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BRAUER’S HEIGHT ZERO CONJECTURE FOR THE 2-BLOCKSOF MAXIMAL DEFECT

GABRIEL NAVARRO AND PHAM HUU TIEP

Abstract. We prove Brauer’s Height Zero Conjecture on blocks of finite groupsfor every 2-block of maximal defect.

1. Introduction

Suppose that G is a finite group, p is a prime and B is a p-block of G with defectgroup D. Richard Brauer’s Height Zero Conjecture, one of the main conjecturesin the Representation Theory of Finite Groups, states that all irreducible complexcharacters in B have height zero if and only if D is abelian. In 1984 Brauer’s HeightZero Conjecture was proved for p-solvable groups by D. Gluck and T. Wolf ([GW1],[GW2]), with the “only if” part being extraordinarily complicated. In 1988 the “if”implication was reduced to a question on quasisimple groups by T. Berger and R.Knorr ([BK]). P. Fong and M. Harris proved the “if” direction of the conjecturefor the principal 2-block in [FH]. (In fact, they proved the Broue Conjecture forthose blocks.) Now, the recent advances on the McKay conjecture in [IMN], togetherwith the recent and powerful results of M. Broue and J. Michel ([BM]) on unionsof `-blocks, of C. Bonnafe and R. Rouquier ([BR]) on Morita equivalences, and ofcourse the Deligne-Lusztig theory [L], allow us to handle the full Brauer’s HeightZero Conjecture for the 2-blocks of maximal defect of arbitrary finite groups.

Theorem A. Let B be a 2-block of G with defect group P ∈ Syl2(G). Then χ(1)is odd for all χ ∈ Irr(B) if and only if P is abelian.

1991 Mathematics Subject Classification: 20C20, 20C15.The research of the first author is partially supported by the Spanish Ministerio de Educacion

y Ciencia proyectos MTM2007-61161 and MTM2010-15296, and a fellowship from Programa deMovilidad. The second author gratefully acknowledges the support of the NSF (grants DMS-0600967and DMS-0901241).

The authors would like to thank the referee for various comments and suggestions which greatlyimproved the exposition of the paper.

Part of the paper was written while both authors visited the MSRI. The authors are grateful tothe Institute for its hospitality, support, and stimulating environment. They are also grateful toP. Fong, M. Isaacs, R. Kessar, and J. Michel for helpful discussions. Special thanks are due to F.Lubeck for providing us with a nice proof of Lemma 6.3.

1

2 GABRIEL NAVARRO AND PHAM HUU TIEP

Almost all of this paper is devoted to prove the “only if” part of Theorem A, andwe use the Classification of Finite Simple Groups.

A natural question is what happens with Theorem A for odd primes p. In §7 belowwe will comment on what exactly needs to be proved in order to apply the techniquesdeveloped in this paper to establish Brauer’s Height Zero Conjecture for p-blocks ofmaximal defect.

Brauer’s Height Zero Conjecture provides a wonderful way to detect from thecharacter table of a finite group G whether G has abelian Sylow p-subgroups.

Corollary B. Suppose that G is a finite group, and let P ∈ Syl2(G). Then P isabelian if and only if for every χ ∈ Irr(G) of even degree, we have that∑

x∈G, o(x) is odd

χ(x) = 0 .

Proof. By Theorem (3.19) and Corollary (3.25) of [N1], we have that χ ∈ Irr(G)belongs to the principal block of G if and only if

∑χ(x) 6= 0 (where the sum is taken

over all elements x ∈ G of odd order). Now, apply Theorem A for the principal blockof G.

2. Groups with Abelian Sylow 2-subgroups

In general, the groups with an abelian Sylow p-subgroup have a very restrictedstructure. (See, for instance, Theorem (2.1) of [KS].) This is even more restrictedwhen p = 2. We take advantage of this fact, but we do remark that the same (butmore elaborate) ideas should apply in the case where p is odd.

We recall that Walter’s classification of groups with abelian Sylow 2-subgroups [W]implies that the only non-abelian simple groups with abelian Sylow 2-subgroups areJ1, SL2(2

n) for n > 1, PSL2(q) for q ≡ ±3(mod 8), and the Ree groups 2G2(32e+1).

We mention now that all these groups have Schur multiplier of order 1 or 2. Also, allthese groups have cyclic outer automorphism group ([GLS, Theorem 2.5.12]).

Let us start with the following.

Lemma 2.1. Let G be a finite group with a Sylow p-subgroup P . Then CG(P ) =Z(P ) × A for some group A of order not divisible by p. Furthermore, the numberof p-blocks of G of maximal defect is the number of NG(P )-orbits on the conjugacyclasses of elements in A.

Proof. Set C = CG(P ). Then C/Z(P ) ∼= CP/P has order not divisible by p. By theSchur-Zassenhaus Theorem, Z(P ) has a complement A in C, which is centralized byZ(P ). It follows that C = Z(P )× A, CP = P × A, and A = Op′(NG(P )). Next, bythe Extended First Main Theorem (9.7) of [N1], it easily follows that the number ofp-blocks of maximal defect of G is the number of NG(P )-orbits on Irr(A). Hence, by

BRAUER’S HEIGHT ZERO CONJECTURE FOR THE 2-BLOCKS OF MAXIMAL DEFECT 3

Corollary (6.33) of [I1], we have that this is the number of NG(P )-orbits on the setof conjugacy classes of A.

We also recall the following well-known fact:

Lemma 2.2. Let S be a finite non-abelian simple group with an abelian Sylow2-subgroup P . Then CS(P ) = P . Also, O2′(N) = Z(N) = 1 for N := NS(P ). 2

The following is the exact structure theorem that we will need.

Theorem 2.3. Suppose that G has an abelian Sylow 2-subgroup P , and assumethat O2′(G) = G and O2′(G) ≤ Z(G). If G is solvable, then G = P . If G is non-solvable, then G = O2(G)×O2(G), where O2(G) is an abelian 2-group, and O2(G) isa direct product of non-abelian simple groups of order divisible by 2. In both cases,O2′(G) = 1 and CG(P ) = P . In particular, the principal block is the only 2-block ofG with maximal defect group.

Proof. Write Z = O2′(G). If G is solvable, then G/Z has a normal Sylow 2-subgroupby the Hall-Higman 1.2.3. Lemma. Then PZ / G has 2′-index and we have thatG = PZ. Hence G = P ×Z, and using again that O2′(G) = G, we have that G = Pis a 2-group. We conclude in this case that Z = 1 and CG(P ) = P .

Suppose that G/Z is non-solvable. By Walter’s theorem (or Theorem (2.1) of[KS]), we can write G/Z = T/Z × Q/Z, where T/Z is a direct product of r non-abelian simple groups Si/Z with non-trivial abelian Sylow 2-subgroups, and Q/Zis a 2-group. By standard group theoretical arguments, we have that [Si, Sj] = 1for i 6= j, T = T ′Z and T ′ =

∏S ′i. We also have that T/Z = O2(G/Z). Now

we have that S ′i/Z ∩ S ′i is a quasisimple group with abelian Sylow 2-subgroups andcenter of odd order. All these groups have Schur multiplier a group of order 1 or 2.Hence Z ∩ S ′i = 1. We then have that the groups S ′i are all simple non-abelian. NowY =

∏S ′i = T ′ is a direct product of non-abelian simple groups S ′i and in particular,

Y ∩ Z = 1. Thus G = Y × Q, and Z = O2′(G) = 1. Clearly, Y = O2(G). Finally,CG(P ) = CS1(P ∩S1)×· · ·×CSr(P ∩Sr)×Q, and therefore the proof of the theoremis complete by Lemma 2.2, except the assertion on principal blocks. This is howevera direct consequence of Lemma 2.1.

We will also need the general structure of the normalizer of a Sylow 2-subgroup inany finite non-abelian simple group, which has been worked out completely in [KM].For the sake of convenience, in what follows we will use the notation (P )SLεn(q) todenote (P )SLn(q) when ε = + and (P )SUn(q) when ε = −. Similarly, Eε

6(q) denotesE6(q) when ε = + and 2E6(q) when ε = −; furthermore, the subscript sc, resp. ad,indicates that we are considering the corresponding finite connected reductive groupof simply connected type, resp. of adjoint type. (For instance, Eε

6(q)sc and Eε6(q)ad

have the same order, but the former is quasisimple and the latter is almost-simple.)


We refer the reader to [C] and [DM] for various facts on the structure and the complexrepresentation theory of finite groups of Lie type.

Proposition 2.4. Let S be a finite non-abelian simple group and let P be a Sylow2-subgroup of S. Then one of the following holds.

