Representation Theory of Finite Groups

Notes (loosely) inspired by a class taught

by Brian Parshall in Fall 2017

George H. Seelinger

1. Introduction

From one perspective, the representation theory of finite groups is merelya special case of the representation theory of associative algebras by consid-ering the fact that a sufficiently nice finite group ring k[G] is semisimple,and thus the results from Artin-Wedderburn theory apply. However, the ex-tra structure of the group provides a rich connection between k[G]-modulesand linear algebra.

A representation of a finite group is a way to induce an action of afinite group on a vector space. Sometimes, such a perspective allows mathe-maticians to see structure or symmetries in groups that may not have beenreadily apparent from a purely group theoretic point of view, much like mod-ules can provide insight into the structure of rings. In fact, a reprsentationof a finite group is fundamentally the same as a module of a finite groupalgebra, but the representation theoretic perspective allows us to leveragetools from linear algebra to gain additional insights.

One of the most oft-cited examples of representation theory in action isBurnside’s pq-theorem (see 11), but representations are also often studied intheir own right. In general, representation theories seem to follow a generalprogram along the following lines.

(a) To what degree do representations decompose into direct sums ofsubrepresentations? The best answer is when all representationsare completely reducible, that is, a representation breaks up into adirect sum of the irreducible subrepresentations. This happens forfinite group representations over C.

(b) Can we determine the irreducible and indecomposable representa-tions of a given object? For finite group representations over C,the irreducible representations are completely determined by thegroup’s “character table.”

(c) Given a tensor product of representations, which irreducible and/orindecomposable representations does the product contain?

These are the questions this monograph will try to answer for representationsof finite groups over C. Over fields where the characteristic of the fielddivides the order of the group, different techniques are often used in an areaof math known as “modular representation theory,” and such a situationwill not be discussed much here.

While the classic texts in representation theory continue to be cited andused often, and contain many details ommitted here, I am eternally gratefulfor [EGH+11] and [Smi10] for being approachable, providing simplifyinginsights, and using more modern language to rephrase classical results. Iam also grateful for [Tel05], out of which I first learned the representationtheory of finite groups, even if I did not understand a lot of it at the time.Much of the writings here are shamelessly borrowed and synthesized fromthese sources.

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2. Group Actions

The following exposition is a summary of [Gro]. The reader is probablyfamiliar with the notion of a group action.

2.1. Definition. A group action on a set X is a homomorphism φ : G→Aut(X) such that φ(e) = IdX .

If the reader is unfamiliar with group actions, most introductory textson group theory will provide more than adequate treatment.

2.2. Definition. A group action is called faithful if φ is injective. Agroup action is called transitive if φ(G) induces only one orbit on X.

When an action is not transitive, we have multiple orbits and we canconsider the group action on each orbit separately. Thus, in a sense, we can“decompose” this group action (on the level of sets). Thus, let us focus ontransitive actions as our basic building blocks of group actions.

2.3. Theorem. Let φ be a transitive group action of G on a set X. Then,consider H = StabG(x). Then, G/H ∼= X as sets with a (left) G-action.

Proof. Given H = StabG(x), consider the correspondence

gH ↔ g.x = φ(g)x

Then, the action of G preserves this correspondence, namely

g′.gH = (g′g)H ↔ (g′g).x = φ(g′g)x = φ(g′)φ(g)x = g′.φ(g)x = g′.(g.x)

�

Thus, we have reduced the classification of transitive G-actions on X tothe study of conjugacy classes of subgroups of G. Thus, the structure of agroup that controls a group action on a set X is the subgroup structure ofG. However, understanding the subgroup structure of a group is, in general,incredibly difficult. For instance, recall Cayley’s theorem.

2.4. Theorem (Cayley’s Theorem). Any finite group G with |G| = n <∞ can be embedded into Sn via the action of G on itself.

Thus, to understand the subgroups of Sn, one must understand all finitegroups G with order less than n. Thus, while group actions are incrediblyuseful and general, they lack structure that can be otherwise useful.

3. Group Algebras

3.1. Definition. The group ring or group algebra of a group G over aring R, denoted R[G] is a free R-module over the set

{eg | g ∈ G}

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with addition being formal sums. Additionally, R[G] has the structure ofa ring where multiplication is given by eg · eh = egh for g, h ∈ G. Thus,1R[G] = e1G

Typically, we will take R to be a field, usually C. Furthermore, we willoften replace eg by g when it is understood that we are working in the groupalgebra. That is, for λ ∈ R, we may rewrite

λeg → λg

Since group algebras are R-algebras, we can ask many mathematical ques-tions about their modules.

3.2. Theorem (Maschke’s Theorem). Let G be a finite group and k bea field such that chark - |G|. Then, k[G] is completely reducible as a moduleover itself.

Proof. Let V be a proper k[G]-submodule of k[G]. Consider the k-linear map map π : k[G]→ B and let φ : k[G]→ V be given by

φ(x) =1

|G|∑s∈G

s.π(s−1.x)

Thus, φ is also a projection since, for v ∈ V ,

φ(v) =1

|G|∑s∈G

s.π(s−1.v) =1

|G|∑s∈G

s.s−1.v = v

and it is k-linear since it is simply the (appropriately scaled) sum of k linearmaps. In fact,

φ(t.x) =1

|G|∑s∈G

s.π(s−1.t.x)

=1

|G|∑s′∈G

(ts′) · π(s′−1.x) (s′−1 = s−1t =⇒ s = ts′)

= t.φ(x)

so, φ is k[G]-linear. Thus, k[G] ∼= V ⊕ kerφ, but V was an arbitrary propersubmodule. Thus, k[G] is completely reducible. �

3.3. Remark. The converse of Maschke’s theorem is also true.

Thus, since k[G] is finite-dimensional and thus also artinian, k[G] is asemisimple ring when chark - |G| and, using artin-wedderburn theory, hasdecomposition as an module over itself

k[G] ∼=k⊕i=1

Mni(Di)

for ni ∈ N and Di a division ring, and these direct summands represent allthe irreducible k[G]-modules. However, we still wish to understand what

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these division rings and ni’s actually are (see [See17]). Thus, we will shiftour perspective to looking at group representations instead of group algebramodules.

4. Representations

In this section, we seek to establish some of the basic ideas of representa-tions of finite groups. Much of this exposition is borrowed from [EGH+11]where the results are presented more generally for R-algebras. Also, manyof these results can be rephrased using the language in [See17].

4.1. Definition. A representation of a finite group G in a (complex)vector space V is a homomorphism ρ : G→ GL(V ).

4.2. Remark. Throughout this monograph, we will often simply say“representation” to mean a representation of a finite group. When we referto a vector space V , we will mean a vector space over C.

4.3. Theorem. There is a natural bijection between C[G]-modules andcomplex representations of G.

Proof. Given a representation of G given by ρ : G→ GL(V ), we get aC[G] action on v ∈ V via

eg.v = ρ(g)v

and extend via linearity. Conversely, given M a C[G]-module, we can con-sider M as a vector space over Ce1 and then view the action of each eg asan invertible linear transformation on this vector space. �

4.4. Definition. We say two representations ρ, ρ′ : G → GL(V ) aresimilar or isomorphic if there is a linear transformation τ : V → V ′ suchthat

τ ◦ ρ(g) = ρ′(g) ◦ τfor all g ∈ G.

4.5. Definition. We say that a subspace W ⊆ V is G-stable if ρ(g)W ⊆W for all g ∈ G.

4.6. Proposition. Let ρ : G → GL(V ) be a linear representation of Gin V and let W be a subspace of V such that ρ(g)W = W for all g ∈ G.Then, there exists a complement W ′ of W in V such that ρ(g)W ′ ⊆W ′ forall g ∈ G

Proof. LetW ′ be a vector space complement ofW in V (not necessarilyG-stable) and let p : V →W be the standard projection. Then, consider themap

p0 :=1

|G|∑t∈G

ρ(t) ◦ p ◦ ρ(t)−1

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Such a map is a projection of V onto a subspace of W . In fact, p0|W = IdWsince, for w ∈W ,

p ◦ ρ(t)−1w = ρ(t)−1w =⇒ ρ(t) ◦ p ◦ ρ(t)−1w = w =⇒ p0w = w

Thus, we seek to show W 0 := ker p0 is stable under the G action via ρ.Indeed, it is easy to check ρ(s)p0ρ(s)−1 = p0 and thus p0◦ρ(s)x = ρ(s)◦p0x =0, that is, ρ(s)x ∈W 0. Thus, V = W ⊕W 0 is a decomposition into G-stablesubspaces. �

4.7. Remark. Note the similarity between this proof and the proof ofMaschke’s theorem above (3.2). Indeed, these proofs are more or less equiv-alent and the proposition above is the same as Maschke’s theorem for k = C.

4.8. Definition. Given representations ρ1 : G → GL(V1) and ρ2 : G →GL(V2), we define ρ := ρ1 ⊕ ρ2 : G→ GL(V1 ⊕ V2) by the map

g 7→(ρ1(g) 0

0 ρ2(g)

)That is, ρ(g)|Vi = ρi(g).

4.9. Definition. We say that a representation ρ : G → GL(V ) is irre-ducible or simple if V 6= 0 and there is no subspace W ⊆ V such that W isstable under the action of G via ρ. By the proposition above, this is equiv-alent to saying that ρ does not break into a direct sum of representations.

