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Recognizing L 2 (p) by its order and one special conjugacy class size

Abstract

In the past thirty years, many authors investigated some quantitative characterizations of finite groups, especially finite simple groups, such as quantitative characterizations by group order and element orders, by the set of lengths of conjugacy classes, by dimensions of irreducible characters, etc. In this article the projective special linear group L 2 (p) is characterized by its order and one special conjugacy class size, where p is a prime. This work implies that Thompson’s conjecture holds for L 2 (p).

MSC:20D08, 20D60.

1 Introduction

All groups considered in this paper are finite, and simple groups are non-Abelian. For convenience, we use π(n) and n p to denote the set of prime divisors and p-part of the nature number n, respectively. For any group G, we also denote by N(G) the set of conjugacy class sizes of G and by π(G)=π(|G|).

In 1970s, a simple graph called prime graph of the group G was introduced: the vertex set of this graph is π(G), two vertices p and q are joined by an edge if and only if G contains an element of order pq (see [1]). Denote the connected components of the prime graph of the group G by T(G)={ π i (G)|1it(G)}, where t(G) is the number of the prime graph components of G. If the order of G is even, we always assume that 2 π 1 (G). Then |G| can be expressed as a product of m 1 , m 2 ,, m t ( G ) , where m i ’s are positive integers with π( m i )= π i (G). These m i ’s are called the order components of G. In particular, if m i is an odd number, then we call it an odd-order component of G.

In 1988, Thompson posed the following conjecture (ref. to [[2], Problem 12.38]).

Thompson’s conjecture Let G be a group with trivial central. If L is a simple group satisfying N(G)=N(L), then GL.

In 1994, Chen proved in his Ph.D. dissertation [3] that Thompson’s conjecture holds for all simple groups with a non-connected prime graph (also ref. to [46]). In 2009, Vasil’ev first dealt with the simple groups with a connected prime graph and proved that Thompson’s conjecture holds for A 10 and L 4 (4) (see [7]). Later on, Ahanjideh in [8] proved that Thompson’s conjecture is true for some projective special linear groups. Recently, Chen and Li contributed their interests on Thompson’s conjecture under a weak condition. They only used group order and one or two special conjugacy class sizes of simple groups and characterized successfully sporadic simple groups (see Li’s Ph.D. dissertation [9]) and simple K 3 -groups (a finite simple group is called a simple K n -group if its order is divisible by exactly n distinct primes), by which they checked Thompson’s conjecture for sporadic simple groups and simple K 3 -groups. Hence, it is an interesting topic to characterize simple groups with their orders and few conjugacy class sizes. In this paper, we characterize the projective special linear group L 2 (p) by its order and one special conjugacy class length, where p is a prime. This work partially generalizes Chen and Ahanjideh’s work [4, 8], which proved that Thompson’s conjecture holds for all the projective special linear group L n (q).

For convenience, we use ϵ to denote ±1. In addition, we denote by G p and Syl p (G) a Sylow p-subgroup of the group G and the set of all of its Sylow p-subgroups for pπ(G), respectively. We also denote by A:B an extension of a normal subgroup A by another subgroup B. The other notation and terminologies in this paper are standard and the reader is referred to [10] and [11] if necessary.

2 Some lemmas

A group G is called 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. Now, we quote some known lemmas which are useful in the sequel.

Lemma 2.1 [[12], Theorem 1]

Suppose that G is a Frobenius group of even order and H, K are the Frobenius kernel and the Frobenius complement of G, respectively. Then t(G)=2, T(G)={π(H),π(K)} and G has one of the following structures:

  1. (i)

    2π(H) and all Sylow subgroups of K are cyclic;

  2. (ii)

    2π(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;

  3. (iii)

    2π(K), H is Abelian, and there exists a subgroup K 0 of K such that

    |K: K 0 |2, K 0 =Z×SL(2,5), ( | Z | , 2 × 3 × 5 ) =1,

and the Sylow subgroups of Z are cyclic.

