Representations of Finite Groups I(Math 240A)

(Robert Boltje, UCSC, Fall 2014)

Contents

1 Representations and Characters 1

2 Orthogonality Relations 13

3 Algebraic Integers 26

4 Burnside’s paqb-Theorem 33

5 The Group Algebra and its Modules 39

6 The Tensor Product 50

7 Induction 58

8 Frobenius Groups 70

9 Elementary Clifford Theory 75

10 Artin’s and Brauer’s Induction Thoerems 82

1 Representations and Characters

Throughout this section, F denotes a field and G a finite group.

1.1 Definition A (matrix) representation of G over F of degree n ∈ N isa group homomorphism ∆: G → GLn(F ). The representation ∆ is calledfaithful if ∆ is injective. Two representations ∆: G→ GLn(F ) and Γ: G→GLm(F ) are called equivalent if n = m and if there exists an invertible matrixS ∈ GLn(F ) such that Γ(g) = S∆(g)S−1 for all g ∈ G. In this case we oftenwrite Γ = S∆S−1 for short.

1.2 Remark In the literature, sometimes a representation of G over F isdefined as a pair (V, ρ) where V is a finite-dimensional F -vector space andρ : G → AutF (V )(= GL(V )) is a group homomorphism into the group ofF -linear automorphisms of V . These two concepts can be translated intoeach other. Choosing an F -basis of V we obtain a group isomorphismAutF (V )

∼→ GLn(F ) (where n = dimF V ) and composing ρ with this iso-morphism we obtain a representation ∆: G → AutF (V )

∼→ GLn(F ). If wechoose another basis then we obtain equivalent representations. Conversely,if ∆: G → GLn(F ) is a representation, we choose any n-dimensional F -vector space V and a basis of V (e.g. V = F n with the canonical basis)and obtain a homomorphism ρ : G→ GLn(F )

∼→ AutF (V ). Representations(V, ρ) and (W,σ) are called equivalent if there exists an F -linear isomorphismϕ : V → W such that σ(g) = ϕ ◦ ρ(g) ◦ ϕ−1 for all g ∈ G. The above con-structions induce mutually inverse bijections between the set of equivalenceclasses of representations (V, ρ) and the set of equivalence classes of matrixrepresentations ∆ of G over F .

If one takes V := F n, the space of column vectors together with thestandard basis (e1, . . . , en) then the matrix representation ∆ corresponds to(F n, ρ) with ρ(g)v = ∆(g)v.

1.3 Examples (a) The trivial representation is the homomorphism ∆: G→GL1(F ), g 7→ 1.

(b) For n > 2 let D2n = 〈σ, τ | σn = 1, τ 2 = 1, τστ = σ−1〉 be the dihedralgroup of order 2n. We define a faithful representation ∆ of D2n over R by

∆(σ) :=

(cosφ − sinφsinφ cosφ

)and ∆(τ) :=

(−1 00 1

),

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where φ = 2π/n. Note that ∆(σ) is the counterclockwise rotation about2π/n and τ is the reflection about the vertical axis in R2.

(c) The map ∆: Sym(n) → GLn(F ), which maps σ ∈ Sym(n) to thepermutation matrix of σ is a faithful representation of Sym(n) of degreen over any field. It is called the natural representation of Sym(n). (Thepermutation matrix of σ has in its i-th colum the canonical basis vector eσ(i),i.e., ∆(σ) maps ei to eσ(i).)

(d) If ∆: G→ GLn(F ) is a representation and H 6 G then the restrictionof ∆ to H,

ResGH(∆): H i qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq G ∆ qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq GLn(F )

is a representation of H over F of the same degree. Here, i : H → G denotesthe inclusion map. More generally, if f : H → G is an arbitrary grouphomomorphism between two arbitrary finite groups then also

Resf (∆): H f qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq G ∆ qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq GLn(F )

is a representation of H over F of degree n.

(e) If G = {g1, . . . , gn} and if we define the homomorphism f : G →Sym(n) by (f(g))(i) = j, when ggi = gj, then

G f qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq Sym(n) ∆ qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq GLn(F ) ,

with ∆ as in (c), is a representation, called the regular representation of G. Ifwe rearrange the ordering of the elements g1, . . . , gn we obtain an equivalentrepresentation.

(f) If ∆1 : G→ GLn1(F ) and ∆2 : GLn2(F ) are representations of G overF then also their direct sum

∆1 ⊕∆2 : G→ GLn1+n2(F ) , g 7→(

∆1(g) 00 ∆2(g)

)

is a representation of G over F . Since(∆2(g) 0

0 ∆1(g)

)=

(0 In2

In1 0

)(∆1(g) 0

0 ∆2(g)

)(0 In1

In2 0

), ,

we see that ∆1 ⊕∆2 and ∆2 ⊕∆1 are equivalent.

(g) Let E/F be a Galois extension of degree n with Galois group G.Then G 6 AutF (E) and we obtain a representaion (E, i) in the sense ofRemark 1.2, with i : G→ AutF (E) the inclusion.

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1.4 Definition Let ∆: G→ GLn(F ) be a representation.

(a) We say that ∆ is decomposable if it is equivalent to ∆1⊕∆2 for somerepresentations ∆1 and ∆2 ofG over F . Otherwise it is called indecomposable.

(b) We say that ∆ is reducible if it is equivalent to a block upper triangularrepresentation, i.e., to a representation of the form

g 7→(A(g) B(g)

0 C(g)

)

where A(g) ∈ Matn1(F ), B(g) ∈ Matn1×n2(F ), C(g) ∈ Matn2(F ), and wheren1, n2 ∈ N are independent of g ∈ G. Otherwise, it is called irreducible.

1.5 Remark Let ∆: G→ GLn(F ) be a representation.

(a) If ∆ is decomposable then ∆ is also reducible. Thus, if ∆ is irreducibleit is also indecomposable.

(b) A subspace U of F n is called invariant under ∆ (or ∆-stable) if∆(g)u ∈ U for all g ∈ G and u ∈ U . More generally, if (V, ρ) is a rep-resentation corresponding to ∆, a subspace U of V is called invariant under∆ (or ∆-stable) if (ρ(g))(u) ∈ U for all g ∈ G and u ∈ U . Note that ∆ isdecomposable if and only if V has a decomposition V = U ⊕ W into twonon-zero ∆-invariant subspaces U and V . And ∆ is reducible if V has a∆-invariant subspace U with {0} 6= U 6= V .

In the following remark we recollect a few results from linear algebra.

1.6 Remark Let V be a finite-dimensional C-vector space.

(a) A map(−,−) : V × V → C

is called a hermitian scalar product on V if the following holds for all x, x′, y, y′ ∈V and λ ∈ C:

(x+ x′, y) = (x, y) + (x′, y) , (x, y + y′) = (x, y) + (x, y′) ,

(λx, y) = λ(x, y) , (x, λy) = λ(x, y) ,

(y, x) = (x, y) , (x, x) ∈ R>0 ,

(x, x) = 0 ⇐⇒ x = 0 .

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(b) Let (−,−) be a hermitian scalar product on V . And let f : V → Vbe an automorphism of V . Recall that f is called unitary with respect to(−,−) if (f(x), f(y)) = (x, y) for all x, y ∈ V . The following results hold:

(i) The vector space V has an orthonormal basis with respect to (−,−),i.e., a basis (v1, . . . , vn) such that (vi, vj) = δi,j for i, j ∈ {1, . . . , n}.

(ii) An automorphism f of V is unitary if and only if the representingmatrix A of f with respect to an orthonormal basis is unitary (i.e., AA∗ = 1,

where A∗ = At).

(c) The complex vector space Cn has the standard hermitian scalar prod-uct

(x, y) = x1y1 + x2y2 + · · ·+ xnyn .

The standard basis (e1, . . . , en) is an orthonormal basis with respect to thisproduct. An automorphism f : Cn → Cn is unitary with respect to (−,−) ifand only if the unique matrix A ∈ GLn(C) with the property f(x) = Ax, forall x ∈ Cn, is a unitary matrix, since A is the representing matrix of f withrespect to (e1, . . . , en).

(d) The set of unitary n× n-matrices form a subgroup U(n) of GLn(C).Every unitary matrix is diagonalizable. Its eigenvalues are complex numbersof absolute value 1.

1.7 Proposition Every complex representation ∆: G → GLn(C) is equiv-alent to a unitary representation, i.e., to a representation Γ: G → GLn(C)with Γ(g) ∈ U(n) for all g ∈ G.

Proof For x, y ∈ Cn define

〈x, y〉 :=∑g∈G

(∆(g)x,∆(g)y) ,

where (−,−) is the standard hermitian scalar product on Cn. It is easy tocheck that < −,− > is again a hermitian scalar product on Cn and we have

〈∆(g)x,∆(g)y〉 =∑h∈G

(∆(h)∆(g)x,∆(h)∆(g)y) =∑k∈G

(∆(k)x,∆(k)y) = 〈x, y〉 ,

for all x, y ∈ Cn and g ∈ G. Thus, ∆(g) is unitary with respect to 〈−,−〉, forevery g ∈ G. Let S ∈ GLn(C) be a matrix such that its columns (v1, . . . , vn)form an orthonormal basis of Cn with respect to 〈−,−〉. Then the represent-ing matrix S−1∆(g)S of ∆(g) with respect to (v1, . . . , vn) is unitary. Thus,the representation Γ = S−1∆S is unitary.

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1.8 Corollary Let ∆: G → GLn(C) be a representation and let g ∈ G bean element of order k. Furthermore, set ζ := e2πi/k ∈ C, a primitive k-throot of unity. Then there exists S ∈ GLn(C) (which can depend on g) suchthat

S∆(g)S−1 = diag(ζ i1 , . . . , ζ in) =

ζ i1

. . .

ζ in

,

the diagonal matrix with diagonal entries ζ i1 , . . . , ζ in , where the exponentsi1, . . . , in are elements in {0, 1, . . . , k − 1}.

Proof By Proposition 1.7, the matrix ∆(g) is conjugate to a unitary matrix.Moreover, by Remark 1.6(d), every unitary matrix is conjugate to a diagonalmatrix. Thus, there exists S ∈ GLn(C) such that S∆(g)S−1 is a diagonalmatrix, say with diagonal entries d1, . . . , dn. Finally, since gk = 1, we have(S∆(g)S−1)k = S∆(g)kS−1 = S∆(gk)S−1 = In. This implies that the entriesdi must satisfy dki = 1 for i = 1, . . . , n. Now the statement of the corollaryfollows immediately.

Note that the previous corollary only says that each matrix ∆(g), g ∈ G,is individually diagonalizable. This does not mean that ∆ as a repre-sentation is diagonalizable, i.e., ∆ is equivalent to a representation withdiagonal matrices as values. Clearly, ∆ as a representation is diagonal-izable if and only if it is equivalent to a direct sum of representations ofdegree 1.

1.9 Theorem Assume that |G| = |G| · 1F is invertible in F . Let ∆: G →GLn(F ) be a representation and let U ⊆ F n be an F -subspace that is ∆-invariant. Then there exists a ∆-invariant F -subspace V ⊆ F n such thatF n = U ⊕ V .

Proof Let u1, . . . , ur be an F -basis of U and extend it to an F -basisu1, . . . , un of F n. Let A ∈ Matn(F ) be defined by Aui = ui for i = 1, . . . , rand Aui = 0 for i = r + 1, . . . , n. We set

A′ :=1

|G|∑g∈G

∆(g−1)A∆(g).

Then we have(i) A′u = u for all u ∈ U .

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(ii) A′x ∈ U for all x ∈ F n.(iii) A′A′x = A′x for all x ∈ F n.(iv) A′∆(h) = ∆(h)A′ for all h ∈ G.

In fact, to see (i), note that

A′u =1

|G|∑g∈G

∆(g−1)A∆(g)u =1

|G|∑g∈G

∆(g−1)∆(g)u =1

|G|∑g∈G

u = u ,

since ∆(g)u ∈ U for all g ∈ G and u ∈ U . To see (ii), note that A∆(g)x ∈ Ufor all x ∈ F n. Now (iii) follows immediately from (i) and (ii). And (iv)follows from the computation

A′∆(h) =1

|G|∑g∈G

∆(g−1)A∆(gh) =1

|G|∑g∈G

∆(hg−1)A∆(g) = ∆(h)A′ ,

where we made the substitution g = gh.Now set V := kerA′. We first show that V is ∆-invariant. For v ∈ V and

g ∈ G we obtain from (iv) that A′∆(g)v = ∆(g)A′v = 0, which means that∆(g)v ∈ kerA′ = V . Next we show that F n = U + V . In fact, let x ∈ F n.Then using (ii) and (iii) we obtain x = A′x+ (x−A′x) ∈ U +V . Finally, weshow that U ∩ V = {0}. So let x ∈ U ∩ V . Then 0 = A′x = x by (i). Thiscompletes the proof.

1.10 Corollary Assume that |G| is invertible in F and let ∆: G→ GLn(F )be a representation. Then one has:

(a) The representation ∆ is equivalent to a direct sum of irreduciblerepresentations.

(b) The representation ∆ is indecomposable if and only if it is irreducible.

Proof (a) We prove this by induction on n. If n = 1, the statement holds,since every representation of degree 1 is irreducible by definition. Assumenow that n > 1. If ∆ is irreducible then we are done. So assume that ∆ isnot irreducible. Then there exists a ∆-invariant F -subspace {0} 6= U 6= F n.By Theorem 1.9, there exists a ∆-invariant F -subspace V of F n such thatF n = U ⊕ V . Therefore, ∆ is equivalent to ∆1 ⊕ ∆2 for representations ofdegree n1, n2 < n. By induction the claim follows.

(b) If ∆ is irreducible then it is also indecomposable for trivial reasons.Now assume that ∆ is not irreducible. Then there exists a ∆-invariant sub-space {0} 6= U 6= F n. By Theorem 1.9, U has a ∆-stable complement V inF n. This implies that ∆ is not indecomposable.

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We recall the following theorem from Linear Algebra without proof.

1.11 Theorem Let A1, . . . , Ak ∈ Matn(F ). Then the following are equiva-lent:

(i) The matrices A1, . . . , Ak are simultaneously diagonalizable over F ,i.e., there exists S ∈ GLn(F ) such SAiS

−1 is a diagonal matrix, for alli = 1, . . . , k.

(ii) The matrices A1, . . . , Ak are individually diagonalizable over F andthey commute, i.e., AiAj = AjAi for all i, j ∈ {1, . . . , k}.

1.12 Corollary (a) Let ∆: G → GLn(C) be a representation satisfying∆(g)∆(h) = ∆(h)∆(g) for all g, h ∈ G. Then ∆ is equivalent to a direct sumof representations of degree 1.

(b) Assume that G is abelian. Then every representation ∆: G →GLn(C) is equivalent to a direct sum of representations of degree 1. In par-ticular, every irreducible representation of G has degree 1. In other words,the irreducible representations of G are just the homomorphisms from G tothe multiplicative group C× of C.

1.13 Definition Let ∆: G → GLn(F ) be a representation. The characterof ∆ is defined as the function

χ := χ∆ : G ∆ qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq GLn(F ) tr qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq F .

The characters of representations of G over F are called the F -characters ofG. If F = C, we just call them characters, or sometimes complex charactersof G. If F = C and ∆ is irreducible then χ is called a (complex) irreduciblecharacter.

In the following remark we derive a few first properties of characters.

1.14 Remark Assume that ∆: G → GLn(F ) is a representation and letχ = χ∆ : G→ F denote its character.

(a) Recall that for two matrices A,B ∈ Matn(F ) one has tr(AB) =tr(BA). In particular, if S ∈ GLn(F ) one obtains tr(SAS−1) = tr(AS−1S) =tr(A). This implies that if ∆ and Γ are equivalent representations thenχ∆ = χΓ. Moreover, by the same property of the trace map we obtain, forany elements g, x ∈ G, that

χ(xgx−1) = tr(∆(xgx−1) = tr(∆(x)∆(g)∆(x)−1) = tr(∆(g)) = χ(g) .

7

This means that the function χ : G→ F is constant on conjugacy classes ofG. Any such function is called a class function or a central function on G.The class functions f : G → F form an F -vector space CF(G,F ) with theusual addition and scalar multiplication. The dimension of CF(G,F ) is equalto the number k(G) of conjugacy classes of G. In fact, the characteristic classfunctions which have value one on one particular conjugacy class and valuezero on all others form an F -basis of CF(G,F ).

(b) Let ∆1 : G → GLn1(F ) and ∆2 : G → GLn2(F ) be two representa-tions. Then, for every g ∈ G, we have

χ∆1⊕∆2(g) = tr

(∆1(g) 0

0 ∆2(g)

)= tr(∆1(g))+tr(∆2(g)) = χ∆1(g)+χ∆2(g) .

Thus, χ∆n1⊕∆n2= χ∆n1

+ χ∆n2in the vector space CF(G,F ).

(c) The F -characters and representations of degree 1 are precisely thegroup homomorphisms in Hom(G,F×). They form an abelian group underthe product defined by (φ · ψ)(g) = φ(g)ψ(g), for φ, ψ ∈ Hom(G,F×) andg ∈ G. Recall that we have an isomorphism of abelian groups

Hom(G/G′, F×)→ Hom(G,F×) , f 7→ f ◦ ν ,

where G′ denotes the commutator subgroup of G, generated by all commuta-tors [x, y] := xyx−1y−1, x, y ∈ G, and where ν : G→ G/G′, g 7→ gG′, denotesthe natural epimorphism.

(d) Assume from now on that F ⊆ C. Note that

χ(1) = n .

Thus, the character of ∆ ‘knows’ the degree of the representation ∆. We callχ(1) the degree of the character χ.

Let g ∈ G and assume that gk = 1. Then, by Corollary 1.8 we have

∆(g) = S

ζ i1

. . .

ζ in

S−1 ,

for some S ∈ GLn(C), with ζ := e2πi/k and i1, . . . , in ∈ Z. Therefore,

χ(g) = ζ i1 + · · ·+ ζ in .

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The triangle inequality implies that

|χ(g)| 6n∑j=1

|ζ ij | = n

with equality if and only if ζ i1 = · · · = ζ in . Thus, we have

|χ(g)| = χ(1) ⇐⇒ ∆(g) is a scalar matrix.

The set of elements g ∈ G with |χ(g)| = χ(1) is called the center of thecharacter χ and is denoted by Z(χ). In particular, we obtain

χ(g) = χ(1) ⇐⇒ ∆(g) = In ⇐⇒ g ∈ ker(∆) .

Thus, the character χ ‘knows’ the kernel of ∆. The set of elements g ∈ Gwith χ(g) = χ(1) is called the kernel of χ and is denoted by ker(χ).

Finally, we will show that

χ(g−1) = χ(g) .

In fact, from above we have

∆(g−1) = ∆(g)−1 = S

ζ i1

. . .

ζ in

−1

S−1 = S

ζ−i1

. . .

ζ−in

S−1 .

Taking the trace yields

χ(g−1) = ζ−i1 + · · ·+ ζ−in = ζ i1 + · · ·+ ζ in = χ(g) .

1.15 Examples Assume that F = C.

(a) The character of the trivial representation is the constant functionG→ C with value 1. It is called the trivial character and it is often denotedby 1G or just 1. More general, any group homomorphism ϕ : G → C× =GL1(C) is a representation of degree 1 and at the same time a character ofG.

(b) For the regular character ρG, the character of the regular represen-tation (cf. Example 1.3(e)), we find that ρG(1) = |G| and ρG(g) = 0 forall g ∈ G with g 6= 1. In fact, in the latter case one has for all elements

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gi ∈ G that ggi 6= gi. This means that all the diagonal entries in the regularrepresentation of g are equal to 0.

(c) Let G = Sym(3). Since we have an isomorphism

Hom(G,C×) ∼= Hom(G/G′,C×) = Hom(G/〈(1, 2, 3)〉,C×) ,

see Remark 1.14, the group G has exactly two representations of degree1, namely, the trivial representation and the sign representation ε : G →{±1} 6 C×. Moreover, from Example 1.3(b), since Sym(3) ∼= D6, we have arepresentation of degree 2 with the property that

(1, 2, 3) 7→(

cos 120◦ − sin 120◦

sin 120◦ cos 120◦

)=

(−1

2−√

32√

32−1

2

),

(1, 2) 7→(−1 00 1

).

Its character χ is given by χ(1) = 2, χ(1, 2, 3) = −1, χ(1, 2) = 0. Sinceevery other element in G is conjugate to (1, 2, 3) or (1, 2), these three valuesdetermine χ. We arrange these values in a table:

1 (1, 2) (1, 2, 3)1 1 1 1ε 1 −1 1χ 2 0 −1

Since χ 6= 1 + 1, χ 6= ε + ε, and χ 6= 1 + ε, we obtain that every represen-tation whose character is equal to χ is an irreducible representation. Herewe used Corollary 1.10 and the property in Remark 1.14(b). In particular, χis an irreducible character. We will see in Section 2 that there are no otherirreducible characters for Sym(3).

(d) Let G = 〈g〉 be a cyclic group of order k and let ζ := e2πi/k. Thenevery homomorphism in Hom(G,C×) is uniquely determined by its value ong and the possible values are the k-th roots of unity, namely 1, ζ, ζ2, . . . , ζk−1.Thus we obtain k characters of degree 1 which we arrange in the following

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table.1 g g2 · · · gj · · · gk−1

1 1 1 1 · · · 1 · · · 1ϕ 1 ζ ζ2 · · · ζj · · · ζk−1

ϕ2 1 ζ2 ζ4 · · · ζ2j · · · ζ2(k−1)

......

......

......

ϕi 1 ζ i ζ2i · · · ζ ij · · · ζ i(k−1)

......

......

......

ϕk−1 1 ζk−1 ζk−2 · · · ζk−j · · · ζ

Note note that Hom(G,C×) is again a cyclic group of order k, generated bythe character ϕ.

Exercises

1. Let D2n = 〈x, y | xn = y2 = 1, xy = yx−1〉 be the dihedral group of order2n.

(a) Show that

∆1 : D2n → GL2(C) , x 7→(ζn 0

0 ζn

), y 7→

(0 11 0

),

with ζn := e2πi/n defines a representation.

(b) Decide if the representation ∆1 is equivalent to the representation

∆2 : D2n → GL2(C) , x 7→(

cos(2π/n) − sin(2π/n)sin(2π/n) cos(2π/n)

), y 7→

(−1 00 1

).

2. Let G = 〈g〉 be a cyclic group of order 3.

(a) Show that

∆: G→ GL2(Q) , g 7→(−1 −11 0

),

defines a representation.

