E-mail: wangj~szxO.math.pku.edu.cn. Abstract In this paper we have completely determined: (1) all almost simple groups which act. 2-transitively on one of their ...
Acta Mathematica Sinica, English Series 1999, Jan., Vol.15, No.l, p. 131-144
Acta Mathematica Sinica, English Series 9 Springer-Vedag1999
Finite Groups Acting 2-Transitively on Their Sylow Subgroups
Caiheng
Li
Department of Mathematics, the University of Western Australia, Nedlands, WA, 6907, Australia E-mail: li~maths.uwa.edu.au Jie
Wang
Department of Mathematics, Peking University, Beijing 100871, P. R. China E-mail: wangj~szxO.math.pku.edu.cn
Abstract In this paper we have completely determined: (1) all almost simple groups which act 2-transitively on one of their sets of Sylow p-subgroups. (2) all non-abelian simple groups T whose automorphism group acts 2-traasitively on one of the sets of Sylow p-subgroups of T. (3) all finite groups which are 2-transitive on all their sets of Sylow subgroups. K e y w o r d s Finite group, Sylow subgroup, 2-transitive action 1 9 9 1 M R S u b j e c t C l a s s i f i c a t i o n 20D20, 20E07
1
Introduction
T h r o u g h o u t this p a p e r the w o r d "group" always m e a n s a finite group.
In [1] B r e w s t e r a n d
W a r d asked, a m o n g others, t h e following questions: (1) W h i c h nonsolvable g r o u p s act 2 - t r a n s i t i v e l y on one of t h e i r sets of Sylow p - s u b g r o u p s ? (2) $4 acts 2 - t r a n s i t i v e l y on all its sets of Sylow p - s u b g r o u p s . W h a t o t h e r g r o u p s have this property? In this p a p e r we t r y to answer these questions. To be precise, we shall d e t e r m i n e : (1) all a l m o s t simple groups which act 2 - t r a n s i t i v e l y on one of t h e i r sets of Sylow p - s u b g r o u p s ( T h e o r e m 1.1). Received April 24, 1995, Accepted November 11, 1997 The first author acknowledges the support of OPR Scholarship of Australia The second author is supported by the National Natural Science Foundation of China. Thanks are also due to the Department of Mathematics, the University of Western Australia, where he did his part of this work for its hospitality
Caiheng Li ~z Jie Wang
132
(2) all non-abelian simple groups T whose automorphism group is 2-transitively on one of the sets of Sylow p-subgroups of T (Theorem 1.2). (3) all finite groups which are 2-transitive on all their sets of Sylow subgroups
Theorem
1.4). Our first two results are stated as follows. T h e o r e m 1.1
Let T ~ G 1) for p = 2; (e) T -- 2G2(32~+1) (a _> 1) and 3 ~ [G/T[ f o r p = 3; (f) G = 2G2(3) ----PSL(2,8).3 forp-= 3. Remark
Notice that PSL(2, 5) ~ A5 and PSL(2, 9) ~- As. Hence A5 and $5 are 2-transitive
on their Sylow 5-subgroups and, for each G: A6 ~ G _< Aut(A6), G is 2-transitive on its Sylow 3-subgroups. T h e o r e m 1.2
Let T be a nonabelian simple group and A = Aut(T). For some prime p ]T],
A acts 2-transitively on the set of Sylow p-subgroups of T by conjugation, if and only if T is one of the following groups: (a) A5 f o r p = 2; (b) P S L ( 2 , p f ) ; (c) PSV(3,p2f); (d) 2B2(22~+1) (a > 1) f o r p = 2; (e) 2G2(32~+1) (a > 1) f o r p = 3; (f) PSL(2,8) f o r p = 3. Remark
PSL(2, 8) is the only case where T itself is not 2-transitive on the set of its Sylow
p-subgroups while its automorphism group is. In order to state the third result, we need a definition. D e f i n i t i o n 1.3
For a group G, let C(G) be the subgroup of G generated by all normal p-
subgroups Hp such that G/CG(Hp)Hp is a p-group for all prime p
]G[.
