MATH 101B: ALGEBRA II PART D: REPRESENTATIONS OF FINITE GROUPSpeople.brandeis.edu/~igusa/Math131b/Math131b_notesD.pdf · 2019-05-02 · MATH 101B: ALGEBRA II PART D: REPRESENTATIONS

MATH 101B: ALGEBRA IIPART D: REPRESENTATIONS OF FINITE GROUPS

For the rest of the semester we will discuss representations of groups, mostly finitegroups. An elementary treatment would start with characters and their computation.But the Wedderburn structure theorem will allow us to start at a higher level whichwill give more meaning to the character tables which we will be constructing. This isfrom Lang, XVIII, §1-7 with additional material from Serre’s “Linear Representations ofFinite Groups” (Springer Graduate Texts in Math 42) and Alperin and Bell’s “Groupsand Representations” (Springer GTM 162).

Contents

1. The group ring KG 21.1. Representations of groups 21.2. Modules over KG 31.3. Semisimplicity of KG 51.4. Idempotents 81.5. Center of CG 92. Characters 102.1. Basic properties 102.2. Irreducible characters 132.3. formula for idempotents 192.4. orthogonality relations 223. Restriction and Induction 273.1. induction 273.2. induced characters 28

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2 MATH 101B: ALGEBRA II, PART D: REPRESENTATIONS OF GROUPS

1. The group ring KG

The main idea is that representations of a group G over a field K are “the same” asmodules over the group ring KG = K[G]. Also, left module are “the same” as rightmodules: KG-mod ∼= mod-KG. First I defined both terms.

1.1. Representations of groups.

Definition 1.1. A representation of a group G over a field K is defined to be a grouphomomorphism

ρ : G→ AutK(V )

where V is a vector space over K.

Here AutK(V ) is the group of K-linear automorphisms of V . This is also written asGLK(V ). This is the group of units of the ring EndK(V ) = HomK(V, V ) which, as Iexplained before, is a ring with addition defined pointwise and multiplication given bycomposition. If dimK(V ) = d then AutK(V ) ∼= AutK(Kd) = GLd(K) which can also bedescribed as the group of units of the ring Matd(K) or as:

GLd(K) = {A ∈ Matd(K) | det(A) 6= 0}

d = dimK(V ) is also called the degree of the representation ρ.

1.1.1. examples.

Example 1.2. The trivial representation. This is defined to be the one dimensionalrepresentation V = K with trivial action of the group G (which can be arbitrary).Trivial action means that ρ(g) = 1 = idV for all σ ∈ G.

In the next example, I pointed out that the group G needs to be written multiplica-tively no matter what.

Example 1.3. Let G = Z/3. Written multiplicatively, the elements are 1, σ, σ2. LetK = R and let V = R2 with ρ(σ) defined to be rotation by 120◦ = 2π/3 and ρ(σ2) isrotation by 240◦ = 4π/3. I.e.,

ρ(σ) =

(−1/2 −

√3/2√

3/2 −1/2

), ρ(σ2) =

(−1/2

√3/2

−√

3/2 −1/2

)Example 1.4. The regular representation is defined to be V = KG with G acting byleft multiplication. The degree of this representation is equal to n = |G|, the order of G.

Example 1.5. Suppose that E is a normal (Galois) field extension of K and G =Gal(E/K). Then G acts on E by K-linear transformations. This gives a representation:

ρ : G ↪→ AutK(E)

Note that this map is an inclusion by definition of “Galois group.” This representationalso has degree n = |G|.

Is this isomorphic to the regular representation?

MATH 101B: ALGEBRA II, PART D: REPRESENTATIONS OF GROUPS 3

1.1.2. axioms. The axioms for a group representation just say what it means for

ρ : G→ AutK(V )

to be a group homomorphism:

(1) ρ(1) = 1 = idV 1v = v ∀v ∈ V(2) ρ(στ) = ρ(σ)ρ(h) ∀σ, τ ∈ G (στ)v = σ(τv) ∀v ∈ V(3) ρ(σ) is K-linear ∀σ ∈ G σ(av + bw) = aσv + bσw ∀v, w ∈ V, a, b ∈ k

The first two conditions say that ρ is an action of G on V . Actions are usually writtenby juxtaposition:

σv := ρ(σ)(v)

Condition (3) says that the action is K-linear. So, together, the axioms say that arepresentation of G is a K-linear action of G on a vector space V .

1.2. Modules over KG. The group ring KG is defined to be the set of all formal finiteK linear combinations of elements of G:

∑agg where ag ∈ k for all g ∈ G and ag = 0

for almost all g. This is a vector space of K with basis G.For example, R[Z/3] is the set of all linear combinations

x+ yσ + zσ2

where x, y, z ∈ R. I.e., R[Z/3] ∼= R3.Multiplication in KG is given by(∑

aσσ)(∑

bττ)

=(∑

cλλ)

where cλ ∈ G can be given in three different ways:

cλ =∑στ=λ

aσbτ =∑σ∈G

aσbσ−1λ =∑τ∈G

aλτ−1bτ

Proposition 1.6. KG is a K-algebra.

This is straightforward and tedious. So, I didn’t prove it. But I did explain what itmeans and why it is important.

Recall that an algebra over K is a ring which contains K in its center. The centerZ(R) of a (noncommutative) ring R is defined to be the set of elements of R whichcommute with all the other elements:

Z(R) := {x ∈ R |xy = yx ∀y ∈ R}Z(R) is a subring of R.

The center is important for the following reason. Suppose that M is a (left) R-module.Then each element r ∈ R acts on M by left multiplication λr

λr : M →M, λr(x) = rx

This is a homomorphism of Z(R)-modules since:

λr(ax) = rax = arx = aλr(x) ∀a ∈ Z(R)

Thus the action of R on M gives a ring homomorphism:

ρ : R→ EndZ(R)(M)


Getting back to KG, suppose that M is a KG-module. Then the action of KG onM is K-linear since K is in the center of KG. So, we get a ring homomorphism

ρ : KG→ EndK(M)

This restricts to a group homomorphism

ρ|G : G→ AutK(M)

In general, any ring homomorphism φ : R → S will induce a group homomorphismU(R) → U(S) where U(R) is the group of units of R. By definition, AutK(M) is thegroup of units of EndK(M). G is contained in the group of units of KG. (An interestingunanswered question is: Which finite groups occur as groups of units of rings?)

