representation theory -...

Representation Theory

Patrick Da Silva

January 9, 2017

Preface

This book currently contains one part, but will be extended over the years. I personally believe this bookshould and will always remain online and available for free on my personal website. One should considerthis book as being “alive”, i.e. frequently edited ; you can find the date of the latest edit on the cover page.

Part I is concerned with the theory of complex representations. A first reader should definitely not beginrepresentation theory over an arbtirary field as it is incredibly more subtle and the generalized notions canbe more easily understood when the simpler notions applied to complex representations are resting on asolid basis.

The plan for the book is to create a second part concerning representations over an arbitrary field, athird section on locally compact groups and a fourth second on modular representations (which is the theoryin characteristic p). For the moment, this book stays with one part. Any collaboration offer is welcome andcan be discussed with me via e-mail.

One can also see that there are no exercises in this book. This is because the author didn’t have the timeto build a seriously interesting database of exercises together with their solutions. The experienced readermight also notice that I heavily relied on Serre’s classical book “Linear representation of finite groups”,where some of the exercises are solved here as a part of the theory for the sake of completeness. It isthe author’s philosophy that parts of the theory a book is built upon should not be given as exercise, tosolidify the student’s comprehension of the theory. The exercises should be parallel to the theory, so that theproblems required to be solved are problems on which the theory applies, not problems which are requiredto be solved for the theory to be completely understood. Before building such an exercise database, thebook should be somehow a bit more complete so that theory appearing in the future was not given as anexercise in the earlier parts. Therefore, an exercise database is a project for the future.

i

Table of Contents

Page

Preface i

I Complex representations of finite groups 1

1 Linear representations 21.1 Generalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.2 Subrepresentations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.3 Operations on representations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71.4 Irreducible representations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11

2 Character theory 162.1 Complex linear algebra review : the spectral theorem . . . . . . . . . . . . . . . . . . . . . 162.2 Row orthogonality relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 182.3 Column orthogonality relations and the canonical decomposition . . . . . . . . . . . . . . 252.4 Character tables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 282.5 Characters of abelian groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 342.6 Character table of a product of two groups . . . . . . . . . . . . . . . . . . . . . . . . . . . 352.7 Lifted and dropped representations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 362.8 Induced representations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

3 Representations as C[G]-modules 463.1 Generalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 463.2 The center of C[G] . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 503.3 Integrality properties of the characters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 513.4 Frobenius reciprocity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 523.5 Group actions and induced representations . . . . . . . . . . . . . . . . . . . . . . . . . . 593.6 Restriction to subgroups ; Mackey’s irreducibility criterion . . . . . . . . . . . . . . . . . . 62

iii

Part I

Complex representations of finite groups

Chapter 1

Linear representations

1.1 Generalities

Definition 1.1. Let V be a complex vector space. Denote by GL(V ) the group of isomorphisms of V , i.e.a ∈ GL(V ) means that a : V → V is a bijective linear map. If (ei)1≤i≤n is a basis of V , there is a bijection

a ∈ GL(V ) (aij) ∈ GLn(C)1:1

Let G be a finite group with 1 ∈ G the neutral element for the multiplication. A linear representation ofG in V is a morphism of groups

ρ : G→ GL(V ), i.e. ∀s, t ∈ G, ρ(st) = ρ(s)ρ(t).

We will often say that V is the representation and write “V is a representation of G” instead of namingthe triple (G,V, ρ) (this is due to the notion of C[G]-module, c.f. Chapter 3). Under this notation, the

corresponding morphism of groups ρ : G→ GL(V ) is often denoted by ρV . We will write ρsdef= ρ(s). The

degree of (G,V, ρ) is the dimension of V as a complex vector space and is denoted by deg ρ = dimV .

Remark 1.2. For any representation V , we have ρ(1G) = idV and ρ(s−1) = ρ(s)−1. This is just because ρis a morphism of groups.

Convention 1.3. When G is a finite group and s ∈ G, we write 1 to denote the identity element of thegroup G and we let ord(G) denote the order or cardinality of G. For s ∈ G, we write ord(s) for the orderof s, namely

ord(s)def= #sn | n ∈ Z.

We restrict ourselves to finite groups represented in finite-dimensional vector spaces.

Definition 1.4. Let ρ, ρ′ be two representations of G in V and V ′, respectively. A G-linear map τ : V → V ′

is a linear map such that the following diagram commutes :

V V ′

V V ′

τ

ρ(s) ρ′(s)

τ

or in other words, τ(ρ(s)) = ρ′(τ(s)). We also say that τ is a map of G-modules (this expression will beused more later since a map of G-modules corresponds to a map of C[G]-modules ; c.f. Chapter 3). Tworepresentations ρ and ρ′ of G are isomorphic if there exists a bijective G-linear map τ : V → V ′, in whichcase the inverse map is automatically G-linear and we write ρ ' ρ′. In particular, ρ and ρ′ have the samedegree.

2


Example 1.5.

- In matrix form, one can choose a basis of V , say V = 〈e1, · · · , en〉C, and write Rs for the matrix

form of ρsdef= ρ(s) ∈ GL(V ). Similarly, write V ′ = 〈e′1, · · · , e′n〉C for a basis of V ′ and R′s the matrix

form of ρ′sdef= ρ′(s). Saying that ρ and ρ′ are isomorphic is equivalent to the existence of a matrix

T ∈ GLn(C) such that∀s ∈ G, TRs = R′sT.

- A representation of degree 1 of G is the same as a morphism ρ : G → C× since we have a naturalisomorphism GL1(C) ' C× (naturality is as functors from fields to groups). Therefore, for any s ∈ G,we have ρord(s)

s = 1. In particular, ρs is a root of unity and has complex norm 1. If ρs = 1 for alls ∈ G, we call this 1-dimensional representation the trivial representation of G.

- Set g = ord(G) and let V be a vector space with dimV = g. Take a basis ett∈G. For every s ∈ G,define

λs : V → V

et 7→ est.

This gives us a map

ρ : G→ GL(V )

s 7→ ρs

called the left regular representation of G and is denoted by C[G]. It has degree g by construction.Since λs(e1) = es, the images of e1 form a basis for V . Conversely, let W be a representation ofG, and suppose that there exists a vector w ∈ W such that λs(w)s∈G is a basis for W . Then weconclude that W is isomorphic to the regular representation of G. (The isomorphism τ : V → W isgiven by es 7→ λs(w).)

Alternatively, one can define the right regular representation of G, denoted by C[G]′ for the mo-

ment : the vector space still equals Vdef= ett∈G and

ρ′s : V → V

eu 7→ ρ′s(eu)def= eus−1 .

The inverse is there to ensure that multiplication is taken in the right order:

ρ′st(eu) = eu(st)−1 = eut−1s−1 = ρ′s(eut−1) = ρ′s(ρ′t(eu)).

Since we will use the left regular representation almost exclusively, the term “regular representation”stands for the left regular representation. Note that the left regular representation and the right regularrepresentation are isomorphic since inversion in G is G-linear : for s, t ∈ G, if Φ : C[G] → C[G]′

sends et to et−1 and is extended by C-linearity, then

Φ(ρs(et)) = Φ(est) = e(st)−1 = et−1s−1 = ρ′s(et−1) = ρ′s(Φ(et)).

- More generally, suppose G acts on the left on a finite set X (we write G

X to denote the action),i.e. for any s ∈ G, there is a map x 7→ s · x ∈ X such that

∀x ∈ X, 1 · x = x

∀s, t ∈ G, ∀x ∈ X, s · (t · x) = (st) · x.

3

Chapter 1

Let V be the free vector space on the set X with basis exx∈X where ex : X → C is the functiondefined by

∀y ∈ X, ex(y)def=

1 if x = y

0 otherwise.

For any s ∈ G, define a map

λs : V → V

ex 7→ es·x.

The induced representation λ : G → GL(V ) is called a permutation representation associated tothe group action G X . Therefore, each left action of G on a finite set X induces a permutationrepresentation of G. The same can be done for right actions too: if X

G is a right action of G onthe set X , for any s ∈ G, define

ρs : V → V

ex 7→ ex·s−1 .

Again, the inverse is present to ensure multiplication is taken in the right order. This is also called apermutation representation associated to the (right) action X

G.

The most basic example of a permutation representation is given by X = 1, · · · , n and by taking Gto be the group of bijections of X onto itself, called the symmetric group on n letters (or simply thesymmetric group), denoted by Sn. The associated permutation representation V = 〈e1, · · · , en〉Cis such that for σ ∈ Sn and i ∈ 1, · · · , n, we have σ · ei

def= eσ(i). We call this representation the

classical representation of Sn.

- The construction of the free complex vector space on the set G is usually given as follows: as a set,

〈G〉Cdef= HomSet(G,C), i.e. 〈G〉C is the set of all functions f : G→ C, which is naturally a complex

vector space under pointwise addition and multiplication. It admits a basis of the form es | s ∈ Gsuch that

es(t)def=

1 if s = t

0 otherwise.

Under this construction, the left regular representation takes the following form : for f ∈ 〈G〉C ands, x ∈ G,

λs(f)(x)def= f(s−1x).

Indeed, for s, t, x ∈ G, ρs(et)(x) = et(s−1x) equals 1 if and only if s−1x = t, i.e. if and only if

st = x, which means that ρs(et) = est. Therefore, this is exactly the same representation of G butwritten under a different notation. Analogously, the right regular representation takes the form

ρs(f)(x)def= f(xs).

This also works for permutation representations associated to left/right actions : if G

X , then

λs(f)(x)def= f(s−1x), and if X

G, then ρs(f)(x)def= f(xs). Note that the left (resp. right) regular

representation is just the permutation representation of the left (resp. right) avction of G on itself,

namely s · t def= st (resp. t · s def

= ts) ; the left-hand side is the action and the right-hand sidecorresponds to multiplication in G.

- Finally, another way to see that the two notations agree is to see what happens when we write afunction f : X → C over the basis exx∈X for V : for s ∈ G, f ∈ V , write f =

∑x∈X axex where

ax = f(x) ∈ C, so that for y ∈ X ,

ρs(f) = ρs

(∑x∈X

axex

)=∑x∈X

axρs(ex) =∑x∈X

axesx =∑x∈X

as−1xex =⇒ ρs(f)(y) = f(s−1y).

4


Theorem 1.6. Let G be a group. The collection of all complex representations of G (which we assumefinite-dimensional) together with G-linear maps as morphisms form a category ; we denote it by RepC(G).For reasons explained later (c.f. Chapter 3), the hom-set between two representations of G, say V andW , isdenoted by HomC[G](V,W ) in this category.

Proof. It is clear that the identity map idV of the vector space V of a complex representation (V, ρ) ofG is G-linear. Given two G-linear maps τ1 : (V1, ρ

1) → (V2, ρ2) and τ2 : (V2, ρ

2) → (V3, ρ3), their

composition τ2 τ1 : (V1, ρ1) → (V3, ρ

3) is a G-linear map since the following diagram commutes forany s ∈ G.

V1 V2 V3

V1 V2 V3

τ1

ρ1s ρ2s

τ2

ρ3s

τ1 τ2

1.2 Subrepresentations

Definition 1.7. Let ρ : G → GL(V ) be a linear representation of G and let W ⊆ V be a vector subspace.We say that W is G-stable under the action of G if

∀w ∈W, ∀s ∈ G, ρs(w) ∈W.

If W ⊆ V is G-stable, the restriction

ρWsdef= ρs|W : W →W

is an isomorphism and ρWst = ρWs ρWt . It follows that ρW : G→ GL(W ) is a representation of G, called asubrepresentation of V .

Example 1.8. Let V be the regular representation of G and write

W = 〈x〉, xdef=∑s∈G

es.

We haveρt(x) =

∑s∈G

ρt(es) =∑s∈G

ets = x,

henceW is aG-stable subspace of V . Therefore, we conclude thatW is a subrepresentation of V isomorphicto the trivial representation (as ρs(x) = x for any s ∈ G).

Definition 1.9. Assume W,W ′ are two vector subspaces of the complex vector space V . We write V =W ⊕W ′ if and only if V = W + W ′ and W ∩W ′ = 0. In other words, any element v ∈ V can bewritten uniquely as v = w + w′ where w ∈W and w′ ∈W ′. In particular, dimV = dimW + dimW ′. Wecall W ′ a complement of W in V . A projection of V is a linear map p : V → V satisfying p p = p ; thecondition is equivalent to p(w) = w for all w ∈ im p.

Remark 1.10. If V = W ⊕ W ′, then we have a linear projection p : V → V with image W given by

v = w + w′ 7→ w. Conversely, given a linear projection p : V → V where Wdef= im p ⊆ V , we see that

V = W ⊕ ker p (write x ∈ V as x = p(x) + (x − p(x)) ∈ W + ker f ; clearly W ∩ ker p = 0). Therefore,we have a bijection

linear projections p : V → V with image W complements W ′ of W in V .1:1

5

Chapter 1

Lemma 1.11. Let V be a linear representation of G and W,W ′ ⊆ V two linear subspaces such thatV = W ⊕W ′. These come with two linear projections p, p′ : V → V with

ker p = W ′, im p = W, ker p′ = W, im p′ = W ′.

The following are equivalent:

(i) The map p is G-linear.

(ii) The map p′ is G-linear.

(iii) The subspaces W and W ′ are G-stable subspaces of V .

Proof. ( (i) ⇒ (iii) ) Let x ∈ W ′ = ker p. For s ∈ G, we have p(ρs(x)) = ρs(p(x))) = ρs(0) = 0, thusρs(w) ∈ ker p and ker p is G-stable. For w ∈W , we have

ρs(w) = ρs(p(w)) = p(ρs(w)) ∈W,

showing that W is also G-stable.( (iii)⇐ (i) ) If v ∈ V and s ∈ G, writing v = w + w′ with w ∈W , w′ ∈W ′ gives

ρs(p(v)) = ρs(w) = p(ρs(w)︸︷︷︸∈W

) + p(ρs(w′))︸︷︷︸

=0

= p(ρs(v)).

( (ii) ⇐⇒ (iii) ) Simply reverse the roles of p and p′ and use (i)⇐⇒ (iii).

Theorem 1.12. (Maschke’s Theorem) Let ρ : G → GL(V ) be a linear representation of G and W ⊆ V aG-stable subspace. Then there exists a G-stable vector subspace W ′ ⊆ V for which V = W ⊕W ′.

Proof. Let W ′ be an arbitrary complement of W in V and let p : V → V be the corresponding linear

projection (i.e. im p = W and ker p = W ′). Set gdef= ord(G). Consider the following linear map (a

so-called “average of p over G”) :

pdef=

1

g

∑t∈G

ρt p ρ−1t .

Moreover, since ρ−1t (w) = ρt−1(w) ∈W for all w ∈W , we notice that

p(ρ−1t (w)) = ρ−1

t (w) =⇒ (ρt p ρ−1t )(w) = w =⇒ p(w) = w.

Since W is G-stable, p maps V onto W (because p does), so this means p : V → V is a projection withimage W . For any s ∈ G,

ρs p ρ−1s =

1

g

∑t∈G

ρs ρt p ρ−1t ρ−1

s =1

g

∑t∈G

ρst p ρ(st)−1 = p,

hence p is a G-linear projection. By Lemma 1.11, we see thatW ′def= ker p is G-stable and V = W⊕W ′.

Remark 1.13. One sees in the proof of Theorem 1.12 that ¯p = p, so that the construction p 7→ p projectsthe set of linear projections p : V → V onto those that are G-linear. This is possible because we canaverage over G, i.e. because G is a finite group and g ∈ C× (when either of these conditions is removed,say by changing the field or working over infinite groups, the situation becomes much more complicated).Because it will prove so useful, Maschke’s theorem is one of the main reasons we restrict our attention tofinite groups.

6


Remark 1.14. Let G be a finite group, gdef= ord(G), let V be a complex vector space and denote (in the

entire document) the complex conjugation map by a bar, i.e. : C → C. A hermitian inner product onV (or simply inner product) is a map (− |−) : V × V → C satisfying the three following properties :

(i) Conjugate symmetry : for all x, y ∈ V , (x | y) = (y |x)

(ii) Sesquilinearity : for all x, y, z, w ∈ V and α, β, γ, δ ∈ C,

(αx+ βy | z) = α (x | z) + β (y | z) , (x | γz + δw) = γ (x | z) + δ (x |w)

(iii) Positive-definiteness : for all x ∈ V , (x |x) ≥ 0 and (x |x) = 0 if and only if x = 0.

A pair (V, (− |−)) is called an inner product space. Given a basis for V , an isomorphism V ' Cn inducesan inner product space on V via the standard inner product on Cn, namely(

n∑i=1

xiei

∣∣∣∣∣n∑i=1

yiei

)def=

n∑i=1

xiyi.

Therefore, any complex vector space can be equipped with an inner product space.

When W,W ′ ≤ V are linear subspaces and v ∈ V , we say that W and W ′ are orthogonal (which wedenote by W ⊥ W ′) if for any w ∈ W , w′ ∈ W ′, we have (w |w)′ = 0 ; if W ′ = 〈v〉C for some v ∈ V , wealso write v ⊥W . Given an inner product on a linear representation V , one says that it is G-invariant if itsatisfies

∀s ∈ G, ∀x, y ∈ V, (ρs(x) | ρs(y)) = (x | y) .

A G-invariant inner product (− |−)′ can always be constructed out of an inner product (− |−) by letting

(− |−) 7→ (− |−)′def=

1

g

∑s∈G

(ρs(−) | ρs(−))

which maps an inner product to an invariant one. Note that if (− |−) is G-invariant, then this constructiondoes not change the G-invariant inner product, i.e. (− |−)′ = (− |−). We deduce that any complexrepresentation admits a G-invariant inner product. The orthogonal complement of W in V is defined as

W⊥def= v ∈ V | v ⊥W.

Under such an inner product, the orthogonal complement W⊥ of a G-stable subspace W of V is also aG-stable subspace since for w⊥ ∈W⊥ and all w ∈W , we have(

w∣∣∣ ρs(w⊥)

)=(ρ−1s (w)

∣∣∣w⊥) = 0 =⇒ ρs(w⊥) ⊥W.

1.3 Operations on representations

Definition 1.15. Keeping the notations of Theorem 1.12, V is a linear representation of G, W,W ′ are G-stable subspaces with V = W ⊕W ′ as linear subspaces so that v ∈ V has a unique expression of the formv = w + w′ where w ∈W , w′ ∈W ′. It follows that

ρs(v) = ρs(w)︸︷︷︸∈W

+ ρs(w′)︸︷︷︸

∈W ′

so that the subrepresentations W and W ′ determine the representation V .

7

Chapter 1

We say that a finite-dimensional vector space V is the direct sum of the two representations W andW ′ when V = W ⊕W ′ and W,W ′ are G-stable subspaces. If the representations W and W ′ are given inmatrix form by the matrices Rs and R′s, then the representation W ⊕W ′ is given in matrix form by

ρVs =⇒

Rs 0

0 R′s

(using block form notation). If W and W ′ were to be linear subspaces of V for which V = W ⊕W ′ only asvector spaces and not as subrepresentations, we would explicitly mention it so that the notation V = W⊕W ′for representations automatically stands for the direct sum of subrepresentations (c.f. Definition 2.65 forinstance).

Remark 1.16. Recall that the tensor product of two complex vector spaces V1, V2 is the vector space V1⊗V2

together with a C-bilinear map ⊗ : V1 × V2 → V1 ⊗ V2 (written (v1, v2) 7→ v1 ⊗ v2) satisfying the followingproperty : for any C-bilinear map L : V1 × V2 → W , there exists a unique linear map L : V1 ⊗ V2 → Wmaking the following diagram commute:

V1 × V2 V1 ⊗ V2

W

L

⊗

L

or in other words, L(v1, v2) = L(v1⊗v2). In particular, this tells us how to define linear maps V1⊗V2 →W ,namely through C-bilinear maps V1 × V2 →W .

In category-theoretical terms, given complex vector spaces V1, V2, the construction which sends a vectorspace W to the set of C-bilinear maps Bil(V1, V2;W ) is functorial ; given a linear map W → W ′, post-composition gives a map Bil(V1, V2;W ) → Bil(V1, V2;W ′). The corresponding functor is represented byV1 ⊗ V2 by the above commutative diagram, which gives us the following natural bijection

Bil(V1, V2;W ) ' HomC(V1 ⊗ V2,W ).

Definition 1.17. A linear representation ρ : G → GL(V ) induces a natural left group action G

V : for

s ∈ G and v ∈ V , we set s · v def= ρs(v). We call this the canonical G-action of the representation. A

G-linear map ϕ : V → W is therefore equivalent to a C-linear map which commutes with the action of G,namely s ·ϕ(v) = ϕ(s · v) for all s ∈ G and v ∈ V . In particular, defining a representation ρ : G→ GL(V )is the same as defining an action G

V where G acts by C-linear endomorphisms of V , which we call aG-linear action on V .

Definition 1.18. Let V1, V2 be representations of the finite group G. Define a representation of G overV1 ⊗ V2 as follows: for s ∈ G,

s · (v1 ⊗ v2)def= (s · v1)⊗ (s · v2).

This gives us a representation ρ : G→ GL(V1⊗V2) called the tensor product of the two representations ρ1

and ρ2 because each s induces the C-bilinear map (v1, v2) 7→ (s · v1, s · v2), so that the above defined map

is a C-linear isomorphism of V1 ⊗ V2. We denote the corresponding morphism of groups by ρdef= ρ1 ⊗ ρ2.

Remark 1.19. If we write ρi in matrix form over the bases e1, · · · , en and f1, · · · , fm, we get matrices

(ri1j1(s)) and (ri2j2(s)) for each s ∈ G. The matrix form of ρdef= ρ1 ⊗ ρ2 is computed as follows : since

8


ei ⊗ fj is a basis for V1 ⊗ V2,

ρ(ej1 ⊗ fj2) = ρ1(ej1)⊗ ρ2(ej2)

=

(n∑

i1=1

ri1j1(s)ei1

)⊗

(m∑i2=1

ri2j2(s)fi2

)

=

n∑i1=1

m∑i2=1

(ri1j1(s)ri2j2(s))ei1 ⊗ fi2 .

It follows that the ((i1, i2), (j1, j2))-coefficient of the matrix representation of ρ1 ⊗ ρ2 is ri1j1(s)ri2j2(s).

Proposition 1.20. Let V1, V2,W be two representations of G. The group G acts on V1 × V2 diagonally,

namely s · (v1, v2)def= (s · v1, s · v2). A C-bilinear map ϕ : V1 × V2 →W is said to preserve the action of

G if for any s ∈ G, v1, v2 ∈ V , we have

ϕ(s · (v1, v2))def= ϕ(s · v1, s · v2) = s · ϕ(v1, v2).

The tensor product representation V1 ⊗ V2 is characterized by the following universal property. By Re-mark 1.16, a C-bilinear map ϕ : V1×V2 →W corresponds to a C-linear map ϕ : V1⊗V2 →W . The tensorproduct representation ρ : G → GL(V1 ⊗ V2) is such that ϕ respects the action of G if and only if ϕ is aG-linear map. Just as the tensor product of vector spaces, the tensor product representation is unique up toa unique G-linear isomorphism respecting ⊗.

Proof. We have a functor RepC(G) → Set which maps a representation W to the set BilG(V1, V2;W )of C-bilinear maps V1 × V2 → W which respect the action of G. It suffices to show that this functor isrepresented by V1⊗V2 with the given G-linear action. Remark 1.16 shows that the functor Bil(V1, V2;−) isalready represented by the vector space V1⊗V2. To see that the tensor product representation representsBilG(V1, V2;W ), it suffices to show that the bijection of Remark 1.16 maps BilG(V1, V2;W ) bijectivelyonto HomC[G](V1 ⊗ V2,W ). If ϕ : V1 × V2 →W respects the action of G, then for s ∈ G,

ϕ(s · (v1 ⊗ v2)) = ϕ((s · v1)⊗ (s · v2)) = ϕ(s · v1, s · v2) = s · ϕ(v1, v2) = s · ϕ(v1 ⊗ v2).

Conversely, if ϕ : V1 ⊗ V2 →W is G-linear, then

ϕ(s · v1, s · v2) = ϕ((s · v1)⊗ (s · v2)) = ϕ(s · (v1 ⊗ v2)) = s · ϕ(v1 ⊗ v2) = s · ϕ(v1, v2).

Example 1.21. Consider a vector space V and the tensor productW = V ⊗V together with a representationρ : G → GL(V ). The C-bilinear map V × V → V × V defined by (v1, v2) 7→ (v2, v1) induces anisomorphism θ : V ⊗ V → V ⊗ V which is characterized by the formula θ(v1 ⊗ v2) = v2 ⊗ v1 for anyv1, v2 ∈ V ; in particular, θ θ = idV⊗V . There are two relevant subspaces of V ⊗ V worth mentioning,namely

Sym2(V )def= z ∈ V ⊗ V | θ(z) = z, Alt2(V ) = z ∈ V ⊗ V | θ(z) = −z.

