Conjugacy Class Lengths of Finite Groups with Prime Graph a Tree

Liguo He^*, Yaping Liu, Jianwei Lu

Dept. of Math., Shenyang University of Technology, Shenyang, PR China

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(Liguo He)

^*Corresponding author

To cite this article:

Liguo He, Yaping Liu, Jianwei Lu. Conjugacy Class Lengths of Finite Groups with Prime Graph a Tree. Mathematics and Computer Science. Vol. 1, No. 1, 2016, pp. 17-20. doi: 10.11648/j.mcs.20160101.14

Received: April 11, 2016; Accepted: May 3, 2016; Published: May 28, 2016

Abstract: For a finite group , we write  to denote the prime divisor set of the various conjugacy class lengths of  and  the maximum number of distinct prime divisors of a single conjugacy class length of . It is a famous open problem that  can be bounded by . Let G be an almost simple group  such that the graph  built on element orders is a tree. By using Lucido’s classification theorem, we prove  except possibly when  is isomorphic to , where  is an odd prime and  is a field automorphism of odd prime order . In the exceptional case, . Combining with our known result, we also prove that for a finite group  with  a forest, the inequality  is true.

Keywords: Prime Graph, Conjugacy Class Length, Almost Simple Group

1. Introduction

Throughout the paper, we only consider finite groups. For a finite group , write  to denote the prime divisor set of the various conjugacy class lengths of  and  the maximum number of different prime divisors of a single conjugacy class length of . And  stands for prime divisor set of the positive integer  and write  for , here  denotes the order of . Considering some similarities between the influence of character degrees and conjugacy class lengths on groups, in 1989, Huppert once asked [1] whether the inequality  holds for every solvable group. It was shown in [2, 4, 6] that this is true when  is at most 3. Specifically, the case  is proven in [4], the case  for the solvable  is in [6], and the cases  for the nonsolvable  and  are finished in [2], respectively. For solvable groups, it is proved  in [7], and an improved version  in [20]. Generally, Casolo proved in [1, Corollary 2] that the inequality is true except possibly when  is  nilpotent with abelian Sylow  subgroups for at least two prime divisors of . And yet Casolo and Dolfi in [3, Example 2] show the inequality is invalid by constructing an infinite family of group examples . Specifically, the quotient  approaches  (from below) as  approaches infinity. The prime numberdivides each . When take , we may obtain an infinity family of counterexamples such that each  is metabelian and super solvable and the constant is 3 in that inequality. Note that the subscripts  in these counterexamples are sufficient large. By observation of these counterexamples, in 1998, Zhang further conjectured [20] the weak version of that inequality should hold by saying that "Now, it seems true that  for any finite solvable group and probably also for any finite group." We attach a prime graph  to a finite group : its vertices are , and any two vertices are adjacent by an edge just when  has an element of order . We use  to denote the connected components of the graph , in particular,  if  is of even order. Furthermore, the graph  is called a tree when it is a connected graph without any loop; and  is called a forest when each connected component is a tree. In this note, we prove the following results.

Theorem A. Suppose that  is an almost simple group such that  is a tree. Then  except possibly when  is isomorphic to , whereis an odd prim and  is a field automorphism of odd prime order . In the exceptional case, .

Theorem B. Suppose that  is a finite group such that is a forest. Then .

In the proof of Theorem A, we apply the classification result due to Lucido (Theorem 2.2). In the process, GAP [6] plays a crucial role. In essence, we use finite simpl egroup classification theorem.

Unless otherwise specified, we adopt the standard notation and terminology as presented in [9].

2. Preliminaries

The following fact is useful and basic on conjugacy class lengths, which will be frequently applied without reference.

Lemma 2.1 ((Lemma 33.2 of [9])). Letbe a normal subgroup of  and . Then

1.  divides  for any , and so.

2.  divides  for any , and thus .

Theorem 2.2. If  is an almost simple group with  a tree, then  is one of the following types of groups.

1.  for the alternating group  of degree 6.

2.  such that  is a prime,  is an odd prime and  is a field automorphism of order .

3. .

4. .

5. , where  is a diagonal automorphism of order 2.

6.  with  is a graph-field automorphism of order 2.