4.10. Proposition. Every representation is a direct sum of irreduciblerepresentations.

Proof. The proof follows from induction on dimV and application ofthe proposition above. �

4.11. Definition. Let ρ1 : G→ GL(V1) and ρ2 : G→ GL(V2) be repre-sentations. Then, we define ρ := ρ1 ⊗ ρ2 : G→ GL(V1 ⊗ V2) by, for s ∈ G,

ρ(s)(v1 ⊗ v2) = ρ1(s)v1 ⊗ ρ2(s)v2

4.12. Remark. The tensor product of two irreducible representations isnot typically irreducible.

4.13. Theorem. Let ρ : G → GL(V ) be a representation. Then, givenρ⊗ ρ→ GL(V ⊗ V ) decomposes as

V ⊗ V ∼=V ⊗ V

(x⊗ y − y ⊗ x)⊕ V ⊗ V

(x⊗ y + y ⊗ x)= Sym2(V )⊕Alt2(V )

Proof. Consider the automorphism of V ⊗V given by τ(ei⊗ej) = ej⊗eifor basis {e1, . . . , en} of V . Then, τ(v ⊗ w) = w ⊗ v for any v, w ∈ V andτ2 = IdV⊗V . Thus, as vector spaces,

V ⊗ V ∼= ker(τ − Id)⊕ im(τ − Id)

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However, ker(τ −Id) = 〈x⊗y | x⊗y−y⊗x = 0〉 ∼= (V ⊗V )/(x⊗y−y⊗x).Now, consider that

(τ − Id)(x⊗ y) + τ(τ − Id)(x⊗ y) = (τ − Id)(x⊗ y) + (Id− τ)(x⊗ y) = 0

and so, for any v ∈ im(τ − Id), we get v + τ(v) = 0. Thus, im(τ − Id) ⊆(V ⊗ V )/(x⊗ y + y ⊗ x). However, by dimension counting, we note that

dim ker(τ − Id) = dim Sym2(V ) =n(n+ 1)

2=⇒

dim im(τ − Id) = n− n(n+ 1)

2=n(n− 1)

2= dim Alt2(V )

Thus, im(τ − Id) = Alt2(V ).

However, we must also check that these spaces are stable under G. How-ever, for ρ′ = ρ⊗ ρ,

ρ′(g) ◦ τ ◦ ρ′(g−1)(v ⊗ w) = ρ′(g) ◦ τ(ρ−1(g)v ⊗ ρ−1(g)w)

= ρ′(g)(ρ−1(g)w ⊗ ρ−1(g)v)

= w ⊗ v= τ(v ⊗ w)

and so the image and kernel of τ − Id is stable under the action of G. �

We now also seek to prove some other useful tools in representationtheory.

4.14. Theorem (Schur’s Lemma). Let V1, V2 be representations of G andlet φ : V1 → V2 be a nontrivial homomorphism of representations. Then,

(a) If V1 is irreducible, then φ is injective.(b) If V2 is irreducible, then φ is surjective.

Proof. Exercise for the reader. �

4.15. Corollary (Schur’s Lemma for algebraically closed fields). LetV be a finite dimensional irreducible representation of a group G over analgebraically closed field k, and φ : V → V a commuting homomorphism (ieρ(g) ◦ φ = φ ◦ ρ(g)). Then, φ = λIdV for some λ ∈ k.

4.16. Remark. (a) We sometimes call φ a “scalar operator” or saythat φ “acts as a scalar” in this situation. In [Ser97], Serre callssuch an operator a “homothey.”

(b) This proposition is false over R.

Proof. Let λ be an eigenvalue of φ, which exists since k is algebraicallyclosed. Then, since φ commutes with ρ(g), so does φ − λId : V → V .However, φ − λId is not an isomorphism since det(φ − λId) = 0. Thus,φ− λId = 0 =⇒ φ = λId. �

4.17. Corollary. Let G be an abelian group. Then, every irreduciblefinite dimensional representation of G is 1-dimensional.

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Proof. Let V be an irreducible finite dimensional representation of G.Then, for g, h ∈ G, v ∈ V

ρ(g)ρ(h)v = ρ(gh)v = ρ(hg)v = ρ(h)ρ(g)v

Thus, ρ(g) commutes with all ρ(h) and so, by the corollary above, ρ(g) = λIdfor some λ ∈ C. Thus, every vector subspace of V is a subrepresentation,but V is irreducible. Thus, dimV = 1. �

4.18. Example. Let G = S3, the smallest order non-abelian group, andlet V be a complex representation of G. Since S3 is generated by σ = (123)and τ = (12), we need only find subrepresentations of V that are stableunder the actions of σ and τ . An important tool for this process will be tofind eigenvectors. Let v be an eigenvector for σ with eigenvalue λ. Then,consider τv. Such a vector must be an eigenvector for σ as well since

στv = τσ2v = τλ2v = λ2τv

Thus, the space span{v, τv} is stable under the actions of σ, τ and must bea subrepresentation of V .

Now, if we wish to find all irreducible representations of G, we knowthat the dimension must be less than or equal to 2. However, we can figureout more. We know that σ3 = Id, and so it must be that v = σ3v = λ3vand so λ = 1, ζ3, or ζ23 , where ζ3 is a primitive 3rd root of unity.

• Let λ 6= 1. Then, λ 6= λ2 and so v and τv have distinct eigen-values and thus they are linearly independent. Thus, span{v, τv}defines an irreducible, 2-dimensional representation. This is calledthe standard representation. Explicitly, it is given by

τ 7→(

0 11 0

), σ 7→

(ζ3 00 ζ23

)• Let λ = 1 and τv = −v. Then, span{v} is an irreducible 1-

dimensional representation, but it is distinct from the trivial repre-sentation. It is called the alternating representation.• Let λ = 1 and τv = v. This is the trivial representation.

Thus, we have found 3 irreducible representations of S3.

4.19. Definition. We say a representation ρ : G→ GL(V ) is faithful ifker ρ = {1}.

4.20. Proposition. Given a representation ρ : G→ GL(V ) and a nor-mal subgroup H E G such that ρ acts trivially on H, (that is, ρ(H) ={IdV }), ρ can descend to a representation of G/H, say ρ, via

ρ(gH) = ρ(g)

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Proof. We must check that such an action is well defined. Let g, g′ ∈gH. Then, g = hg′ so

ρ(gH) = ρ(g) = ρ(hg′) = ρ(h)ρ(g′) = ρ(g′) = ρ(g′H)

�

4.21. Proposition. Given representation ρ : G→ GL(V ), we have thatρ : G/ ker ρ→ GL(V ) is faithful.

4.22. Proposition. Let ρ : G/H → GL(V ) be a representation of G/Hfor H E G. Then, we can lift ρ to a representation of G, say ρ̃, via

ρ̃(g) = ρ(gH)

Furthermore, if ρ is a faitful representation of G/H, then ker ρ̃ = H.

Proof. This construction is immediately well defined. If ρ is faithful,then we note that

ρ̃(g) = Id⇐⇒ ρ(gH) = Id⇐⇒ gH = H

�

5. The Regular Representation

One of the most important representations is the regular representation.Later, we will see that every irreducible representation appears in the reg-ular representation. Furthermore, the regular representation is a somewhatnatural representation to define.

5.1. Definition. Let G be a finite group. Then, the regular represen-tation of G is a homomorphism ρreg : G → GL(V ) where V = C|G| givenby

g 7→ (g : V → V, eh 7→ egh)

Note that the regular representation is equivalent to viewing k[G] as amodule over itself.

5.2. Example. Consider G = S3. Then, the regular representation ofS3 is given by

σ.eσ′ = eσσ′

Now, consider that the element

e(1) + e(12) + e(13) + e(23) + e(123) + e(321)

is a G-invariant subspace of C6 under the regular representation. This sub-representation is isomorphic to the trivial represenation. Similarly, the spacespanned by

e(1) − e(12) − e(13) − e(23) + e(123) + e(321)is G-stable and, as a subrepresentation of the regular representation, is iso-morphic to the alternating representation. Thus, we have

ρreg = ρtrivial ⊕ ρalt ⊕ ρ′

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where ρ′ : G → GL(C6/(Vtrivial ⊕ Valt)). We know from our analysis abovethat ρ′ must break up further, and we will see later that it breaks up as thedirect sum of 2 standard representations.

6. Hom and dual representation

We take a quick detour to view some other methods for constructingnew representations from old ones. These results will be useful later, so thereader may choose to skip them for now and come back later when they areneeded.

6.1. Proposition. Let V,W be representations of a finite group G.Then, the action of G on the vector space HomC(V,W ) given by, for φ ∈HomC(V,W ),

g.φ : V →W

v 7→ g.φ(g−1.v)

makes HomC(V,W ) into a representation of G.

6.2. Corollary. Let V be a representation of a finite group G. Then,V ∗ = HomC(V,C) has a natural representation structure. Explicitly, forbasis f ∈ V ∗,

ρV ∗(g).f = f ◦ ρV (g−1)

6.3. Proposition. Given finite-dimensional representations V,W of G,then there is a natural isomorphism

V ∗ ⊗W∼→ HomC(V,W )

Proof. Consider the C-bilinear map

V ∗ ×W → HomC(V,W )

(f(·), w) 7→ f(·)wThen, by the universal property of tensor products, there is a unique ho-momorphism φ : V ∗ ⊗ W → HomC(V,W ) such that f(·) ⊗ w 7→ f(·)w.Such a map is certainly injective since kerφ = {0}. It is surjective sinceHomC(V,W ) is isomorphic to dimW×dimV matrices, so dim HomC(V,W ) =dimV · dimW = dimV ⊗W = dimV ∗ ⊗W . �

6.4. Remark. The construction of the Hom representation more gener-ally comes from the Hopf algebra structure on C[G].

7. Character Theory

Throughout this section, let G be a finite group and F = C. We wishto extend the notion of the central character above to a general characterχV : G → C that will still carry useful information about the group repre-sentation V .

