OD-characterization of the automorphism groups of simple -groups
© Yan et al.; licensee Springer 2013
Received: 17 December 2012
Accepted: 18 February 2013
Published: 7 March 2013
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 . The group G is called k-fold OD-characterizable if there exist exactly k non-isomorphic groups H satisfying conditions (1) and (2) . 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 -groups are characterized by their orders and degree patterns. In fact, the automorphism groups of simple -groups except and are OD-characterizable. Moreover, is fourfold OD-characterizable and is at least fourfold OD-characterizable.
AMS Subject Classification:20D05.
Keywordsalmost simple group prime graph degree pattern simple -group degree of a vertex
Let G be a finite group, denote the set of element orders of G and the set of all prime divisors of . The prime graph of G was defined by Gruenberg and Kegel (ref. to ), which was denoted by and constructed as follows: The vertex set of this graph is and two distinct vertices p and q are jointed by an edge if and only if . In this case, we say vertices p and q are adjacent and denote this fact by . The number of connected components of is denoted by and the sets of vertices of connected components of are denoted as (). If is even, we always assume that . Set .
Let n be a positive integer, we use to denote the set of all prime divisors of n. If the prime graph of G is known, then can be expressed as a product of , where s are positive integers such that . These 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 ). The set of order components of G is denoted as . We also use the following notations. Given a finite group G, denote by the socle of G which is the subgroup generated by the set of all minimal normal subgroups of G. denotes the set of all Sylow p-subgroups of G, where . And denotes a Sylow r-subgroup of G for . All further unexplained notations are referred to .
Definition 1.1 
Let G be a finite group and , where s are primes and are integers. For , let , called the degree of p. We also define , where . We call the degree pattern of G.
Definition 1.2 
A group M is called k-fold OD-characterizable if there exist exactly k non-isomorphic groups G such that and . 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 for some non-abelian simple group S.
In a series of articles such as [4–7], 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.
All sporadic simple groups and their automorphism groups except and ;
The alternating groups , , and the symmetric groups and , where p is a prime;
All finite simple -groups except ;
The simple groups of Lie type , , , and for certain prime power q;
All finite simple -groups;
The alternating groups , where is a composite number, is a prime and ;
The almost simple groups of , and .
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 . In this paper, we continue this topic and get the following Main Theorem.
If M is one of the following simple -groups: , , , , , and , then . In other words, is OD-characterizable.
If , then G is isomorphic to one of the following groups: , , and . In other words, is fourfold OD-characterizable.
If , then G is isomorphic to one of the following groups: , , and , where for each . In other words, is at least fourfold OD-characterizable.
In this section, we give some results which will be applied to our further investigations.
Lemma 2.1 [, Theorem A]
G is a Frobenius or 2-Frobenius group;
G has a normal series such that H and are -groups and is a non-abelian simple group, where is the prime graph component containing 2, H is a nilpotent group and . Moreover, any odd order component of G is also an odd order component of .
Remark 2.2 A group G is a 2-Frobenius group if there exists a normal series such that K and are Frobenius groups with kernels H and , respectively.
Lemma 2.3 
Finite non-abelian simple groups with
Lemma 2.4 
F is a nilpotent group.
Lemma 2.5 
Let G be a Frobenius group of even order with H and K being its Frobenius kernel and Frobenius complement, respectively. Then and .
Lemma 2.6 
If , G is isomorphic to one of the following simple groups: , , , , , , , , , , , , , , where and .
If , G is isomorphic to one of the following simple groups: , , , , , , HS, , , , , , , , , , , , , , , where and .
If , G is isomorphic to one of the following simple groups: , , , , , , , , , , , , , , , , , , , , , , , , where and .
If , G is isomorphic to one of the following simple groups: , , , , He, , , ( which is a prime, ), , , , , .
Remark 2.7 Let p be a prime. A group G is called a -group if and only if 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 . We say that a finite group G is a -group if and only if .
Let G be a finite group satisfying (1) and (2) . 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 .
Evidently, . In fact, we have and . We first show that G is neither Frobenius nor 2-Frobenius group. Suppose is a Frobenius group with kernel N and complement H, and hence by Lemma 2.5. Since divides , it follows that and . Clearly, this is impossible because cannot act fixed-point-freely for instance on as .
Now assume that G is a 2-Frobenius group with kernels H and , respectively. Since and , it follows that . On the other hand, . Hence , which implies . In this case, we have , 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 such that N and are -groups and is a non-abelian simple group, N is a nilpotent group. Note that one of the components of the prime graph of must be and is a simple group. By Lemma 2.6, can only be isomorphic to one of the following simple groups: , , , , , , , , , , , , and , where and .
Considering the orders of the simple groups, can only be isomorphic to , that is, . Since , we get .
If , and since , we deduce and .
But whether G is isomorphic to or , it always follows that by  (see ATLAS), a contradiction.
