Introduction to Character Theory and the McKay Conjecture

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1 Introduction to Character Theory and the McKay Conjecture Ben West MSRI, July 11-22, 2016 The following are my notes of lectures delivered by Professor Gabriel Navarro at the July 2016 MSRI Summer Graduate School on Character Theory and the McKay Conjecture. Any misprints, mathematical missteps, or general mistakes are mine alone. Contents 1 July 11, July 12, July 13, July 14, July 18, July 20, July 22, Exercises 20 1 July 11, 2016 We use the notation x g = g 1 xg. Theorem 1.1. Let G be a finite group, and p be a prime. Suppose further that P is a p-group that acts by automorphisms on G, and p G. Consider the P -invariant classes of G: cl P (G) = {K : K is a conjugacy class of G and K x = K for all x P } Let C = C G (P ) be the fixed point subgroup of G under the action of P. Then the map is a well-defined bijection. First, a small lemma. cl P (G) cl(c) : K K C Lemma 1.2. Suppose G = N H. Then N N (H) = C N (H). Proof. Clearly C N (H) N N (H). Conversely, take n N N (H), and h H. Then since n 1 h 1 n H since n normalizes H, and since h 1 nh N since N G, we see [n, h] = n 1 h 1 nh N H = 1 1

2 so n and h commute. Thus n C N (H). Proof. (Of Theorem) We first show that K C is never empty, and that K C is a conjugacy class in C. Since P acts on G by automorphisms, let Γ = GP be their semidirect product. Observe [Γ : P ] = GP P = G P G P P = G. Since G is coprime to p, it follows that P is a p-sylow subgroup of Γ. Now let K be a P -invariant class, say K = cl G (x) for some x. If y P, then K y = cl G (x y ) = K = cl G (x). Hence x and x y are conjugate in G, so there exists g G such that x y = x g. Thus x yg 1 = x, which implies z := yg 1 C Γ (x), or y = zg C Γ (x)g. Note that C Γ (x)g is a (sub)group of Γ since G Γ. As y P is arbitrary, we have P C Γ (x)g Γ = GP. But clearly G C Γ (x)g, thus Γ = GP C Γ (x)g Γ. Hence we have Γ = C Γ (x)g. As before, we compute [Γ : C Γ (x)] = G C Γ (x) G which is coprime to p. It follows that if Q Syl p (C Γ (x)), then Q Syl p (Γ), since a maximal p-group in C Γ (x) is still a maximal p-group of Γ. Since P Syl p (Γ), we find that P and Q are conjugate, so there exists z Γ such that P z = Q C Γ (x). As Γ = GP = P G, we may write z = z 1 g for z 1 P, and g G. Thus P z = P z1g = P g. Since P g C Γ (x), we have P C Γ (x g 1 ), since direct computation yields, that x yg = x implies (x g 1 ) y ) = x g 1 for y P. So x g 1 is centralized by P, and is clearly conjugate to x, thus x g 1 K. Hence K C, as it contains x g 1 for instance. We have that K = cl G (x) = cl G (x g 1 ), and x g 1 C G (P ), so without loss of generality, we may replace x by x g 1 to assume that K = cl G (x), where x C G (P ). Lastly, we must show K C is a conjugacy class in C. Since x K C, any other element has form x g for some g G. We must show that x g = x c for some c C, so that x and x g are actually conjugate in C, not merely in G. Take x g K C, so x g C, thus P C Γ (x g ), and in turn P g 1 C Γ (x). But x C G (P ), so P C Γ (x). Thus P and P g 1 are both Sylow subgroups of C Γ (x), and thus must be conjugate in C Γ (x). Take y C Γ (x) such that P g 1 = P y. From Γ = GP, and C Γ (x) G = C G (x), we see C Γ (x) = P C G (x). So we may write y = uv, for u P, v C G (x), and so P g 1 = P y = P uv = P v. It follows that P = P vg, so vg N G (P ) = C G (P ), this equality following from the lemma. So c := vg C G (P ). So finally, x c = x vg = (x v ) g = x g the last equality following since v C G (x). So K C is actually a C-conjugacy class, as was to be shown. Lastly, we show the map is a bijection. Suppose K and L are classes in cl P (G). If K C = L C, these are nonempty, so K L. As the are classes, we must have K = L. For surjectivity, suppose X cl(c), say X = cl c (C). Let K = cl G (c). Since K is P -invariant, c K C, so K C = X. 2

3 Remark 1.3. Professor Isaacs pointed out a quicker route to showing K C is nonempty. It s well known that if a p-group acts on a set Ω, and if Ω 0 is the set of fixed points, then Ω Ω 0 (mod p). Since P is acting on K in this case, K K 0 = K C (mod p). If K = cl G (x), then K = [G : C G (x), so p K, since p G. So p K C, so certainly K C. 2 July 12, 2016 Suppose a group H acts on a group N. If X : N GL n (C) is a representation, then for h H, we get an induced representation X h : N GL n (C) : n X(n h 1 ). If X affords χ, we denote the character afforded by X h to be χ h. One can compute that if α and β are characters, then (α, β) = (α h, β h ) so that α is irreducible iff α h is irreducible. This raises the question, when is Irr H (N) = cl H (N)? The former set is the set of irreducible H- invariant characters, and the latter the H-invariant classes of N. Recall that is x, y G, then [H, K] = [h, k] : h H, k K H, K. Suppose that H acts on N, and M N is an H-invariant normal subgroup. We get a natural induced action of H on N/M. We see that H acts trivially on N/M iff (Mn) h = Mn, iff Mn h = Mn, which holds iff n h n 1 M. Abusing notation in the semidirect product, since [N, H] = n 1 n h : n N, h H, we see that H acts trivially on N/M iff [N, H] M. Let us consider the case where N is abelian. In this case, cl N (n) = {n}, so cl N (n) is H-invariant iff n h = n for all h H, iff n C N (H). Hence it follows cl H (N) = C N (H). Moreover, let λ Irr(N), so λ is linear as N is abelian. We have That is, λ is H-invariant λ h (n) = λ(n) λ(n h 1 ) = λ(n) λ(n h 1 n 1 ) = 1 [N, H] ker λ λ Irr(N/[N, H]). Irr H (N) = Irr(N/[N, H]) = N/[N, H], the last equality following since N/[N, H] is abelian, so there is an irreducible character for every conjugacy class, i.e., every singleton. To recount, if N is abelian, asking if Irr H (N) = cl H (N) is the same as asking if N/[N, H] = C N (H). It turns out this is not generally true. For a counterexample, take N = Z/(3) Z/(3), and let H = h, k S 3 GL 2 (Z/(3)) where h = ( ) 1 0, k = 0 1 ( ) If (a, b) N is a fixed point under H, then (a, b) = (a, b) h = ( a, b), so that a = 0, and (a, b) = (a, b) k = (a, a + b), so b can be arbitrary. Then C N (H) = 3, but one computes N = [N, H], so that N/[N, H] = 1. The equality does hold if H is cyclic. 3

