Steinberg-like characters for finite simple groups
aa r X i v : . [ m a t h . R T ] J a n STEINBERG-LIKE CHARACTERS FOR FINITE SIMPLE GROUPS
GUNTER MALLE AND ALEXANDRE ZALESSKI
Abstract.
Let G be a finite group and, for a prime p , let S be a Sylow p -subgroup of G . Acharacter χ of G is called Syl p -regular if the restriction of χ to S is the character of the regularrepresentation of S . If, in addition, χ vanishes at all elements of order divisible by p , χ is saidto be Steinberg-like. For every finite simple group G we determine all primes p for which G admits a Steinberg-like character, except for alternating groups in characteristic 2. Moreover,we determine all primes for which G has a projective F G -module of dimension | S | , where F isan algebraically closed field of characteristic p . Introduction
Let G be a finite group and, for a prime p , let S be a Sylow p -subgroup of G . A character χ of G is called Syl p - vanishing if χ ( u ) = 0 for every 1 = u ∈ S ; and if, additionally, χ (1) = | S | then we say that χ is Syl p - regular . If χ ( g ) = 0 whenever | g | is divisible by p then χ is called p - vanishing ; and if, additionally, χ (1) = | S | then we say that χ is Steinberg-like . Steinberg-likeand Syl p -regular characters for Chevalley groups in defining characteristic p are studied in [17].Specifically, for all simple groups of Lie type in characteristic p except B n ( q ), n = 3 , ,
5, and D n ( q ), n = 4 ,
5, the Steinberg-like characters for the prime p have been determined in [17].Our main motivation to study this kind of characters is their connection with charactersof projective indecomposable modules. The study of projective indecomposable modules ofdimension | S | was initiated by Malle and Weigel [13]; they obtained a full classification of suchmodules for arbitrary finite simple groups G assuming that the character of the module has thetrivial character 1 G as a constituent. In [22], this restriction was removed for simple groups ofLie type with defining characteristic p . Some parts of the proofs there were valid not only forcharacters of projective modules, but also for Steinberg-like or even Syl p -regular characters.In this paper we complete the classification of projective indecomposable modules of dimension | S | for simple groups G . The first main result is a classification of Steinberg-like characters forsimple groups, with the sole exception of alternating groups for the prime p = 2: Theorem 1.1.
Let G be a finite non-abelian simple group, p a prime dividing | G | and let χ bea Steinberg-like character of G with respect to p . Then one of the following holds: (1) χ is irreducible, and the triple ( G, p, χ (1)) is as in Proposition . ; (2) Sylow p -subgroups of G are cyclic and ( G, p, χ (1)) is as in Proposition . ; (3) G is of Lie type in characteristic p (see [17] ); (4) p = 2 and G = PSL ( q ) with q + 1 = 2 k ; or Date : October 4, 2018.2010
Mathematics Subject Classification.
Key words and phrases. characters of projective indecomposable modules, Steinberg-like characters.The first author gratefully acknowledges financial support by SFB TRR 195. (5) p = 2 and G = A n , n ≥ . In fact, in many instances we even classify all Syl p -regular characters. Examples for case (5)when n = 2 k or 2 k + 1 are presented in Corollaries 6.8 and 6.10. We are not aware of any furtherexamples.Our second main result determines reducible projective modules of simple groups of minimalpossible dimension | G | p . Theorem 1.2.
Let G be a finite non-abelian simple group, p a prime dividing | G | and S a Sylow p -subgroup of G . Then G has a reducible projective F p G -module of dimension | S | if and only ifone of the following holds: (1) G = PSL ( q ) , q > , | S | = q + 1 ; (2) G = PSL n ( q ) , n is an odd prime, n ( q − , | S | = ( q n − / ( q − ; (3) G = A p , | S | = p ≥ ; (4) G = M , | S | = 11 ; or (5) G = M , | S | = 23 . Note that irreducible projective F p G -modules of dimension | G | p are in bijection with irre-ducible characters of defect 0 of that degree, listed in Proposition 3.1 for simple groups.The paper is built up as follows. After some preliminaries we recall the classification ofirreducible Steinberg-like characters in Section 3 (Proposition 3.1). In Section 4 we classifySyl p -regular characters in the case of cyclic Sylow p -subgroups (Proposition 4.4), in Section 5we treat the sporadic groups (Theorem 5.1). The alternating groups are handled in Section 6(Theorem 6.4 for p odd, and in Section 6.2 some partial results for p = 2, see Theorems 6.12and 6.14). The exceptional groups of Lie type are considered in Section 7 (Theorem 7.1). Therest of our paper deals with the classical groups of Lie type. We start off in Section 8 by rulingout the remaining possibilities in defining characteristic from [17]. The case of large Sylow p -subgroups for non-defining primes p is settled in Section 9. In Section 10 we discuss the smallcases when p >
2, while the proof of our main theorems is achieved in Section 11 by treatingthe case when p = 2. 2. Preliminaries
We start off by fixing some notation. Let F q be the finite field of q elements and F q analgebraic closure of F q . The cardinality of a set X is denoted by | X | . The greatest commondivisor of integers m, n is denoted by ( m, n ); if p is a prime then | n | p is the p -part of n , that is, n = | n | p m , where ( m, p ) = 1. If ( m, n ) = m , we write m | n .For a finite group G , Irr( G ) is the set of its irreducible characters and Irr ( G ) is the set of alllinear characters of G (that is, of degree 1). We denote by 1 G the trivial character and by ρ reg G the regular character of G . We write S ∈ Syl p ( G ) to mean that S is a Sylow p -subgroup of G .A group of order coprime to p is called a p ′ -group. Further, Z ( G ), G ′ denote the center and thederived subgroup of G , respectively.If H is a subgroup of G then C G ( H ), N G ( H ) denote the centraliser and normaliser of H in G , respectively. If χ is a character of G then we write χ | H for the restriction of χ to H . The H - level of χ is the maximal integer l ≥ χ | H − l · ρ reg H is a proper character of H . If aprime p is fixed then the p - level l p ( χ ) of χ is the S -level of χ for S ∈ Syl p ( G ). (For quasi-simplegroups with cyclic Sylow p -subgroups irreducible characters of p -level l = 1 , TEINBERG-LIKE CHARACTERS FOR FINITE SIMPLE GROUPS 3 [21, 18], respectively.) The inner product of characters λ, µ of G is denoted by ( λ, µ ), sometimesby ( λ, µ ) G . The character of G induced from a character µ of H is denoted by µ H .Let P ≤ G be finite groups, N a normal subgroup of P and L = P/N . Let F be a field and M an F G -module. Then M N := C M ( N ) becomes an F L -module, which is called generalisedrestriction of M to L and denoted by r GP/N M in [2, § β is the Brauer (or ordinary)character of M then we also write r GP/N β for the Brauer (or ordinary) character of L affordedby M N .Let e = e p ( q ) ( p >
2, ( p, q ) = 1) be the minimal integer i > q i − p . If p = 2 and q is odd then we set e ( q ) = 1 if 4 | ( q −
1) and e ( q ) = 2 if 4 | ( q + 1).The next two lemmas follow from the definitions; here G is a finite group and S ∈ Syl p ( G ). Lemma 2.1.
Let χ be a Syl p -regular character of G . Then every linear character occurs in χ | S with multiplicity . In particular, ( χ | S , S ) = 1 . If S is abelian then χ | S is multiplicity free.Proof. As χ | S = ρ reg S , this follows from the corresponding properties of ρ reg S . (cid:3) Lemma 2.2.
Let G = G × G be a direct product, and let χ , χ be irreducible characters of G , G respectively. Then the p -level of χ ⊗ χ is the product of the p -levels of χ and χ . Lemma 2.3.
Let N be a p ′ -subgroup of G normalised by S . Let χ be a faithful Steinberg-likecharacter of G . Then N is abelian and C G ( S ) = Z ( G ) Z ( S ) .Proof. Let H = N S . Then χ | H is Steinberg-like. As H is p -solvable, every p -vanishing characteris the character of a projective module [15, Lemma 10.16]. As χ (1) = | S | , the module inquestion is indecomposable. Then χ | H is induced from an irreducible character α , say, of N [15,Thm. 10.13]. As α H (1) = α (1) · | H : N | = α (1) · | S | and χ (1) = | S | , it follows that α (1) = 1.Let N ′ be the derived subgroup of N . Then N ′ is normal in H and α ( N ′ ) = 1. Therefore, α H | N ′ = | S | · N ′ , that is, N ′ lies in the kernel of α H . Since χ and hence χ | H = α H is faithful,we have N ′ = 1. So N is abelian as claimed.Note that C G ( S ) = A × Z ( S ), where A is a p ′ -group. Take N = A above, so H = A × S .So now [ N, S ] = 1 and N is abelian. It follows that in any representation afforded by α H , N consists of scalar matrices. As χ is faithful, we have [ N, G ] = 1, as required. (cid:3)
Thus, if G is a simple group then C G ( S ) = Z ( S ) is a necessary condition for G to have aSteinberg-like character. Remark 2.4. A p ′ -subgroup N normalised by a Sylow p -subgroup of G is called a p -signaliser in the theory of finite groups. Thus, Lemma 2.3 tells us that if G admits a faithful Steinberg-likecharacter then every p -signaliser is abelian, and C G ( S ) = Z ( G ) Z ( S ). Lemma 2.5.
Let G be a finite group, P a subgroup with ( | G : P | , p ) = 1 , U a normal p -subgroupof P and let L = P/U . Let
T, S be Sylow p -subgroups of L, G , respectively. Let χ be a characterof G and λ = r GP/U ( χ ) . (a) If χ | S = m · ρ reg S then λ | T = m · ρ reg T . In other words, l p ( χ ) = l p ( λ ) . In particular, if χ is Syl p -regular then so is λ . (b) If χ is a p -vanishing character of G then λ is a p -vanishing character of L . (c) Let K := O p ′ ( L ) . If χ is a Steinberg-like or Syl p -regular character of G then so is thecharacter λ | K of K . GUNTER MALLE AND ALEXANDRE ZALESSKI
Proof.
We can assume that S ≤ P and T = S/U .(a) As χ | S = m · ρ reg S , it follows that λ | T coincides with m · ρ reg T , whence the claim.(b) We have to show that λ vanishes at all p -singular elements of L . Let M be a C G -moduleafforded by χ . Then C M ( U ) = { | U | P u ∈ U ux | x ∈ M } . Observe that if g ∈ P has projectionto L which is not a p ′ -element, then gu is not a p ′ -element for any u ∈ U . Thus, for any suchelement g , it follows that λ ( g ) = | U | P u ∈ U χ ( gu ) = 0 by assumption, whence the claim.(c) Obvious. (cid:3) Lemma 2.6.
Let G = G × G be a direct product. Suppose that l p ( σ ) ≥ k for every non-zero Syl p -vanishing (resp., p -vanishing) character σ of G . Then l p ( χ ) ≥ k for every Syl p -vanishing(resp., p -vanishing) character χ of G .Proof. Let S ∈ Syl p ( G ). Set U = S and P = N G ( U ), so P = N G ( S ) × G . Then L := P/U = L × G , where L = N G ( S ) /S . Let χ be a Syl p -vanishing (resp., p -vanishing)character of G . Let λ = r GP/U ( χ ) be the generalised restriction of χ to L . By Lemma 2.5, λ is aSyl p -vanishing (resp., p -vanishing) character of L and l p ( χ ) = l p ( λ ). Then l p ( λ ) = l p ( λ | G ), as L is a p ′ -group. By assumption, l p ( λ | G ) ≥ k , whence the result. (cid:3) Lemma 2.7.
Let G = G × G , where | G | p > and let χ be a p -vanishing character of G .Then χ = P i η i σ i , where η i ∈ Irr( G ) are all distinct, and σ i are p -vanishing characters of G .In addition, χ := P i l p ( σ i ) η i is a p -vanishing character of G , and l p ( χ ) = l p ( χ ) .Proof. Write χ = P i η i σ i , where η i ∈ Irr( G ) are all distinct, and the σ i ’s are some charactersof G (reducible, in general). Let g ∈ G , and let x ∈ G be p -singular. Then 0 = χ ( gx ) = P i η i ( g ) σ i ( x ). As the characters η i are linearly independent, it follows that σ i ( x ) = 0 for every i , that is, the σ i ’s are p -vanishing.In addition, | G | p P l p ( σ i ) η i = P i η i σ i (1) = χ | G . So P i l p ( σ i ) η i is p -vanishing. Let l p ( χ ) = m ; then χ (1) = m | G | p = m | G | p | G | p = P η i (1) σ i (1) = P η i (1) l p ( σ i ) | G | , whence m | G | p = P η i (1) l p ( σ i ), as required. (cid:3) Corollary 2.8.
Let G = G × G and χ be as in Lemma . , and S i a Sylow p -subgroup of G i , i = 1 , . Let η , . . . , η k be the irreducible constituents of χ | G , and η = η + · · · + η k . Supposethat l p ( σ ) ≥ m for every non-zero p -vanishing character σ of G . Then l p ( χ ) ≥ m · η (1) / | S | .Proof. Let χ = P η i σ i be as in Lemma 2.7. By assumption, σ i | S = m i · ρ reg S , where m i ≥ m .So m · ρ reg S is a subcharacter of σ i | S . Therefore, P i ( η i | S · m · ρ reg S ) = ( P i η i ) | S · m · ρ reg S is asubcharacter of χ | S × S . Now χ (1) ≥ m η (1) | S | = m η (1) | G | p / | S | . As χ (1) is a multiple of | G | p , we have χ (1) = l p ( χ ) | G | p , and the result follows. (cid:3) Proposition 2.9.
Let G be a finite group and N ⊳ G a normal subgroup such that G/N is acyclic p -group. Let χ be a p -vanishing character of G . Then: (a) χ = ψ G for some character ψ of N; (b) if h ∈ N is p -singular and the conjugacy classes of h in G and in N coincide then ψ ( h ) = 0 ; (c) if ψ is G-invariant then ψ is p -vanishing.Proof. (a) Let λ ∈ Irr( G ) be a linear character that generates Irr( G/N ). As all elements of G \ N are p -singular, χ vanishes on G \ N . It follows that λ · χ = χ . Thus, if we write χ = P j a j χ j as TEINBERG-LIKE CHARACTERS FOR FINITE SIMPLE GROUPS 5 a non-negative linear combination of irreducible characters χ j ∈ Irr( G ), then a j is constant onorbits under multiplication with λ . It clearly suffices to show the claim for a single orbit, say χ = P p f i =1 λ i χ ′ with χ ′ ∈ Irr( G ) and f minimal such that λ p f χ ′ = χ ′ .Set M := ker( λ p f ). Then χ ′ | M is irreducible as so is χ ′ , so χ = ( χ ′ | M ) G . Now note that λ p f generates Irr( G/M ), so λ p f ( m ) = 1 for m / ∈ ( M \ N ). Thus, as λ p f χ ′ = χ ′ , it followsthat χ ′ vanishes on M \ N , and hence χ ′ | M = ψ M is induced from some ψ ∈ Irr( N ). Then χ = ( χ ′ | M ) G = ( ψ M ) G = ψ G as claimed.(b) For g ∈ G define the character ψ g of N by ψ g ( x ) = ψ ( gxg − ) ( x ∈ N ). It is well knowthat ψ G | N is a sum of p k characters ψ g for suitable g ∈ G . By assumption, ψ g ( h ) = ψ ( h ), andhence 0 = χ ( h ) = p k ψ ( h ), whence (b).(c) If ψ is G -invariant then ψ g = ψ , and hence χ | N = p k · ψ . It follows that ψ is p -vanishingwhence the result. (cid:3) Remark 2.10.
Let
G, N, p, χ, ψ be as in Proposition 2.9. Then ψ is not necessarily p -vanishing.Indeed, let C = h c i be the cyclic group of order 4, and let ε be a square root of −
1. Define µ i ∈ Irr( C ) ( i = 1 , , ,
4) by µ i ( c ) = ε i . Then P i µ i = ρ reg C , the regular character of C . Let D be the dihedral group of order 8 with normal subgroup C . Then ( P i µ i ) D = ρ reg D . One observesthat µ D = µ D , and hence (2 µ + µ + µ ) D = ρ reg D . However, 2 µ + µ + µ is not a 2-vanishingcharacter of C . Corollary 2.11.
Let
G, N be as in Proposition , and let χ be a Steinberg-like character of G . Suppose that every irreducible character of N of degree at most | N | p is G -invariant. Then χ = ψ G for some Steinberg-like character ψ of N . In particular, if N does not have Steinberg-likecharacters then neither has G .Proof. By Proposition 2.9(a), χ = ψ G for some character ψ of N . Clearly, ψ (1) = χ (1) / | G : N | = | G | p / | G : N | = | N | p , so, by assumption, every irreducible constituent of ψ is G -invariant.Therefore, so is ψ , and the claim follows from Proposition 2.9(c). (cid:3) Lemma 2.12.