This discussion shows that a KG-module M gives, by restriction, a representation ofthe group G on the K-vector space M . Conversely, suppose that

ρ : G→ AutK(V )

is a group representation. Then we can extend ρ to a ring homomorphism

ρ : KG→ EndK(V )

by the simple formula

ρ(∑

agg)

=∑

agρ(g)

When we say that a representation of a group G is “the same” as a KG-module weare talking about this correspondence. The vector space V is also called a G-module.So, it would be more accurate to say that a G-module is the same as a KG-module.

Corollary 1.7. (1) Any group representation ρ : G→ AutK(V ) extends uniquely toa ring homomorphism ρ : KG→ EndK(V ) making V into a KG-module.

(2) For any KG-module M , the action of KG on M restricts to give a group repre-sentation G→ AutK(M).

(3) These two operations are inverse to each other in the sense that ρ is the restrictionof ρ and an action of the ring KG is the unique extension of its restriction to G.

There are some conceptual differences between the group representation and the cor-responding KG-module. For example, the module might not be faithful even if thegroup representation is:

Definition 1.8. A group representation ρ : G→ AutK(V ) is called faithful if only thetrivial element of G acts as the identity on V . I.e., if the kernel of ρ is trivial. AnR-module M is called faithful if the annihilator of M is zero where

ann(M) := {r ∈ R | rx = 0 ∀x ∈M}.These two notions of “faithful” do not agree. For example, take the representation

ρ : Z/3 ↪→ GL2(R)

which we discussed earlier. This is faithful. But the extension to a ring homomorphism

ρ : R[Z/3]→ Mat2(R)

is not a monomorphism since 1 + σ + σ2 is in its kernel.Another thing I did in class:


Proposition 1.9. The categories of left and right KG-modules are isomorphic.

Proof. Given a left KG-module M , the corresponding right KG-module M ′ is the samevector space as M but with right action of G given by

xg := g−1x

This makes M into a right KG module since

(xg)h = (g−1x)h = h−1g−1x = (gh)−1x = x(gh)

It is straightforward to show that this gives an isomorphism of categories:

KG-mod ∼= mod-KG

as claimed. �

1.3. Semisimplicity of KG. The main theorem about KG is the following.

Theorem 1.10 (Maschke). If G is a finite group of order |G| = n and K is a field withchar k - n (or char k = 0) then KG is semisimple.

Instead of saying char k is either 0 or a prime not dividing n, I will say that 1/n ∈ k.By the Wedderburn structure theorem we get the following.

Corollary 1.11. If 1/|G| ∈ K then

KG ∼= Matd1(Di)× · · · ×Matdb(Db)

where Di are finite dimensional division algebras over K.

Example 1.12.R[Z/3] ∼= R× C

In general, if G is abelian, then the numbers di must all be 1 and Di must be finite fieldextensions of K.

1.3.1. homomorphisms. In order to prove Maschke’s theorem, we need to talk abouthomomorphisms of G-modules. We can define these to be the same as homomorphismsof KG-modules. Then the following is a proposition. (Or, we can take the followingas the definition of a G-module homomorphism, in which case the proposition is thatG-module homomorphisms are the same as homomorphisms of KG-modules.)

Proposition 1.13. Suppose that V,W are KG-modules. Then a K-linear mappingφ : V → W is a homomorphism of KG-modules if and only if it commutes with theaction of G. I.e., if

σ(φ(v)) = φ(σv)

for all σ ∈ G.

Proof. Any homomorphism of KG-modules will commute with the action of KG andtherefore with the action of G ⊂ KG. Conversely, if φ : V → W commutes with theaction of G then, for any

∑agσ ∈ KG, we have

φ

(∑σ∈G

agσv

)=∑σ∈G

agφ(σv) =∑σ∈G

agσφ(v) =

(∑σ∈G

agσ

)φ(v)

So, φ is a homomorphism of KG-modules. �


We also have the following Proposition/Definition of a G-submodule.

Proposition 1.14. A subset W of a G-module V over K is a KG-submodule (and wecall it a G-submodule) if and only if

(1) W is a vector subspace of V and(2) W is invariant under the action of G. I.e., σW ⊆ W for all σ ∈ G.

I pointed out that “invariant” is defined as gW ⊆ W for all g ∈ G. This implies thatgW = W since g−1W ⊆ W implies W ⊆ gW .

We also had a discussion about what happens when char K = p and p divides n = |G|.The simplest example is K = Fp and G = Z/p. Then

Fp[Z/p] ∼= Fp[x]/(xp)

To see this, take x = 1−σ where σ is the generator of Z/p. Then xp = 1p−σp = 1−1 = 0.So, Fp is a local ring with maximal ideal m = (x).

In general, the way to handle the “modular” case when p = charK divides |G| isto find a DVR R so that R/m = K (with characteristic p) but QR has characteristiczero. For example, R = Z(p) which has factor ring R/m = Z/p and QR = Q. Since anyring homomorphism R → S induces a functor S-mod → R-mod, the homomorphismsR � K and R ↪→ QR induce functors the other way giving the following diagram:

KG-mod

((

QR[G]-mod

uuRG-mod

This gives a lot of information about the “bad” case of KG-modules in terms of thewell-behaved QR[G]-modules. Perhaps we can discuss this more after we finish with theeasy case when K = C.


Proof of Maschke’s Theorem. Suppose that V is a finitely generated G-module and Wis any G-submodule of V . Then we want to show that W is a direct summand of V .This is one of the characterizations of semisimple modules. This will prove that all f.g.KG-modules are semisimple and therefore KG is a semisimple ring.

Since W is a submodule of V , it is in particular a vector subspace of V . So, there isa linear projection map φ : V → W so that φ|W = idW . If φ is a homomorphism ofG-modules, then V = W ⊕ kerφ and W would split from V . So, we would be done. If φis not a G-homomorphism, we can make it into a G-homomorphism by “averaging overthe group,” i.e., by replacing it with

ψ =1

n

∑g∈G

λg−1 ◦ φ ◦ λg

First, I claim that ψ|W = idW . To see this take any w ∈ W . Since W is G-invariant,gw ∈ W for any g ∈ G. So, φ(gw) = gw and

ψ(w) =1

n

∑g∈G

g−1φ(gw) =1

n

∑g∈G

g−1(gw) = w

Next I claim that ψ is a homomorphism of G-modules. To show this take any h ∈ Gand v ∈ V . Then

ψ(hv) =1

n

∑g∈G

g−1φ(ghv) = h1

n

∑g∈G

h−1g−1φ(ghv) = h1

n

∑k∈G

k−1φ(kv) = hψ(v)

since summing over all g ∈ G is the same as summing over all k = gh ∈ G. So, ψ givesa splitting of V as required. �

1.3.2. R[Z/3]. The isomorphism R[Z/3] ∼= R× C is given by the mapping:

φ : Z/3→ R× Cwhich sends the generator σ to (1, ω) where ω is a primitive third root of unity. Since(1, 0), (1, ω), (1, ω) are linearly independent over R, the linear extension φ of φ is anisomorphism of R-algebras.

