Representations of Finite Groups I (Math 240A)

Home , Character theory, Class function, Elementary group, Faithful representation

(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 is a group homomorphism ∆: G → GLn(F ). The representation ∆ is called faithful 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 matrix −1 S ∈ GLn(F ) such that Γ(g) = S∆(g)S for all g ∈ G. In this case we often write Γ = S∆S−1 for short.

1.2 Remark In the literature, sometimes a representation of G over F is deﬁned as a pair (V, ρ) where V is a ﬁnite-dimensional F -vector space and ρ: G → AutF (V )(= GL(V )) is a group homomorphism into the group of F -linear automorphisms of V . These two concepts can be translated into each other. Choosing an F -basis of V we obtain a group isomorphism ∼ AutF (V ) → GLn(F ) (where n = dimF V ) and composing ρ with this iso- ∼ morphism we obtain a representation ∆: G → AutF (V ) → GLn(F ). If we choose 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 isomorphism ϕ: V → W such that σ(g) = ϕ ◦ ρ(g) ◦ ϕ−1 for all g ∈ G. The above constructions induce mutually inverse bijections between the set of equivalence classes of representations (V, ρ) and the set of equivalence classes of matrix representations ∆ of G over F . If one takes V := F n, the space of column vectors together with the standard 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. n 2 −1 (b) For n > 2 let D2n = hσ, τ | σ = 1, τ = 1, τστ = σ i be the dihedral group of order 2n. We deﬁne a faithful representation ∆ of D2n over R by cos φ − sin φ! −1 0! ∆(σ) := and ∆(τ) := , sin φ cos φ 0 1

1 where φ = 2π/n. Note that ∆(σ) is the counterclockwise rotation about 2π/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 the permutation matrix of σ is a faithful representation of Sym(n) of degree n over any field. It is called the natural representation of Sym(n). (The permutation 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 restriction of ∆ to H, G i ∆ Res (∆): H qqqqqqqqqq G qqqqqqqqqq GL (F ) qqqqq qqqqq H qqqqqqqq qqqqqqqq n qqqqqqqqqq qqqqqqqqqq is a representation of H over F of the same degree. Here, i: H → G denotes the inclusion map. More generally, if f : H → G is an arbitrary group homomorphism between two arbitrary finite groups then also f ∆ Res (∆): H qqqqqqqqqq G qqqqqqqqqq GL (F ) qqqqq qqqqq f qqqqqqqq qqqqqqqq n qqqqqqqqqq qqqqqqqqqq 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 f ∆ G qqqqqqqqqq Sym(n) qqqqqqqqqq GL (F ) , qqqqq qqqqq qqqqqqqq qqqqqqqq n qqqqqqqqqq qqqqqqqqqq with ∆ as in (c), is a representation, called the regular representation of G. If we rearrange the ordering of the elements g1, . . . , gn we obtain an equivalent representation.

(f) If ∆1 : G → GLn1 (F ) and ∆2 : GLn2 (F ) are representations of G over F then also their direct sum ! ∆1(g) 0 ∆1 ⊕ ∆2 : G → GLn1+n2 (F ) , g 7→ 0 ∆2(g) is a representation of G over F . Since ∆ (g) 0 ! 0 I ! ∆ (g) 0 ! 0 I ! 2 = n2 1 n1 ,, 0 ∆1(g) In1 0 0 ∆2(g) 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 of Remark 1.2, with i: G → AutF (E) the inclusion.

2 1.4 Deﬁnition Let ∆: G → GLn(F ) be a representation.

(a) We say that ∆ is decomposable if it is equivalent to ∆1 ⊕ ∆2 for some representations ∆1 and ∆2 of G over F . Otherwise it is called indecomposable. (b) We say that ∆ is reducible if it is equivalent to a block upper triangular representation, i.e., to a representation of the form

A(g) B(g)! g 7→ 0 C(g)

where A(g) ∈ Matn1 (F ), B(g) ∈ Matn1×n2 (F ), C(g) ∈ Matn2 (F ), and where n1, 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 irreducible it 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 representation 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 ∆ is decomposable if and only if V has a decomposition V = U ⊕ W into two non-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, x0, y, y0 ∈ V and λ ∈ C: (x + x0, y) = (x, y) + (x0, y) , (x, y + y0) = (x, y) + (x, y0) , (λx, y) = λ(x, y) , (x, λy) = λ¯(x, y) ,

(y, x) = (x, y) , (x, x) ∈ R>0 , (x, x) = 0 ⇐⇒ x = 0 .

3 (b) Let (−, −) be a hermitian scalar product on V . And let f : V → V be 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 representing matrix 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 product (x, y) = x1y¯1 + x2y¯2 + ··· + xny¯n .

The standard basis (e1, . . . , en) is an orthonormal basis with respect to this product. An automorphism f : Cn → Cn is unitary with respect to (−, −) if and only if the unique matrix A ∈ GLn(C) with the property f(x) = Ax, for all x ∈ Cn, is a unitary matrix, since A is the representing matrix of f with respect 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 numbers of absolute value 1.

1.7 Proposition Every complex representation ∆: G → GLn(C) is equivalent 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 hx, yi := X(∆(g)x, ∆(g)y) , g∈G where (−, −) is the standard hermitian scalar product on Cn. It is easy to check that < −, − > is again a hermitian scalar product on Cn and we have h∆(g)x, ∆(g)yi = X (∆(h)∆(g)x, ∆(h)∆(g)y) = X(∆(k)x, ∆(k)y) = hx, yi , h∈G k∈G for all x, y ∈ Cn and g ∈ G. Thus, ∆(g) is unitary with respect to h−, −i, for every g ∈ G. Let S ∈ GLn(C) be a matrix such that its columns (v1, . . . , vn) form an orthonormal basis of Cn with respect to h−, −i. Then the represent- −1 ing matrix S ∆(g)S of ∆(g) with respect to (v1, . . . , vn) is unitary. Thus, the representation Γ = S−1∆S is unitary.

4 1.8 Corollary Let ∆: G → GLn(C) be a representation and let g ∈ G be an element of order k. Furthermore, set ζ := e2πi/k ∈ C, a primitive k-th root of unity. Then there exists S ∈ GLn(C) (which can depend on g) such that   ζi1 −1 i1 in  .  S∆(g)S = diag(ζ , . . . , ζ ) =  ..  ,   ζin the diagonal matrix with diagonal entries ζi1 , . . . , ζin , where the exponents i1, . . . , 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 diagonal −1 matrix. Thus, there exists S ∈ GLn(C) such that S∆(g)S is a diagonal k matrix, say with diagonal entries d1, . . . , dn. Finally, since g = 1, we have −1 k k −1 k −1 (S∆(g)S ) = S∆(g) S = S∆(g )S = In. This implies that the entries k di must satisfy di = 1 for i = 1, . . . , n. Now the statement of the corollary follows immediately.

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

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

Proof Let u1, . . . , ur be an F -basis of U and extend it to an F -basis n u1, . . . , un of F . Let A ∈ Matn(F ) be deﬁned by Aui = ui for i = 1, . . . , r and Aui = 0 for i = r + 1, . . . , n. We set 1 A0 := X ∆(g−1)A∆(g). |G| g∈G Then we have (i) A0u = u for all u ∈ U.

5 (ii) A0x ∈ U for all x ∈ F n. (iii) A0A0x = A0x for all x ∈ F n. (iv) A0∆(h) = ∆(h)A0 for all h ∈ G. In fact, to see (i), note that 1 1 1 A0u = X ∆(g−1)A∆(g)u = X ∆(g−1)∆(g)u = X u = u , |G| g∈G |G| g∈G |G| g∈G since ∆(g)u ∈ U for all g ∈ G and u ∈ U. To see (ii), note that A∆(g)x ∈ U for all x ∈ F n. Now (iii) follows immediately from (i) and (ii). And (iv) follows from the computation 1 1 A0∆(h) = X ∆(g−1)A∆(gh) = X ∆(hg˜−1)A∆(˜g) = ∆(h)A0 , |G| g∈G |G| g˜∈G where we made the substitutiong ˜ = gh. Now set V := kerA0. We ﬁrst show that V is ∆-invariant. For v ∈ V and g ∈ G we obtain from (iv) that A0∆(g)v = ∆(g)A0v = 0, which means that ∆(g)v ∈ kerA0 = V . Next we show that F n = U + V . In fact, let x ∈ F n. Then using (ii) and (iii) we obtain x = A0x + (x − A0x) ∈ U + V . Finally, we show that U ∩ V = {0}. So let x ∈ U ∩ V . Then 0 = A0x = x by (i). This completes 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 irreducible representations. (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. Assume now that n > 1. If ∆ is irreducible then we are done. So assume that ∆ is not 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 that n F = U ⊕ V . Therefore, ∆ is equivalent to ∆1 ⊕ ∆2 for representations of degree 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 subspace {0} 6= U 6= F n. By Theorem 1.9, U has a ∆-stable complement V in F n. This implies that ∆ is not indecomposable.

6 We recall the following theorem from Linear Algebra without proof.

1.11 Theorem Let A1,...,Ak ∈ Matn(F ). Then the following are equivalent: (i) The matrices A1,...,Ak are simultaneously diagonalizable over F , −1 i.e., there exists S ∈ GLn(F ) such SAiS is a diagonal matrix, for all i = 1, . . . , k. (ii) The matrices A1,...,Ak are individually diagonalizable over F and they 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 sum of 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 particular, every irreducible representation of G has degree 1. In other words, the irreducible representations of G are just the homomorphisms from G to the multiplicative group C× of C.

1.13 Definition Let ∆: G → GLn(F ) be a representation. The character of ∆ is defined as the function ∆ tr χ := χ : G qqqqqqqqqq GL (F ) qqqqqqqqqq F. qqqqq qqqqq ∆ qqqqqqqq n qqqqqqqq qqqqqqqqqq qqqqqqqqqq The characters of representations of G over F are called the F -characters of G. If F = C, we just call them characters, or sometimes complex characters of G. If F = C and ∆ is irreducible then χ is called a (complex) irreducible character. In the following remark we derive a few first properties of characters.

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

(a) Recall that for two matrices A, B ∈ Matn(F ) one has tr(AB) = −1 −1 tr(BA). In particular, if S ∈ GLn(F ) one obtains tr(SAS ) = tr(AS S) = tr(A). This implies that if ∆ and Γ are equivalent representations then χ∆ = χΓ. Moreover, by the same property of the trace map we obtain, for any 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 of G. 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 the usual addition and scalar multiplication. The dimension of CF(G, F ) is equal to the number k(G) of conjugacy classes of G. In fact, the characteristic class functions which have value one on one particular conjugacy class and value zero on all others form an F -basis of CF(G, F ).

(b) Let ∆1 : G → GLn1 (F ) and ∆2 : G → GLn2 (F ) be two representations. Then, for every g ∈ G, we have ! ∆1(g) 0 χ∆1⊕∆2 (g) = tr = tr(∆1(g))+tr(∆2(g)) = χ∆1 (g)+χ∆2 (g) . 0 ∆2(g)

Thus, χ = χ + χ in the vector space CF(G, F ). ∆n1 ⊕∆n2 ∆n1 ∆n2 (c) The F -characters and representations of degree 1 are precisely the group homomorphisms in Hom(G, F ×). They form an abelian group under the product deﬁned by (φ · ψ)(g) = φ(g)ψ(g), for φ, ψ ∈ Hom(G, F ×) and g ∈ G. Recall that we have an isomorphism of abelian groups

Hom(G/G0,F ×) → Hom(G, F ×) , f 7→ f ◦ ν , where G0 denotes the commutator subgroup of G, generated by all commuta- tors [x, y] := xyx−1y−1, x, y ∈ G, and where ν : G → G/G0, g 7→ gG0, denotes the 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 call χ(1) the degree of the character χ. Let g ∈ G and assume that gk = 1. Then, by Corollary 1.8 we have   ζi1  .  −1 ∆(g) = S  ..  S ,   ζin

2πi/k for some S ∈ GLn(C), with ζ := e and i1, . . . , in ∈ Z. Therefore,

χ(g) = ζi1 + ··· + ζin .

8 The triangle inequality implies that

n X i |χ(g)| 6 |ζ j | = n j=1 with equality if and only if ζi1 = ··· = ζin . 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 the character 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 ∈ G with 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

 −1   ζi1 ζ−i1 −1 −1  .  −1  .  −1 ∆(g ) = ∆(g) = S  ..  S = S  ..  S .     ζin ζ−in

Taking the trace yields

−1 −i −i χ(g ) = ζ 1 + ··· + ζ n = ζi1 + ··· + ζin = χ(g) .

1.15 Examples Assume that F = C. (a) The character of the trivial representation is the constant function G → C with value 1. It is called the trivial character and it is often denoted × by 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 of G.

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

9 gi ∈ G that ggi 6= gi. This means that all the diagonal entries in the regular representation of g are equal to 0. (c) Let G = Sym(3). Since we have an isomorphism

× ∼ 0 × × Hom(G, C ) = Hom(G/G , C ) = Hom(G/h(1, 2, 3)i, C ) , see Remark 1.14, the group G has exactly two representations of degree 1, 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 a representation of degree 2 with the property that √ cos 120◦ − sin 120◦! − 1 − 3 ! (1, 2, 3) 7→ = √ 2 2 , sin 120◦ cos 120◦ 3 1 2 − 2 −1 0! (1, 2) 7→ . 0 1

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

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

Since χ 6= 1 + 1, χ 6= ε + ε, and χ 6= 1 + ε, we obtain that every representation whose character is equal to χ is an irreducible representation. Here we used Corollary 1.10 and the property in Remark 1.14(b). In particular, χ is an irreducible character. We will see in Section 2 that there are no other irreducible characters for Sym(3). (d) Let G = hgi be a cyclic group of order k and let ζ := e2πi/k. Then every homomorphism in Hom(G, C×) is uniquely determined by its value on g 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

10 table. 1 g g2 ··· gj ··· gk−1 1 1 1 1 ··· 1 ··· 1 ϕ 1 ζ ζ2 ··· ζj ··· ζk−1 ϕ2 1 ζ2 ζ4 ··· ζ2j ··· ζ2(k−1) ...... ϕi 1 ζi ζ2i ··· ζij ··· ζi(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 by the character ϕ.

Exercises n 2 −1 1. Let D2n = hx, y | x = y = 1, xy = yx i be the dihedral group of order 2n. (a) Show that ζn 0 0 1 ∆1 : D2n → GL2(C) , x 7→ , y 7→ , 0 ζn 1 0

2πi/n with ζn := e deﬁnes a representation.