(b) Show that ∆ is irreducible over Q.

(c) Show that ∆ is decomposable over C.

3. Let ∆: G → GLn(C) be a representation and let χ be its character. Thecenter of ∆ is defined as

Z(∆) := {g ∈ G | ∆(g) = α · 1n for some α ∈ C} ,

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and the center of χ is defined as

Z(χ) := {g ∈ G | |χ(g)| = n}

(a) Show that Z(χ) = Z(∆).

(b) Show that Z(χ) is a normal subgroup of G.

4. Let G = 〈g〉 be a cyclic group of prime order p and let F := Z/pZ.

(a) Show that there exists a unique representation ∆: G→ GL2(F ) with

∆(g) =

(1 10 1

).

(b) Show that the representation ∆ is indecomposable but not irreducible.

5. (a) Let G be a cyclic group of order 2. Try to explicitly decompose theregular representation of G over C into a direct sum of irreducible representa-tions, i.e., find an equivalent representation which is a direct sum of irreduciblerepresentations.

(b) Do the same as in (a) for a cyclic group of order 3.

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2 Orthogonality Relations

Throughout this section, G denotes a finite group. If there is no base fieldspecified, representations are considered to be complex representations.

2.1 Lemma (Schur) Let ∆1 : G → GLm(C) and ∆2 : G → GLn(C) be ir-reducible representations and let A ∈ Matm×n(C) be an intertwining matrix,i.e., a matrix satisfying

∆1(g)A = A∆2(g) for all g ∈ G. (2.1.a)

(a) If ∆1 and ∆2 are not equivalent then A = 0.

(b) If ∆1 = ∆2 then A is a scalar matrix, i.e., there exists λ ∈ C suchthat A = λ · In.

Proof (a) Assume that A 6= 0. From this we will derive a contradiction. LetU := ker(A) ⊆ Cn. Then U is a ∆2-invariant subspace. In fact, for u ∈ U andg ∈ G we have ∆2(g)u ∈ U , since A∆2(g)u = ∆1(g)Au = 0. Since A 6= 0, wehave U 6= Cn. Since ∆2 is irreducible, this implies U = {0}. Thus, the linearmap Cn → Cm, x 7→ Ax is injective. Next consider V := im(A) ⊆ Cm. Thisis a ∆1-invariant subspace. In fact, let v ∈ V and g ∈ G. We need to showthat ∆1(g)v ∈ V . Since v ∈ V , there exists x ∈ Cn with Ax = v. Therefore,∆1(g)v = ∆1(g)Ax = A∆2(g)x ∈ V . Since A 6= 0, we have V 6= {0}. Since∆1 is irreducible, we obtain V = Cm. Therefore, the linear map Cn → Cm,x 7→ Ax, is also surjective, and consequently an isomorphism. Thus, m = nand A ∈ GLn(C). Now Equation (2.1.a) implies ∆1(g) = A∆2(g)A−1 for allg ∈ G. Thus, ∆1 and ∆2 are equivalent. This is a contradiction.

(b) Set ∆ := ∆1 = ∆2. Let λ ∈ C be an eigenvalue of A and setA′ := A − λIn. Then also A′ is an intertwining operator, i.e., it satisfies∆(g)A′ = A′∆(g) for all g ∈ G. Since there exists a non-zero eigenvector forλ, we obtain U ′ := ker(A′) 6= {0}. As in the proof of Part (a) we can seethat U ′ is a ∆-invariant subspace of Cn. Since ∆ is irreducible, we obtainU ′ = Cn. But this means A′ = 0 and A = λ · In.

We derive a sequence of three corollaries from the above lemma, the lastof them showing that the number of irreducible representations of G is finiteand that the sum of the squares of their degrees is bounded by |G|.

13

2.2 Corollary Let ∆1 : G→ GLm(C) and ∆2 : G→ GLn(C) be irreduciblerepresentations and let A ∈ Matm×n(C). Set

A′ :=1

|G|∑g∈G

∆1(g)A∆2(g−1) ∈ Matm×n(C) .

(a) If ∆1 and ∆2 are not equivalent then A′ = 0.

(b) If m = n and ∆1 = ∆2 then A′ = tr(A)n· In.

Proof We first show that A′ satisfies Equation (2.1.a). Let h ∈ G. Then

∆1(h)A′ =1

|G|∑g∈G

∆1(hg)A∆2(g−1) =1

|G|∑g∈G

∆1(g)A∆2(g−1h)

= A′∆2(h) .

Here we made the substitution g = hg.

(a) This follows immediately from Lemma 2.1(a).

(b) Set ∆ := ∆1 = ∆2. By Lemma 2.1(b), there exists λ ∈ C such thatA′ = λ · In. Taking traces, we obtain

nλ = tr(λ · In) =1

|G|∑g∈G

tr(∆(g)A∆(g−1)) = tr(A′) ,

and the desired equation holds.

2.3 Corollary Let ∆1 : G→ GLm(C) and ∆2 : G→ GLn(C) be irreducibleunitary representations and write

∆1(g) =(rij(g)

)16i,j6m

and ∆2(g) =(skl(g)

)16k,l6n

for g ∈ G.

(a) If ∆1 and ∆2 are not equivalent then

1

|G|∑g∈G

rij(g)skl(g) = 0

for all 1 6 i, j 6 m and 1 6 k, l 6 n.

14

(b) If m = n and ∆1 = ∆2 then

1

|G|∑g∈G

rij(g)skl(g) =

1n

if (i, j) = (k, l),

0 if (i, j) 6= (k, l),

for all i, j, k, l ∈ {1, . . . , n}.

Proof Let A := (auv) ∈ Matm×n(C) be arbitrary and let A′ = (a′ik) ∈Matm×n(C) be defined as in Corollary 2.2. Then, since svk(g

−1) = skv(g),the (i, k)-entry of A′ is given by

a′ik =1

|G|

m∑u=1

n∑v=1

∑g∈G

riu(g)auvskv(g) .

(a) By Corollary 2.2(a), we have a′ik = 0 for every choice of A and forevery choice of 1 6 i 6 m and 1 6 k 6 n. We choose A = Ejl, the matrixall of whose entries are 0, except for the (j, l)-entry which is equal to 1. Thisgives the desired equation.

(b) By Corollary 2.2(b), we obtain

a′ik =

tr(A)n

if i = k,

0 if i 6= k.

Now Choose A = Ejl and the result follows.

2.4 Corollary Let G be a group. The number of equivalence classes ofirreducible representations of G is finite. If ∆i : G → GLni(C), i = 1, . . . , k,are representatives of these equivalence classes then one has

∑ki=1 n

2i 6 |G|.

Proof Let ∆t : G → GLnt(C), t ∈ T , be a set of representatives of theequivalence classes of irreducible representations of G. We may choose themto be unitary representations by Proposition 1.7. Write ∆t(g) = (rtij(g)) ∈GLnt(C) for every t ∈ T and g ∈ G. We consider triples (t, i, j) with t ∈ Tand i, j ∈ {1, . . . , nt}. For each such triple we consider the vector (rtij(g)) ∈C|G|. By Corollary 2.3, these vectors are non-zero and pairwise orthogonal inC|G|, and therefore the set of these vectors is linearly independent. Thus, thenumber of triples (t, i, j), namely

∑t∈T n

2t , is bounded by |G|. This implies

that T is finite and the result follows.

15

2.5 Definition On the C-vector space CF(G,C) of class functions from Gto C we introduce the Schur inner product

(f1, f2) :=1

|G|∑g∈G

f1(g)f2(g) .

It is a straightforward verification that (−,−) is a hermitian scalar producton CF(G,C). For characters χ1, χ2 of G we also have

(χ1, χ2) =1

|G|∑g∈G

χ1(g)χ2(g−1) ,

since χ2(g) = χ2(g−1) as observed in Remark 1.14(d). This implies

(χ1, χ2) = (χ2, χ1) ,

i.e., the Schur inner product is symmetric on characters.

2.6 Notation From now on, for the rest of the section, we fix the follow-ing notation. We choose unitary representatives ∆1, . . . ,∆k of the equiv-alence classes of irreducible representations of G. We write n1, . . . , nk fortheir respective degrees and we denote by χ1, . . . , χk their respective char-acters. Then χi(1) = ni for i = 1, . . . , k and

∑ki=1 n

2i 6 |G| by Corol-

lary 2.4. We denote by Irr(G) the set of irreducible characters of G; thus,Irr(G) = {χ1, . . . , χk}. Finally, for t ∈ {1, . . . , k} and g ∈ G, we write

∆t(g) =(rtij)

16i,j6nt.

2.7 Theorem (First orthogonality relation) The irreducible charactersχ1, . . . , χk form a set of orthonormal vectors in CF(G,C) with respect to theSchur inner product. In other words,

(χi, χj) =1

|G|∑g∈G

χi(g)χj(g−1) =

1 if i = j,

0 if i 6= j.

Proof From Corollary 2.3 we obtain, for s, t ∈ {1, . . . , k},

(χs, χt) =1

|G|∑g∈G

( ns∑i=1

rsii(g))( nt∑

k=1

rtkk(g))

=ns∑i=1

nt∑k=1

1

|G|∑g∈G

rsii(g)rtkk(g)

= δstns∑i=1

1

ns= δst

16

and the theorem is proven.

2.8 Corollary The irreducible characters χ1, . . . , χk of G form a set of C-linearly independent vectors in CF(G,C). Moreover, one has k 6 k(G),where k(G) denotes the number of conjugacy classes of G.

Proof The first statement is a general fact about a set of orthogonal vectorsin a C-vector space with hermitian scalar product. The second statementfollows immediately, since dimC CF(G,C) = k(G).

The following theorem is an amazing fact: Although in general it is im-possible to reconstruct a square matrix just from the knowledge of its trace,a representation of G is determined up to equivalence by its character. How-ever, given a character χ of G, it is in practice difficult to construct a repre-sentation ∆ with character χ.

2.9 Theorem Let ∆ and Γ be (complex) representations of G which havethe same character, i.e., χ∆ = χΓ. Then ∆ and Γ are equivalent.

Proof By Corollary 1.10(a), we know that ∆ is equivalent to a direct sumof irreducible representations. Moreover, each of these irreducible represen-tations is equivalent to one of our representatives ∆i, 1 6 i 6 k. Thus,altogether, ∆ is equivalent to a direct sum of the form

k⊕i=1

∆i ⊕ · · · ⊕∆i︸ ︷︷ ︸li summands

. (2.9.a)

Similarly, Γ is equivalent to a direct sum of the form

k⊕i=1

∆i ⊕ · · · ⊕∆i︸ ︷︷ ︸mi summands

. (2.9.b)

Here, li,mi ∈ N0 for i = 1, . . . , k. Remark 1.14(b) implies that

l1χ1 + · · ·+ lkχk = χ∆ = χΓ = m1χ1 + · · ·+mkχk .

The linear independence of χ1, . . . , χk now implies that li = mi for all i. Butthis implies that the representations in (2.9.a) and (2.9.b) coincide. Thus, ∆is equivalent to Γ.

17

2.10 Remark Let ∆: G → GLn(C) be a representation and let χ denoteits character. By the argument in the proof of Theorem 2.9 we see that onecan write χ as

χ = m1χ1 + · · ·+mkχk ,

with unique coefficients m1, . . . ,mk ∈ N0. One can determine the coeffi-cient mi quickly from the knowledge of the character χ and the irreduciblecharacter χi by

mi = (χ, χi) , for i = 1, . . . , k. (2.10.a)

In fact, (χ, χi) =∑kj=1 mi(χj, χi) = mi, by Theorem 2.7. The number mi is

called the multiplicity of χi in χ.

2.11 Proposition For the regular character ρG of G one has

ρG = n1χ1 + · · ·nkχk ,

where ni = χi(1) for i = 1, . . . , k. In particular, evaluating at the identityelement, one obtains

|G| = n21 + · · ·+ n2

k .

Proof Recall from Example 1.15(b) that ρG(1) = |G| and ρG(g) = 0 forg 6= 1. Therefore, the definition of the Schur inner product yields (ρG, χi) =χi(1) = ni. Now equation (2.10.a) implies the formula for ρG. The secondstatement is clear.

2.12 Lemma Let ∆: G → GLn(C) be an irreducible representation withcharacter χ and let f ∈ CF(G,C). Then the matrix

∆f :=∑g∈G

f(g)∆(g−1) ∈ Matn(C)

is a scalar matrix λ · In with λ = |G|n

(f, χ).

Proof For any x ∈ G we have

∆(x)∆f∆(x−1) =∑g∈G

f(g)∆(xg−1x−1) =∑g∈G

f(xgx−1)∆(xg−1x−1)

=∑g∈G

f(g)∆(g−1) = ∆f ,

18

where we substituted g = xgx−1. Therefore we have ∆(x)∆f = ∆f∆(x) forall x ∈ G. Now Corollary 2.2 implies that ∆f = λ · In with

λ =tr(∆f )

n=

1

n

∑g∈G

f(g)χ(g−1) =1

n

∑g∈G

f(g)χ(g) =|G|n

(f, χ) .

2.13 Corollary The set Irr(G) = {χ1, . . . , χk} is an orthonormal basis ofCF(G,C) with respect to the Schur inner product. In particular, |Irr(G)| =dimC CF(G,C) = k(G), the number of conjugacy classes of G. Moreover, forevery class function f ∈ CF(G,C) we have

f =k∑i=1

(f, χi)χi ,

with the Schur inner product (f, χi) ∈ C, for i = 1, . . . , k.

Proof By Theorem 2.7 it suffices to show that Irr(G) generates CF(G,C).For this it suffices to show that the orthogonal complement 〈Irr(G)〉⊥C of〈Irr(G)〉C is equal to 0, since CF(G,C) = 〈Irr(G)〉C ⊕ 〈Irr(G)〉⊥C . So letf ∈ 〈Irr(G)〉⊥C . Then (f, χi) = 0 for all i = 1, . . . , k. Lemma 2.12 impliesthat (∆i)f = 0 for every irreducible representation ∆i of G, i = 1, . . . , k. Set

∆ :=k⊕i=1

∆i ⊕ · · · ⊕∆i︸ ︷︷ ︸ni

and form ∆f as in Lemma 2.12. Then, ∆f is a block diagonal matrix withni blocks equal to (∆i)f . Therefore ∆f = 0. This implies that ∆′f = {0} forevery representation ∆′ which is equivalent to ∆. By Proposition 2.11 andTheorem 2.9, the representation ∆ is equivalent to the regular representationΓ with respect to any ordering g1, . . . , gn of the elements of G. We mayassume that g1 = 1. Thus Γf = 0 and we obtain the following equation inCn, where ei denotes the standard basis vector for i = 1, . . . , n:

0 = Γfe1 =n∑i=1

f(gi)Γ(g−1i )e1 =

n∑i=1

f(g−1i )Γ(gi)e1 =

n∑i=1

f(g−1i )ei .

This implies that f(g−1i ) = 0 for all i = 1, . . . , n. Thus, f = 0 and the proof

is complete.

19

2.14 Remark One arranges the values of the irreducible characters χ1, . . . , χkof G in a quadratic table, the character table of G:

g1 · · · gj · · · gkχ1 χ1(g1) · · · χ1(gj) · · · χ1(gk)...

......

...χi χi(g1) · · · χi(gj) · · · χi(gk)...

......

...χk χk(g1) · · · χk(gj) · · · χk(gk)

Here, g1, . . . , gk are representatives of the conjugacy classes of G. Often onealso writes the cardinality cj of the conjugacy class of gj in a row below gj.Note that cj = [G : CG(gj)] for j = 1, . . . , k. Sometimes It is also usefulto add a row with the orders o(gj) of the group elements gj, j = 1, . . . , k.Usually the elements gj are ordered in ascending element order, and theirreducible characters are ordered according to ascending character degree.Thus, usually g1 = 1. One also usually has χ1 = 1, the trivial character.

2.15 Example For G = Sym(3) (which is isomorphic to D6) we obtain thecharacter table

gj 1 (1, 2) (1, 2, 3)cj 1 3 2

o(gj) 1 2 31 = χ1 1 1 1ε = χ2 1 −1 1χ3 2 0 −1

from our work in in Example 1.15(c), since we now know that |Irr(G)| =k(G) = 3.

2.16 Theorem (Second orthogonality relation) For any g, h ∈ G onehas

k∑i=1

χi(g)χi(h−1) =

|CG(g)| , if g and h are conjugate,

0 , otherwise.

Proof Write fh ∈ CF(G,C) for the class function which is equal to 1 on theclass of h and 0 everywhere else. By Corollary 2.13, we have

fh =k∑i=1

(fh, χi)χi

20

with

(fh, χi) =1

|G|· |G||CG(h)|

· χi(h−1) =1

|CG(h)|· χi(h−1) .

This implies

1

|CG(h)|

k∑i=1

χi(g)χi(h−1) =

k∑i=1

1

|CG(h)|χi(h

−1)χi(g) =( k∑i=1

(fh, χi)χi)(g)

= fh(g) =

1 if g and h are conjugate,

0 otherwise,

and the proof is complete.

The following remark introduces some new operations on representationsand shows how those and the ones that were already established translate tooperations on characters.

2.17 Remark Let F be an arbitrary field.

(a) Tensor product of representations. Let f : V → V ′ and g : W →W ′ be F -linear maps between F -vector spaces and let v1, . . . , vn, v′1, . . . , v

′m,

w1, . . . , ws, and w′1, . . . , w′r be F -bases of V , V ′, W , and W ′, respectively.

Moreover, let A ∈ Matm×n(F ) and B ∈ Matr×s(F ) be the representingmatrices of f and g with respect to these bases, respectively. Then therepresenting matrix of the F -linear map f⊗g : V ⊗FW → V ′⊗FW ′, v⊗w 7→f(v) ⊗ g(w), with respect to the bases vi ⊗ wj of V ⊗F W and v′i ⊗ w′j ofV ′ ⊗F W ′ in lexicographic order is given by the Kronnecker product (alsocalled tensor product) A⊗B of A and B. It is defined by

A⊗B =

a11B · · · a1nB

......

am1B · · · amnB

∈ Matmr×ns(F ) .

Note that if m = n and r = s then

tr(A⊗B) = a11tr(B) + · · · anntr(B) = tr(A) · tr(B) . (2.17.a)

Since the composition of linear maps corresponds to multiplication of ma-trices and since (f1 ⊗ g1) ◦ (f ⊗ g) = (f1 ◦ f) ⊗ (g1 ◦ g), we obtain for anyadditional matrices C ∈ Matl×m(F ) and D ∈ Matq×r(F ) that

(C ⊗D)(A⊗B) = (CA)⊗ (DB) ∈ Matlq×ns(F ) . (2.17.b)

21

If ∆1 : G → GLm(F ) and ∆2 : G → GLn(F ) are representations of G wedefine their tensor product ∆1 ⊗∆2 by

∆1 ⊗∆2 : G→ GLmn(F ) , g 7→ ∆1(g)⊗∆2(g) .

Equation (2.17.b) implies that ∆1 ⊗ ∆2 is in fact a homomorphism, andEquation (2.17.a) implies that

χ∆1⊗∆2(g) = χ∆1(g) · χ∆2(g) ,

for all g ∈ G. If we define the product of class functions f1, f2 ∈ CF(G,C)in the usual way, by (f1 · f2)(g) := f1(g)f2(g) for g ∈ G, then the aboveequation shows that

χ∆1 · χ∆2 = χ∆1⊗∆2 .

In particular, the set of characters of G, as a subset of CF(G,C), is multi-plicatively closed: The product of two characters is again a character.

(b) Contragredient representation. If ∆: G→ GLn(F ) is a representationthen also

∆∗ : G→ GLn(F ) , g 7→ (∆(g)t)−1 = (∆(g)−1)t = ∆(g−1)t ,

is a representation, called the contragredient representation of ∆. One has

χ∆∗(g) = tr((∆(g−1)t) = tr(∆(g−1)) = χ∆(g−1) ,

for all g ∈ G. If F ⊆ C then one also has χ∆∗(g) = χ∆(g). Thus, in thiscase,

χ∆∗ = χ∆ ,

where · : CF(G,C)→ CF(G,C) denotes the function defined by f(g) := f(g),for all g ∈ G. In particular, the set of characters ofG, as a subset of CF(G,C),is closed under complex conjugation. For any character χ of G, we call χ thecontragredient character of χ.

(c) Inflation. Let N /G and let ν : G→ G/N =: G, g 7→ gN , denote thecanonical epimorphism. For any representation Γ: G/N → GLn(F ) we write

InfGG/N(Γ) := Resν(Γ) := Γ ◦ ν : G→ GLn(F ) .

This representation of G is called the inflation of the representation Γ. Forinstance, the trivial representation of G is the inflation of the trivial rep-resentation of G/N . By the fundamental theorem of homomorphisms, the

22

map Γ 7→ InfGG/N(Γ), defines a bijection between the representations Γ ofG/N over F of given degree n and the representations ∆ of G over Fof degree n with N 6 ker(∆). Moreover, since ν is surjective, one hasΓ(G/N) = (Γ ◦ ν)(G) = ∆(G). It follows that two representations Γ1

and Γ2 of G/N are equivalent if and only if their inflations InfGG/N(Γ1) and

InfGG/N(Γ2) are equivalent. Furthermore, a representation Γ of G/N is irre-

ducible (resp. indecomposable) if and only if its inflation InfGG/N(Γ) is irre-ducible (resp. indecomposable). This way, specializing to F = C, inflationinduces a bijection

infGG/N : Irr(G/N)→ {χ ∈ Irr(G) | N 6 ker(χ)} , θ 7→ θ ◦ ν .

(d) The character ring. We set

R(G) := {a1χ1 + · · ·+ akχk | a1, . . . , ak ∈ Z} = 〈Irr(G)〉Z ⊆ CF(G,C) ,

the Z-span of the irreducible characters of G in the C-vector space of classfunctions of G. The elements of R(G) are called generalized characters oralso virtual characters. Every virtual character can be written as a differenceof two characters. Conversely, the difference of two characters is always avirtual character. By Part (a), the product of two virtual characters is againa virtual character. Therefore, R(G) is a subring of the C-algebra of classfunctions CF(G,C). It is called the character ring of G. Note that sinceIrr(G) is a C-linearly independent set in CF(G,C) it is a Z-basis of R(G).Previously defined constructions for representations induce structural mapson character rings and maps between character rings:

direct sum −⊕− → abelian group structure on R(G)tensor product−⊗− → ring structure on R(G)

contragredient −∗ → ring automorphism · : R(G)→ R(G), χ 7→ χ

Resf → ring homom. resf : R(G)→ R(G), χ 7→ χ ◦ fResGH → ring homom. resGH : R(G)→ R(H), χ 7→ χ|H

InfGG/N → ring homom. infGG/N : R(G/N)→ R(G)

Here, f : G → G denotes a group homomorphism, H denotes a subgroup ofG, and N denotes a normal subgroup of G. The functions resGH and infGG/Nare special cases of resf .

