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Article

# Quasirecognition by Prime Graph of the Groups 2D2n(q) Where q < 105

by
1,
2,* and
Ali Iranmanesh
1
1
Department of Mathematics, Tarbiat Modares University, P.O. Box 14115-137, Tehran, Iran
2
School of mathematics, statistics and computer, College of science, University of Tehran, P.O. Box 14155-6455, Tehran, Iran
*
Author to whom correspondence should be addressed.
Mathematics 2018, 6(4), 57; https://doi.org/10.3390/math6040057
Submission received: 8 January 2018 / Revised: 8 March 2018 / Accepted: 14 March 2018 / Published: 9 April 2018

## Abstract

:
Let G be a finite group. The prime graph $Γ ( G )$ of G is defined as follows: The set of vertices of $Γ ( G )$ is the set of prime divisors of $| G |$ and two distinct vertices p and $p ′$ are connected in $Γ ( G )$, whenever G contains an element of order $p p ′$. A non-abelian simple group P is called recognizable by prime graph if for any finite group G with $Γ ( G ) = Γ ( P )$, G has a composition factor isomorphic to P. It is been proved that finite simple groups $2 D n ( q )$, where $n ≠ 4 k$, are quasirecognizable by prime graph. Now in this paper we discuss the quasirecognizability by prime graph of the simple groups $2 D 2 k ( q )$, where $k ≥ 9$ and q is a prime power less than $10 5$.

## 1. Introduction

Let G be a finite group. By $π e ( G )$ we denote the set of element orders of G. For an integer n we define $π ( n )$ as the set of prime divisors of n and we set $π ( G )$ for $π ( | G | )$. The prime graph of the Gruenberg-Kegel graph of G is denoted by $Γ ( G )$ and it is a graph with vertex set $π ( G )$ in which two distinct vertices p and q are adjacent if and only if $p q ∈ π ( G )$, and in this case we will write $p ∼ q$.
A subset of vertices of $Γ ( G )$ is called an independent subset of $Γ ( G )$ if its vertices are pairwise nonadjacent. Denote by $t ( G )$ the maximal number of primes in $π ( G )$ pairwise nonadjacent in $Γ ( G )$. We also denote by $t ( 2 , G )$ the maximal number of vertices in the independent sets of $Γ ( G )$ containing 2. A finite nonabelian simple group P is called quasirecognizable by prime graph if every finite group G with $Γ ( G ) = Γ ( P )$ has a composition factor isomorphic to P. P is called recognizable by prime graph if $Γ ( G ) = Γ ( P )$ implies $G ≅ P$. In addition, finite group P is considered to be recognizable by a set of element orders if the equality $π e ( G ) = π e ( P )$, for each finite group G implies that $G ≅