(b) Decide if the representation ∆1 is equivalent to the representation cos(2π/n) − sin(2π/n) −1 0 ∆ : D → GL ( ) , x 7→ , y 7→ . 2 2n 2 C sin(2π/n) cos(2π/n) 0 1

2. Let G = hgi be a cyclic group of order 3. (a) Show that −1 −1 ∆: G → GL ( ) , g 7→ , 2 Q 1 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. The center of ∆ is deﬁned as

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

11 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 = hgi 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

1 1 ∆(g) = . 0 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 the regular representation of G over C into a direct sum of irreducible representations, i.e., ﬁnd an equivalent representation which is a direct sum of irreducible representations. (b) Do the same as in (a) for a cyclic group of order 3.

12 2 Orthogonality Relations

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

2.1 Lemma (Schur) Let ∆1 : G → GLm(C) and ∆2 : G → GLn(C) be irreducible 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 such that A = λ · In. Proof (a) Assume that A 6= 0. From this we will derive a contradiction. Let n U := ker(A) ⊆ C . Then U is a ∆2-invariant subspace. In fact, for u ∈ U and g ∈ G we have ∆2(g)u ∈ U, since A∆2(g)u = ∆1(g)Au = 0. Since A 6= 0, we n have U 6= C . Since ∆2 is irreducible, this implies U = {0}. Thus, the linear map Cn → Cm, x 7→ Ax is injective. Next consider V := im(A) ⊆ Cm. This is a ∆1-invariant subspace. In fact, let v ∈ V and g ∈ G. We need to show n that ∆1(g)v ∈ V . Since v ∈ V , there exists x ∈ C 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 m n m ∆1 is irreducible, we obtain V = C . Therefore, the linear map C → C , x 7→ Ax, is also surjective, and consequently an isomorphism. Thus, m = n −1 and A ∈ GLn(C). Now Equation (2.1.a) implies ∆1(g) = A∆2(g)A for all g ∈ G. Thus, ∆1 and ∆2 are equivalent. This is a contradiction. (b) Set ∆ := ∆1 = ∆2. Let λ ∈ C be an eigenvalue of A and set 0 0 A := A − λIn. Then also A is an intertwining operator, i.e., it satisﬁes ∆(g)A0 = A0∆(g) for all g ∈ G. Since there exists a non-zero eigenvector for λ, we obtain U 0 := ker(A0) 6= {0}. As in the proof of Part (a) we can see that U 0 is a ∆-invariant subspace of Cn. Since ∆ is irreducible, we obtain 0 n 0 U = C . But this means A = 0 and A = λ · In. We derive a sequence of three corollaries from the above lemma, the last of them showing that the number of irreducible representations of G is ﬁnite and 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 irreducible representations and let A ∈ Matm×n(C). Set

0 1 X −1 A := ∆1(g)A∆2(g ) ∈ Matm×n(C) . |G| g∈G

0 (a) If ∆1 and ∆2 are not equivalent then A = 0. 0 tr(A) (b) If m = n and ∆1 = ∆2 then A = n · In. Proof We ﬁrst show that A0 satisﬁes Equation (2.1.a). Let h ∈ G. Then

0 1 X −1 1 X −1 ∆1(h)A = ∆1(hg)A∆2(g ) = ∆1(˜g)A∆2(˜g h) |G| g∈G |G| g˜∈G 0 = A ∆2(h) .

Here we made the substitutiong ˜ = hg. (a) This follows immediately from Lemma 2.1(a).

(b) Set ∆ := ∆1 = ∆2. By Lemma 2.1(b), there exists λ ∈ C such that 0 A = λ · In. Taking traces, we obtain

1 X −1 0 nλ = tr(λ · In) = tr(∆(g)A∆(g )) = tr(A ) , |G| g∈G and the desired equation holds.

2.3 Corollary Let ∆1 : G → GLm(C) and ∆2 : G → GLn(C) be irreducible unitary representations and write ∆1(g) = rij(g) and ∆2(g) = skl(g) 16i,j6m 16k,l6n for g ∈ G.

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

1 X rij(g)skl(g) = 0 |G| g∈G 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 1 X  n if (i, j) = (k, l), rij(g)skl(g) = |G| g∈G 0 if (i, j) 6= (k, l), for all i, j, k, l ∈ {1, . . . , n}.

0 0 Proof Let A := (auv) ∈ Matm×n(C) be arbitrary and let A = (aik) ∈ −1 Matm×n(C) be deﬁned as in Corollary 2.2. Then, since svk(g ) = skv(g), the (i, k)-entry of A0 is given by

m n 0 1 X X X aik = riu(g)auvskv(g) . |G| u=1 v=1 g∈G

0 (a) By Corollary 2.2(a), we have aik = 0 for every choice of A and for every choice of 1 6 i 6 m and 1 6 k 6 n. We choose A = Ejl, the matrix all of whose entries are 0, except for the (j, l)-entry which is equal to 1. This gives the desired equation. (b) By Corollary 2.2(b), we obtain

 tr(A) 0  n if i = k, aik = 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 of irreducible representations of G is ﬁnite. If ∆i : G → GLni (C), i = 1, . . . , k, Pk 2 are representatives of these equivalence classes then one has i=1 ni 6 |G|.

Proof Let ∆t : G → GLnt (C), t ∈ T , be a set of representatives of the equivalence classes of irreducible representations of G. We may choose them t to be unitary representations by Proposition 1.7. Write ∆t(g) = (rij(g)) ∈

GLnt (C) for every t ∈ T and g ∈ G. We consider triples (t, i, j) with t ∈ T t and i, j ∈ {1, . . . , nt}. For each such triple we consider the vector (rij(g)) ∈ C|G|. By Corollary 2.3, these vectors are non-zero and pairwise orthogonal in C|G|, and therefore the set of these vectors is linearly independent. Thus, the P 2 number of triples (t, i, j), namely t∈T nt , 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 G to C we introduce the Schur inner product 1 X (f1, f2) := f1(g)f2(g) . |G| g∈G It is a straightforward veriﬁcation that (−, −) is a hermitian scalar product on CF(G, C). For characters χ1, χ2 of G we also have

1 X −1 (χ1, χ2) = χ1(g)χ2(g ) , |G| g∈G −1 since χ2(g) = χ2(g ) 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 following notation. We choose unitary representatives ∆1,..., ∆k of the equivalence classes of irreducible representations of G. We write n1, . . . , nk for their respective degrees and we denote by χ1, . . . , χk their respective char- Pk 2 acters. Then χi(1) = ni for i = 1, . . . , k and i=1 ni 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 ∆t(g) = rij . 16i,j6nt 2.7 Theorem (First orthogonality relation) The irreducible characters χ1, . . . , χk form a set of orthonormal vectors in CF(G, C) with respect to the Schur inner product. In other words,  1 X −1 1 if i = j, (χi, χj) = χi(g)χj(g ) = |G| g∈G 0 if i 6= j. Proof From Corollary 2.3 we obtain, for s, t ∈ {1, . . . , k},

ns nt 1 XX s X t (χs, χt) = rii(g) rkk(g) |G| g∈G i=1 k=1 ns nt X X 1 X s t = rii(g)rkk(g) i=1 k=1 |G| g∈G n Xs 1 = δst = δst i=1 ns

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 vectors in a C-vector space with hermitian scalar product. The second statement follows 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 representation ∆ with character χ.

2.9 Theorem Let ∆ and Γ be (complex) representations of G which have the 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 sum of irreducible representations. Moreover, each of these irreducible representations 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 M ∆i ⊕ · · · ⊕ ∆i . (2.9.a) i=1 | {z } li summands Similarly, Γ is equivalent to a direct sum of the form

k M ∆i ⊕ · · · ⊕ ∆i . (2.9.b) i=1 | {z } mi summands

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. But this 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 denote its character. By the argument in the proof of Theorem 2.9 we see that one can 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 irreducible character χi by mi = (χ, χi) , for i = 1, . . . , k. (2.10.a) Pk In fact, (χ, χi) = j=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 identity element, one obtains 2 2 |G| = n1 + ··· + nk .

Proof Recall from Example 1.15(b) that ρG(1) = |G| and ρG(g) = 0 for g 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 second statement is clear.

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

X −1 ∆f := f(g)∆(g ) ∈ Matn(C) g∈G

|G| is a scalar matrix λ · In with λ = n (f, χ). Proof For any x ∈ G we have

−1 X −1 −1 X −1 −1 −1 ∆(x)∆f ∆(x ) = f(g)∆(xg x ) = f(xgx )∆(xg x ) g∈G g∈G X −1 = f(˜g)∆(˜g ) = ∆f , g˜∈G

18 −1 where we substitutedg ˜ = xgx . Therefore we have ∆(x)∆f = ∆f ∆(x) for all x ∈ G. Now Corollary 2.2 implies that ∆f = λ · In with tr(∆ ) 1 1 |G| λ = f = X f(g)χ(g−1) = X f(g)χ(g) = (f, χ) . n n g∈G n g∈G n

2.13 Corollary The set Irr(G) = {χ1, . . . , χk} is an orthonormal basis of CF(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, for every class function f ∈ CF(G, C) we have k X f = (f, χi)χi , i=1 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 hIrr(G)iC of ⊥ hIrr(G)iC is equal to 0, since CF(G, C) = hIrr(G)iC ⊕ hIrr(G)iC . So let ⊥ f ∈ hIrr(G)iC . Then (f, χi) = 0 for all i = 1, . . . , k. Lemma 2.12 implies that (∆i)f = 0 for every irreducible representation ∆i of G, i = 1, . . . , k. Set

k M ∆ := ∆i ⊕ · · · ⊕ ∆i i=1 | {z } ni and form ∆f as in Lemma 2.12. Then, ∆f is a block diagonal matrix with 0 ni blocks equal to (∆i)f . Therefore ∆f = 0. This implies that ∆f = {0} for every representation ∆0 which is equivalent to ∆. By Proposition 2.11 and Theorem 2.9, the representation ∆ is equivalent to the regular representation Γ with respect to any ordering g1, . . . , gn of the elements of G. We may assume that g1 = 1. Thus Γf = 0 and we obtain the following equation in n C , where ei denotes the standard basis vector for i = 1, . . . , n: n n n X −1 X −1 X −1 0 = Γf e1 = f(gi)Γ(gi )e1 = f(gi )Γ(gi)e1 = f(gi )ei . i=1 i=1 i=1 −1 This implies that f(gi ) = 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, . . . , χk of G in a quadratic table, the character table of G:

g1 ··· gj ··· gk χ1 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 one also 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 useful to 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 the irreducible 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 the character table gj 1 (1, 2) (1, 2, 3) cj 1 3 2 o(gj) 1 2 3 1 = χ1 1 1 1 = χ2 1 −1 1 χ3 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 one has k  X −1 |CG(g)| , if g and h are conjugate, χi(g)χi(h ) = i=1 0 , otherwise.

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

k X fh = (fh, χi)χi i=1

20 with 1 |G| −1 1 −1 (fh, χi) = · · χi(h ) = · χi(h ) . |G| |CG(h)| |CG(h)| This implies

k k k 1 X −1 X 1 −1 X χi(g)χi(h ) = χi(h )χi(g) = (fh, χi)χi (g) |CG(h)| i=1 i=1 |CG(h)| i=1  1 if g and h are conjugate, = fh(g) = 0 otherwise, and the proof is complete.

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

2.17 Remark Let F be an arbitrary ﬁeld. (a) Tensor product of representations. Let f : V → V 0 and g : W → 0 0 0 W be F -linear maps between F -vector spaces and let v1, . . . , vn, v1, . . . , vm, 0 0 0 0 w1, . . . , ws, and w1, . . . , wr be F -bases of V , V , W , and W , respectively. Moreover, let A ∈ Matm×n(F ) and B ∈ Matr×s(F ) be the representing matrices of f and g with respect to these bases, respectively. Then the 0 0 representing matrix of the F -linear map f ⊗g : V ⊗F W → V ⊗F W , v⊗w 7→ 0 0 f(v) ⊗ g(w), with respect to the bases vi ⊗ wj of V ⊗F W and vi ⊗ wj of 0 0 V ⊗F W in lexicographic order is given by the Kronnecker product (also called tensor product) A ⊗ B of A and B. It is deﬁned by   a11B ··· a1nB  . .  A ⊗ B =  . .  ∈ Matmr×ns(F ) .  . .  am1B ··· amnB 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 matrices and since (f1 ⊗ g1) ◦ (f ⊗ g) = (f1 ◦ f) ⊗ (g1 ◦ g), we obtain for any additional 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 we deﬁne 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, and Equation (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 above equation 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 multiplicatively closed: The product of two characters is again a character.

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

∗ t −1 −1 t −1 t ∆ : G → GLn(F ) , g 7→ (∆(g) ) = (∆(g) ) = ∆(g ) , is a representation, called the contragredient representation of ∆. One has

−1 t −1 −1 χ∆∗ (g) = tr((∆(g ) ) = tr(∆(g )) = χ∆(g ) , for all g ∈ G. If F ⊆ C then one also has χ∆∗ (g) = χ∆(g). Thus, in this case, 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 of G, as a subset of CF(G, C), is closed under complex conjugation. For any character χ of G, we call χ the contragredient character of χ. (c) Inﬂation. Let N/G and let ν : G → G/N =: G¯, g 7→ gN, denote the canonical epimorphism. For any representation Γ: G/N → GLn(F ) we write

G InfG/N (Γ) := Resν(Γ) := Γ ◦ ν : G → GLn(F ) . This representation of G is called the inﬂation of the representation Γ. For instance, the trivial representation of G is the inﬂation of the trivial representation of G/N. By the fundamental theorem of homomorphisms, the

22 G map Γ 7→ InfG/N (Γ), defines a bijection between the representations Γ of G/N over F of given degree n and the representations ∆ of G over F of degree n with N 6 ker(∆). Moreover, since ν is surjective, one has Γ(G/N) = (Γ ◦ ν)(G) = ∆(G). It follows that two representations Γ1 G and Γ2 of G/N are equivalent if and only if their inflations InfG/N (Γ1) and G InfG/N (Γ2) are equivalent. Furthermore, a representation Γ of G/N is irre- G ducible (resp. indecomposable) if and only if its inflation InfG/N (Γ) is irreducible (resp. indecomposable). This way, specializing to F = C, inflation induces a bijection G infG/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} = hIrr(G)iZ ⊆ CF(G, C) , the Z-span of the irreducible characters of G in the C-vector space of class functions of G. The elements of R(G) are called generalized characters or also virtual characters. Every virtual character can be written as a difference of two characters. Conversely, the difference of two characters is always a virtual character. By Part (a), the product of two virtual characters is again a virtual character. Therefore, R(G) is a subring of the C-algebra of class functions CF(G, C). It is called the character ring of G. Note that since Irr(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 maps on 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 ◦ f G G ResH → ring homom. resH : R(G) → R(H), χ 7→ χ|H G G InfG/N → ring homom. infG/N : R(G/N) → R(G) Here, f : G˜ → G denotes a group homomorphism, H denotes a subgroup of G G G, and N denotes a normal subgroup of G. The functions resH and infG/N are special cases of resf . In the following proposition we give a criterion for when a generalized character χ ∈ R(G) is an irreducible character. Note that χ(1) is always an integer.