1.3.3. group rings over C. We will specialize to the case K = C. In that case, there areno finite dimensional division algebras over C (Part C, Theorem 3.12). So, we get onlymatrix algebras:

Corollary 1.15. If G is any finite group then

CG ∼= Matd1(C)× · · · ×Matdb(C)

In particular, n = |G| =∑d2i .

Example 1.16. If G is a finite abelian group of order n then CG ∼= Cn.

Example 1.17. Take G = S3, the symmetric group on 3 letters. Since this group isnonabelian, the numbers di cannot all be equal to 1. But the only way that 6 can bewritten as a sum of squares, not all 1, is 6 = 1 + 1 + 4. Therefore,

C[S3] ∼= C× C×Mat2(C)


This can be viewed as a subalgebra of Mat4(C) given by∗ 0 0 00 ∗ 0 00 0 ∗ ∗0 0 ∗ ∗

Each star (∗) represents an independent complex variable. In this description, it is easyto visualize what are the simple factors Matdi(C) given by the Wedderburn structuretheorem. But what are the corresponding factors of the group ring CG?

1.4. Idempotents. Suppose that R = R1 × R2 × R3 is a product of three rings. Thenthe unity of R decomposes as 1 = (1, 1, 1). This can be written as a sum of unit vectors:

1 = (1, 0, 0) + (0, 1, 0) + (0, 0, 1) = e1 + e2 + e3

This is a decomposition of unity (1) as a sum of central, orthogonal idempotents ei.Recall that idempotent means that e2i = ei for all i. Also, 0 is not considered to be an

idempotent. Orthogonal means that eiej = 0 if i 6= j. Central means that ei ∈ Z(R).We had a discussion in class about what are these things Ri.

Lemma 1.18. Suppose that I1, I2 are two sided ideals in any ring R so that

(1) I1 ∩ I2 = 0(2) R = I1 + I2

Then we have an isomorphism of rings: R ∼= R/I1 ×R/I2.

Proof. The isomorphism φ : R → R/I1 × R/I2 is given by φ(r) = (r + I1, r + I2). Thekernel of φ is the set of all r so that r+I1 is zero in R/I1 which means r ∈ I1 and r ∈ I2.So, kerφ = I1∩I2 = 0. Also φ is onto: Take any element x = (r+I1, s+I2) ∈ R/I1×R/I2.Since R = I1 ⊕ I2, r = r1 + r2 and s = s1 + s2 where r1, s1 ∈ I1 and r2, s2 ∈ I2. Thenr + I1 = r2 + I1 and s + I2 = s1 + I2. Then φ(s1 + r2) = (s1 + r2 + I1, s1 + r2 + I2) =(r2 + I1, s1 + I2) = x. So, φ is onto. �

R/I1, R/I2 are factor rings of R. But they are isomorphic, as rings without unit tothe ideal I2, I1 respectively and we have the internal direct sum:

R = I1 ⊕ I2of R as direct sum of two ideals.

Theorem 1.19. A ring R can be written as a product of b rings R1, R2, · · · , Rb iff 1 ∈ Rcan be written as a sum of b central, orthogonal idempotents and, in that case, the idealRi = eiR is a ring with unity ei.

Proof. Suppose R ∼= R1 × · · · × Rb. Let φ be the isomorphism. Then 1 = e1 + · · · + ebthere ei = φ−1 of the unity in Ri and these are central orthogonal idempotents of R sincethey come from central orthogonal idempotents of R1 × · · · × Rb. Conversely, supposethe 1 = e1 + · · · eb. Then R is a direct sum of the two sided ideals eiR and the ideal eiRis isomorphic to the factor ring R/

∑j 6=i ejR (the isomorphism is the restriction of the

projection map R→ R/∑

j 6=i ejR to eiR. �


A central idempotent e is called primitive if it cannot be written as a sum of twocentral orthogonal idempotents.

Corollary 1.20. The number of factors Ri = eiR is maximal iff each ei is primitive.

So, the problem is to write unity 1 ∈ CG as a sum of primitive, central (⇒ orthogonal)idempotents. We will derive a formula for this decomposition using characters.

1.5. Center of CG. Before I move on to characters, I want to prove one last thingabout the group ring CG.

Theorem 1.21. The number of factors b in the decomposition

CG ∼=b∏i=1

Matdi(C)

is equal to the number of conjugacy classes of elements of G.

For, example, the group S3 has three conjugacy classes: the identity {1}, the trans-positions {(12), (23), (13)} and the 3-cycles {(123), (132)}.

In order to prove this we note that b is the dimension of the center of the right handside. Any central element of Matdi(C) is a scalar multiple of the unit matrix which weare calling ei (the ith primitive central idempotent). Therefore:

Lemma 1.22. The center of∏b

i=1 Matdi(C) is the vector subspace spanned by the prim-itive central idempotents e1, · · · , eb. In particular it is b-dimensional.

So, it suffices to show that the dimension of the center of CG is equal to the numberof conjugacy classes of elements of G. (If G is abelian, this is clearly true.)

Definition 1.23. A class function on G is a function f : G → X so that f takes thesame value on conjugate elements. I.e.,

f(hgh−1) = f(g)

for all g, h ∈ G. Usually, X = C.

For example, any function on an abelian group is a class function.

Lemma 1.24. For any field K, the center of KG is the set of all∑

g∈G agg so that agis a class function on G. So, Z(KG) ∼= Kc where c is the number of conjugacy classesof elements of G.

Proof. If∑

g∈G agg is central then∑g∈G

agg = h∑g∈G

aggh−1 =

∑g∈G

aghgh−1

The coefficient of hgh−1 on both sides must agree. So

ahgh−1 = ag

I.e., ag is a class function. The converse is also clear. �

These two lemmas clearly imply Theorem 1.21 (which can now be stated as: b = c ifK = C).


2. Characters

If ρ : G → GLd(C) is a representation of G over C then the character of ρ is thefunction

χρ : G→ Cgiven by χρ(g) = Tr(ρ(g)).

