ii) f (f�1(B)) ✓ B.

(iii) In general, equality need not hold in either (i) or (ii).

(iv) G = f�1 (f(G)) for every subset G of X if and only if f is injective (1–1).

(v) f (f�1(H)) = H for every subset H of Y if and only if f is surjective

(onto).

0.3 Take functions f : X ! Y and g : Y ! Z. Prove the following statements.(a) If f and g are both injective, then so is g � f .

(b) If f and g are both surjective, then so is g � f .

(c) If g � f is injective, then so is f , but not necessarily g.

19

Lecture Notes

(d) If g � f is surjective, then so is g, but not necessarily f .

(e) If f and g are bijective, so is g � f .

(f) If g � f is bijective, then neither f nor g need be bijective.

0.4 Show that congruence modulo n is an equivalence relation.

0.5 Show that every equivalence relation on the setX determines a unique partition

of X, and conversely.

20

CHAPTER 1

Groups

First we consider some motivating examples.

Example 1.1. Given a set X, let S(X) be the set of invertible functions from

X to X.

S(X) := {f : X �! X | f is invertible}

We may compose any two functions f, g 2 S(X) as domf = codomg = X. Thuscomposition of functions yields an associative binary operation on S(X) given by

S(X)⇥ S(X) �! S(X), (f, g) 7�! f � g.

Composing f 2 S(X) with the identity function

idX : X �! X, x 7�! x

on either side yields f , that is, f � idX = idX �f = f . By definition, every f 2 S(X)has an inverse f�1 with f � f�1 = f�1 � f = idX .

Example 1.2. Let O be any geometrical object – like for example a regular

polygon in the plane or a tetrahedron in three–dimensional space – and let S be

the set of symmetries of O, that is, the set of rigid motions which map O to itself.

We may view the elements of S as functions from the set of all points of O to itself.

Clearly, any such function which corresponds to a symmetry of O is invertible.

Again, composition of functions yields an associative binary operation on S. The

identity function corresponds to the symmetry ⌘ which does not move anything.

Given � 2 S, we obtain � � ⌘ = ⌘ � � = �. Further, denoting the symmetry whichreverses � by ��1, we have � � ��1 = ��1 � � = ⌘.

Example 1.3. Consider Z together with addition: Given n,m 2 Z we havem + n 2 Z. We also have n + 0 = 0 + n = n for every n 2 Z. Furthermore, forevery n 2 Z there is an element of Z usually donoted by �n such that n+ (�n) =(�n) + n = 0. Finally addition of integers is associative and commutative.

To arrive at the definition of a group we let ourselves be guided by the common

features of these three examples.

21

Lecture Notes

Definition 1.4. A group consists of a pair (G, ⇤), where G is a set and ⇤ is abinary operation on G, that is, a function

⇤ : G⇥G �! G, (a, b) 7�! a ⇤ b,

such that the following three axioms are satisfied:

(G1) (Associativity) For all a, b, c 2 G,

(a ⇤ b) ⇤ c = a ⇤ (b ⇤ c).

(G2) (Existence of Neutral Element) There is an element e of G such that, for

all a 2 G,a ⇤ e = e ⇤ a = a.

(G3) (Existence of Inverses) For each a 2 G there is an element a 2 G suchthat

a ⇤ a = a ⇤ a = e.The group (G, ⇤) is said to be commutative or abelian if and only if the following

axiom is also satisfied.

(G4) (Commutativity) For all a, b 2 G,

a ⇤ b = b ⇤ a.

With this definition the set of symmetries of a geometrical object together

with the composition of symmetries, the set S(X) of Example 1.1 together with

the composition of functions and the set of all integers together with addition all

form groups. In order to see that the composition of symmetries in Example 1.2

is associative we view the geometrical object O as a set of points and the set of

symmetries as a subset of the set of all invertible functions from O to itself. Then

the composition of symmetries corresponds to composition of functions which is

associative.

Note that the groups in 1.1 and 1.2 are not commutative whereas (Z,+) iscommutative.

Proposition 1.5. Let (G, ⇤) be a group. Then the element e postulated in (G2)is unique and the element a postulated in (G3) is uniquely determined by a 2 G.

Proof. Suppose e and e0 are neutral elements for (G, ⇤). Then

e0 = e0 ⇤ e = e.

Now suppose that a and ã are inverses of a. Then

ã(G2)= ã ⇤ e = ã ⇤ (a ⇤ a) (G1)= (ã ⇤ a) ⇤ a = e ⇤ a (G2)= a.

⇤22

Lecture Notes

A set A together with an associative binary operation ⇤ is called a semigroup.A semigroup (G, ⇤) which has a neutral element is called a monoid . Summarising,we have

(G1) semigroup

(G1), (G2) monoid

(G1), (G2), (G3) group

Let (G, ⇤) be a semigroup. An element eR 2 G with a ⇤ eR = a for every a 2 Gis called a right neutral element and an element eL 2 G with eL ⇤ a = a for everya 2 G is called a left neutral element .

Proposition 1.6. If a semigroup (G, ⇤) has both a right and a left neutralelement, then these must be the same.

Proof. Exercise. ⇤Now assume (G, ⇤) is a monoid. Then, given a 2 G, an element aR 2 G with

a ⇤ aR = e is called a right inverse of a and an element aL 2 G with aL ⇤ a = e iscalled a left inverse of a.

Proposition 1.7. Let (G, ⇤) be a monoid. If a 2 G has both a right and a leftinverse, then these must be the same.

Proof. Exercise. ⇤In fact, the existence of a right neutral element and right inverses guarantees

the existence of a left neutral element and left inverses.

Theorem 1.8. Let (G, ⇤) be a semigroup with the following properties.(G2R) There is an element e 2 G such that, for all a 2 G,

a ⇤ e = a.

(G3R) To each a 2 G there is an a 2 G such that

a ⇤ a = e.

Then (G, ⇤) is a group.

