A characterization of automorphism groups of simple <span class="mathjax-formula">${K}_{3}$</span> -groups

A characterization of automorphism groups of simple K

-groups

DAPENGYU(*), JINBAOLI(**), GUIYUNCHEN(***), YANHENGCHEN(***)

ABSTRACT- In this paper, a new characterization of automorphism groups of simple K3-groups is presented.

MATHEMATICSSUBJECTCLASSIFICATION(2010). 20D05, 20D45.

KEYWORDS. SimpleK3-groups; conjugacy classes sizes; characterization.

(*) Indirizzo dell'A.: School of Mathematics and Statistics, Southwest Uni- versity, Chongqing 400715, China; also affiliated with Department of Mathematics, Chongqing University of Arts and Sciences, Chongqing 402160, China.

E-mail: yudapeng0@sina.com

Supported by the Scientific and Technological Research Program of Chongqing Municipal Education Commission (Grant No. KJ111207) and the Scientific Research Foundation of Chongqing University of Arts and Sciences (Grant No. Z2013SC10).

(**) Indirizzo dell'A.: Department of Mathematics, Chongqing University of Arts and Sciences, Chongqing 402160, China.

E-mail: leejinbao25@163.com

Supported by the Scientific Research Foundation of Chongqing Municipal Science and Technology Commission (Grant No. cstc2013jcyjA00034), the Scientific and Technological Research Program of Chongqing Municipal Education Commission (Grant No. KJ131204), the Scientific Research Foundation of Yongchuan Science and Technology Commission (Grant No. Ycstc, 2013nc8006).

(***) Indirizzo dell'A.: School of Mathematics and Statistics, Southwest Uni- versity, Chongqing 400715, China.

E-mail: gychen1963@163.com

Corresponding author and supported by the National Natural Science Founda- tion of China (Grant Nos. 11271301, 11171364).

(***) Indirizzo dell'A.: School of Mathematics and Statistics, Chongqing Three Gorges University, Chongqing 404100, China.

E-mail: math yan@126.com

Supported by the Scientific Research Foundation of Chongqing Municipal Science and Technology Commission (Grant No. cstc2014jcyjA00009), the Scientific and Tech- nologicalResearchProgramofChongqingMunicipalEducationCommission(GrantNo.

KJ1401006) and the Fundamental Research Funds for the Central Universities.

1. Introduction

All groups considered in this paper are finite.

It is known that the setcs(G) of sizes of the conjugacy classes of a finite groupGencodes much information about the structure ofG. Many authors have studied the connection between arithmetical properties ofcs(G) and structural properties ofG. The present paper is a contribution along this line, which is related to Thompson's conjecture (see [11, Problem 12.38]

and [2, 3, 4, 5, 12, 1] for detail).

CONJECTURE1.1. Let Gbe a group with Z(G)ˆ1 andM is a non- abelian simple group satisfying thatcs(G)ˆcs(M), thenG'M.

In this conjecture, M is a non-abelian simple group. Hence it seems interesting to consider the following question.

Let G be a group with Z(G)ˆ1 and M is an almost simple group satisfying that cs(G)ˆcs(M). Then, what can we say about the struc- ture of G?

A groupGis almost simple if there exists a non-abelian simple groupS such that S4G4Aut(S). In [9, 10], the almost sporadic groups, Aut(PSL(2;q)) andPGL(2;p) for some special cases are discussed. In this paper, we investigate the almost simpleK3-groups. A groupGis called aK3- group ifp(G) consists of exactly three distinct primes. Among the known simple groups, there are exactly eight simpleK₃-groups (see [7, p. 12]):

A₅; A₆; L₂(7); L₂(8); L₂(17); L₃(3); U₃(3); U₄(2):

Our main result is as follows.

THEOREM 1.2. Let G be a group with Z(G)ˆ1 and S be one of the simple K₃-groups. Set MˆAut(S).

(1) If S2 fA₅;A₆;L₂(7);L₂(8);L₂(17);L₃(3);U₃(3)gand cs(G)ˆcs(M), then G'M.