In the following proposition we give a criterion for when a generalizedcharacter χ ∈ R(G) is an irreducible character. Note that χ(1) is always aninteger.

23

2.18 Proposition For every χ ∈ R(G) the following are equivalent:

(i) χ ∈ Irr(G).

(ii) (χ, χ) = 1 and χ(1) > 0.

Proof If χ is irreducible then the properties in (ii) are clearly satisfied.Conversely, assume that χ satisfies the properties in (ii) and write χ = a1χ1+· · ·+ akχk as a Z-linear combination of the irreducible characters χ1, . . . , χkof G. Then, by the first orthogonality relation (Theorem 2.7), we have

1 = (χ, χ) = (a1χ1+· · ·+akχk, a1χ1+· · ·+akχk) =k∑i=1

k∑j=1

aiaj(χi, χj) =k∑i=1

a2i .

This implies that there exists a unique index i ∈ {1, . . . , k} such that ai 6= 0and moreover that ai ∈ {1,−1}. In otherwords, χ = ±χi. Now χ(1) > 0implies that χ = χi ∈ Irr(G).

Exercises

1. Let G be a finite group and let χ ∈ Irr(G). Show that Z(G) 6 Z(χ).

2. (a) Determine all the irreducible characters of the dihedral group D8 =〈a, b | a4 = b2 = 1, bab = a−1〉 of order 8.

(b) Determine all the irreducible characters of the quaternion group Q8 =〈x, y | x4 = y4 = 1, x2 = y2, yxy−1 = x3〉 of order 8.

3. Assume that ∆: G→ GLn(F ) is a representation of a finite group G overa field F and let ϕ : G→ F× be a homomorphism.

(a) Show that also ϕ ·∆: G → GLn(F ), g 7→ ϕ(g)∆(g) is a representation ofG over F , the twist of ∆ by ϕ.

(b) Show that ∆ is irreducible (resp. indecomposable) if and only if ϕ · ∆ isirreducible (resp. indecomposable).

(c) Compute the irreducible representations of D10 (up to equivalence) andtheir characters.

4. Determine the irreducible characters of Alt(4).

5. (a) Let f : G→ H be a surjective group homomorphism and let ∆: H →GLn(F ) be a representation of H over an arbitrary field F . Show that ∆ is

24

irreducible (resp. indecomposable) if and only if Resf (∆) is irreducible (resp.indecomposable).

(b) Determine the character table of the symmetric group Sym(4). (Hint: Use(a) with H a factor group of Sym(4).)

6. Let G be a finite group and write Irr(G) = {χ1, . . . , χk}.(a) Let χ and χ′ be characters of G. Show that ker(χ+χ′) = ker(χ)∩ ker(χ′).

(b) Show that⋂ki=1 ker(χi) = {1}.

(c) Let N E G. Show that there exists a subset I ⊆ {1, . . . , k} such that

N =⋂i∈I

ker(χi) .

In particular, one can determine from the character table the partially ordered setof normal subgroups of G.

25

3 Algebraic Integers

Throughout this section, we denote by S a commutative ring and by R ⊆ Sa subring with 1S ∈ R. We introduce and study elements of S which areintegral over R. We specialize to the situation where R = Z and S = Cto obtain consequences for characters. Again, throughout this section, Gdenotes a finite group. The main goal of this section is to show that thedegrees of the irreducible characters of a finite group G are divisors of |G|.

Recall that a polynomial f = anXn + · · · a1X + a0 ∈ R[X] of degree n is

called monic if an = 1. Thus, a monic polynomial is always non-zero.

3.1 Definition An element s ∈ S is called integral over R if there exists amonic polynomial f ∈ R[X] such that f(s) = 0.

3.2 Remark For s ∈ S one denote by R[s] the set of all finite R-linearcombinations of the elements 1, s, s2, . . .. In other words, one has R[s] ={f(s) | f ∈ R[X]} = 〈1, s, s2, . . .〉R. Clearly, R[s] is a subring of S. Moreover,it is the smallest subring of S which contains R and s. In other words,whenever R′ ⊆ S is a subring with R ⊆ R′ and s ∈ R′ then R[s] ⊆ R′.

More generally, for elements s1, . . . , sn ∈ S, one writes R[s1, . . . , sn] forthe set of all R-linear combinations of elements sa11 · · · sann with a1, . . . , an ∈N0. In other words, R[s1, . . . , sn] = {f(s1, . . . , sn) | f ∈ R[X1, . . . , Xn]}.Again it is easy to see that R[s1, . . . , sn] is a subring of S and that it is thesmallest subring of S which contains R and {s1, . . . , sn}. Thus, R[s1, . . . , sn]is independent of the order of the elements s1, . . . , sn. It only depends onthe set {s1, . . . , sn}. Moreover, if also t1, . . . , tm are elements in S one has(R[s1, . . . , sn])[t1, . . . , tm] = R[s1, . . . , sn, t1, . . . , tm].

The following theorem gives useful characterizations of when an elements ∈ S is integral over R. Note that any subring T of S which contains R isalso an R-module.

For the proof of the following theorem recall the following notion and factfrom linear algebra: If B ∈ Matn(R) is a square matrix with entries bij onedefines its adjunct matrix B ∈ Matn(R), also called the cofactor matrix ofB, as the matrix with entries bij, where bij := (−1)i+j det(Bji), where Bji isthe square matrix of size n−1 that arises form deleting the j-th row and thei-th column from B. This matrix B has the property

B ·B = det(B) · In = B · B .

26

3.3 Theorem For s ∈ S the following are equivalent:

(i) The element s is integral over R.

(ii) The R-module R[s] is finitely generated.

(iii) There exists a subring T of S with R[s] ⊆ T ⊆ S such that T isfinitely generated as R-module.

Proof (i)⇒(ii): By definition, there exist n > 1 and r0, . . . , rn−1 ∈ R suchthat sn + rn−1x

n−1 + · · ·+ r1s+ r0 = 0. We will show by induction on k > 1that sk ∈ 〈1, s, s2, . . . , sn−1〉R. For k = 0, . . . , n−1 this is clear and for k = nthis follows from

sn = −r0 · 1− r1s− r2s2 − · · · − rn−1s

n−1 . (3.3.a)

Now assume that k > n and that sk ∈ 〈1, s, s2, . . . , sn−1〉R. Then, by Equa-tion (3.3.a), we obtain

sk+1 = s · sk ∈ s · 〈1, s, s2, . . . , sn−1〉R = 〈s, s2, . . . , sn−1, sn〉R⊆ 〈1, s, s2, . . . , sn−1〉R .

This finishes the proof by induction and we have R[s] = 〈1, s, s2, . . .〉R =〈1, s, s2, . . . , sn−1〉R, which is finitely generated.

(ii)⇒(iii): This is trivial: choose T = R[s].

(iii)⇒(i): Let {t1, . . . , tn} ⊆ T be a generating set of T as R-module.Then we can express sti as an R-linear combination of t1, . . . , tn for everyi = 1, . . . , n. In other words, there exist elements aij ∈ R, i, j ∈ {1, . . . , n},such that

sti =n∑j=1

aijtj for all i = 1, . . . , n.

One can rewrite this as a matrix equation for the matrix A = (aij) ∈Matn(R):

(s · In − A)

t1...tn

=

0...0

in T n.

Therefore, the matrix B := s · In − A ∈ Matn(R[s]) satisfies

det(B)In

t1...tn

= BB

t1...tn

=

0...0

in T n.

27

This implies that det(B)·ti = 0 for all i ∈ {1, . . . , n}. Since t1, . . . , tn generateT as an R-module we obtain det(B) · T = 0 and det(B) = det(B) · 1T =0. But, looking at the definition of B, we see that there exist elementsr0, . . . , rn−1 ∈ R such that det(B) = sn + rn−1s

n−1 + · · · + r1s + r0. Thiscompletes the proof of the theorem.

3.4 Corollary The set R′ of all elements of S that are integral over R is asubring of S containing R.

Proof Clearly, R′ is contained in S. Moreover, it contains R, since, forr ∈ R, the monic polynomial f = X − r ∈ R[X] satisfies f(r) = 0. Inorder to show that R′ is a subring of S, let s, t ∈ R′. We need to showthat s − t and st are elements of R′. Since s − t, st ∈ R[s, t], it suffices toshow that R[s, t] = (R[s])[t] is a finitely generated R-module, cf. Theorem 3.3(iii)⇒(i). Since s is integral over R, Theorem 3.3 implies that the R-moduleR[s] is generated by a finite set {x1, . . . , xm}. Since t is integral over R, itis also integral over R[s]. Therefore, by the same argument, (R[s])[t] has afinite generating set {y1, . . . , yn} as R[s]-module. This implies that (R[s])[t]is generated by the finite set {xiyj | 1 6 i 6 m, 1 6 j 6 n} as R-module.This completes the proof.

3.5 Definition The elements in C that are integral over Z are called alge-braic integers.

3.6 Examples (a) For instance√

2,√

3, 3√

7, (√

2 +√

3 3√

7)2 are algebraicintegers. But

√2/2 is not (see Exercise 1).

(b) Every n-th root of unity ζ ∈ C is an algebraic integer, since it is aroot of the monic polynomial Xn − 1 ∈ Z[X].

(c) Let χ ∈ R(G) and let g ∈ G. Then χ(g) is an algebraic integer. Infact, one can write χ = χ1 − χ2 as difference of two characters χ1 and χ2,and by Remark 1.14(d), χi(g) is a sum of roots of unity for i = 1, 2.

3.7 Proposition Let s ∈ Q and assume that s is integral over Z. Thens ∈ Z.

Proof We can assume that s 6= 0 and write s = ab

with a ∈ Z and b ∈ Nsatisfying gcd(a, b) = 1. Assume that b > 1. Then there exists a prime

28

p which divides b. Since s is an algebraic integer, there exist n > 1 andr0, . . . , rn−1 ∈ Z satisfying(a

b

)n+ rn−1

(ab

)n−1+ · · ·+ r1

a

b+ r0 = 0 .

Multiplication with bn yields

an + rn−1an−1b+ rn−2a

n−2b2 + · · ·+ r1abn−1r0b

n = 0 .

But this implies that an is divisible by p. Therefore, also a is divisible by p.This is a contradiction to gcd(a, b) = 1. Therefore, b = 1 and s ∈ Z.

3.8 Notation Let C1, . . . , Ck denote the conjugacy classes of the group G.For every r ∈ {1, . . . , k} we choose an element zr ∈ Cr, and for r, s, t ∈{1, . . . , k} we set

arst := |{(x, y) ∈ Cr × Cs | xy = zt}| .

Note that the element arst ∈ N0 does not depend on the choice of zr. In fact,if also z′r ∈ Cr then z′r = gzrg

−1, for some g ∈ G, and the maps

{(x, y) ∈ Cr × Cs | xy = zt} ↔ {(x′, y′) ∈ Cr × Cs | x′y′ = z′t}(x, y) 7→ (gxg−1, gyg−1)

(g−1x′g, g−1y′g)←7 (x′, y′)

are mutually inverse bijections.

3.9 Lemma Let C1, . . . , Ck denote the conjugacy classes of G and let gr ∈ Crfor r ∈ {1, . . . , k}. Moreover let ∆: G→ GLn(C) be an irreducible represen-tation and let χ denote the character of ∆. Then, for all r ∈ {1, . . . , k}, onehas ∑

x∈Cr∆(x) = λrIn with λr = |Cr|

χ(gr)

n.

Moreover, the complex numbers λ1, . . . , λk are algebraic integers.

Proof For r ∈ {1, . . . , k} we set Ar :=∑x∈Cr ∆(x) ∈ Matn(C). Then we

obtain, for every g ∈ G, the equation

∆(g)Ar∆(g)−1 =∑x∈Cr

∆(gxg−1) =∑y∈Cr

∆(y) = Ar ,

29

after substituting gxg−1 with y. By Corollary 2.2(b), we obtain Ar = λr · Inwith

λr =tr(Ar)

n=∑x∈Cr

tr(∆(x))

n= |Cr|

χ(gr)

n.

We still need to show that the numbers λ1, . . . , λr are algebraic integers. Forr, s ∈ {1, . . . , k} we have

λrλs · In = ArAs =∑

(x,y)∈Cr×Cs

∆(xy) =k∑t=1

∑z∈Ct

arst∆(z)

=k∑t=1

arstAt =( k∑t=1

arstλt)· In ,

where the integers arst are defined as in 3.8. This implies that λrλs =∑kt=1 arstλt and that the finitely generated Z-module T := 〈λ1, . . . , λk〉Z ⊂ C

is a subring of C. Thus, we have Z[λr] ⊆ T ⊂ C and λr is an algebraicinteger for all r ∈ {1, . . . , k}, by Theorem 3.3(iii)⇒(i).

3.10 Corollary Let ∆: G→ GLn(C) be an irreducible representation. Thenn divides |G|.

Proof Let χ denote the character of ∆. Then, with the notation fromLemma 3.9, we have

k∑r=1

λrχ(g−1r ) =

k∑r=1

|Cr|χ(gr)

nχ(g−1

r ) =1

n

∑x∈G

χ(x)χ(x−1) =|G|n

(χ, χ) =|G|n.

The first term of this equality is an algebraic integer by Lemma 3.9 and byCorollary 3.4. On the other hand the last term in the equality is a rationalnumber. Thus, by Proposition 3.7, we obtain that this number is an integer.This implies that n divides |G|.

3.11 Remark (McKay Conjecture) Let p be a prime and denote by Irrp′(G)the set of irreducible characters χ of G whose degree is not divisible by p.

A longstanding conjecture in character theory, the McKay Conjecture,states the following: Let G be a finite group, let p be a prime, let P ∈ Sylp(G)and set H := NG(P ). Then

|Irrp′(G)| = |Irrp′(H)| . (3.11.a)

30

Note that if p does not divide the order of G then P = 1, H = G and theabove equation holds for trivial reasons. More generally, if G has a normalSylow p-subgroup, Equation (3.11.a) always holds, since H = G.

3.12 Examples (a) Let G = Sym(3). The irreducible characters of G havedegree 1, 1, 2. For p = 2 we obtain P ∼= C2, a cyclic group of order 2, andH := NG(P ) = P ∼= C2. Thus |Irr2′(G)| = 2 = |Irr2′(H)|. For p = 3 weobtain P = Alt(3). Thus, H := NG(P ) = G and the equation (3.11.a) holds.

(b) Let G = Sym(4). The irreducible characters of G have degrees1, 1, 2, 3, 3. For p = 2 we obtain P ∼= D8 and H := NG(P ) = P ∼= D8.Since the character degrees of D8 are 1, 1, 1, 1, 2, we obtain |Irr2′(G)| = 4 =|Irr2′(H)| as predicted. For p = 3, we have P ∼= C3, a cyclic group of order3, for instance we can choose P = Alt(3). Then H := NG(P ) = Sym(3) andwe obtain |Irr3′(G)| = 3 = |Irr3′(H)| as predicted by the conjecture.

Exercises

1. Show that√

2/2 is not an algebraic integer.

2. Let F3 = Z/3Z be the field with 3 elements and let G := SL2(F3).

(a) Compute the order of G.

(b) Compute the conjugacy classes of G.

(c) Compute Z(G) and determine the isomorphism type of G/Z(G).

(d) Compute the character table of G.

3. (a) Determine the character degrees of Alt(5) and verify the McKay Con-jecture for the primes 2, 3, 5.

(b) Compute the character degrees of Sym(5) as follows. Compute the commu-tator subgroup by using that Alt(5) is simple. Show that the natural permutationcharacter decomposes into the trival character and an irreducible character of de-gree 4. Show that Sym(5) has no irreducible character of degree 2 by consideringthe restriction to Alt(5).

(c) Verify the McKay Conjecture for Sym(5) for p = 2, 3, 5.

4. Let G := GL2(F3).

(a) Determine the order of G, the center of G and the commutator subgroupof G.

(b) Determine the character degrees of G and verify the McKay Conjecture forthe primes 2 and 3.

31

(c) Show that G := G/Z ∼= Sym(4). (Hint: Find some set with four elementson which G acts naturally.)

(d) Compute as much as you can of the character table of G.

5. Compute as much as you can of the character table of Alt(5).

6. Let χ be an irreducible character of G. Show that χ(1) divides [G : Z(χ)].

32

4 Burnside’s paqb-Theorem

The goal of this section is to prove a famous Theorem of Burnside which saysthat every finite group of order paqb (with primes p and q and with a, b ∈ N0)is solvable. The proof will use character theory and some basic facts fromfield theory applied to character values. We will recall these facts first.

4.1 Remark (a) Let α be a complex number and consider the unique ringhomomorphism

ε : Q[X]→ Q[α] , X 7→ α .

By definition it is surjective. If it is injective, α is called transcendentalover Q. From now on we assume that ε is not injective (in this case α iscalled algebraic over Q). The image of ε (as a subring of C) is an integraldomain. By the fundamental theorem of homomorphisms, its kernel must bea non-zero prime ideal, thus of the form (f) for a unique monic irreduciblepolynomial f ∈ Q[X]. This polynomial is called the minimal polynomial of αover Q. Since (f) is even a maximal ideal, Q[X]/(f) ∼= Q[α] is a field. We setn := deg(f). Then Q[X]/(f) ∼= Q[α] are Q-vector spaces of dimension n. Wewrite f = (X−α1) · · · (X−αn) with unique complex numbers α = α1, . . . , αn.These numbers are pairwise distinct (every field extension of Q is separable,since Q has characteristic 0). For each i = 1, . . . , n we obtain a unique ringhomomorphism

Q[X]/(f)→ C , X 7→ αi .

Composing these n ring homomorphisms with ε−1 : Q[α] → Q[X]/(f), weobtain n pairwise distinct ring homomorphisms

σi : Q[α]→ C , α 7→ αi , (i = 1, . . . , n .)

There are no other ring homomorphisms from Q[α] to C, since each such ringhomomorphism must take a root of f to a root of f .

(b) If α from above is an algebraic integer then each αi, i = 1, . . . , n, isan algebraic integer as well. In fact, there exist c0, . . . , ck−1 ∈ Z such thatαk+ck−1α

k−1+· · ·+c1α+c0 = 0 and applying σi from Part (a) to this equationyields the equation αki +ck−1α

k−1i +· · ·+c1αi+c0 = 0 as desired. Corollary 3.4

implies that the coefficients of the polynomial f = (X −α1) · · · (X −αn) arealgebraic integers. Since they are also rational numbers, Proposition 3.7implies f ∈ Z[X].

33

(c) Similar considerations as in Part (a) lead quickly to the followingresult: Let Q ⊆ K ⊆ L ⊆ C be fields with dimQ(L) <∞ and let σ : K → Cbe a ring homomorphism. Then there exists a ring homomorphism τ : L→ Cwith τ |K = σ. Simply replace Q with K and make some minor adjustmentswhere necessary. With some more effort one can show that σ can be extendedto L, even when [L : K] is infinite.

(d) Let ζ := e2πi/k ∈ C. Then the complete set of ring homomorphismsQ[ζ] → C can be constructed as follows: For every i ∈ {1, . . . , k} withgcd(i, k) = 1 there is a unique ring homomorphism

σi : Q[ζ]→ C , ζ 7→ ζ i .

One clearly has σi(Q[ζ]) = Q[ζ], and the inverse of σi is equal to σj wherej ∈ {1, . . . , k} is the unique element with ij ≡ 1 mod k. The field extensionQ[ζ]/Q is a Galois extension with Galois group

Gal(Q[ζ]/Q) = {σi | i = 1, . . . k, gcd(i, k) = 1} ∼= (Z/kZ)× .

4.2 Lemma Let G be a finite group, let g ∈ G and let χ be a character ofG. Then 0 6 |χ(g)/χ(1)| 6 1. Moreover, if χ(g)/χ(1) is an algebraic integerthen |χ(g)/χ(1)| = 1 or χ(g) = 0.

Proof Set d := χ(1) for the degree of χ and let k denote the order ofg. By Remark 1.14(d), we know that χ(g) can be expressed as χ(g) =ζ i1 + · · · + ζ id with ζ := e2πi/k and elements i1, . . . , id ∈ {0, . . . , k − 1}. Thetriangle inequality then yields the first assertion:

0 6 |χ(g)/d| =∣∣∣ζ i1d

+ · · ·+ ζ id

d

∣∣∣ 6 ∣∣∣ζ i1d

∣∣∣+ · · ·+ ∣∣∣ζ idd

∣∣∣ =1

d+ · · ·+ 1

d= 1 .

From now on assume that α := χ(g)/d is an algebraic integer. We alreadyknow that 0 6 |α| 6 1. Assume that |α| < 1. It suffices to show that α = 0.Since α is an algebraic integer, it is algebraic over Q and has a minimalpolynomial f = Xn+cn−1X

n−1+· · ·+c1X+c0 ∈ Q[X]. Since α is an algebraicinteger, we even have f ∈ Z[X], by Remark 4.1(b). Note that α ∈ Q[ζ] andtherefore Q[α] ⊆ Q[ζ]. We can write f = (X − α1) · · · (X − αn) ∈ C[X] andthe roots α = α1, . . . , αn ∈ C of f are given by the images σ(α), where σruns through the ring homomorphisms from Q[α] to C, cf. Remark 4.1(a). ByRemark 4.1(c) and (d), each such σ is the restriction of a ring homomorphism

34

σj : Q[ζ]→ C with σj(ζ) = ζj. Thus, each αi, i = 1, . . . , n, is of the form

σj(α) =ζ i1j

d+ · · ·+ ζ idj

d,

where j ∈ {1, . . . , k} with gcd(j, k) = 1. As in the first part of the proofthe triangle inequality implies that |αi| 6 1 for all i = 1, . . . , n. Since|α| = |α1| < 1, we obtain that |α1 · · ·αn| < 1. But, on the other handα1 · · ·αn = ±c0 ∈ Z. This implies that |c0| < 1 and c0 = 0. But thenf = X(c1 + c2X + · · · + Xn−1). Since f is irreducible, this implies that thesecond factor must be a unit in Q[X], i.e., a constant non-zero polynomial.Since f is monic, this implies that f = X. Since α is a root of f , this impliesα = 0 as desired.