Proof. Exercise. ⇤This proposition is significant since we do not insist that a group be commuta-

tive. It is of practical importance since it simplifies the test as to whether a given

pair (G, ⇤) is a group.

23

Lecture Notes

Proposition 1.9. Let (G, ⇤) be a group. Then, for every a, b 2 G,(i) a = a

(ii) a ⇤ b = b ⇤ a.

Proof. (i) Take a 2 G. Then

a(G2)= a ⇤ e (G3)= a ⇤ (a ⇤ a) (G1)= (a ⇤ a) ⇤ a (G3)= e ⇤ a (G2)= a.

(ii) Observe that, for every a, b 2 G,

(a ⇤ b) ⇤ (b ⇤ a) = a ⇤ (b ⇤ (b ⇤ a)) = a ⇤ ((b ⇤ b) ⇤ a)

= a ⇤ (e ⇤ a) = a ⇤ a = e.

Since G is a group, any right inverse is automatically a left inverse. This implies

a ⇤ b = b ⇤ a

⇤

Example 1.10. The set Q+ of positive rational numbers together with multi-plication forms a group.

Example 1.11. The set R⇤ of non–zero real numbers together with multiplica-tion forms a group.

Example 1.12. The set mZ := {km |k 2 Z} of all integers divisible by thenatural number m together with addition forms a group.

Example 1.13. Take m 2 Z. Then the set Zm of residue classes (mod m)together with addition defined by [l] + [k] := [l + k] for l, k 2 Z forms a group.

Example 1.14. Let V be the underlying set of a vector space. Then V together

with addition of vectors forms a group.

Example 1.15. The set M(m;R) of (m ⇥ m)–matrices with real coe�cientstogether with addition of matrices forms a group.

Example 1.16. The set GL(m;R) of invertible (m ⇥ m)–matrices with realcoe�cients together with matrix multiplication forms a group.

Example 1.17. Let K be a field. The set Gm(K) of non–zero elements of K

together with multiplication forms a group called the multiplicative group of K.

Example 1.18. Let K be a field. The set GL(m;K) of invertible (m ⇥ m)–matrices with entries in K together with matrix multiplication forms a group.

24

Lecture Notes

Exercises:

1.1 Let (G, ⇤) be a semigroup. Show that, if there is both a right and a left neutralelement, then these must be the same.

1.2 Let (G, ⇤) be a monoid. Show that, if a 2 G has both a right and a left inverse,then these must be the same.

1.3 Let (G, ⇤) be a semigroup with the following properties.(G2R) There is an element e 2 G such that, for all a 2 G,

a ⇤ e = a.

(G3R) To each a 2 G there is an a 2 G such that

a ⇤ a = e.

Show that (G, ⇤) is a group.1.4 Take m 2 Z and let Zm be the set of residue classes mod m. Show that

+ : Zm ⇥ Zm �! Zm, ([l], [k]) 7�! [l + k]

is a well–defined binary operation, that is, show that [l + k] does not depend on

the choice of representatives l and k of the equivalence classes [k] and [l]. Show

further, that (Zm,+) is a group.

25

CHAPTER 2

Basic Properties

We introduce some convenient notation. Given a group (G, ⇤), we call G theset underlying the group (G, ⇤). If the binary operation ⇤ is given by the context,we permit ourselves to write G instead of (G, ⇤). (This convention is to be usedwith care, as a group is not determined by its underlying set).

Our next notational convention is very di↵erent in nature. Very many examples

of groups, including the most familiar ones, are drawn from our number system.

They thus involve the operations of addition and multiplication. In (Q+,⇥) therôle of the neutral element is played by the number 1 and the inverse of the non–

zero rational number a is just the rational number a�1. Moreover, the product of

a and b is generally written just by juxtaposing, namely as ab. Our convention is

to adopt the multiplicative notation for an arbitrary group (G, ⇤). This means wewrite

ab for a ⇤ b, a, b 2 G;1 for e;

a�1 for ā, a 2 G.

Sometimes we may prefer to retain e for the neutral element and write a · b for ab.We follow the logic of this convention by adopting the notation of exponents.

Thus, for a positive integer n and an element a 2 G, we define an inductively by

a0 := 1, an+1 := ana,

for integers n > 1. We extend the convention to any integer by defining

a�n := (a�1)n.

The group axioms (in particular, the associative law) then immediately validate

the laws of indices:

Proposition 2.1. aman = am+n, (an)m = amn.

If (G, ⇤) is an abelian group, it is customary to adopt additive notation insteadof multiplicative notation. Thus in an abelian group (A, ⇤) we often write

27

Lecture Notes

a+ b for a ⇤ b, a, b 2 A;0 for e;

�a for a, a 2 A;a� b for a+ (�b), a, b 2 A;na for an, a 2 A.

With the last convention the laws of indices become

ma+ na = (m+ n)a, m(na) = (mn)a.

Unless otherwise specified, we will use multiplicative notation and denote the

group and its underlying set by the same symbol.

Our final convention has to do with the associative law. This law asserts that

the two ways of forming the product of three elements a, b, c 2 G in that order,yield the same result. Using the multiplicative notation we have

(ab)c = a(bc).

Thus we may write abc for this product, and the notation is unambiguous. Sim-

ilarly, given k elements a1, a2, . . . , ak in the group G, we have many (how many?)

ways of forming their product in the given order, and by virtue of the associative

law, all these ways yield the same element of G. Thus we may write a1a2 . . . akfor this element. The reader will readily appreciate the simplification produced by

this convention — rewrite the proof of Proposition 1.9 if you need convincing.

When we are working with N,Z,Q,R or C, we can cancel. For example, givenpositive real numbers a, b, c, the equality ab = ac implies b = c, we say that we

can cancel on the left . And the equality ab = cb implies a = c, we say that we

can cancel on the right . While we cannot expect that all the rules applying to

(R,+) or (R+,⇥) extend to an arbitrary group (G, ⇤), cancellation does extend toan arbitrary group.

Proposition 2.2. Let (G, ⇤) be a group. Thenab = ac =) b = c (left cancellation)ab = cb =) a = c (right cancellation).

for every a, b, c 2 G.