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7.1. Proposition. If V is a G-module, then every g ∈ G, viewed asa linear operator gV : V → V , is semisimple, and thus V has a basis ofeigenvectors for gV .

Proof. Since G is a finite group, g is of finite order, so gnV = IdV . Thus,gnV = (D+N)n for diagonalizable D and nilpotent N (by Jordan CanonicalForm since C is algebraically closed) such that DN = ND. Thus,

IdV = gnV = (D+N)n = Dn+nNDn−1+

(n

2

)N2Dn−2+· · ·+nNn−1D+Nn

which thus tells us that N = 0 and Dn = IdV since IdV has no nilpotentpart. �

In general, it is quite annoying to compute the eigenvalues for eachg ∈ G. However, we will see that it suffices to merely compute their sumusing the trace.

7.2. Definition. The character of group representation ρ : G → V ,denoted χρ = χρ, is the function χρ : G→ k defined by χρ(g) = tr(ρ(g)).

7.3. Proposition. For k = C and fixed group representation ρ,

(a) χ(1) = dimV ,

(b) χ(g−1) = χ(g) for all g ∈ G,(c) χ(tst−1) = χ(s) for all s, t ∈ G.

Proof. All these properties follow from standard properties of trace.�

7.4. Proposition. Let V be a group representation and let χV be itscharacter. Then,

χV (g) = χV ∗(g)

Proof. Consider that, by 6.2, χV ∗(g) = χV (g−1). However, since χV (g)is just the sum of eigenvalues of g (which must be roots of unity by the proofof 7.1), we have that

χV ∗(g) = χV (g−1) =∑

λ−1i =∑

λi =∑

λi = χV (g)

�

7.5. Definition. A function f : G → C is called a class function if itis constant on conjugacy classes. Note that Class(G) is the C-algebra of allclass functions on G and dim Class(G) = the number of conjugacy classesin G.

7.6. Proposition. Given a representation of a finite group G and arepresentation ρ, χρ : G→ C is a class function.

Proof. This follows from part (c) of the previous proposition. �

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7.7. Proposition. Let ρi : G → GL(Vi), i = 1, 2 be representations ofG with characters χρi.

(a) χρ1⊕ρ2 = χρ1 + χρ2(b) χρ1⊗ρ2 = χρ1 · χρ2

Proof. This follows from the definition of direct sum and tensor prod-uct of representations and the definition of the character. �

Based on the identity (a) above, we see that the characters of irreduciblerepresentations will be of primary interest.

8. Orthogonality Relations of Irreducible Characters

Inspired by the proof of Maschke’s theorem, we present the followingproposition.

8.1. Proposition. Let h : V1 → V2 be a linear transformation and letρ1 : G → GL(V1), ρ

2 : G → GL(V2) be irreducible representations of finitegroup G. Consider

h0 :=1

|G|∑t∈G

ρ2(t−1)hρ1(t)

h0 is a homomorphism of representations. Furthermore

(a) If ρ1 6∼= ρ2, then h0 = 0.(b) If ρ1 ∼= ρ2 (and thus V1 ∼= V2), then h0 = 1

dimV1tr(h)IdV1.

Proof. � Fill this in

8.2. Corollary. Consider that ρ(t) ∈ GL(V ) and thus we can representρ(t) as a n× n matrix, say ρ1(t) = [ri1,j1(t)] and ρ2(t) = [ri2,j2(t)]. Then,

(a) If ρ1 ∼= ρ2,

1

|G|∑t∈G

ri2,j2(t−1)rj1,i1(t) =1

nδj1,j2δi1,i2

(b) If ρ1 6∼= ρ2,

1

|G|∑t∈G

ri2,j2(t−1)rj1,i1(t) = 0

This provides us motivation for definiting a positive definite Hermitianinner product on the set of class functions on G.

8.3. Definition. For class functions φ, ψ : G→ C, we define inner prod-uct

〈φ, ψ〉 =1

|G|∑g∈G

φ(g−1)ψ(g)

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and also

(φ, ψ) =1

|G|∑g∈G

φ(g)ψ(g)

8.4. Proposition. Let 〈·, ·〉 and (·, ·) be as above.

(a) 〈·, ·〉 is a bilinear form.(b) 〈φ, ψ〉 = 〈ψ, φ〉, that is 〈·, ·〉 is symmetric.(c) (φ, ψ) = 〈φ, ψ ◦ i〉 where i : G→ G takes i(g) = g−1.(d) (·, ·) is a positive definite Hermitian inner form.

8.5. Proposition. If ρ is a representation of G in V and if ψ(t) =tr ρ(t), then

〈φ, ψ〉 = (φ, ψ)

Thus, it is with the form (·, ·) that we can arise at the following

8.6. Theorem. For any representations V,W ,

(χV , χW ) = dim HomG(W,V )

and, if V,W are irreducible,

(χV , χW ) =

{1 V ∼= W

0 V 6∼= W

Proof. We first compute for arbitrary representations V,W ,

(χV , χW ) =1

|G|∑g∈G

χV (g)χW (g)

=1

|G|∑g∈G

χV (g)χW ∗(g) by 7.4

=1

|G|∑g∈G

χV⊗W ∗(g)

=1

|G|∑g∈G

χHomC(W,V )(g) by 6.3

= trHomC(W,V )

1

|G|∑g∈G

g

However, P := 1

|G|∑

g∈G g is a homomorphism of G representations (that

is, P.v is a representation of G) and the image of P is stable under anyaction of G since g.Pv = Pv. However, the only such irreducible repre-sentation is the trivial representation or the zero representation. Thus, forarbitrary representation U , trU (P ) simply counts the number of times the

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trivial representation occurs as a subrepresentation of U . However, consid-ering HomC(W,V ) as a G-representation, the G-invariant subrepresentationis exactly HomG(W,V ). Thus,

(χV , χW ) = trHomC(W,V )

1

|G|∑g∈G

g

= dim HomG(W,V )

Furthermore, if V,W are irreducible, Schur’s lemma tells us that

dim HomG(W,V ) =

{1 V ∼= W

0 else

�

Thus, we have shown that the irreducible characters of a finite group areorthogonal to each other under (·, ·). In fact, we have the following theorem

8.7. Theorem. The irreducible characters of G, {χ1, . . . , χk} form anorthonormal basis for the set of class functions on G.

8.8. Lemma. Let ρ : G → GL(V ) with dimV = n be an irreduciblerepresentation of G and let f be a class function on G. Then, given

ρf : V → V, ρf :=∑g∈G

f(g)ρ(g)

we have ρf = 1n |G|(f, χρ)IdV .

Proof of Lemma. Let h ∈ G. Then,

ρfρ(h) =∑g∈G

f(g)ρ(g)ρ(h)

=∑g∈G

f(g)ρ(gh)

=∑g∈G

ρ(h)ρ(h−1)f(g)ρ(gh)

= ρ(h)∑g∈G

f(g)ρ(h−1gh)

= ρ(h)∑g∈G

f(h−1gh)ρ(h−1gh)

= ρ(h)ρf

Since V is irreducible, it must be that ρf = λ1V by Schur’s lemma. Thus,taking the trace of both sides, we get

nλ = tr(ρf ) =∑g∈G

f(g)χρ(g)

15

and thus

λ =1

n|G|(f, χ)

�

Proof of Theorem. By the theorem above, we know that the irre-ducible characters form an orthonormal system. Thus, we need only showthat they generate Class(G). It will suffice to show that if f ∈ Class(G) has(f, χi) = 0 for all irreducible characters χi, then f ≡ 0. Assume (f, χi) = 0.Then, by the lemma above, ρf = 0 for any irreducible representation ρ.Thus, from the direct sum decomposition, we can conlude that ρf = 0 forany ρ. So, applying this to the regular representation, ρreg, and computingthe image of a basis vector e1, we get

0 = (ρreg)fe1 =∑g∈G

f(g)ρreg(g)e1 =∑g∈G

f(g)eg =⇒ f(g) = 0,∀g ∈ G.

Thus, f = 0. �

8.9. Corollary. The number of irreducible representations of G is equalto the number of conjugacy classes of G.

8.10. Remark. We also know this from Wedderburn’s theorem + Maschke’stheorem.

Proof. Let g ∈ G and fg ∈ Class(G) be defined by

fg(h) :=

{1 if g and h are conjugate

0 otherwise

These fg functions form a natural basis for Class(G) when one is chosen foreach conjugacy class in G. Thus, dim Class(G) is the number of conjugacyclasses of G. On the other hand, using the theorem above, dim Class(G) isequal to the number of irreducible representations of G. Thus, these twoquantities are equal. �

8.11. Theorem. Let g ∈ G and C(g) be the number of elements in theconjugacy class of g. Then,

(a)∑d

i=1 χi(g)χi(g) = |G|C(g)

(b)∑d

i=1 χi(g)χi(h) = 0 if g is not conjugate to h.

8.12. Remark. In particular,∑d

i=1 χi(1)χi(1) = |G|.

