1. Introduction
Let G be a finite group. By we denote the set of element orders of G. For an integer n we define as the set of prime divisors of n and we set for . The prime graph of the Gruenberg-Kegel graph of G is denoted by and it is a graph with vertex set in which two distinct vertices p and q are adjacent if and only if , and in this case we will write .
A subset of vertices of is called an independent subset of if its vertices are pairwise nonadjacent. Denote by the maximal number of primes in pairwise nonadjacent in . We also denote by the maximal number of vertices in the independent sets of containing 2. A finite nonabelian simple group P is called quasirecognizable by prime graph if every finite group G with has a composition factor isomorphic to P. P is called recognizable by prime graph if implies . In addition, finite group P is considered to be recognizable by a set of element orders if the equality , for each finite group G implies that . A finite simple nonabelian group P is considered to be quasirecognizable by the set of element orders if each finite group H with has a composition factor isomorphic to P. If a finite simple group is quasirecognizable (recognizable) by prime graph, then it is quasirecognizable (recognizable) by set of element orders, but the inverse is not true necessarily and proving by prime graph is more difficult.
Hagie determined finite groups
H satisfying
, where
S is a sporadic simple group [
1]. In [
2,
3], finite groups with the same prime graph as
, where
q is a prime power, are determined. Quasirecognizability by prime graph of groups
and
has been proved in [
4]. In [
5,
6,
7], finite groups with the same prime graphs as
,
,
,
and
are obtained. In addition, in [
8], it is proved that if
p is a prime less than 1000, for suitable
n, the finite simple groups
and
are quasirecognizable by prime graph. Now as the main result of this paper, we prove the following theorem:
Main Theorem. The finite simple group , where and , is quasirecognizable by prime graph.
Throughout this paper, all groups are finite and by a simple group we mean a nonabelian simple group. All further unexplained notations are standard and the reader is referred to [
9].
3. Proof of the Main Theorem
Throughout this section, we suppose that where , and G is a finite group such that . We denote a primitive prime divisor of by and a primitive prime divisor of by , where .
By ([
12] Tables 4, 6 and 8), we deduce that
and
. Therefore,
and
. Now by Lemma 1, it follows that there exists a finite nonabelian simple group
S such that
where
K is the maximal normal solvable subgroup of
G. In addition,
and
by Lemma 1. Therefore,
and
. On the other hand, by ([
12] Tables 2 and 9), if
S is isomorphic to a sporadic or an exceptional simple group of Lie type, then
. This implies that
S is not isomorphic to any sporadic or any exceptional simple groups of Lie type.
In the sequel, we consider each possibility for S.
Lemma 7. S is not isomorphic to any alternating group.
Proof. Suppose that , where . Since , Lemma 1, we get that and so {.
Case 1. Let , where . Therefore, we get that or . , so G contains an element , such that , which implies that . Now we have . Hence, according to Remark 1, in S. On the other hand, 47 is not connected to the prime numbers in the interval in the prime graph of and similarly, 59 is not connected to the prime numbers in the interval , in the prime graph of . However, these intervals contain at most 16 prime numbers, and this implies that , is a contradiction.
Case 2. Let , where . Using GAP, we get that:
First let . Then we have and so similarly to Case 1, we get that , is a contradiction.
Let . Since , again similarly to Case 1, we get a contradiction. ☐
Lemma 8. If S is isomorphic to a classical simple group of Lie type over a field of characteristic p, then .
Proof. Let
S be a nonabelian simple group of Lie type over
,
. By the hypothesis,
where
N is the maximal normal solvable subgroup of
G. In the sequel, we denote by
a primitive prime divisor of
and by
a primitive prime divisor of
. We remark that
and
by Lemma 1.
Now we consider the following cases:
Case 1. Let . In addition, let and be two primitive prime divisors of and , respectively. So we may assume that and are and , respectively. This implies that . Thus is a primitive prime divisor of and is a primitive prime divisor of , where and . It follows that and . On the other hand, using Zsigmondy’s theorem, we conclude that and so . Furthermore, since , we have and .
Now we consider each possibility for
S, separately. If
, then using the results in [
12], each
, where
is adjacent to
p in
.
Subcase 1.1. Let
. By [[
12] Proposition 2.6], we see that each prime divisor of
is adjacent to
p, except
and
. Hence
. Therefore,
and
are some primitive prime divisors of
and
. Since
, we conclude that
and
. Hence
and
. Consequently, we get that
and
, that is
. Then
S has a maximal torus of order
, say
T. Obviously,
. Therefore,
in
, whereas
in
, by Lemma 2, which is a contradiction.
Subcase 1.2. Let . If , then and so and . Hence and . Therefore , is a contradiction.
If , then and so and . Hence and . Then and so , is a contradiction.
If , then Therefore, and . Hence and . Now easy computation shows that it is impossible.
If , then similarly to the above discussion, we get a contradiction.
Subcase 1.3. Let
or
. Since
, using [[
12] Table 4], we get that
m is odd. In this case,
. Hence
and
and so
and
, which implies that
, which is a contradiction. Let
, where
m is odd. Since
, we conclude that
and
and so
, which is impossible, since
n is even.
Similarly, we can prove that , where m is even and .
Case 2. Let . Hence . Let and be as and , respectively. Therefore and are primitive prime divisors of and , respectively, where and . Now we conclude that and . On the other hand, using Zsigmondy’s theorem, we conclude that and so . If , then using Zsigmondy’s theorem, we conclude that , which implies that , which is a contradiction. Hence we suppose that .
Subcase 2.1. Let , where . We know that . Hence , , , and . These equations imply that , which is impossible.
Subcase 2.2. Let , where m is odd. We note that and so and and so , which is impossible.
Subcase 2.3. Let , where m is even. Since , we get that and or . If , then we get that , which is impossible. Hence , which implies that and , and so , which is a contradiction, since .
We can use a similar proof for groups , , and and get a contradiction. We omit the proof for convenience. ☐
Lemma 9. If S is isomorphic to a classical simple group of Lie type over a field of characteristic , then .
Proof. Let
S be isomorphic to a classical simple group of Lie type over a field with
elements, where
. Using [[
12] Table 4],
and so Lemma 1 implies that
. On the other hand, by Lemma 2, we deduce that if
, then
. Hence
and so
.
Consider and . By Lemmas 2, 3, 4 and 5 we get that for each classical simple group of Lie type S, we have . On the other hand, using Remark 1, we have and so . We note that . Hence, by Lemma 2, it follows that . Since , we conclude that , where and we have a similar argument for , and .
Hence, according to the above discussion, if
, then the following condition holds:
Using GAP, we get that the above condition holds only for , where . Since , we conclude that . If and , then and so , which contradicts Remark 1. Therefore, by the above argument, we get that S is not isomorphic to any classical simple group of Lie type over a field of characteristic . ☐
Using the Classification Theorem of finite simple groups and Lemmas 7–9, we get that the finite simple group , where and is quasirecognizable by prime graph.