Proof. Take a, b, c 2 G. Then

ab = ac =) aab = aac =) eb = ec =) b = c and

ab = cb =) abb = cbb =) ae = ce =) a = c.

⇤28

Lecture Notes

The next proposition establishes an important criterion for a given set with an

associative binary operation to be a group.

Proposition 2.3. Let G be a nonempty set with a given associative binary

operation. Then G is a group if and only if, for all a, b 2 G, the equations

(16) ax = b, ya = b

have solutions in G.

Proof. Let G be a group. Then

aa�1b = b and ba�1a = b,

so that the equation ax = b has the solution x = a�1b and the equation ya = b has

the solution y = ba�1.

Conversely, suppose that the equations (16) always have solutions in G. Then,

for each a 2 G, there is an element ea 2 G such that aea = a. We show that ea isindependent of a. For, if b 2 G, there is an element y 2 G with ya = b. Then

bea = yaea = ya = b.

Thus e = ea is a right neutral element for G. Furthermore, for every a 2 G thereis a solution a�1 of the equation ax = e. The proof is completed by appealing to

Proposition 1.8. ⇤

Definition 2.4. Let (G, ⇤) be a group. Then the order of (G, ⇤) is defined tobe the number (possibly infinite) of elements of the set G and we denote it by |G|.The group is finite (infinite) if its order is finite (infinite).

Now let G = {g1, g2, . . . , gn} be a finite group of order n. Then we may describeG by its group table or Cayley table:

g1 g2 . . . gj . . . gn

g1 g1g1 g1g2 . . . g1gj . . . g1gng2 g2g1 g2g2 . . . g2gj . . . g2gn...

......

gi gig1 gig2 . . . gigj . . . gign...

......

gn gng1 gng2 . . . gngj . . . gngn

Observation 2.5. The left and right cancellation laws in G imply that every

element of G occurs exactly once in each row and exactly once in each column of

the group table.

Question 2.6. Given a positive integer n, are there any groups of order n?

29

Lecture Notes

One approach to settling this question is to figure out all possible group tables. For

n = 1 G has exactly one element which must be the neutral element. Now consider

n = 2, say G = {a, b}. By Observation 2.5 the only possible tables are given by

a b

a a b

b b a

and

a b

a b a

b a b

It is an exercise to show that both tables satisfy (G1) – (G3). In the table on

the left the element a plays the rôle of the neutral element and in the one on the

right the element b plays the rôle of the neutral element. Note that we obtain one

table from the other by permuting a and b. Next we consider the case n = 3, say

G = {a, b, c}. Observation 2.5 implies that the only possible tables are given by

a b c

a a b c

b b c a

c c a b

,

a b c

a c a b

b a b c

c b c a

and

a b c

a b c a

b c a b

c a b c

Again it is an exercise to show that three tables satisfy (G1) – (G3). And again we

see that we obtain any of the tables above from any other by permuting the elements

of the set G. Thus, “morally,” the three groups obtained should be essentially the

same. This leads to the following

Question: How do we compare groups?

The answer to this question in turn leads to the definition of the notion of homo-

morphism and isomorphism which we discuss in the next chapter.

30

CHAPTER 3

Comparing Groups

Sets are compared by means of functions between them. Vector spaces are

compared by means of linear transformations, that is, transformations which re-

spect the vector space structure. When comparing groups we are only interested

in those functions which respect the group structure. To be more precise, let G

and H be groups and let ' : G �! H be a function from the set underlying G tothe set underlying H. What does it mean for ' to preserve the group structure?

It means that ' maps the neutral element of G to the neutral element of H, maps

the inverse of an element of G to the inverse of the image and maps the product

of two elements to the product of their images.

Proposition 3.1. Let G and H be groups and let ' : G �! H be a functionbetween the sets underlying G and H, respectively, such that '(gg0) = '(g)'(g0)

for every g, g0 2 G. Then(i) '(eG) = eH and

(ii) '(g�1) = ('(g))�1,

for every g 2 G, where eG and eH denote the neutral elements of G and H respec-tively.

Proof. (i) Note that eH'(eG) = '(eG) = '(eGeG) = '(eG)'(eG) by assump-

tion. Cancellation yields eH = '(eG).

(ii) Take g 2 G. Then '(g)'(g�1) = '(gg�1) = '(eG) = eH by (i), whichimplies that '(g�1) is a right inverse of '(g) and hence a left inverse by Proposi-

tion 1.8. As the inverse of an element of a group is uniquely determined, it follows

that '(g�1) = ('(g))�1. ⇤

Structure preserving functions between groups are called group homomorphisms

or simply homomorphisms and we arrive at the following definition.

Definition 3.2. Let G and H be groups. A function ' : G �! H is called ahomomorphism if and only if

'(gg0) = '(g)'(g0)

for every g, g0 2 G.31

Lecture Notes

Unless otherwise specified we will assume from now on that G,H, . . ., are groups

and ', , . . . are homomorphisms.

Example 3.3. Multiplication by a fixed integer m defines a homomorphism

(Z,+) �! (Z,+), k 7�! mk,

as m(k + `) = mk +m` for all k, ` 2 Z.

Example 3.4. The determinant

det : GL(n;R) �! (R+,⇥), A 7�! det(A)

is a homomorphism, as det(AB) = det(A) det(B) for A,B 2 GL(n;R).

Example 3.5. The exponential function

exp : R �! S1 = {z 2 C | |z| = 1}, x 7�! e2⇡ix

is a homomorphism from (R,+) to (S1,⇥), as e2⇡i(x+y) = e2⇡ixe2⇡iy.

Proposition 3.6. Given two homomorphisms ' : G �! H and : H �! Ktheir composition � ' : G �! K is a homomorphism.