Let G be a finite group and N ⊳ G a normal subgroup of p -power index. Supposethat l p ( χ ) ≥ m for some integer m > and every p -vanishing character χ of G . Then l p ( χ ) ≥ m for every p -vanishing character χ of N .Proof. Suppose the contrary. Let χ be a p -vanishing character of N such that l p ( χ ) < m . Thenthe induced character χ G is p -vanishing and l p ( χ G ) = l p ( χ ) < m . This is a contradiction. (cid:3) The following fact is well known.
Lemma 2.13.
Let G be a finite group and N ⊳ G a normal subgroup of p -power index. Let F bean algebraically closed field of characteristic p . Let Φ be a projective indecomposable F G -module.Then
Φ = Ψ G , where Ψ is a projective indecomposable F N -module and l p (Ψ) = l p (Φ) .Proof. It is well-known that induction sends projective modules to projective modules. Further-more, by Green’s indecomposability theorem [4, Thm. 3.8] induction from normal subgroups of p -power index preserves indecomposability. So, if Ψ is an indecomposable direct summand ofΦ | N , then Ψ is projective, Ψ G is projective indecomposable and so Ψ G = Φ. The statement l p (Ψ) = l p (Φ) also follows as | G : N | = | G : N | p by assumption. (cid:3) GUNTER MALLE AND ALEXANDRE ZALESSKI Irreducible Steinberg-like characters for simple groups
Here we complete the list of irreducible characters of simple groups G of degree | G | p . Forthis it suffices to extract the characters of degree | G | p from the list of irreducible characters ofprime-power degree obtained in [14, Thm. 1.1]. This list already appeared in [23, Prop. 2.8],where the case with p = 3, G = F (2) ′ was inadvertently omitted.Note that an irreducible character is Steinberg-like if and only if it is Syl p -regular. Proposition 3.1.
Let G be a non-abelian simple group. Suppose that G has an irreducible Syl p -regular character χ . Then one of the following holds: (1) G is a simple group of Lie type in characteristic p and χ is its Steinberg character; (2) G = PSL ( q ) , q even, and p = χ (1) = q ± , or G = SL (8) , p = 3 and χ (1) = 9;(3) G = PSL ( q ) , q odd, χ (1) = ( q ± / is a p -power for p > , or p = 2 and χ (1) = q ± is a -power; (4) G = PSL n ( q ) , q > , n is an odd prime, ( n, q −
1) = 1 , such that χ (1) = ( q n − / ( q − is a p -power; (5) G = PSU n ( q ) , n is an odd prime, ( n, q + 1) = 1 , such that χ (1) = ( q n + 1) / ( q + 1) is a p -power; (6) G = PSp n ( q ) , n > , q = r k with r an odd prime, kn is a -power such that χ (1) =( q n + 1) / is a p -power; (7) G = PSp n (3) , n > is a prime such that χ (1) = (3 n − / is a p -power; (8) G = A p +1 and χ (1) = p ; (9) G = Sp (2) and χ (1) = 7 ; (10) G ∈ { M , M } and χ (1) = 11 ; (11) G ∈ { M , PSL (3) } and χ (1) = 16 ; (12) G ∈ { M , Co , Co } and χ (1) = 23 ; (13) G = F (2) ′ and χ (1) = 27 ; (14) G = PSU (3) ∼ = G (2) ′ and χ (1) = 32 ; or (15) G = G (3) and χ (1) = 64 . The problem of determining the minimal degree of irreducible characters of p -defect 0 looksmuch more complicated. Remark 3.2.
Let us point out the following cases not explicitly mentioned in Proposition 3.1.SL (2) ∼ = PSL (7) , A ∼ = PSL (9), PSU (2) ∼ = PSp (3), A ∼ = SL (2).4. Cyclic Sylow p -subgroups In this section we determine the reducible Steinberg-like characters for simple groups withcyclic Sylow p -subgroups. Proposition 4.1.
Let G be a finite group with a cyclic TI Sylow p -subgroup S , and assume that N G ( S ) /S is abelian. Then l p ( τ ) = ⌊ τ (1) / | S |⌋ for all τ ∈ Irr( G ) .Proof. Let N := N G ( S ). By assumption, N/S is abelian of order prime to p , so it has | N : S | irreducible p -Brauer characters of degree 1. Hence, each of the corresponding PIMs of N hasdimension | S | . Since the Brauer tree for any p -block of N is a star, all PIMs are uniserial [4,Ch. VII, Cor. 2.22]. But then by [4, Ch. I, Thm. 16.14], any indecomposable F N -module, where
TEINBERG-LIKE CHARACTERS FOR FINITE SIMPLE GROUPS 7 F is a sufficiently large field of characteristic p , is a quotient of a PIM, so has dimension strictlysmaller than | S | if it is not projective.Now let τ ∈ Irr( G ). If τ is of p -defect zero, τ | S is a multiple of ρ reg S , and the claim follows.Else, τ lies in a block of full defect, and there exists an indecomposable F G -module X withlift τ [4, Ch. I, Thm. 17.12]. Then X | F N = Y ⊕ P , where P is projective (and hence ofdimension divisible by | S | ) and Y is the Green correspondent of X , an indecomposable, non-projective F N -module [4, Ch. VII, Lem. 1.5]. Thus, dim
Y < | S | by what we said before, so τ (1) / | S | ≤ l p ( τ ) < τ (1) / | S | + 1, and the claim follows. (cid:3) Lemma 4.2.
Let G be a non-abelian simple group. Let p be a prime such that a Sylow p -subgroup of G is cyclic. Let µ denote the minimal degree of any non-linear irreducible characterof G . Then µ > | G | p , except in the case where G = PSL ( p ) , p ≡ and µ = ( p − / . Proof.
The values of µ = µ ( G ) for every simple group G are either known explicitly or thereis a good lower bound. For the sporadic simple groups one can inspect [1], for the alternatinggroups A n we have µ ( A n ) = n − n >
5, and µ ( A ) = 3, for simple groups G of Lie type thevalues µ ( G ) are listed in [19]. The lemma follows by comparison of these data with | G | p . (cid:3) Proposition 4.3.
Let p be a prime and let G be a non-abelian simple group with a cyclic Sylow p -subgroup S . Let χ be a Syl p -regular character of G . Then one of the following holds: (1) χ is irreducible of degree | G | p ; (2) ( χ, G ) = 1 , τ := χ − G is irreducible and ( τ | S , S ) = 0 ; or (3) G = PSL ( p ) , p ≡ and χ = 1 G + τ + τ , where τ , τ are distinct irreduciblecharacters of degree ( p − / .Proof. Suppose that χ is reducible. The result for G = PSL ( p ) easily follows by computationwith the character table of this group. Suppose G = PSL ( p ). Let τ = 1 G be an irreducibleconstituent of χ . By Lemma 4.2, χ = τ + k · G , where k = | G | p − τ (1). Therefore, 1 G is aconstituent of χ . By Lemma 2.1, k = 1 and ( τ | S , S ) = 0. (cid:3) Proposition 4.4.
Let p be a prime and let G be a non-abelian simple group with a cyclic Sylow p -subgroup S . Then G has a reducible Syl p -regular character χ if and only if one of the followingholds: (1) G = PSL ( q ) , q > even, | S | = q + 1 ; (2) G = PSL ( p ) , | S | = p > ; (3) G = PSL n ( q ) , n is an odd prime, n ( q − , | S | = ( q n − / ( q − ; (4) G = PSU n ( q ) , n is an odd prime, n ( q + 1) , | S | = ( q n + 1) / ( q + 1) ; (5) G = A p , | S | = p ≥ ; (6) G = M , | S | = 11 ; or (7) G = M , | S | = 23 .Furthermore, in each case (1) – (7) , C G ( S ) = S and χ is Steinberg-like. In addition, χ − G isan irreducible character of G , unless possibly when (2) holds, when χ − G may be the sum oftwo irreducible constituents of equal degree.Proof. The additional statement follows from Proposition 4.3. If χ − G is reducible, we havethe case (3) of Proposition 4.3. So we may assume that τ = χ − G is irreducible and thus that( τ | S , S ) = 0. The irreducible characters of G of level 0 are determined in [21, Thm. 1.1], so τ belongs to the list in [21, Thm. 1.1]. If we drop from that list the characters of degree other GUNTER MALLE AND ALEXANDRE ZALESSKI than | S | −
1, the remaining cases are given in the statement of the proposition. (Note that thelist in [21, Thm. 1.1] includes quasi-simple groups so one first needs to delete the representationsnon-trivial on the center. For instance, if G = PSp n ( q ) then | S | = ( q n + 1) / τ (1) = χ (1) − | S | − n ( q ) of evendegree ( q n − / G has no irreducible representation of even degree( q n − /
2. In contrast, there do exist irreducible representations of G = PSL n ( q ) and PSU n ( q )for n odd of degree | S | − G + τ is Syl p -regular, that is, χ | S = ρ reg S . Let Ψ be a representations of G afforded by τ . Let s ∈ S with S = h s i . By [21,Cor. 1.3(2)], the multiplicity of every eigenvalue of Ψ( s ) is 1. As det Ψ( s ) = 1, it follows that 1is not an eigenvalue of Ψ( s ). Therefore, χ | S = ρ reg S , as required.Next, we show that C G ( S ) = S . In cases (6) and (7) this follows by inspection in [1]. The cases(1), (2), (5) are trivial. In cases (3), (4) one can take the preimage T , say, of S in G = SL n ( q ),SU n ( q ), respectively. Then T is irreducible on the natural module for G . The groups C G ( T )are described by Huppert [8, S¨atze 4,5]. It easily follows that T is self-centralising in G . Then C G ( S ) = S unless [ g, T ] ⊆ Z ( G ) for some g ∈ N G ( T ) \ T . By order consideration, S is a Sylow p -subgroup of G , so g is not a p -element. Let t ∈ T . Then [ g, t i ] = [ g i , t ] = 1 for i = | S | , so g | S | ∈ C G ( T ) = T by the above. This is a contradiction as S is a Sylow p -subgroup.It follows that every element of G is either a p - or a p ′ -element. Therefore, χ is Steinberg-likeif and only if χ | S = ρ reg S . (cid:3) Lemma 4.5.
Under the assumptions and in the notation of Proposition . we have: (a) χ is unique unless (2) or (6) holds; (b) χ is the character of a projective module when (1) , (3) , (5) or (7) holds; and (c) χ − G is a proper character, and if m is the minimal degree of a non-linear characterof G then either m = χ (1) − , or (1) holds and m = χ (1) − , or (2) holds and m = ( χ (1) ± / .Proof. (a) Let τ = χ − G . Then τ (1) = | S | − τ is irreducible unless (2) holds. We showthat an irreducible character of this degree is unique unless (2) or (6) holds. If G = M , thisfollows from the character table of this group, for A p this is well known. For G = PSL n ( q ), n >
2, and PSU n ( q ), n >
2, this is observed in [19, Table II].In case (2) the number of characters equals the number of irreducible characters of degree p −
1, which is ( p − / p ≡ p − /
4. If G = M then there are threeSteinberg-like characters, see [1].(b) Recall that the principal projective indecomposable module is the only PIM whose char-acter contains 1 G as a constituent. All the characters χ in Proposition 4.4 contain 1 G as aconstituent. Therefore, if χ is the character of a projective module Φ, say, then Φ is indecom-posable and principal. So we compare the list of characters χ in Proposition 4.4 with the mainresult of [13]. The comparison rules out the case (4) of Proposition 4.4. Furthermore, if G admits at least two Steinberg-like characters then at most one of them can be the character ofa projective module. By (a) this leaves us with cases (1), (5) and (7). As in each of these cases χ is unique, it must be the character of the principal projective indecomposable module listedin [13].(c) This follows by inspection in [19, Table II]. (cid:3) TEINBERG-LIKE CHARACTERS FOR FINITE SIMPLE GROUPS 9
Remark 4.6.
The group G = PSL ( p ) has several Syl p -regular characters, all of them areSteinberg-like, and only one of them is projective.5. Sporadic groups
Theorem 5.1.
Let G be a sporadic simple group. Then G does not have a reducible Syl p -regularcharacter unless one of the following holds: (1) G = M , p = 3 , four characters with constituents of degrees and each, allSteinberg-like; (2) G = M , p = 2 , six characters, none of them Steinberg-like; (3) G = M , | S | = 11 ; or (4) G = M , | S | = 23 .Proof. For most groups and primes, by [1] there is a conjugacy class of non-trivial p -elementstaking strictly positive value at all irreducible characters of degree at most | G | p . In a few cases,like in Co and F i at p = 3, or Co and J at p = 2, one has to solve a little linear systemof equations for non-negative integral solutions. The only cases where such solutions exist arelisted in the statement. Note that the cases (3), (4) occur also in Proposition 4.4. (cid:3) Alternating groups
In this section we consider Steinberg-like characters of alternating groups.6.1.
Alternating groups for p > . For odd primes we give a short proof using a recent resultof Giannelli and Law [5] which replaces our earlier more direct proof.
Lemma 6.1.
Let G = A p , p > , and χ ∈ Irr( G ) . Then l p ( χ ) = ⌊ χ (1) /p ⌋ . In addition, l p ( χ ) = 1 for p > (this fact has also been observed in [18] ).Proof. The first part is just Proposition 4.1. In addition, if p > G has no irreduciblecharacter of degree d for p ≤ d < p . This implies the claim. (cid:3) Lemma 6.2.
Let n = kp , where p > and k < p . Let G = A n and let χ be a p -vanishingcharacter. Then l p ( χ ) ≥ k − , equivalently, χ (1) ≥ k − | G | p .Proof. For k = 1 the lemma is trivial. Let k >
1. Let X ∼ = A p , X ∼ = A n − p be commutingsubgroups of G . Set X = X X . Then χ | X = P η i σ i , where the σ i ’s are p -vanishing charactersof X and the η i ’s are distinct irreducible characters of X (Lemma 2.7). By induction, l p ( σ i ) ≥ k − . If l p ( η i ) ≥ i then l p ( χ ) ≥ l p ( η i σ i ) ≥ k − . By Lemma 6.1, if l p ( η i ) < p > l p ( η i ) = 0, and hence either η i = 1 X or η i is the unique irreducible character ofdegree p −
1. (If p = 7 then we may have η i (1) = 10, see [18].)Suppose the lemma is false and p >
7. Then we can rearrange the above to get χ | X = 1 X · σ + η · σ , where η (1) = p − σ , σ are p -vanishing characters of X . It follows that χ | X , as well as τ | X for every irreducible constituent τ of χ , contains no irreducible constituent distinct from1 X , η . It is well known and easily follows from the branching rule that this implies τ (1) = n − G has a single character of degree n −
1. Therefore, χ = a · G + b · τ , where τ (1) = n −
1. Let x ∈ X be of order p . Then τ ( x ) = n − p − >
0, which implies χ ( x ) > Suppose p = 7. Then η i (1) ∈ { , , } . There are two irreducible characters of X ofdegree 10, let us denote them by η , η ′ . Therefore, assuming the lemma is false, we can write χ | X = 1 X · σ + η σ + η σ + η ′ σ . Let 1 = x ∈ X be a p -element. Then η ( x ) = ε + ε + ε and η ′ ( x ) = ε − + ε − + ε − , where ε is some primitive 7th root of unity. As χ ( x ) , η ( x ) areintegers, so is η ( x ) σ (1) + η ′ ( x ) σ (1). This implies σ (1) = σ (1). Then χ (1) = σ (1) +( p − σ (1) + 20 σ (1) > σ (1), and the lemma follows unless σ (1) = 0. If σ (1) = 0 then χ | X = 1 X · σ + η · σ , and the above argument applies. (cid:3) Lemma 6.3.
Let p ≥ be odd and let λ be a hook partition of n ≥ p . Then the correspondingcharacter χ λ of S n takes a positive value on p -cycles.Proof. It is well known that any hook character χ λ is the m th exterior power, for some 0 ≤ m ≤ n −
1, of the irreducible reflection character ρ n of S n (the constituent of degree n − π n ). Let Y = Y × Y , with Y = S p and Y = S n − p , be aYoung subgroup of S n and g = g ′ × ∈ Y a p -cycle. Clearly ρ n | Y = ρ p ⊠ Y + 1 Y ⊠ ( π n − p ),and Λ i ( ρ p )( g ′ ) = ( − i for i < p , Λ i ( ρ p )( g ′ ) = 0 for i ≥ p . Thus χ λ ( g ) = Λ m ( ρ n )( g ) = m X i =0 Λ i ( ρ p )( g ′ )Λ m − i ( π n − p )(1) = min( p − ,m ) X i =0 ( − i (cid:18) n − pm − i (cid:19) , which clearly is positive for m ≤ ( n − p ) / n = 2 p , since the restrictionof a hook character from S n to S n − only contains hook characters. But for n = 2 p we are donesince by symmetry we may assume that m ≤ p = ( n − p ) / (cid:3) Theorem 6.4.
