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OD-characterization of the automorphism groups of simple K 3 -groups

Abstract

The degree pattern of a finite group G associated to its prime graph has been introduced in (Moghaddamfar et al. in Algebra Colloq. 12(3):431-442, 2005) and denoted by D(G). The group G is called k-fold OD-characterizable if there exist exactly k non-isomorphic groups H satisfying conditions (1) |G|=|H| and (2) D(G)=D(H). Moreover, a one-fold OD-characterizable group is simply called OD-characterizable group. In this problem, those groups with connected prime graphs are somewhat much difficult to be solved. In the present paper, we continue this investigation and show that the automorphism groups of simple K 3 -groups are characterized by their orders and degree patterns. In fact, the automorphism groups of simple K 3 -groups except A 6 and U 4 (2) are OD-characterizable. Moreover, Aut( A 6 ) is fourfold OD-characterizable and Aut( U 4 (2)) is at least fourfold OD-characterizable.

AMS Subject Classification:20D05.

1 Introduction

Let G be a finite group, π e (G) denote the set of element orders of G and π(G) the set of all prime divisors of |G|. The prime graph of G was defined by Gruenberg and Kegel (ref. to [1]), which was denoted by Γ(G) and constructed as follows: The vertex set of this graph is π(G) and two distinct vertices p and q are jointed by an edge if and only if pq π e (G). In this case, we say vertices p and q are adjacent and denote this fact by pq. The number of connected components of Γ(G) is denoted by t(G) and the sets of vertices of connected components of Γ(G) are denoted as π i = π i (G) (i=1,2,,t(G)). If |G| is even, we always assume that 2 π 1 (G). Set T(G)={ π i (G)|i=1,2,,t(G)}.

Let n be a positive integer, we use π(n) to denote the set of all prime divisors of n. If the prime graph of G is known, then |G| can be expressed as a product of m 1 , m 2 ,, m t ( G ) , where m i s are positive integers such that π( m i )= π i . These m i s were called the order components of G by the second author, who proved a lot of finite simple groups can be uniquely determined by their order components (ref. to [2]). The set of order components of G is denoted as OC(G)={ m 1 , m 2 ,, m t ( G ) }. We also use the following notations. Given a finite group G, denote by Soc(G) the socle of G which is the subgroup generated by the set of all minimal normal subgroups of G. Syl p (G) denotes the set of all Sylow p-subgroups of G, where pπ(G). And P r denotes a Sylow r-subgroup of G for rπ(G). All further unexplained notations are referred to [3].

Definition 1.1 [4]

Let G be a finite group and |G|= p 1 α 1 p 2 α 2 p k α k , where p i s are primes and α i are integers. For pπ(G), let deg(p):=|{qπ(G)|pq}|, called the degree of p. We also define D(G):=(deg( p 1 ),deg( p 2 ),,deg( p k )), where p 1 < p 2 << p k . We call D(G) the degree pattern of G.

Definition 1.2 [4]

A group M is called k-fold OD-characterizable if there exist exactly k non-isomorphic groups G such that |G|=|M| and D(G)=D(M). Moreover, a one-fold OD-characterizable group is simply called an OD-characterizable group.

Definition 1.3 A group G is said to be almost simple related to S if and only if SGAut(S) for some non-abelian simple group S.

In a series of articles such as [47], many finite non-abelian simple groups or almost simple groups were shown to be OD-characterizable. For convenience, we recall some of them in the following proposition.

Proposition 1.4 A finite group G is OD-characterizable if G is one of the following groups:

  1. (1)

    All sporadic simple groups and their automorphism groups except Aut( J 2 ) and Aut( M c L);

  2. (2)

    The alternating groups A p , A p + 1 , A p + 2 and the symmetric groups S p and S p + 1 , where p is a prime;

  3. (3)

    All finite simple K 4 -groups except A 10 ;

  4. (4)

    The simple groups of Lie type L 2 (q), L 3 (q), U 3 (q), B 2 2 (q) and G 2 2 (q) for certain prime power q;

  5. (5)

    All finite simple C 2 , 2 -groups;

  6. (6)

    The alternating groups A p + 3 , where p+2 is a composite number, p+4 is a prime and 7pπ(1,000!);

  7. (7)

    The almost simple groups of Aut( O 10 + (2)), Aut( O 10 (2)) and Aut( F 4 (2)).

