NOTES ON REPRESENTATIONS OF FINITE GROUPSaaronlan/assets/representation-theory.pdfNOTES ON...

NOTES ON REPRESENTATIONS OF FINITE GROUPS

AARON LANDESMAN

CONTENTS

1. Introduction 31.1. Acknowledgements 31.2. A first definition 31.3. Examples 41.4. Characters 71.5. Character Tables and strange coincidences 82. Basic Properties of Representations 112.1. Irreducible representations 122.2. Direct sums 143. Desiderata and problems 163.1. Desiderata 163.2. Applications 173.3. Dihedral Groups 173.4. The Quaternion group 183.5. Representations of A4 183.6. Representations of S4 193.7. Representations of A5 193.8. Groups of order p3 203.9. Further Challenge exercises 224. Complete Reducibility of Complex Representations 245. Schur’s Lemma 306. Isotypic Decomposition 326.1. Proving uniqueness of isotypic decomposition 327. Homs and duals and tensors, Oh My! 357.1. Homs of representations 357.2. Duals of representations 357.3. Tensors of representations 367.4. Relations among dual, tensor, and hom 388. Orthogonality of Characters 418.1. Reducing Theorem 8.1 to Proposition 8.6 418.2. Projection operators 43

1

2 AARON LANDESMAN

8.3. Proving Proposition 8.6 449. Orthogonality of character tables 4610. The Sum of Squares Formula 4810.1. The inner product on characters 4810.2. The Regular Representation 5011. The number of irreducible representations 5211.1. Proving characters are independent 5311.2. Proving characters form a basis for class functions 5412. Dimensions of Irreps divide the order of the Group 57Appendix A. Definition and constructions of fields 59A.1. The definition of a field 59A.2. Constructing field extensions by adjoining elements 60Appendix B. A quick intro to field theory 63B.1. Maps of fields 63B.2. Characteristic of a field 64B.3. Basic properties of characteristic 64Appendix C. Algebraic closures 66Appendix D. Existence of algebraic closures 67

NOTES ON REPRESENTATIONS OF FINITE GROUPS 3

1. INTRODUCTION

Loosely speaking, representation theory is the study of groups acting onvector spaces. It is the natural intersection of group theory and linear algebra.In math, representation theory is the building block for subjects like Fourieranalysis, while also the underpinning for abstract areas of number theorylike the Langlands program. It appears crucially in the study of Lie groups,algebraic groups, matrix groups over finite fields, combinatorics, and alge-braic geometry, just to name a few. In addition to great relevance in nearly allfields of mathematics, representation theory has many applications outsideof mathematics. For example, it is used in chemistry to study the states of thehydrogen atom and in quantum mechanics to the simple harmonic oscillator.

To start, I’ll try and describe some examples of representations, and high-light some strange coincidences. Much of our goal through the course willbe to prove that these “coincidences” are actually theorems.

1.1. Acknowledgements. I’d like to thank Noah Snyder for sending mesome homework from a course he taught in representation theory. Many ofhis problems appear in these notes. In turn, some of those problems mayhave been taken from a class Noah took with Richard Taylor.

1.2. A first definition. To start, let’s begin by defining a representation sothat we can hit the ground running. In order to define a representation, we’llneed some preliminary definitions to set up our notation.

Remark 1.1. Throughout these notes, we’ll adapt the following conventions:(1) Our vector spaces will be taken over a field k. You should assume k =

C unless you are familiar with field theory, in particular the notion ofcharacteristic and algebraically closed. (We will make explicit whichassumptions we need on k when they come up. After a certain point,we will work over C.)

(2) All vector spaces will be finite dimensional.

To start, we recall the definition of a group action.

Definition 1.2. For A and B two sets, we let Homsets(A, B) denote the setof maps from A to B and Autsets(A) := Homsets(A, A). For A and B twovector spaces over a field k, we let Homvect(A, B) denote the set of linearmaps from A to B and Autvect(A) := Homvect(A, A).

Remark 1.3. In general, A and B have extra structure, we will often letHom(A, B) denote the set maps preserving that extra structure. In particular,when A and B are vector spaces, we will often use Hom(A, B) in place ofHomvect(A, B) and similarly Aut(A) to denote Autvect(A).

4 AARON LANDESMAN

Definition 1.4. Let G be a group and S be a set. An action of G on S is a mapπ : G → Hom(S, S) so that π(e) = idS and π(g) ◦ π(h) = π(g · h), whereg, h ∈ G, s ∈ S and e ∈ G is the identity.

Joke 1.5. Groups, like men, are judged by their actions.

We can now define a group representation.

Definition 1.6. Let G be a group. A representation of G (also called a G-representation, or just a representation) is a pair (π, V) where V is a vectorspace and π : G → Homvect(V, V) is a group action. I.e., an action on the setV so that for each g ∈ G, π(g) : V → V is a linear map.

Remark 1.7. If you’re having trouble understanding the definition of a repre-sentation, a good way to think about it is an assignment of a matrix to everyelement of the group, in a way compatible with multiplication.

Exercise 1.8 (Easy exercise). If g ∈ G has order n so that gn = 1, andπ : G → Aut(V) is a representation, show that π(g) is a matrix of orderdividing n.

1.3. Examples. Let’s start by giving few examples of representations. Thefirst one is rather trivial.

Example 1.9. Let G be any group. Consider the representation (π, V) whereV is a 1 dimensional vector space and every element of G acts by the matrix(1). Formally, π : G → Aut(V) satisfies π(g) = (1) for every g ∈ G. Saidanother way, G acts as the identity map V → V. This is a representationbecause for any g, h ∈ G,

π(g)π(h) = (1) · (1) = (1) = π(gh).

This is called the trivial representation and we denote it by triv.

Let’s see a couple more examples of representations on a 1-dimensionalvector space.

Remark 1.10. We refer to representations on an n-dimensional vector spaceas n-dimensional representations. Further, for (π, V) a representation, weoften refer to the representation simply as π when or V when clear fromcontext. We call a representation (π, V) a complex representation if V is avector space over the complex numbers. Similarly, we say the representationis a real representation if V is a vector space is over the real numbers.

Example 1.11. Consider the 1-dimensional complex representation (π, V) ofthe group Z/2 = {ι, id} (with ι the nontrivial element ι ∈ Z/2) defined byπ(ι) = ×(−1). In other words, π(ι)(v) = −v.


Exercise 1.12 (Easy exercise). Verify the representation of Example 1.11 isindeed a representation.

We refer to this as the sign representation and denote it by sgn.

Example 1.13. Consider the 1-dimensional representation of Z/3 on thecomplex numbers defined as follows: Let x be a generator for Z/3 so thatZ/3 =

{x, x2, id

}. Define (π, C) by π(x) = (ω) for ω := −1

2 + i√

32 a cube

root of unity. This forces π(x2) = (ω2), and, as always, π(id) = (1). Herewe use parenthesis around a complex number to denote a 1× 1 matrix, whichis the element of Aut(C) given by multiplication by the entry.

There is another representation, which we call given by π(x) = ω2 andπ(x2) = ω.

Exercise 1.14. Generalizing Example 1.13 For n an integer, construct n dif-ferent representations of Z/n on C (i.e., construct action maps π : Z/n→Aut(C)) and prove these are indeed representations. Hint: For each of the npossible roots of unity ζ with ζn = 1, send a generator of Z/n to the lineartransformation C→ C given by x 7→ ζ · x.

The following example is more complicated, but prototypical.

Example 1.15. Let V denote the complex numbers, considered as a 2-dimensionalR-vector space with basis {1, i}. Consider the triangle with vertices atthe three cube roots of unity. Explicitly, vertex 1 is at (1, 0) vertex 2 is at(−1/2,

√3/2) and vertex 3 is at

(−1/2,−

√3/2

). See Figure 1 for a picture.

Then, there is a group action of the symmetric group on three elements, S3,permuting the three vertices of the triangle. In fact, this action can be realizedas the restriction of a representation of S3 on R2 to the set {1, 2, 3}. Indeed,since the vectors corresponding to the vertices 1 and 2 are independent, anylinear transformation is uniquely determined by where it sends 1 and 2. Thisshows the representation is unique. To show existence, we can write downthe representation in terms of matrices. Explicitly, we let the elements of S3act by the following matrices. That is, we consider the map π : S3 → Aut(V)

6 AARON LANDESMAN

FIGURE 1. Picture of a triangle with three labeled vertices,corresponding to the S3 action

given by

id 7→(

1 00 1

)(12) 7→

(−1

2

√3

2√3

212

)

(23) 7→(

1 00 −1

)(13) 7→

(−1

2 −√

32

−√

32

12

)

(123) 7→(−1

2 −√

32√

32 −1

2

)

(132) 7→(−1

2

√3

2−√

32 −1

2

)

To check this is a representation, we need to see that id maps to the identityand π(g) ◦ π(h) = π(g · h). Let’s check this in the case g = (12), h = (23).


Indeed, in this case,

π(12)π(23) =

(−1

2

√3

2√3

212

)(1 00 −1

)

=

(−1

2 −√

32√

32 −1

2

)= π(123).

Visually, this is saying that if one first flips the triangle flips 1 and 2, and thenflips 2 and 3, this composition is the same as rotating 1 to 2.

Exercise 1.16. Verify that this is indeed a representation by checking that forall g, h ∈ S3, π(g)π(h) = π(g · h). (We checked g = (12), h = (23) above.)Hint: It will probably be much easier to check this holds visually in terms ofrotations and reflections than by writing out matrix multiplications.

The moral of this example is that a representation is an assignment of amatrix to each element of a group, in a way that respects composition ofgroup elements.

We call this representation the standard representation of S3 and denoteit by std.

1.4. Characters. One useful piece of information which can be garneredfrom a representation is its character.

Definition 1.17. For M : V → V a matrix, the trace of M, denoted tr(M), isthe sum of the diagonal elements of M.

Definition 1.18. For (π, V) a G-representation, define the character of (π, V)by

χπ : G → kg 7→ tr (π(g)) .

Example 1.19. The character of the standard representation of S3 from Ex-ample 1.15 is given by

id 7→ 2

(12) 7→ 0

(23) 7→ 0

(13) 7→ 0

(123) 7→ −1

(123) 7→ −1

8 AARON LANDESMAN

Warning 1.20. The character is in no way a group homomorphism. This isalready seen in Example 1.19.

Definition 1.21. Two elements g, h ∈ G are conjugate if there is some x ∈ Gwith xgx−1 = h.

Exercise 1.22. Verify that the relation g ∼ h if g is conjugate to h is anequivalence relation.

Definition 1.23. A conjugacy class of a group G is a maximal set of elementsof G all conjugate to each other. That is, it is an equivalence class of G underthe relation of conjugacy.

Lemma 1.24. For (π, V) a G-representation, if g and h are conjugate in G, then

χπ(g) = χπ(h).

Proof. Say h = kgk−1. We wish to verify tr(π(h)) = tr(π(kgk−1)) = tr(π(k)π(g)π(k)−1).So, taking A = π(k) and B = π(g), it suffices to show

tr(B) = tr(ABA−1).

Exercise 1.25. Verify this identity. Hint: Here are two possible approaches.(1) If you’ve seen generalized eigenvalues, use that the trace is the sum ofgeneralized eigenvalues with multiplicity, and note that generalized eigen-values are preserved under conjugation. (2) If you haven’t seen the notion ofgeneralized eigenvalues, reduce to showing tr(CD) = tr(DC) and determinea formula for both of these in terms of the matrix entries of C and D.

�

Remark 1.26. Let Conj(G) denote the set of conjugacy classes of G. Forc ∈ Conj(G) and g a representative for c, and π as G-representation, defineχπ(c) := χπ(g). This is independent of choice of g by Lemma 1.24.

1.5. Character Tables and strange coincidences. As promised, we now es-pouse the amazing coincidences associated with representations of groups.To see some of these coincidences, we look at the character table. Givena group G we can associate to G a character table, where in the rows welist certain representations (the irreducible representations, to be definedprecisely later) and in the columns we list the conjugacy classes of G. In therow associated to (π, V) and column associated to the conjugacy class of g,we put χπ(g) (which is independent of choice of g by Lemma 1.24). Let’s seea few examples.


id ιtriv 1 1sgn 1 −1

TABLE 1. Character table for Z/2

Example 1.27. Let G = Z/2, let ι be the nontrivial element, and let triv andsgn be the representations discussed in Example 1.9 and Example 1.11. Wecan form the character table

Here’s another example:

Example 1.28. Let G = Z/3 with elements x, x2, id, and let π1, π2 be the tworepresentations discussed in Example 1.13 so πi(x) = ωi for ω a cube root ofunity. We can form the character table

id x x2

triv 1 1 1π1 1 ω ω2

π2 1 ω2 ω

TABLE 2. Character table for Z/3

Let’s do one more example:

Example 1.29. Let G = S3 be the symmetric group on three elements. Letstd denote the standard representation of Example 1.15. Let sgn denote the1-dimensional representation where the transpositions (12), (13), (23) actby (−1) and all other elements act by (1) (this is related to Example 1.11by composing the map S3 → Z/2, sending transpositions to the nontrivialelement of Z/2, with the sign representation for Z/2). Then, with some helpfrom Example 1.19, S3 has character table given as follows:

id (12) (123)triv 1 1 1sgn 1 −1 1std 2 0 −1

TABLE 3. Character table for S3

10 AARON LANDESMAN

Note that in the top row we use (12) to denote the conjugacy class of (12).So it really corresponds to the three elements (12), (13), (23). Similarly, (123)denotes the conjugacy class of (123), which is {(123), (132)}.

Let x denote the complex conjugate of x. Recall that the dot product oftwo complex vectors v and w ∈ Cn is by definition 〈v, w〉 := ∑n

i=1 viwi.

Remark 1.30. Let us make some observations regarding the above two char-acter tables:

(1) The number of representations agrees with the number of conjugacyclasses.

(2) The size of the group is equal to the sum of the squares of the dimen-sions of the representations.

(3) The dimensions of all representations divide #G.(4) The dot product of any two distinct columns is equal to 0.(5) The dot product of any column, corresponding to a conjugacy class c,

with itself is equal to #G/#c.(6) Given two rows of a character table, corresponding to represen-

tations π and ρ, define a weighted inner product by 〈χπ, χρ〉 :=∑c∈Conj(G) #cχπ(c)χρ(c). Then, this weighted inner product of anyrow with itself takes value #G and the weighted inner product of anytwo distinct rows is 0.

Exercise 1.31 (Crucial exercise). (1) Verify all claims of Remark 1.30 forthe groups Z/2, Z/3, S3 as in examples Example 1.27, Example 1.28,and Example 1.29.

(2) Write down the character table for the representations of Z/n con-structed in Exercise 1.14. Verify the claims of Remark 1.30 for thoserepresentations.

It turns out these amazing coincidences hold for all finite groups. Much ofthe rest of the course will be devoted to proving this.


2. BASIC PROPERTIES OF REPRESENTATIONS

Having seen a preview of coming attractions yesterday, we’ll begin solidi-fying basic notions today.