Let p be odd and G = A n with n > max { , p + 1 } . Then G has no Syl p -regular character. If n = p + 1 > then every Syl p -regular character of G is irreducible, unless ( n, p ) = (6 , .Proof. If p ≤ n < p the Sylow p -subgroups of G are cyclic and so the claim is in Proposition 4.4.Now assume that n ≥ p and let S be a Sylow p -subgroup of S n . First assume that n = p k forsome k ≥ n >
10 when p = 3. Then by the main result of [5], the restriction of anyirreducible character of S n to S contains the trivial character. A moments thought shows thatthe same is true for the restriction of any irreducible character of A n to S . So by Lemma 2.1any Syl p -regular character of A n is irreducible.Now assume that n = p k for some k ≥
2, and n >
10 when p = 3. Then again by [5, Thm. A]the only irreducible characters of S n whose restriction to S does not contain the trivial characterare the two characters of degree n −
1. So the only irreducible character of A n whose restrictionto S does not contain the trivial character is ψ of degree n −
1. Hence a Syl p -regular character χ of A n has the form χ = aψ + ψ ′ , for some a ≥ ψ ′ ∈ Irr( A n ). Let g ∈ A n bea p k -cycle. Then ψ ( g ) = −
1, and by the Murnaghan–Nakayama rule any irreducible characterof S n takes value 0 or ± g . In particular, if χ is reducible then we have that a = 1 and ψ ′ ( g ) = 0. But then ψ ′ is parametrised by a hook partition, of degree (cid:0) n − m (cid:1) for some m ≤ n .But then χ takes positive values on p -cycles by Lemma 6.3, a contradiction.Finally, the cases when p = 3 and 6 ≤ n ≤
10 can easily be checked individually. For example,all irreducible characters of A of degree at most 81 are non-negative on class 3C, and thosewhich vanish there are positive either on class 3B or 3A. So A has no Syl -regular character.As A has the same Sylow 2-subgroup, this also deals with n = 10. (cid:3) TEINBERG-LIKE CHARACTERS FOR FINITE SIMPLE GROUPS 11
Corollary 6.5.
Let G be a finite group and p > . Suppose that G has a subgroup P containinga Sylow p -subgroup of G and such that P/O p ( P ) ∼ = A n with n > max { , p + 1 } . Then G has noSteinberg-like character.Proof. This follows from Lemma 2.5 and Theorem 6.4. (cid:3)
Alternating groups for p = 2 . The situation is more complicated in the case of p = 2and we don’t have complete results. This is in part due to the existence of an infinite family ofexamples which we now construct.Set Γ = P ni =1 Γ i , where Γ i is the irreducible character of S n corresponding to the partition[ i, n − i ] for i >
1, and [1 n ] for i = 1. So the Young diagram γ i of Γ i is a hook with leg length n − i , and Γ i (1) = n ! n ( n − i )!( i − (cid:18) n − i − (cid:19) so Γ(1) = P ni =1 Γ i (1) = 2 n − . Lemma 6.6.
Let < m < n , where m is even, and g = ch ∈ S m × S n − m ≤ S n , where c isan m -cycle and h fixes all letters moved by c . Let Γ n − mk ∈ Irr( S n − m ) correspond to the hookpartition [ k, n − m − k ] . Then Γ i ( g ) = − Γ n − mi ( h ) if i ≤ m, Γ n − mi − m ( h ) if n − m < i, Γ n − mi − m ( h ) − Γ n − mi ( h ) if m < i ≤ n − m. Proof.
One observes that the restriction of Γ i to S m × S n − m is a sum of irreducible characters στ , where σ , τ are irreducible characters of S m , S n − m , resp., and both σ, τ are hook charactersof the respective groups (see [9, Lemma 21.3]). Next we use [9, Lemma 21.1] which states thatΓ i ( g ) = P j ( − j χ ν j ( h ), where χ ν j ∈ Irr( S n − m ), ν j is the Young diagram of χ ν j , and ν j issuch that γ i \ ν j is a skew hook with leg length j . In our case γ i is a hook, so the rim of γ i is γ i itself. By definition, a skew hook is connected, so it is either a row or a column inour case, and hence j = 0 or j = m −
1. (A column hook of length m has leg length m − m is even.) If j = 0 , m − ν = [ i − m, n − i ] , [ i, n − i − m ], respectively.This is a proper diagram if and only if i > m , resp., n − i ≥ m . So if n − i < m then ν = [ i − m, n − i ], j = 0, and Γ i ( g ) = χ ν ( h ) = Γ n − mi − m ( h ); if i ≤ m then ν = [ i, n − i − m ], j = m − i ( g ) = − χ ν ( h ) = − Γ n − mi ( h ); if m < i ≤ n − m then ν ∈ { [ i − m, n − i ] , [ i, n − i − m ] } andΓ i ( g ) = Γ n − mi − m ( h ) − Γ n − mi ( h ), as claimed. (cid:3) Proposition 6.7.
Suppose that n is even. Then: (a) Γ is a -vanishing character of S n . (b) Γ is Steinberg-like if and only if n = 2 k for some integer k > .Proof. (a) Let g ∈ S n be of even order. Suppose first that g is a cycle of length n . By [9, Lemma21.1], Γ i ( g ) = ( − n − i , so Γ( g ) = 0.Suppose that g is not a cycle of length n . Then we can express g as the product ch of a cycle c of even size m , say, and an element h fixing all letters moved by c . Then g ∈ S m × S n − m . ByLemma 6.6, we haveΓ( g ) = n X i =1 Γ i ( g ) = n X i = m +1 Γ n − mi − m ( h ) − n − m X i =1 Γ n − mi ( h ) = n − m X k =1 Γ n − mk ( h ) − n − m X i =1 Γ n − mi ( h ) = 0 . (b) If n = 2 k then | S n | = 2 · | S n/ | . As | S | = 2, by induction we have | S n | = 2 · (2 k − − ) = 2 k − = 2 n − . Write n = 2 k + l where 0 < l < k . Then | S n | = | S k | · | S l | . By induction, | S l | ≤ l − , so | S n | = 2 k − ·| S l | ≤ (2 k − l − = 2 k + l − = 2 n − . The statement follows as Γ(1) = 2 n − . (cid:3) Corollary 6.8.
Let n be even, and Γ = P n/ i =1 Γ i | A n . Then Γ is a -vanishing character of A n . If n = 2 k then this character is Steinberg-like.Proof. The characters Γ i remain irreducible under restriction to A n and Γ i | A n = Γ n − i +1 | A n . Itfollows that Γ | A n = 2Γ . Therefore, Γ ( g ) = Γ( g ) / g ofeven order. The last claim follows from Proposition 6.7(b). (cid:3) Suppose that n is odd. SetΓ e = ( n − / X i =1 Γ i = Γ + Γ + · · · + Γ n − , andΓ o = ( n +1) / X i =1 Γ i − = Γ + Γ + · · · + Γ n . Observe that Γ i | S n − = Γ n − i + Γ n − i − provided 1 < i < n , and Γ | S n − = Γ n − , Γ n | S n − = Γ n − n − .Therefore, Γ e | S n − = Γ n − + · · · + Γ n − n − = Γ o | S n − . As Γ = Γ e + Γ o we have Γ e (1) = Γ o (1) =Γ(1) / n − . Proposition 6.9.
Suppose that n is odd. Then: (a) Γ e and Γ o are -vanishing characters of S n . (b) Γ e is Steinberg-like if and only if n = 2 k + 1 for some integer k > .Proof. (a) Let g ∈ S n be of even order, and g = ch where c is a cycle of even size m . ByLemma 6.6,Γ e ( g ) = ( n − / X i =1 Γ n − i +1 ( g ) = ( n − − m ) / X i =1 Γ n − mn − m − i +1 ( h ) − ( n − / X i =( m +2) / Γ n − mn − i +1 ( h ) , as γ n − mn − m − i +1 is a proper diagram only for i < ( n − m ) / γ n − mn − i +1 is a proper diagram only for i ≥ ( m + 2) /
2. Set k = i − m/
2. So the second sum can be written as P ( n − − m ) / k =1 Γ n − mn − m − k +1 ( h ),whence Γ e ( g ) = 0.Similarly,Γ o ( g ) = ( n +1) / X i =1 Γ n − i +2 ( g ) = ( n − − m ) / X i =1 Γ n − mn − m − i +2 ( h ) − ( n − / X i =( m +2) / Γ n − mn − i +2 ( h ) , as γ n − mn − m − i +1 is a proper diagram only for i ≤ ( n − m − / γ n − mn − i +2 is a proper dia-gram only for i ≥ ( m + 2) /
2. Set k = i − m/
2. Then the second sum can be written as P ( n − − m ) / k =1 Γ n − mn − m − k +2 ( h ). So Γ o ( g ) = 0 as well. TEINBERG-LIKE CHARACTERS FOR FINITE SIMPLE GROUPS 13 (b) If n = 2 k + 1 then | S n | = | S n − | = 2 n − (see the proof of Proposition 6.7(b)). By theabove, Γ e (1) = Γ o (1) = 2 n − , so both Γ e and Γ o are Steinberg-like. If n − | S n | = | S n − | < n − by Proposition 6.7(b). (cid:3) Let n be odd. Then Γ i | A n = Γ n +1 − i | A n is irreducible for i = ( n + 1) /
2, whereas Γ ( n +1) / | A n isthe sum of two irreducible constituents which we denote by Γ +( n +1) / and Γ − ( n +1) / . If n = 4 l + 1then set Γ ea = ( n − / X i =1 Γ i | A n = (Γ + Γ + · · · + Γ ( n − / ) | A n , andΓ o ± = Γ ± ( n +1) / + ( n − / X i =1 Γ i − | A n = (Γ + Γ + · · · + Γ ( n − / ) | A n + Γ ± ( n +1) / , while for n = 4 l + 3 we setΓ oa = ( n +1) / X i =1 Γ i − | A n = (Γ + Γ + · · · + Γ ( n − / ) | A n , andΓ e ± = Γ ± ( n +1) / + ( n − / X i =1 Γ i | A n = (Γ + Γ + · · · + Γ ( n − / ) | A n + Γ ± ( n +1) / . Corollary 6.10. (a)
Let n = 4 l + 1 . Then Γ ea , Γ o + and Γ o + are -vanishing characters of A n . If n = 2 k + 1 then they are Steinberg-like characters. (b) Let n = 4 l + 3 > . Then Γ e + , Γ e − and Γ oa are -vanishing characters of A n . None ofthem is Steinberg-like.Proof. Let g ∈ A n be of even order.(a) Let i = ( n + 1) /
2. Then Γ i remains irreducible under restriction to A n . As Γ i and Γ n − i +1 coincide under restriction to A n , it follows that Γ e | A n = 2Γ ea , and hence Γ ea is a 2-vanishingcharacter. As Γ ea (1) = 2 n − = | A n | , this is Steinberg-like for n = 2 k + 1.Observe that Γ +( n +1) / ( g ) = Γ − ( n +1) / ( g ) and Γ +( n +1) / ( g ) + Γ − ( n +1) / ( g ) = Γ ( n +1) / ( g ). It followsthat Γ o + ( g ) = Γ o − ( g ) = Γ o ( g ) /
2, and thus Γ o + ( g ) = Γ o − ( g ) = 0 by Proposition 6.9. Therefore,Γ o + and Γ o + are 2-vanishing characters of A n . In addition, suppose that n = 2 k + 1. ThenΓ o + (1) = Γ o − (1) = Γ o (1) / | A n | , so both Γ o + and Γ o − are Steinberg-like.(b) Let i = ( n + 1) /
2. Then as above it follows that Γ o | A n = 2Γ oa , and hence Γ oa is a2-vanishing character. In addition, Γ oa (1) = Γ o (1) / n − . As here we never have n = 2 k + 1,Γ oa is not Steinberg-like.Consider Γ e ± . Observe that Γ +( n +1) / ( g ) = Γ − ( n +1) / ( g ) and Γ +( n +1) / ( g ) + Γ − ( n +1) / ( g ) =Γ ( n +1) / ( g ). It follows that Γ e + ( g ) = Γ e − ( g ) = Γ e ( g ) /
2, and so Γ e + ( g ) = Γ e − ( g ) = 0 byProposition 6.9. Therefore, Γ e + and Γ e − are 2-vanishing characters of A n but not Steinberg-like. (cid:3) Lemma 6.11.
Let ≤ n ≤ . Then in addition to the character Γ when n = 2 k , and thecharacters Γ e and Γ o when n = 2 k + 1 , the only Steinberg-like characters of S n are: (a) if n = 4 the sum of all non-linear irreducible characters; (b) if n = 6 the irreducible character of degree ; (c) if n = 8 the sum of the two irreducible characters of degree .Proof. For n ≤ n = 8 we use acomputer program to go through all possibilities. For S one checks that no character existswith the right restriction to S × S , and similarly for S one considers the restriction to S × S .Finally, the cases n ∈ { , , } are treated by restricting to S n − . (cid:3) Theorem 6.12.
Suppose that the only Steinberg-like character of S k , k ≥ , for p = 2 is theone constructed in Proposition . Then A n does not have Steinberg-like characters for p = 2 for n ≥ unless n or n − is a -power. In the latter case, the only Steinberg-like charactersare those listed in Proposition 6.9.Proof. Let ψ be a Steinberg-like character for p = 2 of A n , with n ≥
10. Then χ := ψ S n is Steinberg-like for S n . We argue by induction on n that S n does not have a Steinberg likecharacter, unless n or n − n is not a power of 2 and write n = 2 a + . . . + 2 a r for distinct exponents a , . . . , a r >
0. By Lemma 6.11 we may assume n = 12, so one of the summands, say 2 a isdifferent from 4 and 8. Then the Young subgroup Y = Y × Y := S a × S n − a of S n containsa Sylow 2-subgroup, so χ | Y is Steinberg-like. Then by Lemma 2.7 we have that χ | Y = P i η i σ i where η i ∈ Irr( Y ) are all distinct, the σ i are 2-vanishing characters of Y , and χ := P i l p ( σ i ) η i is a 2-vanishing character of Y with l ( χ ) = l ( χ ) = 1. Thus by assumption χ is the characterΓ from Proposition 6.7. In particular, χ is multiplicity-free and so l p ( σ i ) = 1 for all i . So the σ i are Steinberg-like as well. This is not possible, unless n − a is a 2-power as well.In the latter case, by Lemma 6.11 we conclude that n ≥
17. The above argument shows that χ | Y = Γ (1) ⊠ Γ (2) , with Γ ( j ) a Steinberg-like character of Y j . So in particular χ | Y and hencealso χ is multiplicity-free. By possibly interchanging a , a we may assume that 2 a >
8. Nowconsider χ | Y = | Y | Γ (1) , a sum of hooks. By the branching rule, any non-hook character of S m restricted to S m − contains a non-hook character (except when m = 4 which is excludedhere). Thus, inductively, χ cannot contain any non-hook constituent. This in turn means thatall constituents of χ | Y are hooks and thus by induction that χ | Y = | Y | Γ (2) . Now observethat by the Littlewood–Richardson rule [9, Lemma 21.3], Γ i | Y contains Γ (1) j ⊠ Γ (2) l if and onlyif j + l ∈ { i, i + 1 } . Thus, on the one hand side, Γ i | Y and Γ i +1 | Y have a common constituent,and so at most every second hook character occurs in χ . On the other hand, every second hookmust indeed occur. Thus either χ = Γ e or χ = Γ o as defined above. If n = 2 a + 1 then ourclaim follows from Proposition 6.9, otherwise the degree of χ is larger than | S n | . (cid:3) Projective characters for p = 2 .Lemma 6.13. Let p = 2 . Then A n has a projective character of degree | A n | if and only if S n has a projective character of degree | S n | .Proof. This follows from Lemma 2.13. (cid:3)
Theorem 6.14.