Lemma 2.2 [[12], Theorem 2]

Let G be a 2-Frobenius group of even order. Then t(G)=2 and G has a normal series 1HKG such that π(K/H)= π 2 (G), π(H)π(G/K)= π 1 (G), the order of G/K divides the order of the automorphism group of K/H, and both G/K and K/H are cyclic. Especially, |G/K|<|K/H| and G is solvable.

Lemma 2.3 Let G be a group of order 2 s ( 2 s 1)( 2 s 1 1), where 2 s 1 is a Mersenne prime and s is a natural number. Assume that G is a 2-Frobenius group with a normal series 1HKG, where H is an elementary Abelian 2-group of order 2 s and K/H is a cyclic group of order 2 s 1. Then G( Z 2 × Z 2 × Z 2 : Z 7 ): Z 3 .

Proof By hypothesis, |G/K|= 2 s 1 1, and so (|K|,|G/K|)=1. By the Zassenhaus theorem, there exists a complement subgroup C of K in G such that G=KC. Let B be a complement subgroup of H in K. Then G=HBC, where H and HB are normal subgroups of G; HB and BC are Frobenius groups with kernels H, B and complements B, C, respectively. Let F=GF( 2 s ) and so H is the additive group of F. Also, |B|= 2 s 1 and so B is the multiplicative group of F. Now, C acts by conjugation on H and similarly C acts by conjugation on B, and this action is faithful. Therefore, C keeps the structure of the field F and so C is isomorphic to a subgroup of the automorphism group of F. Hence, |C|=( 2 s 1 1)||Aut(F)|=s. Since 2 s 1 is a Mersenne prime, s is a prime. It follows that s=2 or 3. If s=2, then G is a group of order 12. By the structures of groups of order 12, we have that any group of order 12 is not a 2-Frobenius group, a contradiction. Hence, s=3, and then G is a group of order 168 that contains eight subgroups of order seven. By [13], we have that the total number of groups of order 168 with eight subgroups of order seven is three, which are ( Z 2 × Z 2 × Z 2 : Z 7 ): Z 3 , ( Z 2 × Z 2 × Z 2 : Z 7 )× Z 3 , and L 2 (7). Since G is a 2-Frobenius group, we obtain that G( Z 2 × Z 2 × Z 2 : Z 7 ): Z 3 . □

Lemma 2.4 [[1], Theorem A]

Let G be a group with more than one prime graph component. Then G is one of the following:

  1. (i)

    a Frobenius or 2-Frobenius group;

  2. (ii)

    G has a normal series 1HKG, where H is a nilpotent π 1 -group, K/H is a simple group and G/K is a π 1 -group such that |G/K| divides the order of the outer automorphism group of K/H. Besides, each odd-order component of G is also an odd-order component of K/H.

Lemma 2.5 [[11], Theorem 4.5.3]

Let G be a p-group with order p n , n1, and d is the number of minimal generators of G. Then |Aut(G)|| p d ( n d ) ( p d 1)( p d p)( p d p d 1 ).

Lemma 2.6 Let G be a simple group with a disconnected prime graph. Then its order components are exhibited in Tables 1-3, where p is an odd prime and q is a prime power.

Table 1 The order components of simple groups G with t(G)=2

Proof By [1, 14, 15] and the definition of order component, we easily get order components of G in Tables 1-3. Note that some mistakes and misprints in [1, 14, 15] are amended in this paper. □

3 Characterization of L 2 (p) by its order and one special conjugacy class size

By Lemma 2.3, we know that L 2 (7) is some special such that we have to choose a different way from other cases to deal with it. In fact, Chen and Li have characterized L 2 (7) by its order and the smallest conjugacy class size larger than one in their unpublished paper which characterized simple K 3 -groups. The following theorem is their one result.

Theorem 3.1 Let G be a group with |G|= 2 3 37. Then G L 2 (7) if and only if G has a conjugacy class size of 21.

Proof The necessity of the theorem can be checked easily, so we only need to prove the sufficiency.