Till now we have proved that if and , that is, is OD-characterizable.
Case 2. To prove the theorem holds for M, one of the following simple groups: , , , and .
If , then ;
If , then ;
If , then ;
If , then ;
If , then .
Hence all the almost simple groups , , , and are OD-characterizable.
Case 3. To prove the theorem holds for .
By Table 2, and . By these facts, we immediately conclude that and . It is evident that the prime graph of G is connected since and . Moreover, it easy to see that . 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 and . By the structure of , C is a -group. By N-C Theorem, . Hence, N is a -group. By the Frattini argument, . This implies that . Since K is solvable, it possesses a Hall -subgroup L of order . Clearly, L is nilpotent, and hence , a contradiction.
Next, we show that K is a 3′-group. Otherwise, let . Again, by the Frattini argument . Hence 5 divides the order of . Then contains a subgroup of order , which leads to a contradiction as before. Therefore K is a 2-group. Since , it follows that G is nonsolvable. This completes the proof of Subcase 3.1.
Subcase 3.2. The quotient group is an almost simple group. In fact, , where S is a non-abelian simple group.
Let and . Then , where s are non-abelian simple groups and . In what follows, we will prove that .
Suppose that . It is easy to see that 5 does not divide the order of S, since otherwise , a contradiction. On the other hand, by the order of G, we obtain that , which is impossible. Therefore and .
Subcase 3.3. and G is isomorphic to one of the following groups: , , and .
By Lemma 2.3 and Subcase 3.1, we may assume that , where . Using Table 1, we see that S can only be isomorphic to the simple group . Thus .
If , then by the order comparison, we obtain that and .
If , then and so or . Now, we divide the proof into two subcases.
Subsubcase 3.3.1. and . By N-C Theorem, we know that the factor group is isomorphic to a subgroup of . Thus , that is, , which implies that . In particular, . On the other hand, we have and hence we obtain . So . Therefore G is a central extension of K by . If G is a non-split extension of K by , then (see ). If G is a split extension over K, we have .
Subsubcase 3.3.2. and . In this case, we have , and so . If , then . Since , we obtain , a contradiction. Therefore and . Furthermore, G is a central extension of by . Obviously, G cannot be a non-split extension central extension of by since the order of Schur multiplier of is 6. If G is a split extension over K, we obtain . This completes the proof of Subcase 3.3 and the case.
Case 4. To prove the theorem if .
In this case, we have and by Table 2. By these hypotheses, we immediately conclude that and . 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 . 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 -group by the same approach as that in Subcase 3.1. We assert that is an almost simple group. And in fact, .
Let and . Then , where s are non-abelian simple groups and . It is easy to see that by Table 2. Therefore .
By Lemma 2.3, we can suppose that , where , . Using Table 1, we see that S can only be isomorphic to one of the following simple groups: , and .
If , then . Hence , a contradiction.
If , then and divides the order of K.
Let for each . By the Frattini argument , 5 divides the order of . Let T be a subgroup of of order 5. By N-C Theorem, the factor group is isomorphic to a subgroup of . Thus , which implies that . Then , a contradiction.
If , then . In this case, , then and . Therefore G is a central extension of K by . If G is a non-split extension of K by , then . If G is a split extension over K, we have . In the latter case , by order comparison, we deduce that and .
Till now we have proved that G is isomorphic to one of the following groups: , and if G is nonsolvable. It is easy to see that the groups , and satisfy the conditions (1) and (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: such that is unity or a 2-group, is a 3-group or 5-group. While is a 5-group, we consider the action a 3-element on , then we see that has an element of order 15, so does G, a contradiction. While is a 3-group, it is enough to consider the action of a 5-element of on , a contradiction appears too if . Hence, and the 5-element of must act fixed-point-freely on . Moreover, the -Hall subgroup H of G is a Frobenius group with kernel and complement , since otherwise there exists an automorphism of of order 5, say ϕ, such that for each . We first show that is an elementary abelian 3-group.
Set . Since char and , then and . By the structure of , G has no elements of order 15, neither does H. Therefore, is an elementary abelian 3-group of order 34, as required.
Let x be an element of of order 3, then we have for every . Now a direct computation shows that , where . Hence . However, the order of ϕ is 5, a contradiction.
We have , a product of and . It is obvious that there exists such a finite group satisfying the following conditions: (1) and (2) . This completes the proof of Main Theorem. □
In 1987, Shi in  put forward the following conjecture:
Conjecture 3.2 Let G be a group and M be a finite simple group. Then if and only if (1) and (2) .
Corollary 3.3 Let M be one of the following simple -groups: , , , , , , and G be a finite group such that and . Then .
Proof If , then G and 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 and , then and . However, G is not isomorphic to . Hence, we put forward the following question:
Question 4.2 Is exactly fourfold OD-characterizable?
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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