Proof. Take g, g0 2 G. As both ' and are homomorphisms, we obtain

�' (gg0) = �

'(gg0)�

= �

'(g)'(g0)�

= �

'(g)�

�

'(g0)�

=�

�' (g)� �

�'(g0)�

,

showing that � ' is a homomorphism. ⇤

Definition 3.7. A homomorphism ' : G �! H is called an isomorphism ifand only if there is a homomorphism : H �! G such that

� ' = idG and ' � = idH .

If there is such a homomorphism it is uniquely determined and called the inverse

of '. (Assume there is a 0 with the same property. Then 0 = 0�idH = 0�'� =idG� = .) Two groups G and H are said to be isomorphic (essentially the same)if and only if there is an isomorphism ' : G �! H. We then write G ⇠= H or' : G ⇠= H or ' : G

⇠=�! H.

Proposition 3.8. The composition of two isomorphisms is again an isomor-

phism.

Proof. Exercise. ⇤

Example 3.9. The groups of order 3 arising from di↵erent tables are all iso-

morphic.

32

Lecture Notes

Proposition 3.10. The homomorphism ' : G �! H is an isomorphism if andonly if it is bijective.

Thus, as in the case of vector spaces, a structure preserving function between

groups which is bijective (and hence invertible as a function) is an isomorphism

— the inverse function is automatically structure preserving. Note that this is not

true in the case of topological spaces and continuous functions — a continuous

bijective function between topological spaces need not have a continuous inverse!

Proof. If ' is an isomorphism it has an inverse as a function from the set

underlying the group G to the set underlying the group H and is thus bijective.

For the converse assume that ' is bijective. Then there is a function : H �!G from the set underlying H to the set underlying G such that � ' = idG and'� = idH . It remains to show that is a homomorphism. But, given h1, h2 2 H,there are g1, g2 2 G such that '(gi) = hi, i = 1, 2, since ' is surjective. Thengi = ( � ')(gi) = ('(gi)) = (hi) for i = 1, 2 and hence

(h1h2) = ('(g1)'(g2)) = ('(g1g2))

= ( � ')(g1g2) = idG(g1g2)

= g1g2 = (h1) (h2).

⇤

Exercises:

3.1 (a) Show that

' : (R,+) �! (R+,⇥), t 7�! et

is an isomorphism of groups.

(b) Show that

exp : R �! S1 = {z 2 C | |z| = 1}, x 7�! e2⇡ix

is a homomorphism, but not an isomorphism.

3.2 Given two isomorphisms, ' : G �! H and : H �! K, show that theircomposition � ' : G �! K is an isomorphism.3.3 Find all possible group tables for G = {a, b, c, d}. Which of these yield isomor-phic groups?

33

CHAPTER 4

Subgroups

By now we have seen what a group is, we have also seen some examples and we

have a means of comparing groups.

Question 4.1. How do we obtain new groups from given groups?

Given a group G, we want to characterise groups “sitting inside” G, that is,

groups whose underlying set is a subset of the set underlying G and whose structure

is compatible with the group structure of G. As groups are sets with additional

structure, we let ourselves be guided by what we know about sets. By Theorem

0.15, the set X is a subset of the set Y if and only if ◆ : X ! Y, x 7! x is a function.For subgroups we insist that the inclusion be a homomorphism.

Definition 4.2. A group H is a subgroup of the group G if and only if

◆ : H �! G, h 7�! h

is a homomorphism. (Note that ◆ is a function if and only if the set underlying H

is a subset of the set underlying G.) We call a subgroup H a proper subgroup if

and only if H 6= G. We call a subgroup H non–trivial if and only if H is not asingleton. We write H G to denote that H is a subgroup of G and H < G todenote that H is a proper subgroup of G.

In practice, mathematicians often study subgroups of a given group in order

to obtain information about the group itself which may be too big or complex to

obtain information directly.

Question 4.3. When does a subset of the set underlying a group G carry the

structure of a subgroup?

In order to answer this question we first assume that H is a subgroup of the

group G. Then the inclusion of H in G is a homomorphism and hence maps the

neutral element of H to the neutral element of G. But this means that the neutral

element of H corresponds to the neutral element of G. The fact that the inclusion

is a homomorphism also implies that for every x, y 2 H, the inverse of x withrespect to the group structure in H corresponds to the inverse with respect to

the group structure in G, and the product of x and y with respect to the group

35

Lecture Notes

structure in H corresponds to their product with respect to the group structure in

G. Thus H must contain the inverse (with respect to the group structure in G) of

any element x 2 H and the product (with respect to the group structure in G) ofany two elements x, y 2 H. The next proposition asserts that these conditions arein fact su�cient.

Proposition 4.4. Let H be a subset of the set underlying the group G. Then

H carries the structure of a subgroup if and only if H contains the neutral element

of G and the inverse x�1 of each x 2 H and is closed under multiplication, that is,xy 2 H for every x, y 2 H.

Proof. Given a subsetH of the set underlying the group G, we already showed

that the conditions are necessary for H to carry the structure of a subgroup.

Now assume that H contains the neutral element and the inverse x�1 for every

x 2 H and is closed under multiplication. We must show that H carries thestructure of a subgroup.

In order for the inclusion ofH in G to be a homomorphism, the binary operation

⇤H defining the group structure onH must be the restriction of the binary operation⇤G defining the group structure on G. So put

⇤H := ⇤G|H⇥H : H ⇥H �! H, (h1, h2) 7�! h1 ⇤G h2.

Since H is closed under multiplication, that is, xy 2 H for every x, y 2 H, thebinary operation ⇤H is well–defined. Further, associativity of ⇤G immediately im-plies associativity of ⇤H . By assumption, the neutral element of G is contained inH and clearly is a neutral element with respect to ⇤H . Also by assumption, theinverse of an element x 2 H with respect to ⇤G is contained in H and is an inverseof x with respect to ⇤H . Hence (H, ⇤H) is a group and by definition the inclusionof H in G is a homomorphism. ⇤

The next proposition provides a condition for a non–empty subset H of the set

underlying the group G to carry the structure of a subgroup which is sometimes

more convenient than Proposition 4.4.