Note that the statement of the following lemma is purely group theoretic.It’s proof, however, uses character theory.

4.3 Lemma Assume that G is a finite group which has a conjugacy classC with pr elements, where p is a prime and r > 1. Then G has a normalsubgroup N with 1 < N < G.

Proof First note that G is not abelian, since it has a conjugacy class withmore than 1 element. Let g ∈ C. Then g 6= 1, since the conjugacy classof 1 has just one element. As before, let χ1, . . . , χk denote the irreduciblecharacters of G and assume that χ1 is the trivial character. By the secondorthogonality relation applied to the elements g and 1 we obtain

0 =k∑i=1

χi(g)χi(1) = 1 +k∑i=2

χi(g)χi(1)

and

−1

p=

k∑i=2

χi(g)χi(1)

p.

Since −1/p is not an algebraic integer (cf. Proposition 3.7), there exists anelement i ∈ {2, . . . , k} such that χi(g)χi(1)/p is not an algebraic integer.To simplify the notation, from now on we set χ := χi and note that χ isnot the trivial character, since i > 2. Since χ(g)χ(1)/p is not an algebraicinteger, we obtain that χ(g) 6= 0 and that p does not divide χ(1). Then

35

gcd(pr, χ(1)) = 1. Thus, there exist a, b ∈ Z such that 1 = apr + bχ(1).Multiplication with χ(g)/χ(1) yields

χ(g)

χ(1)= a|C|χ(g)

χ(1)+ bχ(g) ,

since |C| = pr. By Lemma 3.9, the number |C|χ(g)/χ(1) is an algebraicinteger. Since also χ(g) is an algebraic integer, the right hand side of theabove equation is an algebraic integer and also χ(g)/χ(1) is an algebraicinteger. Now Lemma 4.2 applies. Since χ(g) 6= 0, we obtain |χ(g)/χ(1)| = 1which is equivalent to g ∈ Z(χ), the center of χ, cf. Remark 1.14(d). Sinceg 6= 1, the normal subgroup Z(χ) of G is not trivial, and if Z(χ) < G we aredone. So we assume from now on that Z(χ) = G. In this case we considerthe normal subgroup N := ker(χ) of G. Recall from Remark 1.14(d) thatZ(χ) = G implies that every representation ∆ with character χ takes valuesin scalar matrices. But since χ is irreducible this implies that χ has degree1. Thus, χ is a homomorphism from G to C× and G/N is isomorphic to anabelian group. Now, since G is not abelian, we obtain N 6= 1, and since χis not equal to the trivial character, we obtain N < G. This completes theproof.

Recall that if G is a finite group and N is a normal subgroup of G suchthat N and G/N are solvable then G is solvable.

4.4 Theorem (Burnside’s paqb-Theorem) If G is a finite group of orderpaqb with primes p and q and with a, b ∈ N0 then G is solvable.

Proof We prove the statement by induction on the order of G. If |G| = 1then G is clearly solvable. Now assume that paqb > 1 and that the statementholds for all groups of smaller order. We may assume that G is not abelian,because otherwise we are done. It suffices to show that G has a normalsubgroup different from {1} and G. If a = 0 or b = 0 then G is a p-group (ora q-group) and therefore {1} < Z(G)/G, and we are done. So we can assumethat a, b ∈ N. Choose a Sylow q-subgroup Q of G. Then |Q| = qb > 1 andtherefore Z(Q) > {1}. Choose 1 6= g ∈ Z(Q). Then the conjugacy class ofg has cardinality [G : CG(g)]. Since g ∈ Z(Q), we have Q 6 CG(g) 6 G andtherefore, [G : CG(g)] = pr for some 0 6 r 6 a. If r = 0 then 1 6= 〈g〉 /G andwe are done. If r > 0 then Lemma 4.3 implies that G has a normal subgroupN with {1} < N < G. This completes the proof.

36

Exercises

1. (Galois action on characters) Let G be a finite group of exponent k,let ζ := e2πi/k, set K := Q[ζ], and for i ∈ {1, . . . , k} with gcd(i, k) = 1 letσi : Q[ζ]→ Q[ζ] denote the ring homomorphism given by σi(ζ) = ζi. Note that forany χ ∈ R(G) one has χ(g) ∈ K. Therefore, we can define a function σiχ : G→ Kby g 7→ σi(χ(g)).

(a) Let i ∈ {1, . . . , k} with gcd(i, k) = 1. Show that ( σiχ)(g) = χ(gi) for everyvirtual character χ ∈ R(G) and every g ∈ G. Thus, σiχ = Ψi(χ).

(b) Assume again that i ∈ {1, . . . , k} with gcd(i, k) = 1 and let χ be a characterof G. Show that Ψi(χ) is again a character and show that Ψi(χ) is irreducible ifand only if χ is irreducible. (Thus, Gal(K/Q) acts on Irr(G) by (σi, χ) 7→ σiχ).

(c) Show that all entries of the character table of a symmetric group are inte-gers.

2. Assume that G is a non-abelian finite simple group. Show that G has noabelian subgroup of index pr, where p is a prime and r > 1.

3. (Adams operations) Let G be a finite group. For i ∈ Z and for a classfunction f ∈ CF(G,C) let Ψi(f) denote the function G→ C, g 7→ f(gi).

(a) Show that Ψi(f) is again a class function.

(b) Show that the resulting function Ψi : CF(G,C)→ CF(G,C) is a C-algebrahomomorphism, i.e., a C-linear map and a ring homomorphism. The map Ψi iscalled the i-th Adams operation.

(c) Show that for i, j ∈ Z one has Ψi ◦Ψj = Ψij . Show that Ψi = Ψj wheneveri ≡ j mod exp(G). (Here exp(G) denotes the exponent of G, i.e., the smallestpositive integer n with the property that gn = 1 for all g ∈ G.)

(d) Show that if i ∈ Z with gcd(i, n) = 1 then Ψi is an isometry with respect tothe Schur inner product, i.e., a C-linear isomorphism satisfying (Ψi(f),Ψi(g)) =(f, g) for all f, g ∈ CF(G,C). What is its inverse?

4. (Galois action on characters) Let G be a finite group of exponent n, letζ := e2πi/n, set K := Q[ζ], and for i ∈ {1, . . . , n − 1} with gcd(i, n) = 1 letσi : Q[ζ] → Q[ζ] denote the ring homomorphism given by σi(ζ) = ζi. Note thatfor any χ ∈ R(G) one has χ(g) ∈ K. Therefore, we can define the functionσiχ : G→ K by g 7→ σi(χ(g)).

(a) Let i ∈ {1, . . . , n} with gcd(i, n) = 1. Show that ( σiχ)(g) = χ(gi) for everyvirtual character χ ∈ R(G) and every g ∈ G. Thus, σiχ = Ψi(χ).

37

(b) Assume again that gcd(i, n) = 1 and let χ be a character of G. Show thatΨi(χ) is again a character and show that Ψi(χ) is irreducible if and only if χ isirreducible.

(c) Show that all entries of the character table of a symmetric group are inte-gers.

38

5 The Group Algebra and its Modules

Throughout this section, k denotes a commutative ring and G denotes afinite group. All ring homomorphisms are unitary, i.e., they map the identityelement to the identity element.

5.1 Definition A k-algebra is a pair (R, η) consisting of a ring R togetherwith a ring homomorphism η : k → R satisfying η(k) ⊆ Z(R), i.e., η(α)r =rη(α) for all α ∈ k and r ∈ R. A k-algebra homomorphism between twok-algebras (R, η) and (R′, η′) is a ring homomorphism f : R → R′ such thatf ◦η = η′. Very often it is clear from the context what η is and we only writeR instead of (R, η).

5.2 Remark (a) In general, η need not be injective. If k is a field then η isalways injective and k can be considered as a subring of R via η.

(b) If (R, η) is a k-algebra then R is a k-module via α · r := η(α)r. Forthis module structure one has r(α · s)t = α · (rst), for r, s, t ∈ R and α ∈ k.Conversely, if R is a ring which has also a k-module structure satisfyingr(α · s)t = α · (rst), for r, s, t ∈ R and α ∈ k, then we can define η : k → R,α 7→ α · 1R, and obtain a k-algebra (R, η). These two constructions aremutually inverse.

(c) Let (R, η) and (R′, η′) be k-algebras and let f : R→ R′ be a function.Then, f is a k-algebra homomorphism if and only if f is a ring homomorphismand a k-module homomorphism (with respect to the k-module structuresdefined in (b)).

(d) Z-algebras and Z-algebra homomorphisms are nothing else as ringsand ring homomorphisms. In fact, for every ring R there exists a unique ringhomomorphism η : Z → R and this homomorphism satisfies η(Z) ⊆ Z(R).Moreover, every ring homomorphism f : R→ R′ satisfies f ◦ η = η′ for thesehomomorphism η : Z→ R and η′ : Z→ R′.

5.3 Examples (a) The polynomial ring k[X] is a k-algebra with η mappingan element α ∈ k to the constant polynomial α.

(b) The matrix ring Matn(k) is a k-algebra with η : k → Matn(k), α 7→αIn.

(c) If M is a k-module then the k-endomorphism ring Endk(M) is ak-algebra with η : k → Endk(M), α 7→ lα, where lα : M → M is the leftmultiplication by α, i.e., lα(m) = αm for all m ∈ M . The multiplication onEndk(M) is given by composition.

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5.4 Definition The group algebra kG is defined as the free k-module withbasis G. It is a ring with the following multiplication rule:

(∑g∈G

αgg)(∑h∈G

βhh) =∑g∈Gh∈G

αgβhgh .

In other words, the multiplication on kG is just the k-bilinear extension ofthe multiplication on G. The basis element 1 ∈ G is also the identity elementof kG. Moreover, every basis element g ∈ G is invertible in kG. It’s inverseis the basis element g−1. Clearly, the k-span of 1 is contained in the centerof kG. Therefore, kG is in fact a k-algebra. This construction also works forinfinite groups G.

5.5 Example For G = Sym(3) = {1, σ, σ2, τ, στ, σ2τ} with σ = (1, 2, 3) andτ = (1, 2), one has

kG = k · 1G ⊕ k · σ ⊕ k · σ2 ⊕ k · τ ⊕ k · στ ⊕ k · σ2τ .

For k = Z and for a = 2 · 1G + 3 · σ − τ and b = σ2 − 2 · στ in ZG one has

ab = (2 · 1G + 3 · σ − τ)(σ2 − 2 · στ)

= 2 · σ2 − 4 · στ + 3 · σ3 − 6 · σ2τ − τσ2 + 2 · τστ= 3 · 1G + 4 · σ2 − 5 · στ − 6 · σ2τ ,

since τσ2 = στ and τστ = σ2.

5.6 Remark (a) If f : G → H is a group homomorphism then its k-linearextension

kf : kG→ kH ,∑g∈G

αgg 7→∑g∈G

αgf(g) ,

is a k-algebra homomorphism. In fact, it is k-linear by definition and it ismultiplicative on a k-generating set, namely the set G. One obtains a functork− from the category of (finite) groups to the category of k-algebras (whichare finitely generated as k-modules).

(b) For fixed G and n ∈ N, the maps

{k-alg. hom. Γ: kG→ Matn(k)} ∼←→ {grp. hom. ∆: G→ GLn(k)} ,Γ 7→ Γ|G ,(∑

g∈Gαgg 7→

∑g∈G

αg∆(g))←7 ∆ ,

40

given by restricting Γ to G and k-linearly extending ∆ to kG, are mutuallyinverse bijections. (In category theory language one can show that the functork− : Gr → kAlg is left adjoint to the unit group functor U : kAlg → Gr, andthe above bijection is the special case of this adjunction for a group G and thering Matn(k)) A k-algebra homomorphism Γ: kG → Matn(k) is also calleda matrix representation of the k-algebra kG. Two matrix representationsΓ1 : kG → Matn(k) and Γ2 : kG → Matm(k) are called equivalent if m = nand if there exists S ∈ GLn(k) such that Γ2(a) = SΓ1(a)S−1 for all a ∈ kG.From this it follows immediately that two matrix representations Γ1 and Γ2

of kG are equivalent if and only if the corresponding matrix representations∆1 and ∆2 of G are equivalent. Thus, the above bijections induce mutuallyinverse bijections between equivalence classes:

{k-alg. hom. Γ: kG→ Matn(k)}/ ∼ ∼←→ {grp hom. ∆: G→ GLn(k)}/ ∼ .

(c) Let M be a kG-module. Then, in particular, M is a k-module viaη : k → kG. For every group element g ∈ G, the map m 7→ gm defines an el-ement of Autk(M), the group of k-module automorphisms of M . Altogether,one obtains this way a homomorphism σ : G→ Autk(M). Conversely, if M isa k-module and if σ : G→ Autk(M) is a group homomorphism then we candefine a kG-module structure on M by (

∑g∈G αgg)m :=

∑g∈G αg(σ(g))(m)

for m ∈ M and∑g∈G αgg ∈ kG. These two constructions are mutually in-

verse. Similarly as in (b), using k-linear extensions, one obtains a bijectionbetween the set of pairs (M,σ), where M is a k-module and σ : G→ Autk(M)is a group homomorphism, and the set of pairs (M,ρ), where M is a k-moduleand ρ : kG→ Endk(M) is a k-algebra homomorphism. Thus we have naturalbijections

{kG-modules M} ∼←→ {(M,σ) |M k-module, σ : G→ Autk(M) hom.}∼←→ {(M,ρ) |M k-module, ρ : kG→ Endk(M) hom.} .

We say that two pairs (M,σ) and (M ′, σ′) as above are called equivalent, ifthere exists a k-module isomorphism f : M → M ′ such that σ′(g) = f−1 ◦σ(g) ◦ f for all g ∈ G. Similarly, two pairs (M,ρ) and (M ′, ρ′) are calledequivalent if there exists a k-module isomorphism f : M → M ′ such thatρ′(a) = f−1 ◦ ρ(a) ◦ f for all a ∈ kG. It is now easy to check that twokG-modules M and M ′ are isomorphic if and only if the associated pairs(M,σ) and (M ′, σ′) are equivalent. Moreover, M and M ′ are isomorphic

41

as kG-modules if and only if the associated pairs (M,ρ) and (M ′, ρ′) areequivalent. Consequently, we the above bijections induce bijections betweenisomorphism classes and equivalence classes:

{kG-modules M}/ ∼∼←→ {(M,σ) |M k-module, σ : G→ Autk(M) hom.}/ ∼∼←→ {(M,ρ) |M k-module, ρ : kG→ Endk(M) hom.}/ ∼ .

(d) From now on we assume that k = F is a field and fix n ∈ N. For anypair (M,σ) as in (c), consisting of an n-dimensional F -vector space M anda group homomorphism σ : G → AutF (M), we can define a representation∆: G → GLn(F ) by choosing an F -basis of M and composing σ with theisomorphism AutF (M)

∼→ GLn(F ) determined by this basis. Conversely, if∆: G→ GLn(F ) is a representation of G, we can choose any n-dimensionalF -vector space M and a basis of M , and use the resulting isomorphismGLn(F )

∼→ AutF (M) to define a group homomorphism σ : G → AutF (M)by composing ∆ with the described isomorphism. For each of these con-structions, different choices of M and/or bases, lead to equivalent results.Therefore, altogether we obtain a bijection between equivalence classes,

{(M,σ) |M n-dim. F -vector space, σ : G→ AutF (M) hom.}/ ∼∼←→ {group homomorphisms ∆: G→ GLn(F )}/ ∼ .

Composing this with the previous bijection from (c), we obtain a bijection,for fixed G and n,

{FG-modules M with dimF M = n}/ ∼∼←→ {group homomorphisms ∆: G→ GLn(F )}/ ∼ .

If, under this last bijection, ∆1 and ∆2 correspond to M1 and M2 (for di-mensions n1 and n2), then ∆1 ⊕ ∆2 corresponds to M1 ⊕M2. Moreover, if∆ corresponds to M then we have

M is indecomposable (resp. irreducible)

⇐⇒ ∆ is indecomposable (resp. irreducible).

(e) Assume that FG-modules M1 and M2 of dimension n1 and n2 cor-respond to representations ∆1 : G → GLn1(F ) and ∆2 : G → GLn2(F ), re-spectively. Choosing F -bases for M1 and M2 one obtains an isomorphism of

42

F -vector spaces

HomFG(M1,M2)∼−→ {S ∈ Matn2×n1(F ) | S∆1(g) = ∆2(g)S for all g ∈ G} ,

by mapping a homomorphism f to the representing matrix of f with respectto the chosen bases of M1 and M2. In particular, if M is an FG-moduleand if ∆: G → GLn(F ) is a corresponding representation then one has anisomorphism of F -algebras

EndFG(M)∼−→ {S ∈ Matn(F ) | S∆(g) = ∆(g)S for all g ∈ G} .

(f) For F = C we obtain bijections

{finite dimensional CG-modules}/isom.∼←→ {representations of G over C}/equiv.∼−→ {characters of G}.

If M is a finite-dimensional CG-module, ∆: G→ GLn(C) is a correspondingrepresentation and χ is the corresponding character then we also write χ =χM and say that χ is afforded by M and by ∆. Note that, for g ∈ G, thecharacter value χM(g) is the trace of the C-linear map M →M , m 7→ gm.

The character of the regular CG-module, i.e., the CG-module CG, is equalto the regular character.

The trivial character is afforded by the trivial CG-module, namely themodule C with g · α = α for g ∈ G and α ∈ C.

More generally, if φ : G → C× is a linear character (character of degree1) then φ is afforded by the CG-module Cφ which is defined to be equal to Cas C-space, and whose module structure is given by g ·α := φ(g)α, for g ∈ Gand α ∈ C.

5.7 Lemma (Schur) Let R be a ring and let M be a simple R-module.Then the ring EndR(M) is a division ring, i.e., every non-zero element isinvertible.

Proof Let 0 6= f ∈ EndR(M). We need to show that f is injective andsurjective. Since f 6= 0, we obtain im(f) 6= {0}. Since im(f) is an R-submodule of M and since M is simple, we obtain im(f) = M and f issurjective. Moreover, since f 6= 0, we obtain ker(f) 6= M . Since ker(f) is anR-submodule of M and since M is simple, we obtain ker(f) = {0} and f isinjective. Thus, f is an isomorphism and the proof is complete.

43

5.8 Remark The version of Schur’s lemma in 2.1(b) can be derived from themore general version in 5.7. In fact, let ∆ be an irreducible representationof G over C and let M be a corresponding CG-module. By 5.7, the finite-dimensional C-algebra D := EndCG(M) is a division ring. We can view C as asubfield of D via the structural homomorphism η. The version in 2.1 amountsto the statement C = D. So let d ∈ D. We will show that d ∈ C. In fact,the ring C[d] is commutative and has no zero-divisors. Since dimC C[d] <∞it is also a field. Since C is algebraically closed, every finite-dimensional fieldextension of C is equal to C. Thus, d ∈ C.

5.9 Definition Let R be a ring.

(a) An R-module M is called semisimple, if for every submodule U of Mthere exists a submodule V of M such that U ⊕ V = M .

(b) The ring R is called semisimple, if every R-module is semisimple.

5.10 Theorem (Maschke) Let F be a field. The group algebra FG issemisimple if and only if |G| is invertible in F .

Proof First assume that |G| is invertible in F , let M be an FG-moduleand let U be an FG-submodule of M . We follow the idea of the proof ofTheorem 1.9. Let W be an F -subspace of M such that U ⊕W = M . Wedenote by p : M → U the projection map with respect to the decompositionM = U ⊕W . Define

p′ : M →M , m 7→ 1

|G|∑g∈G

gp(g−1m) .

Then it is easy to see that p′ ∈ HomFG(M,M), p′(M) ⊆ U , p′|U = idU , andp′◦p′ = p′. From this it follows quickly that V := ker(p′) is an FG-submoduleof M satisfying M = U ⊕ V .

Next assume FG is semi simple and that char(F ) = p > 0 and p divides|G|. We will derive a contradiction. Let M := FG be the regular leftFG-module and let u :=

∑g∈G g ∈ M . Then Fu is an FG-submodule of

M . In fact, for every g ∈ G one has gu = u and therefore (∑g∈G αgg)u =

(∑g∈G αg)u. It also follows that u2 = |G|u = 0. Since FG is semisimple,

there exists a submodule V of FG such that U ⊕V = FG. Write 1 = αu+ vwith α ∈ C and v ∈ V . Then u = u1 = αu2 + uv = 0 + uv with uv ∈ V ,since V is an FG-submodule. This implies that u ∈ U ∩ V = {0} and thatu = 0. This is a contradiction.

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Note that, for an arbitrary commutative ring k and an arbitrary kG-module M , one has: M is finitely generated as kG-module if and only if Mis finitely generated as k-module. (See Excersise 3)

5.11 Corollary Let F be a field such that |G| is invertible in F . Everynon-zero finitely generated FG-module M can be decomposed into a directsum of simple submodules.

Proof We prove the result by induction on dimF M . If dimF M = 1 thenM is clearly simple. Now assume that dimF M = n > 1 and that the resultholds for modules of smaller dimension. If M is simple we are done. If Mis not simple then there exists a submodule M1 of M which is different from{0} and M . By Theorem 5.10, there exists a submodule M2 of M such thatM = M1 ⊕M2. By induction, both M1 and M2 can be written as a directsum of simple submodules. Thus, M can be written in this form.

The following theorem is an explicit version of Wedderburn’s Theoremfor the semisimple ring CG.

5.12 Theorem Let G be a finite group, let ∆1, . . . ,∆k be representativesof the equivalence classes of irreducible representations of G over C, and letn1, . . . , nk ∈ N denote their respective degrees. Then the map

Φ: CG→ Matn1(C)× · · · ×Matnk(C) ,∑g∈G

αgg 7→(∑g∈G

αg∆1(g), . . . ,∑g∈G

αg∆k(g))

is an isomorphism of C-algebras with inverse

Ψ: (A1, . . . , Ak) 7→1

|G|∑g∈G

( k∑i=1

nitr(∆i(g−1)Ai)

)g .

Proof Clearly, Φ and Ψ are C-linear maps. Moreover, it is easy to checkthat Φ is multiplicative on the basis elements g ∈ G. Therefore, Φ is a C-algebra homomorphism. Note also that both sides of the map have the samedimension, since |G| = n2

1 + · · · + n2k. To see that Φ is bijective and that Ψ

is its inverse it suffices to show that Ψ(Φ(a)) = a for all a ∈ CG. And itsuffices to show this for the basis elements a = h ∈ G. But, with χi denoting

45

the character of ∆i, we have

Ψ(Φ(h)) = Ψ(∆1(h), . . . ,∆k(h)

)=

1

|G|∑g∈G

( k∑i=1

nitr(∆i(g−1)∆i(h)

)

=1

|G|∑g∈G

( k∑i=1

niχi(g−1h)

)g

with∑ki=1 niχi = ρG, the regular character of G, by Proposition 2.11. Thus,

we obtain Ψ(Φ(h)) = h, since ρG(g−1h) = 0 if g 6= h and ρG(g−1h) = |G| ifg = h.