It is not hard to see that V ⊗ V = Sym2(V )⊕Alt2(V ) since for all α ∈ V ⊗ V , we have

α =α+ θ(α)

2︸︷︷︸∈Sym2(V )

+α− θ(α)

2︸︷︷︸∈Alt2(V )

and if z ∈ Sym2(V ) ∩ Alt2(V ), then z = θ(z) = −z, hence z = 0. The two subspaces Sym2(V ), Alt2(V )come with projections pSym : V ⊗ V → V ⊗ V and pAlt : V ⊗ V → V ⊗ V defined by the above formulas,

9

Chapter 1

namely for α ∈ V ⊗ V ,

pSym(α)def=

α+ θ(α)

2, pAlt(α)

def=

α− θ(α)

2.

Clearly,

ker pSym = Alt2(V ), im pSym = Sym2(V ), ker pAlt = Sym2(V ), im pAlt = Alt2(V ).

With respect to the tensor product representation ρ ⊗ ρ : G → GL(V ⊗ V ), since θ and idV⊗V are bothG-linear endomorphisms of V ⊗ V , so are pSym and pAlt, which means that Sym2(V ) and Alt2(V ) areG-stable subspaces by Lemma 1.11, e.g. subrepresentations of V ⊗ V . We call them the symmetric squareand exterior square of V , respectively.

Constructions of representations where permutations are involved will be studied in more detail insubsequent sections when we investigate properties of the symmetric group Sd for d ≥ 2.

Definition 1.22. Let ρi : Gi → GL(Vi) be a representation of the finite group Gi for i = 1, 2. Define arepresentation ρ1 ρ2 : G1 ×G2 → GL(V1 V2), called the box product of ρ1 and ρ2. As vector spaces,

V1 V2def= V1 ⊗ V2 and we denote simple tensors in V1 V2 by v1 v2. For (s1, s2) ∈ G1 ×G2,

(ρ1 ρ2)(s1,s2)(v1 v2)def= ρ1

s1(v1) ρ2s2(v2).

Remark 1.23. The box product of representations and the tensor product of representations are relatedoperations. If G1 = G2 = G, ρi : G → GL(Vi) are two representations for i = 1, 2 and ∆ : G → G × Gdenotes the diagonal morphism (i.e. ∆(s)

def= (s, s)), then ρ1 ⊗ ρ2 = (ρ1 ρ2) ∆.

Note that in the literature, both operations (⊗ and ) are usually called the tensor product, but wewished to distinguish such operations because even though they are related, they do not live over the samegroups.

Proposition 1.24. Let ρi : Gi → GL(Vi) be representations, i = 1, 2. Given a representation W of thegroup G1 × G2, a C-bilinear map ϕ : V1 × V2 → W is said to respect the action of G1 × G2 if for any(s1, s2) ∈ G1×G2 and (v1, v2) ∈ V1×V2, we have ϕ(s1 · v1, s2 · v2) = (s1, s2) ·ϕ(v1, v2). The box productrepresentation ρ1 ρ2 : G1 × G2 → GL(V1 V2) satisfies the following universal property : a C-bilinearmap ϕ : V1×V2 →W respects the action of G1×G2 if and only if the corresponding map ϕ : V1V2 →Wis G1 ×G2-linear.

Proof. The proof is entirely analogous to Proposition 1.20 ; all we need to show is that the natu-ral bijection Bil(V1, V2;W ) ' HomC(V1 ⊗ V2,W ) establishes a correspondence between the subsetBilG1,G2(V1, V2;W ) of bilinear maps which respects the action of G1 × G2 (which is a functorial con-struction in W , e.g. as a functor from RepC(G1 × G2) to Set) and HomC[G1×G2](V1 V2,W ). Toproceed, we need to show that the bijection of Remark 1.16 maps one subset into the other.If ϕ : V1 × V2 →W respects the action of G1 ×G2, then for (s1, s2) ∈ G1 ×G2 and (v1, v2) ∈ V1 × V2,we have

ϕ((s1, s2) · (v1 v2)) = ϕ((s1 · v1) (s2 · v2))

= ϕ(s1 · v1, s2 · v2)

= (s1, s2) · ϕ(v1, v2)

= (s1, s2) · ϕ(v1 v2).

10


Conversely, if ϕ : V1 V2 →W is G1 ×G2-linear, then

ϕ(s1 · v1, s2 · v2) = ϕ((s1 · v1) (s2 · v2))

= ϕ((s1, s2) · (v1 v2))

= (s1, s2) · ϕ(v1 v2)

= (s1, s2) · ϕ(v1, v2).

1.4 Irreducible representations

Definition 1.25. Let ρ : G → GL(V ) be a linear representation of G. We say that V is irreducible ifV 6= 0 and the set of subrepresentations of V equals 0, V .

Remark 1.26.

- By Maschke’s Theorem, V is irreducible if and only if V is not the direct sum of two proper subrep-resentations (i.e. the only options are V = V ⊕ 0).

- Every representation of degree 1 is irreducible. We will see later (c.f. Theorem 2.53) that any finitenon-abelian group admits an irreducible representation of degree greater than 1, which separates therepresentation theory of abelian groups from that of non-abelian groups.

Theorem 1.27. Every representation of a finite group G is a direct sum of irreducible subrepresentations.

Proof. If dimV = 0, the representation is the empty direct sum. We proceed by induction on dimV ≥ 1.If dimV = 1, then V is irreducible, so we are done. Assume dimV > 1. If V is irreducible, we are againdone ; assume not, so that V is not irreducible. By Maschke’s Theorem, there exists a decompositionV = V ′ ⊕ V ′′ with dimV ′, dimV ′′ < dimV . Since V ′ and V ′′ can be decomposed as a direct sum ofirreducible representations, so does V .

Remark 1.28. In general, this decomposition is not unique. For instance, if ρs = idV for all s ∈ G, thenany subspace of V is a subrepresentation, so for any basis v1, · · · , vn of V where n

def= dimV , we have

V =n⊕i=1

〈vi〉C,

so we have as many possible decompositions for V as we have bases (modulo scalars). However, whatis preserved in any decomposition is the number of irreducible subrepresentations appearing which areisomorphic to a given irreducible representation W (for instance, in the above example, the trivial repre-sentation appears in V precisely n times as a subrepresentation, regardless of the chosen basis). This willbe shown as a consequence of Schur’s Lemma (c.f. Theorem 1.30).

Lemma 1.29. Let ρ1 : G→ GL(V1) and ρ2 : G→ GL(V2) be two representations of G. Then the set

HomC[G](V1, V2)def= f ∈ HomC(V1, V2) | ∀s ∈ G, f ρ1

s = ρ2s f

is a complex vector subspace of HomC(V1, V2).

Proof. One can see HomC[G](V1, V2) as the image of the linear projection

Φ : HomC(V1, V2)→ HomC(V1, V2), f 7→ 1

g

∑s∈G

ρ2s f ρ−1

s ,

11

Chapter 1

so that HomC[G](V1, V2) = Φ(HomC(V1, V2)) is a linear subspace of HomC(V1, V2).

Theorem 1.30. (Schur’s Lemma) Let ρi : G → GL(Vi) be irreducible representations of the finite group Gand f : V1 → V2 be a G-linear map.

(i) The kernel ker f is a subrepresentation of V1 and the image im f is a subrepresentation of V2 (thisdoes not assume V1 or V2 irreducible). In particular, either ρ1 is isomorphic to ρ2 or f = 0.

(ii) If ρ1 is isomorphic to ρ2, let θ : V1 → V2 be a G-linear isomorphism. We get f = λ θ for a uniquelydetermined λ ∈ C.

Proof. (i) The subspace ker f ⊆ V1 is G-stable, since if v1 ∈ ker f , then for s ∈ G,

f(s · v1) = s · f(v1) = s · 0 = 0 =⇒ s · v1 ∈ ker f.

Since ker f is G-stable and V1 is irreducible, ker f ∈ 0, V1.Similarly, if v2 = f(v1) ∈ im f , then for s ∈ G, we have

s · v2 = s · f(v1) = f(s · v1) ∈ im f.

It follows that im f is G-stable ; since V2 is irreducible, im f ∈ 0, V2.If ker f = 0, we cannot have im f = 0 because 0 6= f(V1) ⊆ V2, hence im f = V2 ; if ker f = V1,we must have im f = 0. Therefore f = 0 or f is an isomorphism between ρ1 and ρ2, which endsthe proof.

(ii) Let λ ∈ C be an eigenvalue of θ−1 f , so that (θ−1 f)(v) = λv for some v ∈ V1 \0, or in other

words, f(v) = λ θ(v). Set f ′def= f − λ θ, e.g. f ′ : V1 → V2. By Lemma 1.29, we can apply part (i)

to f ′ so that ker f ′ = V1 because v ∈ ker f ′ 6= 0 and V1 is irreducible. This implies f ′ = 0, or inother words, f = λ θ.

Definition 1.31. Let V,W be complex vector spaces and set V ∗def= HomC(V,C), which we call the dual

vector space of V ; it is the set of C-linear maps from V to C. There is a natural map

Φ : V ∗ ⊗W → HomC(V,W ), ϕ⊗ w 7→ (v 7→ ϕ(v)w).

It is an isomorphism because we assume V and W finite-dimensional. To see this, let v1, · · · , vn be abasis of V and ω1, · · · , ωn be the corresponding dual basis, so that ωi(vj) = δij where δij denotes theKronecker delta. Let w1, · · · , wm be a basis of W . It follows that

Φ(ωj ⊗ wi) 7→ (vk 7→ ωj(vk)wi = δjkwi)

so that Φ maps a basis of V ∗ ⊗W bijectively to a basis of HomC(V,W ) (in matrix form, the linear mapsvk 7→ ωj(vk)wi correspond to matrices with zeros everywhere except the 1 at the (i, j)-coordinate). WhenV = W , there is also a natural evaluation map

ev : V ∗ ⊗ V → C, ϕ⊗ v 7→ ϕ(v).

The map trdef= ev Φ−1 is called the trace map of V . Given a basis v1, · · · , vn of V and writing a linear

map f : V → V in matrix form so that

f(vj) =

n∑i=1

aijvi,

12


over the basis ωj ⊗ vi | i, j = 1, · · · , n of V ∗ ⊗ V , we have

Φ

n∑i,j=1

aijωj ⊗ vi

(vk) =n∑

i,j=1

aijωj(vk)vi =n∑i=1

aikvi = f(vk),

hence Φ(∑n

i,j=1 aijωj ⊗ vi)

= f and we obtain the formula

tr (f) =

n∑i,j=1

ev(aijωj ⊗ vi) =

n∑i,j=1

aijδij =

n∑i=1

aii.

If f, g : V → V , then tr (f g) = tr (g f). It follows from this fact (or the definition itself, since it does notdepend on a choice of basis) that if g is an automorphism of V , then tr

(g f g−1

)= tr (f). Also recall

that the matrix form of the dual map f∗ : V ∗ → V ∗ defined by sending g : V → C to g f is the transposeof the matrix of f when V and V ∗ are given the bases v1, · · · , vn and ω1, · · · , ωn respectively, so thattr (f∗) = tr (f). The trace will be used abundantly when we will have developed character theory.

Remark 1.32. Let V be a complex vector space and f ∈ HomC(V, V ). If v1, · · · , vn is a basis over whichf has Jordan normal form, meaning that f(vi) ∈ λivi, λivi + vi−1 for i = 1, · · · , n, we see by definitionthat tr (f) is the sum of the eigenvalues of f with algebraic multiplicity.

A vector v ∈ V \ 0 is a geometric eigenvector for f if there exists λ ∈ C such that f(v) = λv, inwhich case we call λ a geometric eigenvalue (or simply eigenvalue). An algebraic eigenvector is a vectorv ∈ V in the kernel of the linear map (f − λ idV )k for some k ≥ 1 and λ ∈ C (the kth exponent denotesk-fold composition). When ker(f − λ idV )k 6= 0, we call

⋃k≥0 ker(f − λ idV )k the algebraic eigenspace

of f associated to λ, called an algebraic eigenvalue, whereas when ker(f − λ idV ) 6= 0, ker(f − λ idV ) iscalled the geometric eigenspace. Under these definitions, a complex matrix is diagonalizable if and onlyif its geometric and algebraic eigenspaces are equal (the statement is obvious in Jordan normal form).

Note that in Jordan normal form, the basis vectors corresponding to a Jordan block are precisely thosespanning the algebraic eigenspace of the corresponding algebraic eigenvalue on the diagonal of that block.The algebraic multiplicity of an algebraic eigenvalue is defined as the dimension of its correspondingalgebraic eigenspace, so that the trace is easily seen to be the sum of the algebraic eigenvalues, countedwith algebraic multiplicity.

Corollary 1.33. Let V1,V2 be two irreducible representations of G and h : V1 → V2 be a linear map. Set

gdef= ord(G). Define

hdef=

1

g

∑s∈G

ρ2s h (ρ1

s)−1,

c.f. Lemma 1.29. Then the following holds :

(i) If ρ1 and ρ2 are not isomorphic, we have h = 0.

(ii) If Vdef= V1 = V2 and ρ

def= ρ1 = ρ2, then h = tr(h)

dimV idV . In particular, the map HomC[G](V1, V2)→ Cdefined by h 7→ tr(h)

dimV idV is an isomorphism of vector spaces.

Proof. Since h is G-linear :

ρ2t h (ρ1

t )−1 =

1

g

∑s∈G

ρ2t ρ2

s h (ρ1s)−1 (ρ1

t )−1 =

1

g

∑s∈G

ρ2ts h (ρ1

ts)−1 = h,

13

Chapter 1

(i) holds trivially. For (ii), it suffices to compute the scalar λ. We have

tr(h)

=1

g

∑s∈G

tr(ρs h ρ−1

s

)=

1

g

∑s∈G

tr (h) = tr (h) ,

and since tr(h)

= tr (λ idV ) = λ dimV , we conclude that λ =tr(h)dimV = tr(h)

dimV . The last statementfollows since for a G-linear map h : V1 → V2, we have h = h.

Corollary 1.34. (The matrix representation of Schur’s Lemma) Let V1, V2 be two irreducible representations

of G and set gdef= ord(G). Pick bases e1, · · · , em and f1, · · · , fn of V1 and V2 respectively, so that

R1s = (r1

i1j1(s)) and R2

s = (r2i2j2

(s)).

(i) If V1 and V2 are not isomorphic representations, then for all 1 ≤ i1, j1 ≤ m, 1 ≤ i2, j2 ≤ n, we have

1

g

∑s∈G

r2i2j2(s)r1

i1j1(s−1) = 0.

(ii) In the case where Vdef= V1 = V2 (so that m = n), for all 1 ≤ i1, j1, i2, j2 ≤ n,

1

g

∑s∈G

r2i2j2(s)r1

i1j1(s−1) =1

dimVδi1j2δi2j1 .

Proof. Let h : V1 → V2 be a linear map written as (xij) in matrix form and similarly write h as (x0ij).

Then

x0i2j1 =

1

g

∑s∈G

((r2i2j2(s))(xij)(r

1i1j1(s−1))

)i2j1

=

n∑j2=1

m∑i1=1

(1

g

∑s∈G

r2i2j2(s)r1

i1j1(s−1)

)xj2i1 .

Picking the linear map h specified by xji = δjj2δii1 , since h = 0 in part (i) of Schur’s lemma, we get thatthe double sum reduces to

1

g

∑s∈G

r2i2j2(s)r1

i1j1(s−1) = x0i2j1 = 0,

which proves (i). Using the same map h in part (ii), we obtain x0i2j1

= tr(h)dimV δi2j1 = 1

dimV δi1j2δi2j1 bySchur’s lemma, which completes the proof.

Example 1.35.

- Consider a group G and a representation ρ : G → GL(V ). For s ∈ G, we get a linear mapρs : V → V . When is this map G-linear? In other words, when does the following hold:

∀t ∈ G, ρt ρs = ρs ρt.

If G is abelian, the above condition trivially holds since

ρs ρt = ρst = ρts = ρt ρs.

By Schur’s Lemma, for every irreducible representation ρ : G → GL(V ), ρs = λsidV for someλs ∈ C and each s ∈ G. If W ⊆ V is a non-zero vector subspace, then ρs(W ) ⊆ W (because

14


ρs = λsidV ), hence W is G-stable ; it follows that W = V , i.e. V is one-dimensional. We concludethat the irreducible representations of an abelian group G are elements of its Pontryagin dual group :

Gdef= ρ : G→ S1 | ρ is a group homomorphism .

where S1 ⊆ C× is the subgroup of C× of complex numbers z satisfying |z| = 1.

- Consider the group G = S3def= 〈σ, τ | σ3 = τ2 = 1, στ = τσ2〉. We know two representations of G

of degree 1, namely the trivial representation and the alternating or sign representation, which isjust the homomorphism π : S3 → C× which maps a permutation to its sign, which lands in 1,−1.There is another irreducible representation of S3, which we compute as follows. The group S3 actson the vector space C3 (for which we give the basis 〈e1, e2, e3〉C) by σ · ei = eσ(i). In coordinates,σ(z1, z2, z3) = (zσ−1(1), zσ−1(2), zσ−1(3)). To see this,

x =3∑i=1

ziei =⇒ σ(x) =3∑i=1

ziσ(ei) =3∑i=1

zieσ(i) =3∑i=1

zσ−1(i)ei

(c.f. Example 1.5). The subspaceW = 〈(1, 1, 1)〉C is of course G-stable and of dimension one, so since

the standard inner product of C3 is S3-invariant, the orthogonal complement Vdef= (z1, z2, z3) ∈

C3 | z1 + z2 + z3 = 0 is also G-stable by Remark 1.14 (or by direct computation), leaving us withC3 = V ⊕ W . The representation V is called the standard representation of S3. It is clearlyirreducible, since it has dimension 2 and if it were reducible, there would be a G-stable subspace ofdimension 1 ; if x =

∑3i=1 ziei is the generator, then

3∑i=1

zieσ(i) = σ ·3∑i=1

ziei = λσ∑i=1

ziei,

hence for σ = (ij) we have λσ = ±1 since σ2 = id. It follows that z1 = ±z2 = ±z3 by comparing co-efficients, and since z1+z2+z3 = 0, we reach a contradiction since z1±z1±z1 ∈ −3z1,−z1, z1, 3z1(c.f. Theorem 2.46 for a generalization).

15

Chapter 2

Character theory

2.1 Complex linear algebra review : the spectral theorem

Before we start developing character theory over C, we make a quick review of complex linear algebra andthe spectral theorem for normal matrices.

Definition 2.1. Let n ≥ 1 be an integer and A be an n × n complex matrix. The conjugate transpose

(also called hermitian transpose) of A = (aij) is the matrix A† defined by A†ijdef= aji. The matrix A is

said to be

(i) hermitian if A = A†

(ii) normal if AA† = A†A

(iv) unitary if it is normal and invertible, in which case A−1 = A† since AA† = A†A = I ; equivalently,A has orthonormal rows (resp. columns) under the standard inner product in Cn

(v) diagonal if there exists a basis e1, · · · , en of Cn and complex numbers λ1, · · · , λn such thatAei = λiei for i = 1, · · · , n

(vi) diagonalizable if there exists an n× n matrix P such that PAP−1 is diagonal

(vii) unitarily diagonalizable if there exists an n×n unitary matrix U such that UAU † is diagonalizable(note that UAU † = UAU−1).

(viii) upper triangular (resp. lower triangular) if aij = 0 for i > j (resp. i < j)

(ix) triangular when it is either upper or lower triangular.

Recall that EndC(V )def= HomC(V, V ) is the set of linear endomorphisms of V . We can develop the

analogous concepts for a linear endomorphism f ∈ EndC(V ). The adjoint of f is the linear endomorphismcorresponding to the conjugate transpose of its matrix over a basis, and is defined as the unique linear mapf † such that for any v, w ∈ V ,

(f(v) | w) = (v | f †(w)).

(This determines f † uniquely since we can use the above equation to compute its coefficient matrix, whichturns out to be A† when f has coefficient matrix A.) The endomorphism f is

(i)∗ hermitian (also called autoadjoint) if f = f †, i.e. if for any v, w ∈ V , we have

(f(v) | w) = (v | f(w))

16


(ii)∗ normal if f and f † commute, namely f f † = f † f

(iv)∗ unitary if there exists a basis v1, · · · , vn of V for which the vectors f(v1), · · · , f(vn) form anorthonormal basis of V , namely (f(vi) | f(vj)) = δij

(vii)∗ unitarily diagonalizable if there exists an orthonormal basis of eigenvectors for f .

Remark 2.2. transpose matrix A diagonal hermitian matrix is the same as a real diagonal matrix. Diagonal,hermitian matrices and unitary matrices are all normal. However, there are matrices which are neither ofthose and are still normal: as an example, consider

A =

1 1 00 1 01 0 1

which is normal since

AA† =

2 1 11 2 11 1 2

= A†A

but it is neither hermitian, diagonal nor unitary. Also note that just as for real matrices, we had (AB)> =B>A> where A> denotes the transpose of A, for two complex matrices, we have (AB)† = B†A†.

Lemma 2.3. Let A be an n× n complex matrix. If A is normal and triangular, then A is diagonal.

Proof. Without loss of generality, assume A is upper triangular. The vector Aen is the nth column of Aand A†en is the nth row vector of A on which conjugation has been applied on the coefficients. We have

‖Aen‖2 = (Aen | Aen) = (en | A†Aen) = (en | AA†en) = (A†en, A†en) = ‖A†en‖2.

Since the coefficient in front of en is the only non-zero coefficient of A†en, the coefficent in front of enof Aen also has to be its only non-zero coefficient (because their nth coefficients are conjugates, so thecontrary would imply ‖Aen‖2 > ‖A†‖2). It follows that Aen = λnen and Aei ∈ 〈e1, · · · , en−1〉C fori = 1, · · · , n − 1. The (n − 1) × (n − 1) block submatrix of A is also normal (since A was), so weconclude by induction on n since the case n = 1 is trivial.

Proposition 2.4. Let A be an n×n complex matrix. There exists an upper triangular (resp. lower triangular)matrix T and a unitary matrix U such that A = UTU †. A pair (U, T ) for which this holds is called anupper Schur decomposition for A (resp. lower Schur decomposition ; the name “Schur decomposition”usually stands for an upper Schur decomposition).

Proof. Without loss of generality, we prove the result for upper Schur decompositions ; otherwise, onecan apply the result to A† instead and apply (−)† on the equation A† = UT †U † to recover the lowertriangular case.The goal is to find an orthonormal basis in which the expression of A is upper triangular (orthonormalityguarantees that the base change matrix will be unitary). Let λ be an eigenvalue for A and v a corre-

sponding eigenvector. Set v1def= v and let V2 be the orthogonal subspace to 〈v1〉C in Cn of dimension

n− 1. After choosing some orthonormal basis of V2 and adding v1 to it to obtain an orthonormal basisof Cn, the matrix A has the following form over this orthonormal basis:

λ a12 · · · a1n

0 a22 · · · a2n...

.... . .

...0 an2 · · · ann

17

Chapter 2

so applying the result by induction on n on the block matrix B of rows and columns 2, 3, · · · , n, weobtain an orthonormal basis v2, · · · , vn of V2 such that B is upper triangular when expressed over thatbasis. It follows that A is upper triangular over the basis v1, · · · , vn, and since v1 is orthogonal to V2,the basis v1, · · · , vn is orthonormal.

Remark 2.5. The proof of Proposition 2.4 could have been equally well done in the language of completeflags. A complete flag in V is a filtration (V0, V1, · · · , Vn = V ) of subspaces of V , namely V0 ( V1 ( · · · (Vn = V where dimVi = i for i = 0, · · · , n (when we remove the dimension condition but only ask for strictinclusions, we speak of a flag). In this language, Proposition 2.4 is equivalent to the fact that if f is a linearendomorphism of V , there exists a complete flag (V0, · · · , Vn) which f stabilizes, namely f(Vi) ⊆ Vi. Theproof is equally easy and is left to the interested reader ; use the above proof as a sketch.

Notice that no notion of orthogonality was involved in the proof using complete flags ; to recover theorthonormal basis, just take v1 ∈ V1 \ 0 and build an orthonormal basis of V by picking vi ∈ Vi \ Vi−1

orthogonal to the previously chosen basis vectors v1, · · · , vi−1 (this can be done via the Gram-Schmidtalgorithm). The fact that f stabilizes the flag ensures that the corresponding matrix is upper triangular.

Theorem 2.6. (Spectral theorem) Let A be an n× n complex matrix. Then A is normal if and only if it isunitarily diagonalizable, i.e. if and only if there exists a Schur decomposition (U, T ) where T is diagonal.

Proof. (⇐) If A = UDU † where U is unitary and D is diagonal, since D is normal, so is A:

AA∗ = (UDU †)(UDU †)† = UDU †UD†U † = UDD†U † = UD†DU † = · · · = A∗A.

(⇒) Let (U, T ) be a Schur decomposition. Since A = UTU †, we have UAU † = T . By a computationsimilar as the one proving the direction (⇐), one shows that T is normal ; since it is upper triangular, itis diagonal by Lemma 2.3.

Corollary 2.7. Let V be a complex inner product space and f : V → V a linear endomorphism such thatfor any v, w ∈ V , we have (f(v) | f(w)) = (v | w). There exists an orthonormal basis for V over which fis diagonal.