It is easy to see that C(G) is nilpotent and acts trivially on all the sets of Sylow subgroups of G by conjugation. With this definition we answer question (2) of Brewster and Ward as follows. T h e o r e m 1.4
A finite group G acts 2-transitively on all its sets of Sylow subgroups if and
only if G / C ( G ) --- G1 x G2 X . . . •
such that
(i) for any Sylow subgroup S of G / C ( G ) , there is at most one i : 1 < i < rn such that S N G i
is not normal in G / C ( G ) . (ii) each Gi (1 < i < m) is a soluble 2-transitive group of degree ni and one of the following
holds: (a) ni = pd and Gi = Z~.L, d. where p is prime, L is nilpotent and p ILl. (b) nl = 2d and Gi = zd:(L:Z2t), where 2 t d, 2 t + 1 is a Fermat prime, L is nilpotent and [L, Z2t I = Z2,+1. (c) ni = 3d and Gi = zd:(L:Z3), where 3 d, n is nilpotent and [L, Z3] = Z 2. (d) ni = 2d and Gi = zd:(L:Zs), where 8 d, L is nilpotent and [L, Zs] = Z32.
Finite Groups Acting 2-Transitively on Their Sylow Subgroups
133
(e) ni = 32 and Gi = Z32 :Qa or Z~:(Qs:Z3). The notations and terminology used in this paper are standard (see, for example, [2] and [3]). For two groups K and H, K . H is an arbitrary extension of K by H while K : H stands for a split one. For a positive integer n, Zn stands for a cyclic group of that order while Z d stands for a direct product of d copies of Z,~. Moreover, the notation In] stands for an arbitrary group of order n. For a group G, Sylp(G) stands for the set of all Sylow p-subgroups of G and Op(G) for the maximal normal p-subgroup of G. For a subgroup H of G, H char G means that H is a characteristic subgroup of G. Let ~r be a set of primes. We use rd to denote the complement set of 7r in the set of all prime numbers. For a group G, the notations G~ and G~, stand for the Hall 7r-and 7rr-subgroups of G respectively. Finally we make the following definition for the sake of conciseness.
A group G is said to have the property 7) if it is 2-transitive on all of its sets
D e f i n i t i o n 1.5
of Sylow subgroups. In Section 2 we give some useful lemmas. Sections 3 and 4 are devoted to proving Theorems 1.1, 1.2 and 1.4 respectively.
2
Preliminaries
The following simple lemma enables us to use induction on the group order. L e m m a 2.1
If G is a finite group having the property 79and N is a normal subgroup of G,
then G / N has the property 79. Proof
For any prime p
]G/NI, if a Sylow p-subgroup of G / N is normal, then clearly G / N is
2-transitive on Sylv(G/N). So we assume that the Sylow p-subgroups of G / N are not normal in G / N . Let P1, P2 and ~ be three Sylow p-subgroups of G / N . Then there are three Sylow psubgroups P1, P2, P3 of G such that Pi = P i N / N for i = 1, 2, 3. Since G is 2-transitive on its set of Sylow p-subgroups, there is an z E G such that P~ = P1 and P~' -- P3 and so ( P I N ) x = P~N and (P2N) x = P3N. Thus there is an 5 E G / N such that P-~I~ = Pll and ~22~ = P33. So G / N is 2-transitive on the set of all Sylow p-subgroups of G / N . L e m m a 2.2
Suppose that G = A x B. Then G has the property 79 if and only if, for each
Sylow subgroup S of G, at least one of S N A and S N B is normal in G, and A and B have the property 7). Proof
Assume that G = A x B acts 2-transitively on all of its sets of Sylow subgroups. By
Lemma 2.1, A and B have the property 79. Suppose that there is a Sylow p-subgroup S of G such that neither S A A nor S N B
i s n o r m a l i n G. Let P1 r P2 _< A and Q1 r Q2 _< B be
different Sylow p-subgroups of A and B respectively. Now R1 = P1 x Q1, R2 = P1 x Q2 and R3 = P2 x Q2 are three Sylow p-subgroups of G. However there is no g E G such that Rf = R1 and Rg = R3, a contradiction.
134
Caiheng Li & Jie Wang Conversely, for any three Sylow p-subgroups P~ of G (i -- 1, 2, 3), we have P~ = (P~ n
A) • ( P i n B). Without loss of generality, we may assume that P1 n A _ 1, then by (1) and (2), P { = Pi = Q~. Hence P1 = p [ - 1 = Q1, which
Caiheng Li &: Jie Wang
138
is a contradiction. So we have P~ = / ~
and T~ = 7"1, which implies that Q~ = R1. Therefore
we obtain an element in Aut(T1) which fixes P1 and maps Q1 to R1. It follows that Aut(T1) is 2-transitive on all the sets of Sylow subgroups of T1, contrary to L e m m a 4.1. Therefore, G is soluble. By L e m m a 4.2, in the remainder of this p a p e r we suppose that G is a finite soluble group having the property P. L e m m a 4.3
Let G be a soluble group and N ~ Z d a minimal characteristic subgroup of G.