4 Theorem 2.1. (Brauer) If a cyclic group H acts as automorphisms on N, then Irr H (N) = cl N (H). Proof. Suppose H = h, Irr(N) = {χ 1,..., χ k }, and cl(n) = {K 1,..., K n }, where K i = cl N (n i ). Then put X = (χ i (n j )) as the character table, known to be invertible when viewed as a matrix. Define two k k matrices P = (p ij ) and Q = (q ij ) by the following: p ij = { 0 χ h i = χ j, 1 χ h i = χ i. q ij = { 0 K h i = K j, 1 K h i = K i. We claim P X = XQ. The (i, j)-entry of P X is k p il χ l (n j ) = χ h i (n j ) l=1 since p il = 0 unless χ h i = χ l, in which case it is 1. Likewise, the (i, j)-entry of XQ is k l=1 χ i (n l )q lj = χ i (n h 1 j ) since q lj vanishes unless Kl h = K j, in which case K l = Kj h 1 n h 1 j, so we can evaluate χ i on this represenatative. Since χ i (nj h 1 ) = χ h i (n j) by definition, the claim follows. Since X is invertible X 1 P X = Q, so that tr(p ) = tr(q). However, by the definition of P and Q, tr(p ) is the number of characters fixed by h, hence by H, and tr(q) is the number of classes fixed by h, hence by H. So Irr H (N) = cl H (N). Proposition 2.2. Let A be a finite-dimensional F -algebra, for F a field. If λ 1,..., λ k Hom(A, F ) =: A are distinct, then they are linearly independent over F. In particular, dim F (A ) = dim F (A) k. Proof. The proof essentially follows as Artin s Lemma in Galois Theory. 3 July 13, 2016 Corollary 3.1. Let G be a finite group, F a field. Then the number of distinct algebra homomorphisms Z(F [G]) F is at most Irr(G). Proof. Since the conjugacy class sums form a basis of Z(F [G]), we have dim(z(f [G]) = cl(g) = Irr(G). Since distinct morphisms are linearly independent, there can be at most dim(z(f [G])) = Irr(G) of them. Suppose X : G GL n (C) is an irreducible represenation affording the irreducible character χ. We can extend X linearly to an algebra morphism X : C[G] M n (C), also denoted by X. Since X is irreducible, the extension is surjective onto M n (C). 4

5 Supose z Z(C[G]). Then X(z) is central in Im(X) = M n (C), so necessarily X(z) is a scalar matrix. Thus we can write X(z) = ω(z)i n for some ω(z) C. This value ω(z) depends only on χ, for suppose Y is another representation affording χ. Then Y and X are similar represenations, so for some M GL n (C), Y (z) = M 1 X(z)M = M 1 ω(z)i n M = ω(z)i n. Furthermore, since X is an algebra morphism, we see the above map ω : Z(F [G]) C. For instance, we have for z, w Z(F [G]), ω(zw)i n = X(zw) = X(z)X(w) = ω(z)i n ω(w)i n = ω(z)ω(w)i n so that ω(zw) = ω(z)ω(n). Other morphism properties follow similarly. Since ω depends only on χ, rename it ω χ. Thus for χ Irr(G), we get an associated algebra morphism ω χ : Z(C[G]) C. These ω χ are called the central characters. As observed in Isaacs notes, if K cl(g), say K = cl G (x), and we denote ˆK = k K k, then ωχ( ˆK) K χ(x) = R, χ(1) where R denotes the ring of algebraic integers. Now fix a prime p. Let pr be the generated ideal, and pick a maximal ideal m contains pr. Set F = R/m. Then the image of p in F is zero, so char(f ) = p. We will denote the canonical map R F by. Observe that if n Z, then n = 0 iff p n in Z. For if p n, then up + vn = 1 for some integers u and v. Taking the projection into F, the up + vn vanishes, so that 1 M, contrary to M being a maximal, hence proper, ideal. Take χ Irr(G). Define an algebra morphism by λ χ : Z(F [G]) F : ˆK ω χ ( ˆK) and extending linearly. This makes sense as the class sums ˆK are a basis of Z(F [G]), and ω χ ( ˆK) R. Theorem 3.2. Suppose G is a finite group, and p G. Then {λ χ : χ Irr(G)} = Hom Alg (Z(F [G]), F ). Proof. It is sufficient to show that if χ and ψ are distinct irreducible characters, then λ χ and λ ψ are distinct. This will show that {λ χ : χ Irr(G)} has cardinality Irr(G), which is the maximum number of algebra morphisms Z(F [G]) F by the corollary above. Let f ψ = K cl(g) ψ(x 1 k ) ˆK where x k is a representative of K. Note that ψ(x 1 k ) R since as the value of a character, it is a sum of roots of unity, which are algebraic integers themselves. Thus we set f ψ = K cl(g) ψ(x 1 k ) ˆK which is in Z(F [G]) as the ˆK are central, and sends the coefficients of f ψ into F. By the first orthogonality relation, we have G δ χ,ψ = x G χ(x)ψ(x 1 ) = K cl(g) = χ(1) K χ(x k )ψ(x 1 k ) K cl(g) ω χ ( ˆK)ψ(x 1 k ). 5

6 Since G and χ(1) are integers, and ω( ˆK) and ψ(x 1 k ) are algebraic integers, we can apply, yielding G δχ,ψ = χ(1) ω χ ( ˆK) ψ(x 1 k ). K cl(g) By the definition of the λ χ above, the right side can be rewritten as χ(1) K cl(g) λ χ ( where the equality follows by the F -linearity of λ χ. 1 ˆK)ψ(x k ) = χ(1) λ χ (fψ), Recall p G. So G 0, by our previous observation. Moreover, since χ(1) G, certainly p χ(1), so χ(1) 0. Observe that χ(1) λ χ (f χ) = G 0 so that λ χ (f χ) 0. Suppose λ χ = λ ψ, but χ ψ. Then since δ χ,ψ = 0, a contradiction. 0 = λ χ (f ψ) = λ ψ (f ψ) 0, Now suppose P is a p-group that acts as automorphisms on a group G, where p G. Let C := C G (P ). We wish to find a bijection between Irr P (G) and Irr(C). Suppose χ Irr P (G), so χ z = χ for all z P. We wish to construct a corresponding irreducible character of C, which we have seeen above are in bijection with the morphisms λ χ on Z(F [C]), defined in terms of central characters. So in this case, we must product a corresponding morphism δ χ : Z(F [C]) F. Going back to the first section, we have a bijection cl P (G) cl(c) : K K C. So an arbitrary class sum in Z(F [C]) has form K C for K cl P (G). Thus define by extending linearly. Theorem 3.3. The map δ χ is an algebra morphism. δ χ : Z(F [C]) F : K C λ χ ( ˆK) Proof. Now the action of P on G induces an action of P on cl(g), so we have cl(g) = cl P (G) O(K 1 ) O(K s ) where O(K i ) = {K z i : z P }. Since the K i are the classes which are not P -invariant, O(K i ) = p ai > 1. The orders are p-powers, since they must divide the order of P by the Orbit-Stabilizer Theorem. We need to show δ χ is multiplicative on a basis, as it is linear by construction. That is, we want for K, L cl P (G). First, we have δ χ ( K C L C) = δ χ ( K C)δ χ ( L C) δ χ ( K C)δ χ ( L C) = λ χ ( ˆK)λ χ (ˆL) = λ χ ( ˆK ˆL) by definition of δ χ, and the fact that λ χ is an algebra morphism. Note that ˆK ˆL = M cl(g) a KLM ˆM 6