Proof. Let v1, · · · , vn be an arbitrary orthonormal basis for V and consider the matrix form A of fover this basis. The condition on f states that A is unitary since its columns are orthonormal (becausev1, · · · , vn was chosen orthonormal). Therefore A is normal, hence unitarily diagonalizable by thespectral theorem. The corresponding basis over which f is diagonal has to be orthonormal since the basechange matrix we found was unitary.

We can now turn our attention back to complex linear representations. Note that Corollary 2.7 statesthat when ρ : G → GL(V ) is a representation and (− |−) is a G-invariant inner product on V , thenρ maps each element of G to a unitary diagonalizable automorphism of V , which is why complex linearrepresentations are sometimes called unitary representations.

2.2 Row orthogonality relations

The goal of section is to develop the elementary properties of characters and show that the characterof a representation completely determines a representation and helps understand Theorem 1.27, i.e. thedecompositions of a representation in a direct sum of irreducible subrepresentations. We will prove thatif V =

⊕mi=1Wi is such a decomposition, given an irreducible representation W of G, the number of

1 ≤ i ≤ m such that Wi ' W does not depend on the chosen representation, giving a sense of uniqueness

18


of this decomposition. The main tool to acheive this, as the title of the section says, will be the roworthogonality relations (c.f. Theorem 2.19).

Definition 2.8. Given a complex linear representation ρ : G→ GL(V ), define the character of ρ by

χ : G→ C, s 7→ χ(s)def= tr (ρs) .

A character is called irreducible if it is the character of an irreducible representation. Its degree is defined

as degχdef= χ(1). (It corresponds to the degree of the representation we used to define it.)

Proposition 2.9. If χ is the character of the representation ρ : G→ GL(V ), then

(i) χ(1) = dimV = deg ρ.

(ii) For all s ∈ G, we have χ(s−1) = χ(s).

(iii) For all s, t ∈ G, we have χ(tst−1) = χ(s).

Proof. (i) We have χ(1) = tr (ρ1) = tr (idV ) = dimV = deg ρ.

(ii) Since s ∈ G has finite order, ρns = idV , hence if λ is an eigenvalue of ρs, λn = 1, e.g. |λ| = 1,which implies λ−1 = λ. The eigenvalues of ρs−1 are just λ−1 for λ an eigenvalue of ρs (becauseρs(v) = λv ⇐⇒ λ−1v = ρ−1

s (v)). It follows that

χ(s) = tr (ρs) =∑λ

λ =∑λ

λ−1 = tr (ρs−1) = χ(s−1).

(iii) This follows from properties of the trace map explained in Definition 1.31.

Proposition 2.10. Let ρi : G→ GL(V i) and χi be its character for i = 1, 2. Then

(i) The character χ of Vdef= V1 ⊕ V2 is χ1 + χ2.

(ii) The character ψ of Vdef= V1 ⊗ V2 is χ1 ⊗ χ2 = χ1χ2, where the latter is given by pointwise

multiplication (χ1χ2(s)def= χ1(s)χ2(s)).

Proof. (i) ρ1 and ρ2 are given in matrix form by matrices R1s and R

2s over some bases for V1 and V2,

and in block form, the matrix Rs of the representation ρ = ρ1 ⊕ ρ2 is given by

Rs =

R1s 0

0 R2s

=⇒ tr (Rs) = tr(R1s

)+ tr

(R2s

),

from which we deduce the result.

19

Chapter 2

(ii) Write ρ = ρ1 ⊗ ρ2, Ris is the matrix for ρi and Rs is the matrix for ρ. By Remark 1.19,

χ(s) = tr (Rs) =n∑

i1=1

m∑i2=1

r1i1i1(s)r2

i2i2(s)

=

(n∑

i1=1

r1i1i1(s)

)(m∑i2=1

r2i2i2(s)

)= tr

(R1s

)tr(R2s

)= χ1(s)χ2(s).

Theorem 2.11. Let ρ : G → GL(V ) be a representation of the finite group G and set gdef= ord(G),

ndef= dimV . The endomorphisms ρs are unitarily diagonalizable. The character χ(s) is a sum of n roots

of unity of order g, i.e. complex numbers ω satisfying ωg = 1. In particular, |χ(s)| ≤ n and χ(s) = n holdsif and only if ρs = idV .

Proof. By Remark 1.14, we can choose a G-invariant inner product on V and a basis e1, · · · , en whichis orthonormal with respect to (− | −). The columns of ρs in its matrix form over this basis areρs(e1), · · · , ρs(en), and they are orthogonal under (− | −) by definition since this inner product hasbeen taken G-invariant and our basis is orthonormal. By Corollary 2.7, ρs is unitarily diagonalizable.

Since s ∈ G has finite order, then ρord(s)s = idV . The eigenvalues of ρs are therefore ord(s)th roots of

unity, and summing them up shows that χ(s) is a sum of n roots of unity. Since ord(s) divides g byLagrange’s theorem, the eigenvalues of ρs are gth roots of unity.Write χ(s) = ω1 + · · ·+ ωn where ωi is a gth root of unity. Then by the triangle inequality,

|χ(s)| = |ω1 + · · ·+ ωn| ≤ |ω1|+ · · ·+ |ωn| = 1 + · · ·+ 1 = n.

The statement χ(s) = n is equivalent to ωi = 1 for 1 ≤ i ≤ n (because χ(s) = n implies Re(χ(s)) = n),which means ρs = idV .

Proposition 2.12. Let ρ : G→ GL(V ) be a representation and χ its character. Write χ2σ for the character

of Sym2(V ) and χ2α be the character of Alt2(V ) (c.f. Example 1.21). Then

χ2σ(s) =

1

2

(χ(s)2 + χ(s2)

), χ2

α(s) =1

2

(χ(s)2 − χ(s2)

).

Proof. Let λ1, · · · , λn be the eigenvalues of ρs, so that the eigenvalues of ρs2 = ρ2s are λ2

1, · · · , λ2n.

Choose a basis e1, · · · , en of V which consists of eigenvectors for ρs ; this is always possible byTheorem 2.11. Recall Example 1.21 where we have shown, using the linear endomorphism θ of V ⊗ Vgiven by θ(v ⊗ w) = w ⊗ v, that V ⊗ V = Sym2(V )⊕Alt2(V ). The sets

ei ⊗ ej + ej ⊗ ei | 1 ≤ i ≤ j ≤ n ⊆ Sym2(V ), ei ⊗ ej − ej ⊗ ei | 1 ≤ i < j ≤ n ⊆ Alt2(V )

are obviously linearly independent, and since dimV = n2 = n(n+1)2 + n(n−1)

2 , we see that they generatetheir respective subspaces in which we wrote containment. It follows that they form bases of Sym2(V )and Alt2(V ), respectively. Now

(ρs ⊗ ρs)(ei ⊗ ej) = ρs(ei)⊗ ρs(ej) = (λiei)⊗ (λjej) = λiλj(ei ⊗ ej),

so that ρs ⊗ ρs is a diagonal matrix over the bases given for Sym2(V ) and Alt2(V ) ; in both cases, the

20


eigenvalue corresponding to ei ⊗ ej ± ej ⊗ ei is λiλj . Since

2

∑1≤i≤j≤n

λiλj

−( n∑i=1

λ2i

)=

(n∑i=1

λi

)2

= 2

∑1≤i<j≤n

λiλj

+

(n∑i=1

λ2i

),

we have

χ2σ(s) =

∑1≤i≤j≤n

λiλj =1

2

( n∑i=1

λi

)2

+

(n∑i=1

λ2i

) =χ(s)2 + χ(s2)

2

and

χ2α(s) =

∑1≤i<j≤n

λiλj =1

2

( n∑i=1

λi

)2

−

(n∑i=1

λ2i

) =χ(s)2 − χ(s2)

2.

Theorem 2.13. Let ρ : G→ GL(V ) be a representation of a finite group G. Set

V G def= v ∈ V | ∀t ∈ G, ρt(v) = v.

Set gdef= ord(G) and consider

fdef=

1

g

∑t∈G

ρt ∈ HomC(V, V ).

The map f : V → V is a G-linear projection of V onto V G, meaning that V G is a subrepresentation of Vequal to the sum of all trivial subrepresentations of V .

Proof. By definition, if W ≤ V is a trivial subrepresentation, then W ≤ V G. We know that f is G-linearsince

ρt f = ρt

(1

g

∑s∈G

ρs

)=

1

g

∑s∈G

ρts = f =1

g

∑s∈G

ρst =

(1

g

∑s∈G

ρs

) ρt = f ρt.

It follows that im f ⊆ V G. Conversely, if v ∈ V G, ρs(v) = v for all s ∈ G, hence f(v) = v ; thereforeim f = V G. Since f(v) ∈ V G, f2 = f , so f is a projection onto V G.

Corollary 2.14. Let V be a representation of the finite group G and set gdef= ord(G). Under a decompo-

sition of V as a direct sum of irreducible representations, the number of copies of the trivial representationin V in this decomposition equals dimV G. Let χ be the character of V . Then

dimV G = tr (f) =1

g

∑t∈G

tr (ρt) =1

g

∑t∈G

χ(t).

Definition 2.15. Let ρV : G → GL(V ) and ρW : G → GL(W ) be two representations of the finite groupG. We turn the complex vector space HomC(V,W ) into a representation ρ : G → GL(HomC(V,W )) asfollows: given f : V →W and s ∈ G,

ρs(f)def= ρWs f (ρVs )−1.

In particular, when ρ : G → GL(V ) is a representation of G and W = C is the trivial representation,

we endow V ∗def= HomC(V,C) with this representation ρ∗ : G → GL(V ∗), so that for f ∈ V ∗, ρ∗s(f)

def=

f (ρVs )−1 ; we call it the dual representation .

Proposition 2.16. Let V,W be two representations of the group G.

21

Chapter 2

(i) The natural isomorphism Φ : V ∗ ⊗ W ' HomC(V,W ) of Definition 1.31 is an isomorphism ofrepresentations.

(ii) If χV is the character of V , then χV ∗ = χV .

(iii) If χHom is the character of HomC(V,W ), then χHom = χV χW .

Proof. (i) Let s ∈ G, v ∈ V , ϕ ∈ V ∗ and w ∈W . Then

Φ(s · (ϕ⊗w))(v) = Φ((s ·ϕ)⊗ (s ·w))(v) = ϕ(s−1 · v)︸︷︷︸∈C

(s ·w) = s ·Φ(ϕ⊗w)(s−1 ·v) = (s ·Φ(ϕ⊗w))(v).

(ii) If v1, · · · , vn is an eigenbasis for ρVs , then the dual basis ω1, · · · , ωn is an eigenbasis for ρV∗

s

and if vi is an eigenvector of ρVs with eigenvalue λi, then ωi is an eigenvector of ρV∗

s with eigenvalueλ−1i = λi :

(s · ωi)(vj) = ωi(s−1 · vj) = ωi(λ

−1j vj) = λ−1

j δij = λ−1i δij =⇒ s · ωi = λ−1

i ωi.

Summing over all algebraic eigenvalues of ρVs , we obtain χV ∗ = χV .

(iii) Since HomC(V,W ) and V ∗⊗W are isomorphic representations, they have the same character. ByProposition 2.10 (ii) and part (ii) of this proposition,

χHom = χV ∗⊗W = χV ∗χW = χV χW .

Remark 2.17. The point of this G-action on HomC(V,W ) is that

HomC(V,W )G = f : V →W | ∀s ∈ G, ρWs f (ρVs )−1 = f= f : V →W | ∀s ∈ G, ρWs f = f ρVs = HomC[G](V,W ).

Note that theG-linear projection p : HomC(V,W )→ HomC(V,W ) onto HomC[G](V,W ) = HomC(V,W )G

of Theorem 2.13 is exactly the same map we used in the proofs of Maschke’s theorem and Corollary 1.33.Combining the above equality with Theorem 2.13, we obtain

dim HomC[G](V,W ) = dim HomC(V,W )G =1

g

∑s∈G

χHom(s) =1

g

∑s∈G

χV (s)χW (s).

If V,W are irreducible, then by Schur’s lemma,

dim HomG(V,W ) =

1 if V 'W in RepC(G)

0 otherwise.

Definition 2.18. Consider the complex vector space

Ccl[G]def= f : G→ C | ∀s, t ∈ G, f(sts−1) = f(t),

called the vector space of class functions on G. If gdef= ord(G) and α, β are class functions of G, then

(α, β)def=

1

g

∑s∈G

α(s)β(s),

where β(s)def= β(s), turns Ccl[G] into an inner product space. Note that characters are class functions by

Proposition 2.9.

22


Theorem 2.19. (Row orthogonality relations) Let V,W be representations of G.

(i) In light of Remark 2.17, if V and W are irreducible, we have

(χV , χW ) =1

g

∑t∈G

χV (t)χW (t) = dim HomC[G](V,W ) =

1 if V 'W0 otherwise.

This shows that irreducible representations are determined up to G-linear isomorphism by theircharacter ; furthermore, the irreducible characters of G are pairwise orthogonal in Ccl[G], whichproves that a finite group G admits only finitely many irreducible characters.

(ii) Write V =⊕h

i=1Wi where the Wi are irreducible subrepresentations of V . If W is irreducible, then

(χV , χW ) = dim HomC[G](V,W ) = #1 ≤ i ≤ h |Wi 'W in RepC(G).

In particular, #1 ≤ i ≤ h | Wi ' W in RepC(G) is independent of the choice of decompositionof V as a direct sum of irreducible representations.

(iii) The number (χV , χW ) is a non-negative integer, and so in particular (χV , χW ) = (χW , χV ) =(χW , χV ). This implies that the roles of V and W in (ii) can be reversed, namely, if V is irreducibleand W =

⊕hi=1W

′i is a decomposition of W into irreducible subrepresentations, then

(χV , χW ) = dim HomC[G](V,W ) = #1 ≤ i ≤ h |W ′i ' V in RepC(G)

Proof. (i) The last equality follows from Remark 2.17. By the same remark, since dimensions areintegers, we have

dim HomC[G](V,W ) = dim HomC(V,W )G

=1

g

∑s∈G

χV (s)χW (s)

=1

g

∑s∈G

χV (s)χW (s)

= (χV , χW ).

The number of irreducible characters is thus finite since Ccl[G] is finite-dimensional.

(ii) This follows from part (i) and the fact that χW =∑h

i=1 χWi . The inner product (χV , χW ) doesnot require the direct sum decomposition to be defined, so that the number of irreducible subrep-resentations isomorphic to W appearing in a decomposition is also independent of the choice ofsuch a decomposition.

(iii) Remark 2.17 implies that (χV , χW ) = dim HomC[G](V,W ), so that (χV , χW ) is (the complexconjugate of) an integer, i.e. an integer. This integer has to be non-negative since by writing V andW as a direct sum of irreducible representations, χV and χW are a sum of irreducible characters,so by part (i), it equals a sum of 1’s and 0’s. The rest of the statement follows from part (ii).

Corollary 2.20. Let V be a representation of G. Then V is irreducible if and only if (χV , χV ) = 1. Inparticular, χV is irreducible if and only if χV is irreducible.

23

Chapter 2

Proof. Let V = W1 ⊕ · · · ⊕Wh be a decomposition of V into irreducible representations. Then

(χV , χV ) =h∑

i,j=1

(χWi , χWj ) =h∑

i,j=1

δij = h.

It follows that V is irreducible if and only if (χV , χV ) = h = 1. For the last result, we note that

(χV , χV ) =1

g

∑s∈G

χV (s)χV (s) =1

g

∑s∈G

χV (s)χV (s) = (χV , χV ).

Corollary 2.21. If two representations V1, V2 of G have the same character, they are isomorphic. A characterthus determines a unique isomorphism class of representations of G.

Proof. Write Vj =⊕h

i=1Wi,j for j = 1, 2 where the Wi,j are irreducible subrepresentations of Vj . Givenan irreducible representation W , since (χV1 , χW ) = (χV2 , χW ), the number of Wi,1 isomorphic to Wand the number of Wi,2 isomorphic to W are equal. Considering all irreducible subrepresentations of V1

and V2, we see that V1 and V2 are isomorphic by summing the isomorphisms Wi,1 ' Wσ(i),2 (for somepermutation σ : 1, · · · , h → 1, · · · , h.

Corollary 2.22. Let χ1, · · · , χh be the distinct irreducible characters corresponding to the isomorphismclasses of the irreducible representations W1, · · · ,Wh in RepC(G). If V is a representation of G, then

χV =h∑i=1

(χV , χi)χi.

Proof. If we write V =⊕`

j=1W′j where the W ′j are irreducible, then χW ′j = χi(j) for a unique integer

i(j) ∈ 1, · · · , h, which allows us to re-write V '⊕`

j=1Wi(j). Therefore χV =∑`

j=1 χi(j), and gath-ering the characters of the sum which are equal into a single summand gives the result by Theorem 2.19(ii) since there are precisely (χV , χi) values of j with 1 ≤ j ≤ ` satisfying i(j) = i.

Convention 2.23. Until the end of this chapter, we are given a finite group G with gdef= ord(G) and

we denote by W1, · · · ,Wh the distinct irreducible representations of G with character χi of degree ni =

χi(1) = dimWi. The conjugacy classes of G are denoted by C1, · · · , Ck and if s ∈ Ci, we write csdef= |Ci|.

Definition 2.24. Recall (c.f. Example 1.5) that C[G] is the regular representation of G with basis ett∈Gand ρs(et)

def= est. As a set, C[G] = HomSet(G,C) is the set of complex functions on G, hence we can

equip C[G] with an inner product: for f1, f2 ∈ C[G], let

(f1, f2)def=

1

g

∑s∈G

f1(s)f2(s).

Defined this way, we see that Ccl[G] ⊆ C[G] is a subspace whose inner product is induced from that ofC[G]. Clearly, this inner product is G-invariant:

(s · f1, s · f2) =1

g

∑t∈G

(s · f1)(t)(s · f2)(t) =1

g

∑t∈G

f1(s−1t)f2(s−1t) =1

g

∑t∈G

f1(t)f2(t) = (f1, f2).

Proposition 2.25. For s, t ∈ G, we have (es, et) = 0, hence ett∈G is an orthogonal basis of C[G] ; we

24


can normalize it, i.e. √gett∈G is an orthonormal basis of C[G]. Furthermore,

χC[G](s) =

g if s = 1

0 otherwise.

Proof. For orthogonality of the basis, it suffices to notice that when s 6= t, for any u ∈ G, es(u)et(u) = 0since es(u) or et(u) equals zero. When s = t, we have

(√ges,√ges) =

√g2

g

∑t∈G

es(t)es(t) = 1.

We already know that χC[G](1) = dimC[G] = g by Proposition 2.9. If s ∈ G \ 1, then for all t ∈ G,st 6= t, thus (ρs(et), et) = (est, et) = 0, which means χC[G](s) = tr (ρs) = 0.

Theorem 2.26. Every irreducible representation Wi is contained in C[G] with multiplicity ni = dimWi =χi(1). In particular, χC[G] =

∑hi=1 niχi.

Proof. By Proposition 2.25, we have (χC[G], χi) = 1g

∑s∈G χC[G](s)χi(s) = χi(1) = ni. Use Corol-

lary 2.22.

Corollary 2.27. The degrees ni of the irreducible representations of G satisfies g =∑h

i=1 n2i . If s ∈ G\1,

then∑h

i=1 niχi(s) = 0.

Proof. Substituting s = 1 in Theorem 2.26 gives

g = dimC[G] = χC[G](1) =h∑i=1

niχi(1) =h∑i=1

n2i .

By Proposition 2.25, substituting s ∈ G \ 1 in Theorem 2.26 gives∑h

i=1 niχi(s) = χC[G](s) = 0.

2.3 Column orthogonality relations and the canonical decomposition

Lemma 2.28. Let f be a class function on G and ρ : G → GL(V ) a representation with character χ.Define Pχ : Ccl[G]→ EndC(V ) by

Pχ(f)def=

1

g

∑t∈G

f(t)ρ−1t .

If ρ is irreducible of degree n, then Pχ(f) = (f,χ)n idV . In particular, tr (Pχ(f)) = (f, χ). Note : I chose the

letter Pχ(f) for “projection onto χ” (because after taking the trace, this is the coefficient of the orthogonalprojection of f on the line in Ccl[G] spanned by χ).

Proof. The map Pχ(f) : V → V is G-linear since

ρs Pχ(f) ρ−1s =

1

g

∑t∈G

f(t)ρst−1s−1 =1

g

∑t∈G

f(sts−1)ρ(sts−1)−1 = Pχ(f).

Assume V irreducible. By Schur’s Lemma, there exists a unique λ ∈ C such that Pχ(f) = λ idV . Taking

25

Chapter 2

traces, we get

nλ = tr (λ idV ) = tr (Pχ(f)) =1

g

∑t∈G

f(t)χ(t−1) =1

g

∑t∈G

f(t)χ(t) = (f, χ).

It follows that λ = (f,χ)n .

Remark 2.29. Note that when ψ is a character of G, Pχ(ψ) can also be computed by the formula

Pχ(ψ) =1

g

∑t∈G

ψ(t)ρ−1t =

1

g

∑t∈G

ψ(t−1)ρt =1

g

∑t∈G

ψ(t)ρt.

This formula is perhaps easier to compute (instead of computing the inverse of a group element or of thecorresponding matrix, we only need to compute a complex conjugate).

Remark 2.30. Let G be a finite group and ψ1, ψ2 ∈ C[G]. Compare the definitions

(ψ1, ψ2)def=

1

g

∑t∈G

ψ1(t)ψ2(t), 〈ψ1, ψ2〉Gdef=

1

g

∑t∈G

ψ1(t)ψ2(t−1).

We note that (− |−) defines a G-invariant inner product on C[G]. In the case where we work over the field

of complex numbers (which we have been doing so far), letting ψ(t)def= ψ(t−1), we have

(ψ1 |ψ2) =1

g

∑t∈G

ψ1(t)ψ2(t) =1

g

∑t∈G

ψ1(t)ψ2(t−1) =⟨ψ1, ψ2

⟩G.

The C-bilinear map 〈−,−〉G generalizes (− |−) in the sense that it adapts very well to a representationtheory over fields other than C, since for characters ψ, we have ψ = ψ ; clearly, this relationship still holdswhen ψ is a linear combination of characters with real coefficients (since ψ 7→ ψ is an R-linear involutionof C[G]), so it holds in many useful cases.

This new map 〈−,−〉G shares many properties of (− |−) ; not all of them though, since the relationshipψ = ψ only works in the R-span of the characters, not in the C-span. One sees for instance that Lemma 2.28still holds (and the proof is even shorter, so it sounds like it is closer to being an argument meant to beproved with 〈−,−〉G rather than with (− |−)!).

Remark 2.31. Let G be a finite group with irreducible representations ρ1, · · · , ρh and put each of theserepresentations in matrix form ρk = (rkij) where the matrices are all unitary, so that the relation (ρkt )

−1 =

(ρkt )† leads to rkij(t

−1) = rkji(t) for 1 ≤ i, j ≤ deg ρ, which means rkij = rkji. In particular, by Corollary 1.34,we have (

rk2i2j2

∣∣∣ rk1i1j1) =

⟨rk2i2j2 , r

k1i1j1

⟩G

=⟨rk2i2j2 , r

k1j1i1

⟩G

=1

nk1δi1i2δj1j2δk1k2

(where nk1 = deg ρk1 ) so that the functions rkij : G→ C are orthogonal in C[G]. Note that there are

h∑k=1

(deg ρk)2 = g

such functions according to Corollary 2.27, so since they are orthogonal, they actually span the space of all

functions on G. (Even better, if we normalize them by skijdef=√nkr

kij , then they become orthonormal.)

Theorem 2.32. The distinct irreducible characters χ1, · · · , χh of G form an orthonormal basis for Ccl[G].

26


Proof. We already proved that the irreducible characters are linearly independent in Ccl[G], so it sufficesto prove that they generate Ccl[G]. Let f ∈ Ccl[G] be orthogonal to all χi’s. We know that for eachirreducible representation ρ of G with character χ, Pχ(f) = 1

n(f, χ) = 0. It follows that if we constructPχ(f) starting with any character of G, after writing it as a sum of irreducible characters, we obtainPχ(f) = 0. Letting χ denote the regular representation of G, we obtain

0 = Pχ(f)(e1) =1

g

∑t∈G

f(t)ρt−1(e1) =1

g

∑t∈G

f(t)et−1 =⇒ ∀t ∈ G, f(t) = 0.

Corollary 2.33. The number of distinct irreducible representations of G is the number of conjugacy classesof G.

Proof. Let C1, · · · , Ck be the k conjugacy classes of G. If fi : G→ C is defined by

fi(s) =

1 if s ∈ Ci0 if not.

then f1, · · · , fk is a basis of Ccl[G]. It follows that k = h. (We reserve h for the number of conjugacyclasses/number of irreducible representations of G from now on.)

Theorem 2.34. (Column orthogonality relations) Recall that for s ∈ Ci where 1 ≤ i ≤ h, we defined

csdef= |Ci| and χ1, · · · , χh for the irreducible characters of G. Then

(a)∑h

i=1 χi(s)χi(s−1) =

∑hi=1 χi(s)χi(s) =

∑hi=1 |χi(s)|2 = g

cs.

(b) For t ∈ G which is not conjugate to s, we have

h∑i=1

χi(s)χi(t−1) =

h∑i=1

χi(s)χi(t) = 0.