Proof. Using fg ∈ Class(G) as above, write

fg =k∑i=1

λiχi, λi ∈ C

16

Then, λi = (fg, χi) = C(g)|G| χi(g) where C(g) is the number of elements in the

conjugacy class of g. Therefore,

fg(h) =C(g)

|G|

k∑i=1

χi(g)χi(h)

Thus, by definition of fg(h), the identities in the theorem are proved. �

8.13. Example. Let G = S3. We know that S3 has 3 conjugacy classesgiven by cycle type. Thus, S3 has 3 distinct irreducible characters, denotedχ1, χ2, and θ for trivial, alternating, and standard representations. We thenget the following “character table”

(1) (12) (123)χ1 1 1 1χ2 1 -1 1θ 2 0 -1

where the first 2 rows follow from just computing the trace of the trivial andalternating representations. The third row then follows from our orthogo-nality relations, since

θ(1) = 2

0 = (χ1, θ) =1

6

∑g∈S3

θ(g)

0 = (χ2, θ) =1

6

∑g∈S3

sgn(g)θ(g)

=⇒ 0 =1

3(2 + θ((123)) + θ((321)))

=⇒ θ((123)) = −1

and

0 = χ1((1))χ1((12))+χ2((1))χ2((12))+θ((1))θ((12)) = 1−1+2θ((12)) =⇒ θ((12)) = 0

Finally, returning to the regular representation of S3, we know that χreg(1) =6 and χreg = aχ1 + bχ2 + cθ. However, we also know that χreg(g) = 0 forall non-identity g ∈ S3 since no such elements fix any basis elements. Inparticular, χreg((123)) = 0. Thus, the unique decomposition of χreg is

χreg = χ1 + χ2 + 2θ

From this, we will see that

ρreg = ρtrivial ⊕ ρalt ⊕ ρstd ⊕ ρstd ∼= M1(C)⊕M1(C)⊕M2(C)

17

9. Results of Character Theory

Now, we wish to use character theory to tell us more about representa-tions of a group.

9.1. Proposition. Let H E G and let ρ : G/H → GL(V ) be a repre-sentation. Then, ρ is irreducible if and only if its lifted representation to G,ρ̃, is irreducible. (See 4.22 for a reminder of how a lifted representation isconstructed.)

Proof. We first note that

tr(ρ̃(g)) = tr(ρ(gN))

by definition of ρ̃. Thus, if χ is the character of ρ̃ and χ′ is the character ofρ, then χ(g) = χ′(gH). Now, we compute

(χ′, χ′) =1

|G/H|∑

gH∈G/H

χ′(gH)χ′(gH)

=|H||G|

∑gH∈G/H

χ(g)χ(g)

=1

|G|∑

gH∈G/H

∑h∈H

χ(gh)χ(gh)

=1

|G|∑g∈G

χ(g)χ(g)

= (χ, χ)

�

9.2. Example. Let G = A4. Then, |A4| = 12 and A4 has 4 conjugacyclasses given by

C1 = {(1)}, C2 = {(123), . . .}, C3 = {(132), . . .}, C4 = {(12)(34), . . .}and thus must have 4 irreducible characters, say χ1, χ2, χ3, χ4, and let χ1

be the trivial representation.

Now, consider that C4 is also a normal subgroup of A4 isomorphic toZ/2Z× Z/2Z, so we can find irreducible representations of A4/C4

∼= Z/3Z,we could lift them to irreducible representations of A4 by the propositionabove. Since Z/3Z is abelian, it immediately gives character table

(1) (123) (132)χ1 1 1 1χζ3 1 ζ3 ζ23χζ23 1 ζ23 ζ3

Of course, the trivial representation with character χ1 lifts to a trivial rep-resentation of A4. However, the lifted representation ρ̃ζ3 and ρ̃ζ23 will be

18

irreducible by the proposition above and thus give us irreducible charactersof A4. It is immediate that, up to reordering, χ2 = ρ̃ζ3 and χ3 = ρ̃ζ23 , so we

now fill in the rows of our character table to get

C1 C2 C3 C4

χ1 1 1 1 1χ2 1 ζ3 ζ23 1χ3 1 ζ23 ζ3 1χ4 a b c d

However, we know that

12 + 12 + 12 + a2 = 12 =⇒ a = ±3

by 8.12 and a = χ4(1) = dimV for V the corresponding vector space of ρ4.Thus a > 0 =⇒ a = 3. Furthermore,

12 + 12 + 12 + ad = 0 =⇒ ad = −3 =⇒ d = −1

by 8.11. Now, we can use our orthogonality relations to determine χ4. Usingthe orthongality of rows (8.6), we get

(χ4, χ1) = 3 + 4b+ 4c− 3 · 1 = 0

(χ4, χ2) = 3 + 4bζ23 + 4cζ3 − 3 · 1 = 0

(χ4, χ3) = 3 + 4bζ3 + 4cζ23 − 3 · 1 = 0

=⇒

{b = 0

c = 0

Thus, our completed character table is given by

C1 C2 C3 C4

χ1 1 1 1 1χ2 1 ζ3 ζ23 1χ3 1 ζ23 ζ3 1χ4 3 0 0 −1

Perhaps the most important reason to study character theory is thefollowing theorem.

9.3. Corollary. The multiplicity of an irreducible representation W ina representation V is (χW , χV ).

Proof. Let V ∼= Wm11 ⊕ · · · ⊕Wmk

k be a decomposition of V into irre-ducibles. Then, we have

χV = χWm11 ⊕···⊕Wmk

k= m1χW1 + · · ·+mkχWk

.

Moreover, since the irreducible characters form an orthonormal basis of classfunctions, we can apply (·, χWi) to our equality to get

(χV , χWi) = mi

�

19

9.4. Corollary. Two finite dimensional complex representations of afinite group G, say V,W , are isomorphic if and only if they have the samecharacter. In other words

V ∼= W ⇐⇒ χV = χW

Proof. The forward direction follows by definition of character. Thebackwards direction follows from the fact that the irreducible charactersform a basis of Class(G). Namely, χV = χW implies that χV and χW havethe same unique decomposition into irreducible characters, and thus fromthe corollary above, V,W have the same multiplicity of each irreduciblerepresentation. �

9.5. Corollary. A representation V is irreducible if and only if (χV , χV ) =1.

Proof. The forward direction was proved earlier in 8.6. For the back-wards direction, if χV = m1χV1 + · · · + mkχVk for irreducibles V1, . . . , Vk,then using the fact that the χVi ’s form an orthonormal basis, we get

1 = (χV , χV ) = m21 + · · ·+m2

k

Thus, since each m2i ≥ 0 and are in Z, it must be that exactly one mi = 1

and all others are 0. Thus, V must be irreducible by the corollary above. �

10. The Regular Representation Revisited

10.1. Corollary. The multiplicity of any irreducible representation Vin the regular representation equals its dimension.

Proof. Recall that the dimension of the regular representation is |G|.Consider

(χV , χreg) =1

|G|χV (1)χreg(1) = χV (1) = dimV

�

10.2. Corollary. Let {V1, . . . , Vk} be the set of all irreducible repre-sentations of finite group G. Then,

|G| =k∑i=1

(dimVi)2

Proof. From above, we have χreg =∑k

i=1 dimVi ·χVi , and so we simplytake (χreg, χreg) to get

(χreg, χreg) =∑

1≤i≤k,1≤j≤kdimVi dimVj(χVi , χVj ) =

k∑i=1

(dimVi)2

20

using the orthogonality of the irreducible characters. However, since

χreg(g) =

{|G| g = 1

0 otherwise

we also have (χreg, χreg) = |G|2|G| = |G| by definition of the inner product

(·, ·). �

10.3. Remark. This result should not come as a surprise to a readerfamiliar with Artin-Wedderburn theory. However, the question of what theanalogous decomposition of the regular representation looks like still re-mains.

10.4. Lemma. If ρ1 and ρ2 are irreducible representations of G1 and G2

respectively, then ρ1 ⊗ ρ2 is an irreducible representation of G1 ×G2

Proof. We have, from above, that for characters χ1 of ρ1 and χ2 of ρ2,that

(χ1, χ1) = 1 = (χ2, χ2)

However, χ := χρ1⊗ρ2 has

(χ, χ) =1

|G1||G2|∑

g∈G1,g′∈G2

χ1(g)χ2(g′)χ1(g)χ2(g′)

=1

|G1|

∑g∈G1

χ1(g)χ1(g)

1

|G2|

∑g′∈G2

χ2(g′)χ2(g′)

= (χ1, χ1) · (χ2, χ2)

= 1

Thus, it must be that ρ1 ⊗ ρ2 is irreducible. �

10.5. Remark. The converse, that each irreducibleG1×G2-representationis isomorphic to a tensor product of Gi-representations is also true. The ar-gument is to show that each class function of G1 ×G2 which is orthogonalto characters of the form χ1 · χ2 is zero.

10.6. Proposition. One can realize C[G] as a G × G-module via theaction

(h, k).g = hgk−1

extended linearly.

Proof. Indeed,

(h′, k′).(h, k).g = (h′, k′).hgk−1 = h′hgk−1k′−1 = (h′h, k′k).g

�

21

10.7. Theorem. The regular representation decomposes as a direct sumof G×G representations. More specifically,

C[G] ∼=k⊕i=1

Vi ⊗ V ∗i

for irreducible G-representations {V1, . . . , Vk}.10.8. Remark. Thus, in a sense, we have arrived at the Artin-Wedderburn

theorem in different language.

11. Some Applications of Representation Theory: FrobeniusDivisibility and Burnside

We now seek to discuss some useful applications of representation theorythat would be much more difficult to prove using only group theory. First,we will have to introduce the terminology of algebraic numbers.

11.1. Definition. We say z ∈ R, a commutative ring, is an algebraicinteger or integral over Z if z is a root of a monic polynomial with coefficientsin Z.

11.2. Remark. With this definition, “integer” is a somewhat misleadingword. For instance, if R = C, then i ∈ C is an algebraic integer since it is asolution to the polynomial x2 + 1.

