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
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
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  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 , the case for the solvable is in , and the cases for the nonsolvable and are finished in , respectively. For solvable groups, it is proved in , and an improved version in . 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  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  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 .
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 )). 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 .
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.
with a field automorphism of order . Here is an odd prime.
Proof. This is Lemma 3 of .
Lemma 2.3. Assume that is a finite group with disconnected . Then the inequality is true.
Proof. This is Theorem A of .
Lemma 2.4. Let be a finite group with . Then .
Theorem 2.5. Let be a finite group such that is a tree, then .
Proof. This is Theorem 6 of .
Lemma 2.6. Let be an almost simple group over the nonabelian simple group , then .
Proof. It is known by Theorem 33.4 of . 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  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 , we know is trivial and so . It is known when is an odd prime. As in , 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
By using GAP , 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 , we see for any nontrivial element .
Using Lemmas 1 and 2 of , 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 , we see that for the odd and for .
Lemma 2 of  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.
Project supported by NSF of China (No. 11471054) and NSF of Liaoning Education Department (No. 2014399).