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, . . . , χk of G. Then, by the first orthogonality relation (Theorem 2.7), we have

k k k X X X 2 1 = (χ, χ) = (a1χ1+···+akχk, a1χ1+···+akχk) = aiaj(χi, χj) = ai . i=1 j=1 i=1

This implies that there exists a unique index i ∈ {1, . . . , k} such that ai 6= 0 and moreover that ai ∈ {1, −1}. In otherwords, χ = ±χi. Now χ(1) > 0 implies 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 = ha, b | a4 = b2 = 1, bab = a−1i of order 8.

(b) Determine all the irreducible characters of the quaternion group Q8 = hx, y | x4 = y4 = 1, x2 = y2, yxy−1 = x3i of order 8.

3. Assume that ∆: G → GLn(F ) is a representation of a ﬁnite group G over a 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 of G over F , the twist of ∆ by ϕ. (b) Show that ∆ is irreducible (resp. indecomposable) if and only if ϕ · ∆ is irreducible (resp. indecomposable).

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 χ0 be characters of G. Show that ker(χ + χ0) = ker(χ) ∩ ker(χ0). Tk (b) Show that i=1 ker(χi) = {1}. (c) Let N E G. Show that there exists a subset I ⊆ {1, . . . , k} such that \ N = ker(χi) . i∈I In particular, one can determine from the character table the partially ordered set of normal subgroups of G.

25 3 Algebraic Integers

Throughout this section, we denote by S a commutative ring and by R ⊆ S a subring with 1S ∈ R. We introduce and study elements of S which are integral over R. We specialize to the situation where R = Z and S = C to obtain consequences for characters. Again, throughout this section, G denotes a ﬁnite group. The main goal of this section is to show that the degrees of the irreducible characters of a ﬁnite group G are divisors of |G|. n Recall that a polynomial f = anX + ··· 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 a monic 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-linear combinations of the elements 1, s, s2,.... In other words, one has R[s] = 2 {f(s) | f ∈ R[X]} = h1, s, s ,...iR. 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 R0 ⊆ S is a subring with R ⊆ R0 and s ∈ R0 then R[s] ⊆ R0.

More generally, for elements s1, . . . , sn ∈ S, one writes R[s1, . . . , sn] for a1 an the set of all R-linear combinations of elements s1 ··· sn 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 the smallest 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 on the 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 element s ∈ S is integral over R. Note that any subring T of S which contains R is also an R-module. For the proof of the following theorem recall the following notion and fact from linear algebra: If B ∈ Matn(R) is a square matrix with entries bij one ˜ deﬁnes its adjunct matrix B ∈ Matn(R), also called the cofactor matrix of ˜ ˜ i+j B, as the matrix with entries bij, where bij := (−1) det(Bji), where Bji is the square matrix of size n − 1 that arises form deleting the j-th row and the i-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 is ﬁnitely generated as R-module.

Proof (i)⇒(ii): By deﬁnition, there exist n > 1 and r0, . . . , rn−1 ∈ R such n n−1 that s + rn−1x + ··· + r1s + r0 = 0. We will show by induction on k > 1 k 2 n−1 that s ∈ h1, s, s , . . . , s iR. For k = 0, . . . , n − 1 this is clear and for k = n this follows from

n 2 n−1 s = −r0 · 1 − r1s − r2s − · · · − rn−1s . (3.3.a)

k 2 n−1 Now assume that k > n and that s ∈ h1, s, s , . . . , s iR. Then, by Equa- tion (3.3.a), we obtain

k+1 k 2 n−1 2 n−1 n s = s · s ∈ s · h1, s, s , . . . , s iR = hs, s , . . . , s , s iR 2 n−1 ⊆ h1, s, s , . . . , s iR .

2 This ﬁnishes the proof by induction and we have R[s] = h1, s, s ,...iR = 2 n−1 h1, s, s , . . . , s iR, 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 every i = 1, . . . , n. In other words, there exist elements aij ∈ R, i, j ∈ {1, . . . , n}, such that n X sti = aijtj for all i = 1, . . . , n. j=1

One can rewrite this as a matrix equation for the matrix A = (aij) ∈ Matn(R):     t1 0  .   .  n (s · In − A)  .  =  .  in T .  .   .  tn 0

Therefore, the matrix B := s · In − A ∈ Matn(R[s]) satisﬁes       t1 t1 0  .  ˜  .   .  n det(B)In  .  = BB  .  =  .  in T .  .   .   .  tn tn 0

27 This implies that det(B)·ti = 0 for all i ∈ {1, . . . , n}. Since t1, . . . , tn generate T 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 elements n n−1 r0, . . . , rn−1 ∈ R such that det(B) = s + rn−1s + ··· + r1s + r0. This completes the proof of the theorem.

3.4 Corollary The set R0 of all elements of S that are integral over R is a subring of S containing R. Proof Clearly, R0 is contained in S. Moreover, it contains R, since, for r ∈ R, the monic polynomial f = X − r ∈ R[X] satisfies f(r) = 0. In order to show that R0 is a subring of S, let s, t ∈ R0. We need to show that s − t and st are elements of R0. Since s − t, st ∈ R[s, t], it suffices to show 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-module R[s] is generated by a finite set {x1, . . . , xm}. Since t is integral over R, it is also integral over R[s]. Therefore, by the same argument, (R[s])[t] has a finite 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 algebraic integers. √ √ √ √ √ √ 3 3 2 3.6 Examples√ (a) For instance 2, 3, 7, ( 2 + 3 7) are algebraic integers. But 2/2 is not (see Exercise 1). (b) Every n-th root of unity ζ ∈ C is an algebraic integer, since it is a root of the monic polynomial Xn − 1 ∈ Z[X]. (c) Let χ ∈ R(G) and let g ∈ G. Then χ(g) is an algebraic integer. In fact, 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. Then s ∈ Z. a Proof We can assume that s 6= 0 and write s = b with a ∈ Z and b ∈ N satisfying 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 and r0, . . . , rn−1 ∈ Z satisfying an an−1 a + r + ··· + r + r = 0 . b n−1 b 1 b 0 Multiplication with bn yields

n n−1 n−2 2 n−1 n a + rn−1a b + rn−2a b + ··· + r1ab r0b = 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, 0 0 −1 if also zr ∈ Cr then zr = gzrg , for some g ∈ G, and the maps

0 0 0 0 0 {(x, y) ∈ Cr × Cs | xy = zt} ↔ {(x , y ) ∈ Cr × Cs | x y = zt} (x, y) 7→ (gxg−1, gyg−1) (g−1x0g, g−1y0g) ←7 (x0, y0) are mutually inverse bijections.

3.9 Lemma Let C1,..., Ck denote the conjugacy classes of G and let gr ∈ Cr for r ∈ {1, . . . , k}. Moreover let ∆: G → GLn(C) be an irreducible representation and let χ denote the character of ∆. Then, for all r ∈ {1, . . . , k}, one has χ(g ) X ∆(x) = λ I with λ = |C | r . r n r r n x∈Cr

Moreover, the complex numbers λ1, . . . , λk are algebraic integers. P Proof For r ∈ {1, . . . , k} we set Ar := x∈Cr ∆(x) ∈ Matn(C). Then we obtain, for every g ∈ G, the equation

−1 X −1 X ∆(g)Ar∆(g) = ∆(gxg ) = ∆(y) = Ar , x∈Cr y∈Cr

29 −1 after substituting gxg with y. By Corollary 2.2(b), we obtain Ar = λr · In with tr(A ) tr(∆(x)) χ(g ) λ = r = X = |C | r . r n n r n x∈Cr

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

k X X X λrλs · In = ArAs = ∆(xy) = arst∆(z) (x,y)∈Cr×Cs t=1 z∈Ct k k X X = arstAt = arstλt · In , t=1 t=1 where the integers arst are deﬁned as in 3.8. This implies that λrλs = Pk t=1 arstλt and that the ﬁnitely generated Z-module T := hλ1, . . . , λkiZ ⊂ C is a subring of C. Thus, we have Z[λr] ⊆ T ⊂ C and λr is an algebraic integer for all r ∈ {1, . . . , k}, by Theorem 3.3(iii)⇒(i).

3.10 Corollary Let ∆: G → GLn(C) be an irreducible representation. Then n divides |G|. Proof Let χ denote the character of ∆. Then, with the notation from Lemma 3.9, we have

k k X −1 X χ(gr) −1 1 X −1 |G| |G| λrχ(gr ) = |Cr| χ(gr ) = χ(x)χ(x ) = (χ, χ) = . r=1 r=1 n n x∈G n n The ﬁrst term of this equality is an algebraic integer by Lemma 3.9 and by Corollary 3.4. On the other hand the last term in the equality is a rational number. 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 Irrp0 (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

|Irrp0 (G)| = |Irrp0 (H)| . (3.11.a)

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

3.12 Examples (a) Let G = Sym(3). The irreducible characters of G have ∼ degree 1, 1, 2. For p = 2 we obtain P = C2, a cyclic group of order 2, and ∼ H := NG(P ) = P = C2. Thus |Irr20 (G)| = 2 = |Irr20 (H)|. For p = 3 we obtain 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 degrees ∼ ∼ 1, 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 |Irr20 (G)| = 4 = ∼ |Irr20 (H)| as predicted. For p = 3, we have P = C3, a cyclic group of order 3, for instance we can choose P = Alt(3). Then H := NG(P ) = Sym(3) and we obtain |Irr30 (G)| = 3 = |Irr30 (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 commutator subgroup by using that Alt(5) is simple. Show that the natural permutation character decomposes into the trival character and an irreducible character of degree 4. Show that Sym(5) has no irreducible character of degree 2 by considering the 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 subgroup of G. (b) Determine the character degrees of G and verify the McKay Conjecture for the primes 2 and 3.

31 (c) Show that G¯ := G/Z =∼ Sym(4). (Hint: Find some set with four elements on 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 says a b that every finite group of order p q (with primes p and q and with a, b ∈ N0) is solvable. The proof will use character theory and some basic facts from field 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 ring homomorphism ε: Q[X] → Q[α] ,X 7→ α . By definition it is surjective. If it is injective, α is called transcendental over Q. From now on we assume that ε is not injective (in this case α is called algebraic over Q). The image of ε (as a subring of C) is an integral domain. By the fundamental theorem of homomorphisms, its kernel must be a non-zero prime ideal, thus of the form (f) for a unique monic irreducible polynomial f ∈ Q[X]. This polynomial is called the minimal polynomial of α ∼ over Q. Since (f) is even a maximal ideal, Q[X]/(f) = Q[α] is a field. We set ∼ n := deg(f). Then Q[X]/(f) = Q[α] are Q-vector spaces of dimension n. We write 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 ring homomorphism ¯ Q[X]/(f) → C , X 7→ αi . Composing these n ring homomorphisms withε ¯−1 : Q[α] → Q[X]/(f), we obtain 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 ring homomorphism 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, is an algebraic integer as well. In fact, there exist c0, . . . , ck−1 ∈ Z such that k k−1 α +ck−1α +···+c1α+c0 = 0 and applying σi from Part (a) to this equation k k−1 yields the equation αi +ck−1αi +···+c1αi +c0 = 0 as desired. Corollary 3.4 implies that the coeﬃcients of the polynomial f = (X − α1) ··· (X − αn) are algebraic integers. Since they are also rational numbers, Proposition 3.7 implies f ∈ Z[X].

33 (c) Similar considerations as in Part (a) lead quickly to the following result: Let Q ⊆ K ⊆ L ⊆ C be fields with dimQ(L) < ∞ and let σ : K → C be a ring homomorphism. Then there exists a ring homomorphism τ : L → C with τ|K = σ. Simply replace Q with K and make some minor adjustments where necessary. With some more effort one can show that σ can be extended to L, even when [L : K] is infinite. (d) Let ζ := e2πi/k ∈ C. Then the complete set of ring homomorphisms Q[ζ] → C can be constructed as follows: For every i ∈ {1, . . . , k} with gcd(i, k) = 1 there is a unique ring homomorphism

i σi : Q[ζ] → C , ζ 7→ ζ .

One clearly has σi(Q[ζ]) = Q[ζ], and the inverse of σi is equal to σj where j ∈ {1, . . . , k} is the unique element with ij ≡ 1 mod k. The field extension Q[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 of G. Then 0 6 |χ(g)/χ(1)| 6 1. Moreover, if χ(g)/χ(1) is an algebraic integer then |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 of g. By Remark 1.14(d), we know that χ(g) can be expressed as χ(g) = i1 id 2πi/k ζ + ··· + ζ with ζ := e and elements i1, . . . , id ∈ {0, . . . , k − 1}. The triangle inequality then yields the first assertion:

ζi1 ζid ζi1 ζid 1 1 0 |χ(g)/d| = + ··· + + ··· + = + ··· + = 1 . 6 d d 6 d d d d From now on assume that α := χ(g)/d is an algebraic integer. We already know 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 minimal n n−1 polynomial f = X +cn−1X +···+c1X+c0 ∈ Q[X]. Since α is an algebraic integer, we even have f ∈ Z[X], by Remark 4.1(b). Note that α ∈ Q[ζ] and therefore Q[α] ⊆ Q[ζ]. We can write f = (X − α1) ··· (X − αn) ∈ C[X] and the roots α = α1, . . . , αn ∈ C of f are given by the images σ(α), where σ runs through the ring homomorphisms from Q[α] to C, cf. Remark 4.1(a). By Remark 4.1(c) and (d), each such σ is the restriction of a ring homomorphism

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

ζi1j ζidj σ (α) = + ··· + , j d d where j ∈ {1, . . . , k} with gcd(j, k) = 1. As in the ﬁrst part of the proof the 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 hand α1 ··· αn = ±c0 ∈ Z. This implies that |c0| < 1 and c0 = 0. But then n−1 f = X(c1 + c2X + ··· + X ). Since f is irreducible, this implies that the second 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 implies α = 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 class r C with p elements, where p is a prime and r > 1. Then G has a normal subgroup N with 1 < N < G. Proof First note that G is not abelian, since it has a conjugacy class with more than 1 element. Let g ∈ C. Then g 6= 1, since the conjugacy class of 1 has just one element. As before, let χ1, . . . , χk denote the irreducible characters of G and assume that χ1 is the trivial character. By the second orthogonality relation applied to the elements g and 1 we obtain

k k X X 0 = χi(g)χi(1) = 1 + χi(g)χi(1) i=1 i=2 and k 1 X χi(1) − = χi(g) . p i=2 p Since −1/p is not an algebraic integer (cf. Proposition 3.7), there exists an element 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 is not the trivial character, since i > 2. Since χ(g)χ(1)/p is not an algebraic integer, 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(g) = a|C| + bχ(g) , χ(1) χ(1) since |C| = pr. By Lemma 3.9, the number |C|χ(g)/χ(1) is an algebraic integer. Since also χ(g) is an algebraic integer, the right hand side of the above equation is an algebraic integer and also χ(g)/χ(1) is an algebraic integer. Now Lemma 4.2 applies. Since χ(g) 6= 0, we obtain |χ(g)/χ(1)| = 1 which is equivalent to g ∈ Z(χ), the center of χ, cf. Remark 1.14(d). Since g 6= 1, the normal subgroup Z(χ) of G is not trivial, and if Z(χ) < G we are done. So we assume from now on that Z(χ) = G. In this case we consider the normal subgroup N := ker(χ) of G. Recall from Remark 1.14(d) that Z(χ) = G implies that every representation ∆ with character χ takes values in scalar matrices. But since χ is irreducible this implies that χ has degree 1. Thus, χ is a homomorphism from G to C× and G/N is isomorphic to an abelian group. Now, since G is not abelian, we obtain N 6= 1, and since χ is not equal to the trivial character, we obtain N < G. This completes the proof.