(i) CS(P ) = Z(P ).(ii) S ∼= PSLεn(q) for some odd q.(iii) S = L/Z(L), L = Eε

6(q)sc for some odd q, and CS(P ) = Z(P ) × Cd whered = (q−ε)2′/(3, q−ε). Moreover, if Q is a Sylow 2-subgroup of L, then NL(Q) = Q×Aand CL(Q) = Z(Q)× A with |A| = (q − ε)2′ .

Proof. (i), (ii), and the first part of (iii) follow from Theorem 7 of [KM]. For thesecond part of (iii), we may assume P = QZ/Z for Z = Z(L). Clearly, CL(Q) Zand CL(Q)/Z = CS(P ); also, |Z| = (3, q − ε). It follows that |CG(Q)|2′ = (q − ε)2′ ,and so by Lemma 2.1 we have CG(Q) = Z(Q) × A with |A| = (q − ε)2′ . Next, byTheorem 6(c) of [KM], NS(P ) = P × Cd, whence NS(P ) = QZ/Z × A/Z. SinceNL(Q) ≥ QA and NL(Q)/Z ≤ NS(P ), we obtain NL(Q) = Q× A.

Now we can prove the easy “if” part of our Theorem A.

Theorem 2.5. Suppose that B is a 2-block of a finite group G with abelian defectgroup P ∈ Syl2(G). Then χ(1) is odd for all χ ∈ Irr(B).

Proof. We argue by induction on |G|. If N is a proper normal subgroup of G, and bis a block of N covered by B, by Theorem (9.26) of [N1], we may assume that P ∩Nis a defect group of b. Hence all θ ∈ Irr(N) have odd degree. Now, suppose thatG/N has odd order, and χ ∈ Irr(B). Then if δ ∈ Irr(b) lies over χ, then χ(1)/δ(1)divides |G/N | by Corollary (11.29) of [I1], and we are done in this case.

Suppose now that M / G has odd order. By the Fong-Reynolds Theorem (9.14)of [N1], there is a θ ∈ Irr(M) and a block b of the stabilizer T of θ in G, such that bhas defect group T and induction defines a bijection Irr(b) → Irr(B). Since |G : T |is odd, we may assume that T = G. Now by Theorem (2D) of [F], we may assumethat O2′(G) ≤ Z(G).

Now, by Theorem 2.3, we have that G = U×V , where V is an abelian 2-group andU is the direct product of non-abelian simple groups with abelian Sylow 2-subgroups.By induction, we may assume that G is simple. Now, B is the principal block of G.Now we can either apply [FH] or verify directly that all χ ∈ Irr(B) have odd degreeas follows. If G = SL2(q) with q = 2n ≥ 4, then χ(1) = 1, q ± 1. If G = PSL2(q)with q ≡ ±3(mod 8), then χ(1) = 1, q, (q + ε)/2 with ε = ±1 and q ≡ ε(mod 4), cf.[Bu]. If G = 2G2(q) with q = 32f+1 ≥ 3, then χ(1) = 1, q2−q+1, q3, q(q2−q+1), or(q − 1)m(q + 1± 3m)/2 with m = qf . (See [Wa]). Finally, if G = J1, then χ(1) = 1,77, 133, or 209 ([GAP]).


3. Some block theory

In this and the next sections we will develop several technical tools that we willneed to prove the “only if” part of Theorem A. Basically, we use the notation of [N1]for blocks, and the notation of [I1] for characters.

We start with the following lemma.

Lemma 3.1. Suppose that K / G. Let b be a block of K with defect group Q. IfCG(Q) ≤ K, then bG is defined and is the only block of G covering b.

Proof. We argue by induction on |G|. By definition, we have that b is admissible.Hence, bU is defined for every K ≤ U ≤ G. By Lemma (9.8) of [N1], we also havethat bU covers b for every K ≤ U ≤ G. In fact, if U < G, then bU is the only blockof U covering b, by induction. We have to prove that there is a unique block of Gcovering b.

By the Fong-Reynolds Theorem (9.14) of [N1], we may assume that b is G-invariant.Now, let B be any block covering b. By Theorem (9.26) of [N1], there exists adefect group P of B such that P ∩ K = Q. Then CG(P ) ≤ CG(Q) ≤ K andKP = K(PCG(P )). Now, let b0 be a block of PCG(P ) inducing B (by the ExtendedFirst Main Theorem (9.7) of [N1]). Let b1 = bKP which is defined by Theorem (4.14)of [N1], and notice that bG1 = B (by Problem (4.2) of [N1]). Now, we claim that b1covers b. By Corollary (6.4) of [N1], let ξ ∈ Irr(b1) and χ ∈ Irr(B) be over ξ. Now, ifθ ∈ Irr(K) lies under ξ, then it lies under χ, and therefore θ ∈ Irr(b), because b is theonly block of N covered by B (and using Theorem (9.2) of [N1]). This proves that b1covers b, as claimed. Now, bKP is defined and covers b (by the first paragraph of thisproof). Since KP/K is a p-group, then there is a unique block of KP covering b (byCorollary (9.6) of [N1]), and we deduce that (bKP ) = b1. Then, using again Problem(4.2) of [N1], we see that

bG = (bKP )G = (b1)G = B ,

and bG is the only block covering b by Corollary (9.6) of [N1].

We will need the following theorem on extensions of characters and blocks (whichis our character theoretical version of theorems of Alperin [A] and Dade [D]). Wethank M. Isaacs for the argument using Brauer’s characterization of characters thathandles the case where G/N is not solvable.

Let R be the ring obtained by taking the full ring of algebraic integers in C andlocalizing at a maximal ideal containing the prime p. Let M be the unique maximalideal of R and write α∗ to denote the image of α ∈ R in R/M . Recall that if ψ is

an irreducible character of a group G, then ωψ(K) = ψ(x)|K|/ψ(1) is an algebraic

integer, where K is the class of G containing x ∈ G and K :=∑

y∈K y. Furthermore,


if ψ is in the principal block of G, then (ψ(x)|K|/ψ(1))∗ = |K|∗. In this situation, if|K| is not divisible by p, then 1/|K| lies in R, and we see that ψ(x)/ψ(1) lies in Rand (ψ(x)/ψ(1))∗ = 1∗.

If N / G and θ ∈ Irr(N), then Irr(G|θ) is the set of irreducible characters χ of Gsuch that θ is an irreducible constituent of χN .

Theorem 3.2. Suppose that N / G. Let P ∈ Sylp(G), and let b be the principalblock of N . Suppose that G = NY , where Y is a p′-group centralizing P . If τ ∈ Irr(b)is G-invariant, then there exists a unique τ ∈ Irr(G) in the principal block B of Glying over τ . In fact, τN = τ . Also, if N ≤ H ≤ G, then τH lies in the principal blockof H.

Proof. Since the principal block B of G covers b, by Theorem (9.4) of [N1] thereexists some χ ∈ Irr(G) in B lying over τ . We prove that χ is unique and extends τ .

First of all notice that for all y ∈ Y , we have χ(y)/χ(1) ∈ R and (χ(y)/χ(1))∗ = 1∗.This follows since if y ∈ Y and K is the class of y in G, then |K| is not divisible byp.

Step 1. Assume that τ has an extension σ ∈ Irr(G) such that for all y ∈ Y , wehave σ(y)/σ(1) ∈ R and (σ(y)/σ(1))∗ = 1∗. Then χ = σ. In particular, this holds ifσ lies in the principal block of G.

To prove the step, we have that χ = ασ for some character α ∈ Irr(G/N), byGallagher’s Corollary (6.14) of [I1]. Thus

χ(y)

χ(1)=

(α(y)

α(1)

)(σ(y)

σ(1)

)for y ∈ Y . We know that the left side and the second factor on the right lie in R andare congruent to 1 modulo M . Also, the first factor on the right lies in R since α(1) isnot divisible by p. Applying the homomorphism ( )∗, we deduce that (α(y)/α(1))∗ =1∗. Thus α(y) ≡ α(1) modulo M for all y ∈ Y . Now |Y |[αY , 1Y ] ≡ |Y |α(1) which isnot in M . Since αY is irreducible (because N ≤ ker(α)), we deduce that αY = 1Y ,and thus α = 1G.

Step 2. If G/N is nilpotent, then χ is an extension of τ , and it is the unique memberof Irr(G|τ) in the principal block of G.