5.13 Remark Let ∆1, . . . ,∆k and n1, . . . , nk be as in Theorem 5.12 and letχ1, . . . , χk denote the respective characters of ∆1, . . . ,∆k.

(a) Note that the C-algebra A := Matn1(C)× · · · ×Matnk(C) has centralidempotents εi := (0, . . . , 0, Ini , 0, . . . , 0). Moreover ε1 + · · · + εk = 1A andεiεj = 0 for i 6= j in {1, . . . , k}. Applying the C-algebra isomorphism Ψ fromTheorem 5.12 we obtain central idempotents

ei := Ψ(εi) =1

|G|∑g∈G

(nitr(∆i(g

−1))g =

χi(1)

|G|∑g∈G

χi(g−1)g ∈ CG .

Thus, e2i = ei, eiej = 0, eia = aei and e1+· · ·+ek = 1CG, for all i ∈ {1, . . . , k},

all i 6= j in {1, . . . , k}, and all a ∈ CG. Since eiej = 0 for i 6= j, theelements e1, . . . , ek are C-linearly independent. Since dimC Z(CG) = k (seeExercise 1), these elements form a C-basis of Z(CG). The element ei is calledthe central idempotent associated with χi.

(b) Let M be a finitely generated CG-module. Then the above propertiesof the elements e1, . . . , ek imply immediately that

M = e1M ⊕ · · · ⊕ ekM

is a direct sum decomposition into CG-submodules.(c) Let M be a finitely generated CG-modules. Recall from Remark 5.6(f)

that the character χM of M is the character of any representation of G whichcorresponds to M under the constructions in Remark 5.6(d). In other words,for g ∈ G, χM(g) is defined as the trace of a representing matrix of theC-linear map M →M , m 7→ gm, with respect to any C-basis of M .

46

5.14 Theorem LetM be an irreducible CG-module, let Irr(G) = {χ1, . . . , χk}and let e1, . . . , ek ∈ Z(CG) be the respective associated idempotents. For allm ∈M and i ∈ {1, . . . , k} one has

eim =

m if χM = χi,

0 if χM 6= χi.

Proof Since M is an irreducible CG-module, we have χM = χj for some j ∈{1, . . . , k}. Let m1, . . . ,mnj be a C-basis of M and let ∆j : G→ GLnj(C) de-note the corresponding representation, i.e., ∆j(g) is the representing matrixof the C-linear map M →M , m 7→ gm, for all g ∈ G. Then, χj = χM = χ∆.For m = α1m1 + · · ·+ αnjmnj we have

eim =(χi(1)

|G|∑g∈G

χi(g−1)g

)m

= (m1, . . . ,mnj)(χi(1)

|G|∑g∈G

χi(g−1)∆j(g)

︸ ︷︷ ︸=:Aj∈Matnj (C)

)α1...αnj

.

By the definition of the map Φ in Theorem 5.12, the matrix Aj is equal tothe j-th entry in the tuple Φ(ei) = εi. Thus, Aj = Inj if i = j, and Aj = 0 ifi 6= j. Now the result follows.

5.15 Corollary Let S1, . . . , Sk denote a set of representatives of the isomor-phism classes of simple CG-modules, let χ1, . . . , χk denote their respectivecharacters and e1, . . . , ek ∈ CG their respective central idempotents. LetM be a finitely generated CG-module and let M = M1 ⊕ · · · ⊕ Mr be adecomposition of M into simple submodules. For each i ∈ {1, . . . , k} one has

eiM =⊕

s∈{1,...,k}Ms∼=Si

Ms

and

(χM , χi) =1

χi(1)dimC(eiM) .

In particular, every submodule S of M which is isomorphic to Si is con-tained in eiM . The submodule eiM of M is called the χi-isotypic (or χi-homogeneous) component of M .

47

Proof This follows immediately from Theorem 5.14

Exercises

1. Let R be a commutative ring and let G be a finite group. Moreover,let C1, . . . , Ck denote the conjugacy classes of G and set ci :=

∑g∈Ci g ∈ RG for

i = 1, . . . , k.

(a) Show that (c1, . . . , ck) is a R-basis of Z(RG).

(b) Show that the structure constants of Z(RG) (i.e., the unique elementsaijk ∈ R satisfying cicj =

∑km=1 aijmcm) with respect to the basis c1, . . . , ck are

given byaijm = |{(x, y) ∈ Ci × Cj | xy = z}| ,

where z ∈ Cm is a fixed element.

(c) Let Γ: CG→ Matn(C) be a C-algebra homomorphism whose restriction toG is an irreducible representation. Show that Γ(Z(CG)) = C · 1n.

2. Let K := Q(ζ) with ζ := e2πi/3 and let G := Sym(3). Show that themultiplication by ζ and complex conjugation define a QG-module structure on K.Compute a corresponding representation and its character.

3. Let k be a commutative ring, let G be a finite group, and let M be akG-module. Show that M is finitely generated as kG-module if and only if M isfinitely generated as k-module.

4. Let k be a commutative ring, let G be a finite group and let S be a finiteG-set.

(a) Let M := kS, the free k-module with k-basis S. Construct a kG-modulestructure on M .

(b) Show that the character of M is given by the function

χS : g 7→ |{s ∈ S | gs = s}| ,

Note that this gives a new way to construct characters. Such characters are calledpermutation characters.

(c) Assume that k = C. Show that the scalar product (χS , 1) of χS with thetrivial character equals the number of orbits of S.

5. Let G be a finite group and let M and N be two finitely generated CG-modules. Show that

(χM , χN ) = dimC HomCG(M,N) .

48

(Hint: Reduce to the case that M and N are irreducible.)

6. Let G be a finite group and let M and N be finitely generated CG-modules.

(a) Show that M∗ := HomC(M,C) is a finitely generated CG-module via (g ·f)(m) := f(g−1m), and show that χM∗ = χM . (Taking duals corresponds totaking contragredient representations, and taking complex conjugate characters.)

(b) Show that the C-vector space V := HomC(M,N) becomes a CG-modulevia

(g · f)(m) := gf(g−1m)

for g ∈ G, f ∈ V , m ∈M , and that χV = χM · χN .

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6 The Tensor Product

In this section we first recall the definition and basic properties of the tensorproduct (without proofs). It will be applied at the end of the section toconstruct the irreducible characters of the direct product group G×H fromthe irreducible characters of G and of H. Throughout this section, R, S, andT denote rings.

6.1 Definition Let M be a right R-module, let N be a left R-module, andlet A be an abelian group. A map β : M ×N → A is called biadditive, if

β(m+m′, n) = β(m,n)+β(m′, n) and β(m,n+n′) = β(m,n)+β(m,n′) ,

for all m,m′ ∈M and all n, n′ ∈ N . The map β is called R-balanced if

β(mr, n) = β(m, rn) ,

for all m ∈M , n ∈ N and r ∈ R. (Note that for the definition of ”biadditive”it would have been enough to assume that M and N are abelian groups.)

6.2 Definition Let M be a right R-module and let N be a left R-module.A tensor product of M and N over R is a pair (T, τ) consisting of an abeliangroup T together with an R-balanced, biadditive map τ : M ×N → T satis-fying the following universal property:

If (A, β) is a pair consisting of an abelian group A and an R-balanced,biadditive map β : M ×N → A then there exists a unique group homomor-phism f : T → A such that the diagram

T

���

qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq

τ

M ×N

qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq

f

@@@qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq

β

A

commutes.

6.3 Theorem Let M be a right R-module and let N be a left R-module.

(a) There exists a tensor product (T, τ) of M and N over R.

(b) If both (T, τ) and (T ′, τ ′) are tensor products of M and N over Rthen the unique homomorphism f : T → T ′ such that the diagram

50

T

���

qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq

τ

M ×N

qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqq

f

@@@qqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqqτ ′

T ′

commutes, is an isomorphism. Its inverse is the unique homomorphismg : T ′ → T such that g ◦ τ ′ = τ .

6.4 Notation If (T, τ) is a tensor product of M and N over R then oneusually writes M⊗RN instead of T and m⊗n instead of τ(m,n), for m ∈Mand n ∈ N . With these notations, the following hold for all m,m′ ∈ M ,n, n′ ∈ N and r ∈ R:

(m+m′)⊗ n = (m⊗ n) + (m′ ⊗ n) , m⊗ (n+ n′) = (m⊗ n) + (m⊗ n′) ,0⊗ n = 0 = m⊗ 0 , (−m)⊗ n = −(m⊗ n) = m⊗ (−n) ,

mr ⊗ n = m⊗ rn .

The universal property implies that M⊗RN is generated as abelian groupby elements of the form m⊗ n with m ∈M and n ∈ N . Using the fourth ofthe above rules, one obtains that every element x ∈M ⊗R N can be writtenin the form

x = (m1 ⊗ n1) + · · · (ms ⊗ ns)

with s ∈ N, m1, . . . ,ms ∈ M and n1, . . . , ns ∈ N . Warning: The elementss ∈ N, m1, . . . ,ms ∈ M and n1, . . . , ns ∈ N are not uniquely determined byx.

6.5 Proposition Let M and M ′ be right R-modules, let N and N ′ be left R-modules, and let f : M →M ′ and g : N → N ′ be R-module homomorphisms.Then there exits a unique homomorphism h : M⊗RN →M ′⊗RN ′ such thath(m⊗ n) = f(m)⊗ g(n) for all m ∈ M and n ∈ N . This homomorphism isdenoted by f ⊗ g. Thus, (f ⊗ g)(m⊗ n) = f(m)⊗ g(n) for all m ∈ M andn ∈ N .

6.6 Remark (a) Recall that an (R, S)-bimodule M is an abelian groupwhich is a left R-module and a right S-module such that (rm)s = r(ms)

51

for all r ∈ R, s ∈ S and m ∈ M . If M is an (R, S)-bimodule and N isan (S, T )-bimodule then the abelian group M ⊗S N has a unique (R, T )-bimodule structure satisfying r(m⊗n) = (rm)⊗n and (m⊗n)t = m⊗ (nt)for r ∈ R, t ∈ T , m ∈M and n ∈ N .

(b) Let k be a commutative ring. Every left (resp. right) k-module M isautomatically a (k, k)-bimodule via mα := αm (resp. αm := mα), for α ∈ kand m ∈ M . Thus, if M and N are k-modules then M ⊗k N is again ak-module with the module structure from Part (a).

(c) If (A, ηA) and (B, ηB) are k-algebras then (A ⊗k B, η) is again a k-algebra. The multiplication on A ⊗k B is determined by the property (a ⊗b)(a′ ⊗ b′) = (aa′) ⊗ (bb′), for a, a′ ∈ A and b, b′ ∈ B. Moreover, η(α) :=ηA(α)⊗ 1B = (α · 1A)⊗ 1B = α · (1A ⊗ 1B) = 1A ⊗ (α · 1B) = 1A ⊗ ηB(α) forα ∈ k.

(d) For finite groups G and H, one has an isomorphism kG ⊗k kH ∼=k[G × H] of k-algebras given by g ⊗ h 7→ (g, h). In fact, by the followingproposition, the elements g ⊗ h, with g ∈ G and h ∈ H, form a k-basis ofkG ⊗k kH. Therefore, the above map defines a k-linear isomorphism. It isquickly checked that this is also a ring homomorphism (it suffices to checkthis on the basis elements).

6.7 Proposition (a) Let M be a left R-module (resp. an (R, S)-bimodule).Then

R⊗RM ∼= M

as left R-modules (resp. (R, S)-bimodules) via the map r ⊗m 7→ rm withinverse m 7→ 1R ⊗ m, for r ∈ R and m ∈ M . Similarly, if M is a rightS-module (resp. an (R, S)-bimodule) then

M ⊗S S ∼= M

as right S-modules (resp. (R, S)-bimodules) via the map m ⊗ s 7→ ms withinverse m 7→ m⊗ 1S for m ∈M and s ∈ S.

(b) Let M and M ′ be right S-modules (resp. (R, S)-bimodules) and letN be a left S-module (resp. (S, T )-bimodule). Then

(M ⊕M ′)⊗S N ∼= (M ⊗S N)⊕ (M ′ ⊗S N)

as abelian groups (resp. as (R, T )-bimodules). Similarly, if also N ′ is a leftS-module (resp. an (S, T )-bimodule) then

M ⊗S (N ⊕N ′) ∼= (M ⊗S N)⊕ (M ⊗S N ′)

52

as abelian groups (resp. as (R, T )-bimodules).

(c) Let M be an (R, S)-bmodule, let N be an (S, T )-bimodule and letP be a (T, U)-bimodule for an additional ring U . Then there exists anisomorphism

(M ⊗S N)⊗T P ∼= M ⊗S (N ⊗T P )

of (R,U)-bimodules that maps the element (m⊗ n)⊗ p to the element m⊗(n⊗ p) and whose inverse maps the element m⊗ (n⊗ p) to (m⊗ n)⊗ p, form ∈M , n ∈ N and p ∈ P .

(d) Let M be a right R-module and let N be a left R-module. If M is freewith R-basis m1, . . . ,md, then every element x ∈M ⊗R N can be expressedas

x = (m1 ⊗ n1) + · · ·+ (md ⊗ nd)

with unique elements n1, . . . , nd ∈ N . Similarly, if N is free with R-basisn1, . . . , nd then every element x ∈ M ⊗R N can be expressed as above withuniquely determined elements m1, . . . ,md ∈M .

(e) If k is a commutative ring and if M and N are free k-modules withbases m1, . . . ,md and n1, . . . , ne, respectively, then M⊗kN is a free k-modulewith basis mi ⊗ nj, i = 1, . . . , d, j = 1, . . . , e.

6.8 Remark Let F be a field and let G be a finite group.(a) Let V, V ′,W,W ′ be finite-dimensional F -vector spaces, let f : V →

V ′ and g : W → W ′ be F -linear maps, and let (v1, . . . , vr), (v′1, . . . , v′s),

(w1, . . . , wt), and (w′1, . . . , w′u) be F -bases of V, V ′,W,W ′, respectively. More-

over, let A ∈ Mats×r(F ) and B ∈ Matu×t(F ) be the representing matrices off and g, respectively, with respect to these bases. Then A⊗B ∈ Matsu×rt(F )is the representing matrix of f ⊗ g : V ⊗F W → V ′ ⊗F W ′ with respect tothe bases vi ⊗ wj (1 6 i 6 r, 1 6 j 6 s) and v′k ⊗ w′l (1 6 k 6 t, 1 6 l 6 u)in lexicographic order.

(b) Assume now that V and W are finitely generated FG-modules. Then,V ⊗F W is again an F -vector space. It is even an FG-module with

(∑g∈G

αgg) · (v ⊗ w) =∑g∈G

αg(gv ⊗ gw) .

In fact, one has an F -algebra homomorphism

FG δ qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq FG⊗F FGρ⊗ ρ′

qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq EndF (V )⊗F EndF (W ) qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq EndF (V ⊗W )

53

where δ is the F -linear extension of g 7→ g⊗g, ρ and ρ′ denote the F -algebrahomomorphisms describing the FG-module structures of V and W , and thelast map is the F -algebra homomorphism induced by (f, g) 7→ f ⊗ g. Theabove FG-module structure on V ⊗F W comes from this homomorphism.

Thus, if (v1, . . . , vr) and (w1, . . . , ws) are F -bases of V and W and if∆: G → GLr(F ) and ∆′ : G → GLs(F ) denote the representations corre-sponding to V and W with respect to these bases then ∆⊗∆′ : G→ GLrs(F ),g 7→ ∆(g)⊗∆′(g), corresponds to the FG-module V ⊗F W with respect tothe basis vi⊗wj (1 6 i 6 r, 1 6 j 6 s) in lexicographic order. In particular,since tr(∆(g)⊗∆′(g)) = tr(∆(g))tr(∆′(g)), we obtain χV⊗FW = χV · χW .

In the following definition we construct a representation of G × H froma representation of G and a representation of H.

6.9 Definition Let G and H be finite groups and let F be a field. If V isa finitely generated FG-module and W is a finitely generated FH-modulethen the F -vector space V ⊗F W has an F [G ×H]-module structure whichis uniquely determined by

(g, h) · (v ⊗ w) = gv ⊗ hw ,

for (g, h) ∈ G×H, v ∈ V and w ∈ W . In fact, this module structure comesfrom the F -algebra homomorphism

F [G×H] qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq FG⊗FFH ρ⊗ ρ′ qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq EndF (V )⊗FEndF (W ) qqqqqqqqqqqqqqqqqq qqqqqqqqqqqqqqqqqq EndF (V⊗FW ) ,

where the first map is the F -linear extension of (g, h) 7→ g ⊗ h, the maps ρand ρ′ denote the F -algebra homomorphisms corresponding to the modulestructures of V and W , and the last map is induced by (f, g) 7→ f ⊗ g.

We will often write χV × χW for the character χV⊗FW of G × H. Ifalso V ′ and W ′ are finitely generated modules for FG and FH, respectively,then (χV + χV ′) × χW = χV × χW + χV ′ × χW and χV × (χW + χW ′) =χV × χW + χV × χW ′ . Thus, for F = C, one obtains a biadditive map

−×− : R(G)×R(H)→ R(G×H) , (χ, ψ) 7→ χ× ψ ,

where (f1 × f2)(g, h) := f1(g) · f2(h), for arbitrary class functions f1, f2 ∈CF(G,C) and (g, h) ∈ G×H.

54

6.10 Theorem Let G and H be finite groups, let V be a finitely gener-ated CG-module and let W be a finitely generated CH-module. Then thefollowing hold:

(a) One has (χV × χW )(g, h) = χV (g)χW (h).

(b) The C[G×H]-module V ⊗C W is simple if and only if V and W aresimple modules.

(c) The function Irr(G) × Irr(H) 7→ Irr(G × H), (χ, ψ) 7→ χ × ψ, is abijection. It induces a ring isomorphism R(G)⊗R(H)

∼−→ R(G×H).

Proof Let (v1, . . . , vm) and (w1, . . . , wn) be F -bases of V and W , respec-tively, and let ∆1 : G → GLm(F ) and ∆2 : G → GLn(F ) be the correspond-ing representations. Remark 6.8(a) implies that the representation of theF [G × H]-module V ⊗F W with respect to the basis vi ⊗ wj (1 6 i 6 m,1 6 j 6 n) in lexicographic order is given by (g, h) 7→ ∆1(g)⊗∆2(h). Sincetr(A⊗B) = tr(A)tr(B) for any square matrices A and B, we obtain

(χV × χW )(g, h) = χV⊗FW (g, h) = tr(∆1(g)⊗∆2(h))

= tr(∆1(g))tr(∆2(h)) = χV (g) · χW (h) ,

and Part (a) is proven.Next we show that for characters χ, χ′ of G and ψ, ψ′ of H, one has

(χ× ψ, χ′ × ψ′)G×H = (χ, χ′)G · (ψ, ψ′)H . (6.10.a)

In fact, we have

(χ× ψ, χ′ × ψ′)G×H =1

|G×H|∑

(g,h)∈G×H(χ× ψ)(g, h) · (χ′ × ψ′)(g−1, h−1)

=1

|G| · |H|∑g∈G

∑h∈H

χ(g)ψ(h)χ′(g−1)ψ′(h−1)

=(

1

|G|∑g∈G

χ(g)χ′(g−1))·(

1

|H|∑h∈H

ψ(h)ψ′(h−1))

= (χ, χ′)G · (ψ, ψ′)H .

For arbitrary characters χ of G and ψ of H, the irreducibility criterion inProposition 2.18 and Equation (6.10.a) now imply the following chain of

55

equivalences:

χ× ψ ∈ Irr(G×H) ⇐⇒ (χ× ψ, χ× ψ)G×H = 1

⇐⇒ (χ, χ)G · (ψ, ψ)H = 1

⇐⇒ (χ, χ)G = 1 and (ψ, ψ)H = 1

⇐⇒ χ ∈ Irr(G) and ψ ∈ Irr(H) .

This implies the statement in Part (b). It also shows that the first map inPart (c) takes values in Irr(G×H).

Next we show that the first map in Part (c) is bijective. In order to seethat it is injective, assume that χ, χ′ ∈ Irr(G) and ψ, ψ′ ∈ Irr(H) satisfyχ× ψ = χ′ × ψ′. Then Equation (6.10.a) implies that

1 = (χ× ψ, χ′ × ψ′)G×H = (χ, χ′)G · (ψ, ψ′)H ,

and we obtain (χ, χ′)G 6= 0 and (ψ, ψ′)H 6= 0. This implies χ = χ′ and ψ = ψ′.In order to see that the first map in Part (c) is bijective it suffices now toshow that |Irr(G × H)| = |Irr(G)| · |Irr(H)|. This in turn is equivalent toshowing that the numbers of conjugacy classes of these three groups satisfiesthe equation k(G×H) = k(G)k(H). Now note that two elements (g, h) and(g′, h′) of G×H are conjugate in G×H if and only if g and g′ are conjugate inG and h and h′ are conjugate in H. This implies that if C1, . . . , Ck(G) denotethe conjugacy classes of G and D1, . . . ,Dk(H) denote the conjugacy classes ofH then Ci × Dj, 1 6 i 6 k(G), 1 6 j 6 k(H), are precisely the conjugacyclasses of G×H.

Finally, since the first map in Part (c) is bijective and since the sets inthese maps form Z-bases of the abelian groups in the second map, the secondmap is a group isomorphism. It is now easy to verify, using the formula inPart (a), that it is also multiplicative.

The following theorem is well-know in category theoretic terms as theadjunction between Hom and ⊗. We state it here in plain language.

6.11 Theorem Let R, S, T , and U be rings.

(a) IfM is an (R, S)-bimodule andN is an (R, T )-bimodule then HomR(M,N)is an (S, T )-bimodule via (sft)(m) := f(ms)t for f ∈ HomR(M,N), s ∈ S,t ∈ T and m ∈M .

56

(b) IfM is an (S,R)-bimodule andN is a (T,R)-bimodule then HomR(M,N)is a (T, S)-bimodule via (tfs)(m) := tf(sm) for f ∈ HomR(M,N), s ∈ S,t ∈ T and m ∈M .