7. , with  and  a field automorphism of order 2.

8. .

 with  a field automorphism of order . Here  is an odd prime.

Proof. This is Lemma 3 of [11].

Lemma 2.3. Assume that  is a finite group with disconnected . Then the inequality  is true.

Proof. This is Theorem A of [8].

Lemma 2.4. Let  be a finite group with . Then .

Proof. By [2, 3, 5], the inequality is valid when . Otherwise, we see  and so , as wanted.

Theorem 2.5. Let  be a finite group such that  is a tree, then .

Proof. This is Theorem 6 of [11].

Lemma 2.6. Let  be an almost simple group over the nonabelian simple group , then .

Proof. It is known  by Theorem 33.4 of [9]. If  has a maximal central Hall subgroup , then we can write . As  is a simple group, the intersection  is trivial. It follows that  divides , and so . We obtain  acts trivially on , however, this is a contradiction since . Therefore  is trivial an .

3. Main Results

Theorem 3.1. Suppose that  is an almost simple group such that  is a tree. Then  except possibly whenis isomorphic to , where is an odd prime and is a field automorphism of odd prime order. In the exceptional case, .

Proof. We apply Theorem 2.2 above to prove the claims.

If  is isomorphic to , then we know via GAP [6] that  is trivial and , thus we obtain , as desired.

Assume thatis isomorphic to . Here is an odd prime and  is a field automorphism of order  which is also an odd prime. By Lemma 4.1of [10], we know  is trivial and so . It is known when  is an odd prime. As in [4], denote by  an element of order  in  and by  the element of order 2 in the centre of . Then it is known that the class size of  in  is  (which is the conjugacy class size corresponding to , and so we conclude that  for odd . The field automorphism  indeed leave the class of  invariant by [10, Lemma 4.1]. We further get

for odd . Note that is an odd prime.

Assume now that . Then we see

and

,

By using GAP [6], we know that  has 43 conjugacy class lengths,

 and ; and has 56 conjugacy class lengths,  and . Hence we obtain , as required.

If  is isomorphic to , then the application of GAP yields that  has twenty conjugacy class lengths,  and , and so , as wanted.

Next, consider the case that, where  is a diagonal automorphism of order 2. Using GAP, we reach that  has 20conjugacy class lengths,  and . Its automorphism group  has 61 conjugacy class lengths,  and . As  is an almost simple group, we achieve that  and , thus , as desired.

Now, suppose that  with  is a graph-field automorphism of order 2. As  is of order 2, we attain  is either  or else

. When , by GAP, we know and , and so. When , we also get that  because  is of order 2 which is in .

The next case is , with  and  a field automorphism of order 2. For , we get via GAP that  and . Also since

 this yields . For , by GAP, we attain that  and . Also , this implies .

For , we get by GAP that and . Also , this shows , as desired.

Assume that . By using GAP, it follows that  and . Note that  is an almost simple group with , thus .

Assume finally that  is isomorphic to , where  is a field automorphism of odd prime order . Let be a Sylow 2-subgroup of .

By Proposition 1 of [12], we see  for any nontrivial element .

Using Lemmas 1 and 2 of [12], we may pick the noncentral element , which is indeed a specific form of  (by [12, Theorem 7]). Here neither of  or  can equal 0, 1. Then the centralizer  is properly contained in , and so . If  belongs to , then . Otherwise,  is a Sylow -subgroup. By [7, Theorem 9], we have  and so . The whole proof is complete.

Some remarks on  are made here. By [4], we see that  for the odd  and  for .

Lemma 2 of [11] yields  unless. Some direct calculations by GAP show that ,  and  are all equal to 3, but  and  are 2. Observe that   and , moreover .

Theorem 3.2. Suppose that  is a finite group such that is a forest. Then

.

Proof. If  is disconnected, then Lemma 2.3 yields the result. Otherwise,  is a tree, and so  (by Lemma 2.5). Applying Lemma 2.4, we get the result.

In this note, we show Huppert’s problem has affirmative answer for the finite group whose prime graph a forest, and even has better result when the group is an almost simple group with prime graph a tree.

Acknowledgements

Project supported by NSF of China (No. 11471054) and NSF of Liaoning Education Department (No. 2014399).

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