We argue by induction on |G : N |, and may assume that N < G. Let M / G with|M/N | = q, a prime. Then there exists ψ ∈ Irr(M |τ) in the principal block of M ,and ψ is automatically an extension of τ because M/N is cyclic. By Step 1 (appliedin M) we see that ψ is the unique member of Irr(M |τ) in the principal block, so ψis G-invariant. By the inductive hypothesis, ψ has an extension σ in the principalblock of G, and again by Step 1, we have χ = σ. Since this holds for every possiblechoice of χ ∈ Irr(G|τ) in the principal block of G, the uniqueness follows.


Step 3. Suppose that χ is an extension of τ . Then for N ≤ H ≤ G, the restrictionχH lies in the principal block of H.

Let ψ ∈ Irr(H|τ), where ψ lies in the principal block. We can apply Step 1 with ψin place of χ and χH in place of σ, and we deduce that ψ = χH .

Step 4. In all cases, χ is an extension of τ .

We define a class function θ on G as follows. For g ∈ G, let H = N〈g〉 and by Step2, let σ be the unique extension of τ in the principal block of H. Define θ(g) = σ(g).Clearly, θ is a class function, and we argue that it is a generalized character. For thispurpose, it suffices to show that θE is a character for every subgroup E such thatN ≤ E ≤ G and E/N is nilpotent. By Step 2, there exists ψ ∈ Irr(E|τ) extendingτ and lying in the principal block of E. We argue that θE is a character by showingthat θE = ψ. Suppose g ∈ E and let H = N〈g〉. Then the Step 3 guarantees that ψHis the unique extension of τ to H in the principal block of H, and so by the definitionof θ, we have θ(g) = ψH(g) = ψ(g), as wanted.

We know now that θ is a generalized character by Brauer’s characterization ofcharacters. Also θ(1) > 0. In fact θ is irreducible because the sum of |θ(g)|2 overeach coset of N in G is |N | (Lemma (8.14.c) of [I1]), and thus [θ, θ] = 1. Thus θ is anextension of τ and θ(y)/θ(1) lies in R for all y ∈ Y , and satisfies (θ(y)/θ(1))∗ = 1∗.By Step 1, we have that χ = θ, as wanted.

Lemma 3.3. Suppose that N / G has p′-index. Let P ∈ Sylp(G), and suppose thatNG(P ) = NN(P ) × Y , for some subgroup Y . Let b be the principal p-block of Nand assume that every τ ∈ Irr(b) is G-invariant. By Theorem 3.2, let τ ∈ Irr(G)be the only extension of τ to G in the principal block of G. Then the irreduciblecharacters of the blocks of G covering b are exactly the sets bθ = τ θ | τ ∈ Irr(b),where θ ∈ Irr(G/N). Also, bθ 6= bη, if η 6= θ. Furthermore, if e is the principal blockof NN(P ), then eθNG(P )

is the Brauer first main correspondent of bθ.

Proof. By the Frattini argument, we have that G = NY and N ∩ Y = 1. Supposethat τi ∈ Irr(b) and θi ∈ Irr(G/N) for i = 1, 2. By Theorem 3.2, we know that thereexists a unique extension τi ∈ Irr(G) in the principal block of G. Also τiθi ∈ Irr(G)by Gallagher. Suppose that K is a class of G and that x ∈ K. Using that θi(x)/θi(1)is a local integer, we have that

ωτiθi(K)∗ = ωθi

(K)∗ .

Now, since G/N is a p′-group and θi ∈ Irr(G/N), we have that θ1 and θ2 are in thesame p-block of G if and only if θ1 = θ2 (because the restriction (θi)

0 of θi to p-regularelements belongs to IBr(G) and θi is uniquely by this restriction, for instance). Now,if B is a block of G covering b, then every irreducible character of B lies over someτ ∈ Irr(b), and therefore is of the form τ γ for some γ ∈ Irr(G/N). This proves the


first two parts of the lemma. Now, θ ∈ Irr(bθ) and θNG(P ) ∈ Irr(NG(P )) lies in theblock eθNG(P )

. Now, we apply Theorem (6.6) of [N1].

Lemma 3.4. Suppose that Z is a normal p′-subgroup of G. Suppose that G has anormal p-subgroup Q such that CG(Q) ≤ QZ. If λ ∈ Irr(Z), then Irr(G|λ) = Irr(B)for a unique p-block B of G.

Proof. We have that K = QCG(Q) = QZ is a normal subgroup of G with a normalSylow p-subgroup Q. Now the block of λ is covered by a unique block b0 of K (byCorollary (9.6) of [N1]), which has defect group Q. Now, by Lemma 3.1 we have that(b0)

G is the only block of G that covers b0. Now, let χ ∈ Irr(G) be over λ. Supposethat χ belongs to some block B of G. Hence χK has an irreducible constituent µover λ. Necessarily, µ belongs to b0. Hence B covers b0 (by Theorem (9.2) of [N1]),and B = (b0)

G.

We will also need the following elementary result.

Lemma 3.5. Suppose that α, β ∈ Irr(G) and αβ is irreducible. If α and β are in theprincipal block of G and α(1) or β(1) is not divisible by p, then αβ is in the principalp-block of G.

Proof. Suppose that β(1) is not divisible by p. Let x ∈ K, where K is any conjugacyclass of G. Then β(x)/β(1) is a local integer and

ωαβ(K)∗ =

(α(x)β(x)|K|α(1)β(1)

)∗=

(α(x)|K|α(1)

)∗(β(x)/β(1))∗ =

= |K|∗(β(x)/β(1))∗ = ωβ(K)∗ = |K|∗ ,as desired.

4. Character Correspondences and Character Extensions

Suppose that N / G and θ ∈ Irr(N) is G-invariant. Using the notation and termi-nology of Chapter 11 of [I1], we say in this case that (G,N, θ) is a character triple.It is well-known that θ uniquely determines an element [θ]G/N ∈ H2(G/N,C×) =M(G/N), the Schur multiplier of the group G/N . (See Theorem (11.7) of [I1].) Also,[θ]G/N = [1N ]G/N if and only if θ extends to G. We also mention that if (G,N, θ)and (H,M, γ) are character triples with G/N ∼= H/M , we identify M(G/N) withM(H/M), and [θ]G/N = [γ]H/M then the triples (G,N, θ) and (H,M, γ) are isomor-phic. This implies (see for instance Lemma (11.24) of [I1]) that there is a characterbijection˜: Irr(G|θ)→ Irr(H|γ) satisfying χ(1)/γ(1) = χ(1)/θ(1), among some otherproperties.

The following is straightforward.


Lemma 4.1. Suppose that N / G, and that α, β ∈ Irr(N) are G-invariant withαβ ∈ Irr(N). Then

[αβ]G/N = [α]G/N [β]G/N .

Proof. For γ ∈ α, β, take Yγ be a representation of N affording γ and construct aprojective representation Xγ of G satisfying the three conditions in Theorem (11.2)of [I1]. Now, the tensor product representation Yα ⊗ Yβ affords αβ, and Xα ⊗Xβ isa projective representation of G satisfying the conditions in Theorem (11.2) of [I1].The factor set of Xα⊗Xβ is the product of the factors sets of Xα and Xβ. The proofof the lemma easily follows.

The following is one of the key results that we need for Theorem A. It easily followsfrom our recent reduction of the McKay conjecture in [IMN]. If p is a prime, Irrp′(G)denotes the complex irreducible characters of G of degree not divisible by p.

Theorem 4.2. Suppose that L is a finite group with an abelian Sylow 2-subgroupQ, and assume that L is a direct product of non-abelian simple groups. Suppose alsothat L / G. Then there exists a NG(Q)-equivariant bijection

∗ : Irr2′(L)→ Irr2′(NL(Q))

such that for every θ ∈ Irr2′(L), if T is the stabilizer of θ in G, then

[θ]T/L = [θ∗]NT (Q)/NL(Q) ,

whereH2(T/L) andH2(NT (Q)/NL(Q)) are identified under the natural isomorphismT/L ∼= NT (Q)/NL(Q). In particular, the triples (T, L, θ) and (NT (Q),NL(Q), θ∗)are isomorphic.

Proof. We know that L is a direct product of simple groups S of type J1, SL2(2n),

PSL2(q) where q ≡ 3 or 5 mod 8, or 2G2(32e+1). In Sections 15 and 17 of [IMN],

it is checked that that these simple groups are good for every prime p. Moreover,when the prime p = 2, then the correspondence subgroup in the sense of Sections 10and 11 of [IMN] for these simple groups S, is taken to be the normalizer of a Sylow2-subgroup of the group S, cf. pp. 72, 92, and 99 of [IMN]. The case S = J1 is takencare of by Lemma (10.3) of [IMN]. Now the first part of the proof of this theoremfollows from Theorem (13.1) of [IMN], with H = NL(Q), and the comments after itsstatement. Now, the proof easily follows by using the last paragraph of the proof ofTheorem (13.1) of [IMN] on page 83.