11.3. Lemma. The set of all algebraic integers is a ring.

Proof. Let α be an eigenvalue of some A ∈ Mn(R) with eigenvectorv and β be an eigenvalue of some B ∈ Mm(R) with eigenvector w. Then,α± β is an eigenvalue of

A⊗ Idm ± Idn ⊗Band αβ is an eigenvalue of A⊗B. In both cases, the appropriate eigenvectoris v ⊗ w. Thus, the lemma is proven. �

11.4. Proposition. Let z be an element of a commutative ring R. Thefollowing properties are equivalent.

(a) z is integral over Z.(b) The subring Z[z] ⊆ R generated by z is finitely generated as a Z-

module.(c) There exists a finitely generated sub-module of R considered as a

Z-module containing Z[z]

11.5. Proposition. Let χ be the character of a representation ρ of afinite group G. Then, χ(g) is an algebraic integer for each g ∈ G.

Proof. χ(g) = tr(ρ(g)), so it is a sum of eigenvalues of ρ(g). However,all the eigenvalues are roots of unity, which are clearly algebraic integers.Thus, since the algebraic integers form a ring, this sum is an algebraicinteger. �

22

11.6. Theorem. Given a complex irreducible representation V of G,dimV | |G|.

Proof. Let C1, . . . , Ck be the conjugacy classes of G and set

λi := χV (Ci)|Ci|

dimV

where χV (Ci) = χV (g) for some g ∈ Ci (recall that χV is a class function).Then, all of these λi are algebraic integers. To see this, consider

Pi =∑h∈Ci

h

which is central in Z[G] by a quick computation. Thus, by Schur’s lemma,Pi = µiIdV since V is irreducible. Furthermore, Z[G] is finitely generated,so µ is an algebraic integer. Thus, we compute

tr(Pi) = |Ci|χV (Ci) = µi dimV =⇒ µi =|Ci|χV (Ci)

dimV= λi

Now, to finish our proof, we must show that∑i

λiχV (Ci)

is an algebraic integer. However, we already know that the algebraic integersform a ring and, by above, each λi is an algebraic integer. Moreover, χV (Ci)is an algebraic integer by 11.5. Thus, the sum is an algebraic integer by thering structure on the algebraic integers.

Now, we note, by definition of λi, so we get the following∑i

λiχV (Ci) =1

dimV

∑i

|Ci|χV (Ci)χV (Ci)

=1

dimV

∑g∈G

χV (g)χV (g) since characters are class functions

=1

dimV|G|(χV , χV ) by definition of (·, ·)

=|G|

dimVsince V is irreducible, so (χV , χV ) = 1

Thus, it must be that |G|dimV is an algebraic integer. However, |G|

dimV ∈ Q, so

by Gauss’s lemma, |G|dimV ∈ Z. �

While this result is useful in and of itself, we have something better,namely

11.7. Theorem (Frobenius Divisibility). Given an irreducible represen-tation V of G,

dimV | |G||Z(G)|

.

23

Proof. For g ∈ Z(G), we note that ρ(g) commutes with ρ(h) for allh ∈ G, so ρ(g) = λgIdV for some λ ∈ C. Consider the map from Z(G) toC given by g 7→ λg. Such a map is, in fact, a homomorphism from Z(G) toC×. Now, for some m ≥ 0, consider

ρ⊗m : G×m → GL(V ⊗m)

By 10.4, this is an irreducible representation ofG×m. So, given (g1, . . . , gm) ∈Z(G) × Z(G) × · · · × Z(G) ≤ Z(G×m), it must be that ρ(g1, . . . , gm) =λIdV ⊗m . Now, consider H = {(g1, . . . , gm) ∈ (Z(G))×m | g1 . . . gm = 1G} ≤G. Then, for (h1, . . . , hm) ∈ H,

(h1, . . . , hm).(v1 ⊗ · · · ⊗ vm) = ρ(h1)v1 ⊗ · · · ⊗ ρ(hm)vm

= λh1v1 ⊗ · · · ⊗ λhmvm= (λh1 · · ·λhm)v1 ⊗ · · · ⊗ vm= (h1 · · ·hm)v1 ⊗ · · · ⊗ vm= v1 ⊗ · · · ⊗ vm

so H acts trivially on V ⊗m. Thus, consider that ρ⊗m induces an irreduciblerepresentation onH G×m/H. By the theorem above, (dimV )n | |G×m/H| =|G|m

|Z(G)|m−1 , where |H| = |Z(G)|m−1 since g1 · · · gm = 1 constrains a degree of

freedom. Thus,(

|G||Z(G)|dimV

)m∈ |Z(G)|−1Z for all m and thus |G|

|Z(G)|dimV

is an integer. �

Another famous result is Burnside’s theorem, which can be proved usingrepresentation theory.

11.8. Theorem. Any group G of order paqb for primes p, q and a, b ≥ 0is solvable.

To prove Burnside’s theorem, we will use the following results.

11.9. Theorem. Let V be an irreducible representation of finite groupG and let C be a conjugacy class of G such that gcd(|C|,dimV ) = 1. Then,for any g ∈ C, either χV (g) = 0 or g acts as a scalar on V .

11.10. Lemma. If λ1, λ2, . . . , λn are roots of unity such that 1n(λ1 +λ2 +

· · · + λn) is an algebraic integer, then either λ1 = λ2 = · · · = λn or λ1 +· · ·+ λn = 0.

11.11. Theorem. Let G be a finite group and C be a conjugacy class inG with |C| = pk for p prime and k ∈ Z+. Then, G is not simple.

11.12. Lemma. There exists a non-trivial irreducible representation ofG, say V , with p - dimV such that χV (g) 6= 0.

State and proveBurnside’s theo-rem.

24

12. Induced Representations

Given a representation of a group G, there is a natural way to restrictthe representation to a subgroup H ≤ G.

12.1. Definition. Let ρ : G → GL(V ) be a representation of G andH ≤ G. Then, we define the restricted representation ResGH V to be

ρResGH V = ρ|H

However, we may ask if we can go the other way. That is, given arepresentation V of a subgroup H ≤ G, can we construct a representationof G? We present the following

12.2. Definition. Let ρ : H → GL(V ) be a representation of H ≤ G.Then, we define the induced representation IndGH V to be

IndGH V = {f : G→ V | f(hx) = ρ(h)f(x), ∀x ∈ G, h ∈ H}(g.f)(x) = f(xg),∀g ∈ G

Less abstractly, this says that, IndGH V =⊕

g∈G/H g.V where each g.V ∼= V

and the action is given by, where ggi = gσ(i)hi for {gi} a full set of cosetrepresentatives,

ρIndGH V (g) ·

(k∑i=1

givi

)=

k∑i=1

gσ(i)ρ(hi)vi

12.3. Proposition. Given a representation ρ : H → GL(V ) and H ≤ G,we have that

IndGH V∼= C[G]⊗C[H] V

as C[G]-modules with G-action

g′.(eg ⊗ v) = eg′g ⊗ v = eg′′ ⊗ ρ(h)v

where g′g = g′′h for g′′ ∈ G, h ∈ H.

Proof. This follows from the fact that C[G] ∼=⊕

g∈G/H g.C[H], so

C[G]⊗C[H]V ∼=

⊕g∈G/H

g.C[H]

⊗C[H]V ∼=⊕

g.(C[H]⊗C[H]V ) ∼=⊕

g∈G/H

g.V

�

12.4. Remark. Under this isomorphism, we get IndGH V =⊕

g∈G/H g.V

where V ∼= g.V via v 7→ eg ⊗ v Check that allthese actions areactually correct

12.5. Example. (a) Let H = {1} ≤ G and let W be the trivialrepresentation. Then, since W = C

IndGHW∼= C[G]⊗CW ∼= C[G]⊗C C ∼= C[G]

is the regular representation.

25

(b) Let H ≤ G and W be the trivial representation. Then,

IndGHW∼= C[G]⊗C[H] C ∼= C[G/H]

since g ⊗ λ = g′h⊗ λ = g′ ⊗ h.λ = g′ ⊗ λ for g′ ∈ G, h ∈ H.

12.6. Proposition. Let V, V1, V2 be representations of H ≤ G. Then,

(a) IndGH(V1 ⊕ V2) ∼= IndGH V1 ⊕ IndGH V2.(b) dim IndGH V = |G/H|dimV .(c) For H ≤ K ≤ G, IndGK IndKH V

∼= IndGH V .

Proof. (b) has already been shown in the example above. For (a),

IndGH(V1⊕V2) ∼= C[G]⊗C[H](V1⊕V2) = (C[G]⊗C[H]V1)⊕(C[H]⊗C[H]V2) ∼= IndGH V1⊕IndGH V2

and for (c),

IndGK IndKH V∼= C[G]⊗C[K] (C[K]⊗C[H] V ) ∼= C[G]⊗C[H] V ∼= IndGH V

�

12.7. Definition. Let φ ∈ Class(H) for H ≤ G. Then, we define theinduced class function on G by

IndGH(φ)(g) :=∑

s∈G/H,s−1gs∈H

φ̃(s−1gs) =1

|H|∑t∈G

φ̃(t−1gt)

where

φ̃(g) =

{φ(g) g ∈ H0 else

12.8. Example. LetG = S3 and considerH = Z/3Z ∼= {(1), (123), (132)} ES3. Then, since Z/3Z is abelian, it has character table

(1) (123) (132)χ1 1 1 1χζ 1 ζ ζ2

χζ2 1 ζ2 ζ

where ζ = ζ3, a third root of unity, since g3 = 1 for all g ∈ Z/3Z. We alsohave, from earlier, the irreducible characters of S3 as χ1, χ2, θ (see 8.13).Then, using the fact that H is normal and G/H = {H, (12)H}, we see that

IndGH(χ)(h) = χ(h) + χ((12)h(12)) ∀h ∈ H

= χ(h) + χ(h) since (12)(123)(12) = (132)

IndGH(χ)((12)) = χ̃((12)) + χ̃((12)(12)(12))

= 0

And thus

IndGH(χ1) = χ1 + χ2

IndGH(χζ) = θ

26

IndGH(χζ2) = θ

since ζ + ζ2 = −1.