As is ubiquitous in mathematics, once one has defined the objects ofa category, one should define the maps. For example, in the category ofgroups, the objects are groups and the maps are group homomorphisms.In the category of k-vector spaces, the objects are k-vector spaces and themaps are linear transformations. We next define the maps in the category ofG-representations.

Definition 2.1. Let (π, V) and (ρ, W) be two G-representations. A map ofrepresentations is a linear map of vector spaces T : V → W so that for allg ∈ G,

T ◦ π(g) = ρ(g) ◦ T.

Said another way, the diagram

(2.1)V V

W W

π(g)

T Tρ(g)

commutes.

Definition 2.2. A map of G-representations T : (π, V) → (ρ, W) is an iso-morphism if there is an inverse map of G-representations S : (ρ, W) →(π, V) so that S ◦ T : (π, V) → (π, V) is the identity on V and T ◦ S :(ρ, W) → (ρ, W) is the identity on W. We say two G-representations areisomorphic if there exists an isomorphism between them.

Exercise 2.3 (Optional Exercise). Show that a map of G-representations T isan isomorphism if and only if T : V →W, when viewed simply as a lineartransformation of vector spaces, is an isomorphism. Hint: Show that theinverse map to T is automatically a map of representations.

Let’s see an example of a map of representations.

Example 2.4. Take G = Z/2 and define the 2-dimensional representation(π, V) by

π(ι) =

(0 11 0

)for ι the nontrivial element of Z/2. In other words, ι swaps the basis vectorsof V. Let e1 and e2 denote the two basis vectors of V. Then, we have a map

T : (triv, k)→ (π, V)

12 AARON LANDESMAN

Defined by T(1) = e1 + e2.Let’s check this is a map of representations. We have to check that for all

g ∈ G, v ∈ V, T triv(g)v = π(g)Tv. Pictorially, we want to check

(2.2)k k

V V

id

T Tπ(g)

commutes.In equations, since triv(g) = (1), we only need to check Tv = π(g)Tv. In

other words, we just need to check that all elements g ∈ G fix the image ofT. However, the image of T is the span of e1 + e2. The identity in Z/2 fixeseverything and π(ι)(e1 + e2) = e2 + e1, so it is indeed fixed.

Exercise 2.5. Define a nonzero map of Z/2 representations from (sgn, C) to(π, V) with sgn the sign representation and (π, V) the representation givenin Example 2.4. Show your map of representations is unique up to scaling.

2.1. Irreducible representations. We next introduce irreducible represen-tations, which are essentially the building blocks of representation theory.

Definition 2.6. For (π, V) a representation, a subrepresentation is a sub-space W ⊂ V which is stable under the action of all g ∈ G (i.e., π(g)(W) ⊂Wfor all g ∈ G). Such a W has the structure of a representation given by(π|W , W) where π|W(g) := π(g)|W . I.e., viewing π(g) as a map V → V,π(g)|W is the induced map W → W, which has the correct target becauseπ(g)(W) ⊂W.

Exercise 2.7. Show that for (π, V) a representation and g ∈ G, if W ⊂ Vis subrepresentation then the inclusion π(g)(W) ⊂ W is automatically anequality. (Recall: All representations are finite dimensional.)

Example 2.8. For any representation (π, V), both 0 and V define subrepre-sentations.

Definition 2.9. A representation (π, V) is irreducible if its only subrepresen-tations are the 0 subspace and all of V. If a representation is not irreducible,we say it is reducible. For brevity, we sometimes refer to an irreduciblerepresentation as an irrep.

Exercise 2.10 (Easy exercise). Show that all 1-dimensional representationsare irreducible.

Example 2.11. The standard representation of S3 from Example 1.15 is irre-ducible. To see this, if it were reducible, it would have some 1 dimensional


subrepresentation, spanned by a nonzero vector v. But then, π((123))(v) isa vector not in the span of v, because it is related to v by rotation by 120◦

which is not in the span of v. Therefore,(std, R2) cannot be reducible, so it

is irreducible.

Exercise 2.12. Let Sn denote the symmetric group on n elements. Definethe permutation representation of Sn as the representation (π, V) where Vhas basis e1, . . . , en and σ ∈ Sn acts by π(σ)(ei) = eσ(i). This is called thepermutation representation, which we denote perm. Show that V has twosubrepresentations given by the subspaces W1, W2 where

W1 :=

{n

∑i=1

aiei ∈ V : ∑i

ai = 0

}and

W2 :=

{n

∑i=1

aiei ∈ V : a1 = a2 = · · · = an

}.

For the remainder of this exercise, assume the characteristic of k does notdivide n. If you have not seen the notion of characteristic, feel free to assumek = C.

(1) Determine the dimensions of W1 and W2.(2) Show that W1 and W2 are independent and span V.(3) In the case n = 3, show W1 is isomorphic to std the standard repre-

sentation of Example 1.15 (in fact, this holds for n > 3, but we haven’tyet defined the standard representation for higher n).

(4) Determine which 1-dimensional representation (π|W2 , W2) is isomor-phic to. Hint: It is one we have already seen, and we haven’t seenvery many!

Definition 2.13. We call the representation W1 from Exercise 2.12 the stan-dard representation of Sn, and denote it by std.

Note that Definition 2.13 is not a conflict of notation with our definition ofthe standard representation for S3 by Exercise 2.12(3).

If you are assuming k = C skip the following exercise.

Exercise 2.14 (Optional tricky exercise for those familiar with the character-istic of a field). In the setup of Exercise 2.12, assume n = 3 and char(k) = 3.What is the dimension of W2? Are W1 and W2 independent? Is W1 irre-ducible?

Exercise 2.15 (Important exercise). Let T : (π, V) → (ρ, W) be a map ofrepresentations. Show that ker T ⊂ V is a subrepresentation of V andim T ⊂W is a subrepresentation of W.

14 AARON LANDESMAN

Lemma 2.16. A representation is irreducible if and only if for all nonzero v ∈ V,the set {g · v}g∈G spans V.

Proof. Observe that for any v ∈ V, Span {π(g)v}g∈G is a G-stable subspacebecause for any fixed h ∈ G,

{π(g)v}g∈G = {π(h)π(g)v}g∈G .

Therefore, if the representation is irreducible, Span {π(g)v}g∈G must eitherbe all of V or 0. In other words, if v 6= 0, Span {π(g)v}g∈G = V.

Conversely, if the representation is reducible, let W ⊂ V be a subrepre-sentation and take some nonzero w ∈W. Then {π(g)w}g∈G ⊂W does notspan V, as desired. �

Exercise 2.17. In the setup of Exercise 2.12 (still assuming char(k) - n), showthat V is reducible and W1, W2 are irreducible subrepresentations. Show thatthese are the only nonzero proper subrepresentations. Hint: If W is a thirdsubrepresentation, pick w = ∑n

i=1 aiei ∈ W nonzero and look at the spanof {gw}g∈G. Depending on whether a1 = · · · = an and whether ∑i ai = 0,show that this span is either W1, W2, or all of V.

2.2. Direct sums. We now recall a construction from linear algebra.

Definition 2.18. For V and W two vector spaces, the direct sum, denotedV ⊕W, is defined by

V ⊕W := {(v, w) : v ∈ V, w ∈W} .

Addition is given by (v1, w1) + (v2, w2) = (v1 + v2, w1 + w2) and scalarmultiplication is given by c (v, w) = (cv, cw). This makes V⊕W into a vectorspace. More generally, one can define ⊕n

i=1Vi as the set of tuples (v1, . . . , vn)with vi ∈ Vi, and give it the structure of a vector space analogously.

Remark 2.19. Sometimes, the object defined above as V ⊕W is also referredto as V ×W.

Example 2.20. The vector space R2 = R⊕R. Also, R12 = R7 ⊕R3 ⊕R2.Further, Rn = ⊕n

i=1R.

Just as we often construct new vector spaces by taking direct sums of vectorspaces we have already constructed, we can also construct new representa-tions by taking direct sums of representations we have already constructed.We now formalize this notion.


Definition 2.21. If (π, V) and (ρ, W) are two G-representations, the directsum is the representation (π ⊕ ρ, V ⊕W) defined by

(π ⊕ ρ)(g) : V ⊕W → V ⊕W(v, w) 7→ (π(g)v, ρ(g)w) .

Exercise 2.22. Check that the direct sum of two representations (π, V) and(ρ, W) is indeed a representation. Check that (π, V) and (ρ, W) are bothsubrepresentations of (π ⊕ ρ, V ⊕W). In particular, deduce that the directsum of two nonzero representations is never irreducible. Hint: Choose basesfor V and W, and write out the matrix π(g) as a block matrix with blockscorresponding to the maps V → V and W →W.

Exercise 2.23. Show that the representation (π, V) of Z/2 from Example 2.4can be written as

(π, V) ' (triv, C)⊕ (sgn, C) .

Exercise 2.24. For (π, V) a representation, recall χπ(g) = tr(π(g)). Showthat for (π, V) , (ρ, W) two representations, we have χπ⊕ρ(g) = χπ(g) +χρ(g).

16 AARON LANDESMAN

3. DESIDERATA AND PROBLEMS

In this section, we’ll state the main results we’ll prove later in the course,and use them to compute lots of examples and fun problems. To start,we recall the main things we want to prove about representations of finitegroups, which were essentially already stated in Remark 1.30. In order totemporarily state our desiderata, we need to define certain inner products onthe rows and columns of a character table. We recommend you jump to theexercises in subsection 3.2 and come back to read subsection 3.1 as needed.

3.1. Desiderata. Given a group G and its associated character table (wherewe think of the character table as a matrix with rows corresponding toirreducible representations π, columns corresponding to conjugacy classes cand entries given by tr π(c)) we define the following inner products: Giventwo rows of the character table v and w (thought of as vectors with entriesvc, c ∈ Conj(G),) define

〈v, w〉row := ∑c∈Conj(G)

#c · vcwc.

Let Irrep(G) denote the set of isomorphism classes of complex irreduciblerepresentations of G. Given two columns of the character table x and y(thought of as vectors with entries xπ, π ∈ Irrep(G)) define

〈x, y〉column := ∑π∈Irrep(G)

xπyπ.

(These notations are temporary for this section only, we will re-introduce therelevant inner products later in a broader context.)

Desiderata 3.1. Let G be a finite group. Then, the following hold:Maschke’s theorem Every complex representation is a direct sum of irreducible representations.

Isotypic decomposition The direct sum decomposition of the previous part is unique in the sensethat any two such decompositions have the same multiset of irreduciblecomponents. That is, each irreducible representation which appears in onedecomposition appears in the other with the same multiplicity.

Sum of squares # Irrep(G) is finite, and in fact #G = ∑π∈Irrep(G)(dim π)2.Conjugacy and irreps # Conj(G) = # Irrep(G)Dimensions of irreps For π ∈ Irrep(G), dim π | #G.

Orthogonality of rows For G a group, the rows of the character table are orthogonal in the sense thatfor any two rows v and w, 〈v, w〉row = 0 if v 6= w and 〈v, w〉row = #G ifv = w.

Orthogonality of columns The columns of of the character table are orthogonal in the sense that for anytwo columns x and y, 〈x, y〉column = 0 if x 6= y and 〈x, y〉column = #G

#c forx = y


Multiplicity If (π, V) is a representation with some irreducible representation (W, ρ)appearing with multiplicity n inside π (that is, π ' ρ⊕n ⊕ ρ′ where ρ′ hasno subrepresentations isomorphic to ρ) then ∑c∈Conj(G) #cχπ(c)χρ(c) =#G · n. Recall that, for any g ∈ c, we have defined χπ(c) := tr π(g).

The items above will be proven in Theorem 4.5, Theorem 6.3, Theorem 10.1,Corollary 11.6, Theorem 12.1, Theorem 9.2, and Corollary 10.8 below.

3.2. Applications. We now get to apply our desiderata with great effect. Inwhat follows in this subsection, we will assume Desiderata 3.1 throughout(and we may avoid stating explicitly when we are using it). We also assumeall representations are over the complex numbers so that we can applyDesiderata 3.1 freely.

Exercise 3.2. (1) If you haven’t yet finished Exercise 1.31, (which askedyou to compute the character tables of S3 and Z/n and verify theproperties listed in Desiderata 3.1) now is your chance! Do it!

(2) Show that all the representations you write down are irreducible.(3) Show that every irreducible of the relevant group appears in your list

using Desiderata 3.1[Sum of squares].

Exercise 3.3. Using orthogonality of characters, show that two irreduciblecomplex representations of a finite group G have equal characters if and onlyif they are isomorphic.

3.3. Dihedral Groups. In this section, we work out the character tables ofdihedral groups.

Definition 3.4. The dihedral group of order 2n, denoted D2n is the finitenonabelian group generated by r and s and satisfying the relations rn = s2 =1, sr = r−1s.

Remark 3.5. The dihedral group D2n can be thought of as the symmetries ofthe n-gon with r acting by rotation by 2π/n and s acting as a reflection. Inparticular, D6 ' S3.

Exercise 3.6. (1) If n is odd, construct two distinct 1-dimensional repre-sentations of D2n.

(2) If n is even, construct four distinct 1-dimensional representations ofD2n. Hint: Allow r and s to act by ±1.

Exercise 3.7. (1) If n is odd, construct n−12 distinct 2-dimensional irre-

ducible representations of D2n. Hint: Consider the action on then-gon where r acts by rotation by j · 2π/n for 1 ≤ j < n/2. Computecharacters to show they are distinct.

18 AARON LANDESMAN

(2) If n is even, construct n−22 distinct 2-dimensional irreducible represen-

tations of D2n.

Exercise 3.8. Show that the representations constructed in Exercise 3.6 andExercise 3.7 are all irreducible representations of D2n. Hint: Use Desider-ata 3.1[Sum of squares]

Exercise 3.9. Determine the character table for the group D2n. Check its rowsand columns are orthogonal in the sense of Desiderata 3.1.

3.4. The Quaternion group. In this subsection, we work out the charactertable of the Quaternion group.

Definition 3.10. The Quaternion group, Q is a group of size 8 with elements±1,±i,±j,±k with the relations i2 = j2 = k2 = −1 and ij = k, ji = −k, ki =j, ik = −j, jk = i, kj = −i.

Exercise 3.11. Construct an irreducible 2-dimension complex representationof the Quaternion group and prove it is irreducible. Hint: Consider the2-dimensional vector space V over C with basis vectors 1, j, which is a 4-dimensional R-vector space V with basis 1, i, j, i · j. Identifying k = ij ∈ Q,define an action of Q on {±1,±i,±j,±i · j} ∈ V, and extend this to an actionon V.

Exercise 3.12. Construct 4 1-dimensional representations of Q and provethey are pairwise non-isomorphic. Hint: To construct them, allow i and j toact by either 1 or −1. To show they are distinct, use Exercise 3.3.

Exercise 3.13. Compute a complete character table for Q. It turns out thereare only two order 8 nonabelian groups up to isomorphism: Q and D8. Canyou distinguish these two groups using their character tables?

3.5. Representations of A4. In this section, we determine the character tablefor the alternating group A4.

Exercise 3.14. (1) Show that the Klein-4 subgroup consisting of permuta-tions e, (12)(34), (13)(24), (14)(23) ∈ A4 (isomorphic to Z/2×Z/2)is a normal subgroup of A4.