We will also need the following extension theorem.

Lemma 4.3. Suppose that L/ G is a direct product of groups S1, . . . , Se which arepermuted by G-conjugation. Let L ≤ M / G contained in NG(Si) for all i. Assumethat for 1 ≤ i ≤ e we have characters τi ∈ Irr(L) with Sj ≤ ker(τi) for j 6= i and


some extension τi ∈ Irr(M |τi) such that τi extends to ING(Si)(τi), and (τi)g = τk if

and only if (τi)g = τk for all g ∈ G and i, k. Then τ =

∏i τi extends to IG(τ).

Proof. We argue by induction on |G|. Working in IG(τ), we may assume that τ isG-invariant.

Now, suppose that L = U × V , where U = S1 × · · · × Sd is a direct product ofthe G-orbit, say S1, . . . , Sd, and V = Sd+1 × · · · × Se is the direct product of therest. Hence U, V / G. It is straightforward to check that the hypothesis are satisfiedin G/V and G/U . Hence α =

∏di=1 τi and β =

∏ei=d+1 τi extend to IG(α) and IG(β),

respectively. We claim that both α and β are G-invariant. By the first paragraph,we know that τL =

∏i τi is G-invariant. Now, if g ∈ G and Sgi = Sσ(i) for some

permutation σ of 1, . . . , d, it follows that (τi)g = τσ(i). Hence, (τi)

g = τσ(i) byhypothesis, we see that α, and similarly, β is G-invariant. Hence, α and β bothextend to G, and therefore so does αβ = τ .

So we may assume that the action of G is transitive. Write S = S1. Let H =NG(S). Let x1, . . . , xe be representatives of right cosets of H in G, so that Si = Sxi .Since τL is G-invariant, we may also assume that (τ1)

xi = τi, so that (τ1)xi = τi. Now,

we know that τ1 extends to some γ ∈ Irr(H). Let χ = γ⊗G, which is a character ofG. Now, if m ∈M , by Lemma (4.1) of [GI] we have that

γ⊗G(m) =e∏i=1

γ(ximx−1i ) =

e∏i=1

τ1(ximx−1i ) =

e∏i=1

(τ1)xi(m) = τ(m) ,

as desired.

Finally, we will need the following elementary technical result right at the very endin the proof of Theorem A. We provide a proof of it here for the reader’s convenience.

Lemma 4.4. Suppose that L/ G, where L = S1× · · · ×Se, and the set S1, . . . , Seis permuted by G-conjugation. Suppose that Q ∈ Sylp(L), and let Qi = Q∩Si. ThenNL(Q) = NS1(Q1)×· · ·×NSe(Qe), and the set NS1(Q1), . . . ,NSe(Qe) is permutedby NG(Q)-conjugation. If G = NG(Q) and Si = NSi

(Qi), then

NG(Si)/SiCG(Si)

is isomorphic to a quotient of NG(Si)/SiCG(Si).

Proof. We have that Q = Q1 × · · · × Qe, by Sylow theory. Also, we have thatNL(Q) = NS1(Q1) × · · · ×NSe(Qe). Now, suppose that g ∈ NG(Q). Then we seethat Q1 × · · · × Qe = Q = Qg = (Q1)

g × · · · × (Qe)g. Hence, if (Si)

g = Sj, then(Qi)

g ≤ Sj ∩Q = Qj, and (Qi)g = Qj. In particular, NSi

(Qi)g = NSj

(Qj).To prove the final statement, we certainly may assume that Qi > 1.We claim that X = NG(S1) = NG(Q) ∩ NG(S1). First of all notice that if x

normalizes some 1 < U < S1, then (S1)x ∩ S1 > 1, and x normalizes S1. Hence


X ≤ NG(Q) ∩ NG(S1). Conversely, if y ∈ NG(Q) ∩ NG(S1), then y normalizesQ ∩ S1 = Q1. Hence (S1)

y = NS1(Q1)y = NSy

1(Qy

1) = NS1(Q1), and the claim isproved.

Set C = CG(S1). Next, we claim that X ∩CS1 ≤ Y , where Y = S1CG(S1). Writex = cs ∈ X with c ∈ C and s ∈ S1. Write Q = Q1 × R where R = Q2 × . . . × Qe.Then

Q1 ×R = Q = Qx = Qcs1 ×Rsc = Qs

1 ×Rc.

But notice that Qs1 ≤ S1, and Rc ≤ (

∏ei=2 Si)

c =∏e

i=2 Si. It follows that Qs1 = Q1

and Rc = R. Thus s ∈ NS1(Q1) = S1 ≤ Y . Next, Qc = Qc1 × Rc = Q1 × R = Q, so

c ∈ NG(Q) = G. But c centralizes S1, so c centralizes S1, whence c ∈ CG(S1) ≤ Y .

Therefore, x = sc ∈ S1CG(S1) ≤ Y , as claimed.We have shown that X ∩ CS1 = Y ∩ CS1. Now, G = LNG(Q) by the Frattini ar-

gument. Also, L ≤ S1CG(S1) ≤ NG(S1), and NG(S1) = NG(Q)∩NG(S1). ThereforeX/Y ∼= NG(S1)/S1CG(S1)Y .

5. Proof of the Main Theorem

In this section we prove the “only if” part of Theorem A, assuming some results onquasisimple groups that we will prove in the next section. We should remark that the“only if” part of the Height Zero Conjecture does not exactly reduce to quasisimplegroups.

The following is the main result concerning quasisimple groups that we need toprove Theorem A, and that will be proved in the next section.

Theorem 5.1. Let G be a finite perfect group, with S = G/Z simple, where Z =Z(G) is cyclic of odd order. Let B be a 2-block of G with defect group P ∈ Syl2(G),and assume that χ(1) is odd for all χ ∈ Irr(B). Then one of the following holds.

(a) P is abelian.(b) G ∼= Eε

6(q)sc with ε = ± and q odd, and B is Morita equivalent to a 2-blockwith defect group P of a proper subgroup of G.

We will also use the following deep theorem. It combines Theorem D of [M], withthe Gluck-Wolf Theorem [GW1] for the prime p = 2.

Theorem 5.2. Suppose that N / G and θ ∈ Irr(N) is G-invariant. Let P/N ∈Syl2(G/N). If χ(1)/θ(1) is odd for every χ ∈ Irr(G|θ), then P/N is abelian.

Proof. By using character triples, we may assume that N ≤ Z(G). Now, by TheoremD of [M], we know that G/N is solvable. Then the theorem follows from the mainresult of [GW1].


This theorem deserves a comment. Suppose that N / G, and that θ ∈ Irr(N)is G-invariant. Assume further that there is a prime p such that p does not divideχ(1)/θ(1) for all χ ∈ Irr(G|θ). If p > 2, then it is not necessarily true that G/N isp-solvable. What is a consequence of the Height Zero Conjecture (as can be proved)is that a Sylow p-subgroup P/N of G/N should be abelian. Therefore, any attempt ofproving the Height Zero Conjecture for odd primes, should first address this difficultpurely character theoretical question.

We proceed with the proof of our main result.

Theorem 5.3. Suppose that B is a 2-block of a finite group G with defect groupP ∈ Syl2(G). If χ(1) is odd for all χ ∈ Irr(B), then P is abelian.

Proof. We argue by induction on |G : O2′(G)|. We certainly assume that G is not a2-group. By the Fong-Reynolds Theorem (9.14) of [N1], we may assume that if N / Gand b is covered by B, then b is G-invariant. In particular, there exists a G-invariantθ ∈ Irr(O2′(G)) such that every χ ∈ Irr(B) lies over θ. By Theorem (2D) of [F], wemay assume that Z = O2′(G) is cyclic and central in G, and that B covers somefaithful λ ∈ Irr(Z).

Now let M be a proper normal subgroup of G, and let b be the block covered by B.By Theorem (9.26) of [N1], we have that P ∩M is a defect group of b. If η ∈ Irr(b),then η lies under some χ ∈ Irr(B) by Theorem (9.4) of [N1]. Hence η has odd degree,and we conclude that M has abelian Sylow 2-subgroups by induction. In particular,we may assume that O2′(G) = G.

