A NEW CHARACTERIZATION OF SIMPLE K3-GROUPS

DAI Xue,ZHANG Qing-liang,GONG Lü

(School of Sciences,Nantong University,Nantong 226019,China)

1　Introduction

Let G be a finite group and p a prime.The number np(G)of Sylow p-subgroups of G is an important invariant pertaining to G and we call it the Sylow p-number of G.By a Sylow number for G,we mean an integer which is a Sylow p-number of G for some prime p.The Sylow number was investigated by many authors such as Hall,Brauer,Hall,Zhang,and Moret´o(see for instance[1,3,4,6–9]).Zhang[9]launched a systematic study on the influence of arithmetical properties on the group structure.

We set sn(G)={np(G)|p||G|}.Zhang[9]posed the following problem,namely what can we see about the finite groups G in terms of|sn(G)|?And he made the following claim:it seems true that G is solvable if|sn(G)|=2.The above claim was proved in[7].Now we consider the influence of sn(G)on simple K3-groups.

A finite simple group G is called a simple K3-group if|G|has exactly three distinct prime divisors.We know that|A5|=22·3·5 and sn(A5)={5,2·5,2·3}.So the following problem is interesting:if|G|=p2qr and sn(G)={r,pr,pq},where p ＜ q＜ r are different primes,then G～=A5holds?The answer of the problem is yes.In this paper,we get the following results by using an elementary and skillful method of applying Sylow’s theorem.

Main Theorem(1)Let|G|=p2qr and sn(G)={r,pr,pq},where p＜q＜r are different primes,then G ～=A5.

(2)Let|G|=p3q2r and sn(G)={q2r,pr,p2q2},where p ＜ q＜ r are different primes,then G～=A6.

(3)Let|G|=p3qr and sn(G)={qr,p2r,p3},where p＜ q＜ r are different primes,then G～=L2(7).

(4)Let|G|=p3q2r and sn(G)={q2,p2r,p2q2},where p ＜ q＜ r are different primes,then G～=L2(8).

(5)Let|G|=p4q2r and sn(G)={q2r,p3r,pq2},where p ＜ q＜ r are different primes,then G～=L2(17).

(6)Let|G|=p4q3r and sn(G)={q3r,p2r,p4q2},where p ＜ q＜ r are different primes,then G～=L3(3).

(7)Let|G|=p5q3r and sn(G)={q3r,p2r,p5q2},where p ＜ q＜ r are different primes,then G～=U3(3).

(8)Let|G|=p6q4r and sn(G)={r,pr,pq},where p ＜ q＜ r are different primes,then G～=U4(2).

In this paper,all groups are finite and by simple groups we mean non-abelian simple groups.All further unexplained notations are standard(cf.[2]for example).

2　Preliminaries

We need the following two simple lemmas to show our results.

Lemma 2.1(see[5])If G is a simple K3-group,then G is isomorphic to one of the following groups:A5,A6,L2(7),L2(8),L2(17),L3(3),U3(3)or U4(2).

Lemma 2.2(see[9])Let G be a finite group and M a normal subgroup of G,then the product of np(G)and np(G/M)divides np(G).

3　Proof of Main Theorem

Now we will prove the main theorem case by case.

Proof(1)If G is solvable,then G has an elementary abelian minimal normal subgroup N.Note that sn(G)={r,pr,pq},thus|N|=p and|G/N|=pqr.And it follows that G/N is supersolvable.Therefore G is supersolvable and nr(G)=1,which is a contradiction.And so G is unsolvable and G～=A5by Lemma 2.1.

(2)Assume that G is solvable,then G has a{q,r}-Hall subgroup H and|H|=q2r.By Sylow’s theorem,we know that nr(H)|q2.If nr(H)=q,then q ≡ 1(mod r),which is a contradiction since q＜r.If nr(H)=q2,then q2≡1(mod r),which implies that r|q+1.Consequently q=2 and r=3,a contradiction since p＜q.Thus nr(H)=1.Note that|G:NG(H)||p3,thus nr(G)is at most p3,which is impossible since nr(G)=p2q2.Therefore G is unsolvable.

We obtain that G has a chief factor H/N such that H/N～=A5,A6,L2(7)or L2(8)by Lemma 2.1,where N is a maximal solvable normal subgroup of G.Set:=H/N～=A5,:=G/N,we have

(3)If G is solvable,then G has a maximal subgroup M such that MG and|G:M|is a prime.If|M|=p3q,then nq(G)=nq(M)|p3,a contradiction since nq(G)=p2r.By the same reason|M|/=p3r.Hence|M|=p2qr.Let N be a minimal normal subgroup of M,then|N|=p or p2.If|N|=p,then|M/N|=pqr and so M/N is supersovable.Therefore M is supersolvable,which implies nr(G)=nr(M)=1,a contradiction.If|N|=p2,then N Char M and so NG.Since|G/N|=pqr,we obtain that G/N is supersolvable.Let R∈Sylr(G),then RN/NG/N.Since NG/N(RN/N)=NG(R)N/N=G/N,we have G=NG(R)N.Note that|G|=p3qr and|N|=p2,we get that p||NG(R)|,contradict to nr(G)=p3.Therefore G is unsolvable.