12.9. Proposition. Given φ ∈ Class(H) for H ≤ G, we get IndGH(φ) ∈Class(G).

Proof. Let g, k ∈ G, then

IndGH(φ)(k−1gk) =1

|H|∑t∈G

φ̃(t−1k−1gkt) =1

|H|∑t′∈G

φ(t′−1gt′) = IndGH(φ)(g)

�

12.10. Theorem (The Mackey formula). The character of the inducedrepresentation IndGH V is IndGH(χV ).

Proof. We seek to directly compute the trace of the action of s ∈ G onIndGH V

∼=⊕

g∈G/H g.V . Consider that sgH = gH ⇐⇒ g−1sg = t ∈ H. We

only consider such cosets since all others will not contribute to the trace.On g.V , we get

s.(eg ⊗ v) = esg ⊗ v = egt ⊗ v = eg ⊗ ρ(t)w

so the action of s on g.V corresponds to the action of t = g−1sg on V .Thus, the block corresponding to g.V contributes χV (g−1sg) to the trace ofρIndGH V (s). Thus, summing over all such g, we get

χIndGH ρ(s) =∑

g∈G/H,g−1sg∈H

χV (g−1sg) =∑

g∈G/H

χ̃V (g−1sg) = IndGH(χV )(s)

�

12.11. Example. Thus, from our example above, this actually tells usthe decomposition of the induced representations of S3 from Z/3Z.

13. Frobenius Reciprocity

We seek to find relationships between Ind and Res since Res is mucheasier to compute than Ind. By far the most important one is Frobeniusreciprocity.

13.1. Theorem (Frobenius Reciprocity). Let H ≤ G, a finite group.Then, for φ ∈ Class(G) and ψ ∈ Class(H), we have

(IndGH ψ, φ) = (ψ,ResGH φ)

which is to say, IndGH and ResGH are Hermetian adjoint.

13.2. Lemma. Given M an R-module and N an S-module and ring ho-momorphism f : R→ S for R,S (unital) rings, then

HomR(M,N) ∼= HomS(S ⊗RM,N)

27

where N is considered as an R module via r.n = f(r)n (that is, the restric-tion of scalars).

Proof of Lemma. Given u ∈ HomR(M,N), let F : HomR(M,N) →HomS(S ⊗RM,N) be defined by sending u to the composition

S ⊗RMidS⊗u→ S ⊗R N → N

s⊗ n 7→ sn

Then, F (u) is an S-module homomorphism and so F is well-defined anda homomorphism of abelian groups. We now construct and inverse to F .Let G : HomS(S ⊗R M,N) → HomR(M,N) be defined by sending v ∈HomS(S ⊗RM,N) to the composition

M → R⊗RMf⊗idM→ S ⊗RM

v→ N

m 7→ 1⊗m

One can check that F,G are inverse to each other. �

13.3. Remark. One can also check that this isomorphism depends onlyon the ring homomorphism f . Thus, such an isomorphism is functorial andso the extensions of scalars, S⊗−, is left adjoint to the restriction of scalarsfunctor.

Proof of Frobenius Reciprocity. Frobenius reciprocity ends up be-ing a special case of the lemma. That is,

(IndGH χV , χW ) = dim HomG(C[G]⊗C[H] V,W ) by 8.6

= dim HomH(V,W ) by lemma

= (χV ,ResGH χW )

�

Frobenius reciprocity is useful for computing various representations ofgroups from representations of their subgroups.

13.4. Example. Consider Z/2Z ∼= 〈(12)〉 ≤ S3. Then, Z/2Z hastwo irreducible 1-dimensional characters, let us say χ1 and χ−1. Then,

(IndS3

Z/2Z χ1, χV ) = (χ1,ResS3

Z/2Z χV ) for any irreducible representation V of

S3. Thus,

(IndS3

Z/2Z χ1, χtrivial) = (χ1,ResS3

Z/2Z χtrivial) = (χ1, χ1) = 1

(IndS3

Z/2Z χ1, χalt) = (χ1,ResS3

Z/2Z χalt) = (χ1, χ−1) = 0

(IndS3

Z/2Z χ1, χstd) = (χ1,ResS3

Z/2Z χstd) = (χ1, χ1 + χ−1) = 1

(IndS3

Z/2Z χ−1, χtrivial) = (χ−1,ResS3

Z/2Z χtrivial) = (χ−1, χ1) = 0

(IndS3

Z/2Z χ−1, χalt) = (χ−1,ResS3

Z/2Z χalt) = (χ−1, χ−1) = 1

28

(IndS3

Z/2Z χ−1, χstd) = (χ−1,ResS3

Z/2Z χstd) = (χ−1, χ1 + χ−1) = 1

Thus, IndS3

Z/2Z χ1 = χtrivial + χstd and IndS3

Z/2Z χ−1 = χalt + χstd.

14. Mackey Theory

To gain more insight into induced representations, we wish to understandResGK IndGH V for H,K ≤ G and V a representation of G. To describe ourresults, we will need the language of double cosets.

14.1. Definition. Let H,K ≤ G, a group. The set of double cosetsKnG/H is the set of K × H-orbits on G where (k, h).g = kgh. Often,elements of such a set are denoted KgH for g ∈ G, a representative.

14.2. Remark. If K = H E G, then KnG/H = G/H. Also note thatthere are other definitions of double cosets.

14.3. Definition. Following [Ser97], for H,K ≤ G and s ∈ S, a set ofdouble-coset representatives, let

Hs := sHs−1 ∩K ≤ Kand, given representation ρ : H → GL(V ), let ρs : Hs → GL(Vs) given by

ρs(x) := ρ(s−1xs)

be a representation of Hs. In other words, Vs ∼= s ⊗ V with action of Hs

given by

(shs−1)(s⊗ v) = s⊗ hv

14.4. Remark. Note that Hs is the stabilizer of sH under the actionof K on G/H. Furthermore, if one embeds Hs ↪→ H via x 7→ s−1xs, thenimHs is the stabiliser of Ks under the action of H on HnG.

14.5. Proposition. Given S, a set of double coset representatives forKnG/H, then

ResGK IndGH V∼=⊕s∈S

IndKHsVs

Proof. Let us consider C[G] as a (C[K],C[H])-bimodule. Then,

C[G] ∼=⊕s∈S

C[KsH]

Note, too, this viewpoint gives the isomorphism as (C[K],C[H])-bimodules

C[KsH] ∼= C[K]⊗C[Hs] (s⊗ C[H])

ksh 7→ k ⊗ s⊗ h

Thus, using this decomposition, we get the following isomorphisms of C[K]-modules

ResGK IndGH V∼= ResGK

(C[G]⊗C[H] V

)

29

∼=⊕s∈S

C[KsH]⊗C[H] V

∼=⊕s∈S

C[K]⊗C[Hs] (s⊗ C[H])⊗C[H] V

∼=⊕s∈S

C[K]⊗C[Hs] Vs

∼=⊕s∈S

IndKHsVs

�

14.6. Theorem (Mackey’s Irreducibility Criterion). IndGH V is irreducibleif and only if

(a) V is irreducible and(b) Vs and ResHHs

V share no irreducibles for all s ∈ S, a set of doublecoset representatives of KnG/H.

Proof. We wish to show that (IndGH V, IndGH V ) = 1. We compute

(IndGH V, IndGH V ) = (V,ResGH IndGH V ) by Frobenius Reciprocity

= (V,⊕

s∈HnG/H

IndHHs(ρs)) by proposition above

=∑

s∈HnG/H

(ResHHs(ρ), ρs) by Frobenius reciprocity

Note that the last equality also makes use of the fact

Hom(A,⊕i∈I

Bi) =⊕i∈I

Hom(A,Bi) for finite index set I

Now, when s = 1, (ResHH ρ, ρ1) = (ρ, ρ) ≥ 1. Thus, for our sum to be exactly1, it must be that

(ρ, ρ) = 1 and (ResHHsρ, ρs) = 0 for s 6= 1

This will happen precisely when ρ is irreducible and ResHHsρ and ρs share

no irreducibles. �

14.7. Corollary. If H E G, then IndGH V is irreducible if and only ifV is irreducible and V 6∼= Vs for all s ∈ G/H

Proof. If H E G, then Hs = H and ResHHsρ = ρ, so it must be that ρ

is irreducible and not isomorphic to any ρs, otherwise ρs and ResHHsρ = ρ

will share an irreducible. �

30

14.8. Example. Consider 〈r, s | r4 = s2 = 1, sr = r−1s〉 ∼= D8∼= Z4oZ2.

Then, we know 〈r〉 ∼= Z4 E D8 with character table

1 r r2 r3

χ1 1 1 1 1χ2 1 i −1 −iχ3 1 −1 1 −1χ4 1 −i −1 i

If ρi is the irreducible representation of Z4 corresponding to χi, then we no-tice that tr ρ3s(r) = tr ρ3(srs) = tr ρ3(r−1) = −1 = tr ρ3(r), so IndD8

Z4ρ3 will

not be irreducible. On the other hand, one can check tr ρ2s(r) = tr ρ2(srs) =tr ρ2(r−1) = −i = tr ρ4(r), so ρ2s and ρ2 are orthogonal, thus sharing no

irreducibles. Thus, it must be that IndD8Z4ρ2 is irreducible. In fact, one

can check IndD8Z4ρ2 = IndD8

Z4ρ4 will give the only irreducible 2-dimensional

representation of D8.