Proof. Let fj be defined as in the proof of Corollary 2.33, and pick t ∈ Cj . By Theorem 2.32, we canwrite fj =

∑hi=1 (fj |χi)χi where

(fj |χi) =1

g

∑s∈G

fj(s)χi(s) =ctgχj(s).

It follows that for s ∈ G,

fj(s) =h∑i=1

(ctgχi(t)

)χi(s) =

ctg

h∑i=1

χi(s)χi(t)

which gives (a) and (b) by the definition of fj(s), whether we pick t conjugate to s or not.

Definition 2.35. Let χ1, · · · , χh be the characters of the irreducible representationsW1, · · · ,Wh of G withdegrees n1, · · · , nh and associated morphisms of groups ρi : G → GL(Wi). If we decompose an arbitraryrepresentation into a sum of irreducible representations, we can write mi = (χV , χi) and write V as

V =

h⊕i=1

Vi, Vi 'W⊕mii

27

Chapter 2

where the notation W⊕n denotes the direct sum of n copies of the representation W ; this can be done byproducing an arbitrary direct sum decomposition of V into irreducibles and collecting together those whoare isomorphic. This direct sum decomposition of V is called the canonical decomposition of V and theVi are called the canonical components of V .

Theorem 2.36. The canonical decomposition of the representation ρ : G → GL(V ) (of degree n withcharacter χ) does not depend on the original decomposition into irreducible representations. The G-linearprojection πi : V → V with image imπi = Vi can be computed by

πi =nig

∑t∈G

χi(t)ρ−1t =

nig

∑t∈G

χi(t)ρt.

Proof. We only need to show the first equality ; the second one follows by re-arranging the sum. By

Lemma 2.28 applied to the class function fidef= niχi, if W is an irreducible subrepresentation of V with

character χW of degree nW , we have

nig

∑t∈G

χi(t)ρ−1t |W =

1

g

∑t∈G

fi(t)ρ−1t |W = PχW (fi) =

(fi, χW )

nW=ni(χi, χW )

nW.

Letting V =⊕`

j=1W′j be a decomposition of V into irreducible subrepresentations, we see that the

G-linear map nig

∑t∈G χi(t)ρ

−1t restricted to W ′j equals zero if W ′j 6' Wi and equals idW ′j if W

′j ' Wi.

It follows that imπi equals the sum of those W ′j isomorphic to Wi, and since πi does not depend on thechosen decomposition of V into irreducible, neither does Vi = imπi.

Remark 2.37. If we write C[G] =⊕h

i=1W⊕nii in its canonical decomposition, we see that the class function

fi = niχi is the character of the canonical component associated to χi.

2.4 Character tables

We begin this section with a few results which will serve as lemmas to compute character tables, which wethen define and compute. Character tables form a major invariant of the class of finite groups. Althoughit does not characterize each group uniquely (two groups may have equivalent character tables), it is still avery strong tool and deserves being exploited ; not only can it tell groups apart very often but it also helpsunderstanding representations in general.

The last result of this section tells us how to work with irreducible representations obtained via thecomputation of the character table to explicitly decompose a representation given in matrix form into adirect sum of irreducible representations.

Definition 2.38. We begin by recalling a few concepts concerning group actions. Let G be a finite groupacting on a set X ; we call X equipped with the action G

X a G-set. A function f : X → Y betweentwo G-sets satisfying f(s · x) = s · f(x) is called a morphism of G-sets ; this gives the collection of G-setsthe structure of a category.

The orbit of a subset A ⊆ X is the set OrbG(A)def= GA = s · x | s ∈ G, x ∈ A. In particular, the

orbit of x ∈ X is the setOrbG(x)

def= Gx = s · x | s ∈ G.

Let G\X def= OrbG(x) | x ∈ X denote the set of orbits of G (in particular, if H ≤ G is a subgroup, H\G

denotes the set of right cosets of H in G and the set of orbits under the left action of H on G given by(h, g) 7→ hg. The two concepts agree, which explains why we chose this notation ; we reserve the notation

28


X/G for orbits of a right action X

G ; c.f Section 3.6 for an example of when this is useful. A subsetS ⊆ X which contains precisely one element in each orbit of the action G

X is called a system of orbitrepresentatives. The fixed point set of s ∈ G is defined by

Xs def= x ∈ X | s · x = x.

The stabilizer of x ∈ X is defined as

StabG(x)def= s ∈ G | s · x = x.

More generally, if A ⊆ X , the stabilizer of A is defined as

StabG(A)def= s ∈ G | s(A) = A.

(note that this means that the maps s(−) : A→ A is bijective, but not necessarily equal to the identity mapof A). In particular, StabG(x) = StabG(x).

Note that in general, the subgroup StabG(x) ≤ G is not a normal subgroup since by definition, fors ∈ G and A ⊆ X ,

sStabG(A)s−1 = sts−1 ∈ G | t(A) = A = t ∈ G | s−1ts(A) = (A) = StabG(s(A)).

To define the index |G : H| of a subgroup H of G, note that the map gH 7→ (gH)−1 = Hg−1 is a bijectionbetween the set of left cosets G/H and the set of right cosets H\G. The cardinality of any two of thesesets is the definition of |G : H|. By Lagrange’s theorem, when G is finite, we have |H||G : H| = |G|, hence|G : H| = |G|/|H|.

We say that a group action is transitive if G has only one orbit ; in other words, G is transitive if forevery x, y ∈ X , there exists g ∈ G such that g · x = y.

A group action G on the set X induces a natural diagonal action on X2 def= X ×X via g · (x, x′) def

=(g · x, g · x′). If we denote by ∆ : X → X ×X the map x 7→ (x, x), since g ·∆(x) = ∆(g · (x, x)), we seethat OrbG(∆(x)) = ∆(OrbG(x)). (In particular, if |X| > 1, the action G

X2 is never transitive.) Theaction of G on X is said to be doubly transitive if the action of G on X ×X \∆(X) is transitive, namelyfor every (x, y), (x′, y′) ∈ X ×X with x 6= y and x′ 6= y′, there exists g ∈ G with g · (x, y) = (x′, y′).

More generally, the action of G on X is said to be k-transitive if k ≤ |X| and the diagonal action

on Xk (given by g · (x1, · · · , xk)def= (g · x1, · · · , g · xk)) has the minimal possible number of orbits for an

action on X (Xk has to have some minimal number of orbits ; for example, when k = 2, those are ∆(X)and X2 \ ∆(X), and the minimal number of orbits is always acheived by the permutation group of X ).Phrased differently, G

X is k-transitive if for any 1 ≤ ` ≤ k, any two ordered `-tuples (x1, · · · , x`) and(y1, · · · , y`) of distinct elements of X (i.e. where xi 6= xj and yi 6= yj for 1 ≤ i < j ≤ `) can be mappedonto one another via multiplication by some s ∈ G, namely s · (x1, · · · , x`) = (y1, · · · , y`). It follows thata k-transitive action is also `-transitive for any 1 ≤ ` ≤ k.

Theorem 2.39. (Orbit-Stabilizer Theorem) Let the (not necessarily finite) groupG act on the (not necessarilyfinite) setX and let x ∈ X . Then there is a natural bijection of sets G/StabG(x) ' OrbG(x) (in the categoryof G-sets), which implies |G : StabG(x)| = |OrbG(x)|. In particular, If G is a finite group of order g, then|StabG(x)||OrbG(x)| = g.

Proof. Consider the map ϕ : G → OrbG(x) defined by s 7→ s · x. This map is surjective by definition.If s, t ∈ G satisfy s · x = t · x, then s−1t ∈ StabG(x), so the map ϕ : G/StabG(x) → OrbG(x) definedby the same formula defines a transitive group action G/StabG(x)

OrbG(x), which means that ϕ is abijection, namely |G : StabG(x)| = |G/StabG(x)| = |OrbG(x)|. For naturality, it suffices to see that ϕ isa morphism of G-sets, which is equivalent to the identity s · (t · x) = (st) · x. When G is finite, we have

29

Chapter 2

|G : StabG(x)| = |G|/|StabG(x)| = g/|StabG(x)|, which proves the result.

Lemma 2.40. (Burnside’s Lemma) Let the finite group G of order g act on the finite set X . Then

|G\X| = 1

g

∑s∈G|Xs|.

Proof. We have ∑s∈G|Xs| = |(s, x) ∈ G×X | s · x = x| =

∑x∈X|StabG(x)|.

By the Orbit-Stabilizer theorem, we have |StabG(x)||OrbG(x)| = g, hence by writing X as the disjointunion of its orbits, we have∑

x∈X|StabG(x)| =

∑x∈X

g

|OrbG(x)|

= g∑x∈X

1

|OrbG(x)|

= g∑

OrbG(x)∈G\X

∑y∈OrbG(x)

1

|OrbG(x)|

= g

∑OrbG(x)∈G\X

1

= g|G\X|.

Proposition 2.41. If V is a permutation representation with character χ associated to the action of the finitegroup G on a finite set X , then χ(s) = |Xs| is the number of elements in X fixed by s ∈ G.

Proof. We have ρ : G → GL(V ) where V is the free vector space on the set X with basis exx∈X ands ∈ G satisfies ρs(ex) = es·x for all x ∈ X . Computing the trace over this basis, it follows that x ∈ Xcontributes to either 0 or 1 to χ(s), depending on ex = es·x or not. Summing over all x ∈ X gives theresult.

Proposition 2.42. Let the finite group G act on the finite set X . Consider the corresponding permutationrepresentation V with character χ and denote the trivial representation of G by χ1. Then |G\X| = (χ, χ1)and there exists a unique character θ of G such that θ + |G\X|χ1 = χ and (θ, χ1) = 0. This character isequal to zero (i.e. the trace of the zero representation) if and only if G acts trivially on X (i.e. s · x = x forall (s, x) ∈ G×X ). This character is called the standard representation associated to χ.

Proof. This follows directly from Burnside’s Lemma since χ(s) = |Xs|, hence

(χ, χ1) =1

g

∑s∈G

χ(s)χ1(s) =1

ord(G)

∑s∈G|Xs| = |G\X|

The character θdef= χ− |G\X|χ1 is then well-defined. Writing χ as a sum of irreducible characters, we

see that θ(1) 6= 0 if and ony if |G\X| < |X|, which completes the proof.

Remark 2.43. There is nothing wrong in the unique morphism of groups ρ : G → GL(0) where allmorphisms have zero trace according to Definition 1.31 since the composition HomC(V, V ) ' V ∗ ⊗ V ev→ C

30


defines the trace map on V = 0 as the zero map 0 = V → C. If one is uneasy with dealing with astatement such as “the trace of the zero representation” , one can interpret the statement of Proposition 2.42differently. Namely, one can consider θ as a class function in full generality, and it will become the characterof a positive-dimensional vector space if and only if the action G X is non-trivial, i.e. if there exists atleast one pair (s, x) ∈ G×X such that s · x 6= x.

However, there is a theoretical interest in considering the zero representation as a representation since itturns the set of isomorphism classes of representations of G over C into a commutative monoid under theoperation of direct sum ; the zero representation is the identity of this monoid since 0 ⊕ V ' V ' V ⊕ 0.By Corollary 2.21 and Proposition 2.10, this is isomorphic to the monoid of characters of G under addition.By Theorem 2.32, if RepC(G)/ ∼ denotes this monoid (the / ∼ stands for “isomorphism classes of”) andGr denotes the operation of taking the Grothendieck group of a commutative monoid, then

C⊗ZGr(RepC(G)/ ∼) ' Ccl[G].

(InterpretingRepC(G)/ ∼ as the N-span of the irreducible characters sitting inside Ccl[G], its Grothendieckgroup corresponds to the Z-span of the irreducible characters, making the result trivial.)

Lemma 2.44. Let G

X and let χ be the character corresponding to the associated permutation represen-tation. If G acts diagonally on Xk (c.f. Definition 2.38), then

1

g

∑s∈G

χ(s)k = |G\Xk|.

Proof. If s ∈ G has |Xs| fixed points in X , the corresponding diagonal action G Xk is such that|(Xk)s| = |Xs|k ; this is because (x1, · · · , xk) ∈ (Xk)s if and only if x1, · · · , xk ∈ Xs. Since χ(s) =|Xs| by Proposition 2.41, the result follows by applying Burnside’s Lemma to the diagonal action G

Xk.

Remark 2.45. Lemma 2.44 also follows from the fact that χk is the character of the kth power of χ and thepermutation representation associated toG

Xk happens to be the k-fold tensor product of the permutationrepresentation associated to G

X .

Theorem 2.46. Let X be a finite set with |X| > 1 and let the finite group G act transitively on X (so

that |G\X| = 1). Denote the associated representation by V and its character by χ. Set θdef= χ − χ1 (c.f.

Proposition 2.42). Then the following are equivalent :

(i) G acts doubly transitively on X .

(ii) |G\X2| = 2, i.e. the diagonal action G X2 has the two orbits ∆(X) and X2 \∆(X)

(iii) (χ2, 1) = 2.

(iv) The representation θ is irreducible.

Proof. Statements (i) and (ii) are trivially equivalent since ∆(X) and X2 \ ∆(X) have to be a disjointunion of orbits of the diagonal action G

X2. As for (iii), since χ(s) = |Xs|, it suffices to see that byLemma 2.44,

(χ2, 1) =1

g

∑s∈G

χ(s)2 =1

g

∑s∈G|Xs|2 = |G\X2|.

which proves that (ii) and (iii) are equivalent.For (iv), note that θ is irreducible if and only if (θ, θ) = 1, so we compute : since χ = χ and χ1 = χ1, we

31

Chapter 2

have(θ, θ) = (χ− χ1, χ− χ1) = (χ, χ)− 2(χ, χ1) + (χ1, χ1) = (χ2, 1)− 1.

(Note that by Proposition 2.42, (χ, χ1) = 1 by assumption that G is transitive.) Therefore (θ, θ) = 1 ifand only if (χ2, 1) = 2, i.e. (iii) and (iv) are equivalent.

Corollary 2.47. Let n ≥ 1 be an integer, and consider the standard action of Sn on Xdef= 1, · · · , n.

Denote its character by χ and the trivial representation of Sn by χ1. Then the standard representation

θdef= χ− χ1 is irreducible.

Proof. This follows since Sn acts doubly transitively on 1, · · · , n ; the action Sn

X has only oneorbit since it is transitive, hence |Sn\X| = 1, so that the standard representation χ−χ1 indeed equals θ.(In fact, Sn even acts n-transitively on 1, · · · , n by its very definition since given any two n-tuples(x1, · · · , xn) and (y1, · · · , yn) (i.e. two orderings of the elements of 1, · · · , n), there is a uniquepermutation of 1, · · · , n which maps xi to yi for all i = 1, · · · , n (and it is defined by this formula).)

We now have enough information to begin computing a few character tables, a notion which we nowdefine.

Definition 2.48. Let G be a finite group. Assume the irreducible characters of G are given by χ1, · · · , χhand the conjugacy classes of G are denoted by C1, · · · , Ch. A character table of G is an h × h matrixwhose (i, j)th entry equals χi(s) for some s ∈ Cj (i.e. any s ∈ Cj ). When we want to omit defining s ∈ Cjexplicitly, we write χi(Cj)

def= χi(s). By re-indexing the characters and the conjugacy classes, one obtains

different character tables ; the set of all such possible character tables is denoted by Ch(G). When we areonly interested in listing the irreducible characters of G, we write them down in a table (see Example 2.51for instance) and speak of “the” character table of G.

Remark 2.49. If ϕ is an automorphism of G and ρ : G → GL(V ) is a representation of G, then ρ ϕ :G → GL(V ) is also a representation. This induces a right action of the automorphism group Aut(G) on

RepC(G) via ρϕdef= ρ ϕ (right actions are often denoted using exponential notation in order to set them

on the right of the element they act on). Note that if ρ is irreducible, so is ρϕ ; one can even see thatgiven v ∈ V \ 0, the vector space V is the C-span of ρs(v) | s ∈ G and this set of generators remainsunchanged if we replace ρ by ρϕ (it only permutes that set).

If χ is the corresponding character of the representation ρ, then the character of ρϕ equals χϕdef= χϕ :

χϕ(s) = tr (ρϕs ) = tr(ρϕ(s)

)= χ(ϕ(s)),

hence Aut(G) acts on the right on the set of irreducible characters of G ; since inner automorphisms of

G act trivially on characters, this induces a right action of the outer automorphism group Out(G)def=

Aut(G)/Inn(G) on the set of irreducible characters of G, and therefore on Ch(G) via the formula

(χi(Cj))ϕdef= (χϕi (Cj)).

This action can be helpful to produce new irreducible characters out of already known ones ; in particular,note that abelian groups G (written additively) admit the outer automorphism s 7→ −s (abelian groups areprecisely those groups for which Inn(G) is trivial), and this outer automorphism is non-trivial unless G isan abelian elementary 2-group (every non-zero element has order 2).

Proposition 2.50. Let G be a finite group. Assume χ, χ′ are two irreducible characters of G wheredegχ = 1. Then χχ′ is an irreducible character of G. In particular, the irreducible characters of degree 1form a group under multiplication which acts on the set of irreducible characters of G. (When G is abelian,this group is equal to G, the Pontryagin dual group of G.)

32


Proof. Recall that a character of degree n takes values which are sums of nth roots of unity by Theo-rem 2.11 ; in particular, χ takes values in the set of roots of unity, so for any s ∈ G, |χ(s)| = 1. Therefore,by Corollary 2.20, we see that χχ′ is irreducible since

(χχ′, χχ′) =1

g

∑s∈G|χχ′(s)|2 =

1

g

∑s∈G|χ(s)|2|χ′(s)|2 =

1

g

∑s∈G|χ′(s)|2 = (χ′, χ′) = 1.

Example 2.51. Consider the group S3. One can easily see that S3 admits 3 conjugacy classes, namely theidentity permutation, the even permutations and the odd ones. Therefore, S3 admits precisely 3 irreduciblecharacters. We already know three irreducible representations of S3, namely the trivial and the sign repre-sentation (both of degree 1, hence irreducible) and the standard representation of S3, which is irreducibleof degree 2.

We can use this information to build a character table of S3 : indicating each conjugacy class byrepresenting it via one of its elements, we obtain

cs 1 3 2S3 (1) (12) (123)

trivial 1 1 1

sign 1 −1 1

standard 2 0 −1

Example 2.52. Consider G = S4. Recall that in general, for the group Sn, the conjugacy class of apermutation is completely determined by its cycle structure. More precisely, one can show that the conjugacyclasses are in one-to-one correspondence with partitions of the integer n. This is better understood fromthe following identity : let (a1 a2 · · · am) denote an m-cycle in Sn and σ ∈ Sn. Then

σ(a1 a2 · · · am)σ−1 = (σ(a1)σ(a2) · · · σ(am))

(This could have also quickly determined the conjugacy classes of S3 in Example 2.51.) It follows from thisthat the conjugacy class of a permutation depends only on the length of its disjoint cycles (this is what wecall the cycle structure of a permutation). For instance, we have the following partitions of the integer 4(we write the cycles very explicitly to explain the correspondence): letting C(σ) denote the conjugacy classof the permutation σ ∈ S4, we have

1 + 1 + 1 + 1 ←→ C((1)(2)(3)(4))2 + 1 + 1 ←→ C((12)(3)(4))

3 + 1 ←→ C((123)(1))4 ←→ C((1234))

2 + 2 ←→ C((12)(34))

We can now begin to build a character table. We indicate the number of elements in each conjugacy class ina row above to help in computations. Using Proposition 2.41 and Corollary 2.47, we obtain the first, secondand third irreducible representations as the trivial, sign and standard representations of G, which gives thefollowing:

cs 1 6 8 6 3S4 (1) (12) (123) (1234) (12)(34)

χ1, trivial 1 1 1 1 1

χ2, sign 1 −1 1 −1 1

χ3, standard 3 1 0 −1 −1

χ4

χ5

33

Chapter 2

One sees by Proposition 2.50 that χ4def= χ2χ3 are irreducible. Since there is only one representation

left, we can use the fact that g = 4! = 24 is the sum of the squares of the degrees of χ1, · · · , χ5 (c.f.Corollary 2.27) and the column orthogonality relations (c.f. Theorem 2.34) to fill in the last row. We get thefull table :

cs 1 6 8 6 3S4 (1) (12) (123) (1234) (12)(34)

χ1, trivial 1 1 1 1 1

χ2, sign 1 −1 1 −1 1

χ3, standard 3 1 0 −1 −1

χ4 3 −1 0 1 −1

χ5 2 0 −1 0 2

2.5 Characters of abelian groups

Theorem 2.53. Let G be a finite group of order g. Then G is abelian ⇐⇒ all its irreducible charactershave degree 1.

Proof. Let n1, · · · , nh the degrees of the irreducible representations. Recall that h is the number ofirreducible representations of G and the number of conjugacy classes of G. By Corollary 2.27,

∑hi=1 n

2i =

g, so if ni = 1 for 1 ≤ i ≤ h, this means h = g ; in particular, G has g conjugacy classes, so it must beabelian.Conversely, if G is abelian, h = g, hence

∑hi=1 n

2i = g implies that ni = 1 for each 1 ≤ i ≤ h = g.

Corollary 2.54. Assume A ≤ G is an abelian subgroup. Then every irreducible representation of G hasdegree at most |G : A| = ord(G)

ord(A) .

Proof. Let ρ : G → GL(V ) be an irreducible representation of G, hence giving a representation ρAdef=

ρ|A of the subgroup A. Take an irreducible subrepresentationW of ρA : A→ GL(V ), so that dimW = 1by Theorem 2.53. Put

V ′def=∑s∈G

ρs(W ) ⊆ V,

Since V ′ 6= 0 is G-stable and V is an irreducible representation of G, we must have V ′ = V . For s ∈ Gand t ∈ A, we have

ρst(W ) = ρs(ρt(W )) = ρs(W ),

hence the image of ρst depends only on which coset of A the element st is in G. If we write W = 〈w〉Cand take a set of representatives R for the left cosets of G/A, then V is generated by ρs(w) | s ∈ R,hence dimV ≤ |R| = |G : A|.

Remark 2.55. In the previous proof, we could be tempted to write H = g ∈ G | ρg(W ) = W withA ≤ H ≤ G, take a coset of representatives R for G/H and conclude that since V is spanned by ρr(w) |r ∈ R, that dimV = |R| = |G : H| (which divides |G : A|) and get the stronger result “the degree divides|G : A|”. First of all, this is false since there are counter-examples ; any non-abelian group of order pq wherep, q are distinct primes is a counter-example because they have irreducible representations of degree > 1 byTheorem 2.53, but dividing p and q implies being equal to 1). Second of all, the flaw in the proof is whenwe assume dimV = |R| ; the spaces ρr(W ) are pairwise linearly independent, but not necessarily linearlyindependent. So this argument only works when we can be sure that |G : H| = 2, e.g. when |G : A| = 2 ;but in this case we don’t learn anything new (because dimV dividing |G : A| and dimV ≤ |G : A| are

34


equivalent), so it is not interesting.

Example 2.56. The dihedral group Dn contains an abelian group of index 2 (the cyclic subgroup of ordern), hence all irreducible representations have degree ≤ 2. We will compute the character table of Dn inExample 2.71.

2.6 Character table of a product of two groups

Recall that given two finite groups G1, G2 of order g1 and g2 respectively, one can form their (external) direct

product G1 × G2def= (s1, s2) | si ∈ Gi with the operation (s1, s2)(t1, t2)

def= (s1t1, s2t2). The group

Gdef= G1 × G2 has order g

def= g1g2, we have canonical inclusions Gi −→ G1 × G2 given by s1 7→ (s1, 1)

and s2 7→ (1, s2). If we identify Gi with its image under this inclusion, then the elements of G1 commutewith the elements of G2.

Conversely, given a group G which contains two subgroups G1, G2 satisfying G1G2 = G and G1∩G2 =1 where the elements of G1 commute with the elements of G2, then G ' G1×G2 ; this is because in thiscase, each element of G can be written uniquely in the form s = s1s2 where si ∈ Gi. (In this case, we saythat G is the (internal) direct product of G1 and G2.) Using the results of this section, when we recognizea group G as the internal direct product of two subgroups G1 and G2, we can deduce the character table ofG by the character tables of G1 and G2.

Theorem 2.57. Let G = G1 × G2 and ρi : Gi → GL(Vi) be two representations (i = 1, 2) with characterχi. The box product representation ρ1 ρ2 has character χ1 χ2, called the box product of χ1 and χ2,which is given by

(χ1 χ2)(s1, s2) = χ1(s1)χ2(s2).

The representation ρ1 ρ2 is irreducible if and only if ρ1 and ρ2 are irreducible. Each irreducible represen-tation of G is of the form ρ1 ρ2 for some irreducible representations of G1 and G2. (c.f. Definition 1.22).