By hypothesis, there exists an element x of G such that | x G |=|G: C G (x)|=37. In view of Z(G) C G (x), one has that Z(G) is a proper subgroup of G, and 3,7π(Z(G)). We assert that every minimal normal subgroup of G ¯ =G/Z(G) is non-solvable. Let S be any minimal normal subgroup of G ¯ . Suppose that S is solvable. Then S is an elementary Abelian group, from which we get the preimage T of S in G is a nilpotent group. If |S|= r t , then the Sylow r-subgroup R of T is normal in G. Moreover, R cannot be contained in Z(G). Thus, there exists an element y of R which is not contained in Z(G) such that 1<| y G ||R|<37, a contradiction. Hence, every minimal normal subgroup of G ¯ is non-solvable, as desired. It follows that M G ¯ Aut(M), where M= S 1 ×× S k and S i is a direct product of isomorphic non-Abelian simple groups for i=1,2,,k.

It is clear that 3 and 7 must belong to π(M). Otherwise, M is solvable, a contradiction. Hence, M is a simple K 3 -group and π(M)={2,3,7}. By checking possible order of M, M must be isomorphic to L 2 (7), which implies that G L 2 (7), as claimed. □

Theorem 3.2 Let G be a group. Then G L 2 (p) if and only if |G|=p( p 2 1)/(2,p1) and G has a special conjugacy class size of ( p 2 1)/(2,p1), where p is a prime and not equal to seven.

Proof Since the necessity of the theorem can be checked easily, we only need to prove the sufficiency.

If p=2,3, then G is a group of order six or twelve and not an element of order six. By the structures of the groups of order six and twelve, one has that G S 3 or A 4 , where S 3 is a symmetric group of degree three and A 4 is an alternating group of degree four. Note that L 2 (2) S 3 and L 2 (3) A 4 , as desired.

Let p5 but not equal to seven. By hypothesis, there exists an element x of order p in G such that C G (x)=x and C G (x) is a Sylow p-subgroup of G. By the Sylow theorem, we have that C G (y)=y for any element y in G of order p. So, {p} is a prime graph component of G and t(G)2. Therefore, G has one of the structures in Lemma 2.4. In addition, p is the maximal prime divisor of |G| and an odd-order component of G.

Suppose that G is a Frobenius group with a kernel H and a complement K. Then |K||(|H|1). If pπ(H), then by Lemma 2.1, |H|=p and |K|=( p 2 1)/2. It follows that p 2 1 2 |(p1), and thus p=1, a contradiction. If pπ(K), then |K|=p and |H|=( p 2 1)/2 by Lemma 2.1, and so p| p 2 3 2 . Since p is odd, we have that p|( p 2 3), which implies p=3, a contradiction. Hence, G is not a Frobenius group.

Assume that G is a 2-Frobenius group. By Lemma 2.2, we have that G has a normal series 1HKG such that π(K/H)={p}= π 2 (G), π(H)π(G/K)= π 1 (G), and |G/K||(p1). Then we have that K/H is of order p and π((p+1)/2)π(H).

  1. (a)

    Let p be not a Mersenne number. Then there exists an odd prime rπ((p+1)/2) such that | H r |<p. By Lemma 2.5, (p,|Aut( H r )|)=1, which implies that an element of order p of G can trivially act on H r . In other words, r can be connected to p in the prime graph of G, a contradiction.

  2. (b)

    Let p be a Mersenne number, and p= 2 s 1, s2. Then s is a prime and (p1)/2 is an odd number. Recall that |G/K||(p1).

If |G/K|=p1, then |H|=(p+1)/2= 2 s 1 <p such that (p,|Aut(H)|)=1 by Lemma 2.5. It follows that 2 and p connect in the prime graph of G, a contradiction.

If |G/K|=(p1)/2, then |H|=(p+1)= 2 s . If H is not an elementary Abelian 2-group, then (p,|Aut(H)|)=1 by Lemma 2.5, a contradiction. If H is an elementary Abelian 2-group, then by Lemma 2.3, s=3, and thus p=7, a contradiction.