Proposition 4.5. Let (G, ⇤) be a group and let H be a non–empty subset of G.Then (H, ⇤|H⇥H), where ⇤|H⇥H denotes the restriction of ⇤ to H⇥H, is a subgroupof (G, ⇤) if and only if xy�1 2 H whenever x, y 2 H.

Proof. If (H, ⇤|H⇥H) is a subgroup of (G, ⇤), clearly xy�1 2 H wheneverx, y 2 H.

For the converse, assume that xy�1 2 H whenever x, y 2 H. Now take x 2 H(H is non–empty). Then e = xx�1 2 H and hence x�1 = ex�1 2 H by assumption.

36

Lecture Notes

This also implies that xy = x(y�1)�1 2 H whenever x, y 2 H. Thus H is asubgroup of G by Proposition 4.4. ⇤

Example 4.6. Put M := {1, 2, · · · , n} and let Sn = S(M) be the group con-sisting of the set of permutations of M together with composition of functions. Snis also called the symmetric group on n symbols.

Given a regular polygon with n vertices we label the vertices 1, 2, . . . , n. Then

a symmetry of the polygon corresponds to a permutation of M and we may view

the group of symmetries as a subgroup of the group Sn.

Example 4.7. Given a field, K, recall that the set GL(n;K) of non–zero n⇥n–matrices with entries in K together with matrix multiplication forms a group and

define

SL(n;K) := {A 2 GL(n;K) | detA = 1}.

Clearly, the identity matrix is contained in SL(n;K), so that SL(n;K) is non–

empty. Given A,B 2 SL(n;K), we obtain det(AB�1) = (detA)(detB)�1 = 1⇥1 =1 and hence AB�1 2 SLn(K). Thus, by Proposition 4.5, SL(n;K) is a subgroup ofGL(n;K).

Example 4.8. Given a field, K, the elements of the set

µn(K) := {x 2 K | xn = 1}

are called the n–th roots of unity . Then µ4(R) = {1,�1}, so that |µ4(R)| = 2 andµ4(C) = {1,�1, i,�i}, so that |µ4(C)| = 4. Furthermore µ4(C) is a subgroup of

S1 = {z 2 C | |z| = 1}.

Example 4.9. For m 2 Z, put mZ := {km | k 2 Z}, that is, mZ consists ofall integer multiples of m. Then mZ is a subgroup of Z (with respect to addition).

Example 4.10. Given a group G, the centre, C(G), of G consists of all those

elements x 2 G which commute with every other element of G, that is, gx = xg orgxg�1 = x for every g 2 G.

C(G) := {x 2 G | gxg�1 = x 8 g 2 G}

Clearly eG 2 C(G), so that C(G) is non–empty. Now take x, y 2 C(G). Then, forevery g 2 G,

gxy�1g�1 = gxg�1gyg�1 = gxg�1(gyg�1)�1 = xy.

Thus C(G) is a subgroup of G by Proposition 4.5.

There are further examples of subgroups arising from group homomorphisms.

37

Lecture Notes

Proposition 4.11. Let ' : G �! H be a homomorphism and put

ker' := {g 2 G | '(g) = eH} and

im' := {h 2 H | h = '(g) for some g 2 G}.

Then ker' is a subgroup of G and im' is a subgroup of H.

We also sometimes write '(G) for im'.

Proof. We show that im' is a subgroup of H. First note that im' 6= ; aseH = '(eG) 2 im'. By Proposition 4.5 it is thus enough to show that xy�1 2 im'whenever x, y 2 im'. So take x, y 2 im'. Then there are x0, y0 2 G such thatx = '(x0) and y = '(y0). As ' is a homomorphism,

xy�1 = '(x0)'(y0)�1 = '(x0)'((y0)�1)) = '(x0(y0)�1),

and hence xy�1 2 im'.It is left as an exercise to show that ker' is a subgroup of G. ⇤

Proposition 4.12. The homomorphism ' : G ! H is injective if and only ifker' = {eG}.

Proof. Exercise. ⇤Note that every subgroup H of the group G is the image of the inclusion

◆ : H �! G, h 7�! h.

Thus we see that every subgroup is the image of a homomorphism.

Furthermore, if the subgroup H of the group G is the kernel of the homomor-

phism ' : G �! K, that is H = ker', then

gHg�1 := {ghg�1 | h 2 H} = H

for every g 2 G. To see this, take g 2 G and x 2 gHg�1. Then x = ghg�1, forsome h 2 H and hence

'(x) = '(ghg�1) = '(g)'(h)'(g�1)

= '(g)1('(g))�1 = '(gg�1) = eH ,

which shows that x 2 H and thus gHg�1 ✓ H for every g 2 G. In order toshow that H ✓ gHg�1 for every g 2 G, note that every h 2 H can be written ash = gg�1hgg�1 and g�1hg 2 gH(g�1)�1 ✓ H.

Definition 4.13. The elements g1, g2 of the group G are conjugate in G if and

only if there is an element h 2 G with g1 = hg2h�1. The subgroup H of the groupG is normal if and only if gHg�1 = H for every g 2 G. We write H E G to denotethat H is a normal subgroup of G.

38

Lecture Notes

Example 4.14. The kernel of a homomorphism ' : G �! H is a normalsubgroup of G.

The following example shows that not every subgroup is normal.

Example 4.15. Consider the subset

H := {

1 0

0 1

!

,

1 1

0 1

!

} ✓ SL(2;F2),

where F2 = {0, 1}, the field with two elements. It is not di�cult to see that H isindeed a subgroup of SL(F2). But H is not a normal subgroup as

0 1

1 0

!

1 1

0 1

!

0 1

1 0

!

=

0 1

1 1

!

0 1

1 0

!

=

1 0

1 1

!

/2 H.

Lemma 4.16. Let ' : G ! H be a homomorphism of groups. Then the inverseimage of a normal subgroup of H is normal in G. If ' is surjective, then the image

of a normal subgroup of G is normal in H.