Example 2.1. For ρ the degree 2 representation of Z/3 we have

χ(1) = 2, χ(σ) = −1

2− 1

2= −1 = χ(σ2)

The main property of characters is that they determine the representation uniquelyup to isomorphism. So, once we find all the characters (by constructing the charactertable) we will in some sense know all the representations. We will assume that all groupsG are finite and all representations are finite dimensional over C.

2.1. Basic properties. The basic property of trace is that it is invariant under conju-gation:

Tr(ABA−1) = Tr(B)

Letting A = ρ(g), B = ρ(h) we get

χρ(ghg−1) = Tr(ρ(ghg−1)) = Tr(ρ(g)ρ(h)ρ(g)−1) = Tr(ρ(h)) = χρ(h)

for any representation ρ. So:

Theorem 2.2. Characters are class functions. (They have the same value on conjugateelements.)

If ρ : G → AutC(V ) is a representation of G over C, then the character of ρ, alsocalled the character of V , is defined to be the function

χρ = χV : G→ Cgiven by

χV (g) = Tr(ρ(g)) = Tr(φ ◦ ρ(g) ◦ φ−1)for any linear isomorphism φ : V

≈−→ Cd.


There are three basic formulas that I want to explain. In order of difficulty they are:

(1) The character of a direct sum is the sum of the characters:

χV⊕W = χV + χW

(2) The character of a tensor product is the product of the characters:

χV⊗W = χV χW

(3) The character of the dual representation is the complex conjugate of the originalcharacter:

χV ∗ = χV

2.1.1. direct sum. The trace of a direct sum of matrices is the sum of traces:

Tr(A⊕B) = Tr

(A 00 B

)= Tr(A) + Tr(B)

Theorem 2.3. If V,W are two G-modules then

χV⊕W = χV + χW

Proof. If ρV , ρW , ρV⊕W are the corresponding representations then

ρV⊕W (g) = ρV (g)⊕ ρW (g)

The theorem follows. �

2.1.2. character formula using dual basis. Instead of using traces of matrices, I preferthe following equivalent formula for characters using bases and dual bases.

If V is a G-module, we choose a basis {v1, · · · , vd} for V as a vector space overC. Then recall that the dual basis for V ∗ = HomC(V,C) consists of the dual vectorsv∗1, · · · , v∗d : V → C given by

v∗j

(d∑i=1

aivi

)= aj

I.e., v∗j picks out the coefficient of vj.

Proposition 2.4.

χV (g) =d∑i=1

v∗i (σvi)

Proof. The matrix of the linear transformation ρ(g) has (i, j) entry v∗i (σvj) Therefore,its trace is

∑v∗i (σvi). �

For example, the trace of the identity map is

Tr(idV ) =d∑i=1

v∗i (vi) = d

Theorem 2.5. The value of the character at 1 is the degree of the representation:

χV (1) = d = dimC(V )


2.1.3. tensor product. If V,W are two G-modules then the tensor product V ⊗ W isdefined to be the tensor product over C with the following action of G:

σ(v ⊗ w) = σv ⊗ σw

Theorem 2.6. The character of V ⊗W is the product of the characters of V and W .I.e.,

χV⊗W (g) = χV (g)χW (g)

for all σ ∈ G.

Proof. Choose bases {vi}, {wj} for V,W with dual bases {v∗i }, {w∗j}. Then the tensorproduct V ⊗W has basis elements vi ⊗ wj with dual basis elements v∗i ⊗ w∗j . So, thecharacter is:

χV⊗W (g) =∑i,j

(v∗i ⊗ w∗j )σ(vi ⊗ wj) =∑i,j

(v∗i ⊗ w∗j )(σvi ⊗ σwj)

=∑i,j

v∗i (σvi)w∗j (σwj) =

∑i

v∗i (σvi)∑j

w∗j (σwj) = χV (g)χW (g)

�

2.1.4. dual reprentation. The dual space V ∗ is a right G-module. In order to make it aleft G-module we have to invert the elements of the group. I.e., for all f ∈ V ∗ we define

(gf)(v) := f(g−1v)

Lemma 2.7.

χV ∗(g) = χV (g−1)

Lemma 2.8.

χV (g−1) = χV (g)

Proof. The trace of a matrix A is equal to the sum of its eigenvalues λi. If A has finiteorder: Am = Id then its eigenvalues are roots of unity. Therefore, their inverses areequal to their complex conjugates. So,

Tr(A−1) =∑

λ−1i =∑

λi = Tr(A)

Since G is finite, the lemma follows. �

Theorem 2.9. The character of the dual representation V ∗ is the complex conjugate ofthe character of V :

χV ∗(g) = χV (g)


2.2. Irreducible characters.

Definition 2.10. A representation ρ : G→ AutC(V ) is called irreducible if V is a simpleG-module. The character

χρ = χV : G→ Cof an irreducible representation is called an irreducible character.

Theorem 2.11. Every character is a nonnegative integer linear combination of irre-ducible characters.

Proof. Since CG is semisimple, any G-module V is a direct sum of simple modulesV ∼=

⊕Si. So, the character of V is a sum of the corresponding irreducible characters:

χV =∑χSi

. �

If we collect together multiple copies of the same simple module we get V =⊕

niSiand

χV =r∑i=1

niχi

where χi is the character of Si. This makes sense only if we know that there are onlyfinitely many nonisomorphic simple modules Si and that the corresponding charactersχi are distinct functions G→ C. In fact we will show the following.

Theorem 2.12. (1) There are exactly b (the number of blocks) irreducible represen-tations Si up to isomorphism.

(2) The corresponding characters χi are linearly independent.

This will immediately imply the following.

Corollary 2.13. The irreducible characters χ1, · · · , χb form a basis for the b-dimensionalvector space of all class functions G→ C.


2.2.1. Example: abelian groups. Let’s do the easy case first: when G is abelian. In thatcase c = b = n = |G| since no two elements of G are conjugate. We have:

CG ∼=b∏1

Mat1(C) = Cn

Thus each simple module is one dimensional. This gives n irreducible characters χi :G→ GL1(C) = C∗ which are representations: χi = ρi.

For example, if G = Z/n is cyclic with generator t, then the possible representationsare ρi(t) = ζ i where ζ is a primitive n-th root of unity. For n = 3, 4 we get the following:

Z/3 1 t t2

χ1 1 ω ω2

χ2 1 ω2 ωχ3 1 1 1

Z/4 1 t t2 t3

χ1 1 i −1 −iχ2 1 −1 1 −1χ3 1 −i −1 iχ4 1 1 1 1

where ω = −12

+ i√32

is the third root of unity and i =√−1. These are character tables

with b rows, one for each irreducible character and c columns, one for each conjugacyclass of elements of G. Since b = c the character table is always a square matrix.