Regarding other cases of |G/K|, we always can find an odd prime rπ(H) and r| p 1 2 such that (p,|Aut( H r )|)=1, which implies that G has an element of order pr, a contradiction. Therefore, G is not a 2-Frobenius group either.

Now, G has a normal series 1HKG, where H is a nilpotent π 1 -group, K/H is a simple group, G/K is a π 1 -group such that |G/K| divides the order of the outer automorphism group of K/H and each odd-order component of G is also an odd-order component of K/H. It follows that p is an odd-order component of K/H, and t(K/H)t(G)2, and also K/HG/HAut(K/H). We now proceed with the following proof in five steps by the possible number of the prime graph component of K/H in the Tables 1-3. Note that t(K/H)=2,3,4,5 or 6.

Step 1. Assume that t(K/H)=2. Then K/H is isomorphic to one of simple groups in Table 1. We assert that it is impossible.

1.1. If K/H A n , where 6< n = p , p +1, p +2, and one of n , n 2 is not a prime, then p= p and n ! 2 | p ( p 2 1 ) 2 . If n = p , then 2(p2)<p+1, and thus p<5, a contradiction. The cases n = p +1 or p +2 can be ruled out similarly.

1.2. If K/H A p 1 ( q ), where ( p , q )(3,2),(3,4), then p= q p 1 ( q 1 ) ( p , q 1 ) and q p ( p 1 ) 2 i = 1 p 1 ( q i 1)| p 2 1 2 . Since p is an odd prime and q is a prime power, we have that p 2 = ( q p 1 ) 2 [ ( q 1 ) ( p , q 1 ) ] 2 < q 2 p and q p ( p 1 ) 2 < p 2 1 2 < p 2 . It follows that q p ( p 1 ) 2 < q 2 p , and so p ( p 1 ) 2 <2 p , which implies that p =3. Hence, p 2 = ( q 3 1 ) 2 [ ( q 1 ) ( 3 , q 1 ) ] 2 and q 3 ( q 2 1)( q 1) p 2 1 2 . Since p 2 2 q 3 ( q 2 1) and 4 q 4 = q 2 ( 2 q ) 2 > q 2 ( q + 2 ) 2 > ( q 2 + q + 1 ) 2 ( q 3 1 ) 2 [ ( q 1 ) ( 3 , q 1 ) ] 2 = p 2 , we get that 4 q 4 >2 q 3 ( q 2 1), and so q 2 2 q 1<0. Therefore, q =2 if p =3, a contradiction.

1.3. If K/H A p ( q ), where ( q 1)|( p +1), then p= q p 1 q 1 and q p ( p + 1 ) 2 ( q p + 1 1) i = 1 p 1 ( q i 1)| p 2 1 2 . Since p 2 q 2 p and q p ( p + 1 ) 2 < p 2 , we have that p ( p + 1 ) 2 <2 p , and thus p <3, a contradiction.

1.4. If K/H 2 A p 1 ( q ), then p= q p + 1 ( q + 1 ) ( p , q + 1 ) and q p ( p 1 ) 2 i = 1 p 1 ( q i ( 1 ) i )| p 2 1 2 . Since p 2 < q 2 ( p + 1 ) and q p ( p 1 ) 2 < p 2 , we get that p ( p 1 ) 2 <2( p +1), and thus p 2 5 p 4<0. It follows that p =3 or 5. If p =3, then p 2 = ( q 3 + 1 ) 2 [ ( q + 1 ) ( 3 , q + 1 ) ] 2 and q 3 ( q 2 1)( q +1)| p 2 1 2 . Therefore p 2 ( q 2 q + 1 ) 2 q 4 and q 4 < q 3 ( q +1)< p 2 , a contradiction. Similarly, if p =5, then p 2 q 8 and q 10 < p 2 , a contradiction.