Let p = 2 and G = A n or S n for n > . Then G has no reducible projectivecharacter of degree | G | .Proof. One can inspect the decomposition matrix modulo 2 of G = A n for n ≤ G has no projective character of degree | G | . Analysing the character table of G = A n for TEINBERG-LIKE CHARACTERS FOR FINITE SIMPLE GROUPS 15 < n ≤
15 one observes that G has no Syl -regular characters, and hence no PIM of dimension | G | .One can inspect the decomposition matrix of G = A to observe that the minimal dimensionof a PIM is 320. Analysing the character table of G = A n for 9 < n ≤
15 one observes that G has no Syl -regular characters, and hence no PIM of dimension | G | .Let n = 16. Using the known character table of A one finds that there is a unique Syl -regular character, viz. the character Γ ; it is multiplicity-free with constituents of degrees1 , , , , , , , G as a constituent. However, by [13], the principal PIM is not of degree | G | .Let n = 2 k , where k >
4. Then G has a subgroup Y such that Y /N ∼ = A for some normal2-subgroup N and | Y | = | G | . Indeed, let P , . . . , P be a partition of { , . . . , n } with all partsof size n/
16. If G = S n then N is the direct product of 16 copies of a Sylow 2-subgroup of S n/ .If G = A n then we take N ∩ A n for the subgroup in question. Then Y is a semidirect productof N with S . The latter permutes P , . . . , P in the natural way. One easily observes that | G : Y | is odd.If Φ is a PIM of degree | G | = | Y | then so is Φ | Y . By [22, Lemma 3.8], the generalisedrestriction r GY/N (Φ) of Φ is a PIM of dimension | Y /N | . Such a PIM does not exist as we havejust seen.Let n >
16 be not a 2-power, and write n = 2 k + m , where m < k . Let X = X × X ≤ S n ,where X ∼ = S k and X ∼ = S m . Then the index | S n : X | is odd, so Φ | X is a PIM of degree | X | .Therefore, Φ | X is a direct product Φ × Φ , where Φ i is a PIM for X i for i = 1 ,
2. Obviously,dim Φ i = | X i | . This is a contradiction, as X has no PIM of degree | X | . For G = A n theresult follows from Lemma 6.13. (cid:3) Exceptional groups of Lie type
Theorem 7.1.
Let G be a simple group of Lie type which is not classical. Then G does nothave a Syl p -regular character in non-defining characteristic, except for the group G = F (2) ′ which has two reducible Syl -regular characters and two irreducible Steinberg-like characters for p = 3 .Proof. As in the proof of [13, Thm. 4.1] we compare the maximal order of a Sylow p -subgroup of G , which is bounded above by the order of the normaliser of a maximal torus, with the smallestirreducible character degrees (given for example in [19, Tab. I]). This shows that except for verysmall q there cannot be any examples of Syl p -regular characters. A closer inspection of thefinitely many remaining cases shows that G = F (2) ′ has two reducible Syl -regular charactersand two irreducible Steinberg-like character for p = 3, but no further cases arise. (cid:3) Groups of Lie type in their defining characteristic
It was shown in [17] that simple groups of Lie type of sufficiently large rank don’t haveSteinberg-like characters with respect to the defining characteristic apart from the (irreducible)Steinberg character. More precisely, the Steinberg-like characters were classified except forgroups of types B n with 3 ≤ n ≤ D n with n = 4 , Proposition 8.1.
Let G = Spin n +1 ( q ) , n ∈ { , , } , with q = p f odd. Then G has no reducibleSteinberg-like character with respect to p .Proof. We freely use results and methods from [17]. First assume that n = 3. According to [17,Prop. 6.2] it suffices to consider a group H (coming from an algebraic group with connectedcentre) such that [ H, H ] = Spin ( q ). Let χ be a reducible Steinberg-like character of H . Then χ has a linear constituent by [17, Thm. 8.6]. Multiplying by the inverse of that character, wemay assume that the trivial character occurs in χ (exactly once). By [17, Lemma 2.1], then allconstituents of χ belong to the principal p -block, so we may in fact replace H by H/Z ( H ), thatis, we may assume that H is of adjoint type.Let P ≤ H be a parabolic subgroup of H and U = O p ( P ). Then by Lemma 2.5(c) theHarish-Chandra restriction r HL ( χ ) is a Steinberg-like character of L = P/U . We will show thatthere is no possibility for χ compatible with Harish-Chandra restriction to all Levi subgroups.Clearly, r HL ( χ ) also contains the trivial character, so is again reducible. The reducibleSteinberg-like characters of all proper Levi subgroups of H are known by [17, Lemmas 7.4and 7.8]. In particular we must have 7 | ( q + 1) and for L of type A we have r HL ( χ ) = 1 L + µ with µ ∈ Irr( L ) of degree q −
1. Thus µ lies in the Lusztig series of a regular semisimple element s ∈ L ∗ (the dual group of L ) with centraliser a maximal torus of order ( q − q − χ has to contain a constituent ψ lying in the Lusztig series of s . It is easily seen that thecentraliser of s in G ∗ is either a maximal torus, or of type A ( q ) . ( q − ψ (1) ∈ { ( q − q + 1)( q + 1) , ( q − q + 1) , q ( q − q + 1) } . But the first and the last are bigger than q , so ψ (1) = ( q − q + 1). Now if χ contains anyother constituent apart from 1 G in the principal series, then its generalised restriction to L isnon-zero, contradicting r HL ( χ ) = 1 L + µ .Next, the Harish-Chandra restriction to a Levi subgroup L of type B has the form r HL ( χ ) =1 L + ν + ν + ν with ν (1) = ( q − ( q + 1) and ν (1) = ν (1) = ( q − q + 1). In particular ν lies in the Lusztig series of a regular semisimple element t ∈ L ∗ (of order 7 dividing q + 1)with centraliser a maximal torus of order ( q − q + 1). The centraliser of t in G ∗ then eitheris the same maximal torus, or of type A ( q ) . ( q + 1) . Correspondingly, χ has a constituent ψ in the Lusztig series of t of degree d := ( q − q + 1)( q − / ( q + 1) , d := q ( q − q + 1)( q − / ( q + 1) or ( q − q + 1)( q − . The last one is larger than q − − ψ (1), so ψ (1) ∈ { d , d } . Furthermore, by [17, Lemma 3.1], χ contains at least one regular constituent. This is either ψ of degree d , or, if ψ (1) = d thenone can check from the known list of character degrees of H (which can be found at [12]) that theonly regular character ψ of small enough degree has degree d := ( q − q + 1)( q − / ( q + 1).Observe that d = d + d . So the sum of remaining character degrees is d := q − − ψ (1) − d = q ( q − q − q + 3 q − q + 1) . Now note that χ cannot have further unipotent constituents since they would lead to unipotentconstituents of r HL ( χ ) (as H has no cuspidal unipotent characters). It turns out that all remainingcandidates except for one of degree λ (1) = ( q − q +1)( q −
1) have degree divisible by q − q +1.Now λ (1) ≡ q − q + 1), while d ≡ − q − q + 1). It follows that λ would haveto occur at least q − q times in χ . As ( q − q ) λ (1) > d , this is not possible. This contradictionconcludes the proof for the case n = 3. TEINBERG-LIKE CHARACTERS FOR FINITE SIMPLE GROUPS 17
The cases of Spin ( q ) and Spin ( q ) now follow from the previous one by application of theinductive argument in the proof of [17, Thm. 10.1]. (cid:3) Proposition 8.2.
Let G = Spin +2 n ( q ) , n ∈ { , } , q = p f . Then G has no reducible Steinberg-likecharacter with respect to p .Proof. First consider the case n = 4. As in the previous proof, by [17, Prop. 6.2 and Lemma 2.1]we may work with H of adjoint type. Let χ be a reducible Steinberg-like character of H . Then χ contains 1 H by [17, Thm. 8.6] and hence so does its Harish-Chandra restriction r HL ( χ ) to aLevi subgroup of type A . Then by [17, Lemma 9.1] we have q ≡ − r HL ( χ ) =1 L + µ + µ + µ with µ a cuspidal character labelled by a regular element s ∈ L ∗ in a torus oforder q −
1, of order dividing ( q + 1)( q + 1) in L ∗ /Z ( L ∗ ). But then s is also regular in H ∗ , thatis, χ has a constituent ψ of degree ( q − q − q − A . Comparison of degrees shows that this is not possible.The case of Spin ( q ) again follows from the previous one by application of the argument in theproof of [17, Thm. 10.1]. (cid:3) Classical groups of large rank
As an application of results obtained in Section 6 we show here that classical groups of largerank have no Steinberg like character for p >
2, provided p is not the defining characteristic of G . Throughout p is an odd prime not dividing q and we set e := e p ( q ), the order of q modulo p .We first illustrate our method on the groups GL n ( q ). Lemma 9.1.
Let G = GL n ( q ) , p > , and let S be a Sylow p -subgroup of G . (a) Write n = me + m ′ , where ≤ m ′ < e . Then there exist subgroups U ≤ S ≤ N ≤ G ,where U is an abelian normal p -subgroup of N and N/U ∼ = A m . (b) If m > max { , p + 1 } or m ′ > then G has no Steinberg-like character.Proof. (a) See [20]. (b) If m ′ > G contains a subgroup X such that X ∼ = GL m ′ ( q ) and C G ( X ) contains a Sylow p -subgroup of G . As X is a p ′ -group, the result follows from Lemma 2.3.Let m ′ = 0. Suppose the contrary, and let χ be a Steinberg-like character of G . By Lemma 2.5, A m must have a Steinberg-like character. However, this is false by Theorem 6.4. (cid:3) For other classical groups the argument is similar, but involves more technical details. Let d = e p ( − q ) be the order of − q modulo p , equivalently, d = 2 e if e is odd, d = e/ e ≡ d = e if 4 | e . So d = 1 if and only if e = 2, equivalently, p | ( q + 1). Note that e = 2 e p ( q ) if e is even. Lemma 9.2. [20]
Let G = GU n ( q ) and p > . Then the Sylow p -subgroups of G are isomorphicto those of H , where H ∼ = GL ⌊ n/ ⌋ ( q ) if e is odd, H ∼ = GL ⌊ n/ ⌋ ( q ) if | e and H ∼ = GL n ( q ) if e ≡ . Lemma 9.3.
Let G = GU n ( q ) , p > , and let S be a Sylow p -subgroup of G . Suppose that e ≡ , equivalently, d is odd. (a) Write n = md + m ′ , where ≤ m ′ < d . Then there exist subgroups U ≤ S ≤ N ≤ G ,where U is an abelian normal p -subgroup of N and N/U ∼ = A m . (b) If m > max { , p + 1 } or m ′ > then G has no Steinberg-like character. Proof. (a) Suppose first that e = 2. Let V be the natural F q H -module. Then V is a directsum ⊕ ni =1 V i , where V i ’s are non-degenerate subspaces of dimension 1. Let X be the stabiliserof this decomposition, that is, X = { x ∈ G | xV i = V j for some j = j ( x ) ∈ { , . . . , n }} . Then X ∼ = X · S n (a semidirect product), where X ∼ = (GU ( q ) × · · · × GU ( q )) ( n factors). Let U be the Sylow p -subgroup of X . Then U is normal in X and abelian. It is well known that X contains a Sylow p -subgroup of G . Therefore, N = U A n satisfies the statement.Let e >
2. As d is odd, there is an embedding GU m ( q d ) → GU md ( q ) (see [8, Hilfssatz 1]).Note that e p ( q d ) = 2 and | GU m ( q d ) | p = | GU md ( q ) | p . As GU md ( q ) is isomorphic to a subgroupof G , the result follows.(b) is similar to the proof of Lemma 9.1(b). (cid:3) Lemma 9.4.
Let p > , n = me , where e = e p ( q ) is even, and X = GU m ( q e/ ) . (a) If m is even (resp. odd) then X is isomorphic to a subgroup of GO +2 n ( q ) (resp. GO − n ( q ) ). (b) X is isomorphic to a subgroup of Sp n ( q ) , of GO n +1 ( q ) , of GO +2 n + e ( q ) as well as of GO − n + e ( q ) .In addition, X contains a Sylow p -subgroup of the respective group.Proof. (a) follows from [3, Lemma 6.6] as well as (b) for Sp n ( q ). The second case in (b) followsfrom (a) as the groups GO n +1 ( q ), GO +2 n + e ( q ) and GO − n + e ( q ) contain subgroups isomorphic toGO +2 n ( q ) and GO − n ( q ).The additional statement can be read off from the orders of the groups in question. (Thecases with Sp n ( q ), GO − n ( q ) and GO n +1 ( q ) are considered in [6, Lemmas 3.14 and 3.16], thatof GO +2 n ( q ) is similar.) (cid:3) Lemma 9.5.
Let p > . Let H be one of the following groups: (1) H = GU n ( q ) with n = md + m ′ , where m ′ < d ; (2) H ∈ { Sp n ( q ) , GO n +1 ( q ) , GO +2 n ( q ) , GO − n +1) ( q ) } with n = me + m ′ , where e is odd and m ′ < e ; (3) H ∈ { Sp n ( q ) , GO n +1 ( q ) } with n = me + m ′ where e is even and m ′ < e ; (4) H = GO ± n ( q ) with n = me + m ′ where e is even, m ′ < e , and either m ′ > , or m ′ = 0 and then either H = GO +2 n ( q ) , m is even, or H = GO − n ( q ) , m is odd; (5) let e be even, n = ( m + 1) e and H = GO +2 n ( q ) , m + 1 is odd, or H = GO − n ( q ) and m + 1 is even.Let S denote a Sylow p -subgroup of H . Then there exist subgroups U ≤ S ≤ P ≤ H , where U isan abelian normal p -subgroup of P and P/U ∼ = A m .Proof. (1) The case e ≡ [ n/ ( q ) is isomorphic to a subgroup of G .(2) By Lemma 9.2, | H | p = | GL n ( q ) | p . So the result follows from Lemma 9.1.(3) This follows from Lemmas 9.4 and 9.1. (Note that H = GO n +1 ( q ) contains subgroupsisomorphic to GO +2 n +1 ( q ) and GO − n ( q ) and one of them contains a Sylow p -subgroup of H .)(4) Similar to (3). Note that if m ′ > H contains subgroups isomorphic to GO + me ( q )and GO − me ( q ), and one of them contains a Sylow p -subgroup of H .(5) In this case a subgroup X of H isomorphic to GO +2 n − e ( q ) and GO − n − e ( q ), respectively,contains a Sylow p -subgroup of H . (One can easily check that | H : X | p = 1.) So the resultfollows from (4). (cid:3) TEINBERG-LIKE CHARACTERS FOR FINITE SIMPLE GROUPS 19
Our result for alternating groups (Corollary 6.5) implies the following:
Proposition 9.6.
Let p > and e = e p ( q ) . Let m = max { , p + 2 } . Let G be one of thefollowing groups: (1) PSL n ( q ) and n ≥ em ; (2) PSU n ( q ) and n ≥ dm , where d = e p ( − q ) ; (3) Ω n +1 ( q ) , q odd, or PSp n ( q ) , n > , and n ≥ em if e is odd, otherwise n ≥ em ; (4) PΩ +2 n ( q ) , and n ≥ em if e is odd, otherwise n ≥ em ; (5) PΩ − n ( q ) , and n − ≥ em if e is odd, otherwise n ≥ em .Then G has no Steinberg-like character. This remains true for any group H such that G isnormal in H/Z ( H ) and ( H/Z ( H )) /G is abelian.Proof. Suppose first that H is as in Lemma 9.5. Let S ∈ Syl p ( H ). Then there are subgroups U ≤ S ≤ P ≤ H , where U is normal in P and N/U ∼ = A m with m ≥ m = max { , p + 2 } .So A m is perfect. Let H be the derived subgroup of H . Set P = P ∩ H , S = S ∩ H , U = U ∩ H . Then S ∈ Syl p ( H ), U ≤ S ≤ P ≤ H and P /U ∼ = A m , as A m is perfect. Asimilar statement is true for the quotient of H by a central subgroup. Then the result followsfrom Theorem 6.4 using Lemma 2.5. (cid:3) Minimal characters and Sylow p -subgroups, p > p > G is a simple classical group not satisfying theassumptions in Proposition 9.6, p is not the defining characteristic of G and a Sylow p -subgroup S of G is not cyclic, then G has no Syl p -regular character and hence no Steinberg-like character.Observe that S is cyclic if and only if m = 1, and abelian if and only if m < p , where m is as inLemma 9.5. The case where S is cyclic has been dealt with in Section 4.For a group G , let µ ( G ) = 1 < µ ( G ) < µ ( G ) < · · · denote the sequence of integers suchthat for i > G has an irreducible character of degree µ i ( G ) and no irreducible character ρ with µ i − ( G ) < ρ (1) < µ i ( G ). For universal covering groups of finite classical groups thevalues µ ( G ) , µ ( G ) , µ ( G ) were determined in [19]. In our analysis below these three valuesplay a significant role, but mainly for classical centerless groups G such as PGL n ( q ), PGU n ( q ),PSp n ( q ), PΩ ± n ( q ) and Ω n +1 ( q ). For these groups, mainly for 2 e ≤ n ≤ pe with p >
2, weobserve that | G | p < µ ( G ) and sometimes | G | p < µ ( G ). In the latter case it is immediate toconclude that G has no Syl p -regular character, in the other cases we observe that there existsan element g ∈ G of order p such that ρ ( g ) > ρ of degree atmost µ ( G ). For n > pe we use a different method.Recall that e = e p ( q ) denotes the minimal integer i > q i − p .10.1. The groups GL n ( q ) , n ≥ e . Set d n = ( q n − / ( q − G = SL n ( q ). The min-imal degrees of projective irreducible representations of PSL n ( q ) are given in [19, Table IV].Table 1 is obtained from this by omitting the representations that are not realisable as ordinaryrepresentations of SL n ( q ). Lemma 10.1.