If G has the property ~ and G / C c ( N )
is not a p-group, then the following hold:
(i) for any prime q ~ p, ifq I IG/CG(N)F then N is transitive on Sylq(G). (ii) G is transitive on N \ {1} by conjugation. (iii) / f N =- C G ( N ) then G is a 2-transitive group of degree n and one of the following holds: (a) n = pd, G = Zd:L, where p is prime, L is nilpotent and (p, ILl) = 1. (b) n = 2 d, G = Z~:(L:Z2t) where 2' d, 2 ~ + 1 is a Fermat prime, L is nilpotent and [L, Z2t] = Z2~+I. (c) n = 3 4, G = Zd3:(L:Z3), where 3 d, L is nilpotent and In, Z3] = Z 2. (d) n = 2 d, G = z d : ( n : z s ) where 8 d, L is nilpotent and [L, Zs] = Z 2.
(e) n = 9, G = Z2: Qs or Z2: Qs.3. Proof
Assume that G / C G ( N ) is not a p-group and q r p is a prime divisor of I G / C 6 ( N ) I . I f N
is not transitive on Sylq(G ) by conjugation, then there are two subgroups Q, Q1 c Sylq(G) such that Q1 r Q~ for any n E N. Let Q2 = Q'~ for some n E N \ {1}. Since G acts 2-transitively on Sylq(G), there is an element y e G such t h a t Qy = Q and Q~ = Q2- Let n' = yny -1. Then n' E N and
O ~'~ = Q ~ = ( Q ~ F = Q~ = Q2 = Q~. So Q'r = Q~, a contradiction. Thus N is transitive on Sylq(G) and (i) is true. For a fixed Q E Sylq(G), set K = n ~ e G N c ( Q ) ~ = N ~ e c N c ( Q ~ ) , the kernel of G on Sylq(G). Then K is a characteristic subgroup of G. If N A K r 1, the minimality of N leads to N _< K , contrary to (i). Thus N is faithful on Sylq(G). Notice that N is regular on Sylq(G) since it is abelian. It follows that Sylq(G) = { Q'~ [ n E N }. For any n and n~ E N \ {1}, since G acts 2-transitively on Sylq(G), there is an x E G such that Q~ = Q and (Qn~)~ = Q,~. Thus Q n~x'~-~z-~ = Q. Since n l x n - l x -1 E N and N is regular on Sylq(G), we obtain n l x n - l x -1 = 1 and hence x-~n~x = n. It follows that G is transitive on N \ {1} and (ii) is true. Next suppose that N = C a ( N ) and let Go = G / N .
T h e n Go is a subgroup of A u t ( N )
which acts irreducibly on N \ {1}. Identifying N with the vector space V = V(d, p), we conclude that G ~ V:Go is an affine 2-transitive group of degree n = pd and Go is the stabilizer of the zero vector in G.
Suppose that there is a prime q IG0l such that q r p and t h a t a Sylow
q-subgroup Sq of Go is not normal in Go. Let M be the kerneI of G on Sylq(G). Since N is minimal characteristic in G and N = C a ( N ) , if M r 1 then N < M. Thus (N, Sq) = N x Sq, so C a ( N ) > N x Sq, a contradiction. Hence M = 1 and G is faithful on Sylq(G). However NG(Sq) r Go because Sq is not normal in Go. So G is not 2-transitive on Sylq(G). Hence for
139
Finite Groups Acting 2-Transitively on Their Sylow Subgroups
every prime q [Go [, either q -- p or the Sylow q-subgroup of Go is normal in Go.
(3)
If Go ~ FL~(p d) then, by Theorem 2.5 (i) and (ii), Go is one of the following table: No.
Go = G / N
/
Qs, Qs:Z3, Qs:S3
32
Qs:Z3,
52
o Qs):Z3, (z4 o Qs):S3
qs:S3, (Z3 • Qs):S3
72
(Z5 x Qs):Z3, (Z5 x Qs):S3
112
( Z l l x Q8):$3
232
(Ds o Qs):Zs, (Ds o Qs):(Z5:Z2), (Ds o Qs):(Zs:Z4)
34
It is not hard to verify that Go satisfies condition (3) only when Go = Qs or Qs:Z3 for N -- Z32. So part (e) holds. Therefore we assume t h a t Go 1, then 1 ~ Z W 1 and Go = L:Zp, where L is a H a l l / - s u b g r o u p of Go. Let ~ = Sylp(G0)
and K the kernel of Go on ~. Then G o / K is a faithful 2-transitive group acting on ~. Notice that K normalizes all Sylow p-subgroups of Go. If K N Sp r 1 for some Sp E f~, then K N Sv is a non-trivial normal subgroup of Go, contrary to [2, II3.2]. Thus K is a / - s u b g r o u p of Go. Since L is the unique H a l l / - s u b g r o u p of Go, we have K < L and G o / K = Zq.Zp~, where --
q is a prime divisor of [L[ and Zq"is regular on Sylp(Go/K).
r.