1.5. If K/H 2 A p ( q ), where ( q +1)|( p +1) and ( p , q )(3,3),(5,2), then p= q p + 1 q + 1 and q p ( p + 1 ) 2 ( q p 1) i = 1 p 1 ( q i 1)| p 2 1 2 . Hence, p 2 q 2 ( p + 1 ) and q p ( p + 1 ) 2 < p 2 , and thus q p ( p + 1 ) 2 < q 2 ( p + 1 ) . It follows that p ( p + 1 ) 2 <2( p +1), which means that p =3. Therefore, q 6 < p 2 = ( q 3 + 1 q + 1 ) 2 = ( q 2 q + 1 ) 2 q 4 , a contradiction.

1.6. If K/H B n ( q ), where n = 2 m 4 and q is odd, then p= q n + 1 2 and q n 2 ( q n 1) i = 1 n 1 ( q 2 i 1)| p 2 1 2 . Since p 2 < q 2 ( n + 1 ) and q n 2 < p 2 , we have that n 2 <2( n +1), and thus n 2 2 n 2<0, which implies that n <3, a contradiction.

1.7. If K/H B p (3), then p= 3 p 1 2 and 3 p 2 ( 3 p +1) i = 1 p 1 ( 3 2 i 1)| p 2 1 2 . Thus, p 2 < 3 2 p and 3 p 2 < p 2 , and so p <2, a contradiction.

1.8. If K/H C n ( q ), where n = 2 m 2, then p= q n + 1 ( 2 , q 1 ) and q n 2 ( q n 1) i = 1 n 1 ( q 2 i 1)| p 2 1 2 . Since p 2 < q 2 ( n + 1 ) and q n 2 < p 2 , we have that n 2 <2( n +1), and thus n 2 2 n 2<0, which implies that n =2. Therefore, p 2 ( q 2 + 1 ) 2 <4 q 4 and q 4 ( q 2 1 ) 2 < p 2 , and thus ( q 2 1 ) 2 <4. It follows that q =1, a contradiction.

1.9. If K/H C p ( q ), where q =2 or 3, then p= q p 1 ( 2 , q 1 ) and q p 2 ( q p +1) i = 1 p 1 ( q 2 i 1)| p 2 1 2 . Hence, p 2 < q 2 p and q p 2 < p 2 , and so p <2, a contradiction.

1.10. If K/H D p ( q ), where p 5 and q =2,3, or 5, then p= q p 1 q 1 and q p ( p 1 ) i = 1 p 1 ( q 2 i 1)| p 2 1 2 . Hence, p 2 < q 2 p and q p ( p 1 ) < p 2 , and so p ( p 1)<2 p , which implies that p <3, a contradiction.

1.11. If K/H D p + 1 ( q ), where q =2 or 3, then p= q p 1 ( 2 , q 1 ) and 1 ( 2 , q 1 ) q p ( p + 1 ) ( q p +1)( q p + 1 1) i = 1 p 1 ( q 2 i 1)| p 2 1 2 . Since p 2 < q 2 p and q p ( p + 1 ) < p 2 , we get that p ( p +1)<2 p , and thus 0< p <1, a contradiction.

1.12. If K/H 2 D n ( q ), where n = 2 m 4, then p= q n + 1 ( 2 , q + 1 ) and q n ( n 1 ) i = 1 n 1 ( q 2 i 1)| p 2 1 2 . Therefore, p 2 < q 2 ( n + 1 ) and q n ( n 1 ) < p 2 . Then we have that n ( n 1)<2( n +1), and thus n 2 3 n 2<0, which implies that n =2, a contradiction.

1.13. If K/H 2 D n ( q ), where n = 2 m +15 if q =2 and n 9 if q =3, then p= q n 1 + 1 ( 2 , q 1 ) and 1 ( 2 , q 1 ) q n ( n 1 ) ( q n +1)( q n 1) i = 1 n 2 ( q 2 i 1)| p 2 1 2 . Because p 2 < q 2 n and q n ( n 1 ) < p 2 , we have that n ( n 1)<2 n , and so 0< n <3, a contradiction.