(c) If M is an (R, S)-bimodule, N is an (S, T )-bimodule and P is an(R,U)-bimodule then the maps

HomR(M ⊗S N,P )∼↔ HomS(N,HomR(M,P ))

f 7→(n 7→ (m 7→ f(m⊗ n))

)(m⊗ n 7→ (g(n))(m)

)←7 g

are well-defined mutually inverse isomorphisms of (R,U)-bimodules.

Exercises

1. Let G be a finite group and let M be a finitely generated CG-module.Denote by τ : M ⊗C M → M ⊗C M the map given by m1 ⊗m2 7→ m2 ⊗m1 form1,m2 ∈M .

(a) Consider M ⊗C M as CG-module with the diagonal action g(m1 ⊗m2) =gm1 ⊗ gm2, for g ∈ G, m1,m2 ∈M . Show that the subsets

Ms := {x ∈M ⊗C M | τ(x) = x} and Ma := {x ∈M ⊗C M | τ(x) = −x} ,

the symmetric and alternating square of M , are CG-submodules of M ⊗CM withMs ⊕Ma = M ⊗C M .

(b) Let χ, χs, and χa be the character of M , Ms, and Ma, respectively. Showthat

χs(g) =1

2

(χ2(g) + χ(g2)

)and χa(g) =

1

2

(χ2(g)− χ(g2)

)for all g ∈ G. Again, this gives is a new way to construct characters.

2. Compute the character tables of Sym(5) and Alt(5).

3. Let F be a field and let M and N be finitely generated FG-modules.Construct an FG-module isomorphism between HomF (M,N) and M∗⊗FN . (TheFG-module structures of HomF (M,N) and M∗ are given in Exercise 5.6)

4. Verify the the statements in Theorem 6.11.

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7 Induction

Throughout this section, G denotes a finite group, H denotes a subgroup ofG, and F denotes a field. Note that FG is an (FG,FH)-bimodule.

7.1 Definition Let W be an FH-module. The FG-module

IndGH(W ) := FG⊗FH W

is called the module induced by W from H to G, or also the induction of Wfrom H to G.

7.2 Remark Let W be a finitely generated FH-module and let g1, . . . , gn ∈G denote coset representatives of G/H.

(a) The F -algebra FG is free as right FH-module with basis elementsg1, . . . , gn. In fact, every element

∑g∈G αgg in FG can be written as

∑g∈G

αgg =n∑i=1

∑h∈H

αgihgih =n∑i=1

gi(∑h∈H

αgihh)∈

n∑i=1

gi · FH .

Moreover, if∑ni=1 gi·xi = 0, with xi =

∑h∈H αh,ih ∈ FH, then

∑ni=1

∑h∈H αh,igih =

0. For given i ∈ {1, . . . , n} and h ∈ H, the coefficient of gih in this elementof FG is equal to αh,i. Since gih = gjh

′ implies i = j and h = h′, fori, j ∈ {1, . . . , n} and h, h′ ∈ H, we have αh,i = 0. This implies xi = 0 for alli = 1, . . . , n. Therefore, by Proposition 6.7(d), one has an isomorphism ofF -vector spaces

FG⊗FH W∼↔

n⊕i=1

W ,

n∑i=1

gi ⊗ wi ↔ (w1, . . . , wn) .

This impliesdimF IndGH(W ) = [G : H] dimF W .

(b) Let w1, . . . , wm be an F -basis of W , and let ∆: H → GLm(F ) denotethe corresponding representation ofH. Then, by the isomorphism in Part (a),the elements gi ⊗wj, 1 6 i 6 n, 1 6 j 6 m, form an F -basis of FG⊗FH W .We denote the corresponding representation (using the lexicographic order of

58

the basis) by IndGH(∆). Our next goal is to describe the matrix (IndGH(∆))(g)for a given group element g ∈ G. We have to rewrite the element ggi⊗wj aslinear combination of the elements gk ⊗wl. First note that ggiH = gσ(i)H ∈G/H for some σ(i) ∈ {1, . . . , n}. This defines a permutation σg = σ ∈Sym(n) satisfying ggi ∈ gσ(i)H for every i ∈ {1, . . . , n}. It is the permutationinduced by the action of the element g by left translation on G/H. Thus,there exists an element hi ∈ H (depending on g) such that ggi = gσ(i)hi.With this we obtain

g(gi ⊗ wj) = ggi ⊗ wj = gσ(i)hi ⊗ wj = gσ(i) ⊗ hiwj .

This implies that the matrix (IndGH(∆))(g) consists of n2 blocks of size m×m,and that the m ×m block matrix in the i-th block column and k-th blockrow is equal to 0 unless k = σ(i). In the case k = σ(i) this block matrix isequal to ∆(hi) = ∆(g−1

σ(i)ggi).

(c) Let ψ denote the character of W (or of ∆). With the results fromPart (b), we can compute the character of IndGH(∆) which we will denote byindGH(ψ). With the notation in Part (b), we obtain that

(indGH(ψ)

)(g) =

n∑i=1σ(i)=i

ψ(g−1i ggi) =

n∑i=1

g−1i ggi∈H

ψ(g−1i ggi)

=∑

gi∈G/Hψ(g−1

i ggi) =1

|H|∑x∈G

ψ(x−1gx)

with

ψ : G→ F , g 7→

ψ(g) if g ∈ H,

0 if g /∈ H.

(d) By Proposition 6.7(b), we have

IndGH(W ⊕W ′) ∼= IndGH(W )⊕ IndGH(W ′)

if also W ′ is a finitely generated FH-module. In the case that F = C, thisimplies that we obtain a group homomorphism

indGH : R(H)→ R(G) , χ 7→ indGH(χ) ,

between the character rings.

59

Now let K 6 H 6 G and assume that W is a finitely generated CK-module. Parts (a) and (c) in Proposition 6.7 imply that we have a chain ofisomorphisms of left CG-modules,

IndGH(IndHK(W )) = CG⊗CH (CH ⊗CK W ) ∼= (CG⊗CH CH)⊗CK W

∼= CG⊗CK W = IndGK(W ) .

This transitivity property of induction for modules implies the transitivityproperty of induction for characters:

indGH ◦ indHK = indGK : R(K)→ R(G)

whenever K 6 H 6 G.

7.3 Example Let G = Sym(3) = {1, x, x2, y, xy, x2y} with x = (1, 2, 3) andy = (1, 2) and let H := Alt(3) = {1, x, x2} 6 G. The character table of H isgiven by

1 x x2

ψ1 1 1 1ψ2 1 ζ ζ2

ψ3 1 ζ2 ζ

with ζ := e2πi/3. The elements 1 and y form a set of coset representativesfor G/H and we can compute the character values of indGH(ψ) for the variousψ ∈ Irr(H) using the formula in Remark 7.2(c). Note that y−1xy = x2. Weinclude the computations of these characters in the character table of G:

1 y xχ1 1 1 1χ2 1 −1 1χ3 2 0 −1

indGH(ψ1) 2 0 2indGH(ψ2) 2 0 −1indGH(ψ3) 2 0 −1

Now it is easy to see that

indGH(ψ1) = χ1 + χ2 , indGH(ψ2) = χ3 , indGH(ψ3) = χ3 .

60

7.4 Proposition Let V and W be CG-modules and let χV and χW denotetheir respective characters. Then

(χV , χW )G = dimC HomCG(V,W ) .

Proof Exercise 5.5. Hint: Both sides of the equation additive in V andW with respect to direct sums. Therefore, it suffices to prove the equationfor simple CG-modules V and W . But in this case this follows immediatelyfrom distinguishing the two cases that V is isomorphic to W or not.

7.5 Theorem (Frobenius reciprocity) For ψ ∈ R(H) and χ ∈ R(G) onehas

(indGH(ψ), χ)G = (ψ, resGH(χ))H .

Proof Since (−,−) is additive in both arguments, and since indGH and resGHare group homomorphisms, it suffices to prove the equation for irreduciblecharacters ψ and χ. In this case there exists a CG-module V and a CH-module W such that χ = χV and ψ = χW . By Proposition 7.4, we have

(indGH(ψ), χ)G = dimC HomCG(CG⊗CH W,V )

and(ψ, resGH(χ))H = dimC HomCH(W,ResGH(V )) .

Therefore, it suffices to show that the C-vector spaces HomCG(CG⊗CHW,V )and HomCH(W,ResGH(V )) are isomorphic. Note that one has well-definedmutually inverse isomorphisms

HomCG(CG, V )∼↔ ResGH(V )

f 7→ f(1)

(a 7→ av)←7 v

of CH-modules, where the left CH-module structure of HomCG(CG, V ) comesfrom the right CH-module structure of CG, cf. Theorem 6.11(a). There-fore, it suffices to show that the C-vector spaces HomCG(CG⊗CH W,V ) andHomCH(W,HomCG(CG, V )) are isomorphic. But this is precisely the contentof Theorem 6.11(c).

61

7.6 Example Let G = Sym(4). The character table of G is given by

1 (ab)(cd) (ab) (abc) (abcd)χ1 1 1 1 1 1χ2 1 1 −1 1 −1χ3 2 2 0 −1 0χ4 3 −1 1 0 −1χ5 3 −1 −1 0 1

(a) Let H := Sym(3) 6 G. The character table of H together with therestrictions χi|H of the irreducible characters of G is given below:

1 (ab) (abc)ψ1 1 1 1ψ2 1 −1 1ψ3 2 0 −1

resGH(χ1) 1 1 1resGH(χ2) 1 −1 1resGH(χ3) 2 0 −1resGH(χ4) 3 1 0resGH(χ5) 3 −1 0

It follows immediately that

resGH(χ1) = ψ1

resGH(χ2) = ψ2

resGH(χ3) = ψ3

resGH(χ4) = ψ1 + ψ3

resGH(χ5) = ψ2 + ψ3

Frobenius reciprocity now immediately implies that

indGH(ψ1) = χ1 + χ4

indGH(ψ2) = χ2 + χ5

indGH(ψ3) = χ3 + χ4 + χ5

7.7 Theorem (Frobenius property) For ψ ∈ R(H) and χ ∈ R(G) onehas

indGH(ψ) · ψ = indGH(ψ · resGH(χ)) .

In particular, indGH(R(H)) is an ideal in the commutative ring R(G).

62

Proof For g ∈ G we have

(indGH(ψ) · χ)(g) = (indGH(ψ))(g) · χ(g) =1

|H|∑x∈G

ψ(x−1gx)χ(g)

=1

|H|∑x∈G

ψ(x−1gx)χ(x−1gx)

and

indGH(ψ · resGH(χ))(g) =1

|H|∑x∈G

x−1gx∈H

(ψ · resGH(χ))(x−1gx)

=1

|H|∑x∈G

x−1gx∈H

ψ(x−1gx) · χ(x−1gx)

=1

|H|∑x∈G

ψ(x−1gx)χ(x−1gx) .

This completes the proof of the theorem.

7.8 Definition Let H and U be subgroups of G. A double coset of G withrespect to U and H is an equivalence class for the equivalence relation on Gwhich is defined by

g ∼ g′ :⇐⇒ there exist u ∈ U and h ∈ H such that g′ = ugh.

The double coset of g, i.e., all elements g′ ∈ G which are equivalent to g, isUgH = {ugh | u ∈ U, h ∈ H}. We write U\G/H := {UgH | g ∈ G} for theset of double cosets of G with respect to U and H.

7.9 Remark Let U and H be subgroups of G. For an element g ∈ G we setgH := gHg−1 and Hg := g−1Hg.

(a) In the sequel, by abuse of notation, a summation of the form∑g∈U\G/H

means that the elements g run through a set of double coset representativesof G with respect to U and H.

(b) Every double coset in UgH ∈ U\G/H is a disjoint union of left cosetsin G/H and also a disjoint union of right cosets in U\G. If {g1, . . . , gn} ⊆ Gis a set of representatives of the double cosets of G with respect to U and

63

H then {g−11 , . . . , g−1

n } is a set of representatives of the double cosets of Gwith respect to H and U . In fact, the bijection G→ G, g 7→ g−1, maps thedouble coset UgH to the double coset Hg−1U . Moreover, if U is normal inG then UH is a subgroup of G and UgH = gUH for all g ∈ G. ThereforeU\G/H = G/UH. Similarly, if H is normal in G then U\G/H = UH\G.

(c) Let F be a field. Then

FG =⊕

g∈U\G/HF [UgH]

is a decomposition of the (FU, FH)-bimodule FG into (FU, FH)-subbimodules.

(d) Let F be a field, let W be an FH-module and let g ∈ G. We set

gW := Rescg−1 (W ) ,

the restriction of W along the isomorphism

cg−1 : gH∼−→ H , x 7→ g−1xg .

In other words, the F gH-module gW is equal to W as an F -vector space,and for x ∈ gH and w ∈ gW one has x ·w = (g−1xg)w. If ψ ∈ R(H) denotesthe character of W then the character of gW is denoted by gψ. It is given by

( gψ)(x) = ψ(g−1xg) , for x ∈ gH. (7.9.a)

This induces a ring isomorphism

cg,H : R(H)→ R( gH) , ψ 7→ gψ ,

with gψ defined as in (7.9.a) for arbitrary ψ ∈ R(H). The map cg,H is anisometry with respect to the Schur inner product: One has

( gψ1,gψ2) g

H= (ψ1, ψ2)H ,

for all ψ1, ψ2 ∈ R(H). In particular one has: ψ ∈ Irr(H) if and only ifgψ ∈ Irr( gH).

7.10 Theorem (Mackey decomposition formula) Let U and H be sub-groups of G, let F be a field, and let W be an FH-module. Then one hasan isomorphism

ResGU (IndGH(W )) ∼=⊕

g∈U\G/HIndU

U∩gH

(ResgHU∩

gH

( gW ))

of FU -modules.

64

Proof By the decomposition FG =⊕g∈U\G/H F [UgH] of FG into (FU, FH)-

subbimodules we obtain an isomorphism

ResGU (IndGH(W )) = FG⊗FH W ∼=⊕

g∈U\G/HF [UgH]⊗FH W

of FU -modules. Moreover, for fixed g ∈ G, it is an easy verification that themaps

F [UgH]⊗FH W∼↔ FU ⊗

F [U∩gH]

gW ,

ugh⊗ w 7→ u⊗ (ghg−1) · w ,ug ⊗ w ←7 u⊗ w ,

are well-defined, mutually inverse isomorphisms of FU -modules. Note thatone has to verify that the first map does not depend on the choice of u ∈ Uand h ∈ H when expressing the element ugh ∈ UgH.

7.11 Corollary Let U and H be subgroups of G and let ψ ∈ R(H). Then

resGU (indGH(ψ)) =∑

g∈U\G/HindU

U∩gH

(resgHU∩

gH

( gψ)) .

Proof If ψ is the character of a CH-module W then the equation followsimmediately from Theorem 7.10. If ψ ∈ R(H) is arbitrary then we can writeψ = ψ1−ψ2 with characters ψ1 and ψ2. Since both sides of the equation areadditive in ψ, the result follows.

7.12 Corollary Let U and H be subgroups of G, let ψ ∈ R(H), and letλ ∈ R(U). Then

indGU (λ) · indGH(ψ) =∑

g∈U\G/HindG

U∩gH

(resU

U∩gH

(λ) · resgHU∩

gH

( gψ)).

65

Proof Using the Frobenius property (cf. Theorem 7.7) and Mackey’s de-composition formula (cf. Corollary 7.11), we obtain

indGU (λ) · indGH(ψ) = indGU(λ · resGU (indGH(ψ))

)= indGU

(λ ·

∑g∈U\G/H

indUU∩

gH

(resgHU∩

gH

( gψ)))

= indGU( ∑g∈U\G/H

indUU∩

gH

(resUU∩

gH

(λ) · resgHU∩

gH

( gψ))

=∑

g∈U\G/HindG

U∩gH

(resU

U∩gH

(λ) · resgHU∩

gH

( gψ)),

and the corollary is proved.

7.13 Corollary (Mackey’s irreducibility criterion) Let ψ ∈ Irr(H).

(a) The character indGH(ψ) is irreducible if and only if(resH

H∩gH

(ψ), resgHH∩

gH

( gψ))H∩

gH

= 0 , for all g ∈ GrH.

(b) Assume that H is normal in G. Then indGH(ψ) is irreducible if andonly if

ψ 6= gψ , for all g ∈ GrH.

Proof (a) Since the degree of indGH(ψ) is positive, Proposition 2.18 impliesthat indGH(ψ) is irreducible if and only if (indGH(ψ), indGH(ψ))G = 1. Moreover,by Frobenius reciprocity (cf. Theorem 7.5) and the Mackey decompositionformula (cf. Corollary 7.11), we have(

indGH(ψ), indGH(ψ))G

=(ψ, resGH(indGH(ψ))

)H

=(ψ,

∑g∈H\G/H

indHH∩

gH

(resgHH∩

gH

( gψ)))H

=∑

g∈H\G/H

(resH

H∩gH

(ψ), resgHH∩

gH

( gψ))H∩

gH.

The summand in the last sum which is indexed by g = 1 is equal to (ψ, ψ) =1. Since all the other summands are non-negative integers, the result follows.

(b) This follows immediately from Part (a).

66

Exercises

1. Let G be a finite group and let S be a finite G-set with G-orbits S1, . . . , Sn.For each i = 1, . . . , n, let Hi be the stabilizer in G of some element si ∈ Si. Showthat the permutation character χS introduced in an earlier problem satisfies

χS =

n∑i=1

indGHi1Hi .

2. Compute the character table of Sym(5) (or similarly for Sym(6)) using thefollowing approach. For two partitions λ = (λ1, λ2, . . . , . . .) and µ = (µ1, µ2, . . .)of 5, define the dominance order

λ E µ :⇐⇒j∑i=1

λi 6j∑i=1

µi for all j.

This defines a partial order on all partitions of 5. For every partition λ considerthe corresponding Young subgroup Sλ which is defined as the set of all permu-tations which permute the first λ1 elements among themselves, the second λ2

elements among themselves, etc. The subgroup Sλ of Sym(5) is clearly isomor-phic to Sym(λ1) × Sym(λ2) × · · · . Now start with the unique maximal partition

λ = (5) and compute the permutation character πλ := indSym(5)Sλ

(1). Then go downthe poset of partitions, at every step taking a new λ, maximal among the parti-tions that were not considered yet, compute πλ and check which of the previouslyconstructed irreducible characters are constituents of πλ and compute their multi-plicity in πλ. You should be left with precisely one irreducible summand that hasnot been constructed before. Call this character χλ and proceed down the posetof partitions. (One can prove that this process works for Sym(n) for arbitrary n.)

3. (a) Assume that S is a transitive G-set with |S| > 1, let s ∈ S and setH := stabG(s) < G. By Problem 1 and Frobenius reciprocity, one has (χS , 1) =(indGH(1H), 1G)G = (1H , resGH(1G))H = (1H , 1H)H = 1. Therefore, χS = 1G+ψ forsome character ψ ofG, which does not contain the trivial character as a constituent.Show that the following are equivalent:

(i) The character ψ is irreducible.(ii) The G-set S × S (with g(s1, s2) := (gs1, gs2) for g ∈ G and s1, s2 ∈ S)

has exactly two orbits.(iii) One has (χ2

S , 1) = 2(iv) S is a doubly transitive G-set (i.e., for any two pairs (s1, s2), (s′1, s

′2) ∈

S × S with s1 6= s2 and s′1 6= s′2 there exists g ∈ G with gs1 = s′1 and gs2 = gs′2).

67

(v) One has |H\G/H| = 2.

(b) Let n ∈ N and let λ = (n− 1, 1). Show that πλ (from Exercise 2) is equal

to indSym(n)Sym(n−1)(1) and equal to the permutation character of Sn. Also show that

πλ = 1 + χ for an irreducible character χ of Sym(n).

4. Let G be a finite group, let H be a subgroup of G, and let g1, . . . , gn ∈ Gbe coset representatives for G/H. Moreover, let V be a finitely generated FG-module and let W be a finitely generated FH-module. Show that the followingare equivalent:

(i) The FG-modules V and IndGH(W ) are isomorphic.(ii) There exists an FH-submodule W ′ of V , with W ′ ∼= W , such that V =⊕ni=1 giW

′.

5. Let G be a finite group, U 6 H 6 G, and g ∈ G. Furthermore, letcg,H : R(H) → R( gH), ψ 7→ gψ, be the conjugation map with gψ(x) := ψ(g−1xg)for x ∈ gH := gHg−1.

(a) Show that ch,H is the identity if h ∈ H and that cg′, gH ◦ cg,H = cg′g,H for

g, g′ ∈ G.

(b) Show that cg,U ◦ resHU = resgHgU◦ cg,H and cg,H ◦ indHU = ind

gHgU◦ cg,U .

6. Let G be a finite group and let U,H 6 G. Let g1, . . . , gn ∈ G be a setof representatives of U\G/H and for each i = 1, . . . , n let ui,1, . . . , ui,ki be a setof representatives for U/U ∩ giH. Show that the elements ui,jgi, i = 1, . . . , n,j = 1, . . . , ki form a set of representatives for G/H.

7. (a) Let G be a finite group, g1, . . . , gh a set of representatives for theconjugacy classes of G,

n := |{χ ∈ Irr(G) | χ(g) /∈ R for some g ∈ G}| ,

and m ∈ N with 2m = n. Show that the determinant d of the character tablesatisfies

d2 = (−1)m|CG(g1)| · · · |CG(gh)| .(b) Let p be an odd prime and let ζ := e2πi/p. Show that√

(−1)p−12 p ∈ Q(ζ) .

8. Let H be a subgroup of a finite group G and let ψ ∈ CF(H,F ) be a classfunction of H with values in a field F . For g ∈ G set

indGH(ψ)(g) :=1

|H|∑x∈G

ψ(x−1gx) .

68

Show that this defines an F -linear map

indGH : CF(H,F )→ CF(G,F ) ,

and verify that all the rules for induction of virtual characters is also true inductionof class functions (transitivity, Frobenius Reciprocity, Frobenius property, Mackeyformula, what else?)

69

8 Frobenius Groups

In this section we prove Frobenius’ Theorem, Theorem 8.1, on (what are nowcalled) Frobenius groups. The statement of this theorem is purely in grouptheoretic terms. Its proof involves characters and until now there is nowproof known that avoids representation theory and character theory.

For a large part of this section we assume that G is a finite group andthat H is a subgroup of G which satisfies

H ∩ gH = {1} , for every g ∈ GrH. (8.0.a)

We will prove the following theorem.