15. Representations of Nilpotent Groups

15.1. Theorem. Given an irreducible representation of a nilpotent groupG, it is induced from a one-dimensional representation of some subgroupH ≤ G.

15.2. Remark. This is a specific case of Brauer’s theorem (see 18.2).We will prove this theorem using the program given in [Tel05].

15.3. Lemma. Given finite group G with N E G and irreducible repre-sentation V , then V = IndGHW for some N ≤ H ≤ G and ResHN W

∼= U⊕r

for U an irreducible representation of N , r ∈ N.

Proof. Let ResGN V =⊕

i Vi for Vi irreducible N representations. Ifthere is a single summand, then H = G and we are done. Otherwise, givenρ : N → GL(V ), note that ρs(n) := ρ(s−1ns) for n ∈ N , s ∈ G will permutethe summands of ResGN V since conjugation defines an automorphism of thenormal subgroup N , and so must permute the irreducible characters of N .Thus, this conjugation action gives us an isomorphism of N -representations.Furthermore, this permutation action by G must be transitive since we as-sumed that V has no G-invariant subspaces and so each non-zero Vi has thesame dimension.

Now choose a non-zero block Vk and let H = StabG(Vk). Then, it must

be that dimV = |G||H| dimVk = dim IndGH Vk by the orbit-stabilizer theorem.

Furthermore, we get

dim HomG(IndGH Vk, V ) = dim HomH(Vk,ResGH V ) ≥ 1

31

Thus, since V is irreducible as a G-representation and dimV = dim IndGH Vk,it must be that

(IndGH Vk, V ) = dim HomG(IndGH Vk, V ) = 1

and so V = IndGH Vk. �

15.4. Corollary. If N E G is abelian, then either its action on V is ascalar or V = IndGHW for some proper subgroup H � G.

We will also use the following (standard) group theory fact.

15.5. Lemma. If G is nilpotent, then it contains a characteristic, abelian,non-central subgroup.

Proof of Theorem. Let ρ : G → GL(V ) be an irreducible represen-tation of a nilpotent group G. Then, G/ ker ρ is nilpotent. If G/ ker ρis abelian, then ρ : G/ ker ρ → GL(V ) is one-dimensional and so ρ : G →GL(V ) must be as well. Otherwise, take A E G/ ker ρ to be a characteris-tic, abelian, and non-central subgroup given by the lemma above. Then, fornon-central element a ∈ A, ρ(a) cannot be a scalar.

Thus, from the corollary above, there is a proper subgroup H � G with a

representation ρ′ : H/ ker ρ→ GL(W ) such that ρ = IndG/ ker ρH/ ker ρ ρ

′ = IndGH ρ̃′

where ρ̃′ : H → GL(W ) is a lift of ρ′. Since H is also nilpotent, we canrepeat this argument until we find a 1-dimensional representation ρ′. �

16. Example: Representations of the Symmetric Group

Elements of the symmetric group Sn can be represented in 1-line no-tation as products of disjoint cycles, and to each element, we can assign acycle type in the form of a partition of n.

16.1. Example. Consider (12345)(876)(9, 10) ∈ S10. This cycle hascycle type (5, 3, 2) ` 10.

Partitions are useful combinatorial tools with many applications beyondwhat is described here. It is also useful to realize partitions of n as Youngdiagrams. Once again, we shall define the correspondence via an example.

16.2. Example. We can represent a partition (λ1, λ2, . . .) as a Ferrersdiagram with λi boxes on the ith row. So, consider (5, 3, 2) ` 10. Thecorresponding diagram would be

32

Furthermore, we can fill in the boxes of the diagram to get a Young tableau.For example

1 2 3 4 5

6 7 8

9 10

Finally, Sn can act on a Young tableau by permuting the numbers. So,(12345)(876)(9, 10) ∈ S10 would act on the above diagram to yield

2 3 4 5 1

8 6 7

10 9

Now, using this combinatorial device, we can define the following.

16.3. Definition. Let T be a tableau of shape λ ` n for n ∈ N. Then,we define the row stabalizer subgroup R(T) ≤ Sn to be the subgroup ofpermutations such that every permutation in R(T) preserves the elementsin the rows of T. One analogously defines C(T) to be the column stabalizersubgroup. Finally, one defines Sλ := R(T′) where T′ = {(1, 2, . . . , λ1), (λ1 +1, . . . , λ1 + λ2), . . . , (n− λ` + 1, . . . , n)}.

16.4. Example. In the example above, (12345)(876)(9, 10) is in the rowstabalizer of the tableau. In fact, in that setup R(T) = S{1,2,3,4,5}×S{6,7,8}×S{9,10} ∼= S5 ×S3 ×S2.

Now, given Sn, we automatically know of 2 irreducible, 1-dimensional,complex representations, namely the trivial representation and the sign/alternatingrepresentation. Furthermore, we know that conjugacy classes of Sn are en-coded by cycle type, which are encoded by partitions of n. So, since char-acter tables are square, there must be as many irreducible representationsof Sn as there are partitions of n. Thus, we have

{Partitions of n} 1-1←→ {Irreducible representations of Sn}

Our task, then, is to figure out what these irreducible representations are.A systematic treatment will not be given here, but some of the tools inthe previous chapters will be used to get an idea of some results about therepresentation theory of symmetric groups. For more, see [Jam78], [Ful97].We will follow the program of [Sag91], strategically omitting details andadapting the content to the language in these notes.

16.5. Definition. We say two λ-tableaux T1 and T2 are row equivalentif the corresponding rows of the two tableaux contain the same elements,that is, R(T1) = R(T2). We then define a tabloid of shape λ or a λ-tabloid{T} to be

{T} = {T′ | T′ ∼ T, shapeT = λ}

33

16.6. Example. If

T =1 2

3

then,

{T} =

1 2

3,

2 1

3

=1 2

3

16.7. Proposition. If λ = (λ1, . . . , λ`) ` n, then the number of λ-tableaux in any given λ-tabloid is

λ1!λ2! · · ·λ`! =: λ!

and thus the number of λ-tabloids is n!/λ!.

Proof. This follows immediately since, if T1 ∼ T2, then g.T1 = T2 forsome g ∈ R(T1) = R(T2), that is, T1 ∼ T2 if and only if T1 and T2 are inthe same R(T1)-orbit and |R(T1)| = λ!. �

16.8. Proposition. Given T of shape λ, R(T) ≤ Sn and trivial repre-sentation 1 of R(T), we get that

IndSn

R(T)∼= C[Sn/R(T)] = C[π1R(T), . . . , πkR(T)]

and we can index each πi by a λ-tabloid, say {Ti}.16.9. Definition. Let λ ` n. We define

Mλ := C[{T1}, . . . , {Tk}]where {T1}, . . . , {Tk} is a complete list of λ-tabloids as the permutationmodule corresponding to λ.

16.10. Example. Let λ = (n) = · · · . Then, there is only a singleλ-tabloid consisting of all λ-tableaux. Thus,

M (n) = C[

1 2 · · ·n]∼= C

with a trivial action. Thus, M (n) is the trivial representation.

16.11. Example. Let λ = (1n). Then, each λ-tabloids consists of asingle λ-tableau and thus they are in bijection with Sn. Thus,

M (1n) = C[{T1}, . . . , {Tn!}] ∼= CSn

is the regular representation of Sn.

16.12. Example. Let λ = (n − 1, 1) = · · · . Then, each λ-tabloid

is uniquely determined by the number in the second row, so there are nλ-tabloids. Thus, we get

M (n−1,1) = C[{T1}, . . . , {Tn}]with action σ.Tj = Tσ(j) for σ ∈ Sn. Thus, we have the “standard” or“defining” representation of Sn, that is, the action of Sn on {1, . . . , n}.

34

16.13. Proposition. If λ ` n, then Mλ is cyclicly generated by anygiven λ-tabloid. Furthermore, dimMλ = n!/λ!, that is, the number of λ-tabloids.

Proof. This is a relatively straightforward application of the transi-tivity of the Sn-action on the set of λ-tabloids, which initially came fromcosets of Sn/R(T) for some T of shape λ. The dimension computation isimmediate from the construction as well. �

16.14. Theorem. Let λ ` n and let Sλ correspond to tabloid {Tλ}.Then, CSnSλ

∼= CSn{Tλ} as Sn-modules.

Proof. Define a map by sending πiSλ 7→ {πiTλ} and extend linearly.�

We can also define a partial order on tableau.

16.15. Definition. Given partitions λ = (λ1, λ2, . . .) ` n and µ =(µ1, µ2, . . .), we say that λ E µ if

λ1 ≤ µ1λ1 + λ2 ≤ µ1 + µ2...

λ1 + · · ·+ λk ≤ µ1 + · · ·+ µk...

16.16. Example. Using the notation that (22, 12) = (2, 2, 1, 1), the fol-lowing partial order is induced on the partitions of 6.

(6)

(5, 1)

(4, 2)

(3, 3) (4, 12)

(3, 2, 1)

(3, 13) (23)

(22, 12)

(2, 14)

(16)

35

16.17. Lemma (Dominance Lemma). [Sag91, Lem 2.2.4]. Let T and Sbe tableaux of shape λ and µ, respectively. If, for each index i, the elementsof row i of S are all in different columns in T, then λ D µ.