7 where a KLM is the number of pairs (x, y) K L such that xy = x M, a representative of the class M. Certainly ˆK ˆL is an positive integral linear combination of class sums, and we can compute the coefficients of ˆM by counting the pairs whose product if a fixed representative, because the number of summand of any element in M must be the same, else we would not have an integral coefficient of ˆM. Note for z P, we have a bijection Hence a KLM = a K z L z M z. {(x, y) K L : xy = x M } {(u, v) K z L z : uv = x z M } : (x, y) (x z, y z ). We can further decompose this sum. Taking smaller sums over the class sums in the various orbits listed above, we have ˆK ˆL = a KLM ˆM + a KLX ˆX + + a KLX ˆX. M cl P (G) X O(K 1) X O(K s) However, the classes in O(K i ) are of form K z i for z P. By the above observation, we have a KLK z i = a K z L z K z i = a KLK i =: b i, where the first equality follows since K = K z and L = L z as they are P -invariant. So the coefficient for each class sum in the orbit of a given non-p -invariant class is the same, so we can rewrite the product as ˆK ˆL = a KLM ˆM + b 1 ˆX + + b s ˆM = s a KLM ˆM + ˆX. M cl P (G) X O(K 1) X O(K s) M cl P (G) j=1 b j X O(K j) If M cl P (G), then M C cl(c) is nonempty, so pick x M M C, so that x M is fixed by P. Set Ω = {(x, y) K L : xy = x M }. Then P acts on Ω, since for z P, (x z, y z ) K z L z = K L, and if xy = x M, then (xy) z = x z y z = x z M = x M. First recall that if P acts on a set Ω, we have the congruence Ω Ω 0 (mod p), that is Ω = Ω 0 in F = R/m. Note Ω = a KLM, and the P -fixed points of K and L are K C and L C, respectively, so Ω 0 = {(x, y) K C L C : xy = x M } = a K C L C M C since x M M C, and K C and L C are classes in C. Thus a KLM = a K C L C M C. By hypothesis, χ Irr P (G), so if z P, χ = χ z, so λ χ ( ˆX z ) = λ χ z( ˆX z ) = λ χ ( ˆX zz 1 ) = λ χ ( ˆX). We now compute λ χ ( ˆK ˆL) = λ χ s a KLM ˆM + b j λ χ M cl P (G) j=1 X O(K j) ˆX. But since λ χ is constant on classes, λ χ X O(K j) ˆX = O(K j ) λ χ ( ˆX) = p aj λ χ ( ˆX) = 0 since char(f ) = p, and p aj > 1. 7

8 Thus we simply have λ χ ( ˆK ˆL) = λ χ a KLM ˆM which was to be shown. = λ χ M cl P (G) M cl P (G) a K C L C M C ˆM = a K C L C M C λ χ ˆM M cl P (G) = a K C L C M C δ χ ( M C) M cl P (G) = δ χ M cl P (G) a K C L C M C ( M C) = δ χ ( K C L C) 4 July 14, 2016 Suppose P is a p-group acting by automorphisms on a group G, where p G. In the previous section, we saw that if χ Irr P (G), there is an algebra morphism δ χ : Z(F [G]) F : K C λ χ ( ˆK). Since C has order coprime to p, and we have determined all possible algebra morphisms Z(F [C]) F be a previous theorem, necessarily δ χ = λ χ for a unique χ Irr(C). Suppose c C, K = cl G (c), so K C = cl C (c). From their definitions, we have ( ) ( ) K χ(c) = ω λ ( ˆK) = λ χ ( ˆK) K C χ(c) = δ χ ( K C) = λ χ( K C) = ω χ( K C) =. χ(1) χ(1) Multiplying through by χ(1) χ(1), and using the fact that is a homomorphism, inspecting the leftmost and rightmost terms above yields χ(1) K χ(c) = χ(1) K C χ(c). However, K K C (mod p), so that K = K C 0 (mod p), so cancelling and pulling back to R under, we conclude χ(1)χ(c) χ(1) χ(c) (mod m). Theorem 4.1. (Glauberman) Suppose P is a p-group acting on a group G by automorphisms, and p G. Then the map G : Irr P (G) Irr(C) : χ χ is a bijection. (The character χ is determined by the algebra morphism δ χ in the prior discussion.) Also, χ C = e χ + p, where e ±1 (mod p), and where is a linear combination of irreducible characters. Note that e is not necessarily the multiplicity of χ. Proof. Let χ Irr P (G), and θ Irr(C). Using the equation just derived above, we see χ(1) C (χ C, θ) = χ(1)χ C (c)θ(c 1 ) m χ(1) χ(c)θ(c 1 ) = χ(1)( χ, θ). c C c C 8

9 Thus and since p C, χ(1) C (χ C, θ) χ(1) C ( χ, θ) (mod p) χ(1)(χ C, θ) χ(1)( χ, θ) (mod p). We consider two cases. If χ = θ, χ(1)(χ C, χ) χ(1)( χ, χ) = χ(1) (mod p), and so p (χ C, χ), since p χ(1) as χ(1) G. On the other hand, if χ θ, then ( χ, θ) = 0, so that (χ C, θ) 0 (mod p), as p χ(1). With this information of the coefficients, we have χ C = e χ + pψ, where p e, and Ψ is some linear combination of irreducible characters. Now suppose χ, ϕ Irr P (G). We can write cl(g) = cl P (G) O(K 1 ) O(K s ), where the K i are non-p -invariant classes. Observe taht if K i = cl G (y i ), and z P, then K1 z = cl G (yi z ), and χ(y z i ) = χ z 1 (y i ) = χ(y i ) since χ is P -invariant. Thus χ takes the same value on classes in the same orbit. Since O(K i ) is a p-power greater than 1, it vanishes in F, similar to a previous argument. So we have χ(1) G (χ, ϕ) = χ(1) g G χ(g)ϕ(g 1 ) = χ(1) p χ(1) p χ(1) = χ(1) K cl(g) K cl P (G) K cl P (G) K C cl(c) = χ(1) C (χ C, ϕ C ) K χ(x K )ϕ(x 1 k ) K χ(x K )ϕ(x 1 K ) K C χ(x K )ϕ(x 1 K ) K C χ(x K )ϕ(x 1 K ) where x K K is a class rep, and we choose x K K C, so that x K is P -invariant. The first congruence that appears follows since K vanishes modulo p when K / cl P (G) as noted above. To understand the final two equalities above, recall that from our bijection cl P (G) cl(c) : K K C, we have that K C ranges over cl(c) as K ranges over cl P (G), and since x K C, we can replace χ and ϕ with their restrictions χ C and ϕ C in this sum to get χ(1) G (χ, ϕ) χ(1) C (χ C, ϕ C ) (mod p). Cancelling the χ(1) and using the fact that G C 0 (mod p), since C is the fixed point subgroup, we can justify the first equivalence below: (χ, ϕ) (χ C, ϕ C ) = (e X + p, d ϕ + pψ) ed( χ, ϕ) (mod p), where e, d 0 (mod p). We claim G is injective. If χ = ϕ, then (χ, ϕ) ed 0 (mod p), so since (χ, ϕ) = 0, 1 as they are irreducible, necessarily (χ, ϕ) = 1, forcing χ = ϕ. For surjectivity, take α Irr(C), and suppose towards a contradiction that α χ for any χ Irr P (G). Since χ C = e χ + p, then α is a constituent of instead, so (χ C, α) 0 (mod p). By Frobenius reciprocity, we also find (χ C, α) = (χ, α G ) 0 (mod p). 9