1.14. If K/H 2 D p (3), where p 2 m +1 and p 5, then p= 3 p + 1 4 and 3 p ( p 1 ) i = 1 p 1 ( 3 2 i 1)| p 2 1 2 . Hence, p 2 < 3 2 ( p + 1 ) and 3 p ( p 1 ) < p 2 . It follows that p ( p 1)<2( p +1), and so 0< p <4, a contradiction.

1.15. If K/H G 2 ( q ), where q ϵ(mod3) and q >2, then p= q 2 ϵ q +1 and q 6 ( q 3 ϵ)( q 2 1)( q +ϵ)| p 2 1 2 . It follows that p 2 = ( q 2 ϵ q + 1 ) 2 < q 6 and q 6 < p 2 , a contradiction.

1.16. If K/H 3 D 4 ( q ), then p= q 4 q 2 +1 and q 12 ( q 6 1)( q 2 1)( q 4 + q 2 +1)| p 2 1 2 . Therefore, p 2 = ( q 4 q 2 + 1 ) 2 < q 8 and q 12 < p 2 , a contradiction.

1.17. If K/H F 4 ( q ), where q is odd, then p= q 4 q 2 +1 and q 24 ( q 8 1) ( q 6 1 ) 2 ( q 4 1)| p 2 1 2 . Thus, p 2 < q 8 and q 24 < p 2 , a contradiction.

1.18. If K/H E 6 ( q ), then p= q 6 + q 3 + 1 ( 3 , q 1 ) and q 36 ( q 12 1)( q 8 1)( q 6 1)( q 5 1)( q 3 1)( q 2 1)| p 2 1 2 . Since p 2 ( q 6 + q 3 + 1 ) 2 < q 18 and q 36 < p 2 , a contradiction can be obtained.

1.19. If K/H 2 E 6 ( q ), where q >2, then p= q 6 q 3 + 1 ( 3 , q 1 ) and q 36 ( q 12 1)( q 8 1)( q 6 1)( q 5 +1)( q 3 +1)( q 2 1)| p 2 1 2 . Hence, p 2 ( q 6 q 3 + 1 ) 2 < q 12 and q 36 < p 2 , a contradiction.

1.20. If K/H is isomorphic to one of A 3 2 (2), F 4 2 ( 2 ) , M 12 , J 2 , Ru, He, M c L, Co 1 , Co 3 , F 22 , and HN, then p=5,7,11,13,17,19,23, or 29, and | K / H | 2 | p 2 1 2 . By [10], we have that | K / H | 2 2 6 , but ( p 2 1 2 ) 2 2 4 , a contradiction.

Step 2. Suppose that t(K/H)=3. Then K/H is isomorphic to one of simple groups in Table 2. We assert that it is impossible except A 1 (p), where p5 is an odd prime and not equal to seven.

Table 2 The order components of simple groups G with t(G)=3

2.1. If K/H A p , where p >6 such that p and p 2 are primes, then p= p and p ! 2 | p ( p 2 1 ) 2 . It follows that 2(p2)<p+1, and thus p<5, a contradiction.

2.2. If K/H A 1 ( q ), where q ϵ(mod4) and q >3, then p= q , q + ϵ 2 .

Let p= q . Then K/H A 1 (p)= L 2 (p). Recall that K/HG/H and |G|= p ( p 2 1 ) 2 =| L 2 (p)|, we have that G L 2 (p), as desired.

Let p= q + 1 2 . Then q =2p1 and q ( q 1)| p 2 1 2 , and thus 4(2p1)p+1. Hence we get that p<1, a contradiction.

Let p= q 1 2 . Then q =2p+1 and q ( q +1)| p 2 1 2 . It follows that 4(2p+1)p1, and so it is impossible.

2.3. If K/H A 1 ( q ), where q >4 is even, then p= q +ϵ and q ( q ϵ)| p 2 1 2 , and thus 4(2pϵ)p+ϵ. Therefore, p=5, which means that q =4, a contradiction.