Theorem 2.14 (Artin). Degree one characters (i.e., homomorphisms ρi : G→ C∗) arelinearly independent.

Proof. χ1, · · · , χk are linearly independent by induction on k. If k = 1, the statement iscertainly true. Suppose k ≥ 2. Suppose we have an equation:∑

aiχi = 0

where ai are not all zero. By induction on k, all of the ai must be nonzero. (Otherwisewe have a linear relation among k − 1 characters.) This means that, for all g ∈ G wehave: ∑

aiχi(g) = 0

For any g, h ∈ G we have ∑aiχi(hg) =

∑aiχi(h)χi(g) = 0

Multiply the previous equation by χk(h) and subtract:

k∑i=1

ai(χi(h)− χk(h))χi(g)

Since the i = k term is zero, we get a linear relation among χ1, · · · , χk−1. By inductionon k we have

ai(χi(h)− χk(h)) = 0

for all i < k. Since ai 6= 0 this implies χi(h) = χk(h). Since this holds for any h ∈ G weget χi = χk which is a contradiction. �


2.2.2. regular representation. This is a particularly elementary representation and char-acter which contains all the simple modules.

Definition 2.15. The free module CGCG is called the regular representation of G. Thecorresponding character is called the regular character : χreg = χCG : G→ C.

Theorem 2.16. χreg(g) =

{n = |G| if g = 1

0 if g 6= 1

Proof. I used the basis-dual basis formula for characters. The regular representationV = CG has basis elements g ∈ G and dual basis elements g∗ given by

g∗(∑

ahh)

= ag

I.e., g∗(x) is the coefficient of g in the expansion of x. The regular character is thengiven by

χreg(h) =∑g∈G

g∗(hg)

But this is clearly equal to 0 if h 6= 1 since the coefficient of g in hg is 0. And we alreadyknow that χreg(1) = dimCG = n. �

Lemma 2.17. There are only finitely many isomorphism classes of simple G-modules.

Proof. First choose a decomposition of the regular representation into simple modules:

CG ∼=⊕

Sα

Then I claim that any simple module S is isomorphic to one of the Sα in this decompo-sition. And this will prove the lemma.

To prove the claim, choose any nonzero element x0 ∈ S. Then x0 generates S (thesubmodule generated by x0 is either 0 or S). Therefore, we have an epimorphism

φ : CG� S

given by φ(r) = rx0. When we restrict φ to each simple component Sα, we get ahomomorphism φ|Sα : Sα → S which, by Schur’s lemma, must either be zero or anisomorphism. These restrictions cannot all be zero since φ is an epimorphism. Therefore,one of them is an isomorphism Sα ∼= S. This proves the claim. �

This proof shows more than the lemma states. It proves:

Lemma 2.18. The regular representation CG contains an isomorphic copy of everysimple G-module.

Therefore, in order to find all the irreducible representations, we need to decomposethe regular representation as a sum of simple modules.


2.2.3. decomposition of the regular representation. At this point I used the Wedderburnstructure theorem again:

CG ∼=b∏i=1

Matdi(C) =∏

Ri

where, following Lang, we write Ri = Matdi(C).Let Si = Cdi be the vector space of column vectors. Then Ri acts on the left by matrix

multiplication and it is easy to see that Si is a simple Ri-module since it is generatedby any nonzero element.

Lemma 2.19. If φ : R′ � R is an epimorphism of rings and S is any simple R-module,then S becomes a simple R′-module with the action of R′ induced by φ.

Proof. If x0 ∈ S is any nonzero element then R′x0 = Rx0 = S. So, any nonzero elementof S generates the whole thing as an R′-module. So, it is simple. �

Since CG =∏Ri, we can make Si into a G-module with the ring homomorphism:

ρi : CG πi−→ Ri≈−→ EndC(Si)

Since πi : CG� Ri is an epimorphism, Si becomes a simple G-module. In other words,the corresponding representation is irreducible:

ρi : G→ AutC(Si)

Also, Lang points out that

RjSi = 0

if i 6= j. (And RiSi = Si.) This is the key point. It shows immediately that the G-modules Si are not isomorphic. And it will also show that the characters are linearlyindependent.

In order to show that the characters

χi = χρi = χSi

are linearly independent we will evaluate them on the central idempotents ei corre-sponding to the decomposition CG =

∏Ri. As we discussed earlier, this product

decomposition gives a decomposition of unity:

1 = e1 + · · ·+ eb

where ei is the unity of Ri. (We want to say “ei = 1” but there would be too many 1’s.)We then need to compute χi(ej). But this is not defined since ej is not an element ofG. We need to extend χi to a map on CG.

2.2.4. linear extension of characters. If χ : G→ C is any character, we define the linearextension of χ to CG by the formula

χ(∑

agg)

=∑

agχ(g)

Since the symbol χ is already taken (χ is the complex conjugate of χ), I decided to usethe same symbol χ to denote the linear extension of χ given by the above formula.


The linear extension of χρ is χρ which is the trace of the linear extension of ρ. To seethis let x =

∑agg. Then∑agχρ(g) =

∑ag Tr(ρ(g)) = Tr

(∑agρ(g)

)= Tr(ρ(x))

Lemma 2.20.

χi(ej) =

{0 if i 6= j

di if i = j

Proof. If i 6= j we haveχi(ej) = Tr(ρi(ej)) = 0

since ρi(ej) is the zero matrix (giving the action of ej ∈ Rj on Wi).If i = j then

χi(ei) = dimSi = disince ei is unity in Ri. �

This proves the second part of Theorem 2.12: If∑aiχi = 0 then∑

aiχi(ej) = ajdj = 0

which forces aj = 0 for all j.

Theorem 2.21. The regular representation decomposes as:

CG ∼=b∑i=1

diSi

2.2.5. example. Take G = S3. Then we already saw that

C[S3] ∼= C× C×Mat2(C)

So, there are three simple modules S1, S2, S3

S1 = C is the trivial representation.S2 is the sign representation ρ2(g) = sgn(g) = ±1S3 is a simple 2-dimensional module.Since characters are class functions, their value is the same on conjugate elements.