Proof. Let K be a normal subgroup of H. Take g 2 G and k 2 '�1(K). Then'(gkg�1) = '(g)'(k)'(g)�1 2 K as K is normal in H. Thus gkg�1 2 '�1(K),showing that '�1(K) is normal in G.

Now assume that ' is surjective and let N be a normal subgroup of G. Take

h 2 H and '(n) 2 '(N). As ' is surjective, there is a g 2 G with '(g) = h andwe obtain h'(n)h�1 = '(g)'(n)'(g)�1 = '(gng�1) = '(n0) for some n0 2 N as Nis normal in G. Hence h'(n)h�1 2 '(N), showing that '(N) is nomal in G. ⇤

Exercises:

4.1 Let ' : G �! H be a homorphism. Show that ker' is a subgroup of G.4.2 Show that the homomorphism ' : G ! H is injective if and only if ker' ={eG}.4.3 Given a group G, the centre, C(G), of G is defined by

C(G) := {x 2 G | gxg�1 = x 8 g 2 G}.

Show that C(G) is a normal subgroup of G.

4.4 Let ' : G ! G0 be a homomorphism of groups and let K be a subgroup of G.Show that

K ker' = '�1('(K)).

39

CHAPTER 5

Generators of Groups and Subgroups

First we infer the following important fact from Proposition 4.5.

Proposition 5.1. Let {H↵}↵2A be a family of subgroups of the group G indexedby the set A. Then

T

↵2A H↵ is a subgroup of G.

Proof. Since e 2 H↵ for every ↵ 2 A, we also have e 2T

↵2A H↵. HenceT

↵2A H↵ is non–empty. Further, if a, b 2T

↵2A H↵, then ab�1 2 H↵ for every

↵ 2 A. Thus ab�1 2T

↵2A H↵. By Proposition 4.5,T

↵2A H↵ is a subgroup of

G. ⇤

Now let S be a subset of the group G and consider the collection {H↵}↵2A ofsubgroups of G which contain S. This collection is not empty since S ✓ G. Then

HS :=\

↵2A

H↵

is a subgroup of G by Proposition 5.1. Further, HS contains S and it is the smallest

subgroup of G containing S, as any subgroup H↵ containing S contains HS. We

call HS the subgroup of G generated by S. If HS = G we call S a set of generators

of G. A group is said to be cyclic if and only if it may be generated by a single

element. It is said to be finitely generated if and only if it may be generated by a

finite set of elements.

We may adopt a di↵erent point of view with regard to HS which is often more

useful in practice. We consider the collection W(S) of all elements of HS of theform

a1a2 . . . an,(17)

where n is a counting number and, for each i, ai or a�1i belongs to S. This collection

is non–empty (even if S is empty W(S) contains e), and it is plain from Proposition4.5 that it is a subgroup of HS. Moreover, if H is any subgroup of G containing

S, then H must contain all the elements of the form (17). Hence the collection

W(S) is, in fact, HS. We say that HS is the set of all words formed from the setof symbols {a, a�1 | a 2 S}.

Lemma 5.2. Let G be a group. Then the powers ak, (k 2 N), of an element a ofG are either all distinct or there is a positive integer n such that am = 1 (= eG) if

41

Lecture Notes

and only if n|m. In the former case we say that a is of infinite order, in the lattercase that a is of order n.

Proof. If all powers of a are distinct there is nothing to prove.

Now assume that am = ak with k > m. Then ak�m = 1. Hence

{` 2 N+ | a` = 1} 6= ;.

Let n be the minimum of this set. If n|` for some ` 2 Z, there is an s 2 Z suchthat ` = ns so that

a` = ans = (an)s = 1s = 1.

Now assume a` = 1 for ` 2 Z. By the Euclidean Algorithm there are r, s 2 Z suchthat ` = ns+ r with 0 r < n. Then

1 = a` = ans+r = ansar = (an)sar = (1)sar = ar.

Since r < n this implies r = 0 and hence ` = ns, that is, n|`. ⇤

If S = {x↵ | ↵ 2 A} we also write

HS = hx↵i↵2A

for the subgroup generated by S.

The following lemma shows that the homomorphic image of a generating set

for a group is a generating set for the image of that group.

Lemma 5.3. Given a group homomorphism, ' : G ! H, and a1, a2, . . . , an 2G, n 2 N,

'(ha1, a2, . . . , ani) = h'(a1),'(a2), . . . ,'(an)i.

Proof. Exercise. ⇤

The next lemma completely determines the subgroups of Z.

Lemma 5.4. Let H be a non–trivial subgroup of Z. Then H = hmi for somem 2 Z.

Proof. As Z and its subgroups are commutative, we adopt additive notation.Being non–trivial, H contains a non–zero integer, and being a subgroup, H must

therefore contain a positive integer (n 2 H implies �n 2 H). Let m be the smallestpositive integer in H. Then mZ ✓ H. We show that H ✓ mZ. Take h 2 H. Bythe Euclidean Algorithm, there are r, s 2 Z such that

h = ms+ r with 0 r < m.42

Lecture Notes

Since h,m 2 H and r = h � ms, it follows that r 2 H. Since m is the smallestpositive integer in H it follows that r = 0. Hence

h = ms 2 mZ,

which shows that H ✓ mZ and thus H = mZ. ⇤

The remainder of this chapter is devoted to sets of generators of Sn, the sym-

metric group on n symbols. Recall that Sn is defined to be equal to the set of

permutations (that is, bijections) of the set M = {1, . . . , n} together with compo-sition of functions. Hence

|Sn| = n!

Definition 5.5. A permutation � 2 Sn is called a cycle of length r or r–cycleif and only if there is a subset

{i1, i2, . . . , ir} ✓ {1, · · · , n}

of order r such that

(i) �(ij) = ij+1 for 1 j < r;

(ii) �(ir) = i1;

(iii) �(m) = m for m /2 {i1, . . . , ir}.We then write � = (i1 . . . ir). A cycle, (i1 i2), of length two is also called a trans-

position.

With this definition � = idM = e is the only cycle of length 1.

