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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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).
References