So, we only need their values on representatives 1, (12), (123). The characters χ1, χ2 areeasy to compute. The last irreducible character is determined by the equation

χreg = χ1 + χ2 + 2χ3

So, here is the character table of S3:

1 (12) (123)

χ1 1 1 1χ2 1 −1 1χ3 2 0 −1χreg 6 0 0

All characters of S3 are nonnegative integer linear combinations of χ1, χ2, χ3.


2.2.6. example: Z/2⊕Z/2. Let’s call the elements of the group 1, σ, τ, στ . Since Z/2⊕Z/2 is abelian, all characters are again one dimensional and the values must be squareroots of 1, i.e., they must be ±1. So, we got the following.

1 σ τ στ

χ1 1 1 1 1χ2 1 −1 1 −1χ3 1 −1 −1 1χ4 1 1 −1 −1

Each row is clearly a one-dimensional representation. There are no others because weknow that there are exactly b = 4 such representations. So, this is the complete charactertable.

2.2.7. example: D4. This is the dihedral group of order 8 with presentation:

D4 =⟨σ, τ |σ4, τ 2, στστ

⟩(Replace 4 by any n to get the dihedral group of order 2n.) To find the numbers di wehave to write n = 8 as a sum of squares which are not all 1 (because D4 is nonabelian)and so that there is at least one 1 (since d1 = 1). The solution is:

8 = 1 + 1 + 1 + 1 + 4

Therefore, b = c = 5.The elements of the group are:

D4 = {e, σ, σ2, σ3, τ, στ, σ2τ, σ3τ}Among these, σ, σ3 are conjugate since τστ−1 = σ3, τ, σ2τ = στσ−1 are conjugate andστ, σ3τ = σ(στ)σ−1 are conjugate. There are no other conjugacy relations since we gotit down to 5 classes.

Among the 5 characters, the first 4 are 1-dimensional. And we can find them veryquickly as follows. The center of D4 is the set of elements which are alone in theirconjugacy class. So,

Z(D4) = {1, σ2}This is a normal subgroup of D4 with quotient isomorphic to Z/2⊕ Z/2.

We already have four irreducible representations ρ1, · · · , ρ4 of Z/2 ⊕ Z/2. We cancompose with the projection to get four irreducible representations of D4

D4 � D4/Zρi=χi−−−→ C×

This gives the first four lines in the character table:

1 σ2 σ τ στ

χ1 1 1 1 1 1χ2 1 1 −1 1 −1χ3 1 1 −1 −1 1χ4 1 1 1 −1 −1χ5 2 −2 0 0 0


To get the last line we use the equation:

χreg =∑

diχi = χ1 + χ2 + χ3 + χ4 + 2χ5

2.3. formula for idempotents. Lang gives a formula for the idempotents ei ∈ CG interms of the corresponding irreducible character χi. The key point is that the linearextension

ρi : CG→ EndC(Si) = Matdi(C)

of ρi sends ei to the identity matrix and ej to zero for j 6= i. Therefore,

ρi(eig) = ρi(ei)ρi(g) = ρi(g)

χi(ejg) =

{χi(g) if i = j

0 if i 6= j

Now use the regular character. Since χreg =∑djχj we get:

χreg(eig−1) =

b∑j=1

djχj(eig−1) = diχi(g

−1)

If ei =∑ahh then

χreg(eig−1) = χreg

(∑h∈G

ahhg−1

)= nag

So,

ag =1

nχreg

(eig−1) =

1

n

∑j

djχj(eig−1) =

dinχi(g

−1)

Theorem 2.22.

ei =din

∑g∈G

χi(g−1)g

This formula has an important consequence.

Corollary 2.23. di|n (Each di divides n = |G|.)

Proof. First recall that ei is an idempotent. So,

ei = e2i =din

∑τ∈G

χi(τ−1)τei

Now multiply by n/di to get:

(2.1)n

diei =

∑τ∈G

χi(τ−1)τei

I mentioned earlier that χi(τ−1) =

∑λj is a sum of mth roots of unity where m =

o(τ−1) = o(h). But this number divides n = |G|. So, each λj is a power of ζ = e2πi/n

Let Mi ⊂ CG be the additive subgroup generated by all elements of the form ζjτei (forall j and fixed i). This is a finitely generated torsion free (and thus free) Z-module and


equation (2.1) shows that Mi is invariant under multiplication by the rational numbern/di. Therefore, n/di is integral. Since Z is integrally closed in Q this implies thatn/di ∈ Z. �

2.3.1. kernel of a representation. Looking at the character table, we can determine whichelements of the group lie in the kernel of each representation.

Lemma 2.24. σ ∈ ker ρ ⇐⇒ χρ(g) = d = χρ(1).

Proof. In a d-dimensional representation, χ(g) = λ1 + · · · + λd is a sum of d roots ofunity. This sum is equal to d if and only if every λi = 1 which is equivalent to sayingthat ρ(g) is the identity matrix (since ρ(g) has finite order). �

Using the same argument it follows that:

Proposition 2.25. |χρ(g)| = d if and only if ρ(g) = λId is a scalar multiple of theidentity matrix. Furthermore, λ = χρ(g)/d.

For example, in the last irreducible representation of D4 we have

|χ5(σ2)| = 2 = d5

Therefore, ρ5(σ2) = −I2.

2.3.2. finding all normal subgroups. Finally, I claimed that the character table deter-mines all normal subgroups of the group G. This is based on the trick that we used toconstruct the character table of D4.

Suppose that N is a normal subgroup of G and ρi, i = 1, · · · , r are the irreduciblerepresentations of G/N .

Lemma 2.26.N =

⋂ker(ρi ◦ π)

where π : G� G/N is the quotient map.

Proof. Let K =⋂

ker(ρi ◦ π). Then clearly, N ⊆ K. So, suppose that K is bigger thanN . Then the representations ρi would all factor through the quotient G/K:

ρi : G/Nφ−→ G/K

ψi−→ AutC(Si)

This is not possible because the sum of the squares of the dimensions of these represen-tations add up to the order of G/N :

|G/K| < |G/N | =∑

d2i

So, the ψi are distinct irreducible representations of G/K whose dimensions squared addup to more than the order of the group. This contradiction proves the lemma. �

Combining Lemmas 2.26 and 2.24, we get the following.

Theorem 2.27. The normal subgroups of a finite group G can be determined from itscharacter table as follows.

(1) The kernel of ρi is the union of all conjugacy classes cj for which χi(cj) = di =χi(1).


(2) A collection of conjugacy classes forms a normal subgroup if and only if it is anintersection of kernels of irreducible representations ρi.


