REPRESENTATIONS OF FINITE GROUPS

1. Definition and Introduction 1.1. Definitions. Let V be a vector space over the field C of complex numbers and let GL(V ) be the group of isomorphisms of V onto itself. An element a of GL(V ) is a linear mapping of V into V which has a linear inverse a−1. When V has dimension n and has n a finite basis (ei)i=1, each map a : V → V is defined by a square matrix (aij) of order n. The coefficients aij are complex numbers. They are obtained by expressing the image a(ej) in terms of the basis (ei): X a(ej) = aijei. i Remark 1.1. Saying that a is an isomorphism is equivalent to saying that the determinant det(a) = det(aij) of a is not zero. The group GL(V ) is thus identified with the group of invertible square matrices of order n. Suppose G is a finite group with identity element 1. A representation of a finite group G on a finite-dimensional complex vector space V is a homomorphism ρ : G → GL(V ) of G to the group of automorphisms of V . We say such a map gives V the structure of a G-module. The dimension V is called the degree of ρ. Remark 1.2. When there is little ambiguity of the map ρ, we sometimes call V itself a representation of G. In this vein we often write gv˙ or gv for ρ(g)v. Remark 1.3. Since ρ is a homomorphism, we have equality ρ(st) = ρ(s)ρ(t) for s, t ∈ G. In particular, we have ρ(1) = 1, , ρ(s−1) = ρ(s)−1. A map ϕ between two representations V and W of G is a vector space map ϕ : V → W such that ϕ V −−−−→ W   g g (1.1) y y V −−−−→ W ϕ commutes for all g ∈ G. We will call this a G-linear map when we want to distinguish it from an arbitrary linear map between the vector spaces V and W . We can then define Ker ϕ, Im ϕ, and Coker ϕ, which are also G-modules. (Check this.)

∗These are some notes I wrote for the students in the class Representations of Finite Groups SS2013. The materials are from the following two books. J. P. Serre Linear representations of finite groups GTM42 W. Fulton; J. Haris Representation theory Part 1, GTM129 1 2

1.2. Basic examples. 1.2.1. Characters. A representation of degree 1 of a group G is a homomorphism ρ : × × G → C , where C denotes the multiplicative group of nonzero complex numbers. Since each element of G has finite order, the values ρ(g) of ρ are roots of unity. In particular, |ρ(g)| = 1. Degree 1 representations are called characters. If we take ρ(g) = 1 for all g ∈ G, we obtain a representation of G which is called the trivial representation or the unit representation. 1.2.2. Regular representation. Let |G| be the order of G, and let V be a vector space of dimension |G| with basis (eg)g∈G indexed by the elements g of G. For g ∈ G, let ρ(g) be the linear map of V into V which sends eg to egh. This defines a linear representation, which is called the regular representation of G. Its degree is equal to the order of G. Note that eg = ρg(e1). Hence the images of e1 form a basis of V . Conversely, let W be a representation of G containing a vector w such that ρg(w) for g ∈ G form a basis of W , then W is isomorphic to the regular representation. Define τ : V → W by τ(eg) = ρg(w), then τ gives us such an isomorphism. 1.2.3. Permutation representations. Suppose that G acts on a finite set X. This means that, for each g ∈ G, there is a permutation x 7→ gx of X, satisfying the identities 1x = x; g(hx) = (gh)x if g, h ∈ G, x ∈ X.

Let V be a vector space having a basis (ex)x∈X indexed by the elements of X. For g ∈ G, let ρg be the linear map of V into V which sends ex to egx. The linear representation of G thus obtained is called the permutation representation associated with X. 1.3. Subrepresentations and irreducible representations. 1.3.1. Some linear algebra. Let V be a vector space, and let W and W 0 be two subspaces of V . The space V is said to be the direct sum of W and W 0 if each x ∈ V can be written uniquely in the form x = w + w0, with w ∈ W and w0 ∈ W 0. This amounts to saying that the intersection W ∩ W 0 = 0 and that dim V = dim W + dim W 0. We then write V = W ⊕ W 0 and say that W 0 is a complement of W in V . The mapping p which sends each x ∈ V to its component of w ∈ W is called the projection of V onto W associated with the decomposition V = W ⊕ W 0. The image of p is W and p(x) = x for x ∈ W . Conversely, if p is a linear map of V into itself satisfying these two properties, one checks that V is the direct sum of W and the kernel W 0 of p. 1.3.2. Subrepresentations. A subrepresentation of a representation V is a vector subspace W of V which is invariant under G. A representation V is called irreducible if there is no proper nonzero invariant subspace W of V . Example 1.4. Take V for example the regular representation of G, and let W be the P subspace of dimension 1 of V generated by the element x = g∈G eg. We have ρgx = x for all g ∈ G. Therefore, W is a subrepresentation of V , isomorphic to the unit representation. Later we will determine all the subrepresentations of the regular representation. Theorem 1.5. Let ρ : G → GL(V ) be a linear representation of G in V and let W be a vector subspace of V stable under G. Then there exists a complement W 0 of W in V which stable under G. 3

Proof. We give two proofs of this theorem. (1) Recall a Hermitian product on V is a map ( , ): V × V → C such that • (w, z) = (z, w) • (w + w0, z) = (w, z) + (w0, z) • (aw, bz) =ab ¯ (w, z)

Let H0 be any Hermitian product on V , we define X H(x, y) = H0(gx, gy). g∈G

It is easy to check that H is also a Hermitian product on V and H(gx, gy) = H(x, y). Let W 0 be the orthogonal complement of W in V . Then W 0 is stable under G and we have our theorem. (2) Let W 0 be any complement of W in V and let p be the corresponding projection of V onto W . Form the average p0 of the conjugates of p by the elements of G:

1 X p0 = ρ ◦ p ◦ ρ−1. |G| g g g∈G

0 Since p maps V into W and ρg preserves W we see that p maps V into W . Note that −1 0 0 ρg x ∈ W for x ∈ W , so p x = x. Thus p is a projection of V onto W , corresponding to some complement W 0 of W . It is easy to check that

0 −1 0 ρg ◦ p ◦ ρg = p .

0 0 0 0 0 0 Thus ρg ◦ p = p ◦ ρg. If x ∈ W and g ∈ G, we have p x = 0, so p gx = xp x = 0. That 0 shows that W is stable under G and the theorem follows. 

1.3.3. Irreducible representations. Let V be a representation of G. By definition, V is irreducible if V is not the direct sum of two representations except for the trivial decom- position V = 0 ⊕ V . A representation of degree 1 is evidently irreducible.

Theorem 1.6. Every representation is a direct sum of irreducible representations.

Proof. Let V be a linear representation of G. We proceed by induction on dim(V ). If dim(V ) = 0, there is nothing to prove. Suppose that dim(V ) ≥ 1. If V is irreducible, there is nothing to prove. Otherwise, because of last theorem, V can be decomposed into a direct sum V 0 ⊕ V 00 with dim(V 0) < dim(V ) and dim(V 00) < dim(V ). By the induction hypothesis, V 0 and V 00 are direct sums of irreducible representations, and so the same is true for V . 

Remark 1.7. Let V be a representation of G, and let V = W1 ⊕· · ·⊕Wk be a decomposition of V into a direct sum of irreducible representations. We can ask if this decomposition is unique. The case where all the ρg are equal to 1 shows that this is not true in general. Nevertheless, we will show later that the number of Wi isomorphic to a given irreducible representation does not depend on the chosen decomposition. 4

1.4. Schur’s Lemma. Theorem 1.8. If V and W are irreducible representations of G and ϕ : V → W is a G-module homomorphism, then (1) Either ϕ is an isomorphism or ϕ = 0. (2) If V = W , then ϕ = λ · I for some λ ∈ C, I the identity map. Proof. The first claim follows from the the fact that Ker ϕ and Im ϕ are invariant sub- spaces. For the second, since C is an algebraic closed field, ϕ must have an eigenvalue λ, i.e., for some λ ∈ C, ϕ − λI has a nonzero kernel. By (1), we must have ϕ − λI = 0 and the claim follows.  Proposition 1.9. For any representation V of a finite group G, there is a decomposition

a1 ak V = V1 ⊕ · · · ⊕ Vk , where the Vi are distinct irreducible representations. The decomposition of V into a direct sum of the k factors is unique, as are the Vi that occur and their multiplicities ai. Proof. It follows from Schur’s lemma that if W is another representation of G, with a bj decomposition W = ⊕jWj and ϕ : V → W is a map of representations, then ϕ must map ai bj ∼ the factors Vi into the factors Wj for which Vi = Wj. If we apply this to the identity map of V to V , the stated uniqueness follows.  1.5. Construct new representations from old ones. 1.5.1. General construction. If V and W are representations, the direct sum V ⊕ W and the tensor product are also representations, via g(v ⊕ w) = gv ⊕ gw, g(v ⊗ w) = gv ⊗ gw. For a representation V , the nth tensor power V ⊗n is again a representation of G by this rule, and the exterior power Vn(V ) and symmetric power Symmn(V ) are subrepresen- tations of it. ∗ The dual V = Hom(V, C) of V is also a representation of G via t ρ∗(g) = ρ(g−1): V ∗ → V ∗. Note under this definition, we have hρ∗(g)(v∗), ρ(g)(v)i = hv∗, vi, where h, i is the natural pairing between V and V ∗.

1.5.2. Symmetric square and alternation square. Suppose that the representations V1 and V2 are identical to the same representation V , so that V1 ⊗ V2 = V ⊗ V . If (ei) is a basis of V , let θ be the automorphism of V ⊗ V such that

θ(ei ⊗ ej) = ej ⊗ ei, for all (i, j). It follows that θ(x ⊗ y) = y ⊗ x for all x, y ∈ V , and θ is independent of the chosen basis 2 (ei). Moreover, θ = 1. The space V ⊗ V then decomposes into a direct sum 2 ^ V ⊗ V = Symm2(V ) ⊕ (V ), 5 where Symm2(V ) is the set of elements z ∈ V ⊗ V such that θ(z) = z and V2(V ) is the set of elements of z ∈ V ⊗ V such that θ(z) = −z. The elements (ei ⊗ ej + ej ⊗ ei)i≤j for a 2 V2 basis of Symm (V ), and the elements (ei ⊗ ej − ej ⊗ ei)iabelian group G are thus simply elements of the dual group, that is, homomorphisms × ρ : G → C .