2.4. If K/H 2 D p (3), where p = 2 m +1 and m 2, then p= 3 p 1 + 1 2 or 3 p + 1 4 .

Let p= 3 p 1 + 1 2 . Then 3 p + 1 2 3 p ( p 1 ) ( 3 ( p 1 ) 1) i = 1 p 2 ( 3 2 i 1)| p 2 1 2 . Since p 2 < 3 2 p and 3 p ( p 1 ) < p 2 , we have that p ( p 1)<2 p , and so p <3, a contradiction.

Let p= 3 p + 1 4 . Then ( 3 p 1 +1) 3 p ( p 1 ) ( 3 ( p 1 ) 1) i = 1 p 2 ( 3 2 i 1)| p 2 1 2 . Hence, p 2 < 3 2 ( p + 1 ) and 3 p ( p 1 ) < p 2 , and thus p ( p 1)<2( p +1). It follows that p =3 and p=7, contradicting p 2 < 3 2 ( p + 1 ) .

2.5. If K/H G 2 ( q ), where q 0(mod3), then p= q 2 ϵ q +1 and ( q 2 +ϵ q +1) q 6 ( q 2 1)| p 2 1 2 . It follows that p 2 = ( q 2 ϵ q + 1 ) 2 < q 6 and q 6 < p 2 , a contradiction.

2.6. If K/H 2 G 2 ( q ), where q = 3 2 m + 1 >3, then p= q ϵ 3 q +1 and ( q +ϵ 3 q +1) q 3 ( q 2 1)| p 2 1 2 . Therefore, p 2 = ( q ϵ 3 q + 1 ) 2 < q 4 and q 3 ( q 2 1)< p 2 , and thus q 2 1< q , a contradiction.

2.7. If K/H F 4 ( q ), where q >2 is even, then p= q 4 +1 or q 4 q 2 +1. Assume p= q 4 +1. Then q 24 ( q 6 1 ) 2 ( q 4 1 ) 2 ( q 4 q 2 +1)| p 2 1 2 . It follows that p 2 = ( q 4 + 1 ) 2 < q 10 and q 24 < p 2 , a contradiction. Similarly, if p= q 4 q 2 +1, then q 24 ( q 6 1 ) 2 ( q 4 1 ) 2 ( q 4 +1)| p 2 1 2 , and so p 2 = ( q 4 q 2 + 1 ) 2 < q 8 and q 24 < p 2 , a contradiction.

2.8. If K/H 2 F 4 ( q ), where q = 2 2 n + 1 and q >2, then p= q 2 +ϵ 2 q 3 + q +ϵ 2 q +1 and q 12 ( q 4 1)( q 3 +1)( q 2 +1)( q 1)( q 2 ϵ 2 q 3 + q ϵ 2 q +1)| p 2 1 2 . Hence, p 2 = ( q 2 + ϵ 2 q 3 + q + ϵ 2 q + 1 ) 2 < ( 2 q 2 + 2 q + 2 ) 2 <4 ( q 3 1 ) 2 <4 q 6 < q 8 and q 12 < p 2 , a contradiction.

2.9. If K/H is isomorphic to one of M 11 , M 23 , and J 3 , then p=11,19, or 23 and | K / H | 2 | p 2 1 2 . By [10], we have that 2 7 | K / H | 2 2 4 , but ( p 2 1 2 ) 2 2 3 , a contradiction.

2.10. If K/H is isomorphic to one of A 5 2 (2), E 7 (2), E 7 (3), M 24 , HS, Sz, Co 2 , F 23 , F 2 , and F 3 , then p=11,13,19,23,31,47,127, or 1,093, and | K / H | 2 | p 2 1 2 . By [10], we have that | K / H | 2 > 2 7 , but ( p 2 1 2 ) 2 2 7 , a contradiction.

Step 3. Assume that t(K/H)=4. Then K/H is isomorphic to one of simple groups in Table 3 except E 8 ( q ), q 0,1,4(mod5) and J 4 . We assert that it is impossible.