It follows from part (ii) of the
theorem that Zp, is transitive on Z~ \ {1}. Thus Zp~ is regular on Z~ \ {1}, which implies that qr _ 1 = / .
By [3, IX2.7], one of the following holds:
(1) q = 2, t --- 1, r is a prime and p = 2 r - 1 is a Mersenne prime. (2) p -- 2, r = 1, t = 2 m and q -- 2 t + 1 is a Fermat prime, or (3) q r = 9 a n d p ~ = 8 . Furthermore, assume (1) holds and G o / K = Z~:Zp~. Let H be the preimage of Z~ in Go. Then H is a subgroup of FLI(p d) and hence can be generated by at most 2 elements. It follows that
H / K = Z~ can also be generated by at most 2 elements. Thus r ~ 2 and we get the conclusion that G o / K = Z2:Z3. To complete the proof of (iii) (b)-(d), it is sufficient to prove that [n, Zp~] ~- L / K . Notice that KZp~ = g x Zp, < NGo(Zp~). Moreover, IL__[ = Igl
=
= [ a 0 : NGo(Zp,)l -
ILIp' INco(Zp,)l
It follows that Cco(Zp,) = NGo(Zp,) = K x Zp, and hence CL(Zp*) = NL(Zp*) = K . Since (p, l L [ ) = 1, it is well known that L can be written as L = [L, Zp,]CL(Zp,) = [L, Zp,]K. For any l E L and h 6 Zp,, if the element x = l - l h - l l h
E K N [L, Zp,], then x fixes all Sylow
Caiheng Li & Jie Wang
140
p-subgroups of Go. In particular, x fixes Ztp, which implies that h E Nvo (Zp,)t = K x Z pt. t So h E Z ~ . If Z'v~ = Zp~ then l E Nao(Zp,) = Cao(Zp,). It follows that x = 1. So we assume that Z pt l # Zp, and h q Zp, F3 Zp~,. Now a = Zp, and/3 = Z ptl are two distinct points in ~ and
NGo(Zv~ ) is transitive on ~ \ {a}, which has size pt. Thus the stabilizer of/3 in NGo(Zv~ ) is of order coprime to p and hence p is not a divisor of [NGo(Zp~) n NGo(Z~)I. Notice that NGo(Zp,) N Nao(Z~,) = ( K • Zp,) n ( K • Zpl,) = K • (Zp, N Z~,). Therefore we have Zp, N Z~, = 1. It follows again t h a t x = 1. Hence we get the conclusion that
In, ZW] ~- L / K =" Z~. This completes the proof of L e m m a 4.3. Proof of Theorem 1.4
Now we use induction on the group order IGI to prove Theorem 1.4.
Suppose that Theorem 1.4 holds for all groups of orders less t h a n ]G]. Recall that C(G) is the subgroup of G generated by all normal p-subgroups Hp such that G/CG(Hp)Hp is a p-group for all prime p ] G I. Since G is soluble, every minimal subgroup N of G is elementary abelian. By L e m a n 2.1, G/N and G/Ca(N) also have the property P. If G is decomposable then it is clear that Theorem 1.4 is true by Lemmas 2.1-2.3. Hence we may assume that G is indecomposable. First assume that C(G) # 1. By L e m m a 2.3 (i), C(G) is characteristic in G. So we can take a p-subgroup N of C(G) such that N is a minimal characteristic subgroup of G. Since G is soluble, N - Z d for some positive d. It follows that G/Ca(N) is a p-group by L e m m a 2.3 (iv). Let G = G/N and C < G such that C/N = C(G). Then C is characteristic in G. We are going to prove t h a t C = C(G). By L e m m a 2.3 (ii), C/N = C(G) is nilpotent. Since
C/Cc(N) = C/(C N C a ( N ) ) ~ CCG(N)/CG(N) < G/CG(N), C/Cc(N) is a p-group. Applying L e m a n 2.4 to C for 7r = {p}, we conclude t h a t C is nilpotent. For any Sylow r-subgroup Cr of C, denote Cr = (C,N)/N. T h e n G/(CG(C~)Cr) -~ (G/N)/((CG(C~)Cr)/N) = G/C-~(C~)Cr, because (CG(Cr)C~)/N = C u ( C . ) C , .
(4)
Moreover, we have Cr char C char G since C is
nilpotent. It follows that Cr