Let p > , e > , and G = GL en ( q ) . Suppose that < n ≤ p , and if q = 2 supposethat either n < p or p < e − . Then | G | p < µ ( G ) and G has no Syl p -regular character.Proof. If n < p then | G | p ≤ ( q e − n / ( q − n (as p is coprime to q − µ ( G ) =( q en − q ) / ( q − n, q µ ( G ) µ ( G ) µ ( G ) n = 3 , q > d − d ( q − q − / (3 , q − n = 4 , q > d − d ( q − q − / (2 , q − n = 4 , q = 3 26 39 52 n > , q > , ( n, q ) = (6 , d n − d n d n ( q n − − q ) / ( q − n > , q = 2 , n = 6 d n − d n (2 n − − / d n d n − / n = 6 , q = 2 62 217 588 n = 6 , q = 3 363 364 6318 Table 1.
Minimal degrees of irreducible characters of SL n ( q )Let n = p . If q > | G | p ≤ p ( q e − p ( q − p , while µ ( G ) = q ep − qq − . (As p >
2, the exceptions inTable 1 can be ignored, except for e = 3 and ( n, q ) = (6 ,
2) or (6 , p ( q e − p < ( q − p − ( q ep − q ) as p < ( q − p − for q > q e − p < q ep − q .If q = 2 and p < e − p ≤ (2 e − / | S | ≤ p (2 e − p / p is less than 2 ep − µ ( G )as p < p . (cid:3) Remark 10.2.
Lemma 10.1 does not extend to the case q = 2 with n = p = 2 e − | G | p = p (2 e − p = p p +1 > ( p + 1) p − ep − µ ( G ) . So the case e > q = 2, which we deal with next. Lemma 10.3.
Let e > and G = GL ep (2) . Then | G | p < µ ( G ) and G has no Syl p -regularcharacter.Proof. We have | G | p ≤ p (2 e − p . By Table 1 we have µ ( G ) = 2 ep − µ ( G ) = (2 ep − ep − − / > | G | p , or ep = 3 , ,
6. As p is odd and e >
1, we have ep = 3 ,
4, so in theexceptional cases e = 2, p = 3 where | G | p = 81 < µ ( G ) = 217.Let π be the permutation character of G associated with the action of G on the non-zerovectors of the natural F G -module. Then π = τ + 1 G , where τ is a character of G of degree τ (1) = 2 ep −
2. There is a unique irreducible character of G of degree 2 ep − τ . Let χ be a Syl p -regular character of G . As χ (1) = | G | p , it followsthat the irreducible constituents of χ are either 1 G or τ . As χ ( g ) = 0 for every p -element g ∈ G ,we get a contradiction as soon as we show that τ ( g ) > p -element g ∈ G . This isequivalent to showing that π ( g ) >
1. This can be easily verified. (cid:3)
Lemma 10.4.
Suppose that p > , p | ( q − and let SL n ( q ) ≤ G ≤ GL n ( q ) for < n < p . Then G has no Syl p -regular character.Proof. Let G = G/Z ( G ). Then | G | p ≤ | q − | n − p and µ ( G ) = ( q n − q ) / ( q − > q n − as above. So | G | p < µ ( G ), and G has no Syl p -regular character. Then neither has G byLemma 2.5. (cid:3) The groups GU n ( q ) , n > . In this section we consider the case where p > p -subgroup of GU n ( q ) is abelian or abelian-by-cyclic. This implies n < dp , where d isthe order of − q modulo p , equivalently, d = 2 e if e is odd, d = e/ e ≡ d = e if 4 | e . So d = 1 if and only if e = 2, equivalently, p | ( q + 1). Note that e = 2 e p ( q ) if e is even. TEINBERG-LIKE CHARACTERS FOR FINITE SIMPLE GROUPS 21
Lemma 10.5.
Let p be odd, let S be a Sylow p -subgroup of G = GU n ( q ) . Then S is abelian andnot cyclic if and only if d ≤ n < dp .Proof. If X = GL n ( q ) then Sylow p -subgroups of X are abelian if and only if n < ep . Let S bea Sylow p -subgroup of G . We use Lemma 9.2. If e ≡ | G | p = | GL n ( q ) | p , so S is abelian if and only if n < e p ( q ) p = ep/ dp . If e is odd then | G | p = | GL ⌊ n/ ⌋ ( q ) | p , so S isabelian if and only if ⌊ n/ ⌋ < ep , equivalently, n < ep = dp . If 4 | e then | G | p = | GL ⌊ n/ ⌋ ( q ) | p ,so S is abelian if and only if ⌊ n/ ⌋ < e p ( q ) p , equivalently, n < ep = dp . Similarly, S is cyclic ifand only if n < d . So the lemma follows. (cid:3) Lemma 10.6.
Let p > , d > and SU dp ( q ) ≤ G ≤ GU dp ( q ) . Then | G | p < µ ( G ) and G hasno Syl p -regular character. This remains true if SU n ( q ) ≤ G ≤ GU n ( q ) with d ≤ n < dp .Proof. Note that d > p does not divide q + 1, so | G | p = | SU d ( q ) | p . So it suffices toprove the lemma for G = SU dp ( q ). First assume that d is odd, so e ≡ d = e/ | G | p ≤ p ( q e/ + 1) p / ( q + 1) p = p ( q d + 1) p / ( q + 1) p and µ ( G ) = ( q dp − q ) / ( q + 1).In this case | G | p < µ ( G ). Similarly, if e is odd then d = 2 e , | G | p ≤ p ( q d − p / ( q + 1) p and µ ( G ) = ( q dp − / ( q + 1), so again | G | p < µ ( G ).Finally assume that 4 | e . Then d = e , | G | p ≤ p ( q e − p / ( q − p and µ ( G ) = ( q ep − / ( q + 1).So | G | p < µ ( G ) again. This implies that G has no Syl p -regular character.The proof of the additional statement is similar. (cid:3) Thus, we are left with primes such that p | ( q + 1). We first consider the case where n < p . Lemma 10.7.
Let p | ( q +1) , < n < p , and SU n ( q ) ≤ G ≤ GU n ( q ) . Then G has no Syl p -regularcharacter.Proof. Let G = { g ∈ G | det g is an element of GU ( q ) of p ′ -order } . Then | G : G | = | Z p | ,where Z p is the Sylow p -subgroup of Z ( G ). As Z p ∩ G = 1, it follows that G = G × Z p . ByLemma 2.6, the result for G follows if we show that G has no Syl p -regular character. In turn,this follows from the same result for G ′ = SU n ( q ) as | G : G ′ | is coprime to p . So we deal with G ′ .Suppose the contrary, and let χ be a Syl p -regular character of G ′ .First, let n = 3. Then χ (1) = | G ′ | p = ( q + 1) p for p >
3. By [19, Table V], µ ( G ′ ) = q − q .Let q >
4. Then χ (1) = | G ′ | p = ( q + 1) p < q − q ) = 2 µ ( G ′ ). So χ has a single non-trivialirreducible constituent ρ , and ρ (1) ≤ χ (1). Again by [19, Table V], q − q ≤ ρ (1) ≤ q − q + 1.As χ is Syl p -regular, ( χ, G ′ ) ≤ χ (1) ≤ ρ (1) + 1 ≤ q − q + 2, whichis false. The case with q = 4 can be read off from the character table of G ′ .Let n > V be the natural F q G ′ -module. Let b , . . . , b n be an orthogonal basis in V and let W = h b , b , b i . Then W is a non-degenerate subspace of V of dimension 3. Set X = { h ∈ G ′ | hW = W and hb i ∈ h b i i for i = 4 , . . . , n } and U := O p ( X ). Then U ⊆ Z ( X ) andevery element of U acts scalarly on W . Let X ′ be the derived subgroup of X ′ and P = X ′ U .Then X ′ ∼ = SU ( q ) and P/U ∼ = SU ( q ) (as p > ( q ) has no Syl p -regularcharacter. As P contains a Sylow p -subgroup of G ′ , the result follows from Lemma 2.5. (cid:3) Lemma 10.8.
Let SU (8) ≤ G ≤ GU (8) . Then G has no Syl -regular character.Proof. By Lemma 2.5, it suffices to prove that PSU (8) and PGU (8) have no Syl -regularcharacter. Suppose the contrary, and let χ be a Syl -regular character of any of these groups. As | PGU (8) | = 243, we have χ (1) ≤ ρ be an irreducible constituent of χ . Then ρ (1) ≤ ρ (1) ∈ { , , , } , and the characters of degree 1,56,133 are positive atthe class 9 A , whereas those of degree 57 vanish at this class. It follows that ρ (1) = 57, but then ρ is positive at the class 3 C . This is a contradiction. (cid:3) Lemma 10.9.
Let H = SU p ( q ) , p > , p | ( q + 1) , or H = SL p ( q ) , p > , p | ( q − ), and h = diag(1 , ε, ε , . . . , ε p − ) ∈ H , where ε ∈ F × q is a primitive p -th root of unity. Let χ be anirreducible character of H whose kernel has order prime to p . Then χ ( h ) = 0 .Proof. The element h is written in an orthogonal basis of the underlying vector space in theunitary case. Then h ∈ E , where E ≤ H is an extraspecial group of order p such that Z ( E ) = Z ( H ). The restriction of χ to E is a direct sum of irreducible representations of E non-trivial on Z ( E ). It is well known and can be easily checked that the character of every suchrepresentation vanishes at h . So the claim follows. (cid:3) Let H = GU n ( q ) or GL n ( q ) with n >
2. Weil representations of these groups were studiedby Howe [7] and other authors, and have many applications, mainly due to the fact that theirirreducible constituents (which we call irreducible Weil representations) essentially exhaust theirreducible representations of degree µ ( H ) and µ ( H ). More details are given below for n = p , p odd. Let M be the underlying space of the Weil representation of H . Then M = ⊕ ζ ∈ Irr( Z ( H )) M ζ ,where M ζ = { m ∈ M | zm = ζ ( z ) m for z ∈ Z ( H ) } . In general, H is irreducible on M ζ , exceptfor the case where H = GL p ( q ) and ζ = 1 Z ( H ) . In this case M ζ is a sum of a one-dimensionaland an irreducible H -invariant subspace.So the irreducible Weil representations ρ of H of dimension greater than 1 are parameterisedby their restriction to Z ( H ), and each of them remains irreducible under restriction to H ′ =SU n ( q ) or SL n ( q ). By [19], every irreducible representation of H ′ of degree µ ( H ) and µ ( H )is an irreducible Weil representation. Moreover, every irreducible representation of H of degree µ ( H ) and µ ( H ) is obtained from an irreducible Weil representation by tensoring with a one-dimensional representation. Lemma 10.10.
Let p > , and H = GU p ( q ) , p | ( q + 1) , ( p, q ) = (3 , , or GL p ( q ) , p | ( q − .Let ζ ∈ Irr( Z ( H )) . Let ρ = ρ ζ be the character of an irreducible constituent of the Weilrepresentation ω of H labeled by ζ (where ρ (1) > ). Let h be as in Lemma . Then ρ ( h ) ∈ { , p, p − } , except for the case with G = GL p ( q ) and ζ = 1 Z ( H ) , where ρ ( h ) = p − . Inaddition, ρ ( h ) = 0 if and only if ρ ( z ) = 1 for an element z ∈ Z ( H ) of order p .Proof. We only consider the case H = GU p ( q ), as the case H = GL p ( q ) is similar.Let Z = Z ( H ), ζ ∈ Irr( Z ) and ρ = ρ ζ be the irreducible constituent of ω labeled by ζ . Thismeans that ρ ( z ) = ρ (1) ζ ( z ).Let X = h Z, h i . Let ε i be the character of h h i such that ε i ( h ) = ν i , where ν is a fixed p throot of unity, i = 1 , . . . , p . Then the multiplicity of the eigenvalue ν i of ρ ( h ) equals ( ω | X , ζ · ε i ).Recall that ω ( x ) = − ( − q ) d , where d is the multiplicity of the eigenvalue 1 of x as a matrix inGU p ( q ). Therefore, ω (1) = q n , and if x = zh k and z p = 1 then ω ( x ) = −
1; if z p = 1 , h = 1 then ω ( x ) = q (also for z = 1). TEINBERG-LIKE CHARACTERS FOR FINITE SIMPLE GROUPS 23
We compute | X | · ( ω | X , ζ · ε i ) = P x ∈ X ω ( x ) ζ ( z ) ε i ( h ), where x = zh . Note that ω ( x ) is aninteger, so ω is self-dual. Let Z p be the subgroup of order p in Z , and X p = h Z p , h i . Then X x ∈ X ω ( x ) ζ ( z ) ε i ( h ) = X x ∈ X p ω ( x ) ζ ( z ) ε i ( h ) + X x/ ∈ X p ω ( x ) ζ ( z ) ε i ( h ) . We first show that the second sum equals 0 if i < p . Note that x = zh / ∈ X p is equivalent to z / ∈ Z p . Therefore, d = d ( x ) = 0 for x / ∈ X p , and ω ( x ) = −
1. For z fixed we have a partial sum ζ ( z ) P h ε i ( h ), and P h ε i ( h ) = 0, as claimed.If i = p and ζ = 1 Z then X x/ ∈ X p ω ( x ) ζ ( z ) ε i ( h ) = − p X z / ∈ Z p ζ ( z ) = − p ( X z ∈ Z ζ ( z ) − X z ∈ Z p ζ ( z )) = p as ζ ( z ) = 1 for z ∈ Z p . If i = p and ζ = 1 Z then X x/ ∈ X p ω ( x ) ζ ( z ) ε i ( h ) = − ( | X | − | X p | ) = − p ( q + 1) + p . Next, we compute P x ∈ X p ω ( x ) ζ ( z ) ε i ( h ). Observe that ζ ( z ) = 1 for z ∈ Z p so this sumsimplifies to P zh ∈ X p ω ( zh ) ε i ( h ). Note that d ( zh ) = 1 if h = 1 and any z ∈ Z p . So if h = 1 then ω ( zh ) = q . If h = 1 then d ( zh ) = d ( z ) = 0 for z = 1 so ω ( z ) = −
1. And ω (1) = q p .Therefore, we have X zh ∈ X p ω ( zh ) ε i ( h ) = X z ∈ Z p ,h =1 ω ( zh ) ε i ( h ) + X z ∈ Z p ,z =1 ω ( z ) + q p = X z ∈ Z p ,h =1 q · ε i ( h ) + X z ∈ Z p ,z =1 ( −
1) + q p = pq X h =1 ε i ( h ) − ( p −
1) + q p . (i) Let i = p . Then P h =1 ε i ( h ) = −
1, and the last sum equals − pq − p +1+ q p = q p +1 − p ( q +1).(ii) Let i = p . Then P h =1 ε i ( h ) = p −
1, and the last sum equals pq ( p − − ( p −
1) + q p = q p + 1 + p q − pq − p .Therefore, | X | · ( ω | X , ζ · ε i ) = q p + 1 − p ( q + 1) if i = p , q p + 1 + ( p − p ( q + 1) if i = p , ζ = 1 Z ,and q p + 1 + ( p − p ( q + 1) if i = p , ζ = 1 Z .In particular, the multiplicities of eigenvalue ν i for i = p of h on the module M ζ for fixed ζ are the same. As P i = p ν i = −
1, the trace of h on M ζ for ζ = 1 Z with ζ ( Z p ) = 1 equals(1 / | X | )( q p + 1 + ( p − p ( q + 1) − ( q p + 1 − p ( q + 1)) = p as | X | = p ( q + 1). Similarly, if ζ = 1 Z then the trace in question equals p −
1. In other words,if ω ζ is the character of M ζ and ζ ( Z p ) = 1 then ω ζ ( h ) = p for ζ = 1 Z and p − (cid:3) Lemma 10.11.