Till now a lot of finite simple groups have been shown to be OD-characterizable, and also some finite groups, especially the automorphism groups of some finite simple groups, have been shown not to be OD-characterizable but k-fold OD-characterizable for some k>1. In this paper, we continue this topic and get the following Main Theorem.

Main Theorem Let M be a simple K 3 -group and G be a finite group such that |G|=|Aut(M)| and D(G)=D(Aut(M)).

  1. (1)

    If M is one of the following simple K 3 -groups: A 5 , A 6 , L 2 (7), L 2 (8), U 3 (3), L 3 (3) and L 2 (17), then GAut(M). In other words, Aut(M) is OD-characterizable.

  2. (2)

    If M= A 6 , then G is isomorphic to one of the following groups: Aut( A 6 ), Z 2 × Z 2 × A 6 , Z 2 ×( Z 2 A 6 ) and Z 4 × A 6 . In other words, Aut( A 6 ) is fourfold OD-characterizable.

  3. (3)

    If M= U 4 (2), then G is isomorphic to one of the following groups: Z 2 U 4 (2), Z 2 × U 4 (2), Aut( U 4 (2)) and ( P 3 P 5 ) P 2 , where P r Syl r (G) for each rπ(G). In other words, Aut( U 4 (2)) is at least fourfold OD-characterizable.

2 Preliminaries

In this section, we give some results which will be applied to our further investigations.

Lemma 2.1 [[1], Theorem A]

Let G be a finite group with t(G)2, then G is one of the following groups:

  1. (a)

    G is a Frobenius or 2-Frobenius group;

  2. (b)

    G has a normal series 1HKG such that H and G/K are π 1 -groups and K/H is a non-abelian simple group, where π 1 is the prime graph component containing 2, H is a nilpotent group and |G/H||Aut(K/H)|. Moreover, any odd order component of G is also an odd order component of K/H.

Remark 2.2 A group G is a 2-Frobenius group if there exists a normal series 1HKG such that K and G/H are Frobenius groups with kernels H and K/H, respectively.

Lemma 2.3 [3]

Let S be a finite non-abelian simple group with order having prime divisors at most 17. Then S is isomorphic to one of the simple groups listed in Table 1. In particular, if |Out(G)|1, then π(Out(G)){2,3}.

Table 1 Finite non-abelian simple groups with π(S){2,3,5,7,11,13,17}

Lemma 2.4 [8]

Let G be a Frobenius group with kernel F and complement C. Then the following assertions hold:

  1. (a)

    F is a nilpotent group.

  2. (b)

    |F|1(mod|C|).

Lemma 2.5 [9]

Let G be a Frobenius group of even order with H and K being its Frobenius kernel and Frobenius complement, respectively. Then t(G)=2 and T(G)={π(K),π(H)}.

Lemma 2.6 [10]

Let G be a simple C p , p -group, where p is a prime. Then

  1. (a)

    If p=5, G is isomorphic to one of the following simple groups: A 5 , A 6 , A 7 , M 11 , M 22 , L 3 (4), S 4 (3), S 4 (7), U 4 (3), Sz(8), Sz(32), L 2 (49), L 2 ( 5 m ), L 2 (2 5 m ±1), where mN and 2 5 m ±1P.

  2. (b)

    If p=7, G is isomorphic to one of the following simple groups: A 7 , A 8 , A 9 , M 22 , J 1 , J 2 , HS, L 3 (4), S 6 (2), O 8 + (2), G 2 (3), G 2 (13), U 3 (3), U 3 (5), U 3 (19), U 4 (3), U 6 (2), Sz(8), L 2 (8), L 2 ( 7 m ), L 2 (2 7 m 1), where mN and 2 7 m 1P.