10 Write α G = χ Irr(G) (αg, χ)χ, and as usual, decompose Irr(G) = Irr P (G) O(χ 1 ) O(χ t ). Since induction of characters commutes with conjugation, if z P, we have (α G ) z = (α z ) G = α G as α is P - invariant, so that (α G, χ i ) = ((α G ) z, χ z i ) = (α G, χ z i ) implying that the multiplicity of any character in O(χ i ) as an irreducible constituent of α G is the same. We can write α G = χ Irr P (G) (α G, χ)χ + t (α G, χ i ) j=1 ψ O(χ j) ψ. Evaluating at 1, and using the fact that (α G, χ) 0 (mod p) for χ Irr P (G), and O(χ j ) 0 (mod p) for the remaining characters, we find α G (1) t (α G, χ j ) O(χ j ) χ j (1) 0 (mod p). j=1 So p α G (1). However, from character theory, we know the degree of an induced character is equal to the product of the index of the group from which its induced, and the degree of the original character. In this case, α G (1) = [G : C]α(1). But p α G (1), since p α(1), as α(1) C and p [G : C], a contradiction. So G is surjective. We extend this to a more general case. Theorem 4.2. Suppose A is a solvable group, acting by automorphisms on a group G, and ( A, G ) = 1. Then Irr A (G) = Irr(C G (A)). Proof. We induct on A. Choose B A a proper, normal subgroup of order p e > 1. (Such a B does exist, for example, every minimal normal subgroup of a finite solvable group is elementary abelian, in particular it must be a p-group for some p.) Then B is a p-group of lesser order acting by automorphisms on G, so by the Glauberman correspondence, we have a bijection Irr B (G) Irr(C G (B)). If χ Irr B (G), then χ CG (B) = c χ + p. We have that C G (B) is A invariant, since if a A, so C G (B) a = C G (B a ) = C G (B), as B is normal in A. Also, if g C G (B), then (χ a ) CG (B)(g) = χ a (g) = χ(g a 1 ) = χ CG (B)(g a 1 ) = (χ CG (B)) a (g) from which it follows that χ a = χ a. So A-invariant characters are mapped to A-invariant characters under the Glauberman correspondence, so Irr A (G) = Irr A (C G (B)) But B acts trivially on C G (B), and we get an induced action of A/B on C G (B). By the induction hypothesis, we get the first equality below: IrrA/B (C G (B)) = Irr(CCG (B)(A/B)) = Irr(CG (A)). The second equality follows since the points of C G (B) fixed by A/B are the points of C G (B) fixed by A, since the B-action is trivial, which is to say C CG (B)(A/B) = C G (B) C G (A) = C G (A) since C G (A) C G (B). Combining these two above equations proves the original claim. 5 July 18, 2016 Suppose H is a subgroup of G, and ψ Irr(H). We say ψ extends to G if there exists χ Irr(G) such that χ H = ψ. If H G, then necessarily ψ must be G-invariant. For if χ H = ψ, then as χ g = χ as a class function, we see ψ = χ H = (χ g ) H = (χ H ) g = ψ g. Theorem 5.1. Suppose N G, [G : N] = p, and θ Irr G (N). Then θ extends to G. 10

11 Proof. Consider the induced character θ G, and pick χ to be an irreducible constituent. Then (χ N, θ) = (χ, θ G ) 0 so θ is an irreducible constituent of χ N. From Clifford Theory (cf. Isaacs, the Theorem following Clifford s theorem in the slides), we know in this case that χ N is either irreducible, or the sum of p distinct irreducible characters in the same orbit. Since θ is a G-invariant constituent, it has trivial orbit, so the latter case cannot happen, hence χ N is irreducible, and χ N = θ. We denote by Lin(G) = {α Irr(G) : α(1) = 1} to be the multiplicative group of linear characters on G. Observe that Lin(G) = [G : G ]. For χ Irr(G), denote by o(χ) to be the order of det(χ) in Lin(G). Lemma 5.2. Suppose that N G, θ Irr G (N), θ(1) = 1, o(θ) is a p-power, and P/N Syl p (G/N). Then if θ extends to P, θ further extends to G. Proof. Since θ extends to P, we can find µ Irr(P ) such that µ N = θ. Thus µ(1) = µ N (1) = θ(1) = 1. Since [G/N : P/N] = [G : P ] is coprime to p, we see P is a Sylow subgroup of G, and thus µ G (1) = [G : P ]µ(1) = [G : P ] 0 (mod p). Writing µ G = χ Irr(G) (µg, χ)χ, evaluating at 1 shows p χ(1) for some χ, lest p µ G (1). For this χ, 0 (χ, µ G ) = (χ P, µ), so χ P = eµ +, so that χ N = eµ N +. However, χ N = eθ as χ N is a sum of G-conjugates, but θ is G-invariant. Then χ N (1) = eθ(1) = e, so that e = χ N (1) = χ(1) 0 (mod p). The corresponding representation X N is thus similar to a direct sum of e copies of a linear represenation, each affording θ. It follows that (det χ) N = θ χ(1). By hypothesis o(θ) is a p-power, so (χ(1), o(θ)) = 1. We may find v such that vχ(1) 1 (mod o(θ)). Thus ((det χ) v ) N = ((det χ) N ) v = (θ χ(1) ) v = θ χ(1)v = θ. Lemma 5.3. Suppose N G, θ(1) = 1, and θ Irr G (N). Assume (o(θ), [G : N]) = 1. Then θ extends to G, and there exists a unique µ Irr(G) such that µ N = θ such that o(µ) = o(θ). Proof. Recall from group theory that if g G is of order n = n 1 n 2 where (n 1, n 2 ) = 1, then we can decompose g as g = g 1 g 2, where o(g 1 ) = n 1, o(g 2 ) = n 2, and g 1 and g 2 commute, and are the unique such elements to satisfy these properties. In this case, we can write g = g 1 g 2, where o(g 1 ) = p f, and p o(g 2 ), and g 1 g 2 = g 2 g 1. Now take θ Lin(N). Applying the above, we can write θ = p θ p, where o(θ p ) is a p-power. Assume θ p 1, so that (p, [G : N]) = 1, since p o(θ). By the uniqueness of the decomposition, θ p is G-invariant, since θ = θ g = p θg p. Since p [G : N], there is no proper P N such that P/N Syl p (G/N), for if there were P is a Sylow subgroup of G containing N, implying [G : N] = [G : P ][P : N] so that p [P : N], a contradiction. So by Lemma 5.2, θ p extends to G, so to µ p. Thus (µ p ) N = θ, and µ p (1) = θ(1) = 1, hence µ p is linear. Set µ = p µ p, and thus µ N = p (µ p ) N = p θ p = θ. For uniqueness, since (o(θ), [G : N]) = 1, we can find b such that b[g : N] 1 (mod o(θ)). Set ν = µ b[g:n]. Since µ N = θ, we find ν N = (µ b[g:n] ) N = (µ N ) b[g:n] = θ b[g:n] = θ. We claim that o(ν) = o(µ), and that ν is the unique such extension. 11