Suppose that G/Z is a simple group. If G/Z is abelian, then |G/Z| = 2, and thetheorem is true. Assume that G/Z is non-abelian. If G′ < G, then G′Z = G, andZ ∩ G′ ≤ Z(G′). Since G′ is perfect, by Theorem 2.3 we conclude that Z ∩ G′ =1. Then G = G′ × Z and G has abelian Sylow 2-subgroups (because G′ does byinduction). Hence, we conclude that G is perfect. Then we can apply Theorem 5.1and we are done in the case of the conclusion (a). So suppose that we are in case (b) ofthat theorem. In this case, B is Morita equivalent to some block b with defect group Pof some proper subgroup of G. Hence, there is a bijection Irr(B)→ Irr(b) preservingcharacters of height zero. (See Theoreme 1.5 and Remarque 1 after Theoreme 3.1 of[B].) Hence, all irreducible characters of b have height zero, and we conclude that Pis abelian by induction.

Now, let 1 < N/Z < G/Z be a minimal normal subgroup of G/Z; in particular,N/Z is a direct product of isomorphic simple groups. If N/Z is solvable, then N/Zis a 2-group. Then N = Q×Z, where Q is an elementary abelian normal 2-subgroupof G. In this case, B covers the principal block of Q. If N is non-solvable, thenN = O2′(N) × Z by Theorem 2.3. Hence L = O2′(N) is a direct product of non-abelian simple groups with abelian Sylow 2-subgroups. Then L has a unique blockof maximal defect, namely the principal block of L, by Theorem 2.3. So in both


cases we have that B covers the principal block of X, where N = X ×Z, and X is aminimal normal subgroup of G. Hence, there is a character χ ∈ Irr(B) lying over 1Xby Theorem (9.4) of [N1]. Now, the block B of G/X containing χ is contained in B.Hence we deduce that G/X (equivalently G/N) has abelian Sylow 2-subgroups byinduction. Also, if G/Z has two distinct minimal normal subgroups N and M , thenG/N ∩M has abelian Sylow 2-subgroups. Since M ∩ N = Z, then G has abelianSylow 2-subgroups, and we are done.

We conclude that G/Z has a unique minimal normal subgroup N/Z having abelianSylow 2-subgroups, such that G/N has abelian Sylow 2-subgroups.

Suppose first that N/Z is solvable. Hence, as we already know, N = Z×Q, whereQ is an elementary abelian 2-group. Assume first that CG(Q) = C < G, so thatC has abelian Sylow 2-subgroups by induction. Now consider K/Z = O2′(C/Z),that contains Q/Z. Now, by applying Theorem 2.3 to K/Z, we have that K/Z is adirect product of two normal subgroups of G/Z. However, G/Z has a unique minimalnormal subgroup. Since K/Z contains a nontrivial normal 2-subgroup, we concludethat K/Z is a 2-group. Thus K = R× Z, where R is an abelian normal 2-subgroupof G containing Q. Now R is a normal Sylow 2-subgroup of C and R ≤ CG(R) ≤ C.Hence CG(R) = R× Z = K. By Lemma 3.4, we have that all irreducible charactersof Irr(G|λ) lie in the same block, which should be B (because B covers λ). Now, Ghas abelian Sylow 2-subgroups by Theorem 5.2.

So we may assume that CG(Q) = G, so that |Q| = 2. Hence O2′(G/Q) = ZQ/Q.Also, O2′(G/Q) = G/Q. We know that G/Q has abelian Sylow 2-subgroups. Soif G/Q is solvable, then by Theorem 2.3, G/Q is a 2-group, and G is a 2-group, acontradiction. Hence, by Theorem 2.3, we deduce that G/Q = S/Q × R/Q, whereS/Q is a direct product of non-abelian simple groups Si/Q, and R is abelian (becauseit is a proper normal 2-subgroup of G). Notice that S/Q is perfect. If S ′ < S, thenS = S ′×Q, G = S ′×R, and we are done. Hence, we may assume that S ′ = S. Now,let B1 be a block of S covered by B. Now, B1 covers the principal block of Q and allδ ∈ Irr(B1) have odd degree. Now, if 1 6= λ ∈ Irr(Q) and δ ∈ Irr(B1) lies over λ, wehave that δQ = δ(1)λ. Since S is perfect, we have that 1 = det(δ)Q = λδ(1), and thisis a contradiction.

Hence we may assume that N/Z is non-solvable. By the fourth paragraph of thisproof, we know that N = L × Z, where L = O2′(N) is a minimal (non-solvable)normal subgroup of G. Recall that N/Z is the unique minimal normal subgroup ofG and theferore CG/Z(N/Z) = 1. Thus CG(L) = Z. Also, we showed that B coversthe principal block b of L. Let Q = P ∩ L. We know that CL(Q) = Q by Theorem2.3. Also, the principal block of L is the unique block of maximal defect in L. Hence,all odd degree irreducible characters of L are in the principal block of L, and coveredby characters of B. But also, by Theorem 2.5, the odd degree irreducible charactersof L are exactly the irreducible characters of b.


Assume thatN < M is a proper normal subgroup ofG. HenceN/Z ≤ O2′(M/Z) =V/Z. By Theorem 2.3, we have that V/Z = O2(V/Z)×O2(V/Z), where O2(V/Z) is adirect product of non-abelian simple groups, andN/Z/O2(V/Z). Since CG/Z(N/Z) =

1, we see that N/Z = O2(V/Z) and O2(V/Z) = 1. Hence O2′(M/Z) = N/Z, andM/N is odd.

Now, let us write M = NCG(Q) = LCG(Q). Recall that CL(Q) = Q. HenceNM(Q)/Q = NL(Q)/Q×CG(Q)/Q.

Suppose now that M = G. Then LCG(Q) = G. Since NG(Q)/CG(Q) ∼= NL(Q)/Qis not divisible by 2, we have that P ≤ CG(Q) and P / NL(Q)P because NG(Q)/Q =NL(Q)/Q × CG(Q)/Q. Now, let µ ∈ Irr(P ), and let ε ∈ Irr(Q) be an irreducibleconsituent. Recall that P/Q is abelian because it is isomorphic to a Sylow 2-subgroupof G/N . Now, ε has a canonical extension ε to its stabilizer in NL(Q), by Corollay(8.16) of [I1]. Now consider (ε)NL(Q) which has odd degree (because Q is abelian),and is CG(Q)-invariant (because ε uniquely determines ε). Hence, by Theorem 4.2,we have that (ε)NL(Q) = τ ∗ for some odd degree τ ∈ Irr(L) which is CG(Q)-invariant.Hence τ is G-invariant. Now τ is in the principal block of L which is covered by B. Sothere is an odd degree χ ∈ Irr(B) over it. Then χLP has some odd degree irreducibleconstituent ξ. Since ξ lies over τ and PL/L is a 2-group, we conclude that ξL = τ .Thus, since (G,L, τ) and (NG(Q),NL(Q), τ ∗) are isomorphic by Theorem 4.2, wededuce that τ ∗ extends to NL(Q)P . Let δ be such an extension, which therefore hasodd degree. Since P / NL(Q)P , it follows that the irreducible constituents of δP arelinear. Now, one of them lies over ε, and we conclude that ε extends to P . Since P/Qis abelian, we conclude that µ is linear, and therefore P is abelian. The dispositionof relevant subgroups is displayed in Figure 1.

G

nnnnnnnnPPPPPPPP

M

ssssss

NNNNNNNN NG(Q)pppppp

LKKKK NM(Q)

ppppppOOO

O

NL(Q)

OOOOOOCG(Q)

oooooo LLLL

Q

PPPPPPPPP Y

qqqqqq

1

Figure 1.

We may assume therefore that M = NCG(Q) = LCG(Q) is a proper normalsubgroup of G. Hence M/L has odd order. Thus CG(Q)/Q has odd order, and thus


CG(Q) = Q× Y , where NM(Q) = NL(Q)× Y . We also have that Z ≤ Y . Now, byTheorem 4.2, it follows that all the irreducible characters of the principal block b ofL are CG(Q)-invariant. Let b be the principal block of NL(Q).

Now, let e be a block of M covered by B. By Fong-Reynolds, recall that we areassuming that e is G-invariant. By Lemma 3.3 (and using the notation of Theorem3.2), there exists ρ ∈ Irr(M/L) such that we have that e = τ ρ | τ ∈ Irr(b). Since eis NG(Q)-invariant, we have that ρ is G-invariant (by using Lemma 3.3, for instance).Now, let e be the Brauer First Main correspondent of e. By Lemma 3.3, we havethat Irr(e) = τ × ρ | τ ∈ Irr(b). Since e is NG(Q)-invariant, so is e.

Now, by Theorem (9.28) of [N1], let B be the Harris-Knorr correspondent of B(applied in (G,M)). Hence B is a block of NG(Q) of defect group P covering e.