Let H = GU p ( q ) , p | ( q + 1) , p > , ( p, q ) = (3 , or GL p ( q ) , p | ( q − . Then Hhas no Syl p -regular character. The same is true for H ′ = SU p ( q ) and SL p ( q ) and for all groups X with H ′ ≤ X ≤ H .Proof. Set G = H/O p ( Z ( H )). By Lemma 2.5, it suffices to prove the lemma for G in place of H .Suppose the contrary, and let χ be a Syl p -regular character of G , and let λ be an irre-ducible constituent of χ . We first observe that λ (1) < µ ( G ), and hence by [19], λ (1) ∈{ , µ ( G ) , µ ( G ) } . Indeed, note that | G | p = p | q + 1 | p − p in the unitary case, respectively | G | p = p | q − | p − p in thelinear case. By [19, Table IV], µ ( G ) ≥ µ ( H ′ ) ≥ ( q p +1)( q p − − q )( q +1)( q − , resp., ( q p − q p − − q )( q − q − if p > | G | p . Let p = 3. Then µ ( G ) ≥ µ ( H ′ ) ≥ ( q − q + 1)( q − /
3, resp.,( q − q − / q >
4. Again, | G | < µ ( G ), unless G = PGU (8) or PGL (4). The formercase is settled in Lemma 10.8.Let G = PGL (4). Then | G | = 27. In this case µ ( G ) = 20, µ ( G ) = 35 and µ ( G ) = 45. So λ (1) ≤ | G | p implies λ (1) ≤
20. The character of degree 20 is positive at class 3 A , a contradiction.So | G | p ≤ µ ( G ). As mentioned prior Lemma 10.10, λ is either one-dimensional or can beseen as a character of H obtained from an irreducible Weil character by tensoring with a linearcharacter of H . Let h ∈ H as in Lemma 10.10. Then h ∈ H ′ , so tensoring can be ignored,and we can assume that λ is an irreducible Weil character of H . Then, by Lemma 10.10, λ ( h ) ∈ { , p, p − } in the unitary case and λ ( h ) ∈ { , p, p − } in the linear case. If λ (1) = 1then λ ( h ) = 1. So λ ( h ) ≥
0. As χ is p -vanishing and | h | = p , we have χ ( h ) = 0. So λ ( h ) = 0 forevery irreducible constituent of χ . This is false as λ is trivial on O p ( Z ( H )) by the definition of G ,and hence λ ( h ) = 0 by Lemma 10.10. This is a contradiction. As irreducible Weil representationsof H remain irreducible upon restriction to H ′ , this argument works for intermediate groups X too. (cid:3) Lemma 10.12.
Let p > and let G be a group such that SL n ( q ) ≤ G ≤ GL n ( q ) with e < n ≤ ep , or SU n ( q ) ≤ G ≤ GU n ( q ) with d < n ≤ dp and ( n, q ) = (3 , . Then G and G/O p ( G ) haveno Syl p -regular character.Proof. For the unitary case with d > G is stated in Lemma 10.6. The case with d = 1 and n < p is dealt with in Lemma 10.7, and the remaining case d = 1 and n = p isexamined in Lemma 10.11.Let H = GL n ( q ). The result for e > , q > e > , q = 2 is proved in Lemma 10.3. The result for e = 1 , n = p is stated in Lemma 10.11.The case with e = 1 , n < p is examined in Lemma 10.4.The statement on G/O p ( G ) follows from Lemma 2.5. (cid:3) Lemma 10.13.
For p > let H = GL n ( q ) , H ′ = SL n ( q ) with ep < n < ep , or H = GU n ( q ) , H ′ = SU n ( q ) with dp < n < dp . Let G be a group such that H ′ ≤ G ≤ H . Then G has no Syl p -regular character, unless p = 3 and H = GU (2) .Proof. Suppose the contrary, and let χ be a Syl p -regular character of G .Suppose first that e > , d >
1. Note that G has a subgroup X , say, isomorphic to SL ep ( q ) × SL n − ep ( q ), resp., SU dp ( q ) × SU n − dp ( q ), and | G : X | p = 1. Let S be a Sylow p -subgroup ofSL n − ep ( q ), resp., SU n − dp ( q ). Let Y = S × SL ep ( q ), resp., S × SU dp ( q ). As | G : Y | p = 1,by Lemma 2.5, r GY/S ( χ ) is a Syl p -regular character of Y /S ∼ = SL ep ( q ), resp., SU dp ( q ). Thiscontradicts Lemma 10.12, unless, possibly, if G = SU n (2) and p = 3. As d >
1, this case doesnot occur.Next, suppose that e = d = 1, that is p | ( q −
1) or q + 1. Then we refine the above argument.Set D = GL p ( q ) or GU p ( q ). Then Y /S ∼ = D . Set Y = G ∩ Y . Then Y is normal in Y andhence O p ( Y ) = Y ∩ O p ( Y ). As Y /S ∼ = D , it follows that Y /O p ( Y ) = D/O p ( D ) = D/O p ( Z ( D )),and hence E := Y /O p ( Y ) is a non-central normal subgroup of D/O p ( Z ( D )). By Lemma 2.5, r GY /O p ( Y ) ( χ ) is a Syl p -regular character of E = Y /O p ( Y ). However, by Lemma 10.12, E has no TEINBERG-LIKE CHARACTERS FOR FINITE SIMPLE GROUPS 25
Syl p -regular character, unless D = GU (2) and p = 3. So we are left with the case H = GU n (2), p = 3 and 3 < n < H = GU (2) is excluded by assumption, so consider first H = GU (2). As H = H ′ × Z ( H ), it suffices to deal with G = SU (2). Suppose the contrary, and let χ be aSyl -regular character of G . Then χ (1) = | G | p = 243. Let λ be an irreducible constituent of χ , so λ (1) ≤ g ∈ G is an element from class 3 E then λ ( g ) > λ (1) = 176or 220. As χ ( g ) = 0, there is a constituent λ , say, of χ such that λ (1) ∈ { , } . Then λ (1) ≤
67 = 243 − h ∈ G from the class 3 F . Then λ ( h ) >
0, and if λ (1) ≤
67 then λ ( h ) >
0, unless λ (1) = 10. So χ must have a constituent λ , say, of degree 10. Then λ ( g ) = 4.If λ (1) = 176 then λ ( g ) = −
4, and hence ( χ − λ − λ )( g ) = 0. As λ ( g ) > λ (1) ≤ χ = λ + λ , but then 0 = χ ( h ) = λ ( h ) + λ ( h ) = 3, a contradiction. So λ (1) = 220, and the other constituents are of degree at most 23. As λ ( g ) = − λ ( g ) = 4, wehave ( χ − λ − λ )( g ) = −
1, in particular, for the other constituents λ of χ we have λ ( g ) ≤
1. By[1], this implies λ (1) = 1, and λ must occur with multiplicity 1, whence χ (1) = 220+10+1 = 231,a contradiction.Let H = GU (2). By Lemma 2.5, it suffices to deal with X := PGU (2). Set X ′ = PSU (2).Then | X ′ | = 3 = 729 and the irreducible characters of X ′ of degree less than 616 are positiveon class 3 A . In addition, | X | = 3 = 2187, and the irreducible characters of X of degree lessthan 2187 and not equal to 616 are positive on class 3 A . Let χ be a Syl -regular characterof X or X ′ . Then the irreducible character µ of degree 616 is a constituent of χ . Note that τ (3 A ) = −
14. If χ ∈ Irr( X ′ ) then the sum of the other constituents of χ is at most 113. By [1],they are of degree 1 or 22. The trivial character cannot occur with multiplicity greater than 1,so 113 or 112 must be a multiple of 22, which is false. Let χ ∈ Irr( X ). Note that the multiplicityof µ in χ is at most 3, and if µ occurs with multiplicity 3 then the sum of the other constituentsof χ is at most 2187 − µ are positive at class 3 C . This is a contradiction. Supposethat µ occurs once. Then the sum of the other constituent values at class 3 A is −
14. It followsthat these constituents may only be of degrees 770,252,232,22,1. Inspecting [1], one observesthat all them as well as µ are positive at class 3 C . This is a contradiction. So the multiplicityof µ must be 2. Then the sum of the other constituent values at class 3 A is −
28. Therefore, thedegrees of the other constituents may only be 770,560,385,252,232,22,1. Let ν be the characterof degree 385. Then ν (3 A ) = 25. If ( χ, ν ) > A is −
3. The trivial character is the only one whose value is at most 3. As this cannotoccur twice, we get a contradiction. Therefore, ( χ, ν ) = 0. As above, this contradicts χ (3 C ) = 0.This completes the analysis of the case with n = 6.Let n = 7. Then H ′ = SU (2) contains a subgroup isomorphic to GU (2), which containsa Sylow 3-subgroup of H ′ . So the result for this case follows from n = 6. In addition, H = H ′ · Z ( H ), so we are done by Lemma 2.5.Similarly, the result for n = 8 follows from that with n = 7. (cid:3) Remark 10.14.
The group SU (2) has an irreducible projective character of degree 81 (for p = 3), and hence H = GU (2) = SU (2) × Z ( H ) has a projective character of degree | H | = 243. Theorem 10.15.
Let p > and G be a group such that SL n ( q ) ≤ G ≤ GL n ( q ) , or SU n ( q ) ≤ G ≤ GU n ( q ) . Suppose that Sylow p -subgroups of G/Z ( G ) are not cyclic. Then G has noSteinberg-like character, unless p = 3 and G ∈ { SU (2) , GU (2) , SU (2) , GU (2) } . Proof. If n ≥ ep in the linear case and n ≥ dp in the unitary case then the result follows fromProposition 9.6 for G/O p ( G ) in place of G , and then for G in view of Lemma 2.5.If ep < n < ep in the linear case and dp < n < dp in the unitary case then the result followsfrom Lemma 10.13. If 2 e ≤ n ≤ ep in the linear case and 2 d ≤ n ≤ dp in the unitary case thenthe result follows from Lemma 10.12. If n < e in the linear case and n < d in the unitary casethen Sylow p -subgroups of G/Z ( G ) are cyclic. (cid:3) Remark 10.16.
Proposition 9.6 gives a better bound for n , but this does not yield an essentialadvantage as the cases with n = e ( p + 1) and d ( p + 1) are not covered by Proposition 9.6, andwe have to use Lemma 10.13 anyway.10.3. The symplectic and orthogonal groups for p > .Lemma 10.17. Let G = Sp n ( q ) ( q even, n ≥ , ( n, q ) = (2 , ), G = Spin n +1 ( q ) ( q odd, n ≥ , ( n, q ) = (3 , ), or G = Spin ± n ( q ) ( n ≥ ). Suppose that Sylow p -subgroups of G areabelian. Then | G | p < µ ( G ) .Proof. Let S ∈ Syl p ( G ). As S is abelian, we have p > | S | ≤ ( q + 1) n . If G = Sp n ( q ), q even, n ≥
2, ( n, q ) = (2 , n +1 (3) then µ ( G ) ≥ ( q n − q n − q ) / q + 1) (see[19, Table II]). This is greater than ( q + 1) n . If G = Spin n +1 ( q ), q > n ≥
3, then µ ( G ) ≥ ( q n − / ( q − µ ( G ) > ( q + 1) n , whence the result. The cases with G = Spin ± n ( q ), n ≥
4, are similar, see [19, Thm. 7.6]. (cid:3)
Proposition 10.18.
Let e be odd, p > , and let H = Sp n ( q ) with n > , GO n +1 ( q ) with n > , GO +2 n ( q ) with n > , or GO − n +2 ( q ) with n > . Suppose that e ≤ n < ep . Then H hasno Syl p -regular character.Proof. Let S ∈ Syl p ( H ). By Lemma 9.2(1), S is conjugate to a Sylow p -subgroup of a subgroup H ∼ = GL n ( q ) of H . By Lemmas 10.1, 10.3, 10.12 and 10.13, GL n ( q ) for 2 e ≤ n < ep has noSyl p -regular character, unless possibly when n = 2.Let n = 2, so H = Sp ( q ), e = 1 and p | ( q − | H | p = | q − | p . If q is even then µ ( G ) = q ( q − / q >
2. This is greater than | H | p , whence the result. If q is odd then | H | p ≤ ( q − / µ ( H ) = ( q − /
2. So again | H | p < µ ( H ). (cid:3) Proposition 10.19.
Let e be even, p > , and let H = Sp n ( q ) ( n > , ( n, q ) = (2 , ), GO n +1 ( q ) ( q odd, n > ), or GO ± n ( q ) with n > . Suppose that e ≤ n < ep . Then H hasno Syl p -regular character.Proof. Write 2 n = ek + m with m < e , where k > H contains a subgroup H with ( | H : H | , p ) = 1, where H ∼ = Sp ke ( q ) or GO ke +1 ( q ), respectively, it suffices to provethe lemma for 2 n = ke . Let 2 n = ke . By Lemma 9.4, a Sylow p -subgroup of H is containedin a subgroup isomorphic to GU k ( q e/ ). By Lemma 10.12 for 2 < k ≤ p and Lemma 10.13 for p < k < p (with d = 1 and q e/ in place of q ), the group GU k ( q e/ ) with ( k, q e/ ) = (3 , p -regular character, whence the claim. (The exceptional case H = Sp (2), p = 3 isconsidered below.)Let k = 2. Then H = Sp e ( q ), and p | ( q e/ + 1). Then | H | p = | q e/ + 1 | p . If q is even then µ ( G ) = ( q e − q e − q ) / q + 1) for q >
2. This is greater than | H | p , whence the result. If q isodd then | H | p ≤ ( q e/ + 1) / µ ( H ) = ( q e − /
2. So | H | p < µ ( H ), whence the result. TEINBERG-LIKE CHARACTERS FOR FINITE SIMPLE GROUPS 27
A similar argument works if H = GO ke +1 ( q ) as well as for H = GO − ke ( q ) with k odd, and for H = GO + ke ( q ) with k > H = GO +8 (2) and e = 2.Let H = GO +8 (2) and e = 2 so p = 3. Then | G | = 243 and the irreducible characters ofdegree less than 243 are of degrees 1 , , , , , , χ be a Syl p -vanishing characterof degree 243 and λ an irreducible constituent of χ . By [1], λ (3 E ) > λ (1) ≤ χ (3 E ) = 0.Let k = 2 and H = GO +2 e ( q ). Then | H | p = | q e/ + 1 | p < µ ( H ) = ( q e − q e − − / ( q − q > q = 2 , e >
4. (If q = 2 , e = 4 then p = 5, and | H | = 25 < µ ( H ) = 28.) So theresult follows. (The case e = 2 has been examined above.)Suppose that H = GO − n ( q ) with k even or H = GO +2 n ( q ) with k odd. Then some Sylow p -subgroup of H is contained in a subgroup H isomorphic to GO − n − e ( q ) or GO +2 n − e ( q ), respec-tively. Note that 2 n − e = ( k − e . For the groups H the result has been proven above, exceptfor the cases where k − k − e ≤
6. However, if k − p -subgroupsof H , and hence of H are cyclic, and this case has been examined in Propositions 3.1 and 4.4.Let ( k − e ≤
6. As k − >
1, we have k = 3 , e = 2 as e is even. Then H = GO − ( q )and µ ( H ) = q ( q + 1) [19]. If p > | H | p = | q + 1 | p , otherwise | H | = 3 | q + 1 | . Then | H | p < µ ( H ), unless q = 2.Let q = 2, p = 3. Then | G | = 81. By [1] the irreducible characters of degree less than 81 areof degrees 1 , ,
52. Therefore only these characters can occur as irreducible constituents of aSyl -regular character χ . However, the values of these characters at an element g ∈ G in class9 A are 1 , ,
1, in particular, positive. As χ ( g ) = 0, this is a contradiction.Suppose that G = Sp (2) and p = 3. Then | S | = 81. Let τ be an irreducible constituent of χ .Then τ (1) ≤
81. Let g ∈ G belong to the conjugacy class 3 C in notation of [1]. By inspectionof the character table of G one observes that τ ( g ) ≥ τ (1) ≤
81. Therefore, τ ( g ) = 0for every irreducible constituent τ of χ . This implies τ (1) ∈ { , } , see [1]. However, such acharacter takes positive values at the elements of class 3 A . So this case is ruled out. (cid:3) Remark 10.20.
Let G be the universal covering group of SO +8 (2). One observes that G has aSyl -regular character, and has no Syl -regular characters. If H = Sp (2) then H has Steinberg-like characters for p = 3, both reducible and irreducible.11. Classical groups at p = 2In this section we investigate Syl p -regular and Steinberg-like characters of simple classicalgroups over fields of odd order q > p = 2.11.1. Linear and unitary groups at p = 2 . We first deal with the smallest case:
Proposition 11.1.
Let q > be odd. (a) Let G = PSL ( q ) . Then G has a reducible Syl -regular character if and only if q + 1 = 2 k for some k ≥ or if q = 5 . (b) Let G = SL ( q ) . Then G has a reducible Syl -regular character if and only if q ± is a -power.Proof. (a) The 2-part of | G | is | q − | if q ≡ | q + 1 | else. The smallest non-trivialcharacter degree is ( q + 1) / q − / -regular characters in the first case, unless q = 5. In the second case, it follows from the character table of G that the sum of the trivial and the Steinberg character is Syl -regular when q + 1 is a power of 2, and there are no cases otherwise. If q = 5 then there are two reducibleSyl -regular characters of degree 4 by [1].(b) Let χ be a reducible Syl -regular character. Let 1 = z ∈ Z ( G ); then χ = χ + χ , where χ ( z ) = χ (1) and χ ( z ) = − χ (1). By Lemma 2.5 (with P = G and U = Z ( G )), χ is aSyl -regular character for G/Z ( G ) = PSL ( q ). If χ is irreducible then q ± χ is reducible. So q ± χ , χ such that χ + χ is 2-vanishing of degree 2( q ±
1) = | G | . Indeed, using the character table of G one observes that there exist irreducible characters χ , χ of G that vanish at non-central 2-elements of G , and such that χ ( z ) = χ (1) and χ ( z ) = − χ (1). It follows that χ + χ is a reducible Syl -regular character. (cid:3) We recall that µ ( G ) denotes the third smallest degree of a non-trivial irreducible represen-tation of G . Lemma 11.2.