Theorem 5.6. The r–cycles in Sn satisfy the following relations

(a) (i1 . . . ir) = (i2 . . . iri1);

(b) (i1 . . . ir) = (i1 . . . ij)(ij . . . ir) for 1 < j < r;

(c) The order of (i1 . . . ir) is r;

(d) Given ⌧ 2 Sn, we have ⌧(i1 · · · ir)⌧�1 = (⌧(i1) · · · ⌧(ir)).

Proof. (a), (b) and (c) follow from the definitions.

(d) First note that (b) implies that every r–cycle g may be written as a product

of transpositions, that is,

g = g1 . . . gr�1

43

Lecture Notes

where gi is a transposition for 1 i r � 1. Namely, for g = (j1 . . . jr), putgi = (jiji+1) for 1 i r � 1. Then, given ⌧ 2 Sn,

⌧g⌧�1 = ⌧g1g2 . . . gr�1⌧�1

= ⌧g1⌧�1⌧g2⌧

�1 . . . ⌧gr�1⌧�1.

Hence it su�ces to show that (d) holds for transpositions. But given i1, i2,m 2{1, . . . , n},

�

⌧(i1i2)⌧�1� (m) =

8

>

>

<

>

>

:

⌧(i1) if ⌧�1(m) = i2,

⌧(i2) if ⌧�1(m) = i1,

m otherwise.

=

8

>

>

<

>

>

:

⌧(i1) if m = ⌧(i2),

⌧(i2) if m = ⌧(i1),

m otherwise.

and hence

⌧(i1i2)⌧�1 = (⌧(i1)⌧(i2)).

⇤

Corollary 5.7. Any two r–cycles (i1 . . . ir) and (j1 . . . jr) are conjugate, that

is, there is a � 2 Sn such that

(j1 . . . jr) = �(i1 . . . ir)��1.

Proof. Define � 2 Sn by

�(ik) = jk for 1 k r

�(m) = m for m /2 {i1, . . . , ir}.

Then, by (d), �(i1 . . . ir)��1 = (j1 . . . jr). ⇤

Theorem 5.8. Every � 2 Sn is a product of disjoint cycles and hence a productof transpositions.

Corollary 5.9. Sn is generated by transpositions.

Proof. Given � 2 Sn define an equivalence relation on {1, · · ·n} by

x1 ⇠ x2 if and only if �`(x1) = x2 for some ` 2 Z

and x1, x2 2 {1, . . . , n}. Then the set {Ai}mi=1 of equivalence classes of this equiva-lence relation is a partition of {1, . . . , n}. If |Ai| = ni and ai 2 Ai, then

{ai, �(ai), �2(ai), . . . , �ni�1(ai)} = Ai.44

Lecture Notes

Hence � permutes the elements of Ai cyclicly. Ai is also called the orbit of ai under

�. Now let �i be the ni–cycle which permutes the elements of Ai the way � does,

that is,

�i|Ai = �|Ai and �i|M\Ai = idM\Ai .

Then

� = �1�2 . . . �m.

Since the Ai are disjoint, the �i are disjoint. ⇤

Definition 5.10. The sign "(�) of a permutation � 2 Sn is defined by

"(�) :=Y

i

Lecture Notes

Since every � 2 Sn is a product of transpositions it is now enough to show that"(i1, i2) 2 {±1}. But

"(i1, i2) =i2 � i1i1 � i2

= �1 or "(i1, i2) =i1 � i2i2 � i1

= �1.

⇤The above proof shows that " = 1 if and only if � is the product of an even

number of transpositions.

Definition 5.12. The alternating group An on n symbols is the kernel of ".

An := ker(" : Sn �! {±1}).

Thus An consists of all elements of Sn which are products of an even number of

transpositions. Furthermore, as An is the kernel of a homomorphism, it is normal

in Sn.

Exercises:

5.1 Let ' : G �! H be a homomorphism of groups and take g1, . . . , gn 2 G. Showthat

'(hg1, . . . , gni) = h'(g1), . . . ,'(gn)i.

5.2 Show that Sn is generated by { (1 2), (1 3), . . . , (1n) }.5.3 Show that there are permutations x, y 2 Sn which generate Sn.5.4 Show that every group G of order n is isomorphic to a subgroup of Sn. (This

is also called Cayley’s Theorem.)

46

CHAPTER 6

Cosets, Quotient Groups and Lagrange’s Theorem

Recall that we may construct a new set from a given one by means of an

equivalence relation. To be more precise, given a set S and an equivalence relation

⇠ on S, we obtain the set of all equivalence classes of ⇠, also denoted by S/⇠.In the case of sets with additional structure, like groups or vector spaces, we are

interested in equivalence relations whose quotient sets carry a structure which

reflects the structure of the original object. We then say that this structure is

induced by the original object. Thus we study equivalence relations which respect

the given additional structure.

Given a vector space V with a subspace U , two vectors v, w 2 V are said tobe equivalent (modulo U) if and only if they di↵er by an element of U , that is, if

and only if v � w 2 U . Then the quotient set of the set underlying V modulo thisequivalence relation carries a vector space structure such that the projection of V

onto the quotient set is a linear transformation. The vector space thus obtained is

called the quotient of V modulo U and denoted by V/U .

In Exercise 1.4 we met an example of an equivalence relation defined on a group

such that the quotient set of the set underlying the group modulo the equivalence

relation carries a group structure. The group was the group of integers under

addition and the equivalence relation was equivalence modulo an integer m. Two

integers k and ` are equivalent modulo m if and only if their di↵erence is a multiple

of m. In other words, k and ` are equivalent modulo m if and only if k � ` 2 mZ.Now we consider an arbitrary group G with a subset H. In multiplicative

notation the di↵erence of two elements is replaced by the product of the first times

the inverse of the second and we obtain the following definition

Definition 6.1. Let H be a subset of the set underlying the group G. Then

x, y 2 G are equivalent modulo H if and only if xy�1 2 H. We write x ⇠H y todenote that x is equivalent to y modulo H.