We are going to prove that if δ ∈ Irr(B), then δ has odd degree. Since NG(Q) < G,by induction we will have proved the theorem. Now, δ lies over some µ× ρ for someµ ∈ Irr(NL(Q)) of odd degree. Now, µ = τ ∗ for some τ ∈ Irr(L) of odd degree.Notice that µ = µ× 1Y , where we are using the notation of Theorem 3.2 and Lemma3.3. Let T be the stabilizer of τ in G, which we know contains M . By Theorem 4.2,we know that NT (Q) is the stabilizer of τ ∗ in NG(Q) and that

[τ ]T/L = [τ ∗]NT (Q)/L .

Now, τ has a unique extension τ ∈ Irr(M) in the principal 2-block of M . We claimthat the stabilizer of τ ρ is also T . First of all notice that τ and τ uniquely determineeach other. Also, if x ∈ G fixes τ ρ then it fixes (τ ρ)L = ρ(1)τ and x fixes τ . By

uniqueness, x fixes τ , as wanted. By the same reason, the stabilizer of (τ ∗)ρ is NT (Q).By Lemma 4.1, we have that

[τ ρ]T/M = [τ ]T/M [ρ]T/M ,

and

[(τ ∗)ρ]NT (Q)/M = [(τ ∗)]NT (Q)/M [ρ]NT (Q)/M .

Next, we prove that [τ ]T/M and [(τ ∗)]NT (Q)/M are trivial. Once this is proved, wewill know that

(T,M, τρ) and (NT (Q),NM(Q), (τ ∗)ρ)

are character triple isomorphic. Assume this for the moment. By Lemma 3.1, noticethat B = eG is the only block of G covering e. Hence every χ ∈ Irr(G|τ ρ) lies inB, and therefore all of them have odd degree. In particular, |G : T | is odd by theClifford correspondence for characters (and therefore |NG(Q) : NT (Q)|), and everyγ ∈ Irr(T |τ ρ) has odd degree. By character triples (Lemma (11.24) of [I1]), everyγ ∈ Irr(NT (Q)|µ×ρ) has odd degree. Now, δ = γNG(Q) for some γ ∈ Irr(NT (Q)|µ×ρ),and we conclude that δ has odd degree. Hence, P is abelian, by induction.

Hence, the proof of the theorem will be complete if we show that τ extends to T ,

and (τ ∗) extends to NT (Q). Recall that τ ∈ Irr(L) has odd degree and is Y -invariant.


Now write L = S1×· · ·×Se, where the Si are non-abelian simple groups with abelianSylow 2-subgroup. Recall that Q = P ∩ L. Let Qi = Q ∩ Si ∈ Syl2(Si). WriteH = NG(S1). Then L ≤ H and HNG(Q) = G. Hence, there are xi ∈ NG(Q) suchthat Si = (S1)

xi and Qi = Qxi1 . We also have that NL(Q) = NS1(Q1)×· · ·×NSe(Qe).

If c ∈ Y , then [c,NL(Q)] = 1, and therefore Y normalizes Si for all i.Now, write τ = τ1 · · · τe, where τi ∈ Irr(L) and Sj ≤ ker(τi) for i 6= j. We have that

τi have odd degree, so they lie in the principal block of L. Also, since τ is Y -invariantand Y normalizes Si, we have that τi is Y -invariant. By Theorem 3.2, τi has a uniqueextension τi in the principal 2-block of M , which therefore is uniquely determinedby τi. Hence (τi)

g = τj if and only if (τi)g = τj, using that the principal block of

M is G-invariant. By Lemma 3.5, we have that τ =∏

i τi. By Lemma 4.3, we needto show that τi extends to its stabilizer Ti in NG(Si) to conclude that τ extends toT . Let W the product of all Sj except Si, and let C = CG(Si). Hence C ∩ L = W .Again, we display the disposition of relevant subgroups in Figure 2.

NG(Si)

vvvv

Tizzz

z

V||| SSSSSSSSSS

MPPPPPPPP LC

vvvv

SSSSSSSSSS

U

SSSSSSSSSS CG(Si)|||

L

RRRRRRRRRRR CM(Si)vvv

v

W

Figure 2.

Also, because Out(Si) is cyclic, we have that NG(Si)/LC is cyclic. We have thatLC/W = L/W × C/W . Let U = M ∩ LC. Notice that all subgroups M,L,C arenormal in Hi = NG(Si), so in particular L ≤ U / M . Now, since LC/L ∼= L/W thereis a unique extension τi ∈ Irr(LC/C) of τi. By uniqueness, Ti is also the stabilizerof τi in Hi. It is trivial to check that τi lies in the principal block of LC. By thesame reason, (τi)U lies in the principal block of U . By Theorem 3.2, we have that(τi)U = (τi)U . Since τi and τi determine one another, we also see that Ti is thestabilizer of τi in Hi. Finally, by Corollary (4.2) of [I2], we have that restrictiondefines a bijection Irr(V |τi) → Irr(LC|(τi)U), where V = MC. Hence, there is aunique γ ∈ Irr(V ) over τi such that γLC = τ1. Now, γ extends τi, and therefore itextends τi. Now, Ti/V is cyclic and therefore γ (and hence τi) extends to Ti.


Finally, to show that (τ ∗) extends to its stabilizer, we apply the same argumentas before, but replacing G by G = NG(Q), Si by Si = NSi

(Qi), L by L = NL(Q).

Notice that by Lemma 4.4 we know that NG(Si)/SiCG(Si) is cyclic, and by Lemma

2.2 we know that Z(Si) = 1. This concludes the proof of the theorem.

6. Quasisimple Groups

This entire section is devoted to proving Theorem 5.1. For any finite group X, letB0(X) denote the principal 2-block of X, and we write χ ∈ B0(X) to indicate thatan irreducible, ordinary or Brauer, character χ belongs to B0(X).

Lemma 6.1. Theorem 5.1 holds if S is an alternating group, a sporadic simplegroup, the Tits simple group 2F4(2)′, or one of the following simple groups: SU3(3),PSU4(3), Ω7(3), G2(3).

Proof. Let S = An with n ≥ 8. Since |Z| is odd, we must have Z = 1 and G = S. ByProposition 2.4 and Lemma 2.1, G has only one 2-block of maximal defect, whenceB = B0(G). Note that the irreducible characters α, β of H = Sn, labeled by thepartitions (n−2, 2) and (n−2, 12) both belong to B0(H). Since n ≥ 6, both α and βrestrict irreducibly to G, and their restrictions to G belong to B0(G) by [N1, Theorem9.2]. But β(1) − α(1) = 1, so exactly one of them has even degree, a contradiction.All the remaining groups can be checked directly using [GAP].

Lemma 6.2. Let 2|q and let L = SLεn(q) with n ≥ 3. Then for any irreduciblecharacter λ of Z = Z(L), there is χ ∈ Irr(L|λ) with q|χ(1) and χ not of 2-defect 0.

Proof. It is well known that the Steinberg character of L is the only character of2-defect zero. If λ = 1Z then we can take χ to be the unipotent (Weil) characterof L of degree (qn − εn−1q)/(q − ε). Assume λ 6= 1Z and consider the dual groupH = PGLεn(q). Then the quotient H/[H,H] is cyclic and isomorphic to Z. Givenλ 6= 1Z , one can find a semisimple element s ∈ H represented by the diagonal matrixdiag(1, 1, . . . , 1, α) in GLεn(q) (for a suitable 1 6= α ∈ F×q2 with αq−ε = 1) such that

all the irreducible characters in the Lusztig series E(L, s) lie above λ. Note that theSteinberg character ψ of CH(s) ∼= GLεn−1(q) has degree q(n−1)(n−2)/2 divisible by qsince n ≥ 3. It follows that the character χ ∈ E(L, s) parametrized by ψ (in Lusztig’sclassification of irreducible characters of L) has degree

|H : CH(s)|2′ · ψ(1) =qn − εn

q − ε· q(n−1)(n−2)/2

and lies above λ.Alternatively, one can use the dual pairs technique of [LBST] to show that L has

an irreducible character of degree (qn− εn)(qn− εn−1q)/(q− ε)(q2− 1) lying above λ.We omit the details.


Lemma 6.3. Let L = Eε6(q)sc with even q. Then for any irreducible character λ of

Z = Z(L), there is χ ∈ Irr(L|λ) with 2|χ(1) and χ not of 2-defect zero.

Proof. For brevity, in this proof we denote by Φm the value of the mth cyclotomicpolynomial at q. If λ = 1Z , we can choose χ to be the unique unipotent character ofdegree qΦ8Φ9 for ε = +, and qΦ8Φ18 for ε = − (cf. [Lu]).