Let G be a quasi-simple group such that G/Z ( G ) ∈ { PSL n ( q ) , PSU n ( q ) } with n ≥ , q > odd and | Z ( G ) | = 1 . Then | G | < µ ( G ) .Proof. Let first
G/Z ( G ) = PSL ( q ), q > µ ( G ) = q ( q + 1) by Table 1, while | G | = 2 | q − | if q ≡ | G | = 4 | q + 1 | if q ≡ | G | < µ ( G ) unless q − µ ( G ) = ( q − q − G/Z ( G ) = PSL ( q ) with q > µ ( G ) = ( q − q − /
2, whichis larger than | G | ≤ q − for q ≥
5. Now assume that
G/Z ( G ) = PSL n ( q ), with n ≥ q > µ ( G ) = ( q n − q n − − q )( q − q − (see Table 1) while | G | ≤ ( q − n − n − when q ≡ | G | ≤ ( q + 1) ⌊ n/ ⌋ n − when q ≡ G/Z ( G ) = PSU ( q ), q > µ ( G ) = q ( q −
1) by [19, Table V], while | G | =2 | q + 1 | if q ≡ | G | = 4 | q − | if q ≡ | G | < µ ( G )unless q + 1 is a 2-power. In the latter case, µ ( G ) = ( q − q + 1)( q − G/Z ( G ) = PSU ( q ), q > µ ( G ) = ( q − q +1)( q +1)2 . Suppose first that4 | ( q + 1). We have | G | = | PSU ( q ) | ≤ q + 1) , whereas µ ( G ) = ( q + 1)( q − q + 1) / | G | < µ ( G ). Suppose now that 4 | ( q − | G | ≤ q − which is less than µ ( G ) = ( q − / ( q + 1). Now assume that G/Z ( G ) = PSU n ( q ) with n ≥ µ ( G ) = ( q n +1)( q n − − q )( q − q +1) , while | G | ≤ ( q − ( n − / n − if q ≡ | G | ≤ ( q + 1) n − n − if q ≡ G/Z ( G ) = PSU n ( q )with n ≥ µ ( G ) = ( q n − q n − +1)( q − q +1) , while the 2-part of | G | is bounded above as givenbefore. Again we can conclude. (cid:3) Proposition 11.3.
Let G be quasi-simple with G/Z ( G ) ∈ { PSL n ( q ) , PSU n ( q ) } with n ≥ and q > odd. (a) If n = 3 , then G has no Syl -regular character. (b) If n ≥ then G has no Steinberg-like character for p = 2 .Proof. By Lemma 2.5 (with P = G ), it suffices to prove the result in the case where | Z ( G ) | = 1.So we assume this, and then | G | equals the order of a Sylow 2-subgroup of G/Z ( G ).Let first G/Z ( G ) = PSL n ( q ), q > χ be a Syl -regular character for G . ByLemma 11.2, χ (1) < µ ( G ), and hence the non-trivial irreducible constituents of χ are of degree TEINBERG-LIKE CHARACTERS FOR FINITE SIMPLE GROUPS 29 ( q n − / ( q −
1) or ( q n − q ) / ( q − q n − / ( q −
1) are induced characters λ G , where λ = 1 P is a one-dimensional characterof the stabiliser P of a line of the underlying space for GL n ( q ), while the irreducible characterof degree ( q n − q ) / ( q −
1) is the unique non-trivial constituent τ of the permutation character1 GP = τ + 1 G on P .Let n ≥ g ∈ SL n ( q ) be a block-diagonal matrix, with an ( n − × ( n − F q n − with determinant 1, and a 2 × q + 1. Since g has no eigenvalue in F q , no conjugate of g is contained in P , so all induced characters from P to G vanish on g . In particular λ G ( g ) = 0 and τ ( g ) = − g ∈ G of g has even order, so χ ( g ) = 0 if χ is Steinberg-like. Write χ = x G + x τ + Λ, where Λ is a sum of x induced characters of degree ( q n − / ( q − x i ≥
0. Evaluating on g we see that x = x , but then χ (1) = ( x + x )( q n − / ( q − | G | . This proves part (b) for G/Z ( G ) = PSL n ( q ).Now assume that n = 4. Then an easy estimate shows that when q − | q − | ≤ ( q − /
3, and q = 7, then | G | < µ ( G ). So we may assume that in addition either q = 7 or q − q = 7 let g be the 2-element g = a a − ∈ SL ( q ) , where a ∈ F × q is a 2-element of order q −
1. Observe that again g is not conjugate to an elementof P , thus λ G ( h ) = 0 and τ ( h ) = −
1. As g is a 2-element and χ is Syl -regular, we have χ ( g ) = 0. We may now argue as above to conclude. When q = 7 then the candidate charactershave degrees 1, 399 and 400, while | G | = 2 = 512, so clearly there can be no Syl -regularcharacter.Now consider the case when G/Z ( G ) = PSL ( q ). The proof of Lemma 11.2 shows that µ ( G ) > | G | unless q − χ canhave degrees 1 , q + q, q + q + 1, while | G | = 2( q − . Clearly at most one of the degrees q + q, q + q + 1 can contribute to χ (1), but then necessarily q = 5. But in that case thecharacter table shows that there’s no Syl -regular character. This completes the proof of (a)when G/Z ( G ) = PSL n ( q ).Now let G/Z ( G ) = PSU n ( q ) with q > χ be a Syl -regular character of G .According to Lemma 11.2, χ (1) < µ ( G ), and hence the non-trivial irreducible constituents of χ are of degree ( q n − ( − n ) / ( q + 1) or ( q n + ( − n q ) / ( q + 1), see [19, Table V]. The first ofthese are semisimple characters lying in the Lusztig series of an element s of order q + 1 inthe dual group G ∗ = PGU n ( q ) with centraliser C G ∗ ( s ) ∼ = GU n − ( q ), the second is a unipotentcharacter, τ say, corresponding to the character of the Weyl group S n parametrised by thepartition ( n − , g ∈ SU n ( q ) be a regular element of even order in a maximal torus T oforder ( q − q n − − ( − n − ) / ( q + 1), see [10, Lemma 3.1(a)]. Then no conjugate of the dualmaximal torus T ∗ contains s , so the characters in E ( G, s ) vanish on g (see e.g. [11, Prop. 6.4]).If χ is Steinberg-like, then χ ( g ) = 0. As τ is unipotent, its value on g is (up to sign) the same as ψ ( h ) where ψ ∈ Irr( S n ) is labelled by ( n − ,
1) and h is a permutation of cycle shape ( n − , ψ ( h ) ∈ {± } , so τ ( g ) ∈ {± } . We may now argue as in the first part to conclude that χ cannotbe Steinberg-like, thus completing the proof of (b).Next, assume that G/Z ( G ) = PSU ( q ) with q > q + 1 is not a power of 2 and q = 5 , | G | < µ ( G ), as | q + 1 | ≤ ( q + 1) /
3. So now assume that q + 1 is a power of 2, and hence inparticular q ≡ | G | = 2( q + 1) , while the three smallest character degrees are1 , ( q − / ( q + 1) , ( q + q ) / ( q + 1), with the trivial character occurring at most once. It is easilyseen that there is no non-negative integral solution for a possible decomposition of χ . When q = 5 then the three smallest degrees are 1, 104, 105, while | G | = 128; if q = 9 then the threesmallest degrees are 1, 656, 657 while | G | = 512; so in neither case can there be Syl -regularcharacters either.Finally, when G/Z ( G ) = PSU ( q ) then again the proof of Lemma 11.2 shows that q + 1must be a 2-power. Here the possible constituents of χ have degrees 1 , q − q, q − q + 1, while | G | = 2( q + 1) . Again an easy consideration shows that at most the case q = 7 needs specialattention. But there the existence of Syl -regular characters can be ruled out from the knowncharacter table. (cid:3) We now treat the case q = 3, which is considerably more delicate. Lemma 11.4.
Let G = PSL (3) or PSU (3) . Then G does not have reducible Syl -regularcharacters.Proof. For G = PSL (3) we have | G | = 16, and all irreducible characters of degree less than 16take non-negative values on class 4A, so there are no Syl -regular characters. For G = PSU (3)we have | G | = 32, and all irreducible characters have degree at most that large. Since thesmallest non-trivial character degree is 6, those of degrees 27 and 28 cannot be constituents ofa Syl -regular character χ . Thus, we need to consider the characters of degrees 1, 6, 7, 14, 21.Clearly those of degree 21 cannot occur either. As 32 ≡ (cid:3) Remark 11.5.
PSL (3) and PSU (3) both have irreducible Syl -regular characters, see Propo-sition 3.1. Lemma 11.6.
Let G = PSL (3) or PSU (3) . Let χ be a Syl -vanishing character of G . Then l ( χ ) ≥ .Proof. Suppose the contrary. Note that χ is reducible. As | G | = 128, we have χ (1) ≤ | G | · τ be an irreducible constituent of χ of maximal degree. For all numerical data see [1].Let G = PSL (3). Then τ (1) <
351 (otherwise τ (1) = 351, so ( χ − τ )(1) = 33, and then χ (2 B ) >
0, which is false).Let µ ∈ Irr( G ) and µ (1) < µ (4 B ) ≥ µ (1) = 90, and µ (4 B ) = 0 unless µ (1) = 52 or 260. Let µ (1) = 90. Then ( χ, µ ) >
0. Indeed, otherwise the irreducible constituentsof χ are of degree 52 or 260, which implies χ (2 A ) >
0, a contradiction.It follows that ( χ − µ )(1) ≤ σ ∈ Irr( G ) with σ (1) = 39. The irreducible characters of G of degree at most 294 and distinct from σ are non-negative at 2 A . In addition, σ (2 A ) = − µ (2 A ) = 10. As χ (2 A ) = 0, it follows that ( χ, σ ) ≥
10. Then χ (1) ≥ µ (1) + 10 σ (1) > TEINBERG-LIKE CHARACTERS FOR FINITE SIMPLE GROUPS 31
Let G = PSU (3) and µ ∈ Irr( G ) with ( χ, µ ) >
0. The irreducible characters of degree atmost 384 are of degree at most 315. If τ (1) = 315 then µ (1) ≤
69. Then τ (2 A ) > µ (2 A ) >
0, a contradiction.Suppose that τ (1) = 280. Then µ (1) ≤ A .The same consideration rules out τ (1) = 210.Suppose that τ (1) = 189. It occurs once as otherwise 1 G occurs 6 times, which is false as( χ, G ) ≤
3. Then µ (1) = 140 ,
90, so µ (4 A ) > τ (4 A ) >
0, a contradiction. No more optionexists, as the irreducible characters of degree less than 189 are positive on 2 A . (cid:3) Lemma 11.7.
Let
PSL (3) ≤ G ≤ PGL (3) or PSU (3) ≤ G ≤ PGU (3) . Let χ be a -vanishing character of G . Then l ( χ ) ≥ .Proof. By Lemma 2.9(a), χ = ψ G , where ψ is a proper character of G ′ . By inspection of thecharacter table of G ′ , see [1], it is easily checked that the conjugacy class of any element g ∈ G ′ of 2-power order is G -invariant. Then, by Lemma 2.9(b), ψ ( g ) = 0 for every 2-element g = 1 of G ′ . This means that ψ is Syl -vanishing. By Lemma 11.6, l ( ψ ) ≥
4. Then l ( ψ G ) ≥ (cid:3) Lemma 11.8.
Let G = PGL (3) , and let χ be a -vanishing character of G . Let η , . . . , η k be the irreducible constituents of χ disregarding multiplicities, and η = η + · · · + η k . Then η (1) ≥ | G | .Proof. Let G ′ = PSL (3). Then | G ′ | = 128 and thus 2 · | G | = 512. Suppose the contrary. Then η (1) < η i (1) ≥ η j (1) for 1 ≤ i < j ≤ k . Note that k >
1, otherwise χ = aη for some a , and hence η is a 2-vanishing character of G and η (1) is a multiple of | G | .By [1], G has no character of degree at most 512 with this property.By [1], we have η (1) ≤ G ′ of degree at most 468extend to G = G ′ · χ , χ of degree 260, χ , χ of degree 234, χ , χ of degree 65and χ , χ of degree 26. The corresponding characters of G are of degrees 520, 468, 130 and 52,respectively. Let χ = P a i η i , where a i > η (1) = 468. Then P i> η i (1) ≤
44. Computing χ (4 B ) we get a contradiction(as η (4 B ) = 0 and P i> a i η i (4 B ) > k > η (1) = 416. If η (1) ≤
90 then P i> η i (1) ≤
6, whence k = 3 and η = 1 G .Computing χ (2 B ) we get a contradiction.So η (1) ≤
52. If η (1) = 52 then P i> η i (1) ≤
44. Computing χ (4 B ) we get a contradiction,unless k = 2 and η (4 B ) = 0 (that is, η = χ in [1]). In this case computing χ (4 C ) gives acontradiction. So η i (1) ≤
39 for i >
1. This violates χ (2 B ) = 0.(iii) Let η (1) = 390. Then η (1) ≤
90. If η (1) = 90 then η (1) ≤
32, whence k = 1and η (1) = 1. This conflicts with χ (4 A ) = 0. If η (1) ≤
52 then computing χ (4 B ) yields acontradiction.(iv) Let η (1) = 351 so P i> η i (1) ≤ η (1) = 130 then P i> η i (1) ≤
31, and hence η (1) = 1, which yields a contradiction with χ (20 A ) = 0. Let η (1) ≤
90. Then η i (1) ≤
71 for i >
2. Then η i (2 A ) + η i (2 B ) > i = 1 , . . . , k , which violates χ (2 A ) + χ (2 B ) = 0.(v) Let η (1) = 260. Then η (2 A ) ≥ λ of degree lessthan 260 with negative value at 2 A has λ (1) = 39. It follows that ( χ, λ ) ≥
0, and then η i (1) ≤ i > η i = λ . Note that λ (8 A ) = 1 and η (8 A ) = 0. As χ (8 A ) = 0, it follows that acharacter of degree 130 occurs in χ , which implies η (1) = 130. Then we get contradiction to χ (2 A ) + χ (4 B ) = 0. (vi) Let η (1) ≤ χ (2 A ) + χ (4 B ) leads to a contradiction. (cid:3) Lemma 11.9.
Let G = PGU (3) , and χ a -vanishing character of G . Let η , . . . , η k be theirreducible constituents of χ disregarding multiplicities, and η = η + · · · + η k . Then η (1) ≥ | G | .Proof. Let G ′ = PSU (3). Then | G ′ | = 128, | G | = 512 and 2 · | G | = 1024. Suppose thecontrary. Then η (1) < η i (1) ≥ η j (1) for 1 ≤ i < j ≤ k .Note that χ · τ = χ for every linear character τ of G . Therefore, η i τ is a constituent of η . Let g ∈ G \ G ′ . Then η i τ = η implies η i ( g ) = 0.By [1], if 630 = η i (1) >
420 then η i (4 E ) = 0 or η i (4 G ) = 0; it follows that η must contain atleast 2 representations of the same degree, which contradicts η (1) ≤ η i (1) either equals 630 or η i (1) ≤ η i (2 A ) + η i (4 B ) > η i , unless η i (1) = 210. This violates χ (2 A ) + χ (4 B ) = 0 unless k = 1 and η (1) = 210. Then χ (2 A ) > (cid:3) Lemma 11.10.
Let H = H × · · · × H n , where H ∼ = · · · ∼ = H n ∼ = PGL (3) or PGU (3) . Let χ be a -vanishing character of H. Then χ (1) ≥ n +1 | H | .Proof. By Lemma 11.7, the claim holds for n = 1, so by induction we can assume that it is truefor X := H × · · · × H n . By Lemma 2.7, χ = P i η i σ i , where η i ∈ Irr( H ), σ i are 2-vanishingcharacters of X and χ ′ = P i l ( σ i ) η i is a 2-vanishing character of H . By induction, σ i (1) ≥ n | X | . By Lemmas 11.8 and 11.9 applied to χ ′ , we have P i η i (1) ≥ | H | , so χ (1) ≥ n +1 | H | by Lemma 2.9. (cid:3) Proposition 11.11.