1.6.3. Representations of S3. We consider the simplest nonabelian group G = S3. To begin with, we have two one-dimensional representations: the trivial representation, which we will denote U, and the alternating representation U 0, defined by gv = sgn(g)v, for g ∈ G, v ∈ V . Next, since G comes to us as a permutation group, we have a nat- 3 ural permutation representation, in which G acts on C by permuting the coordinates. Explicitly, if (e1, e2, e3) is the standard basis, then g(ei) = eg(i), or equivalently,

g(z1, z2, z2) = (zg−1(1), zg−1(2), zg−1(3)). This representation is not irreducible. The line spanned by the sum (1, 1, 1) of the basis vectors is invariant, with complementary subspace 3 V = {(z1, z2, z3) ∈ C : z1 + z2 + z3 = 0}. 6

This two-dimensional representation V is easily seen to be irreducible, we call it the standard representation of S3. Now we describing an arbitrary representation of S3. (Later we will develop the so called character theory, which is a wonderful tool for describing arbitrary representation of a group.) We have seen that the representation theory of a finite abelian group is virtually trivial, we start our analysis of an arbitrary representation W of S3 by looking at the action of the abelian subgroup U3 = Z/3 ⊂ S3 on W . Let τ be any generator of U3, the space W is spanned by eigenvectors vi for the action of τ, whose eigenvalues are of course all powers of a cube root of unity ω = e2πi/3. Thus

W = ⊕Vi α where Vi = Cvi and τvi = ω i vi. Next, we ask how the remaining elements of S3 act on W in terms of this decomposition. Let σ be any transposition, so that τ and σ together generate S3, with the relation στσ = τ 2. Let v be an eigenvector of τ with eigenvalue ωi, we have τ(σ(v)) = σ(τ 2(v)) = σ(ω2i · v) = ω2i · σ(v). Therefore, σ(v) is again an eigenvector for τ with eigenvalue ω2i.

Lemma 1.10. Let σ = (12), τ = (123). Then the standard representation of S3 has a basis α = (ω, 1, ω2), β = (1, ω, ω2), with τα = ωα, τβ = ω2β, σα = β, σβ = α. Suppose we start with an eigenvector v for τ. If the eigenvalue of v is ωi 6= 1, then σ(v) is an eigenvector with eigenvalue ω2i 6= ωi, and so is independent of v. Now v and 0 0 σ(v) together span a two-dimensional subspace V of W invariant under S3. In fact, V is isomorphic to the standard representation, which follows from the lemma. If, on the other hand, the eigenvalue of v is 1, then σ(v) may or may not be independent of v. If it is not, then v spans a one-dimensional subrepresentation of W , isomorphic to the trivial representation if σ(v) = v and to the alternating representation if σ(v) = −v. If σ(v) and v are independent, then v + σ(v) and v − σ(v) span one-dimensional representations of W isomorphic to the trivial representation and alternating representation respectively. Thus, we have the following proposition.

Proposition 1.11. For an arbitrary representation W of S3, we can write W = U ⊕a ⊕ (U 0)⊕b ⊕ V ⊕c. Moreover, c is the number of independent eigenvectors for τ with eigenvalue ω, a+c is the multiplicity of 1 as an eigenvalue of σ, and b + c is the multoplicity of −1 as an eigenvalue of σ.

2. Character theory 2.1. The character of a representation. Let V be a vector space of dimension n having a basis (ei). Let A be a linear map of V into itself with matrix (aij)n×n. By the trace of A we mean the scalar X Tr(A) = aii. i 7

It is also the sum of the eigenvalues of A (counted with their multiplicities) and does not depend on the choice of the basis. Let ρ : G → GL(V ) be a linear representation of a finite group G in the vector space V . For each g ∈ G, define

χρ(g) = Tr(ρ(g)).

The complex valued function χρ on G thus obtained is called the character of the repre- sentation ρ. Proposition 2.1. If χ is the character of a representation ρ of degree n, then (1) χ(1) = n. (2) χ(g−1) = χ(g)∗ for g ∈ G. Here ∗ denotes the complex conjugation. (3) χ(hgh−1) = χ(g) for g, h ∈ G.

Proof. Since ρ(q) = In×n, Tr(1) = n. Note G is a finite group, any element g has finite order. Thus ρ(g) has finite order. Consequently, all of its eigenvalues λi have finite order and have absolute value equal to 1. Therefore,

∗ ∗ X ∗ X −1 −1 −1 −1 χ(g) = Tr(ρ(g)) = λi = λi = Tr(ρ(g) ) = Tr(ρ(g )) = χ(g ). (3) follows from the fact that similar matrices have the same eigenvalues. It is also a consequent of the formula Tr(AB) = Tr(BA).  Remark 2.2. A function f on G satisfying identity (3), or what amounts to the same thing, f(uv) = f(vu), is called a class function. Proposition 2.3. Let V and W be two representations of G, then

(1) χV ⊕W = χV + χW . (2) χV ⊗W = χV · χW . ∗ (3) χV ∗ = (χV ) . 1 2 2 (4) χ∧2V (g) = 2 (χV (g) − χV (g )). 1 2 2 (5) χSymm2 V (g) = 2 (χV (g) + χV (g )). Proof. For (1)-(3), we compute the values of these characters on a fixed element g ∈ G. For the action of g, V has eigenvalues λi and W has eigenvalues µi. Then V ⊕ W has ∗ −1 ∗ eigenvalues λi + µj, V ⊗ W has eigenvalues λi · µj, V has eigenvalues λi = λi . The formulas follow. 2 Let ei ∈ V be the eigenvector corresponds to the eigenvalue λi. Recall that ∧ V has a 2 basis ei ⊗ ej − ej ⊗ ei i < j, Symm has a basis ei ⊗ ej + ej ⊗ ei i ≤ j.

(ρV ⊗ ρW )(ei ⊗ ej ± ej ⊗ ei) = λiλj(ei ⊗ ej ± ej ⊗ ei). Therefore

X 1 X 2 X 2 χ 2 (g) = λ λ = (( λ ) − λ ). ∧ V i j 2 i i i

X 1 X 2 X 2 χ 2 (g) = λ λ = (( λ ) + λ ). Symm V i j 2 i i i≤j i i The results follow.  ∼ 2 Remark 2.4. The equality χ∧2V + χSymm2 V = χV ⊗V reflects the fact V ⊗ V = ∧ V ⊕ Symm2 V . 2.2. Some corollaries of Schur’s lemma. Recall the statement of Schur’s Lemma.

Theorem 2.5. Let ρi : G → GL(Vi) be two irreducible representations of the finite group G. Let f : V1 → V2 be a morphism of representations, i.e. ρ2(g)f(v1) = f(ρ1(g)v1) for all g ∈ G, v1 ∈ V1. Then ∼ (1) if V1 =6 V2, then f = 0. ∼ (2) if V1 = V2, then f is a scalar multiple of the identity.

Let us keep the hypothesis that V1 and V2 are irreducible.

Corollary 2.6. Let f be a linear map from V1 toe V2. Define 1 X f 0 = ρ (g)−1fρ (g). |G| 2 1 g∈G Then ∼ 0 (1) If V1 =6 V2, then f = 0. (2) If V =∼ V , then f 0 is a homothety of ratio 1 Tr(f). 1 2 dim V1 Proof. For any h ∈ G, we have the following equation 1 X ρ (h)−1f 0ρ (h) = ρ (h)−1ρ (g)−1fρ (g)ρ (h) 2 1 |G| 2 2 1 1 g∈G (2.1) 1 X = ρ (gh)−1fρ (gh) = f 0. |G| 2 1 g∈G 0 Therefore f is also a G-linear map from V1 to V2. The first statement is clear. Assume now V1 and V2 are isomorphic and f is the scalar λ, then 1 X Tr(f 0) = Tr(ρ (g)−1fρ (g)) = Tr(λ) = dim V · λ, |G| 1 1 1 g∈G the second statement follows.  We may write the above corollary in matrix form. Assume that

ρ1(g) = (ri1j1 (g)), ρ2(g) = (ri2j2 (g)). The linear map f is defined by a matrix (x ) and f 0 is defined by (x0 ). i2i1 i2i1 Corollary 2.7. In case (1), we have 1 X r (g−1)r (g) = 0 |G| i2j2 j1i1 g∈G for any ii, i2, j1, j2. 9

In case (2), we have

1 X −1 1 ri2j2 (g )rj1i1 (g) = δi2i1 δj2j1 . |G| dim V1 g∈G ( 1 if i = j Here δij = 0 otherwise Proof. By definition, 1 X x0 = r (g−1)x r (g). i2i1 |G| i2j2 j2j1 j1i1 g,j1,j2

The right hand side is a linear form with respect to xj2j1 . In case (1), this form vanishes for all systems of values of the xj2j1 . Thus its coefficients are zero. The formula follows. In case (2), we have x0 = λδ , i2i1 i2i1

0 1 1 X f = λ = Tr(λ) = δj2j1 xj2j1 . dim V1 dim V1 Therefore, we have the following equation

1 X −1 1 X ri2j2 (g )xj2j1 rj1i1 (g) = δi2i1 δj2j1 xj2j1 . |G| dim V1 g,j1,j2 j1,j2

Comparing the coefficients of xj2j1 , the equation follows.  Remark 2.8. (1) If φ and ψ are functions on G, set 1 X 1 X hφ, ψi = φ(g−1)ψ(g) = φ(g)ψ(g−1). |G| |G| g∈G g∈G We have hφ, ψi = hψ, φi. Moreover, hφ, ψi is linear in φ and ψ. With this notation, we may rewrite the above corollaries as 1 hri2j2 , rj1i1 i = 0, hri2j2 , rj1i1 i = δi2i1 δj2j1 dim V1 respectively. (2) Suppose that the matrices (rij(g)) are unitary (this can be realized by a suitable −1 ∗ choice of basis). We have then rij(g ) = rji(g) . The above corollaries are just orthogonality relations with respect to the scalar product we define next. 2.3. Orthogonality relations for characters. If φ and ψ are two complex-valued func- tions on G, define 1 X (φ|ψ) = φ(g)ψ(g)∗. |G| g∈G This is a scalar product. i.e. it is linear in φ, semilinear in ψ, and (φ|φ) > 0 for all φ 6= 0. If ψˆ is the function defined by the formula ψˆ(g) = ψ(g−1)∗, then 1 X (φ|ψ) = φ(g)ψ(g)∗ = hφ, ψˆi. |G| g∈G 10

In particular, if χ is the character of a representation of G, we haveχ ˆ = χ, so that (φ|χ) = hφ, χi for all function φ on G. So we can use at will (φ|χ) or hφ, χi, as long as we are concerned with characters. Theorem 2.9. (1) If χ is the character of an irreducible representation, we have (χ|χ) = 1. (2) If χ and χ0 are the characters of two nonisomorphic irreducible representations, we have (χ|χ0) = 0 Proof. Let ρ be an irreducible representation with character χ, given in matrix form ρ(g) = P (rij(g)). We have χ(g) = i rii(g), hence X (χ|χ) = hχ, χi = hrii, rjji. i,j

On the other hand, we know hrii, rjji = δij/ dim ρ, the statement follows. The second statement can be proved in the same way.  A character of an irreducible representation is called an irreducible character. The irreducible characters form an orthonormal system. Theorem 2.10. Let V be a linear representation of G, with character φ, and suppose that V decomposes into a direct sum of irreducible representations

V = W1 ⊕ · · · ⊕ Wk.