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

4.4 Theorem (Burnside’s paqb-Theorem) If G is a finite group of order a b p q 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| = 1 then G is clearly solvable. Now assume that paqb > 1 and that the statement holds 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 normal subgroup different from {1} and G. If a = 0 or b = 0 then G is a p-group (or a q-group) and therefore {1} < Z(G)/G, and we are done. So we can assume that a, b ∈ N. Choose a Sylow q-subgroup Q of G. Then |Q| = qb > 1 and therefore Z(Q) > {1}. Choose 1 6= g ∈ Z(Q). Then the conjugacy class of g has cardinality [G: CG(g)]. Since g ∈ Z(Q), we have Q 6 CG(g) 6 G and r therefore, [G: CG(g)] = p for some 0 6 r 6 a. If r = 0 then 1 6= hgi /G and we are done. If r > 0 then Lemma 4.3 implies that G has a normal subgroup N with {1} < N < G. This completes the proof.

36 Exercises 1. (Galois action on characters) Let G be a ﬁnite group of exponent k, 2πi/k let ζ := e , set K := Q[ζ], and for i ∈ {1, . . . , k} with gcd(i, k) = 1 let i σi : Q[ζ] → Q[ζ] denote the ring homomorphism given by σi(ζ) = ζ . Note that for any χ ∈ R(G) one has χ(g) ∈ K. Therefore, we can deﬁne a function σiχ: G → K by g 7→ σi(χ(g)). (a) Let i ∈ {1, . . . , k} with gcd(i, k) = 1. Show that ( σiχ)(g) = χ(gi) for every virtual 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 character of G. Show that Ψi(χ) is again a character and show that Ψi(χ) is irreducible if σ and only if χ is irreducible. (Thus, Gal(K/Q) acts on Irr(G) by (σi, χ) 7→ iχ). (c) Show that all entries of the character table of a symmetric group are integers.

2. Assume that G is a non-abelian ﬁnite simple group. Show that G has no r abelian subgroup of index p , where p is a prime and r > 1.

3. (Adams operations) Let G be a ﬁnite group. For i ∈ Z and for a class i i function f ∈ CF(G, C) let Ψ (f) denote the function G → C, g 7→ f(g ). (a) Show that Ψi(f) is again a class function. i (b) Show that the resulting function Ψ : CF(G, C) → CF(G, C) is a C-algebra i homomorphism, i.e., a C-linear map and a ring homomorphism. The map Ψ is called the i-th Adams operation. i j ij i j (c) Show that for i, j ∈ Z one has Ψ ◦ Ψ = Ψ . Show that Ψ = Ψ whenever i ≡ j mod exp(G). (Here exp(G) denotes the exponent of G, i.e., the smallest positive integer n with the property that gn = 1 for all g ∈ G.) i (d) Show that if i ∈ Z with gcd(i, n) = 1 then Ψ is an isometry with respect to i i the Schur inner product, i.e., a C-linear isomorphism satisfying (Ψ (f), Ψ (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, let 2πi/n ζ := e , set K := Q[ζ], and for i ∈ {1, . . . , n − 1} with gcd(i, n) = 1 let i σi : Q[ζ] → Q[ζ] denote the ring homomorphism given by σi(ζ) = ζ . Note that for any χ ∈ R(G) one has χ(g) ∈ K. Therefore, we can deﬁne the function σ iχ: 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 every virtual 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 that Ψi(χ) is again a character and show that Ψi(χ) is irreducible if and only if χ is irreducible. (c) Show that all entries of the character table of a symmetric group are integers.

38 5 The Group Algebra and its Modules

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

5.1 Deﬁnition A k-algebra is a pair (R, η) consisting of a ring R together with 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 two k-algebras (R, η) and (R0, η0) is a ring homomorphism f : R → R0 such that f ◦η = η0. Very often it is clear from the context what η is and we only write R instead of (R, η).

5.2 Remark (a) In general, η need not be injective. If k is a field then η is always 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. For this 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 satisfying r(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 are mutually inverse. (c) Let (R, η) and (R0, η0) be k-algebras and let f : R → R0 be a function. Then, f is a k-algebra homomorphism if and only if f is a ring homomorphism and a k-module homomorphism (with respect to the k-module structures defined in (b)). (d) Z-algebras and Z-algebra homomorphisms are nothing else as rings and ring homomorphisms. In fact, for every ring R there exists a unique ring homomorphism η : Z → R and this homomorphism satisfies η(Z) ⊆ Z(R). Moreover, every ring homomorphism f : R → R0 satisfies f ◦ η = η0 for these homomorphism η : Z → R and η0 : Z → R0. 5.3 Examples (a) The polynomial ring k[X] is a k-algebra with η mapping an 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 a k-algebra with η : k → Endk(M), α 7→ lα, where lα : M → M is the left multiplication by α, i.e., lα(m) = αm for all m ∈ M. The multiplication on Endk(M) is given by composition.

39 5.4 Definition The group algebra kG is defined as the free k-module with basis G. It is a ring with the following multiplication rule: X X X ( αgg)( βhh) = αgβhgh . g∈G h∈G g∈G h∈G In other words, the multiplication on kG is just the k-bilinear extension of the multiplication on G. The basis element 1 ∈ G is also the identity element of kG. Moreover, every basis element g ∈ G is invertible in kG. It’s inverse is the basis element g−1. Clearly, the k-span of 1 is contained in the center of kG. Therefore, kG is in fact a k-algebra. This construction also works for infinite groups G.

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

2 2 kG = k · 1G ⊕ k · σ ⊕ k · σ ⊕ k · τ ⊕ k · στ ⊕ k · σ τ .

2 For k = Z and for a = 2 · 1G + 3 · σ − τ and b = σ − 2 · στ in ZG one has 2 ab = (2 · 1G + 3 · σ − τ)(σ − 2 · στ) = 2 · σ2 − 4 · στ + 3 · σ3 − 6 · σ2τ − τσ2 + 2 · τστ 2 2 = 3 · 1G + 4 · σ − 5 · στ − 6 · σ τ , since τσ2 = στ and τστ = σ2.

5.6 Remark (a) If f : G → H is a group homomorphism then its k-linear extension X X kf : kG → kH , αgg 7→ αgf(g) , g∈G g∈G is a k-algebra homomorphism. In fact, it is k-linear by definition and it is multiplicative on a k-generating set, namely the set G. One obtains a functor k− from the category of (finite) groups to the category of k-algebras (which are 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 , X X αgg 7→ αg∆(g) ←7 ∆ , g∈G g∈G

40 given by restricting Γ to G and k-linearly extending ∆ to kG, are mutually inverse bijections. (In category theory language one can show that the functor k−: Gr → kAlg is left adjoint to the unit group functor U : kAlg → Gr, and the above bijection is the special case of this adjunction for a group G and the ring Matn(k)) A k-algebra homomorphism Γ: kG → Matn(k) is also called a matrix representation of the k-algebra kG. Two matrix representations Γ1 : kG → Matn(k) and Γ2 : kG → Matm(k) are called equivalent if m = n −1 and if there exists S ∈ GLn(k) such that Γ2(a) = SΓ1(a)S 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 mutually inverse 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 via η : k → kG. For every group element g ∈ G, the map m 7→ gm deﬁnes an element 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 is a k-module and if σ : G → Autk(M) is a group homomorphism then we can P P deﬁne a kG-module structure on M by ( g∈G αgg)m := g∈G αg(σ(g))(m) P for m ∈ M and g∈G αgg ∈ kG. These two constructions are mutually inverse. Similarly as in (b), using k-linear extensions, one obtains a bijection between 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-module and ρ: kG → Endk(M) is a k-algebra homomorphism. Thus we have natural bijections

∼ {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 0, σ0) as above are called equivalent, if there exists a k-module isomorphism f : M → M 0 such that σ0(g) = f −1 ◦ σ(g) ◦ f for all g ∈ G. Similarly, two pairs (M, ρ) and (M 0, ρ0) are called equivalent if there exists a k-module isomorphism f : M → M 0 such that ρ0(a) = f −1 ◦ ρ(a) ◦ f for all a ∈ kG. It is now easy to check that two kG-modules M and M 0 are isomorphic if and only if the associated pairs (M, σ) and (M 0, σ0) are equivalent. Moreover, M and M 0 are isomorphic

41 as kG-modules if and only if the associated pairs (M, ρ) and (M 0, ρ0) are equivalent. Consequently, we the above bijections induce bijections between isomorphism 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 any pair (M, σ) as in (c), consisting of an n-dimensional F -vector space M and a 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 the ∼ isomorphism AutF (M) → GLn(F ) determined by this basis. Conversely, if ∆: G → GLn(F ) is a representation of G, we can choose any n-dimensional F -vector space M and a basis of M, and use the resulting isomorphism ∼ GLn(F ) → AutF (M) to define a group homomorphism σ : G → AutF (M) by composing ∆ with the described isomorphism. For each of these constructions, 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 correspond to representations ∆1 : G → GLn1 (F ) and ∆2 : G → GLn2 (F ), respectively. 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 respect to the chosen bases of M1 and M2. In particular, if M is an FG-module and if ∆: G → GLn(F ) is a corresponding representation then one has an isomorphism 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 corresponding representation 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, the character 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 equal to the regular character. The trivial character is afforded by the trivial CG-module, namely the module C with g · α = α for g ∈ G and α ∈ C. More generally, if φ: G → C× is a linear character (character of degree 1) then φ is afforded by the CG-module Cφ which is defined to be equal to C as C-space, and whose module structure is given by g · α := φ(g)α, for g ∈ G and 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 is invertible.

Proof Let 0 6= f ∈ EndR(M). We need to show that f is injective and surjective. 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 is surjective. Moreover, since f 6= 0, we obtain ker(f) 6= M. Since ker(f) is an R-submodule of M and since M is simple, we obtain ker(f) = {0} and f is injective. 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 the more general version in 5.7. In fact, let ∆ be an irreducible representation of 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 a subfield of D via the structural homomorphism η. The version in 2.1 amounts to 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 field extension 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 M there 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 is semisimple if and only if |G| is invertible in F . Proof First assume that |G| is invertible in F , let M be an FG-module and let U be an FG-submodule of M. We follow the idea of the proof of Theorem 1.9. Let W be an F -subspace of M such that U ⊕ W = M. We denote by p: M → U the projection map with respect to the decomposition M = U ⊕ W . Deﬁne 1 p0 : M → M , m 7→ X gp(g−1m) . |G| g∈G

0 0 0 Then it is easy to see that p ∈ HomFG(M,M), p (M) ⊆ U, p |U = idU , and p0 ◦p0 = p0. From this it follows quickly that V := ker(p0) is an FG-submodule of 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 left P FG-module and let u := g∈G g ∈ M. Then F u is an FG-submodule of P M. In fact, for every g ∈ G one has gu = u and therefore ( g∈G αgg)u = P 2 ( g∈G αg)u. It also follows that u = |G|u = 0. Since FG is semisimple, there exists a submodule V of FG such that U ⊕ V = FG. Write 1 = αu + v with 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 that u = 0. This is a contradiction.

44 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 M is ﬁnitely generated as k-module. (See Excersise 3)

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

Proof We prove the result by induction on dimF M. If dimF M = 1 then M is clearly simple. Now assume that dimF M = n > 1 and that the result holds for modules of smaller dimension. If M is simple we are done. If M is 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 that M = M1 ⊕ M2. By induction, both M1 and M2 can be written as a direct sum of simple submodules. Thus, M can be written in this form.

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

5.12 Theorem Let G be a ﬁnite group, let ∆1,..., ∆k be representatives of the equivalence classes of irreducible representations of G over C, and let n1, . . . , nk ∈ N denote their respective degrees. Then the map

Φ: CG → Matn1 (C) × · · · × Matnk (C) , X X X αgg 7→ αg∆1(g),..., αg∆k(g) g∈G g∈G g∈G is an isomorphism of C-algebras with inverse

k 1 XX −1 Ψ:(A1,...,Ak) 7→ nitr(∆i(g )Ai) g . |G| g∈G i=1

Proof Clearly, Φ and Ψ are C-linear maps. Moreover, it is easy to check that 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 same 2 2 dimension, since |G| = n1 + ··· + nk. 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 it suﬃces to show this for the basis elements a = h ∈ G. But, with χi denoting

45 the character of ∆i, we have

k 1 XX −1 Ψ(Φ(h)) = Ψ ∆1(h),..., ∆k(h) = nitr(∆i(g )∆i(h) |G| g∈G i=1 k 1 XX −1 = niχi(g h) g |G| g∈G i=1

Pk with i=1 niχi = ρG, the regular character of G, by Proposition 2.11. Thus, −1 −1 we obtain Ψ(Φ(h)) = h, since ρG(g h) = 0 if g 6= h and ρG(g h) = |G| if g = h.