2.4. orthogonality relations. The character table satisfies two orthogonality rela-tions:

(1) row orthogonality(2) column orthogonality

First, I will do row orthogonality. The rows are the characters χi. We want to showthat they are “orthogonal” in some sense.

2.4.1. main theorem and consequences.

Definition 2.28. If f, g : G→ C are class functions then we define 〈f, g〉 ∈ C by

〈f, g〉 = 〈g, f〉 =1

n

∑σ∈G

f(σ)g(σ−1)

(n = |G|). The main theorem is the following.

Theorem 2.29. If V,W are G-modules then

〈χV , χW 〉 = dimC HomG(V,W )

Before I prove this let me explain the consequences.

Corollary 2.30. The rows of the character table are orthonormal in the sense that:

〈χi, χj〉 = δij

Proof. It follows from Schur’s lemma that

〈χi, χj〉 = dimC HomG(Si, Sj) = δij

since HomG(Si, Sj) = 0 for i 6= j and HomC(Si, Si) = C. �

Since only conjugacy classes appear in the character table we have:

〈χi, χj〉 =1

n

b∑k=1

|ck|χi(ck)χj(ck)

For example, for G = S3 we have the character table:

|cj| 1 3 21 (12) (123)

χ1 1 1 1χ2 1 −1 1χ3 2 0 −1

〈χ1, χ2〉 =(1)(1) + 3(1)(−1) + 2(1)(1)

6=

1− 3 + 2

6= 0

This formula also tells us that a representation is determined by its character in thefollowing way.

Corollary 2.31. Suppose that the semisimple decomposition of the G-module V is V =∑niSi. Then

ni = 〈χV , χi〉


Proof. Since χV⊕W = χV + χW , we have: χV =∑njχj. So,

〈χV , χi〉 =⟨∑

njχj, χi

⟩= ni

�

2.4.2. proof of the main theorem. Theorem 2.29 will follow from three lemmas. The firstlemma calculates the dimension of the fixed point set of V .

Definition 2.32. If V is a G-module then the fixed point set of the action of G is givenby

V G := {v ∈ V | gv = v ∀g ∈ G}

Lemma 2.33. The dimension of the fixed point set is equal to the average value of thecorresponding character:

dimC VG =

1

n

∑g∈G

χV (g)

Proof. The projection map

π : V → V G

is given by

π(v) =1

n

∑gv

It is clear that

(1) π(v) ∈ V G since multiplication by any g ∈ G will just permute the summands.(2) π(v) = v if v ∈ V G because, in that case, each gv = v and there are n terms.

Therefore, π is a projection map, i.e., a linear retraction onto V G. Looking at theformula we see that π is multiplication by the idempotent e1 = 1

n

∑g∈G g. (This is the

idempotent corresponding to the trivial representation.) So:

dimV G = Tr(π) = χV (e1) = χV

(1

n

∑σ∈G

σ

)=

1

n

∑σ∈G

χV (g)

Explanations:(1) dimV G = Tr(π) because V ∼= V G ⊕W (W = ker π). So, the matrix of π is:

π =

(1V G 0

0 0W

)making Tr(π) = Tr(1V G) = dimC V

G.(2) Tr(π) = χV (e1) by definition of the character:

χV (e1) := Tr(e1· : V → V )

This is the trace of the mapping V → V given by multiplication by e1. But we arecalling that mapping π. �


Lemma 2.34. If V,W are representations of G then

HomG(V,W ) = HomC(V,W )G

where G acts on HomC(V,W ) by conjugation: σf = σ ◦ f ◦ σ−1 for σ ∈ G, i.e.,

(σf)(v) = σf(σ−1v)

Proof. This is trivial. Given any linear map f : V → W , f is a G-homomorphism iff

σ ◦ f = f ◦ σ ⇐⇒ σ ◦ f ◦ σ−1 = f ⇐⇒ σf = f

iff f ∈ HomC(V,W )G. �

Lemma 2.35. HomC(V,W ) ∼= V ∗ ⊗W as G-modules.

Proof. Let φ : V ∗ ⊗W → HomC(V,W ) be given by

φ(f ⊗ w)(v) = f(v)w

To check that this is a G-homomorphism we need to show that φσ = σφ for any σ ∈ G.So, we compute both sides:

φσ(f ⊗ w) = φ(σf ⊗ σw) = φ(f ◦ σ−1 ⊗ σw)

which sends v ∈ V toφ(f ◦ σ−1 ⊗ σw)(v) = f(σ−1v)σw

On the other side we have:

σφ(f ⊗ w) = σ ◦ φ(f ⊗ w) ◦ σ−1

which also sends v ∈ V to

σ ◦ φ(f ⊗ w) ◦ σ−1v = σ(f(σ−1v)w) = f(σ−1v)σw

This shows that φ commutes with the action of G. The fact that φ is an isomorphism iswell-known: If vi, v

∗i form a basis-dual basis pair for V and wj form a basis for W then

v∗j ⊗ wi form a basis for V ∗ ⊗W and

φ(v∗j ⊗ wi) : v =∑

ajvj 7→ v∗j (v)wi = ajwi

is the mapping whose matrix has ij-entry equal to 1 and all other entries 0. So, thesehomomorphisms form a basis for HomC(V,W ) and φ is an isomorphism. �

Proof of Main Theorem 2.29. Using the three lemmas we get:

dimC HomG(V,W ) =2.34 dimC HomC(V,W )G

=2.35 dimC(V ∗ ⊗W )G

=2.331

n

∑σ∈G

χV ∗⊗W (σ)

=1

n

∑σ

χV ∗(σ)χW (σ)

=1

n

∑σ

χV (σ−1)χW (σ) = 〈χV , χW 〉

�


2.4.3. character table of S4. Using these formulas we can calculate the character tablefor S4. First note that there are five conjugacy classes represented by

1, (12), (123), (12)(34), (1234)

The elements of cycle form (12)(34) form (with 1) a normal subgroup

K = {1, (12)(34), (13)(24), (14)(23)} / S4

called the Klein 4-group. The quotient S4/K is isomorphic to the symmetric group on3 letters. Imitating the case of D4, this allows us to construct the following portion ofthe character table for S4:

|cj| 1 6 8 3 61 (12) (123) (12)(34) (1234)

χ1 1 1 1 1 1χ2 1 −1 1 1 −1χ3 2 0 −1 2 0χ4 3χ5 3

Explanations:

(1) Since (12)(34) ∈ K, the value of the first three characters on this conjugacy classis di, the same as in the first column.