(2) Show that the quotient A4/K4 ' Z/3.(3) Use the projection A4 → Z/3 to define 3 irreducible 1-dimensional

representations of A4, coming from the three irreducible representa-tions of Z/3.

Exercise 3.15. Identify the rotations of the tetrahedron in R3 with the alter-nating group A4. Hint: Here, A4 acts as a subgroup of the permutations ofthe 4-points on the tetrahedron.


Exercise 3.16. Using Exercise 3.15, construct an irreducible 3-dimensionalrepresentation of A4.

Exercise 3.17. Show that the 3 1-dimensional representations of A4 con-structed in Exercise 3.14, together with the 3-dimensional representationconstructed in Exercise 3.16 are all irreducible representations of A4. Com-pute the character table of A4. Hint: After determining these are all therepresentations, to avoid computing angles of rotation of the tetrahedron,you can use the orthogonality properties of the character table to computethe character for the 3-dimensional representation.

3.6. Representations of S4. In this section, we determine the character forS4.

Exercise 3.18. (1) Show that the Klein-4 subgroup consisting of permuta-tions e, (12)(34), (13)(24), (14)(23) ∈ S4 (isomorphic to Z/2×Z/2)is a normal subgroup of S4.

(2) Show that the quotient S4/K4 ' S3.(3) Use the projection S4 → S4/K4 ' S3 construct irreducible representa-

tions of S4 corresponding to the three irreducible representations ofS3 (triv, sgn, and std).

Exercise 3.19. Show that S4 can be realized as the automorphisms of thecube. Hint: Stare at the four long diagonals of the cube.

Exercise 3.20. Use Exercise 3.19 to construct an irreducible 3-dimensionalrepresentation of S4.

Exercise 3.21. (1) Show that altogether, S4 has 5 irreducible representa-tions: 3 were constructed in Exercise 3.18, one was constructed inExercise 3.20, and there is one more, call this last one we have not yetconstructed (π, V).

(2) Determine the dimension of V from the previous part.(3) Even though we have not constructed the representation V, compute

the character table of S4.(4) Using the character table as a guide, can you construct the representa-

tion V explicitly. Hint: Can you figure out the relation between V, thesgn representation (coming from sgn on S3 via Exercise 3.18), and the3-dimensional representation of Exercise 3.20? This is a first exampleof the “tensor product“ of representations, which we will see more oflater.

3.7. Representations of A5. In this section, we compute character table ofA5.

20 AARON LANDESMAN

Exercise 3.22. Compute the conjugacy classes of A5. Hint: There are 5 conju-gacy classes.

Exercise 3.23. Using the inclusion A5 → S5 construct a 4-dimensional irre-ducible representation of A5 (which we again call std for A5) by composing

A5 → S5std−→ Aut(V), where the latter map is the standard representation

(std, V) (as defined in Definition 2.13) of S5.

Exercise 3.24. Identify A5 as the group of rotations of the dodecahedron. Usethis to construct an irreducible 3-dimensional representation of A5. Computethe character of this representation. Hint: To compute the character, findautomorphisms of the dodecahedron of orders 2, 3, and 5. The conjugacyclasses are almost determined by their orders, with the exception that thereare 2 conjugacy classes of order 5. Note that all but one of the conjugacyclasses act by rotation. If they rotate an angle of θ about the third basis vector(which we choose to be the angle of rotation), the action of these elementscan be expressed as cos θ − sin θ 0

sin θ cos θ 00 0 1

.

It may be helpful to know cos 2π5 = 1

4

(−1 +

√5)

.

Exercise 3.25. (1) Construct an automorphism S5 → S5 by sending σ 7→(12)σ(12). Show that this restricts to an automorphism φ : A5 → A5.

(2) Let (π, V) denote the 3-dimensional representation of A5 constructedin Exercise 3.24. Compose the automorphism φ : A5 → A5 con-structed in the previous part with the representation π : A5 →Aut(V) to obtain another 3-dimensional representation (ρ, W) of A5.Show (ρ, W) is irreducible, but is not isomorphic to (π, V). Hint: Toshow it is not isomorphic, show that it flips the actions of the twoconjugacy classes consisting of order 5 elements.

(3) Compute the character of (ρ, W).

Exercise 3.26. Show that A5 has 5 irreducible representations: triv, the 4-dimensional std from Exercise 3.23, the two 3-dimensional representationsfrom Exercise 3.24 and Exercise 3.25, and one more. Compute the dimensionof this last representation.

Exercise 3.27. Compute the character table of A5.

3.8. Groups of order p3. In this subsection, we guide the reader in deter-mining the number of conjugacy classes of any group of order p3 for p aprime. This is carried out in Exercise 3.44


Definition 3.28. Let G be a group. Define the center of G, denoted Z(G) by

Z(G) := {g ∈ G : gh = hg for all g ∈ G} .

Remark 3.29. The center is the set of elements that commute with all otherelements, so G is abelian if and only if G = Z(G).

Exercise 3.30. Show Z(G) ⊂ G is a normal subgroup.

Exercise 3.31. Suppose p is a prime and G is a p-group (meaning #G = pn

for some n). Suppose G acts on a finite set S. Let FixG(S) ⊂ S denote theset of s ∈ S with g · s = s for all g ∈ G. I.e., FixG(S) is the set of fixedpoints under the G-action. Show # FixG(S) ≡ #S mod p. Hint: Use the orbitstabilizer theorem.

Exercise 3.32. Let p be a prime. If G is a p-group show Z(G) is nontrivial.Hint: Consider the conjugation action of G on itself and use Exercise 3.31.

Exercise 3.33. Suppose that G is a group with G/Z(G) ' Z/nZ. Show thatin fact G = Z(G) (so n = 1).

Exercise 3.34. For G a nonabelian group of order p3 for p prime, show#Z(G) = p. Hint: Rule out the possibilities that #Z(G) = 1, p2, p3.

Definition 3.35. For G a group, define the commutator subgroup, denoted[G, G] to be the subgroup of G generated by all elements of the form xyx−1y−1

for x, y ∈ G. That is, [G, G] consists of all products of elements of the formxyx−1y−1.

Exercise 3.36. For G a group, show that [G, G] ⊂ G is a normal subgroup.

Exercise 3.37. Show there are no nonabelian groups of order p2. Hint: UseExercise 3.32 and Exercise 3.33.

Exercise 3.38. Let G be a non-abelian group of order p3 for some prime p.Show [G, G] = Z(G) and both have size p in the following steps

(1) Let q : G → G/[G, G] denote the quotient map. Show that if we havea map f : G → H for H an abelian group then there is a unique mapφ : G/ [G, G]→ H so that φ ◦ q = f .

(2) Show that [G, G] /Z(G) is abelian using Exercise 3.37. Conclude fromthe previous part that if f : G → G/Z is the quotient map, there is aunique map φ : G/[G, G]→ G/Z(G) so that φ ◦ q = f .

(3) Show that for H and K groups, and N ⊂ H a normal subgroup, theset of maps H/N → K are in bijection with maps H → K sending Nto the identity element of K.

(4) Conclude from the previous two parts that [G, G] ⊂ Z(G).(5) Show that [G, G] 6= 1

22 AARON LANDESMAN

(6) Conclude [G, G] = Z(G) and both have size p. Hint: Use Exer-cise 3.34.

Exercise 3.39. Suppose G is a non-abelian group of order p3 for some primep. Show that G/ [G, G] = Z/p×Z/p. Hint: Show G/ [G, G] is abelian oforder p2, but not cyclic. For this, use Exercise 3.37 and Exercise 3.38.

Exercise 3.40. For G any finite group, show that if g ∈ [G, G], then forany 1-dimensional (π, k) of G, π(g) = id. Use this to show there is abijection between the 1-dimensional representations of G and 1-dimensionalrepresentations of G/ [G, G].

Exercise 3.41. (1) Construct p2 distinct irreducible 1-dimensional repre-sentations of Z/p×Z/p and show these are all of them. Hint: Sendgenerators to multiplication by roots of unity, as in Exercise 1.31.

(2) If you are so inclined, generalize this to arbitrary finite abelian groups(which are necessarily a product of cyclic groups by the fundamentaltheorem for finitely generated abelian groups).

Exercise 3.42. Suppose G is a non-abelian group of order p3 for some primep. Find all 1-dimensional representations of G. Hint: Combine Exercise 3.39,Exercise 3.40, and Exercise 3.41.

Exercise 3.43. Let G be a nonabelian group of order p3 for p prime. Showthat G has p2 irreps of dimension 1 and p− 1 irreps of dimension p. Hint:Use that the dimension of any irrep must divide the order of the group,together with the previous exercises.

Exercise 3.44. Let p be a prime and G a nonabelian group of order p3. Showthat G has p2 + p− 1 conjugacy classes. Hint: Use Exercise 3.43.

3.9. Further Challenge exercises.

Exercise 3.45. Consider the regular representation of S3, Reg(S3), definedas the 6-dimensional representation spanned by basis vectors

{eg}

g∈S3with

action given by σ · eg = eσ·g. Calculate the character of the regular representa-tion and use the multiplicity criterion from Desiderata 3.1 to find the numberof copies of the three irreducible representations of S3 (triv, sgn, std) insideReg(S3). (Note, this calculation is done for an arbitrary group in the proofof the sum of squares formula.) Then, describe an explicit decomposition ofReg(S3) as a direct sum of irreducible subrepresentations.

Exercise 3.46 (Tricky exercise). In this exercise, we explore the possible num-ber of even and odd dimensional irreducible representations of a group G. Apriori, there are four possibilities either G has


(1) an even number of odd dimensional irreducible representations andan even number of even dimensional irreducible representations

(2) an odd number of odd dimensional irreducible representations andan even number of even dimensional irreducible representations

(3) an even number of odd dimensional irreducible representations andan odd number of even dimensional irreducible representations

(4) an odd number of odd dimensional irreducible representations andan odd number of even dimensional irreducible representations

For which of (1), (2), (3), and (4) does there exist a finite group G with sucha number of odd and even dimensional irreducible representations? Eithergiven an example of such a G or prove no such G exist.

Here are two problems, which can be solved both by elementary methodsand using representation theory.

Exercise 3.47. Let Sn act on [n] := {1, . . . , n} by permuting the elements. Forg ∈ Sn, let Fix(g) denote the set of x ∈ [n] so that g(x) = x. That is, Fix(g) isthe set of x fixed by g.

(1) Show using elementary methods (without representation theory) that∑g∈Sn # Fix(g) = n!.

(2) Show using representation theory ∑g∈Sn # Fix(g) = n!. Hint: Considerthe permutation representation of Sn as defined in Exercise 2.12, andrecall it decomposes as a direct sum of triv and std, two irreduciblerepresentations, as shown in Exercise 2.12 and Exercise 2.17. Let χpermbe the associated character and relate ∑g∈Sn # Fix(g) to 〈χperm, χtriv〉.

Exercise 3.48. Let Sn act on [n] := {1, . . . , n} by permuting the elements.Show that ∑g∈Sn(# Fix(g))2 = 2n!. See if you can do it both using represen-tation theory, and via elementary means.

Exercise 3.49. Show that G is abelian if and only if all irreducible representa-tions are 1-dimensional.

Exercise 3.50. Fix a group G. Show that G is abelian if and only if it ispossible to reorder the irreps and conjugacy classes of G so that the charactertable of G is symmetric about the diagonal line from the upper left to lowerright.

Exercise 3.51. Let D∞ be the “infinite dihedral group“ generated by r and swith the relations that s2 = 1 and sr = r−1s. (If we imposed rn = 1, we wouldget the usual dihedral group D2n.) Find a 2-dimensional representation of D∞which is not irreducible, but cannot be written as a direct sum of irreduciblerepresentations. Note this does not contradict Maschke’s theorem becauseD∞ is infinite.

24 AARON LANDESMAN

4. COMPLETE REDUCIBILITY OF COMPLEX REPRESENTATIONS

We begin the saga of proving Desiderata 3.1. To start, we next introducea notion of complete reducibility. Just as every vector space breaks up as adirect sum of copies of the base field (by choosing a basis), in many nice cases,every representation breaks up as a direct sum of irreducible representations.

We next codify the notion of complete reducibility:

Definition 4.1. A G-representation (π, V) is completely reducible if V canbe written as a direct sum of irreducible G-representations. That is, one canfind a collection of irreducible subrepresentations Vi ⊂ V which have trivialpairwise intersection and span V.

We’d like to give a criterion for when every G-representation is completelyreducible, so it can always be broken down into irreducible pieces. This isdone in Theorem 4.5. In particular, we will see this is always true over C ormore generally any field of characteristic 0.

To set this up, we define quotient representations. Recall that for W ⊂ V asubspace, the quotient V/W is the quotient of V by the equivalence relationv1 ∼ v2 if v1 − v2 ∈ V. The following exercise is crucial, especially if youhaven’t seen quotients:

Exercise 4.2 (Important Exercise). Let W ⊂ V be a subspace.

(1) Verify that the relation v1 ∼ v2 if v1 − v2 ∈ W is an equivalencerelation.

(2) Give the set V/W the structure of a vector space by declaring c[v] :=[cv] and [v1] + [v2] := [v1 + v2]. Check that this is well defined (i.e.,independent of the choice of representative for v ∈ V.

(3) Check that dim V/W = dim V−dim W. Hint: Choose a basis e1, . . . , emfor W and extend it to a basis e1, . . . , en for V. Check the images[em+1], . . . , [en] in V/W form a basis for V/W.

(4) Let

q : V → V/Wv 7→ [v]

denote the projection map. Verify that there is a map of vector spacess : V/W → V so that q ◦ s = idV/W . Hint: Choose bases as in theprevious part and define s by sending [em+j] ∈ V/W to em+j ∈ V.

(5) Let i : W → V denote the given inclusion. Show that for any suchmap s as in the previous part, the map (i, s) : W ⊕ V/W → V is an


isomorphism. Here, (i, s) denotes the map

(i, s) : W ⊕V/W → V

(w, x) 7→ i(w) + s(x).

Definition 4.3. For (π, V) a G-representation and W ⊂ V a subrepresenta-tion, define the quotient representation (π, V/W) to be the representationwith vector space V/W and action given by π(g)[v] := [π(g)(v)] wherefor x ∈ V, [x] denotes the image under the map V → V/W taking theequivalence class in the quotient.

Exercise 4.4. (1) Check definition Definition 4.3 is well defined. That is,check that if v1, v2 are two representatives for [v] then [π(g)(v1)] =[π(g)(v2)].

(2) Verify that V/W is indeed a G-representation.(3) Show q : V → V/W is a map of representations.

If you are assuming k = C, then ignore the assumption on char(k) every-where that follows, as it is automatically satisfied.

Theorem 4.5 (Maschke’s Theorem). Suppose char(k) - #G. Then every G-representation is completely reducible.

Before continuing with the proof, we include a couple illustrative exercises.