By Lemma 2.1,it follows that G has a chief factor H/N such that H/N～=A5or L2(7).If H/N～=A5,then nr(G)=n5(G)=p3=8,which is a contradiction.If H/N～=L2(7),then by Lemma 2.1,we have N=1 and H=G ～=L2(7)since G=p3qr and|L2(7)|=23·3 ·7.

(4)Suppose that G is solvable,then G has a{q,r}-Hall subgroup H and|G:NG(H)||p3.It is easy to show that nr(H)=1 by Sylow’s theorem.Therefore nr(G)is at most p3,contradict to nr(G)=p2q2.So G is unsolvable.

By Lemma 2.1 G has a chief factor H/N such that H/N～=A5,A6,L2(7)or L2(8).If H/N～=A5,then n2(G)=np(G)=q2=9.By Lemma 2.2,we get that n2(H/N)|n2(G),namely 5|9,which is a contradiction.By the same reason,H/N≇A6,L2(7).If H/N～=L2(8),then by Lemma 2.1 we have N=1 and H=G～=L2(8)since G=p3q2r and|L2(8)|=23·32·7.

(5)Suppose that G is solvable,then G has a maximal subgroup M such that MG and|G:M|is a prime.If|G:M|=q,then the Sylow p-subgroup P of M is also the the Sylow p-subgroup of G.Since np(G)=q2r we have NG(P)=P≤M,which implies that NG(M)=M,contradict to MG.Similarly|G:M|/=r.Consequently|G:M|=p and|M|=p3q2r.Now we consider the{q,r}-Hall subgroup N of M.It is easy to show that nr(N)=1.Therefore nr(M)is at most p3,a contradiction since nr(M)=nr(G)=pq2.So G is unsolvable.

We get that G has a chief factor H/N such that H/N～=A5,A6,L2(7),L2(8)or L2(17)by Lemma 2.1.Similarly to the above,we can show that G～=L2(17).

(6)Suppose that G is solvable,then there exist a maximal subgroup M of G such that MG and|G:M|is a prime.If|G:M|=r,then nq(G)=nq(M)|p4by Sylow’s theorem,contradict to nq(G)=p2r.If|G:M|=q,then np(G)=np(M)=|q2r,a contradiction since np(G)=q3r.If|G:M|=p,then nr(G)=nr(M)|p3q3,which is impossible since nr(G)=p4q2.Therefore G is unsolvable.

We can see from Lemma 2.1 that G has a chief factor H/N such that H/N～=A5,A6,L2(7),L2(8),L2(17)or L3(3).Now similarly to the above,we can show that G～=L3(3).

(7)Assume that G is solvable,then G has a maximal subgroup M such that MG and|G:M|is a prime.If|G:M|=r,then nq(G)=nq(M)|p5by Sylow’s theorem,which is contradict to nq(G)=p2r.If|G:M|=q,then np(G)=np(M)|q2r,a contradiction since np(G)=q3r.If|G:M|=p,then nr(G)=nr(M)|p4q3,which is impossible since nr(G)=p5q2.Therefore G is unsolvable.

It is easy to see that G has a chief factor H/N such that H/N～=A5,A6,L2(7),L2(8),L2(17),L3(3)or U3(3)by Lemma 2.1.Now similarly to the above,we can show that G～=U3(3).

(8)Suppose that G is solvable,then G has a maximal subgroup M such that MG and|G:M|is a prime.If|G:M|=r,then np(G)=np(M)|q4by Sylow’s theorem,which is contradict to np(G)=q3r.If|G:M|=q,then nr(G)=nr(M)|p6q3,a contradiction since nr(G)=p4q4.It follows that|G:M|=p and|M|=p5q4r.Now we consider a{q,r}-Hall subgroup H of M.It is evident that H is also a{q,r}-Hall subgroup of G.Note that nr(G)=p4q4and|G:NG(H)||p5,thus nr(H)=q4by Sylow’s theorem.In fact,if nr(H)≤q3,then nr(G)is at most p5q3,a contradiction since p5q3＜nr(G)=p4q4.Hence r|q4−1=(q2+1)(q2−1).If r|q2−1,then r|q+1 since q＜r.Consequently q=2 and r=3,which is contradict to p＜q＜r.Therefore r|q2+1.Since nr(G)=p4q4,we get that r|p4q4−1.Hence r|p4−1 since r|q4−1.Therefore r|p2+1.And it follows that r|(q2+1)−(p2+1),namely r|(q−p)(q+p),which implies that r|p+q.Now we get a contradiction since p+q＜2r.Therefore G is unsolvable.

By Lemma 2.1,we know that G has a chief factor H/N such that H/N～=A5,A6,L2(7),L2(8),L2(17),L3(3),U3(3)or U4(2).Now similarly to above,we can show that G～=U4(2).

Now the proof of the theorem is complete.

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