Then, if W is an irreducible representation with character χ, the number of Wi isomorphic to W is equal to the scalar product (φ|χ) = hφ, χi. P P Proof. Let χi be the character of Wi, then φ = i χi.Thus (φ|χ) = i(χi|χ). The theorem follows.  Corollary 2.11. The number of Wi isomorphic to W does not depend on the chosen decomposition. This number is called the number of times that W occurs in V . Corollary 2.12. Two representations with the same character are isomorphic. Proof. The above corollary shows that they contain each given irreducible representation the same number of times.  The above results reduce the study of representations to that of their characters. If χ1, . . . , χh are the distinct irreducible characters of G, and if W1,...,Wh denote corre- sponding representations, each representation V is isomorphic to a direct sum

W = m1W1 ⊕ · · · ⊕ mhWh, mi ∈ Z≥0. P The character φ of W is equal to i miχi, and we have mi = (φ|χi). The orthogonality relations imply X 2 (φ|φ) = mi . i We have the following result. Theorem 2.13. If φ is the character of a representation V , (φ|φ) is a positive integer and we have (φ|φ) = 1 if and only if V is irreducible. P 2 Proof. Indeed, i mi = 1 if and only if one of the mi’s is equal to 1 and the others to 0.  11

2.4. Decomposition of the regular representation. We fix some notation. The irre- ducible characters of G are denoted χi, . . . , χh, their degrees are written n1, . . . , nh. Let R be the regular representation of G. It has a basis (eg)g∈G such that ρ(h)(eg) = ehg. If h 6= 1, we have hg 6= g for all g, which shows that the diagonal terms of the matrix ρ(h) are zero. In particular, Tr(ρ(h)) = 0. On the other hand, for h = 1, we have Tr(ρ(1)) = dim R = |G|.

Proposition 2.14. The character rG of the regular representation is given by

rG(1) = |G|, rG(h) = 0 if h 6= 1.

Corollary 2.15. Every irreducible representation Wi is contained in the regular represen- tation with multiplicity equal to its degree ni

Proof. This number is equal to (ρG|χi). By the above proposition 1 X (ρ |χ ) = r (g−1)χ (g) = χ (1) = n . G i |G| G i i 1 g∈G The result holds.  P 2 Corollary 2.16. (1) The degrees ni satisfy the relation n = |G|. P i i (2) If g ∈ G is different from 1, we have i niχi(g) = 0. P Proof. From the above corollary, rG(g) = i niχi(g) for all g ∈ G. The results follow.  The above result can be used in determining the irreducible representations of a group G. Suppose we have constructed some mutually nonisomorphic irreducible representations of degree n1, . . . , nk, in order that they be all the irreducible representations of G, it is P 2 necessary and sufficient that ni = |G|. 2.5. Number of irreducible representations. Recall that a function f on G is called a class function if f(h−1gh) = f(g) for all h, g ∈ G. Proposition 2.17. Let f be a class function on G, and let ρ : G → GL(V ) be a linear irreducible representation of G. Let ρf be the linear map of V into itself defined by X ρf = f(g)ρ(g). g∈G

If V is irreducible of degree n and character χ, then ρf is a homothety of ratio λ given by 1 X |G| λ = f(g)χ(g) = (f|χ∗). n n g∈G Proof. For any h ∈ G, −1 X −1 ρ(h)ρf ρ(h ) = f(g)ρ(h)ρ(g)ρ(h ) g∈G X −1 (2.2) = f(g)ρ(hgh ) g∈G X −1 −1 = f(hgh )ρ(hgh ) = ρf . g∈G 12

By Schur’s Lemma, ρf is a homothety of ratio λ. On one hand, λ has trace nλ. On the other hand, X X ∗ Tr(ρf ) = f(g) Tr(ρ(g)) = f(g)χ(g) = |G|(f|χ ). g∈G g∈G 1 P |G| ∗ Therefore, λ = n g∈G f(g)χ(g) = n (f|χ ).  Let H denote the space of class functions on G. Note that the irreducible characters χ1, . . . , χh belong to H. Theorem 2.18. The characters χ1, . . . , χh form an orthonormal basis of H. Proof. We have proved above that χi form an orthonormal system in H. It remains to show that they generate H. For this, it suffices to show that every element of H orthogonal ∗ to the χi is zero. ∗ Let f ∈ H and assume that f is orthogonal to all χi . For each representation ρ of G, P ∗ put ρf = g∈G f(g)ρ(g). Since f is orthogonal to χi , the above proposition shows that ρf is zero as long as ρ is irreducible. From the direct sum decomposition, we conclude that ρf is always zero. Applying this to the regular representation R and compute the image of e1 under ρf , we have X X 0 = ρf (e1) = f(g)ρ(g)(e1) = f(g)eg. g∈G g∈G Therefore, f(g) = 0 for al g ∈ G and the theorem follows.  Recall that two elements g and g0 of G are conjugate if there exists h ∈ G such that g0 = hgh−1. This is an equivalence relation, which partitions G into conjugacy classes. Theorem 2.19. The number of irreducible representations of G (up to isomorphism) is equal to the number of conjugacy classes of G.

Proof. Let C1,...,Ck be the distinct classes of G. A function f on G is a class function if and only if f is constant on each Ci. Thus f is determined by its values λi on the Ci, and these can be chosen arbitrarily. Therefore, dim H = k. On the other hand, dim H = h, the number of irreducible representations of G. The theorem follows.  Proposition 2.20. Let g ∈ G, and let c(g) be the number of elements in the conjugacy class of g. Then Ph ∗ (1) i=1 χi(g) χi(g) = |G|/c(s). Ph ∗ (2) for h ∈ G not conjugate to g, i=1 χi(g) χi(h) = 0.

Proof. Let fg be the function equal to 1 on the class of g and equal to 0 elsewhere. It is a class function. We may write h X fg = λiχi, i=1 ∗ where λi = (fg|χi) = c(g)χi(g) /|G|. Therefore, for each h ∈ G, we have h c(g) X f (h) = χ (g)∗χ (h). g |G| i i i=1 The results follow.  13

2.6. Canonical decomposition of a representation. Let ρ : G → GL(V ) be a linear representation of G. Let χ1, . . . , χh be the distinct characters of the irreducible repre- sentations W1,...,Wh of G and n1, . . . , nh their degrees. Let V = U1 ⊕ · · · ⊕ Um be a decomposition of V into a direct sum of irreducible representations. For i = 1, . . . , h denote by Vi the direct sum of those of the U1,...,Um which are isomorphic to Wi. We have V = V1 ⊕ · · · ⊕ Vh. In other words, we have decomposed V into a direct sum of irreducible representations and collected together the isomorphic representations.

Theorem 2.21. (1) The decomposition V = V1 ⊕ · · · ⊕ Vh does not depend on the initially chose decomposition of V into irreducible representations. (2) The projection pi of V onto Vi associated with this decomposition is given by the formula ni X p = χ (g)∗ρ(g). i |G| i g∈G

Proof. Assertion (1) follows from (2) because the projections pi determine the Vi. It suffices to prove (2). Let ni X q = χ (g)∗ρ(g). i |G| i g∈G The restriction of q to an irreducible representation W with character χ and of degree n i ( 0 if χ 6= χi is a homothety of ratio (ni/n)(χi|χ) = . In other words, qi is the identity 1 if χ = χi. on an irreducible representation isomorphic to Wi, and is zero on the others. In view of the definition of Vi, it follows that qi is the identity on Vi and is 0 on Vj for j 6= i. This means that qi is equal to the projection pi of V onto Vi.  The decomposition of a representation V can be done in two steps. First the canonical decomposition V1 ⊕ · · · ⊕ Vh is determined. This can be done easily using the formulas giving the projections pi. Next, one chooses a decomposition of Vi into a direct sum of irreducible representations each isomorphic to Wi. Example 2.22. Let G be the group of two elements {1, g} with g2 = 1. This group has + − two irreducible representations of degree 1, W and W , corresponding to ρg = +1 and + − + ρg = −1. The canonical decomposition of a representation V is V = V ⊕ V , where V (resp. V −) consists of the elements x ∈ V which are symmetric (resp. antosymmetric), i.e., which satisfy ρ(g)(x) = x (resp. ρ(g)(x) = −x). The corresponding projections are p+x = (x + ρ(g)x)/2, p−x = (x − ρ(g)x)/2. 2.7. Explicit decomposition of a representation. Keep the notation of the preceding section, let V = V1 ⊕ · · · ⊕ Vh be the canonical decomposition of the given representation. We can determine the i-th component Vi by means of the projection. We now give a method for explicitly construction a decomposition of Vi into a direct sum of subrepresentations isomorphic to Wi. Let Wi be given in matrix form (rαβ(g)) with respect to a basis (e1, . . . , en). We have χi(g) = 14 P α rαα(g) and n = ni = dim Wi. For each pair of integers α, β taken from 1 to n, let pαβ denote the linear map of V into itself defined by n X p = r (g−1)ρ(g). αβ |G| αβ g∈G

Proposition 2.23. (1) The map pαα is a projection. It is zero on the Vj for j 6= i. Its image Vi,α is contained in Vi, and Vi is the direct sum of the Vi,α for 1 ≤ α ≤ n. P We have pi = α pαα. (2) The linear map pαβ is zero on the Vj for j 6= i, as well as on the Vi,γ for γ 6= β. It defines an isomorphism from Vi,β onto Vi,α. (3) Let x1 be an element 6= 0 of Vi,1 and let xα = pα1(x1) ∈ Vi,α. The xα are linearly independent and generate a vector subspace W (x1) stable under G and of dimension n. For each g ∈ G, we have X ρ(g)(xα) = rβα(g)xβ. β

Inparticular, W (x1) is isomorphic to Wi. (1) (m) (4) If (x1 , . . . , x1 ) is a basis of Vi,1, the representation Vi is the direct sum of the (1) (m) subrepresentations W (x1 ),...,W (x1 ). Proof. By definition, n X n X X p (e ) = r (g−1)ρ(g)(e ) = r (g−1)r (g)(e ). αβ γ |G| αβ γ |G| αβ δγ δ g∈G δ g∈G Therefore, ( eα if γ = β pαβ(eγ) = 0 otherwise. P We get from this the fact that α pαα is the identity map of Wi, and the formulas ( pαδ if β = γ pαβ ◦ pγδ = 0 otherwise

X ρ(g) ◦ pαγ = rβα(g)pβγ. β

For Wj with j 6= i, pαβ is zero by similar argument. We decompose V into a direct sum of subrepresentations isomorphic to Wj and apply the preceding to each of these representations. Then (1) and (2) follow. Under the hypothesis of (3), we have X X ρ(g)(xα) = ρ(g) ◦ pα1(x1) = rβα(g)pβ1(x1) = rβα(g)xβ. β β

This proves (3). The last statement follows from (1)-(3).  15

3. Induced representations 3.1. Abelian groups. Let G be a group. We say that G is abelian if gh = hg for all g, h ∈ G. This amounts to saying that each conjugacy class of G consists of a single element, also that each function on G is a class function. Theorem 3.1. The following two properties are equivalent. (1) G is ableian. (2) All the irreducible representations of G have degree 1.

Proof. Let (n1, . . . , nh) be the degrees of the distinct irreducible representations of G. Here h is the number of classes of G. We have 2 2 |G| = n1 + ··· + nh. Hence |G| = h if and only if all the ni are 1. The theorem follows.  Corollary 3.2. Let A be an abelian subgroup of G, let a be its order and let g be that of G. Each irreducible representation of G has degree ≤ g/a. Proof. Let ρ : G → GL(V ) be an irreducible representation of G. Through restriction to the subgroup A, it defines a representation ρA : A → GL(V ) of A. Let W ⊂ V be 0 an irreducible subrepresentation of ρA, then dim W = 1. Let V ⊂ V be the subspace generated by ρ(s)W , where s ranging over G. It is clear that V 0 is stable under G. Since ρ is irreducible, we must have V 0 = V . Note that for t ∈ A, s ∈ G, ρ(st)W = ρ(s)ρ(t)W = ρ(s)W. Therefore the number of distinct ρ(s)W is at most equal to g/a, hence the desired in- equality dim V ≤ g/a, since V is the sum of the ρ(s)W . 