Proof. The formula for the character χ is computed similarly as the character of the tensor product of tworepresentations over G : fix a basis of Vi over which ρi is diagonal (c.f. Theorem 2.11), say vi1, · · · , vini.Then v1

j1 v2

j2| 1 ≤ ji ≤ ni, i = 1, 2 is a basis of V1 V2 and ρi has eigenvalue λij at v

iji. We have

(ρ1 ρ2)(v1j1 v

2j2) = ρ1(v1

j1) ρ2(vj2) = λ1j1λ

2j2(v1

j1 v2j2),

so that ρ is diagonalized over this basis. Summing all the eigenvalues shows that χ1 χ2 is indeed thecharacter of ρ1 ρ2.The second statement holds because

(χ1 χ2, χ1 χ2) =1

g

∑s1∈G1

∑s2∈G2

|(χ1 χ2)(s1, s2)|2

=1

g1g2

∑s1∈G1

∑s2∈G2

|χ1(s1)|2|χ2(s2)|2

=

1

g1

∑s1∈G1

|χ1(s1)|2 1

g2

∑s2∈G2

|χ2(s2)|2

= (χ1, χ1)(χ2, χ2).

For the last statement, it suffices to show that if f ∈ Ccl[G1 × G2] is orthogonal to all characters of theform χ1

i χ2j where χ

1i (resp. χ

2j ) is an irreducible representation of G1 (resp. G2), then f = 0. We have

(f, χ1i χ

2j ) =

1

g

∑s1∈G1

s2∈G2

f(s1, s2)χ1i (s1)χ2

j (s2) = 0.

35

Chapter 2

Define f ′i : G2 → C by

f ′i(s2)def=

1

g1

∑s1∈G1

f(s1, s2)χ1i (s1) ∈ Ccl[G2].

Since f is a class function, f ′i is also ; since the χ2j ’s form a basis of Ccl[G2] and

1

g2

∑s2∈G2

f ′i(s2)χ2j (s2) = (f, χ1

i χ2j ) = 0,

we obtain f ′i = 0 ; since the χ1i ’s form a basis of Ccl[G1], we have

1

g

∑s1∈G1

f(s1, s2)χ1i (s1) = f ′i(s2) = 0 =⇒ ∀(s1, s2) ∈ G1 ×G2, f(s1, s2) = 0.

Definition 2.58. Let A ∈ Matm×m(C) and B ∈ Matn×n(C). The Kronecker product of A and B is thematrix A⊗B ∈ Matmn×mn(C) defined by

A⊗B def=

a11B · · · a1mB...

. . ....

am1B · · · ammB

More explicitly in terms of coefficients, for 1 ≤ i, k ≤ m and 1 ≤ j, ` ≤ n, we have

(A⊗B)((i−1)n+j)((k−1)n+`) = aikbj`.

Proposition 2.59. Let G1 (resp. G2) be a finite group with character table A1 = (χ1i (C1

k)) (resp. A2 =(χ2j (C2

` ))). Then A1 ⊗A2 is a character table of G1 ×G2.

Proof. First note that the conjugacy classes of G1×G2 are of the form C1k×C2

` since for (s1, s2) ∈ G1×G2

and (t1, t2) ∈ C1k × C2

` ,

(s1, s2)(t1, t2)(s1, s2)−1 = (s1t1s−11 , s2t2s

−12 ) ∈ C1

k × C2` .

It follows that

(A1 ⊗A2)((i−1)n+j)((k−1)n+`) = χ1i (C1

k)χ2j (C2

` ) = (χ1i ⊗ χ2

j )(C1k × C2

` ).

The rows of A1 ⊗A2 form all the irreducible characters of G by Theorem 2.57, so we are done.

2.7 Lifted and dropped representations

This section is meant to put into correspondence the representations of G/K with a subset of those of Gwhen K E G is a normal subgroup.

Definition 2.60. Let K E G be a normal subgroup of the finite group G. Given a representation ρ :

G/K → GL(V ), the lift of ρ to G is the representation L(ρ)def= ρ πK : G → GL(V ) defined by

L(ρ)sdef= ρsK . If χ is the character of ρ, we write L(χ) for the character of L(ρ).

Conversely, if η : G → GL(V ) is a representation of G such that for all s ∈ K , we have ηs = idV , weobtain a morphism of groups D(η) : G/K → GL(V ) by passing to the quotient, i.e. a representation ofG/K called the drop of η along K . Analogously, if ψ is the character of η, we write D(ψ) for the characterof D(η).

36


Theorem 2.61. The lift and drop are inverse constructions of each other in the following sense : the mapsL(−) and D(−) are bijections between the sets

RepC(G/K)←→ ρ ∈ RepC(G) | ker ρ ⊇ K.

In other words, D(L(ρ)) = ρ and L(D(η)) = η. Let ρ be a representation of G/K with character χ and ηbe a representation of G with character ψ. For s ∈ G, we have the formulas

L(χ)(s) = χ(sK), D(ψ)(sK) = ψ(s).

Furthermore, L(−) and D(−) restrict to a bijection between the irreducible representations in each set.

Proof. The bijectivity is given by the quotient map correspondence for groups. As for characters, we areevaluating the trace of the same matrices, so it remains unchanged. For irreducibility, if ρ : G/K →GL(V ) is a representation, then V is irreducible if and only if for any v ∈ V \ 0, the set

ρsK(v) | sK ∈ G/K

spans V ; this set is also equal to L(ρ)s(v) | s ∈ G since for t ∈ K , L(ρ)t(v) = ρtK(v) = ρK(v) = v.Since those two sets of generators are always equal, we deduce that ρ is irreducible if and only if L(ρ) is,which is what was required to prove.

Example 2.62. (Products of groups) Let G = G1 × G2 be two groups. One sees that the subgroupsG1, G2 E G are normal, so if ρ1 (resp. ρ2) is a representation of G1 (resp. G2), then

ρ1 ρ2 ' L(ρ1)⊗ L(ρ2)

(note that the first tensor product is the one between representations of G1 and G2, while the second oneis between representations of G since L(ρi) is lifted from Gi to G). In other words, one can produce allirreducible representations of G by lifting those of G1 and G2, and then tensoring them together (had wenot known that we can tensor them together without lifting to begin with).

Remark 2.63. It is easy to detect irreducible lifted representations via the character table when one knowsthe normal subgroups of a group. Normal subgroups are unions of conjugacy classes, so if a character inthe character table χ is such that χ is constant on a union of conjugacy classes forming a subgroup (whichis then normal) and on its cosets, then χ is lifted from D(χ) via that subgroup.

Theorem 2.64. Let G be a finite group of order g with center C of order c. The degrees of the irreduciblerepresentations χ1, · · · , χh of G divide g/c = |G : C|.

Proof. Let ρ : G → GL(V ) be an irreducible representation of degree n. If s ∈ C , then ρs : V → Vcommutes with the action of G by definition of the center, thus ρs = λsidV by Schur’s Lemma. Note thatρ|C : C → GL(V ) is a representation, thus the function s 7→ λs is a morphism of groups C → C∗.Let m ≥ 1 be an integer, set V m

def= V · · · V︸︷︷︸

m times

and form the box product representation

ρmdef= ρ · · · ρ︸︷︷︸

m times

: Gm → GL(V m)

This is an irreducible representation of Gm by Theorem 2.57. The subgroup H ≤ Gm of those(s1, · · · , sm) ∈ Cm satisfying s1s2 · · · sm = 1 lies in the kernel of ρm, thus we can drop it

D(ρm) : Gm/H → GL(V m). Since hdef= ord(H) = cm−1 (because (s1, · · · , sm) ∈ H if and

only if sm = s−1m−1 · · · s

−11 ), by Corollary 3.20, we see that

deg D(ρm) = dimV m = nm | ord(Gm/H) = gm/cm−1,

37

Chapter 2

we see that (g/cn)m lies in the Z-submodule of Q generated by 1c . This holds for all m ≥ 1, so if we

write (g/cn)m = p/q where p, q ∈ Z are coprime integers and q > 0, this implies that qm divides c forall m ≥ 1, e.g. q = 1.(Remark : according to [1], this proof is due to John T. Tate.)

2.8 Induced representations

Given a finite group G and a subgroup H ≤ G, one easily produces representations of H by composing

the morphism of groups ρ : G → GL(V ) with the inclusion map ι : H → G, so that ResGH(ρ)def=

ρ|H : H → GL(V ). The goal of this section is to show that this construction can be reversed, namelygiven a representation θ : H → GL(W ), one can induce a representation IndGH(θ) : G → GL(V ). Wewill illustrate this induction process by formulating it in many different ways and deduce its properties insubsequent sections.

Definition 2.65. Let ρ : G → GL(V ) be a representation of G, H ≤ G a subgroup and ResGH(ρ) : H →GL(V ) is its restriction to the subgroup H ≤ G, i.e. ResGH(ρ)

def= ρ|H . Let W be a subrepresentation of

ResGH(ρ) denoted by θ : H → GL(W ) (θ is again defined by “restriction”, but not at the same place ; it isthe restriction of ρs ∈ GL(V ) to W for each s ∈ H , which is possible because we assume W is H-stable,although not necessarily G-stable). For s ∈ G, the vector subspace ρs(W ) ⊆ V only depends on the leftcoset in which s is, because if s ∈ G, t ∈ H , we have

ρst(W ) = ρs(ρt(W )) = ρs(θt(W )) = ρs(W ).

For σ = sH ∈ G/H a left coset in G/H , we define Wσdef= ρs(W ) (the last argument shows that this is

well-defined, i.e. does not depend on the choice of representative s for σ = sH ). We have∑

σ∈G/HWσ ⊆ V .

We say that ρ : G→ GL(V ) is induced by θ : H → GL(W ) if as vector spaces, we have

V =⊕

σ∈G/H

Wσ.

Equivalently, each x ∈ V can be written uniquely in the form

x =∑

σ∈G/H

xσ, xσ ∈Wσ.

Another reformulation goes as follows. A section of the projection map πH : G → G/H is a map φ :G/H → G such that πH φ = idG/H ; the data of such a map is equivalent to a set R, called a system ofleft coset representatives of G/H , such that each σ ∈ G/H satisfies σ = rH for a unique r ∈ R. Then asvector spaces, ρ : G→ GL(V ) is induced by θ : H → GL(W ) if and only if

V =⊕r∈R

ρr(W ).

In particular, we note that when V is induced by W , we have dimV = |G : H|dimW . It is also clearthat the notion of being induced is independent of the choice of system of left coset representatives R (sinceWσ = ρr(W ) does not depend on the choice of the group element r representing σ ∈ G/H ).

Remark 2.66. Suppose ρ : G → GL(V ) is induced by θ : H → GL(W ) where W ≤ V is an H-

subrepresentation of V . Set gdef= ord(G) and h

def= ord(H). Fix a section of the map πH : G→ G/H , that

is, fix a set of left coset representatives R = r1, · · · , r` of G/H where ` = g/h. Extend the morphism ofgroups θ : H → GL(W ) ⊆ EndC(W ) to all of G as a map of sets by θ(s) = 0 ∈ EndC(W ) if s ∈ G \H .If we let w1, · · · , wn be a basis of W , then

(ρriwβ | 1 ≤ i ≤ `, 1 ≤ β ≤ n) = (ρr1(w1), · · · , ρr1(wn), ρr2(w1), · · · , ρr`wn)

38


is an ordered basis of V and with respect to this ordered basis, in block form, we have

ρs = (θr−1i srj

)ij =

θr−1

1 sr1θr−1

1 sr2· · · θr−1

1 sr`

θr−12 sr1

θr−12 sr2

· · · θr−11 sr`

......

. . ....

θr−1` sr1

θr−1` sr2

· · · θr−1` sr`

.

To see this, first note that for each j there is a unique i for which r−1i srj ∈ H (because srj belongs

to a unique coset riH ). Because s was arbitrary in the previous argument and r−1i srj ∈ H implies

r−1j s−1ri =

(r−1i srj

)−1 ∈ H , we see that for each i, there is also a unique j satisfying r−1i srj ∈ H . This

is reflected in the fact that the action of G on⊕

σ∈G/HWσ permutes the subspaces Wσ according to thecosets σ.

For 1 ≤ β ≤ n and 1 ≤ j ≤ `, set vj,βdef= ρrj (wβ). Let 1 ≤ i ≤ ` be the unique integer satisfying

r−1i srj ∈ H . Letting (Θαβ

ij )αβ denote the matrix form of θr−1i srj

over the basis w1, · · · , wn, so that for1 ≤ β ≤ n,

θr−1i srj

(wβ) =

n∑α=1

Θαβij wα,

we have

ρs(vj,β) = ρsrj (wβ)

= ρrir−1i srj

(wβ)

= ρri(ρr−1i srj

(wβ))

= ρri(θr−1i srj

(wβ))

= ρri

(n∑

α=1

Θαβwα

)

=

n∑α=1

Θijαβρri(wα)

=n∑

α=1

Θijαβvi,α.

Extending the definition of Θijαβ to Θij

αβ = 0 if r−1i srj ∈ G \H , (so that Θij

αβ is the matrix form of θr−1i srj

for all 1 ≤ i, j ≤ ` and s ∈ G), we can write

ρs(vj,β) =∑i=1

n∑α=1

Θijαβvi,α,

which gives the matrix form of ρs over the basis vi,α | 1 ≤ i ≤ `, 1 ≤ α ≤ n.

Proposition 2.67. Let ρ : G → GL(V ) be a representation of the finite group G induced by the represen-tation θ : H → GL(W ) of the subgroup H ≤ G. Let χ denote the character of θ and write IndGH(χ) denotethe character of ρ. Set R to be a system of left coset representatives for G/H . For any s ∈ G, we have

IndGH(χ) (s) =∑r∈R

r−1sr∈H

χ(r−1sr) =1

h

∑t∈G

t−1st∈H

χ(t−1st).

39

Chapter 2

Proof. The first equality is clear after taking the trace in the matrix form of ρs in Remark 2.66. Note thatif t ∈ rH , letting tr be such that t = rtr, we have t−1st = t−1

r r−1srtr ∈ H if and only if r−1sr ∈ H ,in which case χ(t−1st) = χ(t−1

r r−1srtr) = χ(r−1sr). In other words, the whole coset rH contributesto hχ(r−1sr) to the sum over all t ∈ G, so grouping the elements t ∈ G in the second sum according totheir cosets gives the second equality.

Example 2.68. Let V = C[G] be the regular representation of G. It has a basis ett∈G. Consider asubgroup H ⊆ G and let W = 〈ett∈H〉C ⊆ C[G] be the chosen subrepresentation of ResGH(ρ) : H →GL(C[G]) (notice that it is H-stable). The representation θ : H → GL(W ) is the regular representation ofH . One clearly sees that C[G] is induced by W ' C[H] ; this is just saying that G is a disjoint union of itsleft cosets. In particular, the regular representation of G is always induced by the trivial representation ofits trivial subgroup.

To be able to use induced representations to character tables, it is also important to consider therelationship between conjugacy classes and subgroups. The most important properties are summarized inthe following proposition.

Proposition 2.69. Let G be a group and H a subgroup. The group G acts on itself via conjugation, namely

for s, t ∈ G, s · t def= sts−1 defines a left action of G on itself. We also see that H acts on G via conjugation,and therefore also acts on H . (All the actions considered in this proposition will be defined via conjugation.)

The conjugacy classes of G are disjoint union of orbits of the action H

G. If CG is a conjugacy classof G and CG =

⋃i∈I CHi where the CHi are orbits of H

G, pick t ∈ CG and let sii∈I ⊆ G be such thatsi · t ∈ CHi . The maps

sjs−1i (−)sis

−1j : CHi → CHj

are bijections, so that two conjugacy classes of H within the same conjugacy class of G have the samecardinality.

In particular, if H is a normal subgroup of the finite group G, it is a union of conjugacy classes of Gby definition, and those G-conjugacy classes are a disjoint union of conjugacy classes of H . Therefore, ifs ∈ H and cHs (resp. cGs ) denotes the size of the conjugacy class of s in H (resp. in G), then cHs divides cGsand cHs = cHt for all t ∈ cGs .

Proof. Two elements conjugate in H are also conjugate in G, hence for any conjugacy class CH of H ,

we have CH ⊆ OrbG(CH). Write CG def= OrbG(CH) as a union of conjugacy classes of H . Orbits of an

action always partition the set being acted on (this set being CG considered as an H-action), which gives

the first claim. If CHtdef= OrbH(t), then s · (−) : CHt → CHs·t is bijective (with inverse s−1 · (−)), which

proves the second claim. The particular case where H is normal in G and G is finite follows since thesecond claim implies cHs = cHt for all t ∈ cGs and the first claim implies that cGs is partitioned by subsetsof size cHs , hence c

Gs is a multiple of cHs .

Example 2.70. We now consider the computation of a character table ; the one of the alternating groupA4. Recall that A4 is a normal subgroup of S4 since |S4 : A4| = 2 and any subgroup of a group which hasindex 2 is normal. Therefore, we can apply Proposition 2.69 and our knowledge of the conjugacy classesof S4 to compute the conjugacy classes of A4. Also note that by the Orbit-Stabilizer theorem, the size ofthe conjugacy classes have to divide the order of the group. 33 In Example 2.52, we have seen that theconjugacy classes of S4 correspond to permutations of the same cycle structure. Write the elements of S4 incycle notation, e.g. (124) is the permutation 1 7→ 2 7→ 4 7→ 1 and 3 7→ 3. In A4, there are three types ofcycle structures, namely

(1)(2)(3)(4), (12)(34), (123).

The identity is obviously in its own conjugacy class. As for the cycle structure (12)(34), there are threesuch permutations, so the orbits in this cycle structure have order 1 or 3 ; since their cycle structure is left

40


invariant by the action of A4 (for example, conjugation of (12)(34) by (123) gives (23)(14)), this is an orbitof A4 of order 3. For the remaining eight 3-cycles, we can split them down in two rows, namely

(123) (142) (134) (243)(132) (124) (143) (234)

Note that if σ is the permutation in the first row, then σ2 is below it in the second row. Since the S4-orbitcorresponding to this cycle structure has order 8, the A4-orbit of such permutations has to divide 8 and 12,thus has to divide 4. One readily sees that all these permutations have a fixed point and conjugation byelements of A4 moves this fixed point over all of 1, 2, 3, 4, hence an orbit contains at least four distinctelements, and therefore has order 4. One can easily see that the permutations listed in the top row are allconjugate : letting · denote action via conjugation,

(12)(34) · (123) = (214) = (142), (13)(24) · (123) = (341) = (134), (14)(23) · (123) = (432) = (243),

so that the top row is a conjugacy class of A4, and thus so does the bottom row.

Noticing that Kdef= id, (12)(34), (13)(24), (14)(23) is a subgroup which is a union of conjugacy

classes of A4, we see that it is a normal subgroup of A4. Since A4/K ' Z/3Z is abelian, all its irreduciblerepresentations have degree 1, thus lifting them to A4 gives us three irreducible characters of A4 (namelyχ1, χ2 and χ3 in the table below ; since Z/3Z is cyclic, a morphism of groups Z/3Z→ C× is the same as achoice of a root of the equation x3 = 1, hence the formula for the three characters. They are given below:

write ωdef= e2πi/3 = −1+

√3i

2 , so that

cs 1 3 4 4A4 1 x t t2

χ1, trivial 1 1 1 1

χ2 1 1 ω ω2

χ3 1 1 ω2 ω

ψ 3 −1 0 0

where the last row ψ was filled using Corollary 2.27 and the column orthogonality relations (c.f. The-orem 2.34). Another way to obtain ψ is to notice that A4 acts doubly transitively on 1, 2, 3, 4. Toprove this, recall that S4 does, and if S4 uses an odd permutation σ to map α to a and β to b, write1, 2, 3, 4 = a, b, c, d so that the permutation (cd)σ also does the trick and is an even permutation. Thecorresponding standard representation with character ψ = χ− χ1 is irreducible by Theorem 2.46.

Example 2.71. We compute a second character table, the one of the dihedral group

Dn = 〈r, s | rn = s2 = 1, srs−1 = r−1〉

where n ≥ 2. Because there are many possible conjugacy classes, we list the character values as a functionof the elements instead of the conjugacy classes. In other words, any element can be written in the form rk

or srk for 0 ≤ k ≤ n− 1 and we give the character as a function of k. Note that the number of conjugacyclasses will be deduced from the number of irreducible characters.

The subgroup 〈r〉 is abelian and has index 2 (thus is normal), hence every irreducible representation ofDn has degree at most 2 by Corollary 2.54. We begin with the irreducible characters of degree 1. It sufficesto compute what are the possible group homomorphisms ρ : Dn → C×. It suffices to know the image of ρrand ρs ; they must satisfy the relations

ρ2s = ρnr = 1, ρsρrρ

−1s = ρr−1 =⇒ ρr = ρ−1

r =⇒ ρ2r = 1.

41

Chapter 2

It follows that ρr, ρs ∈ −1, 1. Furthermore, if n is odd, ρnr = ρ2r = 1 implies ρr = 1. We must split into

two cases :n even n odd

Dn rk srk

ψ1 1 1

ψ2 1 −1

ψ3 (−1)k (−1)k

ψ4 (−1)k (−1)k+1

Dn rk srk

ψ1 1 1

ψ2 1 −1

Note that our choice of notation is such that ψ1 and ψ2 are given by the same formulas whether n is evenor not. We now turn to degree 2 representations. To proceed, we use the normal subgroup 〈r〉 and induceits irreducible representations to Dn. Since 〈r〉 ' Z/nZ, an irreducible character of Z/nZ is equivalent to

a choice of an nth root of unity. For each 0 ≤ j ≤ n− 1, let ωjdef= e2πij/n and let ρj : Z/nZ→ C× be the

corresponding character, so that

IndDnZ/nZ(ρj)rk

=

[ωkj 0

0 ω−kj

], IndDnZ/nZ

(ρj)srk

=

[0 ω−kjωkj 0

].

To obtain these formulas via computations, use the formula given in Remark 2.66. Their characters areeasily computed since we have them in matrix form. Let χj be the character associated to IndDnZ/nZ

(ρj), so

thatχj(r

k) = ωkj + ω−kj , χj(srk) = 0.

Note that it is not true in general that representations induced by an irreducible representation are irre-ducible (c.f. Example 2.68 for the example of the regular representation), hence we still have to check thatχj is irreducible:

(χj , χj) =1

2n

n−1∑k=0

χj(rk)2 =

1

2n

n−1∑k=0

(ω2kj + ω−2k

j + 2) = 1 +1

n

n−1∑k=0

ω2kj .

To understand the last equality, the substitution k 7→ n− k gives

n−1∑k=0

ω−2kj =

1∑k=n

ω−2(n−k)j =

n∑k=1

ω2kj

and since ω2nj = 1, we can replace the term with k = n by k = 0 so that

n−1∑k=0

ω−2kj =

n−1∑k=0

ω−2kj =

n−1∑k=0

ω2jk1 .

To check if χj is irreducible, it suffices to determine if the latter sum equals zero. Note that χj = χn−j , sowe can restrict our attention to 0 ≤ j ≤ n/2. We can even remove the case j = 0 (since χ0 = ψ1 + ψ2)and j = n/2 when n is even (since χn/2 = ψ3 + ψ4), so in fact, we restrict our attention to 1 ≤ j < n/2.Irreducibility can then be checked in two different method. The first method consists in computing theabove sum explicitly: since 1 ≤ j < n/2, ω2j

1 6= 1, so we set x = ω2j1 in the geometric progression formula

(x− 1)

(n−1∑k=0

xk

)= xn − 1 =⇒

(n−1∑k=0

ω2jk1

)=ω2jn

1 − 1

ω2j1 − 1

= 0.

The second method is by showing directly that no one-dimensional subspace of C2 is Dn-stable under theaction of IndDnZ/nZ

(ρj). To see this, if v ∈ C2 \0, the orbit of v via the action of the subgroup 〈r〉 ≤ Dn

spans a line if and only if v lies on the coordinate axes of C2, but since s ∈ Dn permutes the axes, the orbitof v has to span C2.

To show that there are no more irreducible characters of Dn, we split into two cases:

42


- Assume n is even, so that there are n2 − 1 values of j satisfying 1 ≤ j < n/2. Since

2n = ord(Dn) = 4 · 12 + ((n/2)− 1) · 22,

we have found all the irreducible representations of Dn, hence the following character table:

Dn rk srk

ψ1 1 1

ψ2 1 −1

ψ3 (−1)k (−1)k

ψ4 (−1)k (−1)k+1

χj , 1 ≤ j < n/2 ωkj + ω−kj 0

In particular, there are n/2 − 1 + 4 = n/2 + 3 irreducible characters/conjugacy classes in Dn for neven.

- Assume n is odd, so that there are bn/2c = n−12 values of j satisfying 1 ≤ j < n/2. Since

2n = ord(Dn) = 2 · 12 +n− 1

2· 22,

we have all the irreducible characters of Dn, hence

Dn rk srk

ψ1 1 1

ψ2 1 −1

χj , 1 ≤ j < n/2 ωkj + ω−kj 0

It follows that there are n−12 + 2 = n+3

2 irreducible characters/conjugacy classes in Dn for n odd.

Example 2.72. Consider the alternating group G = A5. We already know of one irreducible character ofH , namely the trivial character which we denote by χ1. The doubly transitive action S5

1, 2, 3, 4, 5 isalso doubly transitive when we restrict it to the subgroup A5 ; to see this, if a, b, c, d denote distinct integers,the following table shows how to find a permutation in A5 which maps a pair of distinct integers to another :[

a 7→ bc 7→ d

](abcd)

[a 7→ ac 7→ d

](ab)(cd)

[a 7→ ac 7→ c

](ac)(bd)

[a 7→ bc 7→ a

](abc).