(2) Since (1234)K = (12)K, these two columns have the same values of χ1, χ2, χ3.(3) Finally, the two unknown characters χ4, χ5 must be 3-dimensional since

24 =∑

d2i = 1 + 1 + 4 + d24 + d25

has only one solution: d4 = d5 = 3.

To figure out the unknown characters we need another representation. The permuta-tion representation P is the 4-dimensional representation of S4 in which the elements ofS4 act by permuting the unit coordinate vectors. For example

ρP (12) =

0 1 0 01 0 0 00 0 1 00 0 0 1

Note that the trace of ρP (g) is equal to the number of letters left fixed by σ. So, χPtakes values 4, 2, 1, 0, 0 as shown:

|cj| 1 6 8 3 61 (12) (123) (12)(34) (1234)

χ1 1 1 1 1 1χ2 1 −1 1 1 −1χ3 2 0 −1 2 0χP 4 2 1 0 0

χV = χP − χ1 3 1 0 −1 −1


The representation P contains one copy of the trivial representation and no copies ofthe other two:

〈χP , χ1〉 =1

24(4 + 6(2) + 8(1)) = 1

〈χP , χ2〉 =1

24(4 + 6(−1)(2) + 8(1)(1)) = 0

〈χP , χ3〉 =1

24((2)(4) + 8(−1)(1)) = 0

So, P ∼= S1⊕V where V is a 3-dimensional module which does not contain S1, S2 or S3.So, V = nS4 ⊕mS5. But S4, S5 are both 3-dimensional. So, V = S4 (or S5).

Using the fact thatχ1 + χ2 + 2χ3 + 3χ4 + 3χ5 = χreg

we can now complete the character table of S4:

|cj| 1 6 8 3 61 (12) (123) (12)(34) (1234)

χ1 1 1 1 1 1χ2 1 −1 1 1 −1χ3 2 0 −1 2 0χ4 3 1 0 −1 −1χ5 3 −1 0 −1 1

From the character table of S4 we can find all normal subgroups. First, the kernels ofthe 5 irreducible representations are:

(1) ker ρ1 = S4.(2) ker ρ2 = A4 containing the conjugacy classes of 1, (123), (12)(34).(3) ker ρ3 = K containing 1, (12)(34) and conjugates.(4) ker ρ4 = 1. I.e., ρ4 is a faithful representation.(5) ker ρ5 = 1. So, ρ5 is also faithful.

Since these subgroups contain each other:

1 < K < A4 < S4

intersecting them will not give any other subgroups. So, these are the only normalsubgroups of S4.

2.4.4. column orthogonality. We skipped this topic.


3. Restriction and Induction

Given any ring homomorphism S → R we get a functor R-mod→ S-mod by “restric-tion of scalars”. Given a subgroup H of a group G, the inclusion map CH → CG is aring homomorphism and therefore induces a functor

ResGH : G-mod→ H-mod.

Given a G-module M , the H-module ResGHM is the same vector space M where weconsider only the action of the elements of H on M . The character of this module isjust the restriction of the character of M :

ResGHχ(h) = χ(h)

for all h ∈ H. Equivalently,

(3.1) ResGHχV = χResGH

for any G-module V .

3.1. induction. Given a representation V of H, the induced representation IndGHV isthe G-module given by

IndGHV := CG⊗CH V.

For example, the regular representation of H induces the regular representation of G:

IndGHCH := CG⊗CH CH = CG

The main theorem about restriction and induction is Frobenius reciprocity:

Theorem 3.1 (Frobenius reciprocity). If V is a representation of H and W is a repre-sentation of G then

HomG(IndGH V,W ) ∼= HomH(V,ResGHW )

This follows from:

Theorem 3.2 (adjunction formula).

HomR(RMS ⊗S SV, RN) ∼= HomS(SV,HomR(MS, N))

And the easy formula:

HomR(R,N) ∼= N

Letting M = R and S ⊆ R, we get the following.

Corollary 3.3. If S is a subring of R, V is an S-module and W is an R-module then

HomR(R⊗S V,W ) ∼= HomS(V,W )

Putting R = CG,S = C[H], this gives Frobenius reciprocity.


3.2. induced characters.

Definition 3.4. Suppose that H ≤ G (H is a subgroup of G) and χ : H → C is acharacter (or any class function). Then the induced character

IndGH χ : G→ Cis the class function on G defined by

IndGH χ(g) =1

|H|∑h∈G

χ(hgh−1)

where χ(g) = 0 if g /∈ H.

The main theorem about the induced character is the following.

Theorem 3.5. If V is any representation of H then the induced character IndGHχV ofV is the character of the induced representation IndGHV :

IndGH χV = χIndGHV

3.2.1. Frobenius reciprocity for characters. If f : G → C is any class function then therestriction of f to H, denoted ResGH f , is the composition of f with the inclusion mapj : H ↪→ G:

ResGH f = f ◦ j : H → CTheorem 3.6 (Frobenius reciprocity). Suppose that g, h are class functions on G,Hrespectively. Then ⟨

IndGH h, g⟩G

=⟨h,ResGH g

⟩H

Proof. Since both sides of the equation are bilinear in g, h and the irreducible charactersspan the vector space of all class functions, it suffices to prove these when g, h areirreducible characters:

(3.2)⟨IndGH χ

′j, χi

⟩G

=⟨χ′j,ResGH χi

⟩H.

But this follows almost immediately from abstract Frobenius reciprocity (the adjunctionformula): ⟨

IndGH χ′j, χi

⟩G

=thm3.5

⟨χIndGH S′j

, χi

⟩G

=thm2.29 dimC HomG(IndGHS′j, Si)

=thm3.1 dimC HomH(S ′j, ResGHSi)

=3.1

⟨χ′j, χResGH Si

⟩H

=thm2.29

⟨χ′j,ResGH χi

⟩H

�

As I pointed out in the instructions for HW 11, the version of Frobenius reciprocitygiven by the character formula (3.2) implies:

Corollary 3.7. The number of times that the simple G-module Si appears in direct sumdecomposition the induced representation IndGHS

′j is equal to the number of times the

number of times that S ′j appears in the direct sum decomposition of ResGHSi.

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MATH 101B: ALGEBRA II PART D: REPRESENTATIONS OF FINITE GROUPSpeople.brandeis.edu/~igusa/Math131b/Math131b_notesD.pdf · 2019-05-02 · MATH 101B: ALGEBRA II PART D: REPRESENTATIONS