3.2. Product of two groups. Let G1 and G2 be two groups, and let G1 × G2 be their product. As a set, G1 × G2 = {(s1, s2): si ∈ Gi}. Define

(s1, s2) · (t1, t2) = (s1t2, s2t2).

This gives a group structure on G1 × G2. Endowed with this structure, we call G1 × G2 the group product of G1 and G2. If Gi has order gi, then G1 ×G2 has order g = g1g2. The group G1 can be identified with the subgroup of G1 × G2 consisting of elements (s1, 1). Similarly, G2 can be identified with the subgroup of G1 ×G2 consisting of elements (1, s2). It is clear that each element of G1 commutes with each element of G2. Conversely, let G be a group containing G1 and G2 as subgroups, and suppose the following two conditions are satisfied:

(1) Each s ∈ G can be written uniquely in the form s = s1s2 with si ∈ Gi. (2) For si ∈ Gi, we have s1s2 = s2s1. Then G is isomorphic to the product G1 × G2. (Check this!) Let ρi : Gi → GL(Vi) be a linear representation of Gi. We define a linear representation ρ1 ⊗ ρ2 of G1 × G2 into V1 ⊗ V2 by setting

(ρ1 ⊗ r2)(s1, s2) = ρ1(s1) ⊗ ρ2(s2).

This representation is called the tensor product of the representations ρ1 and ρ2. If χi is the character of ρi, the character χ of ρ1 ⊗ ρ2 is given by

χ(s1, s2) = χ1(s1) · χ2(s2). 16

Remark 3.3. In the case G1 = G2 = G, the representation ρ1 ⊗ ρ2 defined above is a representation of G × G. When restricted to the diagonal subgroup G < G × G, it gives the representation of G denoted ρ1 ⊗ ρ2 we defined before. Although we have the identity of notations, it is important to distinguish these two representations.

Theorem 3.4. (1) If ρi is irreducible, then ρ1 ⊗ ρ2 is an irreducible representation of G1 × G2. (2) Each irreducible representation of G1×G2 is isomorphic to a representation ρ1⊗ρ2, where ρi is an irreducible representation of Gi.

Proof. If ρi is irreducible, we have

1 X 2 |χi(si)| = 1 for i = 1, 2. gi si∈Gi Taking product, we have 1 X |χ(s , s )|2 = 1. g 1 2 (s1,s2)∈G1×G2

Thus ρ1 ⊗ r2 is irreducible. To prove (2), we show that each class function f on G1 × G2, which is orthogonal to the characters of the form χ1(s1)χ2(s2), is zero. Suppose f is such a function, we have X ∗ ∗ f(s1, s2)χ1(s1) χ2(s2) = 0. s1,s2 Therefore, X X ∗ ∗ ( f(s1, s2)χ1(s1) )χ2(s2) = 0. s2 s1 This tells us X ∗ f(s1, s2)χ1(s1) = 0. s1 So f(s1, s2) = 0. The theorem follows. 

By the above theorem, to understand the representations of G1 × G2, it suffices to understand the representations of G1 and the representations of G2.

3.3. Induced representations.

3.3.1. Cosets of a subgroup. Let H be a subgroup of G. For s ∈ G, we denote by sH the set of products st with t ∈ H, and say that sH is the left coset of H containing s. Two elements s, s0 of G are said to the congruent modulo H if they belong to the same left coset, i.e., if s−1s0 belongs to H. In this case, we write s ≡ s0 (mod H). The set of left cosets of H is denoted by G/H. It is a partition of G. If G has g elements and H has h elements, then G/H has g/h elements. The integer g/h is the index of H in G and is denoted by (G : H). If we choose an element from each left coset of H, we obtain a subset R of G called a system of representatives of G/H. Each s ∈ G can be written uniquely as s = rt with r ∈ R and t ∈ H. 17

3.3.2. Definition of induced representations. Let ρ : G → GL(V ) be a linear representation of G, and let ρH be its restriction to H. Let W be a subrepresentation of ρH , that is, a vector subspace of V stable under the action of ρ(t), t ∈ H. Denote by θ : H → GL(W ) the representation of H in W thus defined. Let s ∈ G, the vector space ρ(s)W depends only on the left coset sH of s. (Indeed, if we replace s by st with t ∈ H, then ρ(st)W = ρ(s)ρ(t)W = ρ(s)(ρ(t)W ) = ρ(s)W.) If σ is a left coset of H, we can thus define a subspace Wσ of V to be ρ(s)W for any s ∈ σ. It is easy to see that the Wσ are permuted among P themselves by the ρ(s) s ∈ G. Their sum σ∈G/H Wσ is thus a subrepresentation of V . Definition 3.5. We say that the representation ρ of G in V is induced by the represen- tation θ of H in W if V is equal to the sum of the Wσ σ ∈ G/H and if this sum is direct (V = ⊕σ∈G/H Wσ). We can reformulate this condition in several ways: P (1) Each x ∈ V can be written uniquely as σ∈G/H xσ with xσ ∈ Wσ. (2) If R is a system of representatives of G/H, the vector space V is the direct sum of the ρ(r)W r ∈ R. P In particular, we have dim V = r∈R dim(ρ(r)W ) = (G : H) · dim W. 3.3.3. Some remarks.

(1) Take for V the regular representation of G. The space V has a basis (et)t∈G such that ρ(s)et = est for s, t ∈ G. Let W be the subspace of V with basis (et)t∈H . Then representation θ of H in W is the regular representation of H, and it is clear that ρ is induced by θ. (2) Take for V a vector space having a basis (eσ) indexed by the elements σ of G/H and define a representation ρ of G in V by ρ(s)eσ = esσ for s ∈ G. We thus obtain a representation of G which is the permutation representation of G associated with G/H. The vector space eH corresponding to the coset H is invariant under H. The representation of H in the subspace CeH is the unit representation of H and it induces the representation ρ of G in V . (3) If ρi is induced by θi for i = 1, 2, then ρ1 ⊕ ρ2 is induced by θ1 ⊕ θ2. (4) If (V, ρ) is induced by (W, θ), and if W1 is a stable subspace of W , the subspace P V1 = r∈R ρ(r)W1 of V is stable under G, and the representation of G in V1 is induced by the representation of H in W1. 0 0 (5) If ρ is induced by θ and ρ is a representation of G, and if ρH is the restriction of 0 0 0 ρ to H, then ρ ⊗ ρ is induced by θ ⊗ ρH . 3.3.4. Existence and uniqueness of induced representations. Lemma 3.6. Suppose that (V, ρ) is induced by (W, θ). Let ρ0 : G → GL(V 0) be a linear representation of G, and let f : W → V 0 be a linear H-equivariant map. Then there exists a unique linear map F : V → V 0 which extends f and is G-equivariant. Proof. If F satisfies these conditions, and if x ∈ ρ(s)W , then ρ(s)−1x ∈ W and −1 0 −1 0 −1 F (x) = F (ρ(s)ρ(s) x) = ρ (s)F (ρs x) = ρ (s)f(ρs x). This formula determines F on ρ(s)W and hence on V . The uniqueness of F follows. Let x ∈ Wσ and choose s ∈ σ. We define F (x) by the formula 0 −1 F (x) = ρ (s)f(ρs x). 18

This is well defined, i.e., it does not depend on the choice of s ∈ σ. Indeed, if we replace s by st with t ∈ H, we have

0 −1 0 0 −1 −1 0 −1 −1 0 −1 ρ (st)f(ρst x) = ρ(s) ρ(t) f(θ(t) ρ(s) x) = ρ(s) f(θ(t)θ(t) ρ(s) x) = ρ (s)f(ρs x).

Since V is the direct sum of the Wσ, the existence of F follows.  Theorem 3.7. Let (W, θ) be a linear representation of H. There exists a linear represen- tation (V, ρ) of G which is induced by (W, θ), and it is unique up to isomorphism.

Proof. By the remark above, we may assume that θ is irreducible. In this case, θ is isomorphic to a subrepresentation of the regular representation of H, which can be induce to the regular representation of G. By the remark again, θ can be induced. We prove now that ρ is unique up to isomorphism. Let (V, ρ) and (V, ρ0) be two repre- sentations induced by (W, θ). Applying the lemma to the injection W,→ V 0, we obtain a map F : V → V 0 which is G-equivariant and is identity on W . Therefore, the image of F contains all ρ0(s)W and hence all V 0. In other words, F is a surjection between two vector spaces with the same dimension, it is an isomorphism. The theorem follows. 

3.3.5. The character of an induced representation. Suppose that (V, ρ) is induced by (W, θ) and let χρ and χθ be the corresponding characters of G and of H. As we have seen, (W, θ) determines (V, ρ) up to isomorphism, we ought to be able to compute χρ from χθ.

Theorem 3.8. Let h be the order of H and let R be a system of representatives of G/H. For each u ∈ G, we have

X 1 X χ (u) = χ (r−1ur) = χ (s−1us). ρ θ h θ r∈R s∈G r−1ur∈H s−1us∈H

In other words, χρ(u) is a linear combination of the values of χθ on the intersection of H with the conjugacy class of u in G.

Proof. The space V is the direct sum of the ρ(r)W r ∈ R. Moreover, ρ(u) permutes the ρ(r)W among themselves. If we write ur in the form rut with ru ∈ R and t ∈ H, we see that ρ(u) sends ρ(r)W into ρ(ru)W . To determine χρ(u) = TrV (ρ(u)), we can use a basis of V which is a union of bases of the ρ(r)W . The indices r such that ru 6= r give zero diagonal terms. The others give the trace of ρ(u) on the ρ(r)W . We thus have X χρ(u) = Trρ(r)W (ρ|ρ(r)W (u)). r∈Ru

Here Ru denotes the set of r ∈ R such that ru = r. Note that r ∈ Ru if and only if ur can be written as rt with t ∈ H, i.e., if r−1ur ∈ H. Now we compute Trρ(r)W (ρ|ρ(r)W (u)). Write ρ(u, r) = ρ|ρ(r)W (u). Note that ρ(r) defines an isomorphism of W onto ρ(r)W and

ρ(r) ◦ θ(t) = ρ(u, r) ◦ ρ(r) with t = r−1ur ∈ H. 19

Indeed, the following diagram commutes. ρ(r) W −−−−→ ρ(r)W   (3.1) −1   r ury yρ(u,r) ρ(r) W −−−−→ ρ(r)W −1 The trace of ρ(u, r) is thus equal to that of θ(t), that is, to χθ(t) = χθ(r ur). Therefore X −1 χρ(u) = χθ(r ur). r∈R r−1ur∈H The second formula follows from the fact that for any s ∈ G and s ∈ rH, we have −1 −1 χθ(s us) = χθ(r ur).  3.4. Frobenius reciprocity law. Theorem 3.9. Let W (resp. V ) be a representation of H (resp. G). Then G G HomH (W, ResH (V )) = HomG(IndH (W ),V ).