  3. (c)

    If p=13, G is isomorphic to one of the following simple groups: A 13 , A 14 , A 15 , Suz, F i 22 , L 3 (3), L 4 (3), O 7 (3), S 4 (5), S 6 (3), O 8 + (3), U 3 (4), U 3 (23), G 2 (4), G 2 (3), F 4 (2), Sz(8), E 6 2 (2), D 4 3 (2), F 4 2 ( 2 ) , L 2 (27), L 2 (25), L 2 ( 13 m ), L 2 (2 13 m 1), where mN and 2 13 m 1P.

  4. (d)

    If p=17, G is isomorphic to one of the following simple groups: A 17 , A 18 , A 19 , J 3 , He, F i 23 , F i 24 , L 2 (q) (q= 2 4 , 17 m ,2 17 m ±1 which is a prime, m1), S 4 (4), S 8 (2), O 8 (2), F 4 (2), E 6 2 (2).

Remark 2.7 Let p be a prime. A group G is called a C p , p -group if and only if pπ(G) and the centralizers of its elements of order p in G are p-groups.

3 Proof of Main Theorem

In this section, we give the proof of Main Theorem.

Remark 3.1 Let n be a positive integer and n>1. We say that a finite group G is a K n -group if and only if |π(G)|=n.

Proof of Main Theorem Let M be a simple K 3 -group, then M is isomorphic to one of the following simple K 3 -groups: A 5 , A 6 , L 2 (7), L 2 (8), U 3 (3), L 3 (3), L 2 (17) and U 4 (2). For convenience, using [3], we have tabulated |Aut(M)|, D(Aut(M)) and |Out(M)| in Table 2.

Table 2 K 3 -groups

Let G be a finite group satisfying (1) |G|=|Aut(M)| and (2) D(G)=D(Aut(M)). We prove the theorem up to choice of M one by one. The proof is written in four cases.

Case 1. To prove the theorem if M= A 5 .

Evidently, t(G)=2. In fact, we have π 1 (G)={2,3} and π 2 (G)={5}. We first show that G is neither Frobenius nor 2-Frobenius group. Suppose G=NH is a Frobenius group with kernel N and complement H, and hence T(G)={π(N),π(H)} by Lemma 2.5. Since |H| divides |N|1, it follows that |N|= 2 3 3 and |H|=5. Clearly, this is impossible because P 5 cannot act fixed-point-freely for instance on P 3 as 5(31).

Now assume that G is a 2-Frobenius group with kernels H and K/H, respectively. Since T(G)={π(H)π(G/K),π(K/H)} and 2π(H)π(G/K), it follows that |K/H|=5. On the other hand, G/KAut(K/H) Z 4 . Hence |G/K|4, which implies {3,5}π(K). In this case, we have 3π(H), so an element of order 5 must act fixed-point-freely on a subgroup of order 3 in H, which is clearly a contradiction by Table 2.

By Lemma 2.1, G has a normal series 1N G 1 G such that N and G/ G 1 are π 1 -groups and G 1 /N is a non-abelian simple group, N is a nilpotent group. Note that one of the components of the prime graph of G 1 /N must be {5} and G 1 /N is a simple C 5 , 5 group. By Lemma 2.6, G 1 /N can only be isomorphic to one of the following simple groups: A 5 , A 6 , A 7 , M 11 , M 22 , L 3 (4), S 4 (3), S 4 (7), U 4 (3), Sz(8), Sz(32), L 2 (49), L 2 ( 5 m ) and L 2 (2 5 m ±1), where mN and 2 5 m ±1P.

Considering the orders of the simple groups, G 1 /N can only be isomorphic to A 5 , that is, G 1 /N A 5 . Since G/NAut( G 1 /N), we get A 5 G/NAut( A 5 ).

If G/NAut( A 5 ), and since |G|=|Aut( A 5 )|, we deduce N=1 and GAut( A 5 ).

If G/N A 5 , then |N|=2 and so NZ(G). Therefore G is a central extension of Z 2 by A 5 and G is isomorphic to one of the following groups:

2 A 5 ( a non-split extension of  Z 2  by  A 5 ) ; 2 : A 5 Z 2 × A 5 ( a split extension of  Z 2  by  A 5 ) .

But whether G is isomorphic to 2 A 5 or 2: A 5 Z 2 × A 5 , it always follows that 15 π e (G) by [3] (see ATLAS), a contradiction.