5.13 Remark Let ∆1,..., ∆k and n1, . . . , nk be as in Theorem 5.12 and let χ1, . . . , χk denote the respective characters of ∆1,..., ∆k.

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

1 X −1 χi(1) X −1 ei := Ψ(εi) = nitr(∆i(g ) g = χi(g )g ∈ CG. |G| g∈G |G| g∈G

2 Thus, ei = 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, the elements e1, . . . , ek are C-linearly independent. Since dimC Z(CG) = k (see Exercise 1), these elements form a C-basis of Z(CG). The element ei is called the central idempotent associated with χi. (b) Let M be a ﬁnitely generated CG-module. Then the above properties of 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 which corresponds 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 the C-linear map M → M, m 7→ gm, with respect to any C-basis of M.

46 5.14 Theorem Let M be an irreducible CG-module, let Irr(G) = {χ1, . . . , χk} and let e1, . . . , ek ∈ Z(CG) be the respective associated idempotents. For all m ∈ M and i ∈ {1, . . . , k} one has  m if χM = χi, eim = 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) denote the corresponding representation, i.e., ∆j(g) is the representing matrix of the C-linear map M → M, m 7→ gm, for all g ∈ G. Then, χj = χM = χ∆.

For m = α1m1 + ··· + αnj mnj we have

χi(1) X −1 eim = χi(g )g m |G| g∈G  α  χ (1) 1 i X −1  .  = (m1, . . . , mnj ) χi(g )∆j(g)  .  . |G|   g∈G α | {z } nj =:Aj ∈Matnj (C)

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

5.15 Corollary Let S1,...,Sk denote a set of representatives of the isomorphism classes of simple CG-modules, let χ1, . . . , χk denote their respective characters and e1, . . . , ek ∈ CG their respective central idempotents. Let M be a ﬁnitely generated CG-module and let M = M1 ⊕ · · · ⊕ Mr be a decomposition of M into simple submodules. For each i ∈ {1, . . . , k} one has M eiM = Ms s∈{1,...,k} Ms=∼Si and 1 (χM , χi) = dimC(eiM) . χi(1)

In particular, every submodule S of M which is isomorphic to Si is contained 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 C ,..., C denote the conjugacy classes of G and set c := P g ∈ RG for 1 k i g∈Ci 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 elements Pk aijk ∈ R satisfying cicj = m=1 aijmcm) with respect to the basis c1, . . . , ck are given by aijm = |{(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 to G is an irreducible representation. Show that Γ(Z(CG)) = C · 1n.

2πi/3 2. Let K := Q(ζ) with ζ := e and let G := Sym(3). Show that the multiplication 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 a kG-module. Show that M is finitely generated as kG-module if and only if M is finitely generated as k-module.

4. Let k be a commutative ring, let G be a ﬁnite group and let S be a ﬁnite G-set. (a) Let M := kS, the free k-module with k-basis S. Construct a kG-module structure 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 called permutation characters. (c) Assume that k = C. Show that the scalar product (χS, 1) of χS with the trivial 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 · −1 f)(m) := f(g m), and show that χM ∗ = χM . (Taking duals corresponds to taking contragredient representations, and taking complex conjugate characters.)

(b) Show that the C-vector space V := HomC(M,N) becomes a CG-module via (g · f)(m) := gf(g−1m) for g ∈ G, f ∈ V , m ∈ M, and that χV = χM · χN .

49 6 The Tensor Product

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

6.1 Deﬁnition Let M be a right R-module, let N be a left R-module, and let A be an abelian group. A map β : M × N → A is called biadditive, if β(m+m0, n) = β(m, n)+β(m0, n) and β(m, n+n0) = β(m, n)+β(m, n0) , for all m, m0 ∈ M and all n, n0 ∈ 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 abelian group T together with an R-balanced, biadditive map τ : M × N → T satisfying 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 homomorphism f : T → A such that the diagram

T qqqqqqqqqqqqqqqqqqq q q τ

M × N f @ @ qqq qqq β qqqqqqqqqq qqqqqqqqqq @ qqqqqqqqq q q qqqqqqqqqqqqqqqqqqq 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 0, τ 0) are tensor products of M and N over R then the unique homomorphism f : T → T 0 such that the diagram

50 T qqqqqqqqqqqqqqqqqqq q q τ

M × N f @ 0@ qqqqqqqq qqqqqqqq τ @ qqqqqqqqqqqqqqqqqqq q q Tqqqqqqqqqqqqqqqqqqq 0 commutes, is an isomorphism. Its inverse is the unique homomorphism g : T 0 → T such that g ◦ τ 0 = τ.

6.4 Notation If (T, τ) is a tensor product of M and N over R then one usually writes M ⊗R N instead of T and m⊗n instead of τ(m, n), for m ∈ M and n ∈ N. With these notations, the following hold for all m, m0 ∈ M, n, n0 ∈ N and r ∈ R:

(m + m0) ⊗ n = (m ⊗ n) + (m0 ⊗ n) , m ⊗ (n + n0) = (m ⊗ n) + (m ⊗ n0) , 0 ⊗ n = 0 = m ⊗ 0 , (−m) ⊗ n = −(m ⊗ n) = m ⊗ (−n) , mr ⊗ n = m ⊗ rn .

The universal property implies that M ⊗R N is generated as abelian group by elements of the form m ⊗ n with m ∈ M and n ∈ N. Using the fourth of the above rules, one obtains that every element x ∈ M ⊗R N can be written in the form x = (m1 ⊗ n1) + ··· (ms ⊗ ns) with s ∈ N, m1, . . . , ms ∈ M and n1, . . . , ns ∈ N. Warning: The elements s ∈ N, m1, . . . , ms ∈ M and n1, . . . , ns ∈ N are not uniquely determined by x.

6.5 Proposition Let M and M 0 be right R-modules, let N and N 0 be left R- modules, and let f : M → M 0 and g : N → N 0 be R-module homomorphisms. 0 0 Then there exits a unique homomorphism h: M ⊗R N → M ⊗R N such that h(m ⊗ n) = f(m) ⊗ g(n) for all m ∈ M and n ∈ N. This homomorphism is denoted by f ⊗ g. Thus, (f ⊗ g)(m ⊗ n) = f(m) ⊗ g(n) for all m ∈ M and n ∈ N.

6.6 Remark (a) Recall that an (R,S)-bimodule M is an abelian group which 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 is an (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 is automatically a (k, k)-bimodule via mα := αm (resp. αm := mα), for α ∈ k and m ∈ M. Thus, if M and N are k-modules then M ⊗k N is again a k-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)(a0 ⊗ b0) = (aa0) ⊗ (bb0), for a, a0 ∈ A and b, b0 ∈ B. Moreover, η(α) := ηA(α) ⊗ 1B = (α · 1A) ⊗ 1B = α · (1A ⊗ 1B) = 1A ⊗ (α · 1B) = 1A ⊗ ηB(α) for α ∈ 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 following proposition, the elements g ⊗ h, with g ∈ G and h ∈ H, form a k-basis of kG ⊗k kH. Therefore, the above map defines a k-linear isomorphism. It is quickly checked that this is also a ring homomorphism (it suffices to check this on the basis elements).

6.7 Proposition (a) Let M be a left R-module (resp. an (R,S)-bimodule). Then ∼ R ⊗R M = M as left R-modules (resp. (R,S)-bimodules) via the map r ⊗ m 7→ rm with inverse m 7→ 1R ⊗ m, for r ∈ R and m ∈ M. Similarly, if M is a right S-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 with inverse m 7→ m ⊗ 1S for m ∈ M and s ∈ S. (b) Let M and M 0 be right S-modules (resp. (R,S)-bimodules) and let N be a left S-module (resp. (S,T )-bimodule). Then 0 ∼ 0 (M ⊕ M ) ⊗S N = (M ⊗S N) ⊕ (M ⊗S N) as abelian groups (resp. as (R,T )-bimodules). Similarly, if also N 0 is a left S-module (resp. an (S,T )-bimodule) then 0 ∼ 0 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 let P be a (T,U)-bimodule for an additional ring U. Then there exists an isomorphism ∼ (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, for m ∈ 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 free with R-basis m1, . . . , md, then every element x ∈ M ⊗R N can be expressed as x = (m1 ⊗ n1) + ··· + (md ⊗ nd) with unique elements n1, . . . , nd ∈ N. Similarly, if N is free with R-basis n1, . . . , nd then every element x ∈ M ⊗R N can be expressed as above with uniquely determined elements m1, . . . , md ∈ M. (e) If k is a commutative ring and if M and N are free k-modules with bases m1, . . . , md and n1, . . . , ne, respectively, then M ⊗k N is a free k-module with 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 0, W, W 0 be finite-dimensional F -vector spaces, let f : V → 0 0 0 0 V and g : W → W be F -linear maps, and let (v1, . . . , vr), (v1, . . . , vs), 0 0 0 0 (w1, . . . , wt), and (w1, . . . , wu) 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 of f and g, respectively, with respect to these bases. Then A⊗B ∈ Matsu×rt(F ) 0 0 is the representing matrix of f ⊗ g : V ⊗F W → V ⊗F W with respect to 0 0 the bases vi ⊗ wj (1 6 i 6 r, 1 6 j 6 s) and vk ⊗ wl (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 X X ( αgg) · (v ⊗ w) = αg(gv ⊗ gw) . g∈G g∈G

In fact, one has an F -algebra homomorphism

δ ρ ⊗ ρ0 FG qqqqqqqqqq FG ⊗ FG qqqqqqqqqq End (V ) ⊗ End (W ) qqqqqqqqqq End (V ⊗ W ) qqqqq qqqqq qqqqq qqqqqqqq F qqqqqqqq F F F qqqqqqqq F qqqqqqqqqq qqqqqqqqqq qqqqqqqqqq

53 where δ is the F -linear extension of g 7→ g ⊗g, ρ and ρ0 denote the F -algebra homomorphisms describing the FG-module structures of V and W , and the last map is the F -algebra homomorphism induced by (f, g) 7→ f ⊗ g. The above 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 0 ∆: G → GLr(F ) and ∆ : G → GLs(F ) denote the representations corre- 0 sponding to V and W with respect to these bases then ∆⊗∆ : G → GLrs(F ), 0 g 7→ ∆(g) ⊗ ∆ (g), corresponds to the FG-module V ⊗F W with respect to the basis vi ⊗ wj (1 6 i 6 r, 1 6 j 6 s) in lexicographic order. In particular, 0 0 since tr(∆(g) ⊗ ∆ (g)) = tr(∆(g))tr(∆ (g)), we obtain χV ⊗F W = χV · χW . In the following deﬁnition we construct a representation of G × H from a 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 is a finitely generated FG-module and W is a finitely generated FH-module then the F -vector space V ⊗F W has an F [G × H]-module structure which is uniquely determined by

(g, h) · (v ⊗ w) = gv ⊗ hw , for (g, h) ∈ G × H, v ∈ V and w ∈ W . In fact, this module structure comes from the F -algebra homomorphism

ρ ⊗ ρ0 F [G×H] qqqqqqqqqq FG⊗ FH qqqqqqqqqq End (V )⊗ End (W ) qqqqqqqqqq End (V ⊗ W ) , qqqqq qqqqq qqqqq qqqqqqqq F qqqqqqqq F F F qqqqqqqq F F qqqqqqqqqq qqqqqqqqqq qqqqqqqqqq where the ﬁrst map is the F -linear extension of (g, h) 7→ g ⊗ h, the maps ρ and ρ0 denote the F -algebra homomorphisms corresponding to the module structures 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 ⊗F W of G × H. If also V 0 and W 0 are ﬁnitely generated modules for FG and FH, respectively, then (χV + χV 0 ) × χW = χV × χW + χV 0 × χW and χV × (χW + χW 0 ) = χV × χW + χV × χW 0 . 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 generated CG-module and let W be a finitely generated CH-module. Then the following 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 are simple modules. (c) The function Irr(G) × Irr(H) 7→ Irr(G × H), (χ, ψ) 7→ χ × ψ, is a bijection. 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 , respectively, and let ∆1 : G → GLm(F ) and ∆2 : G → GLn(F ) be the corresponding representations. Remark 6.8(a) implies that the representation of the F [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). Since tr(A ⊗ B) = tr(A)tr(B) for any square matrices A and B, we obtain

(χV × χW )(g, h) = χV ⊗F W (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, χ0 of G and ψ, ψ0 of H, one has

0 0 0 0 (χ × ψ, χ × ψ )G×H = (χ, χ )G · (ψ, ψ )H . (6.10.a)

In fact, we have

0 0 1 X 0 0 −1 −1 (χ × ψ, χ × ψ )G×H = (χ × ψ)(g, h) · (χ × ψ )(g , h ) |G × H| (g,h)∈G×H 1 = X X χ(g)ψ(h)χ0(g−1)ψ0(h−1) |G| · |H| g∈G h∈H 1 1 = X χ(g)χ0(g−1) · X ψ(h)ψ0(h−1) |G| g∈G |H| h∈H 0 0 = (χ, χ )G · (ψ, ψ )H .

For arbitrary characters χ of G and ψ of H, the irreducibility criterion in Proposition 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 in Part (c) takes values in Irr(G × H). Next we show that the ﬁrst map in Part (c) is bijective. In order to see that it is injective, assume that χ, χ0 ∈ Irr(G) and ψ, ψ0 ∈ Irr(H) satisfy χ × ψ = χ0 × ψ0. Then Equation (6.10.a) implies that

0 0 0 0 1 = (χ × ψ, χ × ψ )G×H = (χ, χ )G · (ψ, ψ )H ,

0 0 0 0 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 to show that |Irr(G × H)| = |Irr(G)| · |Irr(H)|. This in turn is equivalent to showing that the numbers of conjugacy classes of these three groups satisfies the equation k(G × H) = k(G)k(H). Now note that two elements (g, h) and (g0, h0) of G×H are conjugate in G×H if and only if g and g0 are conjugate in 0 G and h and h are conjugate in H. This implies that if C1,..., Ck(G) denote the conjugacy classes of G and D1,..., Dk(H) denote the conjugacy classes of H then Ci × Dj, 1 6 i 6 k(G), 1 6 j 6 k(H), are precisely the conjugacy classes of G × H. Finally, since the first map in Part (c) is bijective and since the sets in these maps form Z-bases of the abelian groups in the second map, the second map is a group isomorphism. It is now easy to verify, using the formula in Part (a), that it is also multiplicative.