Proof. Applying the lemma, for any f ∈ HomH (W, Res(V )), we have a unique element F ∈ HomG(Ind(W ),V ). On the other hand, for any F ∈ HomG(Ind(W ),V ), we may define an element f ∈ HomH (W, Res(V )) simply by letting f(w) = F (w). Check that this is well defined and the two maps are inverse to each other.  4. Group algebra 4.1. Representations and modules. Let G be a group of finite order g and let K be a commutative ring. We denote by K[G] the algebra of G over K. This algebra has a basis indexed by the elements of G. Each element f of K[G] can be uniquely written in the form X f = ass with as ∈ K, s∈G and multiplication in K[G] extends that in G. Let V be a K-module and let ρ : G → GL(V ) be a linear representation of G in V . For s ∈ G and x ∈ V , set sx = ρ(s)x. By linearity this defines fx for f ∈ K[G] and x ∈ V . Thus V is endowed with the structure of a left K[G]-module. Conversely, such a structure defines a linear representation of G in V . To say a ring or an algebra A is semisimple is equivalent to saying that each A-module M is semisimple, i.e., that each submodule M 0 of M is a direct factor in M as an A-module. Proposition 4.1. If K is a field of characteristic zero, the algebra K[G] is semisimple. Proof. Let V be a K[G]-module and V 0 ⊂ V a submodule. Certainly, V 0 is a sub K-vector space of V and is a direct factor as a K-module. Let p : V → V 0 be a K-linear projection, define 1 X P = sps−1. g s∈G Then P is a projection and is K[G]-linear, which implies that V 0 is a direct factor of V as a K[G]-module.  20

Corollary 4.2. If K is a field of characteristic zero, the algebra K[G] is a product of matrix algebras over skew fields of finite degree over K.

4.2. Decomposition of C[G]. Now we take K = C, so that each skew field of finite degree over C is equal to C. By the above corollary, C[G] is a product of matrix algebras Mni (C). More precisely, let ρi : G → GL(Wi) 1 ≤ i ≤ h be the distinct irreducible representations of G, and set ni = dim Wi, so that the ring End(Wi) if endomorphisms of Wi is isomorphic to Mni (C). The map ρi : G → GL(Wi) extends by linearity to an algebra homomorphismρ ˜i : C[G] → End(Wi). We thus obtain a homomorphism h h Y ∼ Y ρ˜ : C[G] → End(Wi) = Mni (C). i=1 i=1 Proposition 4.3. The homomorphism ρ˜ defined above is an isomorphism.

P 2 Qh Proof. Since g = i ni , both C[G] and i=1 Mni (C) have the same dimension. To show thatρ ˜ is an isomorphism, it suffices to show that it is surjective. Suppose otherwise, there Qh would exist a nonzero linear form on i=1 End(Wi) vanishing on the image ofρ ˜. This would induce a nontrivial relation on the coefficients of the representations ρi, which is impossible because of the orthogonality formulas. The proposition follows.  Qh Proposition 4.4 (Fourier inversion formula). Let (ui)1≤i≤h be an element of End(Wi) P i=1 and let u = s∈G u(s)s be the element of C[G] such that ρ˜i(u) = ui for all i. Then the s-th coefficient u(s) of u is given by the formula

h 1 X u(s) = n Tr (ρ (s−1)u ), where n = dim(W ). g i Wi i i i i i=1

Proof. It suffices to show this for u = t ∈ G. Let χi be the irreducible character of G corresponding to Wi. Then the formula is equivalent to h h 1 X 1 X δ = u(s) = n Tr (ρ (s−1)u ) = n χ (s−1t). st g i Wi i i g i i i=1 i=1 The proposition follows.  Proposition 4.5 (Plancherel formula). Let u = P u(s)s and v = P v(s)s be two elements P −1 of C[G] and define hu, vi = g s∈G u(s )v(s). Then we have formula h X hu, vi = ni TrWi (˜ρi(uv)). i=1 Proof. The proof is similar to the proof of last proposition. It suffices to prove that the −1 formula is true for u, v ∈ G. In this case, u(s ) = δus−1 , v(s) = δvs, TrWi (˜ρi(uv)) = χi(uv), the formula is equivalent to

X 2 g = ni .

This is well known.  21

4.3. The center of C[G]. The center of C[G] is the set of elements of C[G] which commute P with all the elements in C[G]. For c a conjugacy class of G, set ec = s∈c s. It is easy to check that ec form a basis for the center of C[G]. Therefore the center has dimension h, where h is the number of conjugacy classes of G. Let

ρi : G → GL(Wi) be an irreducible representation of G with character χi and degree ni, and let

ρ˜i : C[G] → End(Wi) be the corresponding algebra homomorphism.

Proposition 4.6. The homomorphism ρ˜i maps the center of C[G] into the set of homo- theties of Wi and defines an algebra homomorphism

ωi : Cent .C[G] → C. P If u = u(s)s is an element of Cent .C[G], we have 1 1 X ωi(u) = TrWi (˜ρi(u)) = u(s)χi(s). ni ni s∈G

Proposition 4.7. The family (ωi)1≤i≤h defines an isomorphism of Cent .C[G] onto the h algebra C . Q Proof. If we identify C[G] with the product 1≤i≤h End(Wi), the center of C[G] is the product of the centers of the End(Wi). But the center of End(Wi) consists of homotheties. h We then obtain an isomorphism of Cent .C[G] onto C . It is easy to check that this is the one induced by ωi.  4.4. Some algebra. Let R be a commutative ring and let x ∈ R. We say that x is integral over Z if there exists an integer n ≥ 1 and elements a1, . . . , an ∈ Z such that n n−1 x + a1x + ··· + an = 0. A complex number which is integral over Z is called an algebraic integer. Each root of unity is an algebraic integer.

Lemma 4.8. If x ∈ Q is an algebraic integer, then x ∈ Z. Proof. Assume otherwise, we could write x = p/q with p, q ∈ Z, q ≥ 2, and (p, q) = 1. Then by definition, we have n n−1 n p + a1qp + ··· + anq = 0. n Therefore, q|p , which is a contradiction.  Proposition 4.9. Let x be an element of a commutative ring R. The following properties are equivalent: (1) x is integral over Z. (2) The subring Z[x] of R generated by x is finitely generated as a Z-module. (3) There exists a finitely generated sub Z-module of R which contains Z[x]. 22

Proof. Since Z is noetherian, a submodule of a finitely generated Z-module is finitely generated. Therefore (2) and (3) are equivalent. If x is integral over Z, say x satisfying an equation n n−1 x + a1x + ··· + an = 0 with ai ∈ Z, n−1 the sub Z-module of R generated by 1, x, . . . , x is stable under the multiplication by x, and thus coincides with Z[x], which proves (1) implies (2). Conversely, suppose (2) n−1 holds. Denote by Rn the sub Z-module of R generated by 1, x, . . . , x . The Rn form an increasing sequence, and the union is Z[x]. Since Z[x] is finitely generated, we must have n Z[x] = Rn for n sufficiently large. This shows that x is a linear combination with integer n−1 coefficients of 1, x, . . . , x , therefore (1) holds.  Corollary 4.10. If R is a finitely generated Z-module, each element of R is integral over Z. Corollary 4.11. The elements of R which are integral over Z form a subring of R. Proof. Let x, y ∈ R. If x, y are integral over Z, the rings Z[x] and Z[y] are finitely generated over Z. The same is true for the tensor product Z[x] ⊗ Z[y] and of its image Z[x, y] in R. Thus all the elements of Z[x, y] are integral over Z.  4.5. Integrality properties of characters. Proposition 4.12. Let χ be the character of a representation ρ of a finite group G. Then χ(s) is an algebraic integer for each s ∈ G. Proof. Indeed χ(s) is the trace of ρ(s), hence is the sum of eigenvalues of ρ(s), which are roots of unity. The proposition follows.  P Proposition 4.13. Let u = u(s)s be an element of Cent .C[G] such that the u(s) are algebraic integers. Then u is integral over Z. (This statement makes sense because Cent .C[G] is a commutative ring.) Proof. Let c (1 ≤ i ≤ h) be the conjugacy classes of G and put e = P s. For s ∈ c i i s∈ci i i Ph we can write u in the form u = i=1 u(si)ei. Therefore, it suffices to show that ei is integral over Z for any i. Note that eiej is a linear combination with integral coefficients of the ek. The subgroup R = Ze1 ⊕ · · · ⊕ Zeh of Cent .C[G] is thus a subring and it is finitely generated. Every element in R is integral over Z. The proposition follows.  Corollary 4.14. Let ρ be an irreducible representation of G of degree n and character χ. 1 P If u is as above, then the number n s∈G u(s)χ(s) is an algebraic integer. Proof. This number is the image of u under the homomorphism

ω : Cent .C[G] → C associated with ρ we constructed before. As u is integral over Z, the same is true for its image under ω.  Corollary 4.15. The degrees of the irreducible representations of G divide the order of G. 23

Proof. Let g be the order of G. Applying the above corollary to the element u = P −1 s∈G χ(s )s, which is legitimate since χ is a class function and χ(s) are algebraic inte- gers, we see that the number 1 X g g χ(s−1)χ(s) = hχ, χi = n n n s∈G is an algebraic integer. Therefore it is an integer and n divides g.  Proposition 4.16. Let C be the center of G. The degrees of the irreducible representations of G divide (G : C). Proof. Let g be the order of G and c be the order of C. Let ρ : G → GL(W ) be an irreducible representation of G of degree n. If s ∈ C, ρ(s) commutes with all the ρ(t) t ∈ G. By Schur’s lemma, ρ(s) is a homothety. We denote it by λ(s). The map λ : s 7→ λ(s) is a × homomorphism of C into C . Let m be an integer ≥ 0 and form the tensor product ρm : Gm → GL(W ⊗ · · · ⊗ W ) of m copies of the representation ρ. This is an irreducible representation of Gm. The m m image under ρ of an element (s1, . . . , sm) of C is the homothety of ratio λ(s1 . . . sm). m The subgroup H of C consisting of the (s1, . . . , sm) such that s1 ··· sm = 1 acts trivially on W ⊗ · · · ⊗ W . Passing to the quotient, we obtain an irreducible representation of Gm/H. It follows that the degree nm of this representation divides the order gm/cm−1 of m m −1 G /H. Therefore (g/cn) ∈ c Z for all m, which implies that g/cn is an integer. The proposition follows. 

5. Mackey’s criterion 5.1. Induction. Let H be a subgroup of G and R a system of left coset representatives for H. Let V be a C[G]-module and let W be a sub C[H]-module of V . Recall that the module V (or the representation V ) is said to be induce by W if we have V = ⊕s∈RsW . This property can be reformulated in the following way. Let 0 W = C[G] ⊗C[H] W be the C[G]-module obtained from W by scalar extension from C[H] to C[G]. The injection 0 W,→ V extends by linearity to a C[G]-homomorphism i : W → V . Proposition 5.1. In order that V be induced by W , it is necessary and sufficient that the homomorphism

i : C[G] ⊗C[H] W → V be an isomorphism.