This gives a second irreducible character by counting fixed points of the action (c.f. Theorem 2.46) ;we denote this character by χ2 and note that it has degree 4. Recall that conjugacy classes in S5 arecharacterized by the possible cycle structures (c.f. Example 2.52), hence the conjugacy classes of A5 of agiven cycle structure partitition the set of even permutations of that cycle structure. In other words, we haveat least 4 conjugacy classes, namely those corresponding to the even cycle structures 1, (123), (12)(34) and(12345). We now check that only the latter cycle structure splits in two conjugacy classes, namely that of(12345) and of (12354).

Let a, b, c, d, e ∈ 1, 2, 3, 4, 5 be distinct integers and σ ∈ S5 be a permutation such that σ(123)σ−1 =(abc). If σ is even, (123) and (abc) are conjugate, and if σ is odd, then (de)σ(123)σ−1(de)−1 =(de)(abc)(de) = (abc) where (de)σ is even, so the 3-cycles form a conjugacy class in A5. If σ ∈ S5 issuch that σ(12)(34)σ−1 = (ab)(cd) and σ is even, (12)(34) and (ab)(cd) are conjugate ; if it is odd, then

43

Chapter 2

σ(12) sends (12)(34) to (ab)(cd) via conjugation. Finally, if σ(12345)σ−1 = (abcde) and σ is odd, thenσ(45)(12345)(45)−1σ−1 = σ(12354)σ−1 = (abcde), so it remains to see that (12345) and (12354) are not

conjugate. For σ ∈ S5 and a ∈ 1, 2, 3, 4, 5, write σadef= σ(a). If σ(12345)σ−1 = (σ1σ2σ3σ4σ5) = (12354)

and σ(1) = a, then

(σ1, σ2, σ3, σ4, σ5) ∈ (1, 2, 3, 5, 4), (2, 3, 5, 4, 1), (3, 5, 4, 1, 2), (5, 4, 1, 2, 3), (4, 1, 2, 3, 5)

and one can check that the corresponding values of σ are the odd permutations

(45), (1235), (134)(25), (153)(24), (3214).

It follows that we have five conjugacy classes in A5, and therefore we are missing three irreducible charactersχ3, χ4, χ5. Their degrees n3, n4, n5 have to satisfy the equation

12 + 42 + n23 + n2

4 + n25 = |A5| = 60,

e.g. n23 + n2

4 + n25 = 43. We cannot have ni ≥ 6 since 72 > 43 and the equation a2 + b2 = 43 − 62 = 7

has no solution in the integers. Therefore 1 ≤ n3, n4, n5 ≤ 5. There can be at most one of the ni’s equalto 5 (since 50 > 43) and at least one, since if n3, n4, n5 ≤ 4, they cannot all be equal to 4 (48 6= 43) and ifonly one of them is smaller than 4, the resulting sum is either equal to 45 or is less than 40. Therefore wecan set n5 = 5. The equation n2

3 + n24 = 43 − 25 = 18 has a unique solution in positive integers, namely

n3 = n4 = 3.

Using elementary combinatorics, one determines the size of each conjugacy class. We summarize whatwe have up to now in the following table, to which we temporarily adjoin the symmetric and alternatingpowers of χ2 (which are not irreducible) to continue filling in the table (c.f. Proposition 2.12) :

cs 1 20 15 12 12

A5 1 (123) (12)(34) (12345) (12354)

χ1 1 1 1 1 1

χ2 4 1 0 −1 −1

χ3 3

χ4 3

χ5 5

χ22,σ 10 1 2 0 0

χ22,α 6 0 −2 1 1

Computing explicitly the square of a permutation is not necessary to evaluate these characters via the for-mulas of Proposition 2.12 since the conjugacy classes of the first three classes squared are trivially evaluatedand conjugacy class of a 5-cycle squared is again that of a 5-cycle, and χ2 takes equal values on bothconjugacy classes. Note that an element of A5 is conjugate to its inverse ; this is trivial for the first threeconjugacy classes, and we note that

(12345)−1 = (54321) = (15432) = (25)(34)(12345) ((25)(34))−1 .

Therefore, we can compute (χi |χj) = 〈χi, χj〉A5by summing the products in each of the rows cs, χi and

χj over all columns and then dividing by |A5| = 60. We deduce⟨χ2

2,σ, χ1

⟩= 1,

⟨χ2

2,σ, χ2

⟩= 1,

⟨χ2

2,σ − χ1 − χ2, χ22,σ − χ1 − χ2

⟩= 1

which implies that χ5 = χ22,σ − χ1 − χ2. Similarly, we can compute⟨

χ22,α, χ1

⟩=⟨χ2

2,α, χ2

⟩=⟨χ2

2,α, χ5

⟩= 0,

⟨χ2

2,α, χ22,α

⟩= 2,

44


which implies that χ22,α = χ3 + χ4. To see this, write χ2

2,α = aχ3 + bχ4 where a, b are non-negativeintegers. Then

⟨χ2

2,α, χ22,α

⟩= a2 + b2 = 2 implies a = b = 1. We now claim that we have gathered

enough representation-theoretic information to use the orthogonality relations and find the remaining twocharacters. We introduce variables to compute them :

cs 1 20 15 12 12

A5 1 (123) (12)(34) (12345) (12354)

χ1 1 1 1 1 1

χ2 4 1 0 −1 −1

χ3 3 x1 x2 x3 x4

χ4 3 y1 y2 y3 y4

χ5 5 −1 1 0 0

From the relation χ22α = χ3 + χ4, we deduce

x1 + y1 = 0, x2 + y2 = −2, x3 + y3 = 1, x4 + y4 = 1.

From considering the products 〈χi, χ3〉 and 〈χi, χ4〉 for i = 1, 2, 5, we get the matrix of linear relations 20 15 1220 0 −12−20 15 0

x1

x2

x3 + x4

=

−3−15−12

=⇒

x1

x2

x3 + x4

=

0−11

=

y1

y2

y3 + y4.

Since the column vectors associated to the pairs ((12)(34), (12345)) and ((12)(34), (12354)) are orthogonal,we deduce x3 + y3 = 1 = x4 + y4, and in particular, x3 = y4 and x4 = y3. So it remains to determine x3,and for this we use the orthogonality of the two last columns :

2 + 2x3(1− x3) = 1 + (−1)2 + x3x4 + y3y4 = 0 =⇒ x23 − x3 − 1 = 0,

so it follows that x3 ∈

1+√

52 , 1−

√5

2

. Letting x3 = y4 be one of the two possible values and using the

row orthogonality relations, we see that y3 takes the other value, so it does not matter which root of thisequation we take since both solutions are actually necessary. We finally display the completed charactertable :

cs 1 20 15 12 12

A5 1 (123) (12)(34) (12345) (12354)

χ1 1 1 1 1 1

χ2 4 1 0 −1 −1

χ3 3 0 −1 1+√

52

1−√

52

χ4 3 0 −1 1−√

52

1+√

52

χ5 5 −1 1 0 0

Another technique that could have been used instead of linear algebra is to notice that conjugation by(45) is an automorphism ϕ ∈ Aut(S4) which permutes only the two last conjugacy classes. The charactersχ3 and χ4 cannot be equal on the conjugacy classes (12345) and (12354), otherwise the correspondingcolumns would be equal and thus not orthogonal. It follows that χϕ3 = χ4, hence

x1 = y1, x2 = y2, x3 = y4, x4 = y3

and the column orthogonality relations allow computing the remaining coefficients directly.

45

Chapter 3

Representations as C[G]-modules

In this section, we will start assuming that the reader might be familiar with some notions of categorytheory ; if the reader isn’t, ignoring those notions is fine since we will not use them explicitly yet (perhaps insubsequent chapters). The complex group algebra of a finite group will prove very useful to the understand-ing of representations and we dedicate this chapter to this notion. Representations will be seen as modulesover the group algebra, thus allowing the theory of left modules over a non-commutative ring to apply torepresentations.

We still consider exclusively finite-dimensional vector spaces and all groups denoted by G,H,K denotefinite groups unless mentioned otherwise. We apologize for the ambiguity that h is sometimes used for theorder of a group H and for the number of irreducible representations of G ; it is the author’s opinion thatcontext should make obvious what that letter means, and if things seem confusing, precisions will be added.

3.1 Generalities

Definition 3.1. Let G be a finite group and k a non-commutative unital ring. The group algebra of

G over k, denoted by k[G], is constructed as follows. As a set, k[G]def= HomSet(G, k) is the set of all

functions a : G→ k. It is clearly a non-commutative ring with operations given by pointwise addition andmultiplication. It is also a free left k-module with k-basis ess∈G, so that its elements a ∈ k[G] can be

uniquely written in the form a =∑

s∈G ases where asdef= a(s). We take the convention of writing the basis

elements of k[G] as the group elements themselves, i.e. a =∑

s∈G ass. Under this notation, the group isnaturally a subset of its group algebra, i.e. G ⊆ k[G].

If V is a left k[G]-module, the subset G ⊆ k[G] can act on vectors v ∈ V via s · v def= esv. This induces

a k-linear left group action of G on V , i.e. a morphism of groups G→ GLk(V ) where GLk(V ) denotes thegroup of k-linear automorphisms of V (when k = C, this is a complex representation of G).

If V,W are left k[G]-modules, we can take their tensor product as k-modules to form V ⊗k W (c.f.

Remark 1.16), which is then also a left k[G]-module via s · (v ⊗ w)def= (s · v) ⊗ (s · w). We also give

Homk(V,W ) a left k[G]-module structure via (s · f)(v)def= s · f(s−1 · v).

We will constantly work with finitely generated left k[G]-modules, hence we call such an object a C[G]-module and denote the abelian category of (finitely generated left) k[G]-modules by k[G]-Modfin. Whenk is left-noetherian, so is k[G], hence this category is abelian (just as the category of finitely generatedR-modules is when R is a left-noetherian non-commutative ring), so left k[G]-modules inherit the notion ofkernel, cokernel, image and biproducts (which we will call direct sums). Furthermore, the tensor product ofleft k[G]-modules defined above makes k[G]-Modfin into a symmetric monoidal category.

46


Remark 3.2. Let G be a finite group and k be a non-commutative unital ring. If f1 =∑

s∈G ass andf2 =

∑s∈G bss are in k[G], then

f1 + f2 =∑s∈G

(as + bs)s, f1f2 =∑s∈G

∑tu=st,u∈G

atbu

s =∑s∈G

(∑t∈G

ast−1bt

)s.

This turns k[G] into a unital ring (1k[G]def= 1k1G) which is commutative if and only if k is commutative and

G is abelian. One can also see that multiplication can be interpreted as follows: for as, bt ∈ k and s, t ∈ G,define (ass)(btt)

def= (asbt)(st), where s and t are multiplied in G and as and bt are multiplied in k ; extend

this definition by k-bilinearity to obtain the general definition.

Proposition 3.3. Let G be a finite group. There is a canonical bijection

ρ : G→ GL(V ) | ρ is a complex representation 1:1←→ V a finitely generated C[G]-module .

The C[G]-submodules of V are in one-to-one correspondence with G-stable complex vector subspacesW ⊆ V . More precisely, we have an isomorphism of abelian symmetric monoidal categories

RepC(G) ' C[G]-Modfin.

In other words, this isomorphism preserves kernels, cokernels, images, quotients, direct sums and tensorproducts. It also respects the notion of “hom” defined in both categories (i.e. the HomC(−,−) of tworepresentations and the one defined in Definition 3.1).

Proof. Given a representation ρ : G → GL(V ), we can turn the vector space V into a C[G]-module

by letting s · v def= ρs(v) and extend this definition to C[G] by C-linearity. This defines a C[G]-module

structure on V . Conversely, if V is a C[G]-module, then ρ : G → GL(V ) defined by ρs(v)def= s · v

is a morphism of groups. These correspondences are inverse to each other, thus the first part of thestatement.As for the isomorphism of categories, it is clear that this definition respects morphisms of representa-tions / C[G]-modules since “it does nothing on the maps”, i.e. a G-linear map f : V → W gets mappedto exactly the same map of vector spaces f : V → W , except that it is considered as a C[G]-linearmap in C[G]-Modfin. The notions of kernel, cokernel, image, quotient and direct sum agree with theunderlying notions applied to vector spaces, so we have nothing to prove there. As for the tensor product,one sees that the definitions agree on simple tensors, completing the proof. The same argument appliesto the construction (V,W ) 7→ HomC(V,W ) in both categories.

Definition 3.4. Let G be a finite group. A C[G]-module V is called irreducible or simple if the cor-responding representation is irreducible ; in other words, V is irreducible if it is a simple C[G]-module,i.e. its set of C[G]-submodules has precisely two elements, 0 and V (so that V 6= 0). It is called com-pletely reducible if it can be written as a direct sum of irreducible C[G]-submodules. The character of aC[G]-module is the character of the corresponding representation.

Example 3.5. Under the correspondence of Proposition 3.3, note that the C[G]-module C[G] correspondsto the left regular representation of G, which we fortunately also denoted by C[G]. Therefore, we have adecomposition of C[G] as a direct sum of irreducible C[G]-modules.

Theorem 3.6. Every exact sequence of C[G]-modules splits, or equivalently, for every C[G]-module V andC[G]-submoduleW ⊆ V , there exists a C[G]-submoduleW0 of V such that V = W⊕W0 as C[G]-modules.

47

Chapter 3

Proof. See the proof of Maschke’s Theorem, it is the same proof (and the same statement). Note that Vand W do not have to be finitely generated in that proof, so the result also holds for infinite-dimensionalC[G]-modules.

Remark 3.7. In light of Proposition 3.3, we now think of the terms “linear representation” and “C[G]-module” as being equivalent. Since we assume that all our C[G]-modules are finitely generated, our vectorspaces over C corresponding to them are finite-dimensional.

Theorem 3.8. (Artin-Wedderburn theorem) Let G be a finite group.

(i) Every C[G]-module is completely reducible, i.e. can be written as a direct sum of irreducible C[G]-submodules.

(ii) Every C[G]-module is projective and injective, i.e. every exact sequence of C[G]-modules is split.

(iii) Let ρ1 : G→ GL(Cn1), · · · , ρh : G→ GL(Cnh) be the h irreducible representations of G written insome matrix form. Since this turns Cni into a C[G]-module, let

ρ1 : C[G]→ Matn1×n1(C), · · · , ρh : C[G]→ Matnh×nh(C)

be the corresponding morphism of C-algebras defined by

ρi

(∑s∈G

ass

)def=∑s∈G

asρis.

Taking the product map for i = 1, · · · , h, we obtain a map

ρ = (ρ1, · · · , ρh) : C[G]'−−−−→

h∏i=1

Matni×ni(C)

where n1, · · · , nh are the degrees of the irreducible representations of G (i.e. the dimensions of thesimple C[G]-modules).

In other words, the C-algebra C[G] is semisimple (this means one of the equivalent notions of (i), (ii) or(iii) ; that they are equivalent is precisely Artin-Wedderburn’s theorem).

Proof. By Maschke’s theorem, if V is a non-irreducible C[G]-module, we can write V = W ⊕W ′ whereW,W ′ ≤ V are two C[G]-submodules ; the result in (i) then follows by induction on dimC V . Sinceevery irreducible representation is a direct summand of C[G], if V is a C[G]-module and (W1, · · · ,Wh)are irreducible C[G]-submodules such that V '

⊕hi=1Wi, then V is isomorphic to a direct summand

of C[G]⊕h, hence is projective. It follows that every exact sequence of C[G]-modules splits, thus everyC[G]-module is also injective, which proves (ii).Onto (iii). We first show that ρ is injective. Suppose that

∑s∈G ass maps to zero, i.e. for all irreducible

representations of G, we have∑

s∈G asρis = 0. We show that a1 = 0. Notice that for each t ∈ G, we have

0 = ρit

(∑s∈G

asρis

)ρit−1 =

∑s∈G

asρitst−1 =

∑s∈G

at−1stρis

hence by averaging over t ∈ G and taking the trace, we get∑

s∈G

(1g

∑t∈G at−1st

)χi(s) = 0. Now

ψ(s)def= 1

g

∑t∈G at−1s−1t is a class function, and

(ψ, χi) =1

g

∑s∈G

ψ(s)χi(s) =1

g

∑s∈G

ψ(s)χ(s−1) =1

g

∑s∈G

ψ(s−1)χ(s) =1

g

∑s,t∈G

at−1stχi(s) = 0.

48


This means ψ = 0, and in particular a1 = ψ(1) = 0, as desired. Let t ∈ G and define bsdef= ast. Then

since∑

s∈G ass maps to zero, the sum∑

s∈G bss maps to

∑s∈G

bsρis =

∑s∈G

astρis =

∑s∈G

asρist−1 =

(∑s∈G

asρis

)ρit−1 = 0.

Thus b1 = at = 0, so that ρ is injective. On the other hand,

dimCC[G] = g =

h∑i=1

n2i = dimC

h∏i=1

Mni(C),

so ρ is bijective. It is clearly a map of C-algebras, thus an isomorphism of C-algebras.

Remark 3.9. Alternatively, the proof of (iii) could go as follows: assume ρ is not surjective, so that thereexists a non-zero linear form ` :

∏hk=1 Mnk(C) → C which vanishes on im ρ. Let Ekij ∈

∏hk=1 Mnk(C) be

the zero matrix in all components k′ 6= k of this product, and be the matrix Eij in the kth component (i.e.has a 1 in the (i, j)th component of the matrix of size nk). Then by Remark 2.31, the functions rkij : G→ Care linearly independent as functions on G, so that for all s ∈ G,

0 = `(ρs) = `(ρ1s, · · · , ρks) =

h∑k=1

∑i,j

`(rkij(s)Ekij) =

∑i,j,k

rkij(s)`(Ekij) =⇒

∑i,j,k

`(Ekij)rkij = 0

where the last equality is an equality of functions on G. It follows that the coefficients `(Ekij) must all bezero, i.e. ` = 0, a contradiction. So ρ is surjective ; we use a vector space dimension argument again toconclude bijectivity. (See [1].)

Another proof is as follows : assume∑

s∈G ass maps to zero, that is,∑

s∈G ass acts as 0 on each of theWi’s. Therefore it must act as zero on all representations by linearity ; this is not true since it doesn’t act aszero on 1 ∈ C[G]. Therefore ρ is injective, so we can apply a dimension argument again. (See [2].)

More abstractly, one can reformulate (iii) as follows: if ρi : G → GL(Wi), i = 1, · · · , h are the distinctirreducible representations of G, then the construction of Theorem 3.8 (iii) gives an isomorphism of C-algebras

ρ : C[G]→h∏i=1

EndC(Wi), ρsdef= (ρ1

s, · · · , ρhs ).

A natural question now arises. Since ρ is an isomorphism, the inverse map is well-defined. What isit? The answer is provided by Corollary 3.11. To proceed, we study the inner product space C[G] (c.f.Definition 2.24).

Proposition 3.10. (Plancherel’s formula) Let u, v : G→ C be complex functions onG. Write u =∑

s∈G uss.Define

udef=∑s∈G

us−1s =∑s∈G

uss−1.

Note that (−) : C[G] → C[G] is a C-linear map since it equals pre-composition with the inversion map ofG. We have the formula

(u, v) =1

g

h∑i=1

ni tr(ρi(uv)

)=

1

g

h∑i=1

ni tr(ρi(uv)

).

49

Chapter 3

Proof. By the C-bilinearity of both sides of the equality, we reduce to the case where u, v ∈ G. Wededuce that (u, v) = δuv and

1

g

h∑i=1

ni tr(ρi(uv)

)=

1

g

h∑i=1

ni tr(ρiu−1v

)=

1

g

h∑i=1

niχi(u−1v) =

1

g

h∑i=1

χi(1)χi(u−1v).

The result then follows from the column orthogonality relations (c.f. Theorem 2.34). The second equalityis proved along the same lines (the factor χi(u−1v) is changed by χi(uv−1).

Corollary 3.11. (Fourier inversion formula) Let G be a finite group of order g with irreducible representationsρi : G → GL(Wi) of order ni, i = 1, · · · , h. Suppose (ui)1≤i≤h ∈

∏hi=1 EndC(Wi) and u =

∑s∈G uss ∈

C[G] are such that ρ(u) = (ui)1≤i≤h. Then

us =1

g

h∑i=1

ni tr(ρis−1u

i).

In other words,

ρ−1(u1, · · · , uh) =∑s∈G

(1

g

h∑i=1

ni tr(ρis−1u

i))

s.

Proof. Since G ⊆ C[G] is an orthonormal basis, we have u =∑

s∈G(u, s)s. It suffices to set v = s ∈ Gin Plancherel’s formula:

us = (u, s) =1

g

h∑i=1

ni tr(ρi(us−1)

)=

1

g

h∑i=1

ni tr(uiρis−1

)=

1

g

h∑i=1

ni tr(ρis−1u

i).

3.2 The center of C[G]

ConsiderZ(C[G])

def= u ∈ C[G] | ∀u′ ∈ C[G], uu′ = u′u,

called the center of C[G].

Proposition 3.12. The center of C[G] is a free C-vector space with basis given by the following : let theconjugacy classes of G be written as C1, · · · , Ch and the basis z1, · · · , zh of Z(C[G]) by

zidef=∑s∈Ci

s.

In other words, Z(C[G]) = Ccl[G] since zi : G→ C is the function equal to 1 on Ci and zero elsewhere.

Proof. Suppose∑

s∈G ass ∈ Z(C[G]). Then for all t ∈ G, we have

∑s∈G

ass = t

(∑s∈G

ass

)t−1 =

∑s∈G

as(tst−1) =

∑s∈G

at−1sts,

so that a : G→ C which sends s 7→ as is a class function, hence∑

s∈G ass ∈ 〈z1, · · · , zh〉C. Conversely,the elements zi are in Z(C[G]) by definition of a conjugacy class (they commute with elements of G,hence with elements of C[G] by linearity), so Z(C[G]) = 〈z1, · · · , zh〉C. The linear independence of thezi’s is clear since conjugacy classes are disjoint.

50


Corollary 3.13. The dimension of Z(C[G]) is the number of conjugacy classes of G and the number ofirreducible representations of G.

Proposition 3.14. Recall the map ρ : C[G] →∏hi=1 EndC(Wi) from the proof of Theorem 3.8, which was

the product of the homomorphisms ρi : C[G]→ EndC(Wi).

The maps ρi send Z(C[G]) into the set of homotheties of Wi (namely, the set of endomorphisms ofWi of the form λ idWi for λ ∈ C), which defines an algebra homomorphism ωi : Z(C[G]) → C. Ifu =

∑s∈G uss ∈ Z(C[G]), then by Lemma 2.28, since us is a class function on G,

ωi(u) =1

nitr (ρi(u)) .

Furthermore, the family (ωi)1≤i≤h defines a C-algebra isomorphism between Z(C[G]) and Ch (seen asa subset of

∏hi=1 EndC(Wi)) by taking the product of the ωi’s ; the inclusion Ch ⊆

∏hi=1 EndC(W )i is

obtained by considering a vector (λ1, · · · , λh) as the element (λ1 idW1 , · · · , λh idWh).

Proof. Since u ∈ Z(C[G]) maps to a G-linear endomorphism ρi(u) of Wi, by Corollary 1.33 (noting thath = h in the notation of that corollary), we see that ρi(u) = 1

nitr(ρi(u)

)idWi . Note that ρ|Z(C[G]) is

injective since ρ is an isomorphism, so the claim follows from dimZ(C[G]) = dimCh.

3.3 Integrality properties of the characters

Definition 3.15. Let R ⊆ S be a ring extension, i.e. R and S are both commutative unital rings, R is asubring of S and 1R = 1S . The element s ∈ S is said to be integral over R if there exists r1, · · · , rn ∈ Rsuch that

sn + r1sn−1 + · · ·+ rn = 0.

We denote the set of all elements s ∈ S integral over R by RSand call it the integral closure of R in S.

We note that it can be shown that RSis an R-algebra, e.g. that R ⊆ R

S(which is obvious) and it is closed

under addition and multiplication (c.f. Commutative Algebra, Section 11.2).

The element z ∈ C is said to be an algebraic integer if z ∈ ZC, the integral closure of Z in C. This is

equivalent to the existence of a1, · · · , an ∈ Z such that

zn + a1zn−1 + · · ·+ an = 0.

Remark 3.16. We have ZC ∩ Q = Z, e.g. Z is integrally closed. To see this, if z ∈ ZC ∩ Q, write z = p/qwith q ≥ 1, so that pn + a1p

n−1q+ · · ·+ anqn = 0 ; this means pn ≡ 0 (mod q), a contradiction if we take

p and q coprime unless q = 1, i.e. p ∈ Z.

Proposition 3.17. Let χ be a character of the complex representation ρ : G → GL(V ) of the finite groupG. For all s ∈ G, χ(s) is an algebraic integer.

Proof. χ(s) is the trace of the linear map ρs which is diagonalizable with eigenvalues that are roots ofunity, hence algebraic integers ; χ(s) is the sum of these roots, hence χ(s) is an algebraic integer.