(2) If SˆU4(2), cs(G)ˆcs(M) and jGj_pˆ jMj_p, then G'M, where jGj_pdenotes the order of the Sylow p-subgroups of G for p2 f2;3;5g.

2. Preliminaries

Let Gbe a group and construct its prime graph G(G) as follows: the vertices are the primes dividing the order ofG, two verticespandq are

joined by an edge if and only ifGcontains an element of orderpq(see [13]).

We denote the set of all connected components of the graph G(G) by T(G)ˆ fp_i(G)j14i4t(G)g, where t(G) is the number of the connected components of G(G). If the order ofG is even, we always assume that 22p₁(G). Forx2G,x^Gdenotes the conjugacy class inGcontainingxand C_G(x) denotes the centralizer ofxinG. A groupGis called a 2-Frobenius groupifGhas a normal series 1HKGsuch thatK andG=Hare Frobenius groups with kernelsHandK=Hrespectively (see, for example [9, Definition 2.1]).

The other notation and terminologies in this paper are standard and the reader is referred to ATLAS [6] and [8] if necessary.

The following lemma is well-known. It follows from [4, Lemma 1.1] or [12, Lemma 3].

LEMMA 2.1. Let G and M be groups satisfying Z(G)ˆZ(M)ˆ1and cs(G)ˆcs(M). Thenp(G)ˆp(M).

LEMMA 2.2[4, Lemma 1.4]. Let G be a group with Z(G)ˆ1and M a group with t(M)>1. Suppose that cs(G)ˆcs(M). ThenjGj ˆ jMj.

LEMMA 2.3 [4, Lemma 1.5]. Let G and M be groups satisfying that jGj ˆ jMjand cs(G)ˆcs(M). Then t(G)ˆt(M)and T(G)ˆT(M).

LEMMA2.4. Suppose that G is a Frobenius group of even order and H, K are the Frobenius kernel and the Frobenius complement of G, respec- tively. Then t(G)ˆ2, T(G)ˆ fp(H);p(K)gand G has one of the following structures:

(i) 22p(H)and all Sylow subgroups of K are cyclic;

(ii) 22p(K), H is an abelian group, K is a solvable group, the Sylow subgroups of K of odd order are cyclic groups and the Sylow 2-subgroups of K are cyclic or generalized quaternion groups;

(iii) 22p(K), H is abelian, and there exists a subgroup K₀of K such that

jK:K0j42;K0ˆZSL(2;5);(jZj;235)ˆ1;

and the Sylow subgroups of Z are cyclic.

PROOF. This is Lemma 1.6 in [4]. p

LEMMA2.5. Let G be a2-Frobenius group of even order. Then t(G)ˆ2 and G has a normal series 1HKG such that p(K=H)ˆp2, p(H)[p(G=K)ˆp1, the order of G=K divides the order of the auto- morphism group of K=H, and both G=K and K=H are cyclic. Especially, jG=Kj<jK=Hjand G is solvable.

PROOF. This is Lemma 1.7 in [4]. p

LEMMA 2.6. Let G be a group with more than one prime graph com- ponent. Then G is one of the following:

(i) a Frobenius or2-Frobenius group;

(ii) G has a normal series1HKG, where H is a nilpotent p₁-group, K=H is a non-abelian simple group and G=K is ap₁- group such that jG=Kj divides the order of the outer auto- morphism group of K=H. Besides,pi(K=H)ˆpi(G)for i52.

PROOF. It follows straight forward from Lemmas 1-3 in [13], Lemma

1.5 in [3] and Lemma 7 in [5]. p

LEMMA 2.7 [9, Corollary 5.1]. Let G be a group with Z(G)ˆ1 and MˆAut(L2(q)), where G(M) is not connected. If cs(G)ˆcs(M), then G'M.

LEMMA2.8 [12, Lemma 4]. Suppose that G is a group with Z(G)ˆ1 and p is a prime inp(G)such that p² does not dividejx^Gjfor all x in G.

Then a Sylow p-subgroup of G is elementary abelian.