Proof. This is a consequence of the fact that the elements of R form a basis of C[G] considered as a right C[H]-module.  Remark 5.2. (1) This characterization of the representation induced by W makes it obvious that the induced representation exists and is unique. Indeed, it is

C[G] ⊗C[H] W . G From now on, the representation of G induced by W will be denoted by IndH (W ). 24

(2) If V is induced by W and if E is a C[G]-module, we have a canonical isomorphism ∼ HomH (W, E) = HomG(V,E) by the property of tensor product. This is Frobenius reciprocity. (3) Induction is transitive in the following sense. If G is a subgroup of a group K, we have K G ∼ K IndG (IndH (W )) = IndH (W ). The following proposition is clear.

Proposition 5.3. Let V be a C[G]-module which is a direct sum V = ⊕i∈I Wi of vector spaces permuted transitively by G. Let i0 ∈ I, W = Wi0 . Let H be the stabilizer of W in G, i.e., the set of all s ∈ G such that sW = W . Then W is stable under the subgroup H and the C[G]-module V is induced by the C[H]-module W .

When the Wi are of dimension 1, the representation V is said to be monomial.

5.2. The character of an induced representation. If f is a class function on H, consider the function f 0 on G defined by the formula 1 X f 0(s) = f(t−1st) where h = |H|. h t∈G t−1st∈H 0 G We say that f is induced by f and denote it by either IndH (f) or Ind(f). Proposition 5.4. (1) The function Ind(f) is a class function on G. (2) If f is the character of a representation W of H, Ind(f) is the character of the induced representation Ind(W ) of G. Proof. (2) is proved before. Since each class function is a linear combination of character, (1) follows from (2). 

Let g be the order of G. For φ1 and φ2 two class functions on G, we have defined 1 X hφ , φ i = φ (s−1)φ (s). 1 2 g 1 2 s∈G

We write hφ1, φ2iG instead of hφ1, φ2i if we want to be more explicit about the group G. If V1 and V2 are two C[G]-modules, we set

hV1,V2iG = dim(HomG(V1,V2)).

Lemma 5.5. If φi is the character of Vi, i = 1, 2, we have

hφ1, φ2iG = hV1,V2iG.

Proof. If V1 and V2 are irreducible, the lemma follows from the orthogonality formulas for characters. In the general case, we decompose V1 and V2 into direct sums of irreducible modules, the lemma then follows by a simple computation.  If φ (resp. V ) is a function on G (resp. a representation of G), we denote by Res φ (resp. Res V ) its restriction to H. 25

Theorem 5.6 (Frobenius reciprocity). If ψ is a class function on H and φ is a class function on G, we have hψ, Res φiH = hInd ψ, φiG. Proof. Since each class function is a linear combination of characters, we can assume that ψ is the character of C[H]-module W and φ is the character of a C[G]-module E. By the above lemma, it suffices to show that

hW, Res EiH = hInd W, EiG. This is equivalent to

dim(HomH (W, Res E)) = dim(HomG(Ind W, E)), which is proved before.  Remark 5.7. (1) In the language of categories, the functors Res and Ind are adjoints to each other. (2) Instead of the bilinear form h·, ·i, we can use the scalar product (·|·) defined before. Then the formula is

(ψ| Res φ)H = (Ind ψ|φ)G. (3) By the formula Ind(W ) ⊗ E =∼ Ind(W ⊗ Res E), we have Ind(ψ) · φ = Ind(ψ · Res φ). Proposition 5.8. Let W be an irreducible representation of H and E an irreducible representation of G. Then the number of times that W occurs in Res E is equal to the number of times that E occurs in Ind W . 5.3. Restriction to subgroups. Let H and K be two subgroups of G. Let ρ : H → G GL(W ) be a linear representation of H and V = IndH (W ) be the corresponding induced representation of G. In the following, we determine the restriction ResG (V ) of V to K. K ` Choose a set of representatives S for the (H,K) double cosets of G, i.e., G = s∈S HsK. −1 For s ∈ S, let Hs = sHs ∩ K, which is a subgroup of K. We set s −1 ρ (x) = ρ(s xs) for x ∈ Hs, s we obtain a homomorphism ρ : Hs → GL(W ) and hence a linear representation of Hs, K denoted Ws. Since Hs is a subgroup of K, the induced representation IndHs (Ws) is defined. G Proposition 5.9. The representation ResK IndH (W ) is isomorphic to the direct sum of K ∼ the representations IndHs (Ws) for s ∈ S = K\G/H.

Proof. We know that V = ⊕x∈G/H xW . Let s ∈ S and let V (s) be the subspace of V generated by the images xW , for x ∈ KsH. Then V = ⊕s∈SV (s). It is easy to see K that V (s) is stable under K. We show that V (s) is K-isomorphic to IndHs (Ws). The subgroup of K consisting of the elements x such that x(sW ) = sW is equal to Hs and K V (s) = ⊕x∈K/Hs x(sW ). Therefore V (s) = IndHs (sW ). Moreover, s : Ws → sW induces K an Hs-isomorphism between Ws and sW . Thus V (s) = IndHs (Ws). The proposition follows.  26

Note that V (s) depends only on the image of s in K\G/H, thus the representation K IndHs (Ws) depends only on the double coset of s. 5.4. Mackey’s irreducibility criterion. We apply the preceding results to the case −1 K = H. For s ∈ G, we denote by Hs the subgroup sHs ∩ H of H. The representation ρ of H defines a representation Ress(ρ) by restriction to Hs. Two representations V1 and V2 of a group K are said to be disjoint if they have no irreducible component in common, i.e., hV1,V2iK = 0. G Proposition 5.10. In order that the induced representation V = IndH W be irreducible, it is necessary and sufficient that the following two conditions be satisfied: (1) W is irreducible. s (2) For each s ∈ G − H the two representations ρ and Ress(ρ) of Hs are disjoint.

Proof. In order that V be irreducible, it is the same that hV,V iG = 1. We have

hV,V iG = hW, ResH V iH H s = hW, ⊕s∈H\G/H IndHs (ρ )i X = hW, IndH (ρs)i (5.1) Hs H s∈H\G/H X s = hRess(ρ), ρ iHs . s∈H\G/H s For s = 1, we have ds = hRess(ρ), ρ iHs = hρ, ρi ≥ 1. In order that hV,V i = 1, it is necessary and sufficient that d1 = 1 and ds = 0 for s 6= 1. There are exactly conditions (1) and (2).  G Corollary 5.11. Suppose H is normal in G. In order that IndH ρ be irreducible, it is necessary and sufficient that ρ be irreducible and not isomorphic to any of its conjugates ρs for s 6∈ H.

Proof. Indeed, if H is normal, then Hs = H and Ress(ρ) = ρ.  5.5. Degrees of the irreducible representations. A representation is said to be isotypic if it is a direct sum of isomorphic irreducible representations. Proposition 5.12. Let A be a normal subgroup of a group G, and let ρ : G → GL(V ) be an irreducible representation of G. Then (1) either there exists a subgroup H of G, unequal to G and containing A, and an irreducible representation σ of H such that ρ is induced by σ; (2) or else the restriction of ρ to A is isotypic.

Proof. Let V = ⊕Vi be the canonical decomposition of the representation ρ|A into a direct sum of isotypic representations. For s ∈ G, we see that ρ(s) permutes the Vi. Since V is irreducible, G permutes them transitively. Let Vi0 be one of these. If Vi0 is equal to V , we have case (2). Otherwise, let H be the subgroup of G consisting of those s ∈ G such that ρ(s)Vi0 = Vi0 . We have A ⊂ H and H 6= G. Moreover, ρ is induced by the natural representation σ of H in Vi0 , which is case (1).  27

Remark 5.13. If A is abelian, (2) is equivalent to saying that ρ(a) is a homothety for each a ∈ A.

Corollary 5.14. If A is an abelian normal subgroup of G, the degree of each irreducible representation ρ of G divides the index (G : A) of A in G.

Proof. We prove this by induction on the order of G. In case (1), by induction hypothesis, the degree of σ divides (H : A) and by multiplying this relation by (G : H), we see that the degree of ρ divides (G : A). In case (2), let G0 = ρ(G) and A0 = ρ(A). Since G/A → G0/A0 is surjective, we have (G0 : A0)|(G : A). The above remark shows that the elements of A0 are homotheties, thus are contained in the center of G0. Therefore, the degree of ρ divides 0 0 (G : A ). The corollary follows.  Remark 5.15. If A is an abelian subgroup of G (not necessarily normal), it is no longer true that deg(ρ)|(G : A). Nevertheless, we have proved before that deg(ρ) ≤ (G : A).

6. Artin’s theorem

6.1. The ring R(G). Let G be a finite group and let χ1, ··· , χh be its distinct irreducible characters. A class function on G is a character if and only if it is a linear combination + of χi with non-negative integer coefficients. We will denote by R (G) the set of these functions, and by R(G) the group generated by R+(G), i.e., the set of differences of two characters. We have

R(G) = Zχ1 ⊕ · · · ⊕ Zχh. An element of R(G) is called a virtual character. Since the product of two characters is a character, R(G) is a subring of the ring FC(G) of class functions on G with complex values. Since the χi form a basis of FC(G) over C, we see that C ⊗ R(G) can be identified with FC(G). Remark 6.1. We can also review R(G) as the Grothendieck group of the category of finitely generated C[G]-modules. Let H be a subgroup of G. The operation of restriction defines a ring homomorphism G R(G) → R(H), denoted by ResH or Res. Similarly, the operation of induction defines a homomorphism of abelian groups R(H) → G R(G), denoted by IndH or Ind. The homomorphisms Ind and Res are adjoints of each other with respect to the bilinear forms h·, ·iH and h·, ·iG. Moreover, the formula Ind(φ · Res(ψ)) = Ind(φ) · ψ shows that the image of Ind : R(H) → R(G) is an ideal of the ring R(G). If A is a commutative ring, the homomorphism Res and Ind extend by linearity to A-linear maps: A ⊗ Res : A ⊗ R(G) → A ⊗ R(H).

A ⊗ Ind : A ⊗ R(H) → A ⊗ R(G). 28

6.2. Statement of Artin’s theorem.

Theorem 6.2. Let X be a family of subgroups of a finite group G. Let Ind : ⊕H∈X R(H) → G R(G) be the homomorphism defined by the family of IndH H ∈ X. Then the following properties are equivalent: (1) G is the union of the conjugates of the subgroups belonging to X. (2) The cokernel of Ind : ⊕H∈X R(H) → R(G) is finite. Since R(G) is finitely generated as a group, we can rephrase (2) in the following way: (3) For each character χ of G, there exist virtual characters χH ∈ R(H) and an integer d ≥ 1 such that X G dχ = IndH (χH ). H∈X Proof. First, we show that (2) ⇒ (1). Let S be the union of the conjugates of the P G subgroups H belonging to X. Each function of the form IndH (fH ) with fH ∈ R(H) vanishes outside S. If (2) is satisfied, it follows that each class function on G vanishes outside S, which shows that S = G. Hence (1) holds. Conversely, suppose (1) is satisfied. To prove (2), it suffices to show that the Q-linear map Q ⊗ Ind : ⊕H∈X Q ⊗ R(H) → Q ⊗ R(G) is surjective, which is also equivalent to the surjectivity of the C-linear map

C ⊗ Ind : ⊕H∈X C ⊗ R(H) → C ⊗ R(G) By duality this is equivalent to the injectivity of the adjoint map

C ⊗ Res : C ⊗ R(G) → ⊕H∈X C ⊗ R(H). But this injectivity is amounts to saying that if a class function on G restricts to 0 on conjugates of all H ∈ X, then it is zero. This is obvious and the theorem follows.  Note that the family of cyclic subgroups of G satisfies (1). Hence we have the following result. Corollary 6.3. Each character of G is a linear combination with rational coefficients of characters induced by characters of cyclic subgroups of G. 6.3. Another proof. In this section, we give another proof for (1)⇒(2). Let A be a cyclic group and let a be its order. Define a function θA on A by the formula ( a if x generates A θA(x) = 0 otherwise

Proposition 6.4. If G is a finite group of order g, then

X G g = IndA θA, A⊂G where A runs through all the cyclic subgroups of G. 29

0 G Proof. Put θA = IndA θA. For x ∈ G, we have 1 X θ0 (x) = θ (yxy−1) A a A y∈G yxy−1∈A (6.1) 1 X X = a = 1 a y∈G y∈G =A =A Note that for each y ∈ G, yxy−1 generates a unique cyclic subgroup of G. We then have X 0 X θA(x) = 1 = g. A⊂G y∈G The proposition follows. 