8.1 Theorem Let G and H satisfy (8.0.a). Then the subset N of G, definedas

N := Gr( ⋃g∈GrH

gH r {1})),

is a normal complement of H in G, i.e., N is a normal subgroup of G,HN = G and H ∩N = {1}.

8.2 Definition If G and H satisfy (8.0.a) then G is called a Frobenius group,H is called a Frobenius complement and N is called the Frobenius kernel ofG.

8.3 Remark A deep theorem from 1960 due to John Thompson states thatthe subgroup N is nilpotent (i.e., a direct product of p-subgroups) and in factthe largest normal nilpotent subgroup of G with respect to inclusion. Thisjustifies the terminology ‘the Frobenius kernel’ of G. Moreover, one can showthat gcd(|H|, |G/H|) = 1 (i.e., H is a Hall-subgroup of G, see Exercise 1).Thus the Frobenius complement H is uniquely determined up to conjugationin G by a Theorem of Schur and Zassenhaus.

We will prove Theorem 8.1 in three steps. First we establish two lemmas.

8.4 Lemma Assume that G is a finite group, that H is a subgroup of Gsatisfying (8.0.a) and let ψ ∈ R(H) with ψ(1) = 0. Then the following hold:

(a) resGH(indGH(ψ)) = ψ.

(b) For every ψ′ ∈ R(H) one has (ψ, ψ′)H = (indGH(ψ), indGH(ψ′))G.

70

Proof (a) We need to show that

1

|H|∑x∈G

ψ(xhx−1) = ψ(h) ,

for every h ∈ H. If h = 1 then both sides are equal to 0. If h 6= 1 and x ∈ Gthen xhx−1 belongs to H if and only if x ∈ H. This implies that the lefthand side is equal to 1

|H|∑x∈H ψ(xhx−1) = ψ(h), since ψ is a class function

on H.(b) This follows immediately from Part (a) and Frobenius reciprocity

(cf. Theorem 7.5):

(indGH(ψ), indGH(ψ′))G =(resGH(indGH(ψ)), ψ′

)H

= (ψ, ψ′)H .

8.5 Lemma Assume that G, H and N are as in Theorem 8.1 and let ϕ ∈Irr(H). Then there exists an irreducible character ϕ∗ ∈ Irr(G) such thatresGH(ϕ∗) = ϕ and N ⊆ ker(ϕ∗).

Proof We first assume that ϕ = 1H is the trivial character of H. Then thetrivial character ϕ∗ = 1G of G has all the desired properties.

From now on we assume that ϕ 6= 1H and set ψ := ϕ− ϕ(1) · 1H . Then,by Frobenius reciprocity and by evaluation at the identity element of H wehave

(indGH(ψ), 1G)G = (ψ, 1H)H = −ϕ(1) and ψ(1) = 0 .

Thus, the virtual character ϕ∗ := indGH(ψ) + ϕ(1)1G ∈ R(G) satisfies

(ϕ∗, 1G) = 0 and ϕ∗(1) = ϕ(1) .

This implies that (indGH(ψ), indGH(ψ))G = (ϕ∗, ϕ∗)G + ϕ(1)2. On the otherhand, by Lemma 8.4(b), we obtain

(indGH(ψ), indGH(ψ))G = (ψ, ψ)H = (ϕ, ϕ)H + ϕ(1)2 = 1 + ϕ(1)2 .

This implies that (ϕ∗, ϕ∗)G = 1. Since also ϕ∗(1) = ϕ(1) > 0, Propo-sition 2.18 implies that ϕ∗ is an irreducible character of G. Moreover,Lemma 8.4(a) implies that resGH(indGH(ψ)) = ψ, so that

resGH(ϕ∗ − ϕ(1) · 1G) = ϕ− ϕ(1) · 1H .

71

Since resGH(1G) = 1H , we obtain resGH(ϕ∗) = ϕ.Finally, we show that N ⊆ ker(ϕ∗). We need to show that ϕ∗(x) = ϕ∗(1)

for every x ∈ N . So let x ∈ N . We may assume that x 6= 1. Then x isnot conjugate in G to any element of H. Therefore, (indGH(ψ))(x) = 0 andϕ∗(x) = (indGH(ψ) + ϕ(1) · 1G)(x) = ϕ(1) = ϕ∗(1). This completes the proofof the lemma.

Proof of Theorem 8.1. For each ϕ ∈ Irr(H) we choose an element ϕ∗ ∈Irr(G) which satisfies resGH(ϕ∗) = ϕ and N ⊆ ker(ϕ∗), as guaranteed byLemma 8.5. Then we set

M :=⋂

ϕ∈Irr(H)

ker(ϕ∗) .

Since each ker(ϕ∗) is a normal subgroup of G containing N , also M is anormal subgroup of G with N ⊆ M . Moreover, since resGH(ϕ∗) = ϕ, weobtain

M ∩H =⋂

ϕ∈Irr(H)

(ker(ϕ∗) ∩H) =⋂

ϕ∈Irr(H)

ker(ϕ) = {1} . (8.5.a)

The last equation follows immediately from the following line of argument:First observe that ker(χ) ∩ ker(χ′) = ker(χ + χ′) for all characters χ and χ′

of G. This implies⋂ϕ∈Irr(H) ker(ϕ) = ker(

∑ϕ∈Irr(H) ϕ(1)ϕ). But the latter

character is the regular character ρH of H and one has ker(ρH) = {1}, sinceρH(h) = 0 for all 1 6= h ∈ H. This proves the last equation in (8.5.a).Since M is normal in G, the equation M ∩H = {1} also implies M ∩ gH =g(M ∩H) = {1} for every g ∈ G. This implies that M does not contain anynon-trivial element which is conjugate to an element of H. In other words,we have M ⊆ N . Altogether we obtain M = N . Thus, N is a normalsubgroup of G and N ∩H = {1}.

We still need to show that NH = G. Note that the hypothesis in (8.0.a)on H implies that NG(H) = H. Thus, H has precisely [G : NG(H)] = [G :H] conjugate subgroups and any two distinct conjugates of H have trivialintersection. In fact if g1 and g2 are elements in G such that g1H 6= g2H then

g−11 g2 /∈ NG(H) = H and g−1

1 ( g1H ∩ g2H) = (H ∩ g−11 g2H) = {1}. This implies

that g1H ∩ g2H = {1}. Therefore, we obtain

|N | = |G| − |⋃g∈G

( gH r {1})| = |G| − [G : H](|H| − 1) = [G : H] .

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As a consequence we have |HN | = |H||N |/|H ∩ N | = |G|/1 = |G| andHN = G. This completes the proof of the theorem.

The following corollary describes a situation that leads to the hypothesisin (8.0.a). This was the original situation that Frobenius wanted to under-stand.

8.6 Corollary Let G be a finite group and let X be a transitive G-set suchthat no non-identity element of G fixes two or more elements of X. LetN denote the set consisting of the identity element of G together with allelements of G which do not fix a single element of X. Then N is a normalsubgroup of G which still acts transitively on X.

Proof We fix an arbitrary element x of X and set H := stabG(x), thestabilizer of x in G. For every element y ∈ X there exists g ∈ G suchthat y = gx. This implies stabG(y) = gH. Since no non-identity elementof G fixes more than one element of X we obtain H ∩ gH = {1} for allg ∈ G r H. Moreover, an element of G fixes at least one element of X ifand only if it is conjugate to an element in H. This implies that the set Nas defined in Theorem 8.1 is equal to the set N as defined in this corollary.By Theorem 8.1 we obtain that N is a normal subgroup of G and thatNH = G. To show that N acts still transitively on X, note that G = NHimplies X = Gx = NHx = Nx, and the corollary is proved.

Exercises

1. Let G be a finite group and let H be a subgroup of G satisfying H ∩ gH ∩H = {1} for all g ∈ GrH. Show that gcd(|H|, [G : H]) = 1.

2. Let k be a commutative ring.

(a) Show that, for each a ∈ k× and b ∈ k, the map fa,b : k → k, x 7→ ax+ b, isbijective.

(b) Show that the maps in (a) form a group under composition. This group iscalled the affine linear group of k and is denoted by Aff(k).

(c) Show that Aff(k) is isomorphic to the semidirect product k o k×, wherethe unit group k× acts by multiplication on the additive group k.

(d) Show that if k is a finite field then Aff(k) is a Frobenius group.

3. Show that if N1 and N2 are nilpotent normal subgroups of a finite groupG then N1N2 is also a nilpotent normal subgroup of G. In particular there exists

73

a largest nilpotent normal subgroup in every finite group G, called the Fittingsubgroup of G and denoted by F (G).

74

9 Elementary Clifford Theory

Throughout this section, G denotes a finite group and N denotes a normalsubgroup of G.

Recall from Remark 5.13 that for every irreducible character ψ of N one hasan associated idempotent

eψ =ψ(1)

|N |∑x∈N

ψ(x−1)x ∈ Z(CN) .

Recall also from Remark 7.9(d) that G acts on Irr(N) by gψ(x) = ψ(g−1xg),for g ∈ G and x ∈ N . The following lemma shows that, for g ∈ G, theC-algebra isomorphism CN → CN , a 7→ gag−1, permutes the above idem-potents as expected.

9.1 Lemma Let ψ ∈ Irr(N) and let g ∈ G. Then geψg−1 = e g

ψin CN .

Proof We have

geψg−1 = g

(ψ(1)

|N |∑x∈N

ψ(x−1)x)g−1 =

ψ(1)

|N |∑x∈N

ψ(x−1)gxg−1

=gψ(1)

|N |∑y∈N

gψ(y−1)y = e gψ,

after substituting y = gxg−1, and the lemma is proven.

The following theorem describes the pattern that occurs when one re-stricts an irreducible character to a normal subgroup.

9.2 Theorem Let V be an irreducible CG-module and let χ ∈ Irr(G) denoteits character.

(a) If 0 6= W is a CN -submodule of V then∑g∈G gW = V .

(b) Let W ⊆ V be an irreducible CN -submodule of V and let ψ ∈ Irr(N)denote the character of W . Furthermore, let H := stabG(ψ) := {g ∈ G |gψ = ψ} denote the inertia group of ψ in G, and let 1 = g1 . . . , gn ∈ G be aset of representatives of G/H. Finally, for i = 1, . . . , n, set ψi := giψ and letVi := eψi · V be the ψi-homogeneous component of ResGN(V ). Then

V =n⊕i=1

Vi and resGN(χ) = en∑i=1

ψi

75

for some positive integer e, called the ramification index of χ over N .

(c) For each i = 1, . . . , n, the CN -submodule Vi of V is an irreducibleCHi-submodule, where Hi := giHg

−1i . Moreover, IndGHi(Vi)

∼= V for alli = 1, . . . , n.

Proof (a) Set V ′ :=∑g∈G gW . Then V ′ 6= {0}, since W 6= {0}. Moreover,

it is immediate that V ′ is a CG-submodule of V . Since V is a simple CG-module, we obtain V ′ = V as claimed.

(b) First note that, since eψi · eψj = 0 for i 6= j in {1, . . . , n}, the sumV1 + · · ·+ Vn of CN -submodules of V is a direct sum.

Next we will show that if g ∈ G satisfies ggiH = gjH then gVi = Vj. Infact, in this case there exists h ∈ H such that ggi = gjh and we obtain

gVi = geψiV = geψig−1gV = ggieψg

−1i g−1gV = gjheψh

−1g−1j gV = eψjV = Vj ,

since gV = V and since gjheψh−1g−1

j = e gjhψ = eψj by Lemma 9.1. Since

g1 = 1 ∈ H, the CN -submodule V1 contains W . Therefore, V1 6= {0} andPart (a) applied to V1 implies that V =

∑g∈G gV1 = V1 + · · · + Vn. Thus

we have shown that V decomposes into the direct sum V = V1 ⊕ · · · ⊕ Vn ofCN -submodules. This implies that resGN(χ) is the sum of the characters ofthe CN -modules Vi. Since Vi is the ψi-homogeneous component of ResGN(V ),the character of Vi is equal to ni · ψi with ni = dimC(Vi)/ψi(1). But sinceVi = giV1, we have dimC Vi = dimC V1, and since ψi(1) = giψ(1) = ψ(1), thepositive integers ni, i = 1, . . . , n, are all equal. If we denote this number bye we obtain

resGN(χ) = e(ψ1 + · · ·+ ψn)

and part (b) is proven.(c) It follows from the proof of Part (b) that, for i ∈ {1, . . . , n} and every

h ∈ H and vi ∈ Vi, one has

(gihg−1i )vi = (gih)(g−1

i vi) ∈ gihV1 = giV1 = Vi .

This implies that Vi is a CHi-submodule of V . Moreover, the function

CG⊗CHi Vi → V , a⊗ vi 7→ avi ,

is a well-defined CG-module homomorphism. It is surjective, since it con-tains

∑g∈G gVi = V1 + · · · + Vn, by the proof of part (b). Moreover, since

dimC(CG ⊗CHi Vi) = [G : Hi] · dimC Vi = [G : H] · dimC V1 = n dimC V1 =

76

dimC(V1 ⊕ · · · ⊕ Vn), we obtain that the above map is an isomorphism. Fi-nally, since V ∼= IndGHi(Vi) is irreducible, also the CHi-module Vi must beirreducible.

9.3 Example LetG = Sym(4) andN = V4 = {1, (1, 2)(3, 4), (1, 3)(2, 4), (1, 4)(2, 3)}.The character tables of G and N are given by

G 1 (ab)(cd) (ab) (abc) (abcd)χ1 1 1 1 1 1χ2 1 1 −1 1 −1χ3 2 2 0 −1 0χ4 3 −1 1 0 −1χ5 3 −1 −1 0 1

andN 1 (1, 2)(3, 4) (1, 3)(2, 4) (1, 4)(2, 3)ψ1 1 1 1 1ψ2 1 1 −1 −1ψ3 1 −1 1 −1ψ4 1 −1 −1 1

One can read from the character tables that

resGN(χ1) = ψ1 ,

resGN(χ2) = ψ1 ,

resGN(χ3) = 2 · ψ1 ,

resGN(χ4) = ψ2 + ψ3 + ψ4 ,

resGN(χ5) = ψ2 + ψ3 + ψ4 .

One can also read from the character table ofN that Irr(N) has two orbits un-der G-conjugation, namely {ψ1} and {ψ2, ψ3, ψ4}. Moreover, stabG(ψ1) = Gand stabG(ψ2) = H1 := N · 〈(1, 2)〉, stabG(ψ3) = N · 〈(1, 3)〉 and stabG(ψ4) =N ·〈(1, 4)〉. Note that the latter three subgroups of G have order 8. Moreover,Theorem 9.2(c) implies that χ4 = indGH1

(λ) for some λ ∈ Irr(H1). Comparingdegrees, we obtain λ(1) = 1.

9.4 Remark (a) Let K 6 H 6 G, ψ, ψ′ ∈ R(K), χ ∈ R(H) and g ∈ G.Then it is easily verified (see Exercise 7.4) that

gindHK(ψ) = ind

gHgK

( gψ) ,gresHK(χ) = res

gHgK

( gχ) , ( gψ, gψ′) gK

= (ψ, ψ′)K .

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(b) The following notation will be used in the next theorem. For ψ ∈Irr(N) we set

Irr(G|ψ) := {χ ∈ Irr(G) | (resGN(χ), ψ)N > 0} .

Note that by the above, Irr(G|ψ) = Irr(G| gψ) for all g ∈ G. Moreover, byTheorem 9.2, one has Irr(G|ψ) ∩ Irr(G|ψ′) = ∅, whenever ψ and ψ′ are notin the same G-orbit of Irr(N). Moreover, by Frobenius reciprocity, everyχ ∈ Irr(G) belongs to a subset Irr(G|ψ) for some ψ ∈ Irr(N). Altogether,restriction and induction induce mutually inverse bijections between the setof subsets Irr(G|ψ) of Irr(G) and the set of G-orbits of Irr(N).

(c) In the proof of the next theorem we will need the following result. LetH 6 G and let χ and χ′ be characters of G. Then

(χ, χ′)G 6(resGH(χ), resGH(χ′)

)H.

In fact, let V and V ′ be CG-modules with characters χ and χ′, respec-tively. Then (χ, χ′)G = dimC HomCG(V, V ′) and

(resGH(χ), resGH(χ′)

)H

=

dimC HomCH(V, V ′). But, clearly one has an inclusion HomCG(V, V ′) ⊆HomCH(V, V ′).

9.5 Theorem Let ψ ∈ Irr(N) and set H := stabG(ψ). Write

indHN(ψ) = m1θ1 + · · ·+mrθr

with m1, . . . ,mr ∈ N and with pairwise distinct θ1, . . . , θr ∈ Irr(H).

(a) One has Irr(H|ψ) = {θ1, . . . , θr}, resHN(θi) = miψ for i = 1, . . . , r, andm2

1 + · · ·+m2r = [H : N ].

(b) The function θ 7→ indGH(θ) defines a bijection f : Irr(H|ψ)→ Irr(G|ψ).

Proof (a) The first statement follows immediately from Frobenius reci-procity. The second statement is immediate with Frobenius reciprocity andTheorem 9.2(b), since ψ is stable under H. In order to prove the last state-ment in Part (a), consider the equations

resHN(indHN(ψ)) = resHN(m1θ1 +mrθr) = (m21 + · · ·+m2

r)ψ .

Since the degree of the left hand side is equal to [H : N ]ψ(1) and the degreeof the right hand side is equal to (m2

1 + · · ·+m2r)ψ(1), the result follows.

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(b) For i = 1, . . . , r we set χi := indGH(θi). For g ∈ GrH, Remark 9.4(a),(c), and Part (a) imply

0 6(( gθi)|H∩ gH , θi|H∩ gH

)H∩

gH6(( gθi)|N , θi|N

)N

=(g(θi|N) , θi|N

)N

= m2i (

gψ , ψ)N = 0 ,

since gψ 6= ψ. This implies that each term above is equal to 0. By Mackey’sirreducibility criterion for induced characters, cf. Corollary 7.13, we obtainthat χi ∈ Irr(G). Moreover,

(χi|N , ψ)N =(χi, indGN(ψ)

)G

=(χi, indGH(indHN(ψ))

)G

=r∑i=1

mi

(χi, indGH(θi)

)G

>(χi, indGH(θi)

)G

= (χi, χi)G = 1 .

Thus, the map f in the theorem takes values in the set Irr(G|ψ) as claimed.To see that the function f is surjective, let χ ∈ Irr(G|ψ). Then, by Frobeniusreciprocity, χ is a constituent of indGN(ψ) = indGH(indHN(ψ)) = indGH(m1θ1 +· · ·+mrθr) = m1χ1 + · · ·+mrχr and χ = χi = f(θi) for some i ∈ {1, . . . , r}.Finally we will show that the map f is injective. Assume this is not the case.Then, after renumbering if necessary, we obtain that χ1 = χ2. This implies

(indGN(ψ), indGN(ψ)

)G

=r∑

i,j=1

(miχi , mjχj)G

> m21 + 2m1m2 +m2

2 + · · ·+m2r > m2

1 + · · ·+m2r

= [H : N ] .

On the other hand, the Mackey decomposition formula implies(indGN(ψ), indGN(ψ)

)G

=(ψ, resGN(indGN(ψ))

)G

=∑

g∈G/N

(ψ, indN

N∩gN

(resgNN∩

gN

( gψ)))N

=∑

g∈G/N(ψ, gψ)N =

∑g∈H/N

(ψ, gψ)N = [H : N ] .

This is a contradiction, and the theorem is proven.

79

9.6 Corollary Assume that [G : N ] = 2, that ψ ∈ Irr(N) and χ ∈ Irr(G|ψ).

(a) The following are equivalent:

(i) stabG(ψ) = N .

(ii) indGN(ψ) = χ.

(iii) resGN(χ) = ψ + ψ′ for some ψ 6= ψ′ ∈ Irr(N).

(iv) |Irr(G|ψ)| = 1.

In this case, the G-orbit of ψ is equal to {ψ, ψ′} and indGN(ψ′) = χ.

(b) The following are equivalent:

(i) stabG(ψ) = G.

(ii) indGN(ψ) = χ+ χ′ for some χ 6= χ′ ∈ Irr(G).

(iii) resGN(χ) = ψ.

(iv) |Irr(G|ψ)| = 2.

In this case, one has Irr(G|ψ) = {χ, χ′} and resGN(χ′) = ψ.

Proof First note that there are only two possibilities for stabG(ψ), namelyeither stabG(ψ) = N or stabG(ψ) = G. Also note that in any of these twocases the Mackey decomposition formula implies

resGN(indGN(ψ)) = ψ + gψ (9.6.a)

with g ∈ GrN .

(a) Assume that stabG(ψ) = N . Then Theorem 9.5(b) implies (ii) andEquation (9.6.a) implies (iii). Moreover, Theorem 9.5(b) implies (iv). Now,(ii) and Equation (9.6.a) imply that ψ′ = gψ with g ∈ G r N , and Re-mark 9.4(a) implies that indGNψ

′ = indGN( gψ) =gindGN(ψ) = gχ = χ.

(b) Now assume that stabG(ψ) = G. Then Theorem 9.5(a) implies r = 2and m1 = m2 = 1, since these are the only solutions to m2

1 + · · · + m2r =

[G : N ] = 2. This immediately implies (ii) and (iv). Moreover, with Theo-rem 9.5(a) this also implies (iii). Now, since χ′ occurs in indGN(ψ), Frobeniusreciprocity implies that ψ occurs in resGN(χ′), so that χ′ ∈ Irr(G|ψ). Thus, ev-erything that was proved for χ also holds for χ′. In particular, resGN(χ′) = ψ.

Finally, since each of the conditions (ii), (iii), and (iv) in (a) excludes thecorresponding condition in (b) and vice-versa, each of the conditions (ii), (iii)and (iv) also implies (i) in Parts (a) and (b).

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9.7 Example Consider G = Sym(5) and N = Alt(5). In a homework prob-lem we determined the character degrees of N = Alt(5) as 1, 3, 3, 4, 5. Fromthe character table of Alt(5) it is also easy to see that the two irreduciblecharacters of degree 3 are G-conjugate. Since conjugate characters have thesame degree, the characters of degree 1, 4 and 5 form G-orbits of size 1.Using Corollary 9.6 this implies that the character degrees of G = Sym(5)are given by 1, 1, 4, 4, 5, 5 and 6.

Exercises

1. Let G be a finite group, let N be a normal subgroup of G and assumethat χ is an irreducible character of G satisfying (resGN (χ), 1N )N 6= 0. Show thatN 6 ker(χ).