3. Proof of Theorem 1.2

LEMMA3.1. Let G be a group with Z(G)ˆ1and S2 fA5;L2(7);L2(8);L2(17);L3(3);U3(3)g:

Let MˆAut(S)and cs(G)ˆcs(M). Then G'M.

PROOF. By [6], we have thatG(M) is not connected if S2 fA5;L2(7);L2(8);L2(17);L3(3);U3(3)g:

Therefore, by Lemma 2.7,G'Mprovided that S2 fL₂(7);L₂(8);L₂(17)g:

Now, it suffices to discuss the remaining three cases.

First, we suppose thatSˆA5. Then G(M) is not connected. By the hypothesis and lemma 2.2, we obtain thatjGj ˆ jMj. Since p2(M)ˆ f5g by [6], it follows from Lemma 2.3 thatp2(G)ˆ f5g. Since 5 does not di- vide 23, by Lemma 2.4, we haveGis not a Frobenius group. Suppose that G is a 2-Frobenius group. Then, by Lemma 2.5,G has a normal series 1HKGsuch thatjK=Hj ˆ5. LetPbe a Sylow 3-subgroup ofH.

ThenPis a normal subgroup ofGof order 3. Letx2Gsuch thatjxj ˆ5.

Then xacts trivially on P and soG(G) is connected. This contradiction shows that Gis not a 2-Frobenius group either. Now, by lemma 2.6, G has a normal series 1HKG such that K=H is a non-abelian simple group. It is easy to see that K=His isomorphic toA₅. IfH6ˆ1, thenHis of order 2 and soH4Z(G), contrary to our assumption forG.

Hence Hˆ1 and thereforeK'A5. It follows thatG'S5, as desired.

Next, we assume that SˆL₃(3). Since G(M)ˆG(Aut(S)) is not con- nected, Lemma 2.2 together with the hypothesis imply thatjGj ˆ jMj. B y [6], we have thatf13gis a component ofG(M). Hence, by Lemma 2.3,f13gis a component ofG(G). Similar to above discussion, we can show thatGis not a Frobenius group. We assert thatGis not a 2-Frobenius group either. If not, then, by Lemma 2.5, G has a normal series 1HKG such that jK=Hj ˆ13. Let P be a Sylow 2-subgroup ofH andVˆV₁(Z(P)). Since (13;jGL_n(2)j)ˆ1, wherenˆ2;3;4;5, an elementxofGof order 13 act tri- vially onV. ThereforeGhas an element of order 26, a contradiction. ThusGis not a 2-Frobenius group. By Lemma 2.6,Ghas a chief factorK=Hsuch that K=His a simpleK3-group and 132p(K=H). By [6],K=Hmust be isomorphic toL₃(3). Similarlyas above,weobtainthatHˆ1 and soG'MˆAut(L₃(3)).

Finally, we prove that G'MˆAut(U₃(3)). As above, one can show that G is neither a Frobenius group nor a 2-Frobenius group. By [6], p2(M)ˆ f7gand sop2(G)ˆ f7gby Lemma 2.3. It follows from Lemma 2.6 that G has a chief factorK=Hsuch thatK=His a simpleK3-group and 72p(K=H). ThenK=His isomorphic toL₂(7),L₂(8) orU₃(3). LetP be a Sylow 2-subgroup ofH,Qa Sylow 3-subgroup ofHandxan element ofGof order 7. If K=H'L2(7), then jQj ˆ3². Since (7;jAut(Q)j)ˆ1, x acts trivially on Qand so 7 and 3 are joined, a contradiction. IfK=H'L2(8), then P is of order 2³ and so G has an element x such that jx^Gj 7, a contradiction by [6]. Hence K=H'U₃(3). It is easy to see that Hˆ1 since otherwise,His contained inZ(G), a contradiction. Thus,K'U₃(3)

andG'Aut(U₃(3)). p

LEMMA 3.2. Let G be a group with Z(G)ˆ1 and MˆAut(A6). If cs(G)ˆcs(M), then G'M.

PROOF. We proceed the proof by several steps.

(1) By [6],cs(G) consists ofn1ˆ1,n2ˆ3²5,n3ˆ2⁴5,n4ˆ23²5, n5ˆ2⁴3²,n6ˆ235,n7ˆ2⁴35,n8ˆ2²3²,n9ˆ2²3²5.