Table 3 The order components of simple groups G with t(G)4

3.1. If K/H J 1 , then p=19 and 11| p 2 1 2 , but p 2 1 2 =180, a contradiction.

3.2. If K/H is isomorphic to one of A 2 (4), E 6 2 (2), M 22 , ON, Ly, F 24 , and F 1 , then p=7,11,19,29,31,67, or 71 and | K / H | 2 | p 2 1 2 . By [10], we have that | K / H | 2 2 6 , but ( p 2 1 2 ) 2 2 5 , a contradiction.

3.3. If K/H 2 B 2 ( q ), where q = 2 2 m + 1 and q >2, then p= q 1 or q +ϵ 2 q +1.

Let p= q 1. Then q 2 ( q 2 q +1)( q + 2 q +1)| p 2 1 2 . It follows that q 2 < p 2 = ( q 1 ) 2 < q 2 , a contradiction.

Let p= q 2 q +1. Then q 2 ( q 1)( q + 2 q +1)| p 2 1 2 . Therefore, p 2 < q 2 and q 3 < p 2 , a contradiction.

Let p= q + 2 q +1. Then q 2 ( q 1)( q 2 q +1)| p 2 1 2 . Therefore, p= 2 2 m + 1 + 2 m + 1 +1 and p 2 1 2 = 2 m + 1 ( 2 m +1)( 2 2 m + 2 m +1). Since | K / H | 2 | ( p 2 1 2 ) 2 , we have that 2 2 ( 2 m + 1 ) | 2 m + 1 , and thus 2(2 m +1) m +1, a contradiction.

3.4. If K/H E 8 ( q ), where q 2,3(mod5), then p= q 8 q 4 +1, q 8 q 7 + q 5 q 4 + q 2 q +1 or q 8 + q 7 q 5 q 4 q 2 + q +1. Assume that p= q 8 q 4 +1. Then p 2 < q 16 , but q 120 < p 2 , a contradiction. Similarly, for other cases of p, we can always obtain a contradiction.

Step 4. Assume that t(K/H)=5. Then K/H E 8 ( q ) in Table 3, where q 0,1,4(mod5), and thus p= q 8 q 4 +1, q 8 q 6 + q 4 q 2 +1, q 8 q 7 + q 5 q 4 + q 2 q +1 or q 8 + q 7 q 5 q 4 q 2 + q +1. By the way of 3.4, for all cases of p, we can always obtain a contradiction. Hence, we omit it.

Step 5. Assume that t(K/H)=6. Then K/H J 4 in Table 3. It follows that p=43 and 37| p 2 1 2 , but p 2 1 2 = 2 2 3711, a contradiction.

Therefore, we have that G L 2 (p), where p is a prime and not equal to 7, as desired. □

By Theorems 3.1 and 3.2, the following corollary holds.

Corollary 3.3 Thompson’s conjecture holds for the projective special linear group L 2 (p), where p is a prime.

Proof Let G be a group with trivial central and N(G)=N( L 2 (p)). Then it is proved in [4] that |G|=| L 2 (p)|. Hence, the corollary follows from Theorems 3.1 and 3.2. □

Remark 3.4 By 2.2 of Theorem 3.2 and Lemma 2.3, the following conclusion is true.

Let G be a group with |G|= 2 3 37 . Then G has a class size of 24 if and only if G L 2 (7) or ( Z 2 × Z 2 × Z 2 : Z 7 ): Z 3 .

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Acknowledgements

The authors are very grateful to the referee for carefully reading the earlier version of this manuscript and providing valuable suggestions and useful comments. This work was supported by the National Natural Science Foundation of China (Grant Nos. 11271301, 11171364, 11001226), Science and Technology Foundation of Chongqing Education Committee (No. KJ111107), Youth Foundation of Chongqing Three Gorges University (Grant Nos. 12QN-22, 12QN-23).

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Chen, Y., Chen, G. Recognizing L 2 (p) by its order and one special conjugacy class size. J Inequal Appl 2012, 310 (2012). https://doi.org/10.1186/1029-242X-2012-310

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Keywords

  • finite simple groups
  • conjugacy class size
  • prime graph
  • Thompson’s conjecture