12 First we prove the equality of orders. Note so that o(θ) o(ν). Conversely, To see the last equality, observe that θ o(ν) = (ν N ) o(ν) = (ν o(ν) ) N = 1 N ν o(θ) = (µ b[g:n] ) o(θ) = (µ [G:N] ) b o(θ) = 1. (µ o(θ) ) N = (µ N ) o(θ) = θ 0(θ) = 1 N whence µ Lin(G/N). By Lagrange, µ [G:N] = 1 N. Hence o(ν) o(θ), and so o(ν) = o(θ). For uniqueness, suppose we have two characters ν and ɛ such that ν N = θ = ɛ N, with o(ν) = o(θ) = o(ɛ) =: m. Then (ν ɛ) N = ν N ɛ N = θ θ = 1 N as these are all linear characters. So ν ɛ Lin(G/N), and (ν ɛ) m = ν m ɛ m = 1 N. However, (m, [G : N]) = 1 by hypothesis, so necessarily ν ɛ = 1 N. Indeed, writing 1 = sm + t[g : N], we find ν ɛ = (ν ɛ) sm+t[g:n] = 1 N. Lemma 5.4. Suppose N G, θ Irr G (N) extends to G, and (θ(1), [G : N]) = 1. Then the map is a bijection. Ext(G θ) Ext(G det θ) : χ det χ Proof. Let µ Irr(G) such that µ N = θ. By the Gallagher correspondence, the map Irr(G/N) Irr(G θ) : β βµ is a bijection. Also, (βµ) N = β N µ N = β(1)θ since N ker β, so that β N = β(1). Hence we find that (βµ) N = θ β(1)θ = θ β(1) = 1. It follows that Ext(G θ) = Lin(G/N), since every extension of θ to G is an irreducible character lying over θ which corresponds to a character of G/N by the Gallagher correspondence, which is necessarily linear by the above string of equivalences. Observe that if µ N = θ, then (det µ) N = det θ, so that if µ extends θ, then det µ extends det θ. The above argument does not depend on θ, so by the same argument we have Ext(G θ) = Lin(G/N) = Ext(G det θ), so it suffices to show the map in question is injective to get a bijection. Take β 1 µ and β 2 µ with β i Lin(G/N) such that det(β 1 µ) = det(β 2 µ). Since β i simply scales the µ(1) rows of the image of the affording representation, the equation states β µ(1) 1 det µ = β µ(1) 2 det µ. Cancelling det µ yields β µ(1) 1 = β µ(1) 2. But µ(1) = µ N (1) = θ(1), so β θ(1) 1 = β θ(1) 2. Since β i Lin(G/N) and since in general, Lin(H) = [H : H ] H, we see Lin(G/N) [G : N]. Since (θ(1), [G : N]) = 1, we have that (β 1 β 2 ) θ(1) = 1, so that β 1 = β 2. Lemma 5.5. Suppose N G, G/N is solvable, θ Irr G (N), and (θ(1), [G : N]) = 1. Then θ extends to G iff det θ extends to G. Proof. We have already seen the forward direction. So suppose det θ extends to G. We induct on [G : N]. If [G : N] = 1, G = N, so the characters serve as their own extensions. So suppose [G : N] > 1, and take λ Irr(G) which extends det θ. Pick M to be a maximal normal subgroup, so [G : M] = p for some prime p. (To find such an M, suppose G 1 /N is the first term in some subnormal series witnessing G/N being solvable. Then (G/N)/(G 1 /N) G/G 1 is cyclic of prime order. Then there exists a maximal normal subgroup of index p in G/G 1 as p is the smallest divisor of the order of G/G 1, and by the correspondence theorem this 12

13 subgroup pulls back to a normal subgroup of G with index p.) Now det θ extends to M as λ M lies over it, so by the induction hypothesis θ extends to M as well, and we may find ψ Irr(M) such that ψ N = θ. From Lemma 5.4, the map Ext(M θ) Ext(M det θ) : η det η is a bijection. So there exists a unique η Ext(M θ) such that det η = λ M. But λ M is G-invariant as the restiction of a character of G, and thus for g G, det η = λ M = (λ M ) g = (det η) g = det(η g ), so that η = η g, and η is G-invariant. So η is a G-invariant character of M lying over θ, and [G : M] = p. By Theorem 5.1, η extends to a character χ of G. With these pieces, we finally have the following. Theorem 5.6. Suppose N G, G/N is solvable, θ Irr G (N), and (o(θ), θ(1), [G : N]) = 1. Then θ extends to G, and it has a unique extension ˆθ Irr(G) such that o(ˆθ) = o(θ). Proof. By Lemma 5.2, det θ extends to G. Note det(θ)(1) = 1 as a linear character, and o(det θ) is coprime to [G : N] since o(θ) is, and det θ is G-invariant as θ is G-invariant. By Lemma 5.5, θ extends to G. Then by Lemma 5.3, there is a unique extension of det θ with the same order. Then by Lemma 5.4, from the bijection Ext(G θ) Ext(G det θ), we see there is a unique extension of θ with the same order as θ. Corollary 5.7. Suppose N G, G/N is solvable, and N is a Hall subgroup. If θ Irr G (N), then θ extends to G. Proof. We have θ(1) N, so that (θ(1), [G : N]) = 1. Also, o(θ) Lin(N) = [N : N ] N, so that (o(θ), [G : N]) = 1. The above theorem now applies. 6 July 20, 2016 The McKay Conjecture states that if G is a group, p a prime, and P Syl p (G), then where Irr p (G) = {χ Irr(G) : p χ(1)}. Irr p (G) = Irr p (N G (P )) Recall that if K P is a normal Hall subgroup, and G/K is a p-group, then θ Irr(K) extends to G iff and only if θ is G-invariant. In this situation, the G-invariant characters are precisely the P -invariant characters for P Syl p (G). For then G = KP, since KP = K P K P = K P = K [G : K] = G, since p K, so K has index the maximal p-power of G. Taking g = ky, for k K, y Y, then if θ is P -invariant, and automatically K-invariant, θ g = (θ k ) y = θ y = θ so θ is G-invariant. Theorem 6.1. Suppose there exists K G such that p K, and G/K is a p-group. Let P be a complement of K. Then the McKay Conjecture holds, i.e., Irr p (G) = Irr p (N G (P )). Proof. Let θ Irr P (K). As θ is P -invariant, it has an extension ˆθ Irr(G). By the Gallagher correspondence, Ext(G θ) = {β ˆθ : β Lin(G/K)}. We claim Irr p (G) = {β ˆθ : β Lin(G/K)} θ Irr P (K) 13

14 We have ˆθ K = θ, so ˆθ(1) = θ(1) divides K, has is not divisible by p. So so that β ˆθ Irr p (G). (β ˆθ)(1) = β(1)ˆθ(1) = ˆθ(1) 0 (mod p) Conversely, take χ Irr p (G). Let θ be an irreducible constituent of χ K. We know χ(1)/θ(1) [G : K], a p-power. Since p χ(1), as χ(1) K, necessarily χ(1)/θ(1) =, Thus χ K = θ by comparing degrees. So χ K Irr P (G), and is P -invariant as the restriction of a G-invariant character. By the Gallagher correspondence, χ = β ˆθ for some β Lin(G/K), giving the equality. To see the union is disjoint, suppose β ˆθ = γ ˆη. Then β K ˆθK = (β ˆθ) K = (γ ˆη) K = γ K ˆη K. This implies β(1)θ = γ(1)η, or that η = θ. Hence Irr p (G) = {β ˆθ : β Lin(G/K)} = Irr P (K) Lin(G/K). θ Irr P (K) However, P is a Sylow subgroup of N G (P ), with complement K N G (P ) = N K (P ) = C K (P ), this last equality following from an earlier lemma. Applying the above equation to this group, where the second equality follows since Irr p (N G (P )) = Irr P (N K (P )) Lin(N G (P )/N K (P )) = Irr P (C K (P )) Lin(G/K) = Irr(C K (P )) Lin(G/K) = Irr(C K (P )) Lin(G/K) = Irr P (K) Lin(G/K) N G (P ) N K (P ) = N G(P ) N G (P ) K N G(P )K K = G/K, the third equality following since any character on C is automatically P -invariant, and the last equality follows by the Glauberman correspondence. Consider the case where P is self-normalizing, that is, P = N G (P ). As the p -degree characters of P are precisely the linear ones, McKay posits that Irr p (G) = Irr p (N G (P )) = Irr p (P ) = Lin(P ) = [P : P ]. Lemma 6.2. Suppose N G, P Syl p (G), and χ Irr p (G). Then χ N has a P -invariant irreducible constituent θ, and any two P -invariant irreducible constituents are conjugate in N G (P ). Proof. Suppose η Irr(N) is a constituent of χ N. Put T = I G (η) to be the stabilizer group. By the Clifford correspondence, we find ψ Irr(T η) such that ψ G = χ. Observe that χ(1) = ψ G (1) = [G : T ]ψ(1), so p [G : T ] since p χ(1). Thus any p-sylow subgroup of T is a p-sylow subgroup of G. Thus P g T for some g G. Hence P T g 1 = I G (η) 1 = I G (η 1 ). Let θ = η 1. As η is a constituent of χ N, Clifford s Theorem implies any G-conjugate of η is a constituent as well, so θ is a constituent of χ N, which is P -invariant. Now suppose γ is another P -invariant irreducible constituent of χ N. By Clifford s Theorem, there exists x G such that γ x = θ, so T g 1 = I G (θ) = I G (γ x ) = I G (γ) x. 14