(2) By Lemma 2.1, we have thatp(G)ˆp(M)ˆ f2;3;5g.

(3) The Sylow 5-subgroups ofGare of order 5.

LetPbe a Sylow 5-subgroup ofG. Then, by Lemma 2.8,Pis elementary abelian since 5²does not divide any element incs(G). We assert thatPis of order 5. Assume that 5² divide the order of G. Then, the centralizer of every element ofGcontains an element of order 5. Let y2Gsuch that jy^Gj ˆ2⁴35 andxan element of C_G(y) of order 5. Since the Sylow 5- subgroup ofGis elementary abelian, we have that 5 does not dividejx^Gj and consequently x^Gˆ2⁴3² or 2²3². If 5 does not divide jyj, then C_G(xy)ˆC_G(x)\C_G(y), from which we can deduce that 2⁶3³5 divide (xy)^G, which is impossible. If 5 dividesjyj, thenyˆy₁y₂withjy₁j ˆ5 and (jy₁j;jy₂j)ˆ1. It follows thatjy^G₁jshould dividejy^Gj, a contradiction. Hence the Sylow 5-subgroup ofGis of order 5.

(4)O22⁰(G)ˆO2(G).

WriteKˆO2(G) and GˆG=K. Suppose that the statement is false.

Then there is r2 f3;5g such that Or(G)6ˆ1. If PˆO5(G)6ˆ1, then jPj ˆ5. Let Qbe a Sylow 3-subgroup ofG andx an element ofZ(Q) of order 3. Thenjx^Gj ˆn₃ˆ2⁴5 and so 5 does not dividejC_G(x)jby Step 1.

Since (3;5 1)ˆ1, we see thatxacts trivially onP. Thus 5 dividesC_G(x).

Since (3;jKj)ˆ1, we have thatC_G(x)ˆC_G(x)K=Kand so 5 dividesjC_G(x)j, a contradiction. Now, we assume that O3(G)6ˆ1. Put V ˆV1(Z(O3(G))).

ThenV is a nontrivial normal subgroup ofG. Letybe an element ofGof order 5. ThenV ˆ[V; y]C_V(y). By Step 1,jV :C_V(y)jis at most 3². It follows that [[V;y];y]ˆ1, which implies that [V;y]ˆ1. Therefore V ˆCV(y). LetQbe a Sylow 3-subgroup ofG. ThenV\Z(Q)6ˆ1. Letzbe an element ofV\Q of order 3. Then y zˆz y. Since (3;jKj)ˆ1, there exists a preimagezofzinGsuch thatzis contained in the center a Sylow 3- subgroup ofGand so 5 dividesjC_G(z)j. But, by Step 1, this is impossible.

(5)G'Aut(A6).

SetKˆO2(G) andGˆG=K. ThenGis insoluble by Step 4. Further- more, we have that M4G4Aut(M), where MˆS1S2 S_k is a direct product of non-abelian simplesS₁,S₂,. . .,S_k. Sincep(G)ˆ f2;3;5g and the order of the Sylow 5-subgroups ofGis 5, we obtain thatkˆ1, that is,Mis a non-abelian simpleK₃-group. HenceMis isomorphic toA₅,A₆or U4(2).

IfM'A5, thenG=K'A5orAut(A5) and so 3²does not dividejGj, a contradiction.

Suppose that M'A6. Assume first that G'A6 or S6. Let x be an element of G of order 5 such that n8ˆ jx^Gj ˆ2²3². Then jx^Gj divides 2²3². By [6], we have thatjx^Gj ˆ1, which implies that x4Z(G). There- forexˆ1 for thatZ(G) is trivial and sox2K, a contradiction.

Assume that G'PGL₂(9). Pick x2G with jxj ˆ5 such that jx^Gj ˆ2²3². Then, by [6],jx^Gj ˆ1 or 2²3². As above, it is impossible that jx^Gj ˆ1. Ifjx^Gj ˆ2²3², thenxcentralizesK. IfK6ˆ1, thenxcentralizes an element ofGwhich lies in the center of a Sylow 2-subgroup ofG, con- trary to Step 1. IfK ˆ1, thenG'PGL2(9), butcs(PGL2(9))6ˆcs(G) by [6], a contradiction.

