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 number
divides 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
, where
is 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 when
is 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