Till now we have proved that GAut( A 5 ) if |G|=|Aut( A 5 )| and D(G)=D(Aut( A 5 )), that is, Aut( A 5 ) is OD-characterizable.

Case 2. To prove the theorem holds for M, one of the following simple groups: L 2 (7), L 2 (8), U 3 (3), L 3 (3) and L 2 (17).

Since |G|=|Aut(M)| and D(G)=D(Aut(M)), we have to discuss the following five cases. The method used below is the same as Case 1, so the detailed processes are omitted.

  1. (a)

    If M= L 2 (7), then GAut( L 2 (7));

  2. (b)

    If M= L 2 (8), then GAut( L 2 (8));

  3. (c)

    If M= U 3 (3), then GAut( U 3 (3));

  4. (d)

    If M= L 3 (3), then GAut( L 3 (3));

  5. (e)

    If M= L 2 (17), then GAut( L 2 (17)).

Hence all the almost simple groups Aut( L 2 (7)), Aut( L 2 (8)), Aut( U 3 (3)), Aut( L 3 (3)) and Aut( L 2 (17)) are OD-characterizable.

Case 3. To prove the theorem holds for M= A 6 .

By Table 2, |G|=|Aut( A 6 )|= 2 5 3 2 5 and D(G)=D(Aut( A 6 ))=(2,1,1). By these facts, we immediately conclude that {2,3,5,6,10} π e (G) and 15 π e (G). It is evident that the prime graph of G is connected since deg(2)=2 and |π(G)|=3. Moreover, it easy to see that Γ(G)=Γ(Aut( A 6 )). We break up the proof into a sequence of subcases.

Subcase 3.1. Let K be a maximal normal solvable subgroup of G. Then K is a 2-group. In particular, G is nonsolvable.

We first prove that K is a 5-group. Assume the contrary, then K possesses an element x of order 5. Set C= C G (x) and N= N G (x). By the structure of D(G), C is a {2,5}-group. By N-C Theorem, N/CAut(x) Z 4 . Hence, N is a {2,5}-group. By the Frattini argument, G=KN. This implies that {3,5}π(K). Since K is solvable, it possesses a Hall {3,5}-subgroup L of order 3 2 5. Clearly, L is nilpotent, and hence 15 π e (G), a contradiction.

Next, we show that K is a 3-group. Otherwise, let P 3 Syl 3 (K). Again, by the Frattini argument G=K N G ( P 3 ). Hence 5 divides the order of N G ( P 3 ). Then N G ( P 3 ) contains a subgroup of order 3 2 5, which leads to a contradiction as before. Therefore K is a 2-group. Since KG, it follows that G is nonsolvable. This completes the proof of Subcase 3.1.

Subcase 3.2. The quotient group G/K is an almost simple group. In fact, SG/KAut(S), where S is a non-abelian simple group.

Let G ¯ :=G/K and S:=Soc( G ¯ ). Then S= B 1 × B 2 ×× B m , where B i s are non-abelian simple groups and S G ¯ Aut(S). In what follows, we will prove that m=1.

Suppose that m2. It is easy to see that 5 does not divide the order of S, since otherwise 15 π e (G), a contradiction. On the other hand, by the order of G, we obtain that π(S){2,3}, which is impossible. Therefore m=1 and S= B 1 .

Subcase 3.3. S A 6 and G is isomorphic to one of the following groups: Aut( A 6 ), Z 2 × Z 2 × A 6 , Z 2 ×( Z 2 A 6 ) and Z 4 × A 6 .

By Lemma 2.3 and Subcase 3.1, we may assume that |S|= 2 a 3 2 5, where 2a5. Using Table 1, we see that S can only be isomorphic to the simple group A 6 . Thus A 6 G/KAut( A 6 ).

If G/KAut( A 6 ), then by the order comparison, we obtain that K=1 and GAut( A 6 ).

If G/K A 6 , then |K|=4 and so K Z 2 × Z 2 or K Z 4 . Now, we divide the proof into two subcases.