Now we may assume that q ≡ ε(mod 3) and λ 6= 1Z . We thank F. Lubeck forproviding us with the following explicit argument in this case. Note that S = L/Zhas index 3 in H = Eε

6(q)ad, the corresponding finite group of adjoint type. We willconsider the collection X of all irreducible characters of L of certain even degree D(not divisible by q36, so no α ∈ X has defect zero). Arguing by contradiction, assumethat every α ∈ X is trivial on Z. Then α may be viewed as a character of S, andinducing it up to H we get either an irreducible character of degree 3D, or threedistinct irreducible characters of degree D. Thus Irr(H) either contains a characterof degree 3D, or at least 3|X | characters of degree D. Now we choose

D = qΦ41Φ

22Φ

24Φ5Φ8Φ9Φ12

if ε = + (so q ≡ 4(mod 6)), and

D = qΦ23Φ

26Φ10Φ12Φ18

if ε = − (whence q ≡ 2(mod 6). All the irreducible character degrees of L and Hhave been computed by Lubeck (and posted online on his website, seehttp://www.math.rwth-aachen.de/∼Frank.Luebeck/chev/DegMult/index.html). One canreadily check that |X | = (q2 + q − 2)/3 if ε = + and |X | = q if ε = −. However,Irr(H) contains no characters of degree 3D, and |X |, resp. |X | − 2, characters ofdegree D, a contradiction.

Proposition 6.4. Theorem 5.1 holds if S is a simple finite group of Lie-type incharacteristic 2.

Proof. We will assume S 6∼= SL2(2n), Sp4(2)′, G2(2)′, 2F4(2)′, noting that the cases

Sp4(2)′ ∼= A6, G2(2)′ ∼= SU3(3), and 2F4(2)′ are covered by Lemma 6.1. In each ofthe remaining cases, we will show that B contains an irreducible complex characterof even degree.

(i) First we consider the case Z = 1 and B = B0(G). We can view G as [H,H],where H is a Lie-type group of adjoint type in characteristic 2. Since Z(H) = 1, Hhas exactly one 2-block of maximal defect which is B0(H), and all the other 2-blocksof H have defect 0, cf. [Hu]. Then using [C, §§13.8, 13.9] for instance we can find a(unipotent) character χ ∈ Irr(H), of even degree and not of defect 0. It follows thatχ ∈ B0(H). Under our assumptions |G : H| is odd. Hence any irreducible constituentof χG also has even degree, and it belongs to B0(G) by [N1, Theorem 9.2].

(ii) Now we consider the general case. View Q = PZ/Z as a Sylow 2-subgroup ofS = G/Z. Since |Z| is odd, CG(P )/Z = CS(Q). Next, CS(Q) = Z(Q) by Proposition


2.4 (and by our assumption on S). It follows that CG(P ) = Z(P ) × Z, and so thenumber of 2-blocks of maximal defect of G is exactly |Z| by Lemma 2.1. Assumefurthermore that S 6∼= PSLεn(q) with 2|q, or Eε

6(q) with q = 2f ≡ ε(mod 3). Thenthe Schur multiplier of S is a 2-group (cf. for instance [KL, Theorem 5.1.4]). In thiscase, Z = 1, B = B0(G), and so we are done by (i).

In the remaining cases, G is a quotient of L = SLεn(q) or Eε6(q)sc. The above

argument applied to L implies that L has exactly |Z(L)| 2-blocks of maximal defect;in fact, there is one for each irreducible character of Z(L) since |Z(L)| is odd (justlook at the central character at the central elements of L). Aside from these blocks,by [Hu], L has one more 2-block, which has defect 0 and consists of the Steinbergcharacter St of L. It follows that Irr(L) \ St partitions into |Z(L)| 2-blocks ofmaximal defect, and this partition is completely determined by the restriction toZ(L). Now for the given block B of G, we get a character of even degree by Lemmas6.2 and 6.3.

Lemma 6.5. Theorem 5.1 holds if S ∼= PSLεn(q) for some odd q.

Proof. Since |Z| and q are odd, G is a quotient of SLεn(q), except for S = PSL2(9) ∼=A6 and PSU4(3) which are covered by Lemma 6.1. In all the other cases, the state-ment follows from the main result of [BE].

Proposition 6.6. Theorem 5.1 holds if S is a finite classical group in any charac-teristic p > 2.

Proof. By Lemma 6.5 we may assume that S 6∼= PSLεn(pf ) for any ε = ±. ThusS is one of the following groups: PSp2n(q) with n ≥ 2, Ω2n+1(q) with n ≥ 3, orPΩε

2n(q) with n ≥ 4 and ε = ±, where q = pf . In particular, the Schur multiplierof S is a 2-group, except for the case S = Ω7(3) which is again covered by Lemma6.1. Hence we may assume Z = 1 and G = S. By Proposition 2.4 and Lemma2.1, G has only one 2-block of maximal defect, and so B = B0(G). Now we viewG = L/Z(L), where L = Sp2n(q), Ω2n+1(q), or Ωε

2n(q), respectively. Recall that L actsas a rank 3 permutation group on the set of singular 1-spaces in the natural modulefor L, with Z(L) acting trivially, see e.g. [ST]. Hence we can view the correspondingpermutation character ρ as a G-character, and ρ = 1G + ϕ + ψ, with ϕ, ψ ∈ Irr(G)and ϕ(1)p ≤ ψ(1)p, again cf. [ST]. Now observe that |ϕ(1)− ψ(1)| = qn in the casesL ∈ Sp2n(q),Ω2n+1(q), and |qϕ(1)−ψ(1)| = qn in all the other cases. In particular,exactly one of ϕ, ψ has even degree. This yields a contradiction, since both ϕ andψ belong to B0(G) (cf. for instance the main results of [LST] and [ST], or [E, Prop.1.5]).

To handle the remaining simple groups of Lie type in odd characteristic, we usethe following fact:


Lemma 6.7. Let q be any odd prime power, d = 2 or 4, and L a finite groupcontaining a subgroup H ∼= Spd(q). Then there is a 2-element s ∈ L such that ddivides |L : CL(s)|.

Proof. Assume for instance that d = 4. It suffices to show that 4 divides |NL(〈s〉) :CL(s)|. Note that H > K ∼= Sp2(q

2) : 〈σ〉 where σ is the field automorphism ofS = Sp2(q

2) raising any matrix entry of elements in S to its qth power. Chooseθ ∈ F×q2 of order (q2 − 1)2 and consider s = diag(θ, θ−1) ∈ S. Then s and s−1 are

conjugate in S and moreover σsσ−1 = sq. It follows that the four distinct generatorss±1, s±q of 〈s〉 are conjugate in K, and so we are done. The case d = 2 is similar.

Theorem 6.8. Let G be a perfect quotient of L = Eε6(q)sc with ε = ± and q = pf

odd. Let B be a non-principal 2-block of G with defect group P ∈ Syl2(G), andassume that χ(1) is odd for all χ ∈ Irr(B). Then B is Morita equivalent to a 2-blockwith defect group P of a proper subgroup of G.

Proof. (i) Denote a = (q − ε)2′ and e = (3, q − ε). By [N1, Theorem (9.9)], wemay identify B with a block of L, and so without loss we may assume G = L. ByProposition 2.4(iii) and Lemma 2.1, G has exactly a 2-blocks of maximal defect. Nowwe need a fundamental result of Broue and Michel [BM]. Let G be the underlyingalgebraic group for G so that G = GF for a Frobenius map F on G. Let the pair(H, F ) be dual to (G, F ), whence H = HF ∼= Eε

6(q)ad. To each H-conjugacy class[x] of semisimple elements in H one associates a rational series E(G, x) of irreduciblecharacters of G, cf. [DM, p. 136]. Next, for each semisimple 2′-element s ∈ H,consider

E2(G, s) =⋃

t∈CH(s), t 2−element

E(G, st).

Then the result of Broue and Michel states that each such E2(G, s) is a union of2-blocks, and moreover, one can find s such that B is one of these 2-blocks.

(ii) On the other hand, as shown in [Lu], H has (q − ε) − 1 conjugacy classes ofsemisimple elements y such that |H : CH(y)|p′ = D = (q8 + q4 + 1)(q9 − ε)/(q − ε),which is odd. In fact, each such y has connected centralizer (of type D5+T1, where T1

denotes a one-dimensional torus) in H, and CH(y) contains a central subgroup T oforder q− ε. Take any 1 6= t ∈ T . Then t is semisimple and CH(t) ≥ CH(y). Lusztig’sclassification of irreducible characters of G implies that there is ρ ∈ E(G, t) with1 < ρ(1) = |H : CH(t)|p′ ≤ D. But the degree of any nontrivial irreducible characterof G is either divisible by p or at least D by [Lu]. It follows that |H : CH(t)|p′ = D,and t belongs to one of the (q − ε)− 1 aforementioned conjugacy classes in H.