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

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

(a) If M is an (R,S)-bimodule and N 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) If M is an (S,R)-bimodule and N 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 for m1, 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 ⊗C M with Ms ⊕ Ma = M ⊗C M. (b) Let χ, χs, and χa be the character of M, Ms, and Ma, respectively. Show that 1 1 χ (g) = χ2(g) + χ(g2) and χ (g) = χ2(g) − χ(g2) s 2 a 2 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 ⊗F N. (The ∗ FG-module structures of HomF (M,N) and M are given in Exercise 5.6)

4. Verify the the statements in Theorem 6.11.

57 7 Induction

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

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

G IndH (W ) := FG ⊗FH W is called the module induced by W from H to G, or also the induction of W from 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 elements P g1, . . . , gn. In fact, every element g∈G αgg in FG can be written as

n n n X X X X X X αgg = αgihgih = gi αgihh ∈ gi · FH. g∈G i=1 h∈H i=1 h∈H i=1

Pn P Pn P Moreover, if i=1 gi·xi = 0, with xi = h∈H αh,ih ∈ FH, then i=1 h∈H αh,igih = 0. For given i ∈ {1, . . . , n} and h ∈ H, the coeﬃcient of gih in this element 0 0 of FG is equal to αh,i. Since gih = gjh implies i = j and h = h , for 0 i, j ∈ {1, . . . , n} and h, h ∈ H, we have αh,i = 0. This implies xi = 0 for all i = 1, . . . , n. Therefore, by Proposition 6.7(d), one has an isomorphism of F -vector spaces

n ∼ M FG ⊗FH W ↔ W, i=1 n X gi ⊗ wi ↔ (w1, . . . , wn) . i=1 This implies G dimF IndH (W ) = [G: H] dimF W.

(b) Let w1, . . . , wm be an F -basis of W , and let ∆: H → GLm(F ) denote the corresponding representation of H. 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 G G the basis) by IndH (∆). Our next goal is to describe the matrix (IndH (∆))(g) for a given group element g ∈ G. We have to rewrite the element ggi ⊗ wj as linear 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 permutation induced 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 .

G 2 This implies that the matrix (IndH (∆))(g) consists of n blocks of size m×m, and that the m × m block matrix in the i-th block column and k-th block row is equal to 0 unless k = σ(i). In the case k = σ(i) this block matrix is −1 equal to ∆(hi) = ∆(gσ(i)ggi). (c) Let ψ denote the character of W (or of ∆). With the results from G Part (b), we can compute the character of IndH (∆) which we will denote by G indH (ψ). With the notation in Part (b), we obtain that

n n G X −1 X −1 indH (ψ) (g) = ψ(gi ggi) = ψ(gi ggi) i=1 i=1 σ(i)=i −1 gi ggi∈H 1 = X ψ˙(g−1gg ) = X ψ˙(x−1gx) i i |H| gi∈G/H x∈G with  ψ(g) if g ∈ H, ψ˙ : G → F , g 7→ 0 if g∈ / H.

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

G 0 ∼ G G 0 IndH (W ⊕ W ) = IndH (W ) ⊕ IndH (W ) if also W 0 is a ﬁnitely generated FH-module. In the case that F = C, this implies that we obtain a group homomorphism

G G indH : R(H) → R(G) , χ 7→ indH (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 of isomorphisms of left CG-modules, G H ∼ IndH (IndK (W )) = CG ⊗CH (CH ⊗CK W ) = (CG ⊗CH CH) ⊗CK W ∼ G = CG ⊗CK W = IndK (W ) . This transitivity property of induction for modules implies the transitivity property of induction for characters:

G H G indH ◦ indK = indK : 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) and 2 y = (1, 2) and let H := Alt(3) = {1, x, x } 6 G. The character table of H is given by 1 x x2 ψ1 1 1 1 2 ψ2 1 ζ ζ 2 ψ3 1 ζ ζ with ζ := e2πi/3. The elements 1 and y form a set of coset representatives G for G/H and we can compute the character values of indH (ψ) for the various ψ ∈ Irr(H) using the formula in Remark 7.2(c). Note that y−1xy = x2. We include the computations of these characters in the character table of G:

1 y x χ1 1 1 1 χ2 1 −1 1 χ3 2 0 −1 G indH (ψ1) 2 0 2 G indH (ψ2) 2 0 −1 G indH (ψ3) 2 0 −1 Now it is easy to see that

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

60 7.4 Proposition Let V and W be CG-modules and let χV and χW denote their respective characters. Then

(χV , χW )G = dimC HomCG(V,W ) . Proof Exercise 5.5. Hint: Both sides of the equation additive in V and W with respect to direct sums. Therefore, it suﬃces to prove the equation for simple CG-modules V and W . But in this case this follows immediately from 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) one has G G (indH (ψ), χ)G = (ψ, resH (χ))H . G G Proof Since (−, −) is additive in both arguments, and since indH and resH are group homomorphisms, it suﬃces to prove the equation for irreducible characters 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

G (indH (ψ), χ)G = dimC HomCG(CG ⊗CH W, V ) and G G (ψ, resH (χ))H = dimC HomCH (W, ResH (V )) .

Therefore, it suﬃces to show that the C-vector spaces HomCG(CG⊗CH W, V ) G and HomCH (W, ResH (V )) are isomorphic. Note that one has well-deﬁned mutually inverse isomorphisms

∼ G HomCG(CG, V ) ↔ ResH (V ) f 7→ f(1) (a 7→ av) ←7 v of CH-modules, where the left CH-module structure of HomCG(CG, V ) comes from 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 ) and HomCH (W, HomCG(CG, V )) are isomorphic. But this is precisely the content of 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 1 χ2 1 1 −1 1 −1 χ3 2 2 0 −1 0 χ4 3 −1 1 0 −1 χ5 3 −1 −1 0 1 (a) Let H := Sym(3) 6 G. The character table of H together with the restrictions χi|H of the irreducible characters of G is given below: 1 (ab)(abc) ψ1 1 1 1 ψ2 1 −1 1 ψ3 2 0 −1 G resH (χ1) 1 1 1 G resH (χ2) 1 −1 1 G resH (χ3) 2 0 −1 G resH (χ4) 3 1 0 G resH (χ5) 3 −1 0 It follows immediately that G resH (χ1) = ψ1 G resH (χ2) = ψ2 G resH (χ3) = ψ3 G resH (χ4) = ψ1 + ψ3 G resH (χ5) = ψ2 + ψ3 Frobenius reciprocity now immediately implies that G indH (ψ1) = χ1 + χ4 G indH (ψ2) = χ2 + χ5 G indH (ψ3) = χ3 + χ4 + χ5 7.7 Theorem (Frobenius property) For ψ ∈ R(H) and χ ∈ R(G) one has G G G indH (ψ) · ψ = indH (ψ · resH (χ)) . G In particular, indH (R(H)) is an ideal in the commutative ring R(G).

62 Proof For g ∈ G we have

G G 1 X ˙ −1 (indH (ψ) · χ)(g) = (indH (ψ))(g) · χ(g) = ψ(x gx)χ(g) |H| x∈G 1 = X ψ˙(x−1gx)χ(x−1gx) |H| x∈G and

G G 1 X G −1 indH (ψ · resH (χ))(g) = (ψ · resH (χ))(x gx) |H| x∈G x−1gx∈H 1 = X ψ(x−1gx) · χ(x−1gx) |H| x∈G x−1gx∈H 1 = X ψ˙(x−1gx)χ(x−1gx) . |H| x∈G

This completes the proof of the theorem.

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

g ∼ g0 : ⇐⇒ there exist u ∈ U and h ∈ H such that g0 = ugh.

The double coset of g, i.e., all elements g0 ∈ G which are equivalent to g, is UgH = {ugh | u ∈ U, h ∈ H}. We write U\G/H := {UgH | g ∈ G} for the set 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 set gH := gHg−1 and Hg := g−1Hg. P (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 representatives of G with respect to U and H. (b) Every double coset in UgH ∈ U\G/H is a disjoint union of left cosets in G/H and also a disjoint union of right cosets in U\G. If {g1, . . . , gn} ⊆ G is a set of representatives of the double cosets of G with respect to U and

63 −1 −1 H then {g1 , . . . , gn } is a set of representatives of the double cosets of G with respect to H and U. In fact, the bijection G → G, g 7→ g−1, maps the double coset UgH to the double coset Hg−1U. Moreover, if U is normal in G then UH is a subgroup of G and UgH = gUH for all g ∈ G. Therefore U\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 = M F [UgH] g∈U\G/H is a decomposition of the (F U, F H)-bimodule FG into (F U, F H)-subbimodules. (d) Let F be a ﬁeld, let W be an FH-module and let g ∈ G. We set gW := Res (W ) , cg−1 the restriction of W along the isomorphism

g ∼ −1 cg−1 : H −→ H , x 7→ g xg . 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) denotes the 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

g g cg,H : R(H) → R( H) , ψ 7→ ψ , g with ψ deﬁned as in (7.9.a) for arbitrary ψ ∈ R(H). The map cg,H is an isometry with respect to the Schur inner product: One has

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

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

g G G M U H g Res (Ind (W )) ∼ Ind g (Res g ( W )) U H = U∩ H U∩ H g∈U\G/H of FU-modules.

64 L Proof By the decomposition FG = g∈U\G/H F [UgH] of FG into (F U, F H)- subbimodules we obtain an isomorphism

G G ∼ M ResU (IndH (W )) = FG ⊗FH W = F [UgH] ⊗FH W g∈U\G/H of FU-modules. Moreover, for ﬁxed g ∈ G, it is an easy veriﬁcation that the maps

∼ g g F [UgH] ⊗FH W ↔ FU ⊗F [U∩ H] W, ugh ⊗ w 7→ u ⊗ (ghg−1) · w , ug ⊗ w ←7 u ⊗ w , are well-deﬁned, mutually inverse isomorphisms of FU-modules. Note that one has to verify that the ﬁrst map does not depend on the choice of u ∈ U and 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

g G G X U H g res (ind (ψ)) = ind g (res g ( ψ)) . U H U∩ H U∩ H g∈U\G/H

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

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

g G G X G U H g ind (λ) · ind (ψ) = ind g res g (λ) · res g ( ψ) . U H U∩ H U∩ H U∩ H g∈U\G/H

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

G G G G G indU (λ) · indH (ψ) = indU λ · resU (indH (ψ)) g G X U H g = ind λ · ind g (res g ( ψ)) U U∩ H U∩ H g∈U\G/H g G X U U H g = ind ind g (res g (λ) · res g ( ψ) U U∩ H U∩ H U∩ H g∈U\G/H g X G U H g = ind g res g (λ) · res g ( ψ) , U∩ H U∩ H U∩ H g∈U\G/H and the corollary is proved.

7.13 Corollary (Mackey’s irreducibility criterion) Let ψ ∈ Irr(H). G (a) The character indH (ψ) is irreducible if and only if

g H H g res g (ψ), res g ( ψ) g = 0 , for all g ∈ G H. H∩ H H∩ H H∩ H r

G (b) Assume that H is normal in G. Then indH (ψ) is irreducible if and only if g ψ 6= ψ , for all g ∈ G r H. G Proof (a) Since the degree of indH (ψ) is positive, Proposition 2.18 implies G G G that indH (ψ) is irreducible if and only if (indH (ψ), indH (ψ))G = 1. Moreover, by Frobenius reciprocity (cf. Theorem 7.5) and the Mackey decomposition formula (cf. Corollary 7.11), we have

indG (ψ), indG (ψ) = ψ, resG (indG (ψ)) H H G H H H g X H H g = ψ, ind g (res g ( ψ)) H∩ H H∩ H H g∈H\G/H g X H H g = res g (ψ), res g ( ψ) g . H∩ H H∩ H H∩ H g∈H\G/H

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. Show that the permutation character χS introduced in an earlier problem satisfies

n X G χS = indHi 1Hi . i=1

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

j j X X λ E µ : ⇐⇒ λi 6 µi for all j. i=1 i=1 This defines a partial order on all partitions of 5. For every partition λ consider the 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 isomorphic to Sym(λ1) × Sym(λ2) × · · · . Now start with the unique maximal partition λ = (5) and compute the permutation character πλ := indSym(5)(1). Then go down Sλ the poset of partitions, at every step taking a new λ, maximal among the partitions that were not considered yet, compute πλ and check which of the previously constructed irreducible characters are constituents of πλ and compute their multiplicity in πλ. You should be left with precisely one irreducible summand that has not been constructed before. Call this character χλ and proceed down the poset of 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 set H := stabG(s) < G. By Problem 1 and Frobenius reciprocity, one has (χS, 1) = G G (indH (1H ), 1G)G = (1H , resH (1G))H = (1H , 1H )H = 1. Therefore, χS = 1G + ψ for some character ψ of G, 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. 2 (iii) One has (χS, 1) = 2 0 0 (iv) S is a doubly transitive G-set (i.e., for any two pairs (s1, s2), (s1, s2) ∈ 0 0 0 0 S × S with s1 6= s2 and s1 6= s2 there exists g ∈ G with gs1 = s1 and gs2 = gs2).

67 (v) One has |H\G/H| = 2. λ (b) Let n ∈ N and let λ = (n − 1, 1). Show that π (from Exercise 2) is equal Sym(n) to indSym(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 ∈ G be 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 following are equivalent: G (i) The FG-modules V and IndH (W ) are isomorphic. (ii) There exists an FH-submodule W 0 of V , with W 0 =∼ W , such that V = Ln 0 i=1 giW .

5. Let G be a ﬁnite group, U 6 H 6 G, and g ∈ G. Furthermore, let g g g −1 cg,H : R(H) → R( H), ψ 7→ ψ, be the conjugation map with ψ(x) := ψ(g xg) for x ∈ gH := gHg−1. g (a) Show that ch,H is the identity if h ∈ H and that cg0, H ◦ cg,H = cg0g,H for g, g0 ∈ G. g g H H H H (b) Show that c ◦ res = res g ◦ c and c ◦ ind = ind g ◦ c . g,U U U g,H g,H U U g,U

6. Let G be a ﬁnite group and let U, H 6 G. Let g1, . . . , gn ∈ G be a set of representatives of U\G/H and for each i = 1, . . . , n let ui,1, . . . , ui,ki be a set gi of representatives for U/U ∩ H. 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 the conjugacy 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 table satisﬁes 2 m d = (−1) |CG(g1)| · · · |CG(gh)| . (b) Let p be an odd prime and let ζ := e2πi/p. Show that

q p−1 (−1) 2 p ∈ Q(ζ) .

8. Let H be a subgroup of a ﬁnite group G and let ψ ∈ CF(H,F ) be a class function of H with values in a ﬁeld F . For g ∈ G set 1 X indG (ψ)(g) := ψ˙(x−1gx) . H |H| x∈G

68 Show that this deﬁnes an F -linear map

G indH : CF(H,F ) → CF(G, F ) , and verify that all the rules for induction of virtual characters is also true induction of class functions (transitivity, Frobenius Reciprocity, Frobenius property, Mackey formula, what else?)