15 Since γ is P -invariant, P I G (γ), so P x T g 1. Thus P and P x are both p-sylow subgroups of T g 1, thus there is t T g 1 such that P xt = P, so that xt N G (P ). Then γ xt = θ t = θ, since t T g 1 = I G (θ). So γ and θ are conjugate in N G (P ). Lemma 6.3. Suppose N G, N is a p-group, and P Syl p (G) is Sylow subgroup containing N. Let Ω = {λ Lin(N) : λ extends to P }. Note that N G (P ) acts on Ω by conjugation, that is, if n N G (P ), then λ n Lin(N n ) = Lin(N), and λ n extends to P n = P. Decompose Ω into orbits as Ω = k j=1 O(λ k). Set T j = I G (λ j ). We claim k Irr p (G) = Irr p (T j /N). j=1 Proof. We show k Irr p (G) = Irr p (G λ j ). j=1 Take χ Irr p (G). The restriction χ P must have an irreducible constituent of degree coprime to p, else evaluating χ(1) = chi P (1) shows p χ(1), a contradiction. Let µ be such a constituent. Since µ(1) P, necessarily µ is linear. So µ N Irr(N), and in fact µ N Ω, so µ N is N G (P )-conjugate to some λ i. As the irreducible constituents of χ N form an orbit, λ i is also a constituent of χ N, so χ lies over λ i. Hence χ Irr p (G λ i ). The reverse containment is obvious, so it remains to show the union is disjoint. Suppose χ Irr p (G λ i ) Irr p (G λ j ) for i j. Then λ i and λ j are irreducible constituents of χ N, hence N G (P ) conjugate by the previous lemma, a contradiction as represents of distinct orbits. Hence we conclude Irr p (G) = By the Clifford correspondence, we have a bijection k Irr p (G λ j ). j=1 Irr(T j λ j ) Irr(G λ j ) : ψ ψ G. Since λ j extends to P, it is P -invariant, so P T j, hence p [G : T ]. From ψ G (1) = [G : T ]ψ(1), we see that p ψ(1) iff p ψ G (1), so that in fact Thus Irr p (G) = Irr p (T j λ j ) Irr p (G λ j ). k Irr p (G λ j ) = j=1 k Irr p (T j λ j ). By assumption, λ j extends to P, is T j -invariant by definition, and since o(λ j ) and λ j (1) divide N hence are p-powers, we see (o(λ j ), [T j : P ]) = 1 = (λ(j), [T j : P ]). By the lemma above, λ j extends to T j, denote this extension ν j Irr(T j ). By Gallagher s correspondence, we have an isomorphism j=1 Irr(T j /N) Irr(T j λ j ) : β βν j. Note ν j (1) = 1 as the extension of a linear character, so β is p -degree iff βν j is p -degree. Thus Irr p (T j /N) = Irr p (T j λ j ), and we get the formula. Corollary 6.4. A minimal counterexample to the McKay conjecture mush have O p (G) = 1. (Here O p (G) denotes the largest, normal p-subgroup of G.) Setting T j = I N G (P )(λ j ), the Clifford correspondence gives a bijection Irr p (T j λ j) Irr p (T j λ), so the formula applied to G and N G (P ) gives the result. 15

16 Theorem 6.5. Suppose G is p-solvable, P Syl p (G), and P = N G (P ). If χ Irr p (G), then χ P = λ +, where λ Lin(P ), and = 0, or the constituents of have degree divisible by p. Furthermore, the map is a bijection. Irr p (G) Irr(P/P ) : χ λ Proof. We consider several cases. First suppose there exists K G such that p K, and K > 1. Pick χ Irr p (G). As we have seen before, there exists θ Irr p (K) a constituent of χ, and since p K and K G, P acts on K by conjugation, so the Glauberman correspondence applies and Irr P (K) = Irr(C K (P )). But C K (P ) N G (P ) = P, so C K (P ) P K = 1. Thus Irr P (K) = 1, and necessarily θ = 1 K. So χ K has the principal character 1 K as a constituent, and any other constituent is a conjugate of 1 K. Since 1 K is invariant, 1 K is the only constituent, so K ker χ. So χ Irr p (G/K). Observe that (G/K)/(P K/K) = G/P K, and G/P K = G / P K, so that P K/K is a p-sylow subgroup of G/K, and N G/K (P K/K) = N G (P )K/K = P K/K. This show that if P is self-normalizing in G, then P K/K is self-normalizing in G/K. By the induction hypothesis, χ P K/K = λ+ where λ Lin(P K/K) and the constituents of have degree divisible by p. Since K ker χ, and K ker λ, we can view this as a character on P, so χ P = λ P + P, and we get the result. Now suppose no such K exists. Let N = O p (G). As N is a p-group, it is contained in some Sylow subgroup P, and let θ Irr P (N) lie under χ. If T = I G (θ) is the inertia group, by the Clifford correspondence, there exists ψ Irr p (T θ) such that ψ G = χ. Also by the Clifford correspondence, χ T = ψ + Ξ, where no irreducible constituent of Ξ lies over θ. Then χ P = ψ P + Ξ P. Suppose T < G is proper. Then N T (P ) = N G (P ) T = P T = P, as P T, as θ is P -invariant. By the induction hypothesis, considering the tower of group T/P/N, we have ψ P = λ +, where λ Lin(P ), and the degrees of the constituents of are divisble by p. Thus χ P = ψ P + Ξ P = λ + + Ξ P and it remains to show that Ξ P has all constituents with degrees divisble by p. Towards a contradiction, suppose Ξ P has a constituent µ Irr(P ) such that 1 µ(1). Since µ(1) P, necessarily µ(1) = 1. Note µ is P -invariant as it is a class function on P, so η := µ N is P -invariant and lies under χ. But P -invariant characters of normal subgroups are N G (P )-conjugate, So η and θ are N G (P ) conjugate, but since N G (P ) = P and they are P -invariant, η = θ. This is a contradiction since no constituent of Ξ lies over θ. Lastly, suppose T = G. Then χ P has p -degree since χ does, but every irreducible constituent of χ P must have degree dividing P. Hence there must exist a constituent ɛ which is linear. So ɛ N and θ are P -invariant constituents of χ N, so as we have seen before ɛ N = θ, and thus θ is linear. Thus θ extends to P, and o(θ) is a p-power since θ Lin(N), and Lin(N) = [N : N ] is a p-power. To reiterate, we have a P -invariant linear character of p-power order which extends to P, hence to G, by a previous theorem above. Call this extension ν Irr(G). Consider χ ν Irr p (G), since χ is irreducible, and ν is linear. Then χ N = χ(1)θ, since θ is the only constituent of χ N, and so (χ ν) N = χ(1)θ ν N = χ(1)θ θ = χ(1)1 N. So N ker χ ν, and thus χ ν Irr p (G/N). By the induction hypothesis, (χ ν) P = δ + Ψ where δ is linear, and the constituents of ψ have degrees divisible by p. So χ P = δν p + Ψν P = δɛ + Ψɛ. Note that δɛ is still linear, and the degrees of the constituents of Ψɛ are unchanged as ɛ is linear. It remains to show the map is a bijection. Both sets in question have the same cardinality. If K exists as above, then we make take the quotient by K as it is in the kernel of every character involved to reduce the problem to a smaller group. Otherwise, suppose we are in the case where nontrivial N = O p (G) exists. In this case, the T j /N are smaller groups with self-normalizing p-sylow subgroup P/N, so the McKay conjecture holds. Also, N Tj/N(P/N) = N Tj (P )/N = N G (P ) T j /N = I NG (P )(λ j )/N 16