Proposition 3.18. Suppose u =∑

s∈G uss ∈ Z(C[G]) is such that us is an algebraic integer (noticethat Z(C[G]) is a commutative unital ring, hence Z(C[G])/Z is a ring extension in the sense of Defini-

tion 3.15). Then u ∈ ZZ(C[G]), i.e. u is integral over Z, so there exists a polynomial with integer coefficients

a1, · · · , an ∈ Z such thatun + a1u

n−1 + · · ·+ an = 0.

51

Chapter 3

Proof. Let z1, · · · , zh be the C-basis given for Z(C[G]) in Proposition 3.12. The elements z1, · · · , zh ∈Z(C[G]) are integral over Z since the Z-module Z[z1, · · · , zh] is finitely generated ; to see this, note thatzizj ∈ Z[G]∩Z(C[G]) ⊆ 〈z1, · · · , zh〉Z ; c.f. Commutative Algebra, Proposition 11.17 and Corollary 11.18.Since u is then a sum of products of integral elements (because s 7→ us is a class function), we are done,

because ZZ(C[G])is a subring of Z(C[G]).

Corollary 3.19. Let G be a finite group of order g and ρ be an irreducible representation of G of degreen with character χ. If u =

∑hi=1 uss ∈ Z(C[G]) (i.e. if s 7→ us is a class function) and us is an algebraic

integer for all s ∈ G, then1

n

∑s∈G

usχ(s)

is an algebraic integer.

Proof. The number written above is the image under ωi : Z(C[G])→ C of u, so since ωi is a C-algebraisomorphism (where ωi corresponds to χ = χi, c.f. Proposition 3.14), it is also a Z-algebra isomorphism ;thus it maps algebraic integers to algebraic integers.

Corollary 3.20. Let G be a finite group of order g and χ1, · · · , χh be the irreducible representations of Gwith degrees n1, · · · , nh respectively. Then ni divides g.

Proof. Apply the previous corollary to the element u =∑

s∈G χi(s−1)s ∈ Z(C[G]), which is possible

because s 7→ χi(s−1) is a class function which maps s to algebraic integers. Since 〈χi, χi〉 = (χi, χi) = 1

by Remark 2.30, we get

g

ni=g〈χi, χi〉

ni=

1

ni

∑s∈G

χi(s)χi(s−1) =

1

ni

∑s∈G

χi(s−1)χi(s) =

1

nitr(ρi(u)

)∈ ZC ∩Q = Z,

hence ni divides g.

3.4 Frobenius reciprocity

Let H ≤ G be a subgroup of the finite group G and R a system of left coset representatives for G/H .Let V be a C[G]-module, so that V is also a C[H]-module via the inclusion C[H] ⊆ C[G]. Let W be aC[H]-submodule of V . Recall that V is induced by W if as C-vector spaces, we have the following directsum :

V =⊕r∈R

r(W )

Note that the r(W ) are not C[H]-submodules of V in general (c.f. Theorem 3.38 which explains that r(W )is a C[rHr−1]-module ; in particular, this is a decomposition of C[H]-modules if and only if H E G is anormal subgroup). So we insist that this direct sum is merely in the sense of vector spaces. Since we have aC[H]-linear inclusion i : W → V , we can extend it to a map i : C[G]⊗C[H] W → V .

Remark 3.21. Note that this tensor product is in the sense of non-commutative ring theory, namely thatC[G] is seen as a C[G]-bimodule (both the left and right module action are defined by multiplication inthe ring), and thus a (C[G],C[H])-bimodule. Since W is a C[H]-module, this makes C[G] ⊗C[H] W aleft C[G]-module where for s ∈ G, t ∈ H and w ∈ W , we have st ⊗ w = s ⊗ ρWt (w). This is not thesame as taking the tensor product of the two left C[H]-modules C[G] and W since in this case, we havet · (s ⊗ w) = ts ⊗ ρWt (w) (c.f. Definition 3.1), where as in the left C[G]-module C[G] ⊗C[H] W , the tensorproduct satisfies t · (s⊗ w) = ts⊗ w.

52


Proposition 3.22. Let G be a finite group, H a subgroup, V a representation of G and W an H-subrepresentation of V . The following are equivalent :

(i) V is induced by W

(ii) There exists an isomorphism of C[G]-modules V ' C[G]⊗C[H] W

(iii) The C[G]-linear map i is an isomorphism.

Proof. The set R is a basis of C[G] as a C[H]-module, which makes the result clear.

Definition 3.23. Let H be a subgroup of the finite group G andW be a representation of H . The induced

representation of G by W is defined as IndGH(W )def= C[G] ⊗C[H] W . By the properties of the tensor

product, this induces a functor IndGH(−) : RepC(H)→ RepC(G) which we call the induction functor.

On the other hand, if V is a C[G]-module, one can view it as a C[H]-module. This induces a secondfunctor ResGH(−) : RepC(G)→ RepC(H) which is forgetful, called the restriction functor.

Remark 3.24.

- Induction is transitive : if H ≤ K ≤ G are groups and W is a C[H]-module, then we have naturalisomorphisms

IndGH(W ) ' IndGK(IndKH(W )

).

This is because as C[G]-modules, we have a natural isomorphism given by the properties of the tensorproduct

C[G]⊗C[K] (C[K]⊗C[H] W ) ' (C[G]⊗C[K] C[K])⊗C[H] W ' C[G]⊗C[H] W.

- Since we constructed an explicit representation of V which is induced by W , the induced represen-tation exists and is unique up to isomorphism by the previous proposition. Proposition 3.22 showsmore than proving the existence of the induced representation ; it shows that it is unique up toa unique isomorphism which is natural in W , namely i. This suggests that the induction func-tor must represent something; in fact, Theorem 3.25 shows that IndGH(W ) represents the functorHomC[H](−,ResGH(V )) : RepC(H)→ Set.

Theorem 3.25. Let H be a subgroup of the finite group G, W be a representation of H and V be arepresentation of G. We have a natural isomorphism of complex vector spaces

HomC[G](IndGH(W ) , V ) ' HomC[H](W,ResGH(V ))

given by ϕ 7→ (w 7→ ϕ(1⊗w)). In fact, since H acts on both hom-sets trivially, this is also an isomorphismof C[H]-modules (c.f. Remark 2.17).

Proof. By the tensor-hom adjunction for bimodules over non-commutative rings (see Commutative Alge-bra, Theorem 7.20 for a proof in the commutative case ; in the non-commutative case, the proof is thesame, except that we must act on the appropriate side (e.g. left or right)), since ResGH(V ) = V[C[H]] is theC[G]-module V seen as a C[H]-module, we obtain the natural isomorphisms

HomC[G](C[G]⊗C[H] W,V ) ' HomC[H](C[H],HomC[H](W,V[C[H]])) ' HomC[H](W,ResGH(V )).

Note that the first isomorphism sends a C[G]-linear map ϕ : C[G]⊗C[H]W → V to the map ψ : C[H]→HomC[H](W,V ) defined by 1 7→ (w 7→ ϕ(1 ⊗ w)), which we extend to all of C[H] by C[H]-linearity ofϕ (note that ϕ is C[G]-linear). The second isomorphism sends a map ψ : C[H]→ HomC[H](W,V ) to itsvalue at 1 ∈ C[H], so the composition of those two isomorphisms sends ϕ to the map w 7→ ϕ(1⊗ w).

53

Chapter 3

Corollary 3.26. (Universal property of the induced representation) Let H be a subgroup of the finite groupG, W a representation of H and V a representation of G. The induced representation is characterized upto a unique isomorphism by the following property : an H-linear map ϕ : W → ResGH(V ) lifts uniquely toa G-linear map ϕ : IndGH(W )→ V . Explicitly, ϕ is given by the following formula for s ∈ G and w ∈ W :if ρV : G→ GL(V ) is the morphism of groups corresponding to the representation V , then

ϕ(s⊗ w)def= ρVs (ϕ(w)).

Proof. Since ϕ ∈ HomC[H](W,ResGH(V )), the map ϕ is just the image of ϕ in HomC[G](IndGH(W ) , V )given by Theorem 3.25 ; to see this, it suffices to notice that ϕ is G-linear and agrees with ϕ onC[H]⊗C[H] W (since ϕ is C[H]-linear). The result follows since ϕ 7→ ϕ is bijective.

Definition 3.27. Let H be a subgroup of the finite group G of order g. It is not hard to see that if V is a

C[G]-module with character χ, then ResGH(V ) has character ResGH(χ)def= χ|H , i.e. the function χ : G→ C

restricted to H . The formula in Proposition 2.67 defines IndGH(ψ) when ψ is the character of a C[H]-moduleW . In general, for a function ψ : H → C, we set

IndGH(ψ) (s)def=

1

g

∑t∈G

t−1st∈H

ψ(t−1st).

When ψ is the character of a C[H]-module W , IndGH(ψ) is the character of IndGH(W ) ; also, when ψ ∈Ccl[G] is a class function, we recover the formula in Proposition 2.67 for IndGH(ψ) in terms of a system ofleft coset representatives R. For ϕ1, ϕ2 ∈ C[G] two functions on G, define

〈ϕ1, ϕ2〉Gdef=

1

g

∑s∈G

ϕ1(s)ϕ2(s−1).

(c.f. Remark 2.30).

Theorem 3.28. (Frobenius reciprocity) Let H be a subgroup of the finite group G. If ψ ∈ Ccl[H] andϕ ∈ Ccl[G], then ⟨

IndGH(ψ) , ϕ⟩G

=⟨ψ,ResGH(ϕ)

⟩H.

Proof. Since both ResGH and IndGH are C-linear maps, both sides are C-bilinear, hence we can restrict tothe case where ψ and ϕ are characters by Theorem 2.32. Let ψ be the character of the C[H]-module Wand ϕ be the character of the C[G]-module V . The result follows from the row orthogonality relations,Remark 2.30 and Theorem 3.25 :⟨

IndGH(ψ) , ϕ⟩G

= dim HomC[G](IndGH(W ) , V ) = dim HomC[H](W,ResGH(V )) =⟨ψ,ResGH(ϕ)

⟩H.

Proposition 3.29. Let H be a subgroup of the finite group G. Let W be a representation of H withcharacter χW and V a representation of G with character χV .

(i) We have a natural isomorphism of C[G]-modules

IndGH(W )⊗C V ' IndGH(W ⊗C ResGH(V )

)(note that on both sides, this is a tensor product of representations, c.f. Remark 1.16) together withthe corresponding identity on characters

IndGH(χW )χV = IndGH(χWResGH(χV )

)which can be extended to class functions by C-bilinearity.

54


(ii) Assume W and V are irreducible. Let hG (resp. hH ) denote the number of conjugacy classes of G(resp. H ). We decompose the following representations as a sum of irreducible subrepresentations :

IndGH(W ) 'hG⊕i=1

V ⊕mii , ResGH(V ) 'hH⊕j=1

W⊕njj .

Pick the two subscripts i0, j0 such that Vi0 = V and Wj0 = W . Then

mi0 = 〈IndGH(χW ) , χV 〉G = 〈χW ,ResGH(χV )〉H = nj0 .

Proof. (i) An easy proof of this statement is done by computing the characters on both sides, althoughthis proof does not make the isomorphism natural : for s ∈ G,

χIndGH(W )⊗V (s) = χIndGH(W )(s)χV (s)

=

1

h

∑t∈G

t−1st∈H

χW (t−1st)

χV (s)

=1

h

∑t∈G

t−1st∈H

χW (t−1st)χV |H(t−1st)

=1

h

∑t∈G

t−1st∈H

χW⊗ResGH(V )(t−1st)

= χIndGH(W⊗ResGH(V ))(s).

For a natural isomorphism, recall Corollary 3.26 so that the canonical map

ϕ : W ⊗ ResGH(V )→ IndGH(W )⊗ V, w ⊗ v 7→ (1⊗ w)⊗ v

lifts to a C[G]-linear map ϕ : IndGH(W ⊗ ResGH(V )

)→ IndGH(W ) ⊗ V . Explicitly, if ρV : G →

GL(V ) denotes the representation of V , for s ∈ G, w ∈W and v ∈ V , we have

ϕ(s⊗ (w ⊗ v))def= s · ((1⊗ w)⊗ v) = (s⊗ w)⊗ ρVs (v).

The map ϕ is surjective since ϕ(s ⊗ (w ⊗ ρVs−1(v)) = (s ⊗ w) ⊗ v. Since both vector spaces havethe same dimension, we conclude that ϕ is a C[G]-linear isomorphism.

We finish this section with a generalization of the notion of induction representation.

Definition 3.30. Let α : H → G be a morphism of groups, ρW : H → GL(W ) a representation of H andρV : G→ GL(V ) a representation of G.

(i) The restriction of V along α is denoted by ResαH(V ). As a vector space, ResαH(V )def= V . The

morphism of groups ResαH(ρV)

: H → GL(V ) is defined by ResαH(ρV) def

= ρV α.

(ii) The induced representation of W along α is denoted by IndGα (W ). The group algebra C[G]becomes a C[H]-algebra via the morphism of rings

C[α] : C[H]→ C[G], C[α]

(∑h∈H

ahh

)def=∑h∈H

ahα(h).

55

Chapter 3

The induced representation is then defined by extension of scalars of the C[H]-module W , namely

IndGα (W )def= C[G]⊗C[H] W.

It becomes a C[G]-module via the definition s · (t ⊗ w)def= (st) ⊗ w, which for h ∈ H satisfies

α(h)⊗ w = 1⊗ ρWh (w).

Theorem 3.31. Let α : H → G be a morphism of groups. Let W a C[H]-module with character χW andV a C[G]-module with character χV .

(i) We have a natural isomorphism

HomC[G](IndGα (W ) , V ) ' HomC[H](W,ResαH(V )).

The map is given by sending ϕ : IndGα (W ) → V to the map w 7→ ϕ(1 ⊗ w) and extending byC[G]-linearity.

(ii) (Universal property of the induced representation along α) The isomorphism given in (i) satisfies thefollowing universal property : if ϕ : W → V is a C-linear map satisfying ϕ(h · w) = α(h) · ϕ(w) forall h ∈ H (e.g. ϕ : W → ResαH(V ) is a morphism of C[H]-modules), there exists a unique morphismof C[G]-modules ϕ : IndGα (W )→ V satisfying ϕ(h⊗ w) = α(h)ϕ(w) for all h ∈ H and w ∈W .

Proof. This is exactly the same proof as in Theorem 3.25, the only difference being that the morphismC[α] : C[H]→ C[G] is not necessarily injective anymore (but that changes nothing to the proof).

Corollary 3.32. Let Kα−→ H

β−→ G be two morphism of groups. We have the identities of functors

ResαK ResβH = ResβαK , IndGβ IndHα ' IndGβα.

Proof. The first equality follows from definition. For the second, given representations ρV : G→ GL(V )and ρW : K → GL(W ), since we have a natural bijection :

HomC[G](IndGβ(IndHα (W )

), V ) ' HomC[H](IndHα (W ) ,ResβH(V ))

' HomC[K](W,ResαK

(ResβH(V )

))

= HomC[K](W,ResβαK (V )),

the universal property of the induced representation along β α gives the natural isomorphism.

Remark 3.33. There are three particular cases to distinguish about these two notions of restriction andinduction along a morphism α : H → G :

(i) when α : H → G is the inclusion map of a subgroup, in which case we recover the classical notion ;

(ii) when αdef= πK : G → G/K is the projection modulo a normal subgroup. Recall Section 2.7

concerning lifts and drops of representations. In this case, if ρV : G/K → GL(V ) and ρW : G →GL(W ) are representations, then ResπKG

(ρV)

= L(ρV ) equals the lift of ρV to G/K along πK . The

left-adjoint to ResπKG , namely IndG/KπK , is not equal to the drop in general (c.f. Section 2.7). To

understand what it is, let WK ≤ W be the vector subspace of elements w ∈ W invariant underthe action of K , i.e. ρs(w) = w for all s ∈ K (c.f. Theorem 2.13). Given w ∈ W , we see that in

56


IndG/KπK (W ) = C[G/K]⊗C[G] W ,

K ⊗ w =1

ord(K)

∑s∈K

sK ⊗ w = 1⊗

(1

ord(K)

∑s∈K

ρs(w)

),

so that any simple tensor admits an equivalent expression with w ∈ WK . Since K acts trivially onthe subspace WK and K is a normal subgroup of G, we deduce that WK is a G-subrepresentation

of W , for if s ∈ G, t ∈ K and w ∈WK , letting t′def= s−1ts ∈ K so that ts = st′, we have

ρt(ρs(w)) = ρs(ρt′(w)) = ρs(w) =⇒ ρs(w) ∈WK .

The map WK → IndG/KπK (W ) sending w to 1 ⊗ w thus maps a basis to a basis and is G/K-linear,

hence is an isomorphism of C[G/K]-modules.

In particular, ifW is an irreducible representation of G, sinceWK is a G-subrepresentation, either K

acts trivially onW (in which case IndG/KπK (W ) = D(W ) equals the drop) orWK = Ind

G/KπK (W ) = 0.

So in this sense, the functor IndG/KπK is the correct functorial generalization of the drop D(−) to

cover the cases where K acts non-trivially ; if we decompose W as a direct sum of irreducible

K-subrepresentations, IndG/KπK kills those on which K acts non-trivially and applies D(−) on the

remaining ones.

It is also worth noting that a morphism of groups α : H → G can be factored as the surjectivemorphism α : H → imα followed by the inclusion imα ≤ G, so that understanding these two casesallows understanding the general case.

(iii) when H ≤ G is a subgroup, s ∈ G, Hs def= sHs−1 and α : Hs → H is the map equal to conjugationby s−1, e.g. α(shs−1) = s−1(shs−1)s = h. Given a representation ρ : H → GL(W ) and s ∈ G, thes-twist of ρ is defined by ρs

def= ResαHs(ρ) : Hs → GL(W s) where as vector spaces, W s def= W and

for w ∈W , h ∈ H , we have

ρsh(w)def= ρα(h)(w) = ρs−1hs(w).

By Corollary 3.32, it is clear that ResαHs and Resα−1

H are inverse of each other since ResidHH and

ResidHsHs are the identity functors by definition ; therefore, their adjoints IndH

s

α and IndHα−1 are alsoinverse of each other. What is perhaps more interesting is that we have natural isomorphisms ResαHs 'IndHα (and analogously Resα

−1

H ' IndHs

α−1 ) since if ρ : H → GL(W ) is a representation of H and

ResαHs(ρ)def= ρs : Hs → GL(W s) is the corresponding s-twist, then IndHα (ρs) = C[H]⊗C[Hs] W

s isa C[H]-module satisfying the following for h ∈ H and w ∈W :

h⊗ w = α(α−1(h))⊗ w = 1⊗ ρsα−1(h)(w) = 1⊗ ρα(α−1(h))(w) = 1⊗ ρh(w),

so that we obtain a natural isomorphism C[H]⊗C[Hs] Ws 'W , hence

IndHα ResαHs ' idRepC(H) ' Resα−1

H ResαHs =⇒ IndHα ' Resα−1

H ,

which leads to the following definition.

Definition 3.34. Let H be a subgroup of the finite group G and s ∈ G. If ρ : H → GL(W ) is a

representation of H , its s-twist is the representation of Hs def= sHs−1 given by TwsH(ρ) : Hs → GL(W s)

where TwsH(ρ)

def= ResαHs(ρ) ' IndH

s

α−1(ρ), where α : Hs → H is the map which conjugates by s−1, e.g.

h 7→ s−1hs, andWsdef= W is the underlying vector space which is turned into a C[Hs]-module via Tws

H(ρ).If W has character χ, the character of the twist is obviously given by the formula Tws

H(χ) (t) = χ(s−1ts)for t ∈ Hs (note that χ(s−1ts) 6= χ(t) in general since it is possible that t /∈ H , hence χ(t) is not evendefined).

57

Chapter 3

Proposition 3.35. Let H be a subgroup of the finite group G and s ∈ G. If V is a representation of Ginduced by the H-subrepresentation W , then Tws

G(V ) is induced by the Hs-subrepresentation TwsH(W ).

Proof. Let α : G→ G denote conjugation by s−1 (namely t 7→ s−1ts), so that α|H : Hs → H is the mapof Definition 3.34. Let ι : H → G be inclusion map of G and ιs : Hs → G be the inclusion map of Hs.We have a commutative diagram

G G

Hs H

α

ιs

α|Hι

which by Corollary 3.32 implies

IndGHs TwsH ' IndGHs IndH

s

α−1 ' IndGα−1 IndGH ' TwsG IndGH .

Theorem 3.36. Let α : H → G be a morphism of groups. Let W a C[H]-module with character χ and Va C[G]-module with character ψ. The characters ResαH(ψ) of the representation ResαH(V ) and IndGα (χ) ofthe representation IndGα (W ) satisfy the following formulas :

(i) For t ∈ H , we have ResαH(ψ) (t) = ψ(α(t)).

(ii) If α is injective, letting hdef= ord(H) and α−1 : imα→ H be the inverse map,

IndGα (χ) (s) =1

h

∑t∈G

t−1st∈α(H)

χ(α−1(u)).

(iii) If α is surjective, let Ndef= kerα and n

def= ord(N), so that

IndGα (χ) (s) =1

n

∑z∈Hα(z)=s

χ(z).

(iv) For α : H → G arbitrary, following the notations of (ii) and (iii), we obtain

IndGα (χ) (s) =1

h

∑t∈G,u∈Ht−1st=α(u)

χ(u).

Proof. (i) In the case of restriction, since ρV : G → GL(V ) satisfies ResαH(ρV)

(t) = ρVα(t), we cantake the trace and obtain the formula.

(ii) This is just a reformulation of the formula given in Proposition 2.67 to adapt to the case where α

is not necessarily the inclusion map of a subgroup since letting Kdef= imα (which is isomorphic to

H via α : H → K), we have

IndGα (χ) (s) = IndGK(IndKα (χ)

)(s) =

1

h

∑t∈G

t−1st∈K

IndKα (χ) (t−1st) =1

h

∑t∈G

t−1st∈K

χ(α−1(t−1st))

by Remark 3.33.

58


(iii) Without loss of generality, assume α : G → G/K is the projection map and K E G is a normal

subgroup (so that ndef= ord(K)). In light of Remark 3.33, we consider the G-subrepresentationWK

of W isomorphic to IndGα (W ), so that χWK = IndGα (χ). By Theorem 2.13, the map 1n

∑t∈K ρ

Wt :

W → W is a projection with image WK . Forming a basis of W by taking the union of a basis ofWK and a basis of kerπ, we can evaluate the following trace for s ∈ G as

χWK (sK) = tr(ρWs |Wk

)= tr

(ρWs π

)=

1

n

∑t∈K

tr(ρWs ρWt

)=

1

n

∑t∈K

χ(st) =1

n

∑z∈Gα(z)=s

χ(z).

(iv) Let Ndef= kerα and n

def= ord(N), so we can write α as the composition

H H/N GπN α

where α is injective. Let hdef= ord(H), so that in both cases, we have the formula for the induced

character by (ii) and (iii), hence we evaluate : for s ∈ G,

IndGα (χ) (s) = IndGα

(IndH/NπN

(χ))

(s)

=1

ord(H/N)

∑t∈G,uN∈H/Nt−1st=α(uN)

IndH/NπN(χ) (uN)

=n

h

∑t∈G,uN∈H/Nt−1st=α(u)

1

n

∑z∈H

πN (z)=uN

χ(z)

=

1

h

∑t∈G,u∈Ht−1st=α(u)

χ(u).

Theorem 3.37. (Frobenius reciprocity along α) Let α : H → G be a morphism of groups, ϕ ∈ Ccl[H] andψ ∈ Ccl[G]. Then ⟨

IndGα (ϕ) , ψ⟩G

= 〈ϕ,ResαH(ψ)〉H .

Proof. By C-bilinearity, we can restrict to the case of characters ; the result then follows from Theo-rem 3.31 by taking dimensions.

3.5 Group actions and induced representations

Theorem 3.38. Suppose V is a C[G]-module such that

Vdef=⊕i∈I

Wi

where the Wi’s are vector subspaces which are permuted by G, so that for any s ∈ G and i ∈ I , there existsa unique i′ ∈ I satisfying ρs(Wi) = Wi′ . Let G

I be defined as the unique action characterized by theequation ρs(Wi) = Ws·i for all i ∈ I .

(i) For i ∈ I , letHidef= StabG(Wi). IfG acts transitively on I , then V is induced by the subrepresentation

59

Chapter 3

Wi of Hi. Moreover, if s ∈ G and j = s · i, then Hj = sHis−1.

(ii) If we do not assume that the action G I is transitive, recall that G \ I denotes the set of orbits of I .Given i ∈ J ∈ G \ I (so that J = OrbG(i)), set

VJdef=⊕j∈J

Wj

as vector spaces. By construction, since G J is transitive, VJ is a G-subrepresentation of V inducedby the Hi-subrepresentation Wi, e.g. VJ = IndGHi(Wi). As representations of G, we obtain

V =⊕J∈G\I

VJ .

Proof. (i) Since Hi = StabG(Wi) = StabG(i), for s ∈ G, any t ∈ Hi gives ρst(Wi) = ρs(Wt·i) =ρs(Wi), so that if R is a system of left coset representatives for G/H , we have

V =⊕j∈I

Wj =⊕r∈R

r(Wi).