Lemma 6.2. Let H be a subset of the set underlying the group G. Then equiv-

alence modulo H is an equivalence relation on G if and only if H is a subgroup of

G.

Proof. First let H be a subset of the set underlying the group G such that

the relation defined by x ⇠H y if and only if xy�1 2 H is an equivalence relation.47

Lecture Notes

Reflexivity implies that, for every x 2 G, x ⇠H x which implies e = xx�1 2 H.Hence H is non–empty. Further, take x, y 2 H. Then xe�1, ye�1 2 H and hencex ⇠H e and y ⇠H e so that e ⇠H y by symmetry. Transitivity guarantees thatx ⇠H y and therefore xy�1 2 H. Thus, by Proposition 4.5, H is a subgroup of G.

Now suppose that H is a subgroup of G. For x 2 G we obtain e = xx�1 2 Hand hence x 'H x. Given x, y 2 G with x 'H y, we obtain xy�1 2 H and henceyx�1 = (xy�1)�1 2 H, so that y 'H x. Finally, given x, y, z 2 G with x 'H y andy 'H z, we obtain xy�1, yz�1 2 H. Thus xz�1 = xy�1yz�1 2 H, showing thatx 'H z. ⇤

What are the equivalence classes? Given x, y 2 G, we have x ⇠H y if and onlyif xy�1 2 H if and only if there is some h 2 H such that xy�1 = h if and only ifthere is some h 2 H such that x = hy. Hence the equivalence class [y] of y moduloH is the set of elements of G of the form hy. We denote this set by Hy and call it

the right coset of y modulo H.

Hy := {hy | h 2 H} ✓ G

Observation 6.3. Similarly, we may define x ⇠H y for x, y 2 G if and onlyif x�1y 2 H, for any subset H of the set underlying the group G. Again, 'H isan equivalence relation if and only if H is a subgroup of G. Here the equivalence

classes are the left cosets

xH := {xh | h 2 H} ✓ G.

In the case of a commutative group we have x�1y = (xy�1)�1, and hence x�1y 2 Hif and only if xy�1 2 H. Thus the two equivalence relations coincide.

Lemma 6.4. Let H be a subgroup of the group G and take x, y, z 2 G. Then

(i) xH = H if and only if x 2 H;

(ii) (xy)H = x(yH) so that we may write xyH to denote this set;

(iii) xyH = xzH if and only if yH = zH.

Proof. (i) If xH = H then x 2 H, for x = xe and e 2 H. For the conversesuppose that x 2 H. Then

xH = {xh | h 2 H} ✓ H

as xh 2 H for every h 2 H (H is a subgroup). Every h0 2 H can be written ash0 = xx�1h0 2 xH and hence H ✓ xH.

48

Lecture Notes

(ii) Given x, y 2 G we obtain

(xy)H = {(xy)h | h 2 H} = {x(yh) | h 2 H}

= {xy0 | y0 2 yH} = x(yH).

(iii) Take x, y, z 2 G. Clearly, yH = zH implies xyH = xzH. Now supposethat xyH = xzH. Then yH = eyH = x�1xyH = x�1xzH = zH. ⇤

We obtain the corresponding properties for right cosets in a similar fashion.

Lemma 6.5. Let H be a subgroup of the group G and take x, y, z 2 G. Then(i) Hx = H if and only if x 2 H;

(ii) H(xy) = (Hx)y so that we may write Hxy to denote this set;

(iii) Hxy = Hzy if and only if Hx = Hz.

Proof. Exercise. ⇤

Lemma 6.6. Let H be a subgroup of the group G and take x, y 2 G. Then thefollowing statements are equivalent:

(i) x ⇠H y;

(ii) Hx = Hy;

(iii) x 2 Hy;

(iv) y 2 Hx;

(v) xy�1 2 H;

(vi) yx�1 2 H

Proof. First note that (i) and (v) are equivalent by definition. As H is a

subgroup, (v) is equivalent to (vi) which is equivalent to (iv) by definition. Further,

(v) is equivalent to xy�1 = h for some h 2 H () x = hy for some h 2 H whichis equivalent to (iii). Similarly, (vi) is equivalent to (iv). ⇤

49

Lecture Notes

Question 6.7. Given a subgroup H of the group G, when does the quotient setG/⇠H of the set underlying G modulo the equivalence relation ⇠H admit a groupstructure such that the canonical projection

⌘ : G �! G/⇠His a homomorphism?

In the following we restrict attention to the equivalence modulo H whose equiv-

alence classes [x] are the right cosets Hx. For the canonical projection ⌘ to be a

homomorphism we must have

(18) HxHy = [x][y] = ⌘(x)⌘(y) = ⌘(xy) = [xy] = Hxy.

Observe that (18) yields a well–defined binary operation if and only if it is inde-

pendent of the choice of representatives, that is, for x, x0, y, y0 2 G with x ⇠H x0

and y ⇠H y0 we always have Hx0y0 = Hxy.So take x, x0, y, y0 2 G with x ⇠H x0 and y ⇠H y0. Then there are h, h0 2 H

such that x0 = hx and y0 = h0y and hence Hx0y0 = Hhxh0y = Hxh0y. Thus

Hx0y0 = Hxy () Hxh0y = Hxy for all h0 2 H() Hxh0 = Hx for all h0 2 H() Hxh0x�1 = H for all h0 2 H() xh0x�1 2 H for all h0 2 H.

Lemma 6.8. Given a subgroup H of the group G the following are equivalent.

(i) xhx�1 2 H for all x 2 G and h 2 H;

(ii) xHx�1 = H for all x 2 G;

(iii) xH = Hx for all x 2 G.

Thus H is normal if and only if one of the conditions (i)–(iii) holds. Observe

that (iii) states that the left cosets and the right cosets coincide.

Proof. Take x 2 G and h 2 H.

(i) =) (ii): (i) implies xHx�1 ✓ H. As h = x(x�1hx)x�1 2 xHx�1, weconclud