L2(q)的一個物征性質

趙俊英, 韓章家

(1.天津商業大學理學院,天津300134;2.成都信息工程學院數學學院,四川成都610225)

1 Introduction

All groups considered in this paper are finite groups.The reader may refer to[1]for the notations of simple groups.

In group theory it is an usual way to get the information on the structure of a group G by studying it's subgroups.It was proved in[2]that if G is one of PSL(2,2n),Sz(22m+1),An(n≤10),K3-groups,Mathieu groups and Janko groups,then G is uniquely determined by the set of orders of it's maximal abelian subgroups.In[3],it is proved that alternative groups whose prime graph has three connected components can be uniquely determined by the set of orders of its maximal abelian subgroups.In this paper we will show that the group L2(q),q≡3,5(mod 8)can be determined up to isomorphism by the set of orders of it's maximal abelian subgroups.

Notations：If G is a finite group then Γ(G)denotes the prime graph of G;t(Γ(G))denotes the number of prime graph components of G;πi(1≤i≤t(Γ(G)))denote the set of vertices of prime graph components of G.If G isof even order,then π1always denotes the even prime graph component of G.M(G)denotes the set of ordersof maximal abelian subgroups of G.

Assume that π1,π2,…,πtare all prime graph components of G.Then|G|=m1m2…mt,where π(mi)=πi,i=1,2,…,t.Positive numbers m1,m2,…,mtare called the order components of G[4].The order components of finite simple groups with non-connected prime graphs are listed in Tables 1-4 given by Chen[5].

The following lemma follows from the definition of order component,Theorem A and Lemma 3 in[6].

Lemma1 Let G be a finite group with disconnected prime graph.Then one of the following statements holds：

(i)G is a Frobenius group or a 2-Frobenius group;

Lemma2 Let G be a Frobenius or a 2-Frobenius group,then t(Γ(G))=2[7].

The following result is important for the proof in our theorem.

Lemma3 Let M(G)=M(L2(q))and q≡3,5(mod 8),where q=tm,then

(i)G has a maximal abelian subgroup of order tmand the order of any maximal abelian 2-subgroup is four;

(ii)If 2? q-1,then G has no abelian subgroup of order q-1,and G has no abelian subgroup of order tr,where(t,r)=1.

Proof.The lemma follows from Theorem 3.6.25 in[8]and Lemma 2.

Lemma4 Let t(Γ(G))≥2 andIf N is a π1-group and a1,a2,…,arare order components of G,then every a1,a2,…,aris a divisor of|H|-1[5].

2 Main result

Now we can show that the group L2(q),q≡3,5(mod 8)is characterized by it's sets of orders of maximal abelian subgroups.

Theorem1 Let G be a finite group.If M(G)=M(M),where M=L2(q),q≡3,5(mod 8),then G?M.

Proof.Since t(Γ(M))=3,it follows by Lemma 1 and Lemma 2 that G has a normal seriessatisfying t(Γ(K/H))≥3 and some odd order components of K/H are equal to the odd order components of G.

If t(Γ(K/H))=6,then K/H ?J4by[6]andis equal to one of 23,29,31,37 or 43.It is impossible.

If t(Γ(K/H))=6,then K/H ?E8(q′)by[6],where q′≡0,1,4(mod 5).

We get a system of equations：

It can be easily shown that this system of equations has no solution,so K/H ?E8(q′).where q′0,1,4(mod 5).By the same way we can easily have that K/H ?E8(q′),q≡2,3(mod 5).Hence,if t(Γ(K/H))=4,then K/H is isomorphic to one of the following groups：2B2(22m+1),A2(4),2E6(2),M22,J1,ON,Ly,F′24and M.

If K/H ?2B2(22m+1),then We get a system of equations：

It iseasy to check that there are no solutions for all above equations.Hence,K/H?2B2(22m+1).Similarly,we have that K/H?A2(4),and2E6(2).If K/H is isomorphic to any sporadic simple group,then by way of calculating we have that the only possibility is K/H?M22.In this case,q=11 and Lemma 3 implies that the order of any Sylow 2-subgroup is 8 at most,which is a contradiction.Thisproves that t(Γ(K/H))≠4.If t(Γ(K/H))=3,then by[6],K/H is isomorphic to one of the following groups：

Ap(p and p-2 prime),A1(q′),G2(q′)(3|q′),2G2(32m+1),2Dp(3)(p=2n+1,n ≥2),2Dp+1(2)(p=2n-1,n ≥2),F4(q′)(2|q′),E7(2),E7(3),A2(2),2A5(2),2F4(22m+1)(m ≥1),M11,M23,M24,J3,HS,Suz,Co2,F23,B and Th.

Obviously,K/H can not isomorphic to of the groups E7(2),E7(3),A2(2),2A5(2),J3,HS,Suz,F23,B,Th,M23and M24.If K/H ?G2(q′)(3|q′),then,we have：

Thus q′=3,q=13.In this case,the orders of some maximal 2-subgroups are greater than 4,which contradicts Lemma 3.By the same reason we have K/H ?2G2(q′).

If K/H ?2Dp(3)(p=2n+1,n≥2),then the following equations hold：

It iseasy to check that all these equationshave no solutions,hence K/H?2Dp(3)(p=2n+1,n≥2).By using the same argument,we know that K/H can not be isomorphic to F4(q′)(2|q′),2F4(q′)(q′=22m+1,m ≥1)and2Dp+1(2),p=2n-1,n≥2.If K/H?Ap,then we can get p=q=5.The theorem follows by[11].Hence the only possibility is K/H=A1(q′),which forces that q=q′and then K/H=A1(q).We now claim that H=1.Otherwise the center Z(T)of a Sylow t-subgroup T of H is normal in G.Consequently,if 4|(q+1),we haveby Lemma 4,this implies that G has an abelian subgroup of order greater than q,which contradicts Lemma 3.This proves our claim,hence K?L2(q).Since CG(K)=1,we have(q)).Note that the outer automorphisms of L2(q)are field automorphisms or diagonal automorphisms of order two and every field automorphism ? centralizes the prime field,so the order of ? is a power of two.If G ≠L2(q),then G has a cyclic subgroup of order 2kt(k≠0),which contradicts Lemma 3.IfG has nontrivial diagonal automorphisms,then there is an element of order q-1 in G,which contradicts Lemma 3 too.Hence we obtain that G?L2(q).The proof is complete.

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