Exercise 4.6. Let (perm, V) be the permutation representation for S3, as dis-cussed in Exercise 2.12. Recall this was the 3-dimensional representationwith basis e1, e2, e3 so that perm(σ)(ei) = eσ(i). Let (triv, W) ⊂ (perm, V)denote the 1-dimensional subrepresentation spanned by e1 + e2 + e3. Find aT ∈ Hom(perm, perm) with T2 = T and im T = W. Compute the subrepre-sentation ker T. Hint: You might guess what T and ker T have to be via yoursolution to Exercise 2.12.

Exercise 4.7. (1) Let (π, V) denote the 2-dimensional Z/2 representa-tion defined in Example 2.4 where the nontrivial element ι acts byπ(ι)(e0) = e1, π(ι)(e1) = e0. Write (π, V) as a direct sum of irre-ducible representations. Hint: We already did this in Exercise 2.23.

(2) Similarly, for this part, let (π, V) denote the 3-dimensional repre-sentation of Z/3 = {0, 1, 2} with basis e1, e2, e3 satisfying π(j)ei =ei+j mod 3. Write (π, V) as a direct sum of irreducible Z/3 representa-tions.

(3) Can you generalize the previous part replacing Z/3 with Z/n?

Proof of Theorem 4.5. Let (π, V) be a G-representation. Let i : (π|W , W) ⊂(π, V) be a nonzero proper subrepresentation and (π, V/W) be the quotient

26 AARON LANDESMAN

representation defined in Definition 4.3. Let q : (π, V)→ (π, V/W) denotethe quotient map.

We claim (π, V) ' (π|W , W)⊕ (π, V/W). Indeed, once we show this, byinduction on the dimension of V, we will be done, because π|W and π aresmaller dimensional representations. Therefore, by induction on the dimen-sion, we can write (π|W , W) =

⊕ni=1(πi, Wi) and (π, V/W) =

⊕mj=1(π j, V′j )

for irreducible πi and π j. Then,

(π, V) ' (π|W , W)⊕ (π, V/W) '(

n⊕i=1

(πi, Wi)

)⊕ m⊕j=1

(π j, V′j )

.

So, to conclude, it suffices to show (π, V) ' (π|W , W)⊕ (π, V/W). Toprove this, we claim we claim there is a map of representations ι : (π, V/W)→(π, V) so that q ◦ ι = idV/W . This is shown next in Proposition 4.9. Let’s seewhy this concludes the proof, and then we’ll come back to prove Proposi-tion 4.9.

To conclude the proof, by Exercise 4.2(5) the resulting map (i, ι) : W ⊕V/W → V is an isomorphism of vector spaces. Further, since ι and i areboth maps of representations, it is also a map of representations.

Exercise 4.8 (Easy exercise). Check for yourself that (i, ι) is indeed a map ofrepresentations.

Hence (i, ι) is an isomorphism of representations, which is what we neededto check. �

So, to conclude the proof of Theorem 4.5, we only need to prove thefollowing proposition.

Proposition 4.9. Let char(k) - #G. Let (π, V) be a representation, W ⊂ V asubrepresentation, V/W the corresponding quotient representation with quotientmap q : V → V/W. Then there is a map of representations ι : V/W → V so thatq ◦ ι = idV/W .

Proof. First, we know there exists a map f : V/W → V of vector spaces sothat q ◦ f = idV/W , as shown in Exercise 4.2(4). However, this may not bea map of representations (as there is no reason that f ◦ π(g) should equalπ(g) ◦ f ). We rectify this by an averaging trick: Define

ι : V/W → V

x 7→ 1#G ∑

g∈Gπ(g) f (π(g−1)x).


We claim ι is a map of representations and satisfies q ◦ ι = idV/W . Oncewe check these two facts, we will be done. First, let’s check ι is a map ofrepresentations. Pictorially, this means we have to check

(4.1)V/W V/W

V V

π(h)

ι ι

π(h)

commutes.Indeed, for x ∈ V/W,

ι(π(h)x) =1

#G ∑g∈G

π(g) f (π(g−1)π(h)x)

=1

#G ∑g∈G

π(hh−1g) f (π(g−1h)x)

=1

#G ∑g∈G

π(h)π(h−1g) f(

π

((h−1g

)−1)

x)

= π(h)1

#G ∑h−1g∈G

π(h−1g) f(

π

((h−1g

)−1)

x)

= π(h)ι(x)

as desired.

28 AARON LANDESMAN

To conclude, we check q ◦ ι = idV/W . Indeed, for x ∈ V/W,

q ◦ ι(x) = q

(1

#G ∑g∈G

π(g) f (π(g−1)x)

)

=1

#G ∑g∈G

q(

π(g) f (π(g−1)x))

=1

#G ∑g∈G

π(g)q(

f (π(g−1)x))

=1

#G ∑g∈G

π(g)π(g−1)x

=1

#G ∑g∈G

π(gg−1)x

=1

#G ∑g∈G

π(e)x

=1

#G ∑g∈G

x

= x.

�

This concludes the proof of Maschke’s theorem.We note that the assumption that char(k) - #G is crucial to Theorem 4.5.

(Again, ignore this if you are assuming k = C.) We give an exercise exhibitingof a representation that is not completely reducible when char(k) | #G. Thisis essentially spelling out the solution to Exercise 2.14. Again, skip this if youare assuming k = C.

Exercise 4.10 (Optional exercise for those familiar with characteristic). Let kbe a field of characteristic 2. Consider the action of Z/2 on the two dimen-sional vector space V with basis e1, e2 given by

π(ι)(e1) := e1,

π(ι)(e2) := e1 + e2.

(1) Verify that (π, V) is a representation. Hint: You will need to usechar(k) = 2.

(2) Check that Span(e1) is a subrepresentation, so V is reducible.(3) Check that Span(e1) is the only subrepresentation, so V is not com-

pletely reducible. Hint: Show that for any v ∈ V − Span(e1), v andπ(ι)(v) are independent and use Lemma 2.16.


Remark 4.11. Theorem 4.5 will be psychologically useful for us, as it lets usthink of any representation as a sum of irreducible constituents. As we shallsee next, these irreducible constituents (and their multiplicities) are in factuniquely determined.

30 AARON LANDESMAN

5. SCHUR’S LEMMA

Our next main result is proving isotypic decomposition. In order to showthis, we will need Schur’s lemma, which we prove in this section. Thisis one of the most important results toward setting up the basic theory ofrepresentations of finite groups. Recall that a field k is algebraically closed ifevery monic polynomial with coefficients in k has a root. (Again, feel free toassume k = C if it is helpful.) Here is the statement for Schur’s lemma.

Theorem 5.1 (Schur’s lemma). If (π, V) and (ρ, W) are two G-representationsover an algebraically closed field,

dim Hom(π, ρ) =

{1 if π ' ρ

0 if π 6' ρ.

To prove Schur’s lemma, we will need two lemmas.

Lemma 5.2. Let T : (π, V)→ (ρ, W) be a nonzero map of representations.(1) If (π, V) is irreducible, T is injective.(2) If (ρ, W) is irreducible, T is surjective.(3) If (π, V) and (ρ, W) are both irreducible, T is an isomorphism.

Proof. We prove the parts in order.(1) Consider ker T ⊂ V. Since V is irreducible, and ker T is a subrepre-

sentation by Exercise 2.15, we either have ker T = V or ker T = 0. Theformer cannot occur because T 6= 0. So, ker V = 0 and T is injective.

(2) Consider im T ⊂ W. Since W is irreducible, and im T is a subrepre-sentation by Exercise 2.15, either im T ⊂W = 0 or im T = W. But wecannot have im T = 0 because T 6= 0, so im T = W and T is surjective.

(3) This follows from the first two parts.�

Lemma 5.3. Suppose k is algebraically closed and V is a k vector space with (π, V)a finite dimensional irreducible G-representation. Then, any map of representationsT : (π, V)→ (π, V) is necessarily multiplication by a scalar. That is, T = c · idfor c ∈ k.Proof. Let T be some such map. Note that T has an eigenvalue since eigenval-ues are the same as roots of the characteristic polynomial, and by assumptionk is algebraically closed so the characteristic polynomial has a root in k. (Ifyou are pretending k = C, this is just saying that every polynomial over C

has a root, which is the fundamental theorem of algebra.)Let λ denote the resulting eigenvalue and x the corresponding eigenvector,

so Tx = λx. Let

W := {w ∈ V : Tw = λw} .


Observe that W is nonzero since 0 6= x ∈W (and by definition of eigenvector,x 6= 0). To conclude, it suffices to show W = V, because that will implyTv = λv for all v ∈ V so T = λ · id.

Because V is irreducible, in order to show W = V, it suffices to show Wis a subrepresentation. To check this, we just need to know that for g ∈ G,T(W) ⊂ W. But indeed, for any w ∈ W, Tw = λw ∈ Span(w) ⊂ W, asdesired. �

Using the above two results, we can easily prove Schur’s lemma.

Proof of Theorem 5.1. From Lemma 5.2, we know that if ρ 6' π then the onlymap between ρ and π is 0. So, it suffices to show that if π ' ρ then,dim Hom(π, ρ) = 1. This is precisely the content of Lemma 5.3 �

Exercise 5.4 (Irreducible representations of abelian groups). Let G be anabelian group and (π, V) an irreducible representation over an algebraicallyclosed field k. For this exercise, you are not allowed to use Desiderata 3.1,and

(1) Show that π(g) : (π, V) → (π, V) defines a map of representations.Hint: You need to show π(g) commutes with π(h) for any h ∈ G.

(2) Show that for every g ∈ G, there is some nonzero c ∈ k such thatπ(g) = c · id Hint: Schur’s lemma!

(3) Conclude that V is 1-dimensional.

If you are working over C, skip the following exercise.

Exercise 5.5 (Optional exercise for those familiar with algebraically closedfields). Let G be an abelian group and (ρ, V) an irreducible representation.Show that if k is not algebraically closed then V may have dimension morethan 1 via the following example: Take k = R, G = Z/3 and let ρ be therepresentation on the 2-dimensional vector space R2 so that the generator ofZ/3 acts by rotation by 2π/3. Show that this representation is irreducible.

32 AARON LANDESMAN

6. ISOTYPIC DECOMPOSITION

We next introduce a notion of isotypic decomposition of a representation.This essentially says that a representation breaks into irreducibles in analmost unique way.

First, we introduce notation for repeated direct sums:

Definition 6.1. For (ρ, W) a G-representation, define

(ρ, W)⊕n := (ρ, W)⊕ · · · ⊕ (ρ, W)︸︷︷︸n

.

We shall often denote (ρ, W) simply as ρ, in which case we let ρ⊕n denote(ρ, W)⊕n.

Definition 6.2. Let (π, V) be a G-representation. An isotypic decomposi-tion for π is an isomorphism (π, V) ' ⊕n

i=1(πi, Vi)⊕ni so that each (πi, Vi)

is irreducible and for any i 6= j, πi 6' πj.

The main result is that isotypic decomposition is essentially unique:

Theorem 6.3. Suppose (π, V) is a G-representation and (π, V) ' ⊕ni=1(πi, Vi)

⊕ni

and (π, V) ' ⊕mj=1(ρj, Wj)

⊕mj are two isotypic decompositions. Then isotypicdecomposition is unique in the following sense:

(1) m = n(2) For each i there is a unique j so that (πi, Vi) ' (ρj, Wj) and ni = mj.

Remark 6.4. Note that this does not say there is a unique isomorphismbetween two isotypic decompositions. For example, if (π, V) = (triv, C)⊕2

then any invertible 2 × 2 matrix will define an isomorphism between Vand itself. In general, isotypic decomposition says we can determine theirreducible subrepresentations and their respective multiplicities.

Before proving Theorem 6.3, let’s deduce a quick but useful corollary.

Corollary 6.5. Suppose (π, V) is a G-representation over a field with # - char(k).Then (π, V) has a unique isotypic decomposition (where by unique, we mean uniquein the sense of Theorem 6.3).

Proof. The existence is precisely the content of Theorem 4.5. Uniqueness isprecisely Theorem 6.3. �

6.1. Proving uniqueness of isotypic decomposition. We next aim towardproving Theorem 6.3. The following lemma is key.

Lemma 6.6. Let π ' ⊕ni=1 πi and ρ ' ⊕m

i=1 ρj be two G-representations. ThenHom (π, ρ) ' ⊕n

i=1⊕m

j=1 Hom(πi, ρj

).


Proof. We define maps both ways, and check that these define a bijection.First, we define a map

f : Hom (π, ρ)→n⊕

i=1

m⊕j=1

Hom(πi, ρj

).

This is given by sending a map φ : π → ρ, to the tuple of composite maps

φij : πi ↪→ πφ−→ ρ � ρj.

Similarly, we define a map

g :n⊕

i=1

m⊕j=1

Hom(πi, ρj

)→ Hom (π, ρ)

By sending a tuple of maps(φij)

1≤i≤n,1≤j≤m to the map φ defined as follows.If v = ∑n

i=1 vi with vi in the vector space Vi on which πi acts, then

φ(v) =n

∑i=1

m

∑j=1

φij(vi).

Exercise 6.7. Verify that the maps f and g above are maps of G-representationsand are inverse of each other, completing the proof of Lemma 6.6.

�

From Lemma 6.6, we can easily deduce the following:

Lemma 6.8. If π and ρ1, . . . , ρm are two irreducible G-representations with π not

isomorphic to ρj for any j, Hom(π⊕a,⊕mj=1ρ

⊕bjj ) = 0.

Proof. Lemma 6.6 allows us to reduce to the case m = 1 and then further tothe case a = b1 = 1. This case is then settled by Lemma 5.2. �

Using Lemma 6.8, we can now prove Theorem 6.3.

Proof of Theorem 6.3. Pick some i, 1 ≤ i ≤ n. If there is no ρj with πi ' ρj,then by Lemma 6.8, any map

n⊕i=1

π⊕nii →

m⊕j=1

ρ⊕mjj(6.1)

must be 0 on Vi, and hence cannot be an isomorphism. Therefore, there issome j so that πi ' ρj. Furthermore, j is unique because by definition ofisotypic decomposition, no two ρj are isomorphic. By reasoning similarly

34 AARON LANDESMAN

with ρj, we find that for each j there is a unique i with ρj ' πi. Therefore,there is a bijection between the set of πi and the set of ρj, so m = n.

It only remains to show that when πi ' ρj, ni = mj. We will show ni ≤ mj,and the reverse inequality will follow by symmetry (interchanging the rolesof the πi and ρj).

To conclude, we show ni ≤ mj. Composing the isomorphism (6.1) withthe inclusion π

⊕nii → ⊕n

i=1 π⊕nii we obtain an injection

π⊕nii →

m⊕t=1

ρ⊕mtt .

If πi ' ρj, then by Lemma 6.8, we know the composite map π⊕nii →⊕m

t=1 ρ⊕mtt → ρ⊕ns

s must be 0 whenever for s 6= j. Therefore, we evenhave an injection V⊕ni

i →W⊕mjj . Therefore,

ni · dim Vi = dim V⊕nii ≤ dim W

⊕mjj = mj · dim Wj.