Proposition 6.5. If A is a cyclic group, then θA ∈ R(A). Proof. We prove this by induction on the order a of A. The case a = 1 is trivial. We have X A X A a = IndB(θB) = θA + IndB(θB). B⊂A B6=A A The induction hypothesis gives θB ∈ R(B) for B 6= A, hence IndB(θB) ∈ R(A) for B 6= A. Therefore θA ∈ R(A) since it is clear that a ∈ R(A).  0 G Proof of (1)⇒(2). First note that, if A is contained in a conjugate of A, the image IndA0 G is contained in that of IndA. Hence we can assume that X is the family of all cyclic subgroups of G. Then we have X G g = IndA(θA) with θA ∈ R(A). A∈X Thus the element g belongs to the image of Ind. Since this image is an ideal of R(G), it contains every element of the form gχ with χ ∈ R(G). This proves (3), hence proves (2). 

7. Brauer’s theorem In this section, the letter p denotes a prime number.

7.1. Some group theory. Definition 7.1. One says that G is solvable if there exists a sequence

{1} = G0 ⊂ G1 ⊂ · · · ⊂ Gn = G of subgroups of G, with Gi−1 normal in Gi and Gi/Gi−1 abelian for 1 ≤ i ≤ n. (Equivalent definition: G is obtained from the group {1} by a finite number of extensions with abelian kernels.) One says that G is supersolvable if there exists a sequence

{1} = G0 ⊂ G1 ⊂ · · · ⊂ Gn = G of subgroups of G, with Gi normal in G and Gi/Gi−1 cyclic for 1 ≤ i ≤ n. 30

One says that G is nilpotent if there exists a sequence

{1} = G0 ⊂ G1 ⊂ · · · ⊂ Gn = G of subgroups of G, with Gi−1 normal in Gi and Gi/Gi−1 in the center of G/Gi−1 for 1 ≤ i ≤ n. Let p be a prime number. A group G whose order is a power of p is called a p-group. Remark 7.2. nilpotent⇒supersolvable⇒solvoble. Lemma 7.3. Let G be a p-group acting on a finite set X, and let XG be the set of elements of X fixed by G. We have |X| ≡ |XG| (mod p). Proof. Indeed X −XG is a union of nontrivial orbits of G. The cardinality of each of these α G orbits is a power p of p with α ≥ 1. Hence |X − X | is divisible by p.  Let us now apply this lemma to the case X = G with G acting by inner automorphism. The set XG is just the center C of G. Thus |C| ≡ |G| ≡ 0 (mod p), hence C 6= {1}. We thus proved the following result. Theorem 7.4. Every p-group is nilpotent (thus supersolvable). Another useful application of the lemma is the following. Proposition 7.5. Let V be a vector space 6= 0 over a finite field k of characteristic p and let ρ : G → GL(V ) be a linear representation of a p-group G in V . Then there exists a nonzero element of V which is fixed by all ρ(s) s ∈ G. Proof. Let x be a nonzero element of V and let X be the subgroup of V generated by the ρ(s)x s ∈ G. Applying the lemma to X, we have p||XG| since X is finite and of order a power of p. The result follows.  Corollary 7.6. The only irreducible representation of a p-group in characteristic p is the trivial representation. Definition 7.7. Let p be a prime number, and let G be a group of order g = pnm, where m is prime to p. A subgroup of G of order pn is called a Sylow p-subgroup of G. Theorem 7.8. (1) There exist Sylow p-subgroups. (2) The are conugate by inner automorphism. (3) Each p-subgroup of G is contained in a Sylow p-subgoup. Proof. To prove (1), we use induction on the order of G. We may assume that n ≥ 1, i.e., p||G|. Let C be the center of G. If |C| is divisible of p, then C contains a subgroup D cyclic of order p. By induction hypothesis, G/D has a Syplow p-subgroup, and the inverse image of this subgroup in G is a Sylow p-subgroup of G. If p - |C|, let G act on G − C by inner automorphism. This gives a partition of G − C into orbits (conjugacy classes). As |G − C| 6≡ 0 (mod p), one of these orbits has a cardinality prime to p. It follows that there is a subgroup H unequal to G such that (G : H) 6≡ 0 (mod p). (Indeed, −1 take cx = {gxg |g ∈ G} be such a conjugacy class, we may take H = {g ∈ G|gx = xg}.) 31

The order of H is thus divisible by pn and the inductio hypothesis shows that H contains a subgroup of order pn. Now let P be a Sylow p-subgroup of G and Q a p-subgroup of G. The p-group Q acts on X = G/P by left translations. Then we have |XQ| ≡ |X| 6≡ 0 (mod p). Thus XQ is not empty. There exists an element x ∈ G such that QxP = xP . That is Q ⊂ xP x−1, which proves (3). If in addition |Q| = pn, then we must have Q = xP x−1, which proves (2).  Lemma 7.9. Let G be a nonabelian supersolvable group. Then there exists a normal abelian subgroup of G which is not contained in the center of G. Proof. Let C be the center of G. The quotient H = G/C is supersolvable, thus has a conposition series in which the first nontrivial term H1 is a cyclic normal subgroup of H. The inverse image of H1 in G has the required properties.  Theorem 7.10. Let G be a supersolvable group. Then each irreducible representation of G is induced by a representation of degree 1 of a subgroup of G (i.e., is monomial). Proof. We prove the theorem by induction on the order of G. Consequently we may consider only those irreducible representations ρ which are faithful, i.e., such that Ker(ρ) = {1}. If G is abelian, such a ρ is of degree 1 and there is nothing to prove. Suppose G is not abelian, and let A be a normal abelian subgroup of G which is not contained in the center of G. Since ρ is faithful, this implies that ρ(A) is not contained in the center of ρ(G). Thus there exists a ∈ A such that ρ(a) is not a homothety. The restriction to A is thus not isotypic. This implies that ρ is induced by an irreducible representation of a subgroup of G which is unequal to G. The theorem follows by applying induction to H.  7.2. p-regular elements and p-elementary subgroups. Let x be an element of a finite group G. We say that x is a p-element (or is p-unipotent) if x has order a power of p. We say that x is a p0-element (or is p-regular) if its order is prime to p. Each x ∈ G can be written in a unique way x = xuxr where xu is p-unipotent, xr is p-regular, and xu and xr commute. Moreover, xu and xr are powers of x. (Indeed, assume that x has order m = pαn with (n, p) = 1. Since (m, m + 1) = 1, we may write m + 1 = apα + bn for some integer a, b such that (a, n) = (p, b) = 1. Then x = xm+1 = apα bn bn apα x · x . It is easy to check we may take xu = x and xr = x . In other words, we decompose the cyclic subgroup generated by x as a direct product of its p-component and its p0-component.) A group H is said to be a p-group if its order is a power of p, is said to be p-elementary if it is the direct product of a cyclic group C of order prime to p with a p-group P . Such a group is nilpotent and its decomposition C × P is unique: C is the set of p0-elements of H and P is the set of p-elements. Let x be a p0-element of a finite group G, let C be the cyclic subgroup generated by x, and let Z(x) be the centralizer of x. If P is a Sylow p-subgroup of Z(x), the group H = CP˙ is a p-elementary subgroup of G, which is said to be associated with x. It is unique up to conjugation in Z(x). In the following two subsections, we prove the following result. 32

Theorem 7.11. Let G be a finite group and let Vp be the subgroup R(G) generated by characters induced from those of p-elementary subgroups of G. Then the index of Vp in R(G) is finite and prime to p. 7.3. Induced characters arising from p-elementary subgroups. Let X(p) be the family of p-elementary subgroups of G. The group Vp is the image of the homomorphism

Ind : ⊕H∈X(p)R(H) → R(G) G defined by the induction homomorphism IndH H ∈ X(p). Then Vp is an ideal of R(G). To prove the theorem it is enough to show that there exists an integer m, prime to p, such that m ∈ Vp. In fact, we prove the following more precise result. n Theorem 7.12. Let g = p l be the order of G with (p, l) = 1. Then l ∈ Vp. Let A be the subring of C generated by the g-th roots of unity. This ring is free and finitely generated as a Z-module. Its elements are algebraic integers. We have Q ∩ A = Z since the elements of this intersection are simultaneously rational numbers and algebraic integers. The quotient A/Z is finitely generated and torsion free, hence free. It follows that A has a basis {1, α1, . . . , αc} containing the element 1. The homomorphism Ind defines, by tensoring with A, an A-linear map

A ⊗ Ind : ⊕H∈X(p)A ⊗ R(H) → A ⊗ R(G).

The existence of the basis {1, α1, . . . , αc} implies the following lemma.

Lemma 7.13. The image of A ⊗ Ind is A ⊗ Vp. Moreover we have

(A ⊗ Vp) ∩ R(G) = Vp.