Let R = O2′(T ), which has order a, and Z = Z(G). Then S = G/Z can beembedded as a normal subgroup of index e in H, and then Q = PZ/Z can beembedded as a Sylow 2-subgroup of both H and CH(y). In particular, we see thatCH(Q) ≥ Z(Q) × R. Recall that CS(Q) = Z(Q) × Ca/e and NS(Q) = Q × Ca/e,


cf. [KM]. By the Frattini argument, NS(Q) has index e in NH(Q), so |NH(Q)| =a|Q|. But R < CH(Q), whence NH(Q) = Q × R. Also, since |CH(Q) : CS(Q)|divides |H/S| = e, we must have CH(Q) = Z(Q) × R. By Burnside’s theorem onNH(Q) controlling fusion of elements in CH(Q), the a elements of R are pairwisenon-conjugate in H.

(iii) We have shown that the a elements of R are semisimple 2′-elements, eachcentralizing a Sylow 2-subgroup of H, and moreover they are pairwise non-conjugatein H. Also, if 1 6= t ∈ R then CH(t) has type D5 + T1 and semisimple rank 5; inparticular, t is non-isolated (i.e. CH(t) is contained in an F -stable Levi subgroupof H). Claim that the element s such that Irr(B) ⊆ E2(G, s) must be conjugate tosome nontrivial t ∈ R. Indeed, the a sets E2(G, t) for t ∈ R are disjoint. Each ofthem contains the semisimple character χt corresponding to t, which has odd degree|H : CH(t)|p′ = D. Clearly, χt must belong to some 2-block Bt of maximal defect.Since G has exactly a 2-blocks of maximal defect, we see that B = Bt for some t ∈ R.Now, B1 contains χ1 = 1G and so B1 = B0(G). Since B 6= B0(G), we conclude thatt 6= 1 as stated.

(iv) As the main result of (ii) and (iii), we can say that Irr(B) ⊆ E2(G, s) with snon-isolated: CH(s) ≤ L∗ for some F -stable Levi subgroup L∗ of H. Now, anotherfundamental result, this time of Bonnafe and Rouquier [BR, Theorem 11.8], impliesthat the union of 2-blocks (over a “large enough” discrete valuation ring Λ) in E2(G, s)is Morita equivalent to the union of 2-blocks (over Λ) of C = LF in E2(C, s). (Here, Lis an F -stable Levi subgroup in G in duality with L∗ in H.) Since Morita equivalencepreserves the decomposition matrix and the defect of the block, we see that B isMorita equivalent to a 2-block b of C with the same defect. But B has maximaldefect and C is a subgroup of G, hence we may assume that b has defect group P .Finally, C < G (as it has a composition factor of type Dε

5(q)).

Finally, we prove Theorem 5.1, thus completing the proof of Theorem A.

Theorem 6.9. Let G be a finite perfect group, with G/Z simple, where Z = Z(G)is cyclic of odd order. Let B be a 2-block of G with defect group P ∈ Syl2(G), andassume that χ(1) is odd for all χ ∈ Irr(B). Then one of the following holds.

(a) P is abelian.(b) G ∼= Eε

6(q)sc with ε = ± and q odd, and B is Morita equivalent to a 2-blockwith defect group P of a proper subgroup of G.

Proof. By virtue of Lemma 6.1, and Propositions 6.4, 6.6, we may assume that S isone of the following simple groups: G2(q),

3D4(q), F4(q), E6(q),2E6(q), E7(q), E8(q),

or 2G2(q), where q = pf and p > 2 is a prime. Notice that Sylow 2-subgroups of2G2(q) are abelian. We will either obtain a contradiction or arrive at the conclusion(b) in all the other cases.


(i) First we consider the case G = S and B = B0(G). View S as the derived group[H,H], where H = GF for a simple algebraic group of adjoint type G and a Frobeniusmap F on G. Let the pair (G∗, F ∗) be dual to (G, F ). We claim that it suffices tofind a 2-element s ∈ L = (G∗)F ∗ such that |L : CL(s)|2 > |G : S|2. Indeed, since Gis of adjoint type, s has connected centralizer in G∗, cf. [DM, Remark 13.15]. By theDeligne-Lusztig theory (see e.g. [DM, Chapter 14]), the conjugacy class [s] in L of sdefines an irreducible (semisimple) character χ of H, of degree |L : CL(s)|p′ . Basedon a result of Hiss [H], it is shown in [BNOT, Lemma 3.1] that every irreducibleconstituent α of χS belongs to B0(S). Now the condition |L : CL(s)|2 > |G : S|2ensures that α(1) is even, yielding the desired contradiction.

In the first two cases listed above, we have that |G : S|2 = 1 and L ≥ K ∼= Sp2(q).In the next five cases listed above, we have that |G : S|2 ≤ 2 and L ≥ K ∼=Spin5(q) ∼= Sp4(q). Now the existence of s follows from Lemma 6.7. (In fact, thesame argument can be used to handle principal blocks of simple classical groupsPSp2n(q) and Ω2n+1(q).)

(ii) Now we consider the general case. By Lemma 6.1 we may assume that S 6∼=G2(3). Assume in addition that S 6∼= Eε

6(q) with ε = ±. Under these assumptions,the Schur multiplier of S is a 2-group, cf. for instance [KL, Theorem 5.1.4]. It followsthat Z = 1 and G = S. By Proposition 2.4, CG(P ) = Z(P ). It follows by Lemma2.1 that B = B0(G), and so we are done by (i).

Finally, consider the case S = Eε6(q) with ε = ±. Assume B = B0(G). Since |Z| is

odd, every complex character in B0(S) can be viewed as a character of G belongingto B0(G), and so we are again done by (i). If B 6= B0(G), we can apply Theorem6.8.

7. Future direction: The p odd case

In this final section, we comment on what has to be shown in order to proveTheorem A for odd primes.

To make our long proof shorter, we have used in several instances in Theorem 5.3special facts on finite groups with abelian Sylow 2-subgroups and their blocks. Allthese steps can easily be carried out with some extra work in the p odd case.

What we would like to mention now are the following three deep results that needto be proved to establish Brauer’s Height Zero Conjecture for p-blocks of maximaldefect when p is odd (following our approach).

(A) Obtain the Gluck-Wolf Theorem [GW1] for arbitrary finite groups for theprime p.

That is to prove that: If N is normal in G, p is a prime, θ ∈ Irr(N) is G-invariant,and p does not divide χ(1)/θ(1) for all χ ∈ Irr(G|θ), then G/N has abelian Sylowp-subgroups.


As we have mentioned in §5, this is the first major obstacle towards provingBrauer’s Height Zero Conjecture for p-blocks of maximal defect. There is an on-going project on this by the authors and their collaborators, which we expect to becompleted soon.

(B) Prove Brauer’s Height Zero Conjecture for quasisimple groups for the prime p.

Or at least, as we have done in this paper (see for instance Theorem 5.1), prove thatif B is a block of a quasisimple group, then either the conjecture is true for B or B isMorita equivalent to a block B∗ of another group H with |H/Op′(H)| < |G/Op′(G)|.

(C) Define what is a good simple group for the blocks of maximal defect (inthe Isaacs-Malle-Navarro sense), and prove that simple groups with abelian Sylowp-subgroups are good.

In [IMN], it was defined what a good simple group is for the McKay conjecture.If now we deal with blocks, then it has to be defined what a good simple group isfor the Alperin-McKay conjecture, or at least, for the blocks of maximal defect. Asa consequence of (C), in order to prove Brauer’s Height Zero Conjecture for oddprimes and blocks of maximal defect (following our techniques), a strong form ofthe Alperin-McKay conjecture has to be established for blocks with abelian maximaldefect group. Hence (if our approach is followed), we have that the Alperin-McKayconjecture for abelian defect groups needs to be proved before a proof of the Brauer’sHeight Zero conjecture can be achieved.

Let us finally mention that a relationship between the Alperin-McKay Conjectureand the “only if” part of the Height Zero Conjecture for principal blocks was observedin [Mu]: if the Alperin-McKay conjecture holds for the principal blocks of all finitegroups and if the “only if” part of the Height Zero Conjecture is true for the principalblocks of all finite simple groups, then this part is in fact true for the principal blocksof arbitrary finite groups.

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Departament d’Algebra, Universitat de Valencia, 46100 Burjassot, Valencia, SpainE-mail address: [email protected]

Department of Mathematics, University of Arizona, Tucson, AZ 85721, USAE-mail address: [email protected]

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