69 8 Frobenius Groups

In this section we prove Frobenius’ Theorem, Theorem 8.1, on (what are now called) Frobenius groups. The statement of this theorem is purely in group theoretic terms. Its proof involves characters and until now there is now proof known that avoids representation theory and character theory. For a large part of this section we assume that G is a ﬁnite group and that H is a subgroup of G which satisﬁes

g H ∩ H = {1} , for every g ∈ G r H. (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, deﬁned as [ g N := G r H r {1}) , g∈GrH 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 of G.

8.3 Remark A deep theorem from 1960 due to John Thompson states that the subgroup N is nilpotent (i.e., a direct product of p-subgroups) and in fact the largest normal nilpotent subgroup of G with respect to inclusion. This justiﬁes the terminology ‘the Frobenius kernel’ of G. Moreover, one can show that 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 conjugation in 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 G satisfying (8.0.a) and let ψ ∈ R(H) with ψ(1) = 0. Then the following hold: G G (a) resH (indH (ψ)) = ψ. 0 0 G G 0 (b) For every ψ ∈ R(H) one has (ψ, ψ )H = (indH (ψ), indH (ψ ))G.

70 Proof (a) We need to show that 1 X ψ˙(xhx−1) = ψ(h) , |H| x∈G for every h ∈ H. If h = 1 then both sides are equal to 0. If h 6= 1 and x ∈ G then xhx−1 belongs to H if and only if x ∈ H. This implies that the left 1 P −1 hand side is equal to |H| x∈H ψ(xhx ) = ψ(h), since ψ is a class function on H. (b) This follows immediately from Part (a) and Frobenius reciprocity (cf. Theorem 7.5):

G G 0 G G 0 0 (ind (ψ), ind (ψ ))G = res (ind (ψ)), ψ = (ψ, ψ )H . H H H H 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 that G ∗ ∗ resH (ϕ ) = ϕ and N ⊆ ker(ϕ ).

Proof We ﬁrst assume that ϕ = 1H is the trivial character of H. Then the ∗ trivial 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 we have G (indH (ψ), 1G)G = (ψ, 1H )H = −ϕ(1) and ψ(1) = 0 . ∗ G Thus, the virtual character ϕ := indH (ψ) + ϕ(1)1G ∈ R(G) satisﬁes

∗ ∗ (ϕ , 1G) = 0 and ϕ (1) = ϕ(1) .

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

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

∗ ∗ ∗ 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, G G Lemma 8.4(a) implies that resH (indH (ψ)) = ψ, so that

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

71 G G ∗ Since resH (1G) = 1H , we obtain resH (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 is G not conjugate in G to any element of H. Therefore, (indH (ψ))(x) = 0 and ∗ G ∗ ϕ (x) = (indH (ψ) + ϕ(1) · 1G)(x) = ϕ(1) = ϕ (1). This completes the proof of the lemma.

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

M := \ ker(ϕ∗) . ϕ∈Irr(H)

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

M ∩ H = \ (ker(ϕ∗) ∩ H) = \ ker(ϕ) = {1} . (8.5.a) ϕ∈Irr(H) ϕ∈Irr(H)

The last equation follows immediately from the following line of argument: First observe that ker(χ) ∩ ker(χ0) = ker(χ + χ0) for all characters χ and χ0 T P 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}, since ρH (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 any non-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 normal subgroup 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 trivial g1 g2 intersection. In fact if g1 and g2 are elements in G such that H 6= H then g−1 −1 −1 1 g1 g2 g1 g2 g1 g2 ∈/ NG(H) = H and ( H ∩ H) = (H ∩ H) = {1}. This implies that g1H ∩ g2H = {1}. Therefore, we obtain

[ g |N| = |G| − | ( H r {1})| = |G| − [G : H](|H| − 1) = [G : H] . g∈G

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

The following corollary describes a situation that leads to the hypothesis in (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 such that no non-identity element of G fixes two or more elements of X. Let N denote the set consisting of the identity element of G together with all elements of G which do not fix a single element of X. Then N is a normal subgroup of G which still acts transitively on X.

Proof We fix an arbitrary element x of X and set H := stabG(x), the stabilizer of x in G. For every element y ∈ X there exists g ∈ G such g that y = gx. This implies stabG(y) = H. Since no non-identity element of G fixes more than one element of X we obtain H ∩ gH = {1} for all g ∈ G r H. Moreover, an element of G fixes at least one element of X if and only if it is conjugate to an element in H. This implies that the set N as 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 that NH = G. To show that N acts still transitively on X, note that G = NH implies 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 ∈ G r H. 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, is bijective. (b) Show that the maps in (a) form a group under composition. This group is called 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 , where the 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 group G 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 Fitting subgroup of G and denoted by F (G).

74 9 Elementary Cliﬀord Theory

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

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

ψ(1) X −1 eψ = ψ(x )x ∈ Z(CN) . |N| x∈N 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, the C-algebra isomorphism CN → CN, a 7→ gag−1, permutes the above idempotents as expected.

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

−1 ψ(1) X −1 −1 ψ(1) X −1 −1 geψg = g ψ(x )x g = ψ(x )gxg |N| x∈N |N| x∈N g ψ(1) X g −1 g = ψ(y )y = e ψ , |N| y∈N 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) denote its character. P (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 | g ψ = ψ} denote the inertia group of ψ in G, and let 1 = g1 . . . , gn ∈ G be a gi set of representatives of G/H. Finally, for i = 1, . . . , n, set ψi := ψ and let G Vi := eψi · V be the ψi-homogeneous component of ResN (V ). Then n n M G X V = Vi and resN (χ) = e ψi i=1 i=1

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 irreducible H -submodule, where H := g Hg−1. Moreover, IndG (V ) ∼ V for all C i i i i Hi i = i = 1, . . . , n.

0 P 0 Proof (a) Set V := g∈G gW . Then V 6= {0}, since W 6= {0}. Moreover, it is immediate that V 0 is a CG-submodule of V . Since V is a simple CG- module, we obtain V 0 = V as claimed.

(b) First note that, since eψi · eψj = 0 for i 6= j in {1, . . . , n}, the sum V1 + ··· + Vn of CN-submodules of V is a direct sum. Next we will show that if g ∈ G satisﬁes ggiH = gjH then gVi = Vj. In fact, in this case there exists h ∈ H such that ggi = gjh and we obtain

−1 −1 −1 −1 −1 gVi = geψi V = geψi g gV = ggieψgi g gV = gjheψh gj gV = eψj V = Vj ,

−1 −1 since gV = V and since gjheψh g = e gj h = eψ by Lemma 9.1. Since j ψ j g1 = 1 ∈ H, the CN-submodule V1 contains W . Therefore, V1 6= {0} and P Part (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 of G CN-submodules. This implies that resN (χ) is the sum of the characters of G the CN-modules Vi. Since Vi is the ψi-homogeneous component of ResN (V ), the character of Vi is equal to ni · ψi with ni = dimC(Vi)/ψi(1). But since gi Vi = giV1, we have dimC Vi = dimC V1, and since ψi(1) = ψ(1) = ψ(1), the positive integers ni, i = 1, . . . , n, are all equal. If we denote this number by e we obtain G resN (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

−1 −1 (gihgi )vi = (gih)(gi 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- P 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 ∼ IndG (V ) is irreducible, also the H -module V must be = Hi i C i i irreducible.

9.3 Example Let G = Sym(4) and N = 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 1 χ2 1 1 −1 1 −1 χ3 2 2 0 −1 0 χ4 3 −1 1 0 −1 χ5 3 −1 −1 0 1 and N 1 (1, 2)(3, 4) (1, 3)(2, 4) (1, 4)(2, 3) ψ1 1 1 1 1 ψ2 1 1 −1 −1 ψ3 1 −1 1 −1 ψ4 1 −1 −1 1 One can read from the character tables that

G resN (χ1) = ψ1 , G resN (χ2) = ψ1 , G resN (χ3) = 2 · ψ1 , G resN (χ4) = ψ2 + ψ3 + ψ4 , G resN (χ5) = ψ2 + ψ3 + ψ4 . One can also read from the character table of N that Irr(N) has two orbits under G-conjugation, namely {ψ1} and {ψ2, ψ3, ψ4}. Moreover, stabG(ψ1) = G and stabG(ψ2) = H1 := N · h(1, 2)i, stabG(ψ3) = N · h(1, 3)i and stabG(ψ4) = N ·h(1, 4)i. Note that the latter three subgroups of G have order 8. Moreover, G Theorem 9.2(c) implies that χ4 = indH1 (λ) for some λ ∈ Irr(H1). Comparing degrees, we obtain λ(1) = 1.

0 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

g g g H H g g H H g g g 0 0 ind (ψ) = ind g ( ψ) , res (χ) = res g ( χ) , ( ψ, ψ ) g = (ψ, ψ ) . K K K K K K

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

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

Note that by the above, Irr(G|ψ) = Irr(G| gψ) for all g ∈ G. Moreover, by Theorem 9.2, one has Irr(G|ψ) ∩ Irr(G|ψ0) = ∅, whenever ψ and ψ0 are not in the same G-orbit of Irr(N). Moreover, by Frobenius reciprocity, every χ ∈ Irr(G) belongs to a subset Irr(G|ψ) for some ψ ∈ Irr(N). Altogether, restriction and induction induce mutually inverse bijections between the set of 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. Let 0 H 6 G and let χ and χ be characters of G. Then

0 G G 0 (χ, χ )G res (χ), res (χ ) . 6 H H H

In fact, let V and V 0 be CG-modules with characters χ and χ0, respec- 0 0 G G 0 tively. Then (χ, χ )G = dim Hom G(V,V ) and res (χ), res (χ ) = C C H H H 0 0 dimC HomCH (V,V ). But, clearly one has an inclusion HomCG(V,V ) ⊆ 0 HomCH (V,V ).

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

H indN (ψ) = m1θ1 + ··· + mrθr with m1, . . . , mr ∈ N and with pairwise distinct θ1, . . . , θr ∈ Irr(H). H (a) One has Irr(H|ψ) = {θ1, . . . , θr}, resN (θi) = miψ for i = 1, . . . , r, and 2 2 m1 + ··· + mr = [H : N]. G (b) The function θ 7→ indH (θ) deﬁnes a bijection f : Irr(H|ψ) → Irr(G|ψ). Proof (a) The ﬁrst statement follows immediately from Frobenius reciprocity. The second statement is immediate with Frobenius reciprocity and Theorem 9.2(b), since ψ is stable under H. In order to prove the last statement in Part (a), consider the equations

H H H 2 2 resN (indN (ψ)) = resN (m1θ1 + mrθr) = (m1 + ··· + mr)ψ .

Since the degree of the left hand side is equal to [H : N]ψ(1) and the degree 2 2 of the right hand side is equal to (m1 + ··· + mr)ψ(1), the result follows.

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

g g 0 ( θi)| g , θi| g g ( θi)|N , θi|N 6 H∩ H H∩ H H∩ H 6 N g 2 g = (θi|N ) , θi|N = m ( ψ , ψ)N = 0 , N i since gψ 6= ψ. This implies that each term above is equal to 0. By Mackey’s irreducibility criterion for induced characters, cf. Corollary 7.13, we obtain that χi ∈ Irr(G). Moreover,

r G G H X G (χi|N , ψ)N = χi, ind (ψ) = χi, ind (ind (ψ)) = mi χi, ind (θi) N G H N G H G i=1 G χi, ind (θi) = (χi, χi)G = 1 . > H G 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 Frobenius G G H G reciprocity, χ is a constituent of indN (ψ) = indH (indN (ψ)) = indH (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

r G G X ind (ψ), ind (ψ) = (miχi , mjχj)G N N G i,j=1 2 2 2 2 2 > m1 + 2m1m2 + m2 + ··· + mr > m1 + ··· + mr = [H : N] .

On the other hand, the Mackey decomposition formula implies indG (ψ), indG (ψ) = ψ, resG (indG (ψ)) N N G N N G g X N N g = ψ, ind g (res g ( ψ)) N∩ N N∩ N N g∈G/N X g X g = (ψ, ψ)N = (ψ, ψ)N = [H : N] . g∈G/N g∈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. G (ii) indN (ψ) = χ. G 0 0 (iii) resN (χ) = ψ + ψ for some ψ 6= ψ ∈ Irr(N). (iv) |Irr(G|ψ)| = 1. 0 G 0 In this case, the G-orbit of ψ is equal to {ψ, ψ } and indN (ψ ) = χ. (b) The following are equivalent:

(i) stabG(ψ) = G. G 0 0 (ii) indN (ψ) = χ + χ for some χ 6= χ ∈ Irr(G). G (iii) resN (χ) = ψ. (iv) |Irr(G|ψ)| = 2. 0 G 0 In this case, one has Irr(G|ψ) = {χ, χ } and resN (χ ) = ψ.

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

G G g resN (indN (ψ)) = ψ + ψ (9.6.a) with g ∈ G r N. (a) Assume that stabG(ψ) = N. Then Theorem 9.5(b) implies (ii) and Equation (9.6.a) implies (iii). Moreover, Theorem 9.5(b) implies (iv). Now, g (ii) and Equation (9.6.a) imply that ψ0 = ψ with g ∈ G r N, and Re- G 0 G g g G g mark 9.4(a) implies that indN ψ = indN ( ψ) = indN (ψ) = χ = χ.

(b) Now assume that stabG(ψ) = G. Then Theorem 9.5(a) implies r = 2 2 2 and m1 = m2 = 1, since these are the only solutions to m1 + ··· + mr = [G : N] = 2. This immediately implies (ii) and (iv). Moreover, with Theo- 0 G rem 9.5(a) this also implies (iii). Now, since χ occurs in indN (ψ), Frobenius G 0 0 reciprocity implies that ψ occurs in resN (χ ), so that χ ∈ Irr(G|ψ). Thus, ev- 0 G 0 erything that was proved for χ also holds for χ . In particular, resN (χ ) = ψ. Finally, since each of the conditions (ii), (iii), and (iv) in (a) excludes the corresponding condition in (b) and vice-versa, each of the conditions (ii), (iii) and (iv) also implies (i) in Parts (a) and (b).

80 9.7 Example Consider G = Sym(5) and N = Alt(5). In a homework problem we determined the character degrees of N = Alt(5) as 1, 3, 3, 4, 5. From the character table of Alt(5) it is also easy to see that the two irreducible characters of degree 3 are G-conjugate. Since conjugate characters have the same 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 assume G that χ is an irreducible character of G satisfying (resN (χ), 1N )N 6= 0. Show that N 6 ker(χ).