17 so that Irr p (G) = Irr p (T j /N) = Irrp N Tj/N(P/N) = Irr p (I NG (P )(λ j )/N) = Irr p (N G (P )) = Irr(P/P ) where the second equality follows by the induction hypothesis, the third by the preceding observation, the fourth by the lemma, and the last since N G (P ) = P, so the p -degree characters are precisely the linear ones of P. Take λ Irr(P/P ). Then λ G has p -degree since λ G (1) = [G : P ]λ(1) = [G : P ]. So some constituent of χ G has p -degree, which is to say (λ G, χ) 0 for some χ Irr p (G). By reciprocity, (λ, χ P ) 0. Then λ is a constituent of χ P = ɛ +, where ɛ is linear and has constituents with degrees divisible by p. So necessarily λ = ɛ. This shows the map is surjective, hence bijective for cardinality reasons. 7 July 22, 2016 Recall that if A is an algebra over a field F, then the Jacobson radical is J(A) = ann A (V ), where the intersection is taken over the simple right A-modules V. The algebra A is said to be semisimple if J(A) = 0, and it follows that A/J(A) is semisimple. Remark 7.1. Suppose A is a semisimple algebra. We have the following facts: There exist a finite number B 1,..., B n of distinct minimal ideals of A, and A = n i=1 B i. Let I j be a minimal right ideal in B j. Then the I j are the simple A-modules up to isomorphism. Let X i denote the corresponding irreducible representations. Let r j : A End F (I j ) : a r j a, where r j a is the map acting by right multiplication by a on I j. Since B i B j B i B j = 0, it follows ker r j = i j B j, and the restriction r j Bj is an isomorphism of algebras onto its image. The B i M nj (F ) for some n j, so A is a direct sum of matrix algebras. If χ i is the trace of X i, then {χ 1,..., χ n } is linearly independent. Proof. This is standard. For the last part, by Wedderburn, the irreducible represenations X i are surjective, so we can choose x i B such that X i (x i ) = diag(1, 0,..., 0). It follows that χ i (x j ) = δ ij, and linear independence follows. Now, suppose G is a finite abelian group, and F is a field of characteristic p, and X : G GL n (F ) an irreducible representation. Then X(g)X(h) = X(h)X(g) for all g, h G, and thus X(g) commutes with the image of X, so by Schur, which still applies, X(g) = λ g I n for some λ g. Since X is irreducible, necessarily n = 1. Let R be the ring of algebraic integers, and p a prime. Pick a maximal ideal m such that pr m R, and set F = R/m. We then have a canonical projection and char(f ) = p. : R F : r r = r + m Denote by U = {ζ C : ζ m = 1, p m}, which is a subset of R. Lemma 7.2. The map U F : u u is a group isomorphism, and F is the algebraic closure of Z = Z/pZ. 17

18 Proof. Suppose ζ 1 is an element of U, so that o(ζ) = m for some p m. We have the identity Evalutating at X = 1 yields 1 + X + + X m 1 = Xm 1 1 X 1 = m 1 m = (1 ζ) (1 ζ i ), i=2 (X ζ i ). and so 1 ζ m in R. If ζ = 1, then m = 0 in F, implying p m, a contradiction. So the map has trivial kernel, and is injective. Furthermore, the extension F/Z is algebraic. Take f F, so f = r for some r R. As r is an algebraic integer, it satisfies some relation m 1 i=1 r n + a n 1 r n a 1 r + a 0 = 0 where a i Z. Applying the canonical surjection gives so that f is algebraic over Z. f n + a n 1f n a 1f + a 0 = 0, Suppose K/F is an algebraic extension. I claim K U. Take k K. Since K/F and F/Z are algebraic, k is algebraic over Z. Thus L := Z (k) = Z [k] is a finite field, as a finite extension of a finite field. Thus k L = 1, so o(k) L 1 (mod p), since L is a p-power. Thus the order of k is not divisible by p, and if L = m, then k is a root of X m 1, so k = ζ for some ζ a root of X m 1. Thus k U. Since K U, we see F U, so that U = F. This shows the map is surjective. It also implies K F, so that K = F, and thus F is algebraically closed. Let G = {g G : p o(g)} denote the set of p-regular elements. Let X : G GL n (F ) be a representation. Since F is algebraically closed, X(g) has n-eigenvalues, denote them λ 1,..., λ n, all nonzero since X(g) is invertible. By the above isomorphism, λ i = u i for a unique u i U. Define a function ϕ: G R : g n u i. This ϕ depends on the choice of X, and is the Brauer character afforded by X. This is a close analogue to the trace function, as ϕ(g) = u i = λ = tr(x(g)). Let IBr(G) denote the set of irreducible Brauer characters, i.e., the characters corresponding to irreducible representations. This set is finite as there only finitely many irreducible representations by Wedderburn. As F [G]/J(F [G]) is semisimple, it decomposes as a sum of matrix algebras of sizes n i, so dim(f [G]/J(F [G])) = n i = ϕ(1) 2. i=1 ϕ IBr(G) Also, ϕ(1) is the degree of the representation. Lemma 7.3. Suppose X is an F -representation of G with trace ψ. If g G decomposes as the unique commuting product g = g p g p, where o(g p ) is a p-power, o(g p ) a p -power, then ψ(g) = ψ(g p ). Proof. Without loss of generality, we may assume G = g, as we only care about a property of g. Since X(g) is similar to an upper triangular matrix with irreducible representations on the diagonal, it suffices to prove the claim for irreducible representations of an abelian group, i.e., linear representations. Suppose X i : G F is such a representation. Since g p has order p f, say, then X i (g p ) pf = X i (g pf p ) = 1 18