Suppose thatG'M10. Letxbe an element ofGof order 5 such that jx^Gj ˆ2⁴3². Thenjx^Gj ˆ1 or 2⁴3²by [6]. Similar to the above case, we can derive a contradiction.

IfG'Aut(A₆), then, similar to the forgoing argument, one can show thatKˆ1 and soG'Aut(A₆), as desired.

If G'U4(2), then G has an element x such that jx^Gj ˆ2⁷3²5. It follows thatjx^Gjis divisible by 2⁷3²5, which is impossible by Step 1.

Thus, the proof is complete. p

LEMMA 3.3. Let G be a group with Z(G)ˆ1 and MˆAut(U4(2)).

Suppose that cs(G)ˆcs(M)andjGj_pˆ jMj_p, wherejGj_pdenotes the order of the Sylow p-subgroups of G for p2 f2;3;5g. Then G'M.

PROOF. We proceed the proof by the following steps.

(1) cs(G) consists of n1ˆ1, n2ˆ3²5, n3ˆ23³5, n4ˆ2⁴5, n5ˆ2⁴35, n6ˆ2⁵35, n7ˆ2²3³5, n8 ˆ2³3⁴5, n9 ˆ2⁶3⁴, n10 ˆ2⁴3²5, n11 ˆ2⁵3²5, n12 ˆ2⁴3³5, n13 ˆ2⁷3²5, n14 ˆ 2⁵3³5,n₁₅ˆ2²3²,n₁₆ ˆ2²3⁴5,n₁₇ˆ2⁴3⁴5.

This follows from [6].

(2)p(G)ˆp(M)ˆ f2;3;5g.

This follows from [6] and Lemma 2.1.

(3)O22⁰(G)ˆO2(G).

Write KˆO₂(G) and GˆG=K. Suppose that the assertion is false.

Then O_r(G)6ˆ1, where r2 f3;5g. Assume first that O₅(G)6ˆ1. Then, jO₅(G)j ˆ5. Letx2Z(Q) withjxj>1, whereQis a Sylow 3-subgroup ofG.

Then jx^Gj ˆ2⁴5 by (1). Sincex acts trivially onO5(G), we have that 5 divides the order ofC_G(x), by which we conclude that 5 divides the order of C_G(x). This is impossible by (1). Now assume that O₃(G)6ˆ1 and set V ˆV₁(Z(O₃(G))). Letybe an element ofGof order 5. Thenjy^Gj ˆ2⁶3⁴ or 2²3². Since V ˆ[V;y]C_V(y), we have that jV :C_V(y)j43⁴. If

jV :CV(y)j43³, then [[V;y];y]ˆ1 and so [V;y]ˆ1. Discussing as in Lemma 3.2, we see that this is impossible. If jV :CV(y)j ˆ3⁴, then the Sylow 3-subgroups of G is elementary abelian, which contradicts that jx^Gj ˆn₁₃ˆ2⁷3²5 for some 3-elementx2G.

(4)G'M.

B y (3),Gis insoluble. Hence we have thatM4G4Aut(M), whereM is a direct product of non-abelian simple groups. Since, by (2) p(G)ˆ p(M)ˆ f2;3;5g, we know thatMis simpleK3-groups by (3). Thus,Mis isomorphic to A₅, A₆ or U₄(2). If M is isomorphic to A₅ or A₆, then 3³ does not divide the order of G, a contradiction by (1). Suppose that M'U₄(2). Then G'U₄(2) or U₄(2):2. If G'U₄(2), then Z(G) is non- trivial, a contradiction. If G'U4(2):2, then G is isomorphic to Aut(U4(2)), as desired.

Proof of Theorem 1.2.

It follows from Lemmas 3.1-3.3. p

Acknowledgments. The authors would like to express their sincere thanks for the referee's helpful comments.

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Manoscritto pervenuto in redazione il 2 Settembre 2013.