2. Let G be a ﬁnite group, let N be a normal subgroup, let ψ be an irreducible character 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-subgroup of G. Assume that χ(1) is a power of p for every χ ∈ Irr(G). Show that P has a normal abelian complement, i.e., that there exists a normal abelian subgroup N of G such that P ∩ N = {1} and PN = G. (Hint: Use the equation |G| = [G : 0 P 2 G ]+ χIrr(G),χ(1)6=1 χ(1) 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 G and let χ be an irreducible character of G. Then χ(1) divides [G : A]. (Hint: Use Cliﬀord 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.

e1 er 10.1 Deﬁnition Let 2 6 n ∈ N and let n = p1 ··· pr the prime decomposition of n, with pairwise distinct primes p1, . . . , pr and positive integers e1, . . . , er. We deﬁne  0 if e 2 for some i ∈ {1, . . . , r}; µ(n) := i > r (−1) if ei = 1 for all i =∈ {1, . . . , r}.

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

10.2 Lemma For every positive integer n one has  1 if n = 1; X µ(d) = d|n 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 > 1 e1 er and let n = p1 ··· pr be its prime decomposition. Then, r > 1 and

X X f1 fr X f1 fr µ(d) = µ(p1 ··· pr ) = µ(p1 ··· pr ) d|n (f1,...,fr) (f1,...,fr) 06fi6ei 06fi61 r! = X (−1)k = (1 − 1)r = 0 . k 06k6r r Here, we counted tuples (f1, . . . , fr) ∈ {0, 1} by first fixing the number k of entries that are equal to 1 and then determining the number of such tuples r as the binomial coefficient k . The explicit formula in the next theorem is due to Richard Brauer.

10.3 Theorem For every χ ∈ R(G) one has 1 χ = X |K| · µ([H : K]) · indG (resG (χ)) . (10.3.a) |G| K K K6H6G H cyclic

82 In particular, one has 1 R(G) ⊆ X indG (R(H)) and R(G) = X indG ( R(H)) , |G| H Q H Q H6G H6G H cyclic H cyclic 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, the trivial character of G. The right hand side of the equation evaluated at the element x ∈ G is equal to 1 1 X |K| · µ([H : K]) · X 1 |G| |K| K6H6G g∈G H cyclic g−1xg∈K 1 = X X X µ([H : K]) |G| H6G g∈G g−1hxig6K6H H cyclic 1 = X X X µ(d) |G| H6G g∈G d|[H:g−1hxig] −1 H cyclic g hxig6H 1 X X = δ −1 |G| H,g hxig H6G g∈G −1 H cyclic g hxig6H 1 = X 1 = 1 , |G| g∈G and the equation holds in this case. For the second equality we used that a cyclic group of order n has precisely one subgroup of order d, for every divisor 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 character and the Frobenius property in Theorem 7.7, we obtain

χ = χ · 1G X G G = χ · |K| · µ([H : K]) · indK (resK (1G)) K6H6G H cyclic X G = |K| · µ([H : K]) · (χ · indK (1K )) K6H6G H cyclic X G G = |K| · µ([H : K]) · indK (resK (χ) · 1K ) K6H6G H cyclic X G G = |K| · µ([H : K]) · indK (resK (χ)) , K6H6G H cyclic proving Equation (10.3.a) in the general case. The last two statements in the 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 induction theorem. It was ﬁrst proved by Emil Artin in a diﬀerent way, before Brauer found the explicit formula in the above theorem.

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

n χ = X a · indG (ϕ ) . i Hi i i=1 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 normal series 1 = G0 6 G1 6 ··· 6 Gn = G, (Gi E G) such that Gi/Gi−1 is cyclic for all i = 1, . . . , n.

84 10.6 Remark (a) Every ﬁnite nilpotent group (i.e., direct product of p- groups) is supersolvable. On the other hand, every supersolvable group is solvable. (b) Subgroups and factor groups of supersolvable groups are again supersolvable. This is proved in the same way as the analogous result for solvable groups. (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- G ten as a sum of characters of the form indH (ϕ) with H 6 G and ϕ ∈ Irr(H) of degree 1. This is equivalent to requiring that there exists a representation of G with character χ which takes values in the subgroup of monomial matrices, i.e., matrices which have precisely one non-zero entry in each row and each 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 is G of the form indH (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 (since it has an irreducible character of degree 3, but no subgroup of order 20. In general, one can show that every monomial group is solvable, but we will not use this result.

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

G G/N G H infG/N (indH/N (ψ)) = indH (infH/N (ψ)) .

This is an easy veriﬁcation, noting that if g1, . . . , gn ∈ G is a set of representatives 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 a normal 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 in Z.

85 (b) We prove the statement by induction on |G|. If |G| = 1 then G is 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 that G χ = indK (ϕ) for some K 6 G and ϕ ∈ Irr(K) with ϕ(1) = 1. By Remark 10.8 we 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 in Z(G). Let a ∈ A r Z(G). Since a∈ / Z(G) and since ∆ is injective, we have ∆(a) ∈/ Z(∆(G)). This implies that ∆(a) is not a scalar matrix. By G P g Theorem 9.2(b), we have resA(χ) = e g∈H/A ψ, for some e ∈ N and some ψ ∈ 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 < G G is a proper subgroup. Theorem 9.2(c) implies that χ = indH (θ) for some H θ ∈ Irr(H). By induction, θ = indK (ϕ) for some K 6 H and ϕ ∈ Irr(K) with G G H G ϕ(1) = 1. Altogether, we obtain χ = indH (θ) = indH (indK (ϕ)) = indK (ϕ). This completes the proof.

10.10 Remark (a) In general, subgroups of monomial groups are not monomial, 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 second implication can be found in [I, Theorem 5.12 and Corollary 5.13]. We will not 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} \ G −1 = (resH ) (R(H)) H∈H and

G X G I(G, H) := hindH (ψ) | H 6 G, ψ ∈ R(H)iZ = indH (R(H)) . H∈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).

Proof (a) Clearly, the trivial character 1G is contained in R(G, H). More- G over, since resH : CF(G, C) → CF(H, C) is a ring homomorphism for all H 6 G −1 G and since R(H) is a subring of CF(H, C), the preimage (resH ) (R(H)) is a subring of CF(G, C). Since intersections of subrings of CF(G, C) are again subrings, also R(G, H) is a subring of CF(G, C). P G (b) Let χ ∈ R(G, H) and let H∈H aH indH (ψH ) ∈ I(G, H) with aH ∈ Z and ψH ∈ R(H) for H ∈ H. Then the Frobenius Property, cf. Theorem 7.7, implies

X G X G G χ · aH · indH (ψH ) = aH · indH (resH (χ) · ψH ) ∈ I(G, H) , H∈H H∈H

G since resH (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 p0-group if its order is not divisible by p. A finite group E is called p-elementary if E = C × P is the direct produce of a cyclic p0-subgroup C of E and a p-subgroup P of E. A group is called elementary if it is p-elementary for some 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 nilpotent and therefore supersolvable and monomial. Note also that subgroups and factor 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

n χ = X a · indG (ϕ ) i Ei i i=1

87 with ai ∈ Z, Ei ∈ E(G) (not necessarily distinct), and ϕi ∈ Irr(Ei) of degree 1, for i = 1, . . . , n. (c) A class function f ∈ CF(G, C) is a virtual character of G if and only if, for every elementary subgroup E of G, its restriction f|E is a virtual character of E.

10.15 Remark Part (b) in the above theorem (usually referred to as Brauer’s induction 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 above theorem. In order to prove Part (a) it suﬃces, by Lemma 10.12, to show that the trivial character 1G of G is contained in I(G, E(G)). The remainder of this section is devoted to prove this. The eventual proof will proceed by induction on |G|.

10.16 Lemma (Banaschewski) Let X be a ﬁnite non-empty set and let R 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 the constant function 1X with value 1). Then there exists an element x ∈ X and a 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 then there exists a prime p such that Ix ⊆ pZ and we are done. So assume that Ix = Z for all x ∈ X. We will derive a contradiction. In fact, in this case, for every x ∈ X, there exists a function fx ∈ R such that fx(x) = 1. Thus, for every x ∈ X, the function fx − 1X vanishes at x. This implies that the the Q function 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 of the elements fx ∈ R. Since R is multiplicatively closed and also an additive subgroup, we obtain 1X ∈ R. This is a contradiction, and the proof of the lemma is complete.

10.17 Deﬁnition We set

G P (G) := hindH (1H ) | H 6 GiZ ⊆ R(G) .

88 The set P (G) is called the ring of virtual permutation characters of G. (It is the subgroup of R(G) generated by characters of permutation representations, i.e., representations which take values in permutation matrices.) More generally, if H is a set of subgroups of G we set

G P (G, H) := hindH (1H ) | H ∈ HiZ ⊆ P (G) .

G Note that, for each H 6 G, the character indH (1H ) takes values in Z. Thus P (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 taking subgroups, i.e., if H ∈ H and K 6 H then K ∈ H. Then P (G, H) is an ideal of P (G).

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

G G X G g g indH (1H ) · indK (1K ) = indH∩ K (1H∩ K ) . g∈H\G/K

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 p0-subgroup C such that H/C is a p- group. Note that this implies that H = C o P for any Sylow p-subgroup P of H. Moreover, H is called quasi-elementary if H is p-quasi-elementary for some prime p. We denote by QE p(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 are again p-quasi-elementary. Also note that every p-elementary group is p- quasi-elementary. Thus, Ep(G) ⊆ QE p(G) and E(G) ⊆ QE(G).

10.20 Lemma Let x ∈ G and let p be a prime. Then there exists H ∈ G QE p(G) such that p - (indH (1))(x). Proof We can write hxi = C × D, with C a cyclic p0-subgroup and D a cyclic p-subgroup of hxi. Set N := NG(C) and note that hxi 6 N. Since hxi/C ∼= D is a p-group, we may choose a Sylow p-subgroup H/C of N/C containing hxi/C. Then hxi 6 H and H ∈ QE p(G).

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

G −1 (indH (1H ))(x) = |{gH ∈ G/H | g xg ∈ H}| . But, for g ∈ G, we have

−1 −1 −1 xgH = gH ⇐⇒ g xg ∈ H ⇐⇒ g hxig 6 H =⇒ g Cg 6 H =⇒ g−1Cg = C =⇒ g ∈ N =⇒ gH ∈ N/H .

G Therefore, the character value (indH (1H ))(x) is equal to the number of fixed points of the hxi-set N/H under the usual action by left multiplication. For c ∈ 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 hxi-fixed points of N/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 by p, and the proof is complete.

10.21 Proposition (L. Solomon) The trivial character 1G of G is contained 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 that p | χ(x) for all χ ∈ P (G, QE(G)). But Lemma 10.20 implies that for this prime p and this element x there exists H ∈ QE p(G) such that p does not di- G G vide (indH (1H ))(x). Since indH (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 product decomposition G = N o P with P ∈ Sylp(G). Assume further that ϕ ∈ Irr(N) with ϕ(1) = 1 and gϕ = ϕ for all g ∈ G. Finally assume that CN (P ) 6 ker(ϕ). Then ϕ = 1N , the trivial character of N. Proof Set K := ker(ϕ) E N. Then ϕ induces an injective homomorphism ϕ¯: N/K → C×. Let x ∈ N. We will show that x ∈ K. g −1 Since ϕ = ϕ for all g ∈ G, we have K E G and obtainϕ ¯(g xgK) = ϕ¯(g−1xKg) = ϕ(x) =ϕ ¯(xK). Thus, xK = g−1xgK = g−1xKg for every

90 g ∈ G. This implies that G and by restriction also P acts on the coset xK by 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 not divisible by p, xK has at least one ﬁxed point. Thus, xK ∩ CN (P ) 6= ∅. But CN (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 transitivity of induction together with Lemma 10.12(c), it suﬃces to show that 1G is a G Z-linear combination of characters of the form indH (ψ) with ψ ∈ Irr(H) and H < 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 p0-subgroup N of G such that 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 are done. Hence, we may assume that Z < N. We choose P ∈ Sylp(G) and set E := ZP < G. Note that E is elementary and that NE = G, since P 6 E.. G We can write indE(1E) = 1G + θ with some character 0 6= θ of G and we let χ ∈ Irr(G) denote an irreducible constituent of θ. It suﬃces to show that χ is G induced from a proper subgroup, since then 1G = indE(1E) − θ ∈ I(G, E(G)) by previous arguments. This is what we show in this last paragraph. We have G G G G 1N + resN (θ) = resN (1G + θ) = resN (indE(1E)) g X N E g = ind g (res g ( 1 )) N∩ E N∩ E E g∈N\G/E N N = indN∩E(1N∩E) = indZ (1Z ) , since N ∩ E = N ∩ PZ = (N ∩ P )Z = Z. Therefore, G N N (1N + resN (θ), 1N )N = (indZ 1Z , 1N )N = (1Z , resZ (1N ))Z = (1Z , 1Z )Z = 1 . G G This implies that (1N , resN (θ))N = 0 and also that (1N , resN (χ))N = 0. Now G let ϕ ∈ Irr(N) be an irreducible constituent of resN (χ). Then ϕ 6= 1N . We claim that Z 6 ker(χ). In fact, G G X Z g g resZ (indE(1E)) = indZ∩ E(1Z∩ E) = |Z\G/E| · 1Z = |G/E| · 1Z , g∈Z\G/E

91 since Z ∩ gE = gZ ∩ gE = g(Z ∩ E) = gZ = Z, for every g ∈ G, and since G Z is normal in G and Z 6 E. But since χ is a constituent of indE(1E), also G resZ (χ) is a multiple of 1Z . This proves the claim that Z 6 ker(χ). Now, by Lemma 10.22, ϕ cannot be invariant under G and we have U := stabG(ϕ) < G G. This implies that χ = indU (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 supersolvable.

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 are again p-elementary. (b) Show that subgroups and factor groups of p-quasi-elementary groups are again p-quasi-elementary.

4. Let ∆: G → GLn(C) be a representation of a ﬁnite group G. Show that the following are equivalent. (i) The representation ∆ is equivalent to a representation with values in permutation matrices. (ii) There exists a G-set X such that ∆ corresponds to the CG-module CX. (iii) The character of ∆ is of the form Pr indG (1), for some r ∈ and i=1 Hi N subgroups H1,...,Hr of G.

References

[I] M. Isaacs: Character theory of ﬁnite groups. Dover 1994.