Since Vi ' Wj, dim Vi = dim Wj, so ni · dim Vi ≤ mj · dim Vi which impliesni ≤ mj as claimed. �


7. HOMS AND DUALS AND TENSORS, OH MY!

In order to prove orthogonality of characters, we will need to introduceseveral new constructions of representations from old ones. Namely homs,duals, and tensors For each of homs duals and tensor products, we firstexplain them in the context of linear algebra, and then upgrade them torepresentation theoretic constructions.

7.1. Homs of representations. We next define the notions of duals, tensorproducts, and homs in the setting of linear algebra.

Definition 7.1 (homs). Let V and W be two vector spaces. Then Hom(V, W)denotes the set of linear maps T : V → W. Then, Hom(V, W) can be giventhe structure of a vector space as follows: If T, S : V →W are linear maps andc ∈ k, define (T + S)(v) := T(v) + S(v) and define (c · T)(v) := c · (T(v)).

Definition 7.2 (homs of representations). Let (π, V) and (ρ, W) be tworepresentations. This can be given the structure of a G-representation(Homπ,ρ, Hom (V, W)

)by declaring

Homπ,ρ(g) : Hom(V, W)→ Hom(V, W)

T 7→ ρ(g)Tπ(g−1).

Lemma 7.3. For (π, V) and (ρ, W) two representations, the object(Homπ,ρ, Hom (V, W)

)constructed in Definition 7.2 is indeed a G-representation.

Proof. The crux of the matter is to check that for g, h ∈ G, we have Homπ,ρ(g) ◦Homπ,ρ(h) = Homπ,ρ(gh). Indeed,

(Homπ,ρ(g) ◦Homπ,ρ(h))T = Homπ,ρ(g)(Homπ,ρ(h)T

)= Homπ,ρ(g)

(ρ(h)Tπ(h−1)

)= ρ(g)ρ(h)Tπ(h−1)π(g−1)

= ρ(gh)Tπ((gh)−1)

= Homπ,ρ(gh)T.

�

7.2. Duals of representations.

Definition 7.4 (Duals). For V a k-vector space, let V∨, the dual of V, denotethe set of linear maps V → k. Note that V∨ = Hom(V, k), so this can begiven the structure of a vector space as in Definition 7.1.

36 AARON LANDESMAN

Exercise 7.5. If e1, . . . , en is a basis for V then we can form a dual basise∨1 , . . . , e∨n for V∨ defined by

e∨i : V → kej 7→ δij,

where δij = 1 if i = j and 0 if i 6= j. Check this is a basis for V∨. In particular,conclude dim V∨ = dim V. Hint: Show any linear map V → k is determinedby where it sends the basis elements e1, . . . , en. Show that a linear map V → kcan be expressed uniquely as a linear combination of the e∨i .

Definition 7.6 (Duals of representations). For (π, V) a G-representation, let(π∨, V∨) denote the dual representation of π, which is defined as(

π∨, V∨)

:= (Homπ,triv, Hom (V, C))

using the notation from Definition 7.2.

Exercise 7.7. Verify that for (π, V) a representation, and g ∈ G, ξ ∈ V∨, v ∈V, we have (π∨(g)ξ)(v) = ξ(π(g−1)v). Hint: Thinking of ξ as a mapξ : V → k, use that Homπ,triv(g)(ξ) = triv(g) ◦ ξ ◦ π(g−1).

Exercise 7.8. Let (π, C) be the representation of Z/3 ={

e, x, x2} with xacting by ω, for ω3 = 1, ω 6= 1. Compute (π∨, C∨). Hint: We know (π∨, C∨)is a 1-dimensional representation of Z/3, and it is determined by whetherπ∨(x) = 1, ω, or ω2.

Remark 7.9. We will see later in Lemma 8.11 that over the complex numbers,the dual representation essentially corresponds to taking the complex conju-gate. Technically, Lemma 8.11 only makes a statement about the character ofthe dual representation being the conjugate of the character of the represen-tation. This is true more generally on the level of representations (not just oncharacters). If you’re interested, try to define what is meant by the conjugateof a complex representation, and show that the conjugate representation isisomorphic to the dual representation.

7.3. Tensors of representations.

Definition 7.10 (Tensors). For V and W two finite dimensional k-vectorspaces, let e1, . . . , en be a basis of V and f1, . . . , fm be a basis of W. Then thetensor product of V and W, denoted V ⊗k W (often denoted V ⊗W whenk is understood) is the k-vector space with basis

{ei ⊗ f j

}1≤i≤n,1≤j≤m. For

v = ∑i aiei ∈ V, w = ∑j bj f j ∈ W let v ⊗ w ∈ V ⊗W denote the simpletensor given by

v⊗ w = ∑i,j

ai · bj(ei ⊗ f j

).


Remark 7.11. Here the symbols ei ⊗ f j are merely formal symbols to keeptrack of indexing, and V ⊗k W is a vector space of dimension dim V · dim W.If one thinks of direct sum of vector spaces as adding vector spaces, one maythink of tensor product as multiplying them.

Remark 7.12. One can alternatively define the tensor product of V and Was the vector space generated by all pairs v⊗ w for v ∈ V, w ∈W, subject tothe bilinearity relations that

(v + v′)⊗ w = v⊗ w + v′ ⊗ w,

v⊗ (w + w′) = v⊗ w + v⊗ w′,(av)⊗ w = a (v⊗ w) = v⊗ (aw).

One issue with this is that it is not obvious from this definition that this isfinite dimensional. The way to prove it is finite dimensional is to show itagrees with the definition given in Definition 7.10. There is a further (inmany ways “better”) definition given in terms of a universal property, whichwe do not go into here.

Exercise 7.13. For V any k-vector space, verify V ' V ⊗k k with the isomor-phism given by v 7→ v⊗ 1.

Example 7.14. Consider the vector space V⊗k W where V has basis e1, e2 andW has basis f1, f2. Then, V ⊗W is a 4-dimensional vector space with basise1 ⊗ f1, e2 ⊗ f1, e1 ⊗ f2, e2 ⊗ f2. In particular, viewing C as a 2-dimensionalvector space over R with basis {1, i}, conclude that C⊗R C has basis givenby 1⊗ 1, 1⊗ i, i⊗ 1, i⊗ i.

Exercise 7.15 (Not all tensors are simple). Consider the vector space V ⊗k Wwhere V has basis e1, e2 and W has basis f1, f2. Show that the elemente1 ⊗ f2 + e2 ⊗ f1 ∈ V ⊗k W cannot be expressed as a simple tensor a⊗ b fora ∈ V, b ∈ W. Hint: Expand a general element of the form (ae1 + be2) ⊗(c f1 + d f2).

Exercise 7.16. Show that for vector spaces U, V, and W, a linear map T :V ⊗k W → U is uniquely determined by its values on simple tensors. That isT is determined by its values on elements of the form v⊗ w. That is, showthat if we have two such maps T and S where T(v⊗ w) = S(v⊗ w) for allsimple tensors v⊗ w then T = S.

Exercise 7.17. For V, W two finite dimensional vector spaces, show(1) dim Hom (V, W) = dim V · dim W,(2) dim V∨ = dim V,(3) dim V ⊗W = dim V · dim W,(4) dim V∨ ⊗W = dim Hom(V, W).

38 AARON LANDESMAN

Definition 7.18 (Tensors of representations). For (π, V) and (ρ, W) two G-representations, Define the tensor representation (π ⊗ ρ, V ⊗W) by

(π ⊗ ρ)(g) : V ⊗W → V ⊗Wv⊗ w 7→ π(g)v⊗ ρ(g)w.

Example 7.19. Let G = Z/2 and consider the representation (sgn, k) asdefined in Example 1.11. This is the unique nontrivial representation of Z/2.When one tensors two 1-dimensional representations, one multiplies thecorresponding characters. In particular, we claim that the representationsgn⊗ sgn is isomorphic to the trivial representation because −1 · −1 =1, 1 · 1 = 1. That is, we have an isomorphism

φ : (triv, k) ' (sgn⊗ sgn, k⊗ k)

given by sending a 7→ a⊗ 1 for a ∈ k. We have seen in Exercise 7.13 thatthis defines an isomorphism of vector spaces, so we only need to checkthis is a map of representations. That is, we need to check that for v ∈ k,φ ◦ triv(g)v = ((sgn⊗ sgn)(g) ◦ φ)v. As always, this holds for g = idtautologically, so we only need to check it in the case g is the nontrivialelement of Z/2. Indeed, in this case, we have φ ◦ triv(g)v = φ(v) = v⊗ 1.So, we really just need to check ((sgn⊗ sgn)(g) ◦ φ)v = v⊗ 1. Indeed,

((sgn⊗ sgn)(g) ◦ φ)v = (sgn⊗ sgn) (g)(v⊗ 1)

= (−v)⊗ (−1)

= (−1)2 · (v⊗ 1)= v⊗ 1.

Remark 7.20. Example 7.19 illustrates a general procedure for computingtensors of a representation with a 1-dimensional representation. The rea-son sgn⊗ sgn ' triv is because (−1) · (−1) = 1. In general, the tensorproduct of two 1-dimensional representations acts as the product of the two1-dimensional representations. The situation is similar when only one ofthe representations is 1-dimensional (you multiply the higher dimensionalrepresentation by the scalar coming from the 1-dimensional representation).

Exercise 7.21. For (π, V) a G-representation, verify that that there is anisomorphism of representations (π, V) ' (π ⊗ triv, V ⊗k k) defined on theunderlying vector spaces by sending v 7→ v⊗ 1.

7.4. Relations among dual, tensor, and hom. Having defined dual, tensor,and hom, both as vector spaces and representations, we next explain howthey relate to each other. Let’s start with their relation within linear algebraand then upgrade it to a statement in representation theory.


Lemma 7.22. For V and W k-vector spaces, the map

φ : V∨ ⊗W → Hom (V, W)

ξ ⊗ w 7→ (v 7→ ξ(v) · w) .

is an isomorphism.

Proof. The map φ is a map between two vector spaces of the same dimensionby Exercise 7.17(4). So, to show it is an isomorphism, it suffices to show it isinjective.

Let e1, . . . , en denote a basis for V and let e∨1 , . . . , e∨n denote the dual basisgiven by e∨i (ej) := δij, see Exercise 7.5. Here δij = 1 if i = j and 0 if i 6= j. Letf1, . . . , fm be a basis for W. Then, observe that φ is given by

φ

(∑i,j

aije∨i ⊗ f j

)=

(v 7→∑

i,jaije∨i (v) f j

).

To show φ is injective, we only need to check that if not all aij are 0, thenthe map v 7→ ∑i,j aije∨i (v) f j is nonzero. Indeed, if some ak` is nonzero, thenek 7→ ∑i,j aije∨i (ek) f j = ∑j akj f j 6= 0 since the f j form a basis. Hence, the lineartransformation takes ek to something nonzero, so is nonzero. �

We now give a representation theoretic upgrade of the previous lemma.

Lemma 7.23. If (π, V) and (ρ, W) are two G-representations, then the map φ fromLemma 7.22 gives is an isomorphism of G-representations(

π∨ ⊗ ρ, V∨ ⊗W)'(Homπ,ρ, Hom (V, W)

).

Proof. We know from Lemma 7.22 that the underlying vector spaces areisomorphic via the map φ. So it suffices to show this map φ is in fact a mapof representations. This is now a routine calculation. For g ∈ G we need tocheck (φ ◦ (π∨ ⊗ ρ))(g) = Homπ,ρ(g) ◦ φ. That is, we need the followingdiagram to commute:

(7.1)V∨ ⊗W V∨ ⊗W

Hom(V, W) Hom(V, W).

(π∨⊗ρ)(g)

φ φ

Homπ,ρ(g)

40 AARON LANDESMAN

Indeed,

(φ ◦ (π∨)⊗ ρ(g))(ξ ⊗ w) = φ(π∨ ⊗ ρ(g)(ξ ⊗ w))

= φ(π∨(g)(ξ)⊗ ρ(g)(w)

)=(x 7→ (π∨(g)(ξ))(x)⊗ ρ(g)w

).

On the other hand, we also see

Homπ,ρ(g) ◦ φ(ξ ⊗ w) = Homπ,ρ(g) (x 7→ ξ(x)⊗ w)

= ρ(g) ◦ (x 7→ ξ(x)ρ(g)w) ◦ π(g−1)

= (x 7→ ξ(x)ρ(g)w) ◦ π(g−1)

=(

x 7→ ξ(π(g−1)x)ρ(g)w)

=(x 7→ (π∨(g)(ξ))(x)ρ(g)w

)Exercise 7.7.

Therefore,

(φ ◦ (π∨)⊗ ρ(g)) = Homπ,ρ(g) ◦ φ

�


8. ORTHOGONALITY OF CHARACTERS

In this section, we work over the field C. We will assume this in whatfollows without comment

Recall from Definition 1.18 that if (π, V) is a complex G-representationthen χπ : G → C is the function sending g 7→ tr (π(g)). Let x denote thecomplex conjugate of x ∈ C. The main result is the following:

Theorem 8.1. Suppose (π, V) and (ρ, W) are two irreducible complex G-representations.Then

∑g∈G

χπ(g) · χρ(g) =

{#G if π ' ρ

0 if π 6' ρ

The idea of the proof will be to realize ∑g∈G χπ · χρ as the character of therepresentation Homπ,ρ, to reduce it to the Proposition 8.6. In order to stateProposition 8.6, we need the following definition:

Definition 8.2. For (π, V) a G-representation, the G-invariants, denoted VG,are defined as

VG := {v ∈ V : π(g)v = v for all g ∈ G} .

In other words, VG is the set of G-invariant vectors in V.

Exercise 8.3. Verify that VG ⊂ V is a subspace.

Definition 8.4. For (π, V) and (ρ, W) two G-representations, let Hom(π, ρ)denote the vector space of maps or representations.

Exercise 8.5. Suppose (π, V) and (ρ, W) are two G-representations and let(Homπ,ρ, Hom(V, W)

)be the corresponding hom representation. Show that

the G-invariants of Homπ,ρ on the vector space Hom(V, W) are exactly themaps of G-representations Hom(π, ρ).

Proposition 8.6. For (π, V) an irreducible G-representation over C,

∑g∈G

χπ(g) = #G · dim VG.

8.1. Reducing Theorem 8.1 to Proposition 8.6. Let’s first see why Proposi-tion 8.6 implies Theorem 8.1.

For this, we will need to understand how characters behave with respectto tensor products and duals.

Lemma 8.7. For (π, V) and (ρ, W) two representations,

χπ⊗ρ(g) = χπ(g) · χρ(g)

42 AARON LANDESMAN

Proof. Choosing a basis ei for V and f j for W, we see that ei ⊗ f j is a basis for

V ⊗W. Therefore, if π(g)(ei) = ∑k aikek and ρ(g)( f j) = ∑` bj

` f`, it followsfrom by the definition of the tensor of two representations that

((π ⊗ ρ)(g))(ei ⊗ f j) = ∑k,`

aikbj

`(ek ⊗ f`).

In particular, the diagonal element of the matrix corresponding to π ⊗ ρ

acting on the basis vector ei ⊗ f j is aiib

jj. It follows that

tr((π ⊗ ρ)(g)) = ∑i,j

aiib

jj

=

(∑

iai

i

)(∑

jbj

j

)= tr π(g) · tr ρ(g)

as desired. �

Exercise 8.8 (Important exercise). For (π, V) a representation, show χπ∨(g) =χπ(g−1). Hint: Let ei be a basis for V and e∨i be the dual basis for V∨ (meaninge∨i (ej) = δij).

(1) Check that π∨(g−1)(e∨i ) is the functional sending

ej 7→ e∨i(

χπ(g−1)ej

).

(2) Verify tr π(g) = ∑dim Vi=1 e∨i (π(g)ei)

(3) Use the previous two parts to deduce the result.

Exercise 8.9. For (π, V) a representation, show that every eigenvalue of π(g)has complex norm 1. Hint: Use that G is finite.

Exercise 8.10. Verify that if M is a matrix of finite order over C, then M isdiagonalizable. Hint: Say Mn = id. Then the characteristic polynomial ofM divides xn − 1, and use that the latter splits as a product of distinct linearfactors over C to conclude that V has a basis of eigenvectors for M, so M isdiagonalizable.

Lemma 8.11. For (π, V) a representation, we have χπ(g) = χπ(g−1).

Proof. By Exercise 8.10, π(g) is diagonalizable. Say its eigenvalues areλ1, . . . , λn. Then the eigenvalues of π(g−1) are 1

λ1, . . . , 1

λn. By Exercise 8.9,

(using that x = x−1 for x ∈ C of norm 1), we obtain1λi

= λi.


It follows that the eigenvalues of π(g−1) are λ1, . . . , λn so

tr π(g−1) =n

∑i=1

λi =n

∑i=1

λi = tr π(g)

�

We can combine the above observations to show the following:

Corollary 8.12. For (π, V) and (ρ, W) two representations, we have

χHomπ,ρ = χπ · χρ

Proof. Indeed,

χHomπ,ρ(g) = χπ∨⊗ρ(g) Lemma 7.23

= χπ∨(g) · χρ(g) Lemma 8.7

= χπ∨(g) · χρ(g) Lemma 8.7

= χπ(g−1) · χρ(g) Exercise 8.8

= χπ(g) · χρ(g) Lemma 8.11.

�

We’re finally ready to show why Proposition 8.6 implies Theorem 8.1.

Proof that Proposition 8.6 implies Theorem 8.1. Let (π, V) and (ρ, W) be as inTheorem 8.1. Apply Proposition 8.6 to the representation Hom(π, ρ). Wefind

∑g∈G

χρ(g)χπ(g) = ∑g∈G

χHom(π,ρ)(g) Corollary 8.12

= #G · dim Hom(V, W)G Proposition 8.6

= #G · dim Hom (π, ρ) Exercise 8.5

=

{#G if π ' ρ

0 if π 6' ρLemma 5.2.

�

8.2. Projection operators. We next want to prove Proposition 8.6. For this,we first review some linear algebra involving projection operators.

Definition 8.13. Let V be a vector space and W ⊂ V a subspace. A mapT : V → V is a projection onto W if T|W = id and T(V) ⊂W.

Exercise 8.14. Show that T : V → V is a projection onto im T if and only ifT ◦ T = T, viewed as linear maps V → V.

44 AARON LANDESMAN

Exercise 8.15. If T : V → V is a projection operator onto W, show tr(T) =dim W. Hint: Write T as a block matrix with blocks corresponding to W anda complement of W. Show one of the diagonal blocks is id and the other is 0.

The point of introducing projection operators is to prove the following,which will be used to relate the trace of the averaging operator to dim VG inthe statement of Proposition 8.6 by Exercise 8.15.

Lemma 8.16. Let (π, V) be a representation. The averaging map Avπ := 1#G ∑g∈G π(g) :

V → V is a projection onto VG.

Proof. First, if v ∈ VG then then

Avπ(v) =1

#G ∑g∈G

π(g)v

=1

#G ∑g∈G

v

=1

#G#Gv

= v,

so Avπ |VG = id. In particular, we see VG ⊂ im Avπ.So, to show Avπ is a projection operator, we only need to check VG ⊃

im Avπ . To this end, let v ∈ im Avπ. We want to show v ∈ VG, meaningthat for any h ∈ G, π(h)v = v. By the assumption v ∈ im Avπ, we can writev = 1

#G ∑g∈G π(g)w for some w ∈ V. Then, we see

π(h)v = π(h)1

#G ∑g∈G

π(g)w

=1

#G ∑g∈G

π(h)π(g)w

=1

#G ∑g∈G

π(hg)w

=1

#G ∑t∈G

π(t)w setting t := hg

= v.

�

8.3. Proving Proposition 8.6. We are now equipped to prove Proposition 8.6which will also complete the proof of Theorem 8.1.


Proof of Proposition 8.6. By Lemma 8.16, we know Avπ = 1#G ∑g∈G π(g) is

a projection onto VG. It follows from Exercise 8.15 that tr Avπ = dim VG.Therefore,

dim VG = tr Avπ

= tr

(1

#G ∑g∈G

π(g)

)

=1

#G ∑g∈G

tr π(g)

=1

#G ∑g∈G

χπ(g)

as we wished to show. �

46 AARON LANDESMAN

9. ORTHOGONALITY OF CHARACTER TABLES

We discuss how the rows of a character table are, in an appropriate sense,orthogonal. We then discuss how the columns are also orthogonal. Again,in this section we work over C. We state and prove this now, but the proofdepends on knowing that the number of conjugacy classes equals the numberof irreducible representations, as shown later in Corollary 11.6.

Definition 9.1. Let Irrep(G) denote the set of all irreducible complex repre-sentations of G, up to isomorphism.

Theorem 9.2 (Orthogonality of character tables). For G a group, the rows of thecharacter table are orthogonal in the sense that, for π, ρ ∈ Irrep(G),

∑c∈Conj(G)

#cχπ(c)χρ(c) =

{#G if π ' ρ

0 if π 6' ρ

(where χπ(c) denotes the value of χπ on a conjugacy class, well defined by Lemma 1.24).The columns of of the character table are orthogonal in the sense that for c, d any twoconjugacy classes

∑π∈Irrep(G)

χπ(c)χπ(d) =

{#G#c if c = d0 if c 6= d

Proof assuming Corollary 11.6. The orthogonality claim on the rows is the con-tent of Theorem 9.2.

It remains to prove the column orthogonality. We deduce this from roworthogonality via a clever trick of changing the order of matrix multiplica-tion. Let A denote the matrix indexed by pairs (π, c) ∈ Irrep(G)×Conj(G)defined by

Aπ,c :=

√#c√#G

χπ(c).

Let A denote the entry-wise conjugate matrix (so Aπ,c = Aπ,c).

Exercise 9.3. Verify that column orthogonality is precisely the statement thatA · At = id, where for M a matrix, Mt denotes the transpose of M.

To conclude column orthogonality, we need the following elementary factfrom linear algebra.

Exercise 9.4. For M, N two n× n matrices with MN = id, verify that NM =id as well. Hint: Observe that M(NM− id) = MNM−M = 0. Use thatM : V → V is bijective to conclude NM− id = 0.


So, knowing A · At = id, it follows from Exercise 9.4 that At · A = id. Herewe are using that A is a square matrix, which follows since the number ofconjugacy classes equals the number of irreps (proved later in Corollary 11.6).However, expanding out this product, we see that multiplying the row of At

indexed by a conjugacy class c containing g and the column of A indexed bya conjugacy class d containing h, we have that

(At · A)c,d = ∑π∈Irrep(G)

√#c√#G

√#d√#G

χπ(g)χπ(h).

On the other hand, since At · A = id, it follows that the right hand sideexpression is 0 if c 6= d and 1 if c = d. In the latter case, we obtain

1 = ∑π∈Irrep(G)

#c#G

χπ(g)χπ(h).

Therefore,#G#c

= ∑π∈Irrep(G)

χπ(g)χπ(h),

as claimed. �

48 AARON LANDESMAN

10. THE SUM OF SQUARES FORMULA

In this section, we work over C, so in particular we may apply Schur’slemma, Maschke’s theorem, and orthogonality of characters. The main goalis to prove that the sum of the squares of the dimensions of all irreduciblerepresentations of G is equal to #G. Along the way, we will need to introducethe inner product between characters and the regular representation.

Theorem 10.1 (Sum of Squares Formula). For G any finite group,

∑ρ∈Irrep(G)

(dim ρ)2 = #G.

Remark 10.2. Note that this follows immediately from Theorem 9.2 appliedto the column corresponding to the conjugacy class of the identity, sincedim ρ = χρ(e). However, this proof relies on the fact the the number ofconjugacy classes equals the number of irreducible representations, whichtakes quite some work as we will later see. We now give an alternate proof,as it gives us the opportunity to introduce the regular representation and seemore representation theory techniques in action.

Exercise 10.3. Assuming Theorem 10.1, show that there are only finitelymany isomorphism classes of irreducible representations. Further verify thatif (π, V) is an irreducible G-representation, dim π ≤

√#G with equality if

and only if #G = 1.

10.1. The inner product on characters. We define an “inner product” oncharacters of representations. For now this is just notation as we have notyet described the space on which it is an inner product. It will turn out thatcharacters are actually a basis for the space of class functions, to be definedlater in Definition 11.1.

Definition 10.4. For (π, V) , (ρ, W) two complex representations, define

〈χπ, χρ〉 :=1

#G ∑g∈G

χπ(g)χρ(g).

Lemma 10.5. If (π, V) and (ρ, W) are irreducible then

〈χπ, χρ〉 ={

1 if ρ ' π

0 if ρ 6' π

Furthermore, the number above is equal to dim Hom (π, ρ).

Proof. The first claim is simply a rephrasing of the statement of Theorem 8.1.The second claim follows from Schur’s lemma Theorem 5.1. �


We wish to generalize Lemma 10.5 which computes the dimension of mapsof representations to arbitrary representations, instead of just irreducibleones. For this, we will need the following result on how characters commutewith direct sums:

Exercise 10.6. Verify that for π ' ⊕ni=1 πi, we have

χπ =n

∑i=1

χπi .

Hint: Recall Exercise 2.24 which showed this in the case n = 2.

Proposition 10.7. For (π, V) , (ρ, W) two complex G-representations (not neces-sarily irreducible) we have

〈χπ, χρ〉 = dim Hom(π, ρ).

Proof. We have already proven this holds in the case ρ and π are irreduciblein Lemma 10.5. In general, by Maschke’s theorem Theorem 4.5, we can writeρ and π as π ' ⊕n

i=1 πi, ρ ' ⊕mj=1 ρj. for πi, ρj all irreducible representations.

Then, by Lemma 6.6, we find

Hom (π, ρ) =n⊕

i=1

m⊕j=1

Hom(πi, ρj

).

Taking dimensions of both sides we find

dim Hom (π, ρ) = dim

n⊕i=1

m⊕j=1

Hom(πi, ρj

).

=

n

∑i=1

m

∑j=1

dim Hom(πi, ρj

)=

n

∑i=1

m

∑j=1〈χπi , χρj〉 Lemma 10.5

= 〈n

∑i=1

χπi ,m

∑j=1

χρj〉

= 〈χ⊕ni=1πi , χ⊕m

j=1ρj〉 Exercise 10.6

= 〈χπ, χρ〉.�

The following corollary is crucial for proving Theorem 10.1. Note this isessentially Desiderata 3.1[Multiplicity].

50 AARON LANDESMAN

Corollary 10.8. Let (π, V) be a representation and π ' ⊕ni=1π

nii an isotypic

decomposition. Then

〈χπi , χπ〉 = ni.

Proof. We know from Proposition 10.7 that

〈χπi , χπ〉 = dim Hom (πi, π) .

Therefore,

〈χπi , χπ〉 = dim Hom (πi, π)

= dim Hom

πi,n⊕

j=1

π⊕njj

= dim⊕n

i=1Hom(

πi, π⊕njj

)Lemma 6.6

=n

∑j=1

dim Hom(

πi, π⊕njj

)=

n

∑j=1

nj · dim Hom(πi, πj

)Lemma 6.6

=n

∑j=1

nj · δij Theorem 5.1

= ni.

�

10.2. The Regular Representation. We will apply the above inner productof Definition 10.4 to deduce Theorem 10.1. We will apply it to the regularrepresentation, which we now define. It may be helpful to note that youhave already encountered the regular representation for Z/n in Exercise 4.7.

Definition 10.9. For G a group define the regular representation (Reg(G), V)as follows: Take V to be a vector space with basis

{eg}

g∈G (so dim V = #G).Let Reg(G) act on V by (Reg(G)(h))(eg) := ehg.

Exercise 10.10 (Easy Exercise). Verify Reg(G) is indeed a G-representation.

Theorem 10.11. We have an isotypic decomposition of Reg(G) given by

Reg(G) ' ⊕π∈Irrep(G)π⊕dim π.(10.1)

Before proving Theorem 10.11, let’s see why it implies Theorem 10.1.


Theorem 10.11 implies Theorem 10.1. Simply compute dimensions of both sidesof Equation 10.1. By construction dim Reg(G) = #G. On the other hand,

dim(⊕π∈Irrep(G)π

⊕dim π)= ∑

π∈Irrep Gdim(π⊕dim π)

= ∑π∈Irrep G

(dim π)2 .

Therefore, #G = ∑π∈Irrep G (dim π)2, as desired. �

It remains only to prove Theorem 10.11. For this, we will need the fol-lowing concrete description of the character of the regular representation.

Lemma 10.12. We have

χReg(G)(g) =

{#G if g = id0 if g 6= id

Proof. Choose the basis of the vector space underlying the regular represen-tation given by

{eg}

g∈G. Then, since (Reg(G)(h))(eg) = egh, observe thattr Reg(G)(h) is precisely {g ∈ G : gh = g} (make sure you understand whyfor yourself!). But if gh = g then necessarily h = id. So, this number is 0 ifh 6= id. If h = id, then Reg(G)(id) = id. Since the vector space underlyingReg(G) is #G dimensional, we obtain tr Reg(G)(id) = #G. �

Proof of Theorem 10.11. By Corollary 10.8, it suffices to prove that for (π, V)any irreducible representation, we have

〈χReg(G), χπ〉 = dim πi.

To see this, recall that for any representation π, χπ(id) = dim π, and so

〈χReg(G), χπ〉 =1

#G ∑g∈G

χReg(G)(g) · χπ(g)

=1

#G

(χReg(G)(id) · χπ(id) + ∑

g 6=idχReg(G)(g) · χπ(g)

)

=1

#G

(#G · dim π + ∑

g 6=id0 · χπ(g)

)Lemma 10.12

=1

#G(#G · dim π)

= dim π.

�

52 AARON LANDESMAN

11. THE NUMBER OF IRREDUCIBLE REPRESENTATIONS

Throughout this section, we work over C. In this section, among otherthings, we show that the number of irreducible representations is equal tothe number of conjugacy classes. In order to do this, we will need to define anotion of class function, which are functions constant on conjugacy classes.It will be evident that the dimension of the space of class functions is equal tothe number of conjugacy classes. We will show that characters of irreduciblerepresentations form a basis for the space of class functions, and thereforethe number of irreducible representations equals the number of conjugacyclasses.

Definition 11.1. A class function is a map of sets f : G → C so that f (g) =f (h) if g and h are conjugate in G. Let Class(G) denote the set of classfunctions on G.

Remark 11.2. By Lemma 1.24 we see that for any G-representation (ρ, V),the character χρ ∈ Class(G).

Remark 11.3. Because addition and scalar multiplication preserves classfunctions, Class(G) can be given the structure of a C vector space. Explicitly,if f and h ∈ Class(G), a ∈ C and g ∈ G, then ( f + h)(g) := f (g) + h(g) and(a · f )(g) = a · ( f (g)).

Exercise 11.4. For c ∈ Conj(G) a conjugacy class, define

fc : G → C

g 7→{

1 if g ∈ c0 if g /∈ c.

Show the set of functions { fc}c∈Conj G form a basis for Class(G). Concludedim Class(G) = # Conj(G).

As mentioned above, the main result we will show in this section is thefollowing:

Theorem 11.5. For G a finite group, the set of class functions {χπ}π∈Irrep(G) forma basis for Class(G).

Proof. Since a set is a basis if and only if it is independent and a spanning set,this follows from Lemma 11.10 and Proposition 11.11 below. �

Let’s deduce our motivating consequence.

Corollary 11.6. For G a finite group,

# Irrep(G) = # Conj(G).


Proof. By Theorem 11.5, the set of characters or irreducible representationsform a basis for Class(G). Therefore, dim Class(G) = # Irrep(G).

However, by Exercise 11.4, we know dim Class(G) = # Conj(G). Hence,# Irrep(G) = dim Class(G) = # Conj(G). �

11.1. Proving characters are independent. We prove Theorem 11.5 by show-ing characters of irreducible representations are independent and spanClass(G). Let’s start by showing they are independent. For this, we be-gin by upgrading the temporarily defined pairing in Definition 10.4 to abona fide pairing on Class(G).

Definition 11.7. Define an inner product on the space of class functions asfollows: If f , h ∈ Class(G) define the pairing

〈 f , h〉 :=1

#G ∑g∈G

f (g)h(g).

Remark 11.8. Note that in particular if we take f = χπ, h = χρ, then 〈 f , h〉 de-fined as in Definition 11.7 agrees with 〈χπ, χρ〉 as defined in Definition 10.4.

Exercise 11.9 (Properties of the inner product on class functions). For f , g, h ∈Class(G) and a ∈ C, check the following properties of the inner product onclass functions (usually a map V ×V → C is called an inner product on V ifit satisfies these properties):

(1) 〈 f , g + h〉 = 〈 f , h〉+ 〈 f , g〉(2) 〈 f + g, h〉 = 〈 f , h〉+ 〈g, h〉(3) 〈a f , h〉 = a〈 f , h〉(4) 〈 f , ah〉 = a〈 f , h〉(5) 〈 f , f 〉 ∈ R≥0(6) 〈 f , f 〉 = 0 ⇐⇒ f = 0.

Lemma 11.10. The set of class functions {χπ}π∈Irrep(G) are independent.

Proof. Suppose there are some constants aπ ∈ C so that ∑π∈Irrep(G) aπχπ = 0.We want to show that aπ = 0 for all π ∈ Irrep(G). In order to pick outthe coefficient aρ, we take the inner product with χρ. More precisely, for

54 AARON LANDESMAN

ρ ∈ Irrep(G),

0 = 〈0, χρ〉

=

⟨∑

π∈Irrep(G)

aπχπ, χρ

⟩= ∑

π∈Irrep(G)

aπ〈χπ, χρ〉 Exercise 11.9

= ∑π∈Irrep(G)

aπδπρ Lemma 10.5

= aρ.

So aπ = 0 for all π ∈ Irrep(G), as we wanted to show. �

11.2. Proving characters form a basis for class functions. In order to provethe set of characters of irreps form a basis for the space of class functions, weonly need to show they span all class functions.

Proposition 11.11. The set of class functions {χπ}π∈Irrep(G) span the space of allclass functionsAssuming Proposition 11.13. Suppose f : G → C is some class function. Tosee that f lies in the span of the χπ. We are free to modify it by linearcombinations of the χπ. Define f̃ := f − ∑π∈Irrep(G)〈 f , χπ〉χπ. We claim〈 f̃ , χπ〉 = 0 for all π ∈ IrrepG. Once we verify this, it will follow fromProposition 11.13 below that f̃ = 0 and therefore

f = ∑π∈Irrep(G)

〈 f , χπ〉χπ ∈ Span (χπ) .

So, conditional on Proposition 11.13, we only need to check 〈 f̃ , χπ〉 = 0.

Exercise 11.12. Verify this, completing the proof. Hint: expand out thedefinition of f̃ and use Exercise 11.9.

�

To complete the proof of Theorem 11.5, we have reduced to showing thefollowing:

Proposition 11.13. If f is a class function so that 〈 f , χπ〉 = 0 for all π ∈ Irrep(G)then f = 0.Proof. The proof of this is a bit tricky, so we subdivide it into steps. For (π, V)a G-representation, we consider the Tf ,π := ∑g∈G f (g)π(g−1) : V → V. Themotivation for considering this is that it is closely related to the inner product〈 f , χπ〉, which we are assuming is 0.


11.2.1. Step 1: Tf ,π is a map of representations. We first claim Tf ,π : (π, V)→(π, V) is a map of representations. That is, we want to check

(11.1)V V

V V

π(h)

Tf ,π Tf ,π

π(h)

commutes.This follows from the following computation:

π(h)Tf ,π = π(h)

(∑

g∈Gf (g)π(g−1)

)= ∑

g∈Gf (g)π(hg−1)

= ∑α∈G

f (h−1αh)π(

h(

h−1αh)−1

)setting g := h−1αh

= ∑α∈G

f (α)π(hh−1α−1h) since f is a class function

= ∑α∈G

f (α)π(α−1h)

=

(∑

α∈Gf (α)π(α−1)

)π(h)

= Tf ,π ◦ π(h)

11.2.2. Step 2: Tf ,π = 0 for π irreducible. We saw in the previous step Tf ,πdefines a map of representations. If π is irreducible, it follows from Schur’slemma Theorem 5.1 that Tf ,π must be a · id for some scalar a ∈ C. Therefore,it only remains to prove a = 0. Since tr Tf ,π = tr(c · id) = dim π · a, it sufficesto show tr Tf ,π = 0. Indeed, by our assumption on f ,

tr Tf ,π = tr

(∑

g∈Gf (g)π(g−1)

)= ∑

g∈Gf (g) tr π(g−1)

= 〈 f , χπ〉= 0

56 AARON LANDESMAN

11.2.3. Step 3: Tf ,π = 0 for all representations π. By Maschke’s theorem The-orem 4.5, we can write π ' ⊕n

i=1 πi for πi irreducible representations. Itfollows that Tf ,π =

⊕ni=1 Tf ,πi . Here, if Ti : Vi → Vi are linear maps,

n⊕i=1

Ti :⊕

i

Vi →⊕

i

Vi

denotes the “block diagonal” linear map defined by(n⊕

i=1

Ti

)(v1, . . . , vn) = (T1(v1), . . . , Tn(vn)) .

It follows from the previous step that each Tf ,πi = 0 and hence Tf ,π = 0.

11.2.4. Step 4: Concluding the proof. We apply the previous step in the casethat π = Reg(G). We have Tf ,Reg(G) = 0, meaning that for V the underlyingvector space of Reg(G) with basis

{eg}

g∈G, the function

∑g∈G

f (g)π(g−1) : V → V

is the 0 map. We wish to show this means f (g) = 0 for all g ∈ G. Indeed,

0 =

(∑

g∈Gf (g)π(g−1)

)(eid)

= ∑g∈G

f (g)eg−1 .

Since the set {eg−1}g∈G form a basis of V, and ∑g∈G f (g)eg−1 is a dependencerelation between these basis elements, we must have f (g) = 0 for all g ∈ G,as desired. �


12. DIMENSIONS OF IRREPS DIVIDE THE ORDER OF THE GROUP

In this section, we work over C. Via a series of exercises, we prove that thedimension of any irreducible representation divides the order of the group.This section assumes familiarity with the notion of integral extensions ofrings and algebraic integers, or at least the willingness to take on faith that asum and product of integral elements is integral. For the remainder of thissection, we fix a group G.

The main result is the following, whose proof is concluded below inExercise 12.10.

Theorem 12.1. Let ρ be an irreducible complex representation of G. Then dim ρ |#G.

To prove this, we will need the notion a ring element being integral. Wenow recall this.

Definition 12.2. Let R→ S be a map of rings. An element s ∈ S is integralover R if s is a root of some monic polynomial ∑n

i=1 rixi with ri ∈ R (wheremonic means rn = 1). We say S is integral over R if every s ∈ S is integralover R.

We now state the main result on integral elements which we will need, butdo not prove it.

Theorem 12.3. If R→ S is map of rings and a, b ∈ S are integral then so are a + band a · b.

We now outline how to prove Theorem 12.1 using Theorem 12.3.

Exercise 12.4. Fix an irreducible complex representation (ρ, V) of G. Thepurpose of this exercise is to show χρ(g) is an integral over Z.

(1) Show that all eigenvalues of ρ(g) are integral over Z. Hint: UseExercise 8.10, that ρ(g) is a diagonalizable matrix. (Really, you onlyneed that ρ(g) is conjugate to an upper triangular matrix, and in factevery matrix over C is conjugate to an upper triangular matrix.)

(2) Conclude using Theorem 12.3 that χρ(g) is an integral over Z forevery g ∈ G.

Exercise 12.5. Let (ρ, V) be an irreducible complex G-representation. Thepurpose of this exercise is to show that if c ⊂ G is a conjugacy class andg ∈ c, then #cχρ(g)/χρ(1) is integral over Z.

(1) Show that every linear combination of elements ρ(g) for g ∈ G is inte-gral over Z. Here the map R→ S of Definition 12.2 is the map fromthe integers to dim ρ× dim ρ matrices with integer entries, sendingn 7→ n · idV .

58 AARON LANDESMAN

(2) Let c ⊂ G be a conjugacy class. Show that ∑g∈c ρ(g) commutes withρ(h) for all h ∈ G.

(3) Use Schur’s lemma to show

∑g∈c

ρ(g) =#c · χρ(g)

χρ(1)· idV .

(4) Conclude that #c · χρ(g)/χρ(1) is integral over Z.

Exercise 12.6. Show that #c · χρ(g)χρ(g)/χρ(1) is integral over Z for anyg ∈ c. Hint: Use Exercise 12.4, Exercise 12.5, and χρ = χρ∨ .

Exercise 12.7. Show that1

χρ(1)∑

c∈Conj(G)

#cχρ(c)χρ(c)

is integral over Z.

Exercise 12.8. Show that 1χρ(1)

#G is integral over Z. Hint: Use Exercise 12.7and Theorem 9.2 (orthogonality of character tables).

Exercise 12.9. Show that any x ∈ Q which is integral over Z in fact lies in Z.Hint: If x = a/b in reduced form satisfies some monic polynomial of degreemore than 1, expand the polynomial to show there are no primes dividing b.

Exercise 12.10. Prove Theorem 12.1 by showing #Gχρ(1)

is both integral over Z

and a rational number, hence an integer.


APPENDIX A. DEFINITION AND CONSTRUCTIONS OF FIELDS

To this end, we first define fields. After defining fields, if we have one fieldK, we give a way to construct many fields from K by adjoining elements.

A.1. The definition of a field. A field is a special type of ring. So, we firstdefine a ring:Definition A.1. A commutative ring with unit is a set R together with twooperations (+, ·) satisfying the following properties:

(1) Associativity: a + (b + c) = (a + b) + c, a · (b · c) = (a · b) · c(2) Commutativity: a + b = b + a, a · b = b · a(3) Additive identity: there exists 0 ∈ R so that a + 0 = a(4) Multiplicative identity: there exists 1 6= 0 ∈ R so that 1 · a = a(5) Additive inverses: For every a ∈ R, there is a additive inverse, de-

noted −a satisfying a + (−a) = 0(6) Distributivity of multiplication over addition: a · (b + c) = (a · b) +

(a · c) .Remark A.2. Any mention of “ring” in what follows implicitly means “com-mutative ring with unit.” There will be no non-commutative rings or ringswithout units.Definition A.3. A field is a ring K such that every nonzero element has amultiplicative inverse. That is, for each a ∈ K with a 6= 0, there is somea−1 ∈ K so that a · a−1 = 1.Definition A.4. A finite field is simply a field whose underlying set is finite.Example A.5. Given any prime number p, the set Z/pZ forms a field underaddition and multiplication. This field is denoted Fp. Nearly all the axiomsare immediate, except possibly for the existence of multiplicative inverses.Exercise A.6. Verify that every nonzero element has a multiplicative inversein two ways:

(1) Use the Euclidean algorithm to show that for any a < p there existssome b with ab ≡ 1 mod p and conclude that b is an inverse for a.Hint: Use that gcd(a, p) = 1.

(2) Show that ap−1 = 1, so ap−2 is an inverse for a. This is also knownas “Fermat’s Little theorem,” not to be confused with “Fermat’s Lasttheorem,” which is much more difficult. Hint: Show that the powersof any element form a subgroup of (Z/pZ)× := Z/pZ− {0} undermultiplication. Use Lagrange’s theorem (i.e., the order of a subgroupdivides the order of the ambient group) to deduce that this subgroupgenerated by a has order dividing #(Z/pZ)× = p− 1. Conclude thatam = 1 for some m dividing p− 1 and hence ap−1 = 1.

60 AARON LANDESMAN

A.2. Constructing field extensions by adjoining elements. We now ex-plain how to construct extensions of fields by adjoining elements. Hereis a prototypical example:

Example A.7. Consider the field Q(√

2)

. How should we interpret this?The elements of this field are of the form

Q(√

2)={

a + b√

2 : a, b ∈ Q}

.

Multiplication works by(a + b

√2) (

c + d√

2)= (ac + 2bd) + (ad + bc)

√2.

Here is another perspective on this field: What is√

2? It is simply a root ofthe polynomial x2 − 2. Therefore, we could instead consider the field

Q[x]/(x2 − 2),

where this means the ring where we adjoin a root of the polynomial x2 − 2.Concretely, Q[x] means polynomials with coefficients in Q, and the notationQ[x]/(x2− 2) means that in any polynomial f (x), we can replace x2 by 2. Sofor example, if we had the polynomial x3 + 2x2 + 3 this would be consideredequivalent to (x2) · x + 2 · (x2) + 3 = 2x + 4 + 3 = 2x + 7. In this way, wecan replace any polynomial with a polynomial of degree 1 of the form a + bx.Identifying x with

√2 gives the isomorphism of this ring with the above

field Q(√

2)

.

Exercise A.8. Describe the elements of the fields K as in Example A.7 for Kone of the following fields

(1) K = Q(√

3)

,

(2) K = Q(71/5),

(3) K = Q (ζ3), for ζ3 a primitive cube root of unity.In each of the above cases, write K = Q[x]/ f (x) for an appropriate polyno-mial f . In each of the above cases, what is the dimension of K over Q, whenK is viewed as a Q vector space?