Subsubcase 3.3.1. G/K A 6 and K Z 2 × Z 2 . By N-C Theorem, we know that the factor group G/ C G (K) is isomorphic to a subgroup of Aut(K). Thus |G/ C G (K)|( 2 2 1)( 2 2 2), that is, |G/ C G (K)|6, which implies that 5| C G (K)|. In particular, K< C G (K). On the other hand, we have C G (K)/KG/K A 6 and hence we obtain G= C G (K). So KZ(G). Therefore G is a central extension of K by A 6 . If G is a non-split extension of K by A 6 , then G Z 2 ×( Z 2 A 6 ) (see [3]). If G is a split extension over K, we have G Z 2 × Z 2 × A 6 .

Subsubcase 3.3.2. G/K A 6 and K Z 4 . In this case, we have G/ C G (K)Aut( Z 4 ) Z 2 , and so |G/ C G (K)|=1 or 2. If |G/ C G (K)|=2, then K< C G (K). Since C G (K)/KG/K A 6 , we obtain G= C G (K), a contradiction. Therefore |G/ C G (K)|=1 and KZ(G). Furthermore, G is a central extension of Z 4 by A 6 . Obviously, G cannot be a non-split extension central extension of Z 4 by A 6 since the order of Schur multiplier of A 6 is 6. If G is a split extension over K, we obtain G Z 4 × A 6 . This completes the proof of Subcase 3.3 and the case.

Case 4. To prove the theorem if M= U 4 (2).

In this case, we have |G|=|Aut( U 4 (2))|= 2 7 3 4 5 and D(G)=D(Aut( U 4 (2)))=(2,1,1) by Table 2. By these hypotheses, we immediately conclude that {2,3,5,6,10} π e (G) and 15 π e (G). Clearly, the prime graph of G is connected, because the vertex 2 is adjacent to all other vertices. Moreover, it is easy to see that Γ(G)=Γ(Aut( U 4 (2))). We separate the proof into a sequence of subcases.

Subcase 4.1. To prove the theorem holds while G is nonsolvable.

Let K be the maximal normal solvable subgroup of G. Then K is a {2,3}-group by the same approach as that in Subcase 3.1. We assert that G/K is an almost simple group. And in fact, SG/KAut(S).

Let G ¯ :=G/K and S:=Soc( G ¯ ). Then S= B 1 × B 2 ×× B m , where B i s are non-abelian simple groups and S G ¯ Aut(S). It is easy to see that m=1 by Table 2. Therefore S= B 1 .

By Lemma 2.3, we can suppose that |S|= 2 a 3 2 5, where 2a7, 1b4. Using Table 1, we see that S can only be isomorphic to one of the following simple groups: A 5 , A 6 and U 4 (2).

If S A 5 , then A 5 G/KAut( A 5 ) S 5 . Hence 25 π e (G) π e ( S 5 ), a contradiction.

If S A 6 , then A 6 G/KAut( A 6 ) and 2 2 3 2 divides the order of K.

Let P r Syl r (K) for each rπ(G). By the Frattini argument G=K N G ( P 3 ), 5 divides the order of N G ( P 3 ). Let T be a subgroup of N G ( P 3 ) of order 5. By N-C Theorem, the factor group N G ( P 3 )/ C G ( P 3 ) is isomorphic to a subgroup of Aut( P 3 ). Thus |G/ C G (K)|( 3 2 1)( 3 2 3), which implies that T C G (K). Then 15 π e (G), a contradiction.

If S U 4 (2), then U 4 (2)G/KAut( U 4 (2)). In this case, G/K U 4 (2), then |K|=2 and KZ(G). Therefore G is a central extension of K by U 4 (2). If G is a non-split extension of K by U 4 (2), then G Z 2 U 4 (2). If G is a split extension over K, we have G Z 2 × U 4 (2). In the latter case G/KAut( U 4 (2)), by order comparison, we deduce that K=1 and GAut( U 4 (2)).

Till now we have proved that G is isomorphic to one of the following groups: Z 2 U 4 (2), Z 2 × U 4 (2) and Aut( U 4 (2)) if G is nonsolvable. It is easy to see that the groups Z 2 U 4 (2), Z 2 × U 4 (2) and Aut( U 4 (2)) satisfy the conditions (1) |G|=|Aut( U 4 (2))| and (2) D(G)=D(Aut( U 4 (2))) (see ATLAS).