19 so that X i (g p ) is a root of the polynomial Z pf 1 = (Z 1) pf over F [Z], so that X i (g p ) = 1. Thus as required. X i (g) = X i (g p )X i (g p ) = X i (g p ), From these two lemmas, it is immediate that if ϕ IBr(G), and we define ϕ by ϕ (g) = ϕ(g p ), then {ϕ : ϕ IBr(G)} = {tr(x) : X Irr(G)} It turns out the irreducible Brauer characters are linearly independent over C. We will need a fact from algebraic number theory. Lemma 7.4. Suppose I R is a proper ideal, and α 1,..., α n are algebraic complex numbers, not all 0. Then there exists β Q(α 1,..., α n ) such that βα i R for all i, but there exists at least one βα i / I. Theorem 7.5. The set IBr(G) is linearly independent over C, and cf(g ) are a C-space of dimension equal to the number of p-regular conjugacy classes of G. Proof. Since the irreducible Brauer characters take values in R Q, it is enough to show they are Q-linearly independent. Viewing the characters as elements in Q n, the matrix whose columns are the coordinate vectors has nonzero determinant, and this determinant does not change if we view it as a matrix in C. So suppose ϕ IBr(G) a ϕ ϕ = 0 where the a ϕ Q are not all 0. Since m R is a proper ideal, by the above lemma, we can find β such that βα ϕ R for all ϕ, but there exists µ such that βµ / m. Scaling by β yields (βaϕ )ϕ = 0 and since the coefficients are in R, applying the canonical surjection gives (βaϕ ) ϕ = 0, but since the ϕ are the traces of the irreducible representations, hence F -linear independent, we must have (βα ϕ ) = 0, or βα ϕ m for all ϕ IBr(G), a contradiction. We can extend to the localization R m by R m F : α β α (β ) 1. Now suppose X : G GL n (R m ) affords χ. The map R m F induces a map GL n (R m ) GL n (F ), so by postcomposition we get an induced representation X : G GL n (F ). Thus a C-irreducible character defines an F -representation in characteristic p, which thus has a Brauer character. Observe X affords χ := χ G. Every Brauer character comes from an F -representation, which decomposes into irreducible representations, so χ = d χϕ ϕ ϕ IBr(G) where the d χϕ are nonnegative integers known as decomposition numbers. Theorem 7.6. (Brauer) The set IBr(G) is a basis of cf(g ). conjugacy classes of G of elements of p order. In particular, IBr(G) is the number of Proof. Take δ cf(g ), and extend it to cf(g) by zero on the p-singular classes. Call this extension η. Then η = a χ χ = η = δ = a χ χ = d χϕ ϕ. χ Irr(G) χ Irr(G) χ Irr(G) a χ ϕ IBr(G) 19

20 8 Exercises Exercise 8.1. Let K be an arbitrary field, and let z be the sum of all elements of a finite group G in the group algebra K[G]. Show that the one-dimensional subspace Kz of K[G] is an ideal. Also, prove that z is nilpotent iff the characteristic of K divides G. Proof. It is clear Kz is an additive subgroup. Observe that if g G, then gz = gz = z, since left or right multiplication simply permutes the summands of z. If c i g i K[G] is arbitrary, then ( ) ci g i z = c i g i z = ( ) c i z = ci z Kz. So Kz is a left ideal, and it follows easily that Kz is a two-sided ideal of K[G]. Observe z 2 = ( g i G g i)z = g i G z = G z, and inductively, we see zn = G n 1 z. So suppose z is nilpotent, say z n = 0. Then 0 = z n = G n 1 z. It follows that G n 1 = 0, so that char(k) G n 1. Since char(k) is necessarily prime in this case, Euclid s theorem gives char(k) G. On the other hand, suppose char(k) G. Then G = 0, so in particular z 2 = G z = 0. So z is nilpotent, and z 2 = 0 even. Exercise 8.2. With the notation as above, show that K[G] has a proper ideal I such that Kz + I = K[G] iff the characteristic of K does not divide G. Proof. First, suppose there is a proper I such that Kz + I = K[G]. Observe G = dim(k[g]) = dim(kz + I) = dim(kz) + dim(i) dim(kz I) = 1 + dim(i) since Kz I = 0, as Kz is one dimensional, and not in I. So in fact Kz I = K[G]. Towards a contradiction, suppose char(k) G, so that z 2 = 0 by the previous exercise. Since 1 Kz I, we can write 1 = λz + x for x I, necessarily x = 1 λz. Since I is an ideal, implying I = K[G], a contradiction. (1 λz)z = z λz 2 = z I Now suppose char(k) G. Then the claim follows immediately from Maschke s Theorem. More directly, We have a surjective morphism ϕ: K[G] K : a i g i = a i. Then I := ker ϕ is a proper ideal of K[G], and by rank-nullity, dim I = dim(k[g]) dim K = G 1. Since ϕ(z) = G 0, it follows from dimension considerations that Kz I = K[G], so Kz has a complement. Exercise 8.3. If χ is a character of G and λ is a linear character, define λχ to be the function defined by (λχ)(g) = λ(g)χ(g). Show that λχ is a character, and that it is irreducible iff χ is irreducible. Proof. Suppose the representation ρ affords χ. Define a map ρ: G GL n (K) by ρ(g) = ρ(g)λ(g). Since λ is linear, it takes values in C, so it is easy to check ρ is a homomorphism. Furthermore, so that λχ is a character. tr( ρ(g)) = tr(ρ(g)λ(g)) = λ(g) tr(ρ(g)) = λ(g)χ(g) = (λχ)(g) 20

21 Also, if λχ is irreducible, then χ is irreducible, for if χ = α + β, then λχ = λα + λβ. Conversely, if χ is irreducible, then (χ, χ) = 1. Since λ is linear, it takes roots of unity as it s values. Then we find that so that λχ is irreducible. (λχ, λχ) = 1 G = 1 G = 1 G λ(g)χ(g)λ(g)χ(g) g G λ(g) 2 χ(g) 2 g χ(g) 2 = (χ, χ) = 1 g Exercise 8.4. Let N G and ψ Irr(G/N). Define χ: G C by setting χ(g) = ψ(ng). Show that χ Irr(G) and that N ker χ. Also, so that the map ψ χ from Irr(G/N) into Irr(G) defines a bijection from Irr(G/N) to the set {χ Irr(G) : N ker χ}. Proof. Let R: G/N Aut(G/N) be the regular represenation of G/N acting on itself by multiplication. If χ i is an irreducible character, put N i = ker χ i. Then ker R = {Ng G/N : Ng acts as the identity } = {Ng : g N} = N/N. Since N acts trivially, we get an induced representation R: G Aut(G/N) given by R(g) = R(Ng). Let χ be the character afforded by R, so χ(g) = tr( R(g)) = tr(r(ng)). I claim that N = ker χ. Let n N. Then so N ker χ. Conversely, suppose g ker χ. So Thus χ(g) = G/N = tr(r(ng)), so that g N. χ(n) = tr(r(nn)) = G/N = tr(r(n)) χ(1) = tr( R(1)) = tr(r(n)) = G/N. Also, observe that if χ and ψ are as in the problem statement, (χ, χ) = 1 G g G χ(g)χ(g) = 1 G g G from which we see irreducibility is preserved. ψ(ng)ψ(ng) = N G Ng G/N ψ(ng)ψ(ng) = (ψ, ψ) Exercise 8.5. Show that {ker χ : χ Irr(G)} = 1, the trivial subgroup of G. Proof. We can number the irreducible characters as Irr(G) = {χ 1,..., χ k }. If χ is a character, then we can write k χ = m i χ i i=1 for m i 0 integers. I claim ker χ = i:m i>0 ker χ i. Recall that χ i (g) χ i (1). If g ker χ, then χ(g) = mi χ i (g) = m i χ i (1) = χ(1), so necessarily χ i (g) = χ i (1), and thus g i:m i>0 ker χ i. Conversely, if g i:m i>0 ker χ i, then χ(g) = m i χ i (g) = m i χ i (1) = χ(1) 21