Let n > and G = GL n (3) or GU n (3) . Let χ be a -vanishing characterof G . Then l ( χ ) ≥ .Proof. Let X be the direct product of n copies of GL (3) or GU (3). Let χ be a 2-vanishingcharacter of X . By Lemmas 2.5 and 11.10, χ (1) ≥ n +1 | X | .Let Y = X · S n , the semidirect product, where S n acts on X by permuting the factors. Then Y contains a Sylow 2-subgroup of G . Let M = X · S , where S ∈ Syl ( S n ), so the index | G : M | is odd. Note that | G | = | X | · | S n | . As | S n | ≤ n − (see the proof of Proposition 6.7), theresult follows for these groups. (cid:3) Theorem 11.12.
Let p = 2 , m > and G one of GL m (3) , SL m (3) , PSL m (3) , GU m (3) , SU m (3) , PSU m (3) . Then G has no Steinberg-like character. Moreover, if χ is a -vanishing character of G then l ( χ ) ≥ .Proof. Let first G = GL m (3) or GU m (3). For m ≡ m = 4 n + l , where 1 ≤ l <
4, and H = GL n (3) or GU n (3). Let S be a Sylow2-subgroup of GL l (3) or GU l (3) and set U = H × S . Then U contains a Sylow 2-subgroup of G .Therefore, l ( χ ) = l ( χ | U ). By Lemma 2.5, if ν is a 2-vanishing character of U then l ( ν ) = l ( µ )for some 2-vanishing character µ of H . So l ( ν ) ≥ G = SL m (3) or SU m (3) the result follows from the above and Lemma 2.12. For G =PSL m (3) or PSU m (3) the statement follows from the above and Lemma 2.5. (cid:3) Orthogonal and symplectic groups at p = 2 . Let V be the natural module for H =Sp n ( q ), q odd, and for g ∈ G let d ( g ) be the dimension of the fixed point subspace of g on V . Let TEINBERG-LIKE CHARACTERS FOR FINITE SIMPLE GROUPS 33 ω n denote the Weil character of H . By Howe [7, Prop. 2], | ω n ( g ) | = q d ( g ) / . Let ω n = ω ′ n + ω ′′ n ,where ω ′ n , ω ′′ n ∈ Irr( H ) and ω ′ n ( z ) = − ω ′ n (1) for 1 = z ∈ Z ( H ). Lemma 11.13. (a)
Let h ∈ H be semisimple such that h and zh fix no non-zero vector on V.Then | ω ′′ n ( h ) | ≤ . (b) Let V = V ⊕ V , where V is a non-degenerate subspace of dimension , let g ∈ H be anelement such that gV i = V i , i = 1 , , g | V = − Id , and g and zg fix no non-zero vector on V .Then | ω ′′ n ( g ) | ≥ ( q − / . (c) Let q > . Then ω ′′ n is not constant on the -singular elements of PSp n ( q ) .Proof. (a) We have ω n ( h ) = ω ′ n ( h ) + ω ′′ n ( h ) and ω n ( zh ) = − ω ′ n ( h ) + ω ′′ n ( h ). Therefore, by [7,Prop. 2], 2 ≥ | ω n ( h ) + ω n ( zh ) | = | ω ′′ n ( h ) | , whence the claim.(b) We have | ω n ( g ) | = 1 and | ω n ( zg ) | = q by [7, Prop. 2]. Then q − ≤ | ω n ( g ) + ω n ( zg ) | = | ω ′′ n ( g ) | , whence the claim.(c) Choose g as in (b) and h to be an element stabilising V , V such that h coincides with g on V and the matrix of h on V is similar to (cid:18) − (cid:19) . Then h is a 2-singular elementsatisfying (a), and hence | ω ′′ n ( h ) | ≤
1. Let h and g be the images of h, g in H/Z ( H ). Then h and g are 2-singular elements of PSp n ( q ). As Z ( H ) is in the kernel of ω ′′ n , this can be viewedas a character of H/Z ( H ). As ( q − / q >
3, (c) follows. (cid:3)
Proposition 11.14.
Let G = PSp n ( q ) , with q > odd and n ≥ . Then G has no Syl -regularcharacters.Proof. We have | PSp n ( q ) | = ( | q − | n · n − · | n ! | if 4 divides q − , | q + 1 | n · n − · | n ! | if 4 divides q + 1 . Let k be minimal with n ≤ k , then as | n ! | ≤ | k ! | = 2 k − we have | PSp n ( q ) | ≤ ( | q − | n · n − if 4 | ( q − , | q + 1 | n · n − if 4 | ( q + 1) . On the other hand µ ( G ) = ( q n − q n − q ) / (2( q + 1)) by [19, Thm. 5.2], and this is larger than | G | , unless n = 2 and q = 5 ,
7. Let’s set aside these cases for a moment. Then otherwise if χ isSyl -regular, the constituents of χ are either Weil characters or the trivial character. Now notethat a Weil character of Sp n ( q ) of degree ( q n ± / χ have degree ( q n − / q ≡ n is odd, and ( q n + 1) / χ . As ( q n ± / n ≥ q (consider a Zsigmondyprime divisor), the trivial character must occur exactly once. Let ψ , ψ denote the two Weilcharacters of G , interchanged by the outer diagonal automorphism γ of G . Observe that γ isinduced by an element of GL n ( q ) and thus fixes all involution classes of G . Let g ∈ G be aninvolution and write a := ψ ( g ) = ψ ( g ). Then χ ( g ) = ma + 1, where m is the number ofnon-trivial constituents of χ . As necessarily m > χ ( g ) = 0,so χ is not Syl -regular. We now discuss the two exceptions. For G = PSp (5), | G | = 2 = 64 and all irreduciblecharacters of degree at most 64 take non-negative values on class 2B, so there is no Syl -regularcharacter. For G = PSp (7), | G | = 2 = 256 and all irreducible characters of degree at most 256take positive values on class 8A, except for one of degree 175 which takes value −
1, and one ofdegree 224 which takes value 0. As at most one of those latter two characters could occur, andat most once, there can be no Syl -regular character for p = 2. (cid:3) Proposition 11.15.
Let G = Ω n +1 ( q ) with q > odd and n ≥ . Then G has no Syl -regularcharacters.Proof. According to [19, Thm. 6.1] we have µ ( G ) = ( q n − / ( q − | G | unless either n = 3 and q = 7, or n = 4 and q = 5 , G = Ω (7) the only non-trivial character of degree less than | G | = 2 is the semisimplecharacter of degree 2451 (see [19]). Since the trivial character can occur at most once in a Syl -regular character, we see that no example can arise here. For G = Ω (5) the only non-trivialcharacter of degree less than | G | = 2 is the semisimple character of degree 16276 (see [19]).Again, this does not lead to an example. For G = Ω (7) the only non-trivial character of degreeless than | G | = 2 is the character of degree 120100, and we conclude as before. (cid:3) Proposition 11.16.
Let G = PΩ ± n ( q ) with q > odd and n ≥ . Then G has no Syl -regularcharacters.Proof. The second smallest non-trivial character degree of G = PΩ +2 n ( q ) is given by µ ( G ) =( q n − q n − − / ( q + 1) / | G | unless ( n, q ) = (4 , -regular character of G is a multipleof the smallest non-trivial character, of degree ( q n − q n − + q ) / ( q − G = PΩ +8 (7) the constituents of a Syl -regular charactercould have degree 1, 17500, or 51300. No non-negative integral linear combination of these threedegrees, with 1 G appearing at most once, adds up to | G | = 2 = 65536.The second smallest non-trivial character degree of G = PΩ − n ( q ) is µ ( G ) = ( q n + 1)( q n − +1) / ( q + 1) / | G | . We conclude as before. (cid:3) Again, we are left with the case that p = 2, q = 3. Lemma 11.17.
Let G = PSp (3) , Ω (3) or PΩ − (3) . Then G has no Steinberg-like character.Proof. For G = PSp (3) we have | G | = 2 , and all irreducible characters of G of degree atmost 512 take positive value on the class 4 A , see [1].Let G = Ω (3). Then | G | = 2 . All irreducible characters of G of at most that degree arepositive at the elements of conjugacy class 2 B [1].Let G = PΩ − (3). By [1], G has 8 irreducible characters of degree at most | G | = 2 . All ofthem take positive values on class 2A. So the result follows.This implies the result. (cid:3) Lemma 11.18.
Let G = PGO +8 (3) or PSp (3) . Let χ be a -vanishing character of G , and η , . . . , η k the irreducible constituents of χ disregarding their multiplicities. Set η = η + · · · + η k .Then η (1) ≥ | G | . TEINBERG-LIKE CHARACTERS FOR FINITE SIMPLE GROUPS 35
Proof.
Suppose first that G = PGO +8 (3). Note that | G ′ | = 2 , so 2 · | G | = 2 = 32768.Suppose the contrary. Then η (1) < G ′ of degree less than 32768, the maximal degree among them is 29120.There is only one irreducible character of G ′ of degree less that 32768 that is negative at 4 A (this is of degree 9450), while all other are positive. So it must be a constituent of η | G ′ . Thischaracter extends to G , so the other constituents of η are of degrees at most 32768 − η i (1) ≤ η i (1) < η i (1) = 18200 and η j (1) = 9450for some i = j then η l for l = i, j are of degree at most 23318 − A (as well as those of degree 18200 and 9450). This violates χ (2 A ) = 0.Thus, η i (1) < η i (1) ≤ G by a program in the computer package GAP, one observes that there are 4 distinct irreduciblecharacters of degree 17550, and only one irreducible character of this degree for G ′ . It follows thatthese 4 characters differ from each other by multiplication by a linear character. As | G/G ′ | = 4,one observes that χ · λ = χ for every linear character λ of G . Therefore, η i · λ must be aconstituent of χ . So, if η i (1) = 17550 then there are 3 more constituents of η of this degree,which contradicts the inequality η (1) < η i (1) < i = 1 , . . . , k . By [1], all such irreducible characters of G ′ , and henceof G , are positive at 4 A , which contradicts χ (4 A ) = 0.Let G = PSp (3). Note that | G | = 2 , so 2 · | G | = 2 = 32768. Suppose the contrary.Then η (1) < A , which violates χ (4 A ) = 0. (cid:3) Lemma 11.19. (a)
Let H = H × · · · × H n , where H ∼ = · · · ∼ = H n ∼ = PGO +8 (3) or PSp (3) .Let χ be a -vanishing character of H . Then χ (1) ≥ n | H | . (b) Let G = G × · · · × G n , where G ∼ = · · · ∼ = G n ∼ = GO +8 (3) or Sp (3) . Let χ be a -vanishingcharacter of G. Then χ (1) ≥ n | G | .Proof. (a) If n = 1 then the result is contained in Lemma 11.18. So by induction we can assumethat it is true for X := H × · · · × H n . By Lemma 2.7, χ = P i η i σ i , where η i ∈ Irr( H ), σ i are2-vanishing characters of X and χ ′ = P l ( σ i ) η i is a 2-vanishing character of H . By induction, σ i (1) ≥ n − | X | . By Lemma 11.18 applied to χ ′ , we have P i η i (1) ≥ | H | , so χ (1) ≥ n | H | by Lemma 2.9.(b) This follows from (a) and Lemma 2.5, as Z ( G ) is a 2-group. (cid:3) Lemma 11.20.
Let G = GO +8 n (3) , Ω +8 n (3) , PΩ +8 n (3) , Sp n (3) or PSp n (3) . Then G has noSteinberg-like character for p = 2 .Proof. Let X be the direct product of n copies of GO +8 (3) or Sp (3). Let ν be a 2-vanishingcharacter of X . By Lemma 11.10, ν (1) ≥ n | X | .Let Y = X · S n , the semidirect product, where S n acts on X by permuting the factors. Then Y contains a Sylow 2-subgroup of G . Let M = X · S , where S ∈ Syl ( S n ), so the index | G : M | is odd. Note that | G | = | X | · | S n | . As | S n | ≤ n − (see the proof of Proposition 6.7), theresult follows for the groups GO +8 n (3) and Sp n (3).For G = Ω +8 n (3) the result follows from the above and Lemma 2.12. For G = PΩ +8 n (3) orPSp n (3) the statement follows from the above and Lemma 2.5. (cid:3) Proposition 11.21.
Let m ≥ and G = GO +2 m (3) or Sp m (3) . Then G has no Steinberg-likecharacter for p = 2 . Proof.
For m ≡ m = 4 n + l , where 1 ≤ l < H = GO +8 n (3) or Sp n (3). Let S be a Sylow 2-subgroup of GO +2 l (3) or Sp l (3). Set U = H × S . Then U contains a Sylow 2-subgroup of G . Let χ be a 2-vanishing character of G .Therefore, l ( χ ) = l ( χ | U ). By Lemma 2.5, if ν is a 2-vanishing character of U then l ( ν ) = l ( µ )for some 2-vanishing character µ of H . By Lemma 11.20, l ( µ ) ≥
2. So l ( ν ) ≥
2, and the resultfollows. (cid:3)
Proposition 11.22.
Let G = GO − m (3) , m ≥ . Then G has no Steinberg-like character for p = 2 .Proof. Let m = 4 n + l , where 1 ≤ l ≤
4, and let H = GO +8 n (3). Then G contains a subgroup D isomorphic to H × GO − l (3). Then one concludes that D contains a Sylow 2-subgroup of G . Let S be a Sylow 2-subgroup of GO − l (3). Set U = H × S . Then U contains a Sylow 2-subgroup of G . By Lemma 2.5, if ν is a 2-vanishing character of U then l ( ν ) = l ( µ ) for some 2-vanishingcharacter µ of H . So l ( ν ) ≥ (cid:3) Proposition 11.23.
Let G = GO m +1 (3) , m ≥ . Then G has no Steinberg-like character for p = 2 .Proof. The case m = 3 is dealt with in Lemma 11.17. So we assume that m >
3, that is,2 m + 1 ≥
9. Let m = 4 n + l , where 0 ≤ l ≤
3, and let H = GO +8 n (3). Then G contains asubgroup D isomorphic to H × GO l +1 (3). Then D contains a Sylow 2-subgroup of G . Set U = H × S , where S is a Sylow 2-subgroup of GO l +1 (3), so U contains a Sylow 2-subgroupsof G . By Lemma 2.5, if ν is a 2-vanishing character of U then l ( ν ) = l ( µ ) for some 2-vanishingcharacter µ of H . So l ( ν ) ≥ (cid:3) Theorem 11.24.
Let G = GO m +1 (3) with m ≥ , GO ± m (3) with m ≥ , or Sp m (3) with m ≥ , and let G ′ be the derived group of G. Let H be a group such that G ′ ≤ H ≤ G . Then H and H/Z ( H ) have no Steinberg-like character for p = 2 .Proof. For H the statement follows from Lemma 11.17, Propositions 11.21, 11.22 and 11.23using Lemma 2.12, and for H/Z ( H ) from Lemma 2.5. (cid:3) We now collect our results to prove our main theorems from the introduction.
Proof of Theorem . . Assume that G is a finite non-abelian simple group possessing a Steinberg-like character χ with respect to a prime p . The cases when χ is irreducible have been recalledin Proposition 3.1. If Sylow p -subgroups of G are cyclic, then ( G, p, χ ) is as in Proposition 4 . p -subgroups of G are non-cyclic. For G alternating and p odd there are no cases by Theorem 6.4 except for A ∼ = PSL (9) with p = 3. The Steinberg-likecharacters of sporadic groups are listed in Theorem 5.1.Thus G is of Lie type. The case when p is the defining prime was handled in [17] andPropositions 8.1 and 8.2, respectively. So now assume p is not the defining prime for G . Groupsof exceptional Lie type were handled in Theorem 7.1. For classical groups of large rank with p odd, our result is contained in Proposition 9.6, the cases for PSL n ( q ) and PSU n ( q ) with p > p = 2 are covered by Proposition 11.1 for G = PSL ( q ),Proposition 11.3 for PSL n ( q ) and PSU n ( q ) with q = 3, Theorem 11.12 for PSL n (3) and PSU n (3),Propositions 11.14, 11.15 and 11.16 for classical groups with q = 3, and Theorem 11.24 for thecase that q = 3. (cid:3) TEINBERG-LIKE CHARACTERS FOR FINITE SIMPLE GROUPS 37
Proof of Theorem . . The characters of projective F p G -modules of dimension | G | p are in par-ticular Steinberg-like, so in order to prove this result we need to go through the list given inTheorem 1.1(2)–(5). When Sylow p -subgroups of G are cyclic, the possibilities are given inLemma 4.5(b). For G of Lie type in characteristic p , see [22, Thm. 1.1]. The case (4) of Theo-rem 1.1 is subsumed in statement (1), and finally the alternating groups for p = 2 are discussedin Theorem 6.14. (cid:3) References [1]
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