16.18. Definition. Let λ = (λ1, . . . , λk) abd µ = (µ1, . . . , µm) be par-titions of n. Then, λ ≤ ν if, for some index i,

λj = µj for j < i and λi < µi

This order is called the lexicographic order.

16.19. Proposition. (a) The lexicographic order is a total order.(b) If λ, µ ` n with λ D µ, then λ ≥ µ.

16.20. Definition. We define the row symmetrizer of a tableau T ofshape λ to be

P (T) :=∑

α∈R(T)

α

and the column symmetrizer to be

Q(T) :=∑

α∈C(T)

sgn(α)α

16.21. Definition. Let T be a tableau of shape λ. Then, we define

eT = Q(T) · {T} ∈ CSn/ ∼

to be the polytabloid associated with T.

16.22. Remark. For the knowledgeable, eT is the image of the Youngsymmetrizer

y(T ) =Q(T)P (T)

hλ

under the quotient map induced by ∼ where hλ is the hook length formula. Is this actuallytrue?

16.23. Definition. For any partition λ, the corresponding Specht mod-ule, Sλ, is the submodule of Mλ spanned by the polytabloids eT where T isof shape λ.

Now, given the information we have above, we list some important re-sults without proofs about the irreducible Sn modules.

16.24. Proposition. We have the following fundamental results.

(a) The Sλ are cyclic modules generated by any given polytabloid.(b) Let U be a submodule of Mλ. Then, U ⊇ Sλ or U ⊆ Sλ⊥. Thus,

over C, the Sλ’s are irreducible.(c) The Sλ for λ ` n form a complete list of irreducible Sn-modules

over C.

36

(d) (Branching Rule) If λ ` n, then

ResSnSn−1

(Sλ) ∼=⊕λ−

Sλ−

IndSn+1

Sn

∼=⊕λ+

Sλ+

where λ− is the set of all Ferrers diagrams with a removable cornerof λ revmoved and λ+ is the set of all Ferrers diagrams with anaddable corner of λ added.

(e) (Young’s Rule) Over an algebraically closed field, the permutationmodules decompose as

Mµ =⊕λ

Kλ,µSλ

where Kλ,µ is the number of “semistandard Young tableaux” ofshape λ and content µ, referred to as the Kostka number.

16.25. Example. Let λ = (n). Then, e ··· = 1 2 · · · n and so

the trivial representation is a subrepresentation of S(n). Therefore, S(n) is

the trivial representation, which was the only possibility anyways since M (n)

was also trivial.

16.26. Example. Let λ = (1n). Then, for any T of shape λ, eT is thesigned sum of all n! permutations viewed as tabloids and eσT = σeT =(sgnσ)eT. Thus, every polytabloid is a scalar multiple of eT and thus

S(1n) = C{eT} with action given by σeT = (sgnσ)eT, thus giving the signrepresentation.

16.27. Example. For λ = (n − 1, 1), then we have polytabloids of theform

{T} =i · · · jk

=⇒ Q(T) = 1− (i, k) =⇒ eT =i · · · jk

− k · · · ji

The span of all these eT is given by

S(n−1,1) = {c11+c22+· · ·+cnn | c1+c2+· · ·+cn = 0} where k := i · · · jk

Thus, dimS(n−1,1) = n−1 and we can pick a basis {2−1,3−1, . . . ,n−1}}with character χS(n−1,1)(σ) to be one less than the number of fixed points ofσ ∈ Sn.

17. Artin’s Theorem

Artin’s theorem allows us to gain more insight into how induced charac-ters or representations may be related to the irreducible representations ofa finite group.

17.1. Definition. Let G be a finite group. A virtual representation ofG is an integer linear combination of irreducible representations of G.

37

17.2. Proposition. Let V be an virtual representation with characterχV . If (χV , χV ) = 1 and χV (1) > 0, then χV is a character of an irreduciblerepresentation of G.

Proof. Let V =∑niVi for irreducible representations Vi and ni ∈ Z.

Then, using the fact that the irreducible characters form an orthonomalbasis, we get 1 = (χV , χV ) =

∑n2i . Thus, ni = ±1 for one i and 0 otherwise.

However, χV (1) > 0 by assumption, so ni = 1. �

17.3. Theorem (Artin’s theorem). Let X be a conjugation-invariantfamily of subgroups of a finite group G. Then, the following are equivalent.

(a) Any element of G belongs to a subgroup H ∈ X, that is, G =⋃H∈X H.

(b) The character of any irreducible representation of G belongs to theQ-span of characters of induced representations IndGH V where H ∈X and V is an irreducible representation of H.

17.4. Remark. Part (b) is equivalent to the statement that coker Ind isa finite group where Ind is defined by

Ind:⊕H∈X

R(H)→ R(G)

(θH)H∈X 7→∑H∈X

IndGH θH

For R(H) the ring of virtual characters of H and R(G) similarly defined. Infact, this map can be represented by a matrix

Ind =[(Wi, IndGH VH,j)

]where Wi ranges over all irreducible representations of G and VH,j rangesover all irreducible representations of H for al H ∈ X. Thus, the cokernel isa finite group if and only if the rows of this matrix are linearly independent.

Proof. For (b) =⇒ (a), let S =⋃H∈X H. Then, take g ∈ G \ S. Since

S is conjugation invariant, no conjugate of g is in S. So, by the Mackeyformula, χIndGH V (g) = 0 for all H ∈ X and V . Thus, by assumption, for

any irreducible representation W of G is a Q-span of characters of inducedrepresentations, so χW (g) = 0. However, irreducible characters also spanthe class functions, so this would mean that any class function would vanishof g, a contradiction.

For (a) =⇒ (b), let U be a virtual representation ofG such that (χU , χIndGH V ) =

0 for all H and V . Then, by Frobenius reciprocity, we get (χResGH U , χV ) = 0.

Thus, χU vanishes on H for any H ∈ X and so χU ≡ 0. � Expand on this

17.5. Corollary. Any irreducible character of a finite group is a ratio-nal linear combination of induced characters from its cyclic subgroups.

38

17.6. Example. LetG = A4. Then, |A4| = 12. LetH = 〈1, (123), (321)〉,K = 〈1, (12)(34)〉. Then, given characters of A4, χ1, χζ , χζ2 , and θ, charac-ters of H, χ1, χζ , χζ2 , and characters of K, χ1, χ−1, one can use Frobeniusreciprocity to check

IndGH χ1 = χ1 + θ

IndGH χζ = χζ + θ

IndGH χζ2 = χζ2 + θ

IndGK χ1 = χ1 + χζ + χζ2 + θ

IndGK χ−1 = 2θ

and so

χ1 = IndGH χ1 −1

2IndGK χ−1

χζ = IndGH χζ −1

2IndGK χ−1

χζ2 = IndGH χζ2 −1

2IndGK χ−1

θ =1

2IndGK χ−1

Thus, A4 =⋃g∈A4

gHg−1 ∪⋃g∈A4

gKg−1.

18. Brauer’s Theorem

While Artin’s theorem gives us a way to write characters as a rationallinear combination of characters induced from cyclic subgroups, the use ofthe rational numbers leaves something to be desired. We would rather haveintegral linear combinations. In fact, if we are willing to look a little morebroadly than cyclic groups, we can arrive at such a decomposition.

18.1. Definition. Let p be a prime. A group G is called p-elementaryif

G ∼= Z × Pwhere Z is a cyclic group of order prime to p, (that is gcd(p, |Z|) = 1), andP is a p-group (i.e. |P | = pr). G is elementary if it is p-elementary for someprime p.

18.2. Theorem (Brauer’s Theorem). Every complex character is a Z-linear combination of linear characters of elementary subgroups P inducedto G, that is

χ =∑P

cP IndGP ξp, cP ∈ Z

where ξP is a linear character of P .

However, the proof of Brauer’s theorem is not so straight-forward. Itrequires use of the following theorem.

39

18.3. Theorem. Let G be a finite group and let Vp be the subgroup ofR(G) generated by characters induced from those of p-elementary subgroupsof G. Then, [R(G) : Vp] <∞ and gcd([R(G) : Vp], p) = 1.

Actually proveBrauer’s theo-rem.

40

Bibliography

[Ben95] D. J. Benson, Representations and Cohomology I, 1995.[CR62] C. W. Curtis and I. Reiner, Representation Theory of Finite Groups and As-

sociative Algebras, 1962.[EGH+11] P. Etingof, O. Golberg, S. Hensel, T. Liu, A. Schwendner, D. Vaintrob, and E.

Yudovina, Introduction to Representation Theory (2011). http://math.mit.edu/~etingof/replect.pdf.

[Ful97] W. Fulton, Young Tableaux, 1997.[Gro] I. Gronjnowski, Representation Theory, The Princeton Companion to Mathe-

matics, 419–431.[Jam78] G. D. James, The Representation Theory of the Symmetric Groups, 1978.[Sag91] B. E. Sagan, The Symmetric Group: Representations, Combinatorical Algo-

rithms, and Symmetric Functions, 1991. 2nd Edition, 2001.[See17] G. H. Seelinger, Artin-Wedderburn Theory (2017). See https://ghseeli.

github.io/grad-school-writings/class-notes/artin-wedderburn-notes.

pdf.[Ser97] J.-P. Serre, Linear Representations of Finite Groups, 1997. Translated from

the French by Leonard L. Scott.[Smi10] K. Smith, Groups and their Representations (2010). http://www.math.lsa.

umich.edu/~kesmith/rep.pdf.[Tel05] C. Teleman, Representation Theory (2005). https://math.berkeley.edu/

~teleman/math/RepThry.pdf.

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