Subcase 4.2. To prove the theorem holds while G is solvable.

Since G is solvable, we may take a normal series of G: 1 N 1 N 2 G such that N 1 is unity or a 2-group, N 2 / N 1 is a 3-group or 5-group. While N 2 / N 1 is a 5-group, we consider the action a 3-element x N 1 on N 2 / N 1 , then we see that G/ N 1 has an element of order 15, so does G, a contradiction. While N 2 / N 1 is a 3-group, it is enough to consider the action of a 5-element of G/ N 1 on N 2 / N 1 , a contradiction appears too if | N 2 / N 1 | 3 3 . Hence, | N 2 / N 1 |= 3 4 and the 5-element of G/ N 1 must act fixed-point-freely on N 2 / N 1 . Moreover, the {3,5}-Hall subgroup H of G is a Frobenius group with kernel P 3 and complement P 5 , since otherwise there exists an automorphism of P 3 of order 5, say ϕ, such that ϕ(x)=x for each x P 3 . We first show that P 3 is an elementary abelian 3-group.

Set H= P 3 P 5 . Since Z(Ω( P 3 )) char P 3 H and | P 3 |= 3 4 , then Z(Ω( P 3 ))H and |Z(Ω( P 3 ))| 3 4 . By the structure of D(G), G has no elements of order 15, neither does H. Therefore, Z(Ω( P 3 )) is an elementary abelian 3-group of order 34, as required.

Let x be an element of P 3 of order 3, then we have ϕ(gx)=gx for every g P 3 . Now a direct computation shows that ϕ(g)=g x i , where i=0,1,2. Hence ϕ 3 (g)=g x 3 i =g. However, the order of ϕ is 5, a contradiction.

We have G=( P 3 P 5 ) P 2 , a product of P 2 and P 3 P 5 . It is obvious that there exists such a finite group satisfying the following conditions: (1) |G|=|Aut(M)| and (2) D(G)=D(Aut(M)). This completes the proof of Main Theorem. □

In 1987, Shi in [11] put forward the following conjecture:

Conjecture 3.2 Let G be a group and M be a finite simple group. Then GM if and only if (1) |G|=|M| and (2) π e (G)= π e (M).

Corollary 3.3 Let M be one of the following simple K 3 -groups: A 5 , A 6 , L 2 (7), L 2 (8), U 3 (3), L 3 (3), L 2 (17) and G be a finite group such that |G|=|Aut(M)| and π e (G)= π e (Aut(M)). Then GAut(M).

Proof If π e (G)= π e (Aut(M)), then G and Aut(M) have the same degree pattern. Hence the result follows from Main Theorem. □

4 An example and a question

Example 4.1 According to Main Theorem, let G=( P 3 P 5 )× P 2 and M= U 4 (2), then |G|=|Aut(M)| and D(G)=D(Aut(M)). However, G is not isomorphic to Aut(M). Hence, we put forward the following question:

Question 4.2 Is Aut( U 4 (2)) exactly fourfold OD-characterizable?

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Acknowledgements

This work was supported by the Natural Science Foundation of China (Grant Nos. 11171364; 11271301); by ‘The Fundamental Research Funds for the Central Universities’ (Grant No. XDJK2012D004) and Natural Science Foundation Project of CQ CSTC (Grant No. cstc2011jjA00020); by the Fundamental Research Funds for the Central Universities (Grant No. XDJK2009C074) and Graduate-Innovation Funds of Science of SWU (ky2009013).

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Correspondence to Guiyun Chen.

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YY carried out the study of the alternating group of degree 5. HX carried out the study of the alternating group of degree 6. GC and LH carried out the study of the group U 4 (2). All authors read and approved the final manuscript.

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Yan, Y., Xu, H., Chen, G. et al. OD-characterization of the automorphism groups of simple K 3 -groups. J Inequal Appl 2013, 95 (2013). https://doi.org/10.1186/1029-242X-2013-95

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Keywords

  • almost simple group
  • prime graph
  • degree pattern
  • simple K 3 -group
  • degree of a vertex