Thus, to prove that the constant function l belongs to Vp, it is enough to prove that l P G belongs to the image of A ⊗ Ind, or in other words, that l is of the form H aH IndH (fH ) with aH ∈ A and fH ∈ R(H). 7.4. Proof of the theorem. Lemma 7.14. Each class function on G with integer values divisible by g is an A-linear combination of characters induced from characters of cyclic subgroups of G. Proof. Let f be such a function, and write it in the form gχ, where χ is a class function with integer values. If C is a cyclic subgroup of G, let θA be the element of R(C) defined ( |A| if a generates A as θA(a) = . We have 0 otherwise

X G g = IndC (θC ). C Therefore X G X G G f = gχ = IndC (θC )χ = IndC (θC · ResC χ). C C G Note the values of χC = θC · ResC χ are divisible by the order of C, so if ψ is a character of C, we have hχC , ψi ∈ A, which shows that χC is an A-linear combination of characters of C. Hence χC ∈ A ⊗ R(C) and the lemmma follows.  33

Lemma 7.15. Let x be an element of A ⊗ R(G) with integer values. Let x ∈ G and xr be the p0-component of x. Then

χ(x) ≡ χ(xr) (mod p). Proof. By restriction, we may assume that G is cyclic and generated by x. Assume that P χ = aiχi, where ai ∈ A and χi running over the distinct characters of degree 1 of G. If q q q q q is a sufficiently large power of p, we have x = xr and thus χi(x) = χi(xr) for all i. Hence q X q X q q χ(x) = ( aiχi(x)) ≡ ai χi(x) (7.1) X q q q ≡ ai χi(xr) ≡ χ(xr) (mod pA). Since pA ∩ Z = pZ, this implies q q χ(x) ≡ χ(xr) (mod p). q This implies χ(x) ≡ χ(xr) (mod p) since z ≡ z (mod p) for all z ∈ Z.  Lemma 7.16. Let x be a p0-element of G. Let H be a p-elementary subgroup of G associated with x. Then there exists a function ψ ∈ A ⊗ R(H) with integer values, such 0 G that the induced function ψ = IndH ψ has the following properties: (1) ψ0(x) 6≡ 0 (mod p). (2) ψ0(s) = 0 for each p0-element of G which is not conjugate to x. Proof. Let C be the cyclic subgroup of G generated by x. Let Z(x) be the centralizer of x in G. We have H = C × P , where P is a Sylow p-subgroup of Z(x). Let c be the order a of C and p the order of P . Let ψC be the function defined on C by

ψC (x) = c and ψC (y) = 0 if y 6= x. P −1 We have ψC = χ χ(x )χ, where χ runs through the set of irreducible characters of C. It follows that χC belongs to A ⊗ R(C). Let ψ be the function on H = C × P defined by ψ(xy) = ψC (x) for x ∈ C and y ∈ P . This is the inverse image of ψC under the projection H → C. So we have ψ ∈ A ⊗ R(H). We show that ψ satisfies the conditions of the lemma. If s is a p0-element of G and if y ∈ G, ysy−1 is a p0-element. If ysy−1 belongs to H, it belongs to C and we have ψ(ysy−1) = 0 whenever ysy−1 6= x. It follows that ψ0(s) = 0 if s is not conjugate to x. Moreover, 1 X 1 X |Z(x)| ψ0(x) = ψ(x) = 1 = cpa pa pa yxy−1=x yxy−1=x 0 Thus ψ (x) 6≡ 0 (mod p). 

Lemma 7.17. There exists an element ψ of A⊗Vp with integer values, such that ψ(x) 6≡ 0 (mod p) for each x ∈ G.

Proof. Let (xi)i∈I be a system of representatives of the p-regular classes. The above lemma gives us an element ψi of A ⊗ Vp with integer values, such that

ψi(xi) 6≡ 0 (mod p) and ψi(xj) ≡ 0 (mod p) for i 6= j. 34 P Put ψ = ψi. It is clear that ψ belongs to A ⊗ Vp and has integer values. For x ∈ G, the 0 p -component of x is conjugate to a unique xi and we have

ψ(x) ≡ ψ(xi) ≡ ψi(xi) 6≡ 0 (mod p). The lemma follows.  Proof of the theorem. Let g = pnl be the order of G with (p, l) = 1. It suffices to show that l belongs to A ⊗ Vp. Let ψ be an element of A ⊗ Vp satisfying the conditions in the above lemma. The values of ψ are 6≡ 0 (mod p). Let N = φ(pn) be the order of n × N n the group (A/p Z) , so that λ ≡ 1 (mod p ) for each integer λ prime to p. Hence ψ(x)N ≡ 1 (mod pn) for all x ∈ G and the function l(ψN − 1) has integer values divisible by lpn = g. This function is therefore an A-linear combination of characters induced form N cyclic subgoups of G. Since each cyclic group is p-elementary, we have l(ψ − 1) ∈ A ⊗ Vp. N But A ⊗ Vp is an ideal of A ⊗ R(G), whence lψ ∈ A ⊗ Vp. Therefore l ∈ A ⊗ Vp and the theorem follows.  7.5. Brauer’s theorem. We say that a subgroup fo G is elementary if it is p-elementary for at least one prime number p. Theorem 7.18. Each character of G is a linear combination with integer coefficients of characters induced from characters of elementary subgroups.

Proof. Let Vp be the subgroup of R(G) defined before. It suffices to show that the sum V of the Vp for p prime, is equal to R(G). Now V contains Vp, so the index of V in R(G) divides that of Vp, hence is prime to p. Since this is true for all p, the index must be 1, which proves the theorem.  Theorem 7.19. Each character of G is a linear combination with integer coefficients of monomial chracters. Proof. Since an elementary group is nilpotent, each character of an elementary group is monomial. The theorem follows. 

8. The representations of GL2(Fq) Let q be a prime power and odd. In this section, we study the representations of GL2(Fq), which is the group of invertible 2 × 2 matrices with entries in the finite field Fq × with q elements. The quotient PGL2(Fq) = GL2(Fq)/Fq is the automorphism group of 1 the finite projective line P (Fq). There are several key subgroups of G. a b 1 b G ⊃ B = { } ⊃ N = { }. 0 d 0 1 1 Since G acts transitively on the projective line P (Fq), with B the isotropy group of the point (1 : 0), we have 1 2 |G| = |B| · |P (Fq)| = (q − 1) q(q + 1). Question: What is the cardinality of GLn(Fq)? Let a 0 D = { } = × × ×. 0 b Fq Fq 35

0 We write F for Fq. Let F = Fq2 be the extension of F of degree two, unique up to isomor- phism. We can identify GL2(F) as the group of all F-linear invertible endomorphisms of 0 0 × F . This makes evident a large cyclic subgroup K = (F ) of G. Choosing a generator  × √ 0 √ for the cyclic group F and choosing a square root  in F . Then 1 and  form a basis 0 for F as a vector space over F, so we can make the identifications x y K = { } =∼ ( 0)× y x F (8.1) x y √ ↔ ζ = x + y  y x The conjugacy classes in G are the following. Representatives Number of elements in class Number of classes x 0 a = 1 q − 1 x 0 x x 1 b = q2 − 1 q − 1 x 0 x x 0 c = , x 6= y q2 + q (q − 1)(q − 2)/2 x,y 0 y x y d = , y 6= 0 q2 − q (q − 1)q/2 x,y y x  0 1 Here c and c are conjugate by , and d and d are conjugate by any x,y y,x −1 0 x,y x,−y a −c . To count the number of elements in the conjugacy class of b , look at the c −a x a b action of G on this class by conjugation. The isotropy group is { }, so the number 0 a of elements in this class is the index of this group in G, which is q2 − 1. Similarly, the isotropy group for cx,y is D, for dx,y is K. To see that the classes are disjoint, consider the eigenvalues and the Jordan canonical forms. Since they account for |G| elements, the list is complete. There are q2 − 1 conjugacy classes, so we must find the same number of irreducible 1 representations. Consider first the permutation representation of G on P (F), which has dimension q + 1. It contains the trivial representation. Let V be the complementary q-dimensional representation. Lemma 8.1. The values of the chracter χ of V on the four types of conjugacy classes are χ(ax) = q, χ(bx) = 0, χ(cx,y) = 1, χ(dx,y) = −1. We display as V = (q, 0, 1, −1). Moreover, (χ|χ) = 1. Thus V is irreducible. 1 Proof. Let (a : b) be an element in P (F) and e(a:b) the corresponding basis in the (q + 1)- dimensional permutation representation. We have

(1) ax(e(a:b)) = e(a:b)

(2) bx(e(a:b)) = e(ax+b:bx). Therefore bx(e(a:b)) = e(a:b) if and only if b = 0. 36

(3) cx,y(e(a:b)) = e(ax:by). Therefore cx,y(e(a:b)) = e(a:b) if and only if a = 0 or b = 0.

2 2 (4) dx,y(e(a:b)) = e(xa+yb:ya+xb). Therefore dx,y(e(a:b)) = e(a:b) if and only if b = a , which is impossible since  is not a square in F. The values of χ then follow. Therefore (χ|χ) = ((q − 1)q2 + (q2 + q)(q − 1)(q − 2)/2 + (q2 − q)q(q − 1)/2)/|G| = 1. The lemma follows.  × × × For each of the q − 1 characters α : F → C of F , we have a one-dimensional representation Uα of G defined by Uα(g) = α(det(g)). We also have the representations Vα = V ⊗ Uα. We may write these representations as 2 2 2 2 Uα = (α(x) , α(x) , α(x)α(y), α(x − y )). 2 2 2 Vα = (qα(x) , 0, α(x)α(y), −α(x − y )). 0 × Note that if we identify K with (F ) as before, we have   2 2 x y q q+1 x − y = det = Norm 0 (ζ) = ζ · ζ = ζ . y x F /F The next place to look for representations is at those that are induced from large × subgroups. For each pair α, β of characters of F , there is a character of the subgroup B as follows. a b 7→ α(a)β(d). 0 d

Let Wα,β be the representation induced from B to G by this character. This is a represen- tation of dimension [G : B]. By the formula for the character of induced representations, we have the following.

Wα,β = ((q + 1)α(x)β(x), α(x)β(x), α(x)β(y) + α(y)β(x), 0).

Lemma 8.2. Wα,β = Wβ,α. Moreover, Wα,α = Vα ⊕Uα and Wα,β is irreducible for α 6= β. Proof. By computation.  Note now there are q(q − 1)/2 irreducible representations to be found. A natural way to find new characters is to induce characters from the cyclic subgroup K. For a repre- sentation 0 × × φ : K = (F ) → C , the character values of the induced representation of dimension [G : K] = (q2 − 1) are Ind φ = (q(q − 1)φ(x), 0, 0, φ(ζ) + φ(ζ)q). 0 × ∼ q Here again we identify K with (F ) as before. Note that Ind φ = Ind φ , so the rep- resentations Ind φ for φq 6= φ give q(q − 1)/2 different representations. However, these representations are not irreducible. Lemma 8.3. Let χ be the character of the representation Ind φ, then ( q − 1 if φ 6= φq (χ|χ) = q otherwise.

Proof. By computation.  37

Another attempt to find more representations is to look inside the tensor products of representations we know. We have Vα ⊗ Uγ = Vαγ and Wα,β ⊗ Uγ = Wαγ,βγ. Therefore, tensoring with Uγ does not give us new representations. Now, we look at tensor products of Vα and Wα,β. For example, we have

V ⊗ Wα,1 = (q(q + 1)α(x), 0, α(x) + α(y), 0). Lemma 8.4. We have the following formulas.

(1) (χV ⊗Wα,1 , χWα,1 ) = 2. (2) (χInd φ, χWα,1 ) = 1 if φ|F× = α. (3) (χV ⊗Wα,1 , χV ⊗Wα,1 ) = q + 3. (4) (χV ⊗Wα,1 , χInd φ) = q if φ|F× = α. Proof. By computation. 

Comparing with formula (χInd φ, χInd φ) = q − 1, one deduce that V ⊗ Wα,1 and Ind φ contain many of the same representations. Consider the virtual character

χφ = χV ⊗Wα,1 − χWα,1 − χInd φ It takes values ((q − 1)α(x), −α(x), 0, −(φ(ζ) + φ(ζ)q)).

Lemma 8.5. We have (χφ|χφ) = 1 and χφ(1) = q − 1 > 0. Therefore, χφ is the character of an irreducible subrepresentation of V ⊗ Wα,1 of dimension q − 1. We denote this representation by Xφ. There gives q(q − 1)/2 different representations. Proof. By computation. 

All the irreducible representations of GL2(Fq) are Uα, Vα, Wα,β, and Xφ.