2. Let G be a finite group, let N be a normal subgroup, let ψ be an irreduciblecharacter of N and let χ ∈ Irr(G|ψ). Show that ψ(1) divides χ(1).

3. Let G be a finite group, let p be a prime, and let P be a Sylow p-subgroupof G. Assume that χ(1) is a power of p for every χ ∈ Irr(G). Show that P has anormal abelian complement, i.e., that there exists a normal abelian subgroup Nof G such that P ∩ N = {1} and PN = G. (Hint: Use the equation |G| = [G :G′]+

∑χIrr(G),χ(1)6=1 χ(1)2 to show that G has a normal subgroup of index p. Then

use induction on |G|.)

4. (Ito’s Theorem) Let A be an abelian normal subgroup of a finite group Gand let χ be an irreducible character of G. Then χ(1) divides [G : A]. (Hint: UseClifford Theory to reduce the problem to Exercise 3.6.)

81

10 Artin’s and Brauer’s Induction Thoerems

Throughout this section, G denotes a finite group.

10.1 Definition Let 2 6 n ∈ N and let n = pe11 · · · perr the prime decom-position of n, with pairwise distinct primes p1, . . . , pr and positive integerse1, . . . , er. We define

µ(n) :=

0 if ei > 2 for some i ∈ {1, . . . , r};(−1)r if ei = 1 for all i =∈ {1, . . . , r}.

For completeness, one defines µ(1) := 1. Altogether we have defined a func-tion µ : N→ {−1, 0, 1}. This function is called the (number theoretic) Mobiusfunction.

10.2 Lemma For every positive integer n one has

∑d|nµ(d) =

1 if n = 1;

0 if n > 1.

Here, the sum runs through the set of positive integers d which divide n.

Proof If n = 1 the assertion is true by inspection. Assume now that n > 1and let n = pe11 · · · perr be its prime decomposition. Then, r > 1 and∑

d|nµ(d) =

∑(f1,...,fr)06fi6ei

µ(pf11 · · · pfrr ) =∑

(f1,...,fr)06fi61

µ(pf11 · · · pfrr )

=∑

06k6r

(rk

)(−1)k = (1− 1)r = 0 .

Here, we counted tuples (f1, . . . , fr) ∈ {0, 1}r by first fixing the number k ofentries that are equal to 1 and then determining the number of such tuplesas the binomial coefficient

(rk

).

The explicit formula in the next theorem is due to Richard Brauer.

10.3 Theorem For every χ ∈ R(G) one has

χ =1

|G|∑

K6H6GH cyclic

|K| · µ([H : K]) · indGK(resGK(χ)) . (10.3.a)

82

In particular, one has

R(G) ⊆ 1

|G|∑H6GH cyclic

indGH(R(H)) and QR(G) =∑H6GH cyclic

indGH(QR(H)) ,

where QR(G) denotes the Q-span of Irr(G).

Proof First we prove the Equation (10.3.a) in the case where χ = 1G, thetrivial character of G. The right hand side of the equation evaluated at theelement x ∈ G is equal to

1

|G|∑

K6H6GH cyclic

|K| · µ([H : K]) · 1

|K|∑g∈G

g−1xg∈K

1

=1

|G|∑H6GH cyclic

∑g∈G

∑g−1〈x〉g6K6H

µ([H : K])

=1

|G|∑H6GH cyclic

∑g∈G

g−1〈x〉g6H

∑d|[H:g−1〈x〉g]

µ(d)

=1

|G|∑H6GH cyclic

∑g∈G

g−1〈x〉g6H

δH,g−1〈x〉g

=1

|G|∑g∈G

1 = 1 ,

and the equation holds in this case. For the second equality we used thata cyclic group of order n has precisely one subgroup of order d, for everydivisor d of n. For the third equality we used Lemma 10.2.

83

Next, let χ ∈ R(G) be arbitrary. Using the formula for trivial characterand the Frobenius property in Theorem 7.7, we obtain

χ = χ · 1G= χ ·

∑K6H6GH cyclic

|K| · µ([H : K]) · indGK(resGK(1G))

=∑

K6H6GH cyclic

|K| · µ([H : K]) · (χ · indGK(1K))

=∑

K6H6GH cyclic

|K| · µ([H : K]) · indGK(resGK(χ) · 1K)

=∑

K6H6GH cyclic

|K| · µ([H : K]) · indGK(resGK(χ)) ,

proving Equation (10.3.a) in the general case. The last two statements inthe Theorem are immediate consequences of the formula for χ in Equa-tion (10.3.a).

The following immediate corollary is often referred to as Artin’s inductiontheorem. It was first proved by Emil Artin in a different way, before Brauerfound the explicit formula in the above theorem.

10.4 Corollary (Artin’s induction theorem) For every χ ∈ R(G) thereexist cyclic subgroups H1, . . . , Hn of G (not necessarily distinct), rationalnumbers a1, . . . , an, and irreducible characters ϕi ∈ Irr(Hi), i = 1, . . . , n,such that χ can be expressed as

χ =n∑i=1

ai · indGHi(ϕi) .

Proof This follows immediately from the explicit formula for χ in Equa-tion (10.3.a).

10.5 Definition A finite group G is called supersolvable if it has a normalseries

1 = G0 6 G1 6 · · · 6 Gn = G , (Gi E G)

such that Gi/Gi−1 is cyclic for all i = 1, . . . , n.

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10.6 Remark (a) Every finite nilpotent group (i.e., direct product of p-groups) is supersolvable. On the other hand, every supersolvable group issolvable.

(b) Subgroups and factor groups of supersolvable groups are again super-solvable. This is proved in the same way as the analogous result for solvablegroups.

(c) The group Alt(4) is solvable but not supersolvable. In fact, Alt(4)does not have a normal cyclic subgroup.

10.7 Definition (a) A character χ of G is called monomial if χ can be writ-ten as a sum of characters of the form indGH(ϕ) with H 6 G and ϕ ∈ Irr(H)of degree 1. This is equivalent to requiring that there exists a representationof G with character χ which takes values in the subgroup of monomial ma-trices, i.e., matrices which have precisely one non-zero entry in each row andeach column.

(b) The group G is called monomial if every character of G is monomial.This is equivalent to the condition that every irreducible character of G isof the form indGH(ϕ) with H 6 G and ϕ ∈ Irr(H) of degree 1. Obviously,abelian groups are monomial. But for instance, Alt(5) is not monomial (sinceit has an irreducible character of degree 3, but no subgroup of order 20. Ingeneral, one can show that every monomial group is solvable, but we will notuse this result.

10.8 Remark Let N 6 H 6 G with N E G and let ψ ∈ R(H/N). Then

infGG/N(indG/NH/N(ψ)) = indGH(infHH/N(ψ)) .

This is an easy verification, noting that if g1, . . . , gn ∈ G is a set of repre-sentatives of G/H then g1N, . . . , gnN ∈ G/N is a set of representatives of(G/N)/(H/N).

10.9 Proposition (a) If G is supersolvable and not abelian then G has anormal abelian subgroup A which is not contained in Z(G).

(b) If G is supersolvable then G is monomial.

Proof (a) Set Z := Z(G). Then G/Z is a non-trivial supersolvable group.Therefore, it has a non-trivial normal cyclic subgroup A/Z. Since Z 6 Z(A)and A/Z is cyclic, the group A is abelian, and it is clearly not contained inZ.

85

(b) We prove the statement by induction on |G|. If |G| = 1 then Gis monomial. Assume that |G| > 1. We may assume that G is not abelian,since otherwise G is clearly monomial. Let χ ∈ Irr(G). We need to show thatχ = indGK(ϕ) for someK 6 G and ϕ ∈ Irr(K) with ϕ(1) = 1. By Remark 10.8we may also assume that χ is faithful, i.e., ker(χ) = 1. Let ∆: G→ GLn(C)be a representation with character χ. Then ∆ is injective. By Part (a)there exists a normal abelian subgroup A of G which is not contained inZ(G). Let a ∈ A r Z(G). Since a /∈ Z(G) and since ∆ is injective, wehave ∆(a) /∈ Z(∆(G)). This implies that ∆(a) is not a scalar matrix. ByTheorem 9.2(b), we have resGA(χ) = e

∑g∈H/A

gψ, for some e ∈ N and someψ ∈ Irr(A), where H := stabG(ψ). Note that ψ is a character of degree 1,since A is abelian. Since ∆(a) is not a scalar matrix, we obtain that H < Gis a proper subgroup. Theorem 9.2(c) implies that χ = indGH(θ) for someθ ∈ Irr(H). By induction, θ = indHK(ϕ) for some K 6 H and ϕ ∈ Irr(K) withϕ(1) = 1. Altogether, we obtain χ = indGH(θ) = indGH(indHK(ϕ)) = indGK(ϕ).This completes the proof.

10.10 Remark (a) In general, subgroups of monomial groups are not mono-mial, but factor groups of monomial groups are again monomial.

(b) One has a sequence of implications:

G supersolvable ⇒ G monomial ⇒ G solvable.

In fact, the first implication is the content of Proposition 10.9(b). The secondimplication can be found in [I, Theorem 5.12 and Corollary 5.13]. We willnot use the second one.

10.11 Definition Let H be a set of subgroups of G. We set

R(G,H) := {f ∈ CF(G,C) | f |H ∈ R(H) for all H ∈ H}=

⋂H∈H

(resGH)−1(R(H))

and

I(G,H) := 〈indGH(ψ) | H 6 G,ψ ∈ R(H)〉Z =∑H∈H

indGH(R(H)) .

Obviously, these two sets are additive subgroups of CF(G,C) and one has

I(G,H) ⊆ R(G) ⊆ R(G,H) . (10.11.a)

86

10.12 Lemma Let H be a set of subgroups of G.

(a) The set R(G,H) is a subring of the ring CF (G,C).

(b) The set I(G,H) is an ideal of the ring R(G,H).

(c) One has 1G ∈ I(G,H) if and only if I(G,H) = R(G) = R(G,H).

Proof (a) Clearly, the trivial character 1G is contained in R(G,H). More-over, since resGH : CF(G,C)→ CF(H,C) is a ring homomorphism for all H 6G and since R(H) is a subring of CF(H,C), the preimage (resGH)−1(R(H)) isa subring of CF(G,C). Since intersections of subrings of CF(G,C) are againsubrings, also R(G,H) is a subring of CF(G,C).

(b) Let χ ∈ R(G,H) and let∑H∈H aH indGH(ψH) ∈ I(G,H) with aH ∈ Z

and ψH ∈ R(H) for H ∈ H. Then the Frobenius Property, cf. Theorem 7.7,implies

χ ·∑H∈H

aH · indGH(ψH) =∑H∈H

aH · indGH(resGH(χ) · ψH) ∈ I(G,H) ,

since resGH(χ) ∈ R(H).

(c) This follows immediately from Part (b) and the chain of inclusions in(10.11.a).

10.13 Definition Let p be a prime. A finite group is called a p′-group ifits order is not divisible by p. A finite group E is called p-elementary ifE = C × P is the direct produce of a cyclic p′-subgroup C of E and ap-subgroup P of E. A group is called elementary if it is p-elementary forsome prime p. We write Ep(G), resp. E(G), for the set of p-elementary,resp. elementary, subgroups of G. Note that elementary groups are nilpotentand therefore supersolvable and monomial. Note also that subgroups andfactor groups of p-elementary groups are again p-elementary.

10.14 Theorem (Brauer’s induction theorem, 1947) (a) One has

I(G, E(G)) = R(G) = R(G, E(G)) .

(b) Every virtual character χ ∈ R(G) can be written in the form

χ =n∑i=1

ai · indGEi(ϕi)

87

with ai ∈ Z, Ei ∈ E(G) (not necessarily distinct), and ϕi ∈ Irr(Ei) of degree1, for i = 1, . . . , n.

(c) A class function f ∈ CF(G,C) is a virtual character of G if and onlyif, for every elementary subgroup E of G, its restriction f |E is a virtualcharacter of E.

10.15 Remark Part (b) in the above theorem (usually referred to as Brauer’sinduction theorem) follows immediately from Part (a) and Proposition 10.9(b).Part (c) in the above theorem (usually referred to as Brauer’s characteriza-tion of virtual characters) is just a reformulation of the equation R(G) =R(G, E(G)) in Part (a). Therefore, it suffices to prove Part (a) of the abovetheorem. In order to prove Part (a) it suffices, by Lemma 10.12, to showthat the trivial character 1G of G is contained in I(G, E(G)). The remainderof this section is devoted to prove this. The eventual proof will proceed byinduction on |G|.

10.16 Lemma (Banaschewski) Let X be a finite non-empty set and letR be a non-unitary subring of the ring F (X,Z) of functions from X to Z(i.e., a multiplicatively closed additive subgroup which does not contain theconstant function 1X with value 1). Then there exists an element x ∈ X anda prime p such that p | f(x) for all f ∈ R.

Proof For each x ∈ X, consider the additive subgroup

Ix := {f(x) | f ∈ R}

of Z. If, for some x ∈ X, the subgroup Ix is a proper subgroup of Z thenthere exists a prime p such that Ix ⊆ pZ and we are done. So assume thatIx = Z for all x ∈ X. We will derive a contradiction. In fact, in this case, forevery x ∈ X, there exists a function fx ∈ R such that fx(x) = 1. Thus, forevery x ∈ X, the function fx − 1X vanishes at x. This implies that the thefunction

∏x∈X(fx − 1X) is the 0-element in R. But expanding this product,

yields an expression of 1X as a Z-linear combination of products of some ofthe elements fx ∈ R. Since R is multiplicatively closed and also an additivesubgroup, we obtain 1X ∈ R. This is a contradiction, and the proof of thelemma is complete.

10.17 Definition We set

P (G) := 〈indGH(1H) | H 6 G〉Z ⊆ R(G) .

88

The set P (G) is called the ring of virtual permutation characters of G. (Itis the subgroup of R(G) generated by characters of permutation representa-tions, i.e., representations which take values in permutation matrices.) Moregenerally, if H is a set of subgroups of G we set

P (G,H) := 〈indGH(1H) | H ∈ H〉Z ⊆ P (G) .

Note that, for each H 6 G, the character indGH(1H) takes values in Z. ThusP (G) ⊆ F (G,Z). Note also that P (G,H) ⊆ I(G,H).

10.18 Lemma (a) The subgroup P (G) is a unitary subring of R(G).

(b) Assume that H is a set of subgroups of G which is closed under takingsubgroups, i.e., if H ∈ H and K 6 H then K ∈ H. Then P (G,H) is an idealof P (G).

Proof First note that 1G = indGG(1G) ∈ P (G). Moreover, for any subgroupsH and K of G, by Corollary 7.12, we have

indGH(1H) · indGK(1K) =∑

g∈H\G/KindG

H∩gK

(1H∩

gK

) .

This proves both Part (a) and Part (b) of the Lemma.

10.19 Definition Let p be a prime. A finite group H is called p-quasi-elementary if H has a normal cyclic p′-subgroup C such that H/C is a p-group. Note that this implies that H = C o P for any Sylow p-subgroupP of H. Moreover, H is called quasi-elementary if H is p-quasi-elementaryfor some prime p. We denote by QEp(G) and QE(G) the set of p-quasi-elementary and the set of quasi-elementary subgroups of G, respectively.Note that subgroups and factor groups of p-quasi-elementary groups areagain p-quasi-elementary. Also note that every p-elementary group is p-quasi-elementary. Thus, Ep(G) ⊆ QEp(G) and E(G) ⊆ QE(G).

10.20 Lemma Let x ∈ G and let p be a prime. Then there exists H ∈QEp(G) such that p - (indGH(1))(x).

Proof We can write 〈x〉 = C × D, with C a cyclic p′-subgroup and D acyclic p-subgroup of 〈x〉. Set N := NG(C) and note that 〈x〉 6 N . Since〈x〉/C ∼= D is a p-group, we may choose a Sylow p-subgroup H/C of N/Ccontaining 〈x〉/C. Then 〈x〉 6 H and H ∈ QEp(G).

89

We will show that p does not divide the character value (indGH(1H))(x).This character value is given by

(indGH(1H))(x) = |{gH ∈ G/H | g−1xg ∈ H}| .

But, for g ∈ G, we have

xgH = gH ⇐⇒ g−1xg ∈ H ⇐⇒ g−1〈x〉g 6 H =⇒ g−1Cg 6 H

=⇒ g−1Cg = C =⇒ g ∈ N =⇒ gH ∈ N/H .

Therefore, the character value (indGH(1H))(x) is equal to the number of fixedpoints of the 〈x〉-set N/H under the usual action by left multiplication. Forc ∈ C and n ∈ N we have cnH = n(n−1cn)H = nH, since n−1cn ∈ C. Thus,every element of N/H is fixed by C, and the number of 〈x〉-fixed points ofN/H is equal to the number of D-fixed points of N/H. Since D is a p-group,the number of D-fixed points of N/H is congruent to |N/H| modulo p. Since|N/H| is not divisible by p, the number of D-fixed points is not divisible byp, and the proof is complete.

10.21 Proposition (L. Solomon) The trivial character 1G of G is con-tained in P (G,QE(G)).

Proof The additive subgroup P (G,QE(G)) of the ring of functions F (G,Z)is multiplicatively closed by Lemma 10.18(b). Assume that 1G /∈ P (G,QE(G)).Then, Lemma 10.16 implies that there exists x ∈ G and a prime p such thatp | χ(x) for all χ ∈ P (G,QE(G)). But Lemma 10.20 implies that for thisprime p and this element x there exists H ∈ QEp(G) such that p does not di-vide (indGH(1H))(x). Since indGH(1H) ∈ P (G,QE(G)), this is a contradiction,and the proof of the proposition is complete.

10.22 Lemma Assume that G is p-nilpotent, i.e., G has a semidirect prod-uct decomposition G = N o P with P ∈ Sylp(G). Assume further thatϕ ∈ Irr(N) with ϕ(1) = 1 and gϕ = ϕ for all g ∈ G. Finally assume thatCN(P ) 6 ker(ϕ). Then ϕ = 1N , the trivial character of N .

Proof Set K := ker(ϕ) E N . Then ϕ induces an injective homomorphismϕ : N/K → C×. Let x ∈ N . We will show that x ∈ K.

Since gϕ = ϕ for all g ∈ G, we have K E G and obtain ϕ(g−1xgK) =ϕ(g−1xKg) = ϕ(x) = ϕ(xK). Thus, xK = g−1xgK = g−1xKg for every

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g ∈ G. This implies that G and by restriction also P acts on the coset xKby conjugation, for every x ∈ N . Since P is a p-group, the number of non-fixed points of xK under this action is divisible by p. Since |xK| = |K| is notdivisible by p, xK has at least one fixed point. Thus, xK ∩CN(P ) 6= ∅. ButCN(P ) 6 K by assumption. Thus, xK ∩K 6= ∅. This implies that xK = K.Since x ∈ N was arbitrary, we have K = N and ϕ = 1N .

Proof of Theorem 10.14. By Remark 10.15 it suffices to show that 1G ∈I(G, E(G)). We will show this by induction on |G|. Using the transitivityof induction together with Lemma 10.12(c), it suffices to show that 1G is aZ-linear combination of characters of the form indGH(ψ) with ψ ∈ Irr(H) andH < G. If G is not quasi-elementary, we are done by Proposition 10.21.Therefore, we may assume that G is quasi-elementary.

Then there exists a prime p and a cyclic normal p′-subgroup N of G suchthat G = NP and N ∩ P = {1} for all P ∈ Sylp(G). Set

Z := CN(P ) = {x ∈ N | xy = yx for all y ∈ P} = Z(G) ∩N.

And note that Z is normal in G. If Z = N then G is elementary and we aredone. Hence, we may assume that Z < N . We choose P ∈ Sylp(G) and setE := ZP < G. Note that E is elementary and that NE = G, since P 6 E..We can write indGE(1E) = 1G + θ with some character 0 6= θ of G and we letχ ∈ Irr(G) denote an irreducible constituent of θ. It suffices to show that χ isinduced from a proper subgroup, since then 1G = indGE(1E)− θ ∈ I(G, E(G))by previous arguments.

This is what we show in this last paragraph. We have

1N + resGN(θ) = resGN(1G + θ) = resGN(indGE(1E))

=∑

g∈N\G/EindN

N∩gE

(resgEN∩

gE

( g1E))

= indNN∩E(1N∩E) = indNZ (1Z) ,

since N ∩ E = N ∩ PZ = (N ∩ P )Z = Z. Therefore,

(1N + resGN(θ), 1N)N = (indNZ 1Z , 1N)N = (1Z , resNZ (1N))Z = (1Z , 1Z)Z = 1 .

This implies that (1N , resGN(θ))N = 0 and also that (1N , resGN(χ))N = 0. Nowlet ϕ ∈ Irr(N) be an irreducible constituent of resGN(χ). Then ϕ 6= 1N . Weclaim that Z 6 ker(χ). In fact,

resGZ(indGE(1E)) =∑

g∈Z\G/EindZ

Z∩gE

(1Z∩

gE

) = |Z\G/E| · 1Z = |G/E| · 1Z ,

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since Z ∩ gE = gZ ∩ gE = g(Z ∩ E) = gZ = Z, for every g ∈ G, and sinceZ is normal in G and Z 6 E. But since χ is a constituent of indGE(1E), alsoresGZ(χ) is a multiple of 1Z . This proves the claim that Z 6 ker(χ). Now, byLemma 10.22, ϕ cannot be invariant under G and we have U := stabG(ϕ) <G. This implies that χ = indGU (ψ) for some ψ ∈ Irr(U) by Clifford theory,cf. Theorem 9.5. This completes the proof of Brauer’s induction theorem.

Exercises

1. (a) Show that every finite nilpotent group is supersolvable.

(b) Show that subgroups and factor groups of supersolvable groups are super-solvable.

2. (a) Show that SL2(F3) is not monomial. Is SL2(F3) solvable?

(b) Decide which of the irreducible characters of Alt(5) is monomial.

3. (a) Show that subgroups and factor groups of p-elementary groups areagain p-elementary.

(b) Show that subgroups and factor groups of p-quasi-elementary groups areagain p-quasi-elementary.

4. Let ∆: G → GLn(C) be a representation of a finite group G. Show thatthe following are equivalent.

(i) The representation ∆ is equivalent to a representation with values in per-mutation matrices.

(ii) There exists a G-set X such that ∆ corresponds to the CG-module CX.

(iii) The character of ∆ is of the form∑r

i=1 indGHi(1), for some r ∈ N andsubgroups H1, . . . ,Hr of G.

References

[I] M. Isaacs: Character theory of finite groups. Dover 1994.

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