From Definition 2.38, we deduce

Hj = StabG(j) = StabG(s · i) = sStabG(i)s−1 = sHis−1.

It is clear that

(ii) This follows from (i) ; we simply recorded the statement.

Definition 3.39. Let Λ be a partition of 1, · · · , n, i.e. Λ is a set of non-empty disjoint subsets of1, · · · , n whose union equals 1, · · · , n. Given an n × n matrix A = (aij), the matrix block corre-sponding to (σ, σ′) ∈ Λ×Λ is the matrix formed of the coefficients aij where (i, j) ∈ σ×σ′ and is denotedby Aσ,σ

′. If f is the linear endomorphism of Cn associated to A, then the linear map fσ,σ

′(which we also

call a matrix block) corresponding to the matrix Aσ,σ′is the map

fσ,σ′

: 〈ejj∈σ′〉C → 〈eii∈σ〉C

defined by the coefficients of Aσ,σ′.

Suppose ρ : G → GL(Cn) is a complex matrix representation and let Λ be a partition of the set1, · · · , n. An element of Λ × Λ is called a block. We say that ρ is in block form with respect to Λ iffor each σ ∈ Λ and each s ∈ G, there are unique σ′, σ′′ ∈ Λ such that the matrix blocks ρσ,σ

′s , ρσ

′′,σs are

non-zero (in other words, one non-zero block per row and column). Note that it suffices to make the checkon rows (resp. columns) since ρs is an isomorphism, thus has maximal rank (and so a row or column ofzeros is impossible).

Saying that ρ is in block form means that if e1, · · · , en is the standard basis of Cn, we have a directsum decomposition

Cn =⊕σ∈Λ

Wσ, Wσdef= 〈eii∈σ〉C

and for every s ∈ G and σ′ ∈ Λ, ρs(Wσ′) ⊆ Wσ where σ ∈ Λ is unique with the property that ρσ,σ′

s 6= 0.This suggests that we can define an action G Λ by setting s · σ′ = σ if ρs(Wσ′) ⊆ Wσ . We call this theblock form action corresponding to ρ : G→ GL(Cn).

60


Corollary 3.40. We reformulate Theorem 3.38 in block form using Definition 3.39. Let Λ be a partition of1, · · · , n and V = Cn be the representation corresponding to the morphism of groups ρ : G → GL(V )which is in block form with respect to Λ.

(i) Suppose the corresponding action G Λ is transitive. Let Hσ = StabG(σ) and consider the Hσ-

subrepresentation θ of Cn given by the matrices θsdef= ρσ,σs . Then ρ is induced by θ and if σ, σ′ satisfy

σ = s · σ′, then Hσ = sHσ′s−1. Furthermore, for any σ, σ′ ∈ Λ, we have |σ| = |σ′|.

(ii) Given σ′ ∈ Ω ∈ G \ Λ (so that Ω = OrbG(σ′)), set

VΩdef=⊕σ∈Ω

Wσ ⊆ V.

as vector spaces. By construction, since the restricted action G Ω is transitive, VΩ is a representationof G induced by the Hσ′-subrepresentation Wσ′ , e.g. VΩ = IndGHσ′ (Wσ′). As representations of G,we obtain

V =⊕

Ω∈G\Λ

VΩ.

In other words, to decompose a representation in block form into subrepresentations of G, one canlook at the orbits of the block form action and decompose as a direct sum of the underlying vectorspaces spanned by the blocks, which will be subrepresentations ; furthermore, these subrepresentationsare induced from the subgroups fixing a block in an orbit.

Proof. (i) The only thing we have not proven yet is that |σ| = |σ′|, which is obvious since Wσ ' Wσ′

via ρs where s · σ′ = σ, so that

|σ| = dimWσ = dimWσ′ = |σ′|.

(ii) This entirely follows from interpreting Theorem 3.38.

Remark 3.41. Let G X be a group action (both sets are finite). Consider the associated permutationrepresentation ρ : G → GL(V ) where V = 〈exx∈X〉C (c.f. Example 1.5). By definition, ρ satisfies the

hypotheses of Theorem 3.38 with V =⊕

x∈XWx whereWxdef= 〈ex〉C since ρs(Wx) = Ws·x. It follows that

(i) if G

X is transitive andHxdef= StabG(x) for x ∈ X , then V is induced by theHx-subrepresentation

Wx and for s ∈ G, Hs·x = sHxs−1

(ii) given A ∈ G\X and setting

VAdef= 〈exx∈A〉C,

we have a direct sum decomposition of V '⊕

A∈G\X VA into G-subrepresentations and for eachx ∈ A ∈ G\X , the G-subrepresentation VA is induced by the trivial representation of Hx (since Wx

is 1-dimensional and Hx acts trivially on Wx by definition).

Example 3.42. Consider the classical representation ρ : Sn → GL(V ) associated to the transitive groupaction Sn

1, · · · , n. We show that it is induced by the trivial representation of Sn−1 (c.f. Example 1.5),where Sn−1 is seen as the stabilizer of the element n ∈ 1, · · · , n. Writing V = 〈e1, · · · , en〉C and

Widef= 〈ei〉C for 1 ≤ i ≤ n, we have V =

⊕ni=1Wi and ρ permutes the Wi transitively (because Sn acts

transitively), so that Sn−1 ' Hn = StabSn(n) ≤ Sn is the subgroup inducing V via its trivial representation.

As for its character χC (C stands for classical), via the formula for the induced character, we see that if

R = id1,··· ,n, (1n), (2n), · · · , ((n− 1)n) = (in) ∈ Sn | i ∈ 1, · · · , n

61

Chapter 3

is our chosen system of left coset representatives for Sn/Sn−1, for σ ∈ Sn, since χC is induced by the trivialcharacter of Sn−1, we have

χC(σ) =∑π∈R

π−1σπ∈Sn−1

1

= #i ∈ 1, · · · , n | (σ (in))(n) = (in)(n)= #i ∈ 1, · · · , n | σ(i) = i,

as predicted by the formula given in Proposition 2.41.

Example 3.43. Let H ≤ G be a subgroup of the finite group G. Since the group action G

G/H istransitive, the corresponding permutation representation C[G/H] can be decomposed as

C[G/H] =⊕

σ∈G/H

Wσ, Wσdef= 〈eσ〉C

where s ∈ G permutes theWσ transitively since ρs(Wσ) = Ws·σ by definition. It is clear that StabG(WH) =H , showing that C[G/H] is induced by the trivial representation of H (because H acts trivially on WH ) byTheorem 3.38. We had seen this directly in Example 2.68.

Example 3.44. Recall Example 2.71, where we computed the irreducible characters of the dihedral groupDn. Suppose we do not know what the irreducible representation ρ : Dn → GL(C2) could look like andwe want to construct it (note that n > 2, since for n = 2, D2 is abelian). The matrix ρr is diagonalizable byTheorem 2.11, so we can already write

ρr =

[ωi1 00 ωi2

]where ω

def= e2πi/n is a primitive nth root of unity and i1, i2 ∈ 1, n − 1 (since these are the only two

generators of the cyclic group Z/nZ). We know that ρs cannot be diagonalizable over the eigenbasis ofρr, otherwise ρ would be reducible. Since ρsrs−1 = ρr−1 , letting v1, v2 be the eigenbasis of ρr satisfyingρr(vj) = ωij , we can write

α1ω−i1v1 + α2ω

−i2v2 = ρr−1(α1v1 + α2v2) = ρr(ρs(v1)) = ρs(ρr(v1)) = ρs(ωi1v1) = ωi1(α1v1 + α2v2)

which implies α1(ωi1 − ω−i1) = 0 and α2(ωi1 − ω−i2) = 0. The first equation implies α1 = 0 and α2 6= 0(because ρs(v1) = α1v1 + α2v2 6= 0, so that the second equation implies ωi1 = ω−i2 , i.e. i2 = n − i1. Itfollows that ρs(v1) = ±v2, and by applying ρs on that equation, we deduce that ρs(v2) = ∓v1. Lettingi1 = 1 and choosing the + sign for ρs gives the representation

ρr =

[ωi1 00 ωi2

], ρs =

[0 11 0

]which we can see as being induced by 〈r〉 ' Z/nZ by Corollary 3.40 (where Λ = 1, 2).

3.6 Restriction to subgroups ; Mackey’s irreducibility criterion

In this section, three groups are constantly involved, so for simplicitly, the letters k and h will denote groupelements instead of group orders or number of conjugacy classes of G.

Let H,K ≤ G be two subgroups. Then we can partition G using the (H,K)-double cosets of G :

KsHdef= ksh | k ∈ K,h ∈ H.

62


This is because we have a left action K ×H

G given by (k, h)

sdef= ksh−1, which we call the double-

coset action of the pair (K,H). The subset KsH is simply the orbit of s ∈ G under the action of K ×H ;we call such an orbit a (K,H)-double coset. Let S ⊆ G denote a fixed set of representatives of (K,H)-double cosets for K\G/H . This notation makes sense since this orbit space is also the orbit space of theaction K G/H given by multiplication on the left and the orbit space of the right action K \G H givenby multiplication on the right (in the above, we artificially turned this action into a left action).

Theorem 3.45. Let H,K ≤ G be two subgroups of the finite group G, S a system of (K,H)-double coset

representatives for K\G/H and ρ : H → GL(W ) a representation. Set HsK

def= Hs ∩K = sHs−1 ∩K .

We have an isomorphism of C[K]-modules :

ResGK(IndGH(W )

)'⊕s∈S

IndKHsK

(ResH

s

HsK

(TwsH(W ))

).

Proof. Let Vdef= IndGH(W ) with corresponding representation ρV : G → GL(V ). Note that for w ∈ W

and h ∈ H , ρVh (w) = ρh(w) since W is an H-subrepresentation of V by definition. Consider thecanonical decomposition of the K-representation ResGK(V ) (c.f. Definition 2.35) :

ResGK(V ) =⊕s∈S

Vs, Vsdef=∑k∈K

ρVks(W ).

To see why this sum is direct, remember that V is induced by W , thus V is the direct sum of the ρr(W )with r ranging over a system R of left coset representatives for G/H . By definition of S, each r ∈ Rsatisfies r = ks for a unique s ∈ S and some k ∈ K . The vector space Vs is then the sum over those rwhich correspond to (k, s) for some k ∈ K , e.g. the sum over all k of the ρVks(W ).

It remains to show that Vs ' IndKHsK

(ResH

s

HsK

(TwsH(W ))

). Clearly, the subgroup of those k ∈ K

satisfying ρVk (ρVs (W )) = ρs(W ) equals HsK = sHs−1 ∩K , hence Vs is induced by ResH

s

HsK

(ρs(W )). A

C[Hs]-isomorphism TwsH(W ) ' ρs(W ) is given by Φs

def= ρVs : W → ρVs (W ) since for t ∈ Hs, we have

Φs(t · w) = ρVs (TwsH(ρ)t (w)) = ρVs (ρs−1ts(w)) = ρVt (ρVs (w)) = t · Φs(w).

In Theorem 3.45, we can set Kdef= H ; it follows that Hs

def= Hs

H

def= sHs−1 ∩ H = Hs ∩ H for all

s ∈ G. Write ρ : H → GL(W ). For s ∈ S where S is a set of (H,H)-double coset representatives forH\G/H , we get two representations of Hs, namely ResH

s

Hs (TwsH(ρ)) and ResHHs(ρ). In general, these two

are not isomorphic.

Definition 3.46. Let V1, V2 be two representations of G with characters χ1, χ2, respectively. We say thatthey are linearly disjoint if 〈χ1, χ2〉G = 0, i.e. if V1 and V2 have no irreducible components in common(equivalently, if the inner product in C[G] satisfies (χ1 |χ2) = 0).

Corollary 3.47. (Mackey’s criterion) Let H ≤ G and W a representation of H . Let S be a set of (H,H)-

double cosets representatives of G, and without loss of generality assume 1 ∈ S. Then V def= IndGH(W ) is

irreducible if and only if W is irreducible and for all s ∈ G\H , the Hs-representations ResHs

Hs (TwsH(ρ))

and ResHHs(ρ) are linearly disjoint.

Proof. Let χW denote the character of W . We know that V is irreducible if and only if its characterχV = IndGH(χW ) satisfies 〈χV , χV 〉G = 1. Using Frobenius reciprocity (twice, once on the pair (G,H)

63

Chapter 3

and then on (H,Hs)), recalling that 〈−,−〉 is commutative,

〈χV , χV 〉G =⟨χW ,ResGH(χV )

⟩H

=∑s∈S

⟨χW , IndHHs

(ResH

s

Hs (TwsH(χW ))

)⟩H

=∑s∈S

⟨ResHHs(χW ) ,ResH

s

Hs (TwsH(χW ))

⟩Hs

= 〈χW , χW 〉H +∑

s∈S\1

⟨ResHHs(χW ) ,ResH

s

Hs (TwsH(χW ))

⟩Hs.

Since the numbers involved at both ends are non-negative integers and 〈χW , χW 〉H > 0, the resultfollows.

Corollary 3.48. Suppose H E G is a normal subgroup and W is a representation of H . Then Vdef=

IndGH(W ) is irreducible if and only ifW is irreducible and ρ is not isomorphic to TwsH(ρ) for all s ∈ G\H .

Remark 3.49. Let H ≤ G be a subgroup of index |G : H| = 2 ; in particular H E G. Pick s ∈ G \ Hso that G = H ∪ sH . Let ρ : H → GL(W ) be a representation of H . There are two possibilities : eitherρ ' Tws

H(ρ) or ρ 6' TwsH(ρ).

If ρ ' TwsH(ρ), then ResGH

(IndGH(W )

)= W ⊕ IndHH(Ws) ' W ⊕W ; in particular, IndGH(W ) is a

reducible representation of G by Corollary 3.48. As an example, takeW to be an irreducible representation

of H and let G = H×Z/2Z ; write Z/2Z = 1, s, h def= (h, 1) and s

def= (1, s) so that ResGH

(IndGH(W )

)'

W⊕sW is the decomposition given by Theorem 3.45 ; furthermore, we see that as C[H]-modules, sW 'W .We also see that IndGH(W ) is a reducible representation of G since

IndGH(W ) ' C[H × Z/2Z]⊗C[H] W ' (1 + s)W ⊕ (1− s)W

and these two representations are the box products (c.f. Definition 1.22) of the representation W with theirreducible representations of G/H ' Z/2Z. We note that the trivial representation and the alternatingrepresentation are not isomorphic, hence so is their tensor product with W (c.f. Theorem 2.57). Worthnoting is that when V is reducible, the decomposition of Theorem Theorem 3.45 is in general not thedecomposition of V into irreducible subrepresentations.

If ρ 6' ρs, then IndGH(ρ) is an irreducible representation of G by Corollary 3.48. For example, if weinduce the representations of the cyclic group Cn defined by ψh(rk) = ωkh to Dn (c.f. Example 2.71),since |Dn : Cn| = 2, we are in the right context. The characters satisfy Tws

Cn(ψh) = ψh if and only ifωh = ω−1

h , i.e. if and only if h is divisible by n/2, which coincides again with our previous results (note thatrepresentations of degree 1 are their own characters).

Example 3.50. Let G = S5 and H = A5, so that |G : H| = 2. Recall the computation of the charactertable of A5 made in Example 2.72 :

cs 1 20 15 12 12

A5 1 (123) (12)(34) (12345) (12354)

χ1 1 1 1 1 1

χ2 4 1 0 −1 −1

χ3 3 0 −1 1+√

52

1−√

52

χ4 3 0 −1 1−√

52

1+√

52

χ5 5 −1 1 0 0

64


Let sdef= (45) be the chosen representative for the non-trivial (A5, A5)-double coset of S5. For i =

1, 2, 5, we see that χi(s−1σs) = χi(σ) for all σ ∈ S5, hence it follows that χi = TwsA5

(χi), so IndS5A5

(χi)

is reducible by Corollary 3.48. For i = 3, 4, we see that χi(s−1(12345)s) = χi((12354)) 6= χi((12345)),hence the characters

ψ5(σ)def= IndS5

A5(χ3) (σ) =

χ3(σ) + χ3(s−1σs) if σ ∈ A5

0 otherwise

ψ′5(σ)def= IndS5

A5(χ4) (σ) =

χ4(σ) + χ4(s−1σs) if σ ∈ A5

0 otherwise

are irreducible by Corollary 3.48. We note that ψ5 = ψ′5. As a sidenote, one easily checks (using thecompleted character table) that

IndS5A5

(χ1) = 2ψ1

ResS5A5

(ψ1) = χ1

IndS5A5

(χ2) = 2ψ3

ResS5A5

(ψ3) = χ2

IndS5A5

(χ5) = ψ6 + ψ7

ResS5A5

(ψ6) = ResS5A5

(ψ7) = χ5.

We can compute the character table of S5 by using the trivial/alternating/standard representation forψ1/ψ2/ψ3, multiplying ψ4 = ψ2ψ3 and using the induced representations IndS5

A5(χ3) = ψ5 = IndS5

A5(χ4)

(note that the characters χ3 and χ4 both induce ψ5 since the only conjugacy classes of A5 on which χ3

and χ4 differ form a single conjugacy class on S5). Since there are 7 conjugacy classes, we are missing 2irreducible characters. Since

120 = |S5| =7∑i=1

n2i = 1 + 1 + 16 + 16 + 36 + n2

6 + n27 =⇒ n2

6 + n27 = 50,

there are two possibilities for (n6, n7) up to permutation, namely (5, 5) and (7, 1). Using the group theoryof the symmetric group, since characters of degree 1 are morphisms to an abelian group, they have tocontain [S5, S5] = A5 in their kernel, hence correspond to characters of Z/2Z. This means S5 has only twoirreducible characters of degree 1, e.g. (n6, n7) = (5, 5).

We now make the assumption that at least one of the values χ6((12)), χ6((123)(45)) or χ6((1234))is non-zero (which will turn out to be true), which implies the relationship ψ7 = ψ6ψ2 ; the hypothesis isnecessary to ensure that ψ6ψ2 6= ψ6, so that it is indeed equal to our remaining character of degree 5. Wehave the following unknowns in the character table :

cs 1 10 20 15 20 24 30

A5 1 (12) (123) (12)(34) (123)(45) (12345) (1234)

ψ1 1 1 1 1 1 1 1

ψ2 1 −1 1 1 −1 1 −1

ψ3 4 2 1 0 −1 −1 0

ψ4 4 −2 1 0 1 −1 0

ψ5 6 0 0 −2 0 1 0

ψ6 5 x1 x2 x3 x4 x5 x6

ψ7 5 −x1 x2 x3 −x4 x5 −x6

The orthogonality of the pairs of columns (1, (123)), (1, (12)(34)) and (1, (12345)) shows that x2 =−1, x3 = 1 and x5 = 0 ; that of the columns ((12), (123)(45)), ((12), (1234)) and ((123)(45), (1234)) givesus the equations x1x4 = 1, x1x6 = −1 and x4x6 = −1, from which we can deduce x1 = x4 = −x6 andx2

1 = 1. Giving x1 the values ±1 gives us the two remaining characters (which we can confirm that they are

65

Chapter 3

irreducible characters by computing 〈χ6, χ6〉 = 1 = 〈χ7, χ7〉, so here is the completed character table ofS5 :

cs 1 10 20 15 20 24 30

A5 1 (12) (123) (12)(34) (123)(45) (12345) (1234)

ψ1 1 1 1 1 1 1 1

ψ2 1 −1 1 1 −1 1 −1

ψ3 4 2 1 0 −1 −1 0

ψ4 4 −2 1 0 1 −1 0

ψ5 6 0 0 −2 0 1 0

ψ6 5 1 −1 1 1 0 −1

ψ7 5 −1 −1 1 −1 0 1

Fun fact : one is welcome to check that ψ23,α = ψ5, e.g. the second exterior power of the standard

representation ψ3 is the irreducible representation ψ5.

66

Bibliography

[1] Jean-Pierre Serre, Linear representations of finite groups, Springer-Verlag, New York, 2nd edition, 1977.

[2] William Fulton, Joe Harris, Representation Theory : A First Course, Springer-Verlag, New York, 1991.

67

Notation index

A†, f † conjugate transpose, 16Aσ,σ

′block form of A w.r.t. (σ, σ′), 60

Dn dihedral group of order 2n, 35, 41, 62G

X group action of G on X , 28G\X orbit space of G X , 28G/H left coset space, 28Gx stabilizer subgroup of x, 28H\G right coset space, 28HsK (H,K)-double coset of s, 62Pχ(f) projection onto χ, 25, 26R system of left coset representatives, 38Sn symmetric group, 3V G subspace of fixed elements, 21W⊥ orthogonal complement of W , 7Xs set of fixed points of s ∈ G, 28Z(C[G]) center of C[G], 50∆ : X → X ×X diagonal map, 28 box tensor product, 10χ(s) = tr (ρs) character, 19χC classical representation of Sn, 61χ1 χ2 box product of characters, 35χ1χ2 product of characters, 19χ1 ⊗ χ2 tensor product of characters, 19χ2α exterior square of χ, 20χ2σ symmetric square of χ, 20‖v‖ norm of a vector v, 17H\G/K double coset space, 62〈S〉C C-linear span of S, 2〈−,−〉G bracket product, 26, 54S1 circle group, 14RepC(G) category of representations of G, 5Ci ith conjugacy class of G, 24D(η) drop of η, 36L(ρ) lift of ρ, 36

Aut(G) automorphism group, 32Bil(V1, V2;W ) bilinear maps V1 × V2 →W , 8EndC(V ) endomorphism ring of V , 16GL(V ) automorphism group of V , 2GLn(C) general linear group, 2HomC[G](V,W ) hom-set of V and W , 5

IndGH(χ) induced character, 39IndGH induction functor, 53IndGα induction along α, 55Inn(G) inner automorphism group, 32OrbG(x) orbit of x ∈ X , 28Out(G) outer automorphism group, 32ResGH restriction functor, 38, 53ResαH restriction along α, 55StabG(x) stabilizer subgroup of x, 28Tws

H twist functor, 57ord(G) order of the group G, 2ord(s) order of the element s ∈ G, 2tr trace, 12C field of complex numbers, 2C[G] complex group algebra of G, 46C× unit group of C, 3Ccl[G] space of class functions, 22⊕ direct sum, 7RS

integral closure of R in S, 51degχ degree of χ, 19deg ρ degree of ρ, 2, 26ρ : G→ GL(V ) representation, 2ρs image of s ∈ G under ρ, 2θ = χ− χ1 standard representation, 30, 31fσ,σ

′block form of f w.r.t. (σ, σ′), 60

rij(s) matrix form of a representation, 8k[G]-Modfin category of f.g. k[G]-modules, 46

68

Terminology index

A>, 17G-invariant inner product, 7G-linear map, 2G-set, 28G-stable subspace, 5k[G]-module, 46s-twist, 57

actionk-transitive, 28doubly transitive, 28, 31group, 28transitive, 28

adjoint linear map, 16algebraic eigenspace, 13algebraic eigenvalue, 13algebraic integer, 51autoadjoint, 16automorphism group of a vector space, 2

block, 60block form, 60box tensor product, 10Burnside’s Lemma, 30

canonical G-action, 8canonical component, 27canonical decomposition, 27character, 19

box product, 35degree, 19exterior square, 20induced, 39irreducible, 19symmetric square, 20table, 32

classical representation of Sn, 61column orthogonality relations, 27complement, 5complete flag, 18

completely reducible, 47conjugate transpose, 16cycle structure, 33

diagonal matrix, 16diagonalizable matrix, 16double cosets, 62

evaluation map, 12

flag, 18Fourier inversion formula, 50free vector space, 3Frobenius reciprocity, 54Frobenius reciprocity along α, 59

general linear group, 2geometric eigenspace, 13geometric eigenvalue, 13group

alternating, 40, 43circle, 14dihedral, 35, 41, 62Pontryagin dual, 14symmetric, 3

group algebra, 46group order, 2

hermitian matrix, 16hermitian transpose, 16

induction along α, 55, 58inner product, 7inner product space, 7integral over R, 51internal direct product, 35irreducible k[G]-module, 47irreducible representation, 11

Jordan normal form, 13

Kronecker product, 36

Maschke’s Theorem, 6

69

matrix block, 60

normal matrix, 16

orbit, 28orbit-stabilizer theorem, 29orthogonal, 7orthogonal complement, 7

partition, 60Plancherel’s formula, 49projection, 5

representation, 2alternating, 14degree, 2drop, 36dual, 21exterior square, 9induced, 38left regular, 3lift, 36permutation , 3right regular, 3sign, 14standard, 14, 31, 40

symmetric square, 9trivial, 3unitary, 18

restriction along α, 55, 58row orthogonality relations, 23

Schur decomposition, 17Schur’s Lemma, 12semisimple, 48simple k[G]-module, 47spectral theorem, 18stabilizer, 28subgroup index, 28subrepresentation, 5

tensor product of characters, 19tensor product of vector spaces, 8tensor product representation, 8trace, 12triangular matrix, 16

unitarily diagonalizable, 16unitary matrix, 16universal property, 54, 56universal property of the tensor product, 9upper/lower triangular matrix, 16

70

representation theory -...

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