Definition A.9. Let K be a field. Define the polynomial ring

K[x] :=

{n

∑i=1

aixi : ai ∈ K

}.

For f ∈ K[x], define

K[x]/( f ) := K[x]/ ∼


where ∼ is the equivalence relation defined by g ∼ h if f | g− h.

Exercise A.10. Show that K[x]/(x) ' K, where the map is given by sendinga polynomial to its constant coefficient.

Lemma A.11. Let K be a field and let f ∈ K[x] be a monic irreducible polynomial.Then K[x]/( f ) is a field.

Proof. Note that K[x]/( f ) is a ring as it inherits multiplication and additionand all the resulting properties of a ring from K[x]. (Check this!) Therefore,it suffices to check that if f is monic and irreducible, then every element hasan inverse. In other words, given any g ∈ K[x]/( f ), we need to show thereis some h with gh = 1. We can consider g ∈ K[x] as a polynomial of degreeless than f . Since f is irreducible, and deg g < deg f , it follows that the twopolynomials share no common factors. Then, by the Euclidean algorithmfor polynomials (if you have only seen the Euclidean algorithm over theintegers, check that the natural analog to the Euclidean algorithm for theintegers works equally well in polynomial rings over arbitrary fields, wherethe remainder is then a polynomial of degree less than the polynomial youare dividing by) we obtain some h, k ∈ K[x] with gh + f k = 1 as elements ofK[x]. It follows that gh ∼ 1 in K[x]/( f ) because gh− 1 = f k in K[x]. �

Exercise A.12. Let K be a field and f ∈ K[x] a monic irreducible polynomial.Suppose L = K[x]/( f ). Show that dimK L = deg f , where deg f denotes thedegree of the polynomial f and dimK L denotes the dimension of L as a Kvector space.

Example A.13. Consider the field F2[x]/(x2 + x + 1). We claim this is afinite field of order 4. Indeed, this holds because the polynomial x2 + x + 1is irreducible. To check this, we only need to check it has no linear factors.It has a linear factor if and only if x2 + x + 1 has a root in F2. But, when weevaluate it at 0 we get 1 mod 2 and when we evaluate it at 1, we get 1 mod 2.So it has no roots, and the claim follows from Lemma A.11.

Exercise A.14. For any p > 2, show that there are exactly p+12 elements

x ∈ Fp with x = y2 for some y ∈ Fp. We call such x squares. Conclude thatthere is some x ∈ Fp which is not a square whenever p > 2. Hint: Show thatif x = y2 then we also have x = (−y)2 and further that there y and −y arethe only two elements of Fp squaring to x.

Example A.15. Let p > 2 be a prime and let ε ∈ Fp be an element which isnot a square (which exists by Exercise A.14). Then,

Fp[x]/(x2 − ε)

62 AARON LANDESMAN

is a finite field of order p2. It is order p2 because it is a two dimensionalvector space over Fp spanned by the basis 1 and x. It is a field because x2− εis irreducible in Fp[x]. Indeed, to see this, note that if it were not irreducible,it would factor as a product of two linear factors, which means it wouldhave a root. But, if it had a root, there would be some y ∈ Fp so that y2 = ε.However, we chose ε not to be a square, and so no root exists.


APPENDIX B. A QUICK INTRO TO FIELD THEORY

In this section, we discuss maps of fields and the characteristic of a field.

B.1. Maps of fields.

Definition B.1. Given two fields K and L, a map φ : K → L is a mapof sets sending 1 7→ 1, 0 7→ 0 such that φ(a +K b) = φ(a) +L φ(b) andφ(a ·K b) = φ(a) ·L φ(b).

Remark B.2. Sometimes, a map of fields is referred to as a homomorphism orextension. Whenever we have a map of fields, it is required to be compatiblewith the addition and multiplication operations, as defined above. If we donot wish to require such compatibility, we will call the map “a map of sets”

Remark B.3. We shall typically drop the subscript +K, ·K on addition andmultiplication when it is clear from context.

Exercise B.4. Verify from the definition of map that

φ(a−1) = φ(a)−1

and

φ(−a) = −φ(a).

We next prove that maps of fields are injective. If you have not workedmuch with the notion of injectivity before, you may want to try the followingexercise first.

Exercise B.5. Show that a map of rings is injective (using the definition thatf : R → S is injective if f (a) = f (b) implies a = b) if and only if the onlyelement mapping to 0 is 0. Hint: Consider f (a− b).

Lemma B.6. Any map of fields is injective.

Proof. By Exercise B.5 it suffices to show that any c 6= 0 does not satisfyφ(c) = 0. Suppose there is some such c. But note that 1 = φ(1) = φ(cc−1) =φ(c)φ(c−1) = 0φ(c−1) = 0, a contradiction. Therefore, every nonzero ele-ment does not map to 0 and the map is injective. �

Remark B.7. Because of Lemma B.6, a map of fields is also typically calledan extension of fields or a field extension.

Remark B.8. The property that maps of fields are injective is very special tofields. Indeed, it is not true for groups. For example, the map Z → {1} isnot injective!

64 AARON LANDESMAN

Remark B.9. Using Lemma B.6, whenever we have a map of fields φ :K → L, we can consider L as a vector space over K. The map K × L → Lcorresponding to scalar multiplication is given by

K× L→ L

(a, b) 7→ φ(a) · b

B.2. Characteristic of a field.

Definition B.10. Let K be a field. If there is some n so that

(B.1) n := 1 + 1 + · · ·+ 1︸︷︷︸n

is equal to 0 in K, the the minimal such n is defined to be the characteris-tic of K, denoted char(K). If no such n ∈ Z≥0 exists, then we say K hascharacteristic 0.

Example B.11. The rational numbers Q has characteristic 0, but the field Fphas characteristic p.

Exercise B.12 (Important exercise). Let p be a prime number and suppose Kis a field of characteristic p. Show that for any x, y ∈ K, we have

(x + y)p = xp + yp.

Hint: Expand the left hand side using binomial coefficients, and show that pdivides nearly all of the binomial coefficients.

B.3. Basic properties of characteristic.

Lemma B.13. The characteristic of any field is either 0 or prime.

Proof. Note that the characteristic cannot be 1 because 1 6= 0. So, we have toshow that the characteristic is never composite.

Let n be a composite number with n = f g for f , g > 1 two factors of n.

Exercise B.14. Suppose a, b ∈ K with ab = 0. Then show either a = 0 orb = 0.

By the above exercise, if n = f g = 0, then either f = 0 or g = 0. Say f = 0.But then, we obtain that f < n, and so K does not have characteristic n. �

Definition B.15. For K a field, we say a subset K′ ⊂ K is a subfield if it is afield and the inclusion K′ ⊂ K is a map of fields (meaning 1 7→ 1, 0 7→ 0 andthe multiplication and addition are compatible).


Exercise B.16. Verify similarly that any field of characteristic 0 contains Q asa subfield. Hint: Define a map of fields

φ : Q→ Kab7→ ab−1.

Use that b ∈ K is nonzero by the assumption that K has characteristic 0 toshow this is well defined.

Lemma B.17. Any field K of characteristic p > 0 (for p a prime) contains Fp as asubfield.

Proof. Inside K, consider the subset {0, 1, 2, . . . , p− 1}. These form p distinctelements because char(K) = p. By definition, of n = 1 + 1 + · · ·+ 1︸︷︷︸

n

, the

elements 0, 1, . . . , p− 1 satisfy the same addition and multiplication rules asFp ' (Z/pZ). Therefore, when we restrict the multiplication and additionfrom K to {0, 1, 2, . . . , p− 1}, we realize Fp as this subfield. �

66 AARON LANDESMAN

APPENDIX C. ALGEBRAIC CLOSURES

In this section, we discuss “algebraic closures.”

Definition C.1. An extension of fields φ : K → L is finite if φ makes L into afinite dimensional vector space over K. An extension of fields φ : K → L isalgebraic if for every a ∈ L, there is a finite extension K → La with La ⊂ L asubfield containing a.

It is not too difficult to show algebraic closures exists, but to jump to theinteresting stuff, we will defer it for later:

Definition C.2. A field K is algebraically closed if any finite field extensionK → L is an isomorphism.

Exercise C.3. Show that the real numbers are not algebraically closed. Showthat the rational numbers are not algebraically closed.

Lemma C.4. Let K be a field. The following are equivalent.(1) K is algebraically closed.(2) Every monic irreducible polynomial over K has a root.(3) Every monic irreducible polynomial over K factors as a product of linear

polynomials.

Proof. For (1) =⇒ (2), we suppose K is algebraically closed and showevery monic irreducible polynomial over K has a root. Let f be any monicirreducible polynomial over K. Then, K[x]/( f ) is a field extension of K.Because K is algebraically closed, the natural map K → K[x]/( f ) is anisomorphism. Therefore, dimK K[x]/( f ) = 1 and so f has degree 1 byExercise A.12 (which says deg f = dimK K[x]/( f )), and hence has a root.

Next, if (2) holds, one can prove (3) by induction on the degree of thepolynomial.

Finally, for (3) =⇒ (1), suppose K is not algebraically closed. We wantto show there is some irreducible polynomial over K which does not factorcompletely. Let L be a finite extension of K with the inclusion K → L not anisomorphism. Since K → L is an injection it is not a surjection, so we maytake some y ∈ L \ K. We claim there is some monic irreducible polynomialf ∈ K[x] with f (y) = 0. Indeed, this is the content of the following exercise.

Exercise C.5. Let K → L be an algebraic extension. Show that any elementx ∈ L satisfies some monic irreducible polynomial f (x) = xn + kn−1xn−1 +· · · + k0, for ki ∈ K. Hint: By definition of an algebraic extension, showthat the powers of x satisfy some linear dependence relation, and obtain themonic irreducible polynomial from this relation.


Note that since y /∈ K, the polynomial f with f (y) = 0 has degree morethan 1. Since f is irreducible and has degree more than 1, f does not have aroot in K, as we wanted to show. �

Exercise C.6. Show that the complex numbers are algebraically closed (youmay assume that every polynomial over the complex numbers has a root).

Definition C.7. A field extension K → K is an algebraic closure if(1) K → K is algebraic and(2) K is algebraically closed.

Exercise C.8. Let K → L be an algebraic extension and let L denote analgebraic closure of L. Show that L is also an algebraic closure of K.

Theorem C.9 (Existence of algebraic closures). Let K be a field.(1) K has an algebraic closure.(2) Any two algebraic closures of K are isomorphic as field extensions (meaning

that for two algebraic closures K, K′, with K as a subfield via the mapsφ : K → K, φ′ : K → K′, there is an isomorphism f : K → K′ so thatf ◦ φ = φ′).

APPENDIX D. EXISTENCE OF ALGEBRAIC CLOSURES

We guide the reader through a proof of the existence of algebraic closuresin series of exercises.

We first prove the existence of an algebraic closure Theorem C.9(1), andthen show it is unique up to (non-unique) isomorphism. The key to provingthe existence of an algebraic closure will be Zorn’s lemma, which we nowrecall:

Lemma D.1. Suppose I is a partially ordered set. Suppose any totally orderedsubset I′ ⊂ I has a maximum element, i.e., there is some i ∈ I with i ≥ j for allj ∈ I′. Then I contains a maximal element, i.e., there is some i ∈ I so that for anyj ∈ I, j 6> i.

Remark D.2. Zorn’s lemma is not a lemma in the conventional sense becauseit is equivalent to the axiom of choice. Therefore, we will not prove it, butrather take it as an axiom.

We next aim to prove existence of algebraic closures. Logically, if you’dlike, you can skip directly to Exercise D.5. However, it may help yourunderstanding of that exercise if you do the prior exercises first.

Exercise D.3. We now prove some basic properties about cardinalities offield extensions.

68 AARON LANDESMAN

(1) Show that if L is an algebraic extension of a finite field K, then |L| ≤|Z|. Here |S| denotes the set-theoretic cardinality of a set S.

(2) Show that if L is an algebraic extension of an infinite field K, then|L| = |K|. Hint: Show that K has the same cardinality as K[x] anddefined a map of sets L→ K[x] by sending an element to its minimalpolynomial. Show that there are only finitely many elements with agiven minimal polynomial and deduce |K| = |L|.

(3) Conclude that for any infinite field K, if T is a set with |T| > |K| thenfor any algebraic extension L of K, we have |T| > |L|.

(4) Conclude that for any field K if T is an infinite set with |T| > |K|,then |T| ≥ |L| for any algebraic extension L of K. (By the above, theonly interesting case is the case that K is finite.)

Exercise D.4. Assume K is an infinite field. Using Exercise D.3, solve aslightly simplified version of Exercise D.5 with the modification that S is anyset so that |S| > |K| (so that there is no intermediate set T in the picture).Therefore, the addition of T is only needed to deal with finite fields.

Exercise D.5 (Difficult exercise). Use Zorn’s lemma to show an algebraicclosure of a field K exists as follows: Let T be an infinite set with |T| > |K|and let S be a set with |S| > |T|.

(1) Consider the partially ordered set

R := {(L, φ) : L is an algebraic extension of K and φ : L ↪→ S is a subset }Check that one can define a partial ordering on R by declaring(L1, φ1) ≤ (L2, φ2) if i : L1 → L2 is an algebraic extension, andφ2 ◦ i = φ1.

(2) Use Zorn’s lemma, Lemma D.1, to show that R has a maximal element,call it (M, φ).

(3) Show that M is algebraically closed by showing that if i : M → Nis any algebraic extension then there is a map ψ : N → S withψ ◦ i(x) = φ(x). Hint: Use that |N| ≤ |M| ≤ |T| < S and |S−M| =|S| > |N −M|.

Exercise D.6. Suppose we have an algebraic extension K ⊂ L and K ⊂ Kwith K algebraically closed. Show that there is a map of extensions L→ K inthe following steps:

(1) Consider the partially ordered set I of pairs (M, φ) with K ⊂ M ⊂ Land φ : M→ K a map of fields. Check that the relation

(M1, φ1) ≤ (M2, φ2)

if M1 ⊂ M2 and φ2|M1 = φ1 defines a partial ordering on such pairs(M, φ).


(2) Show that any totally ordered subset I′ ⊂ I corresponding to acollection {(Mi, φi)}i∈I′ has a maximum element given by taking(∪i Mi,∪iφi), with ∪iφi interpreted suitably.

(3) Using Zorn’s lemma obtain a maximal element (M, φ) of I.(4) Verify that the maximum element (M, φ) has M = L and conclude

there is a map L → K Hint: Suppose L 6= M. Then there is somex ∈ L − M. Show that x satisfies some minimal polynomial overL. Deduce there is a map M(x) → K restricting to the given mapφ : M→ K, and hence (M, φ) was not maximal.

Exercise D.7. Prove Theorem C.9(2) using Exercise D.6 as follows:(1) Show that for any two algebraic closures K1, K2 of the same field K

there is an injective map between φ : K1 → K2.(2) Show that the injective map φ is an algebraic extension.(3) Conclude that the map produced K1 → K2 is an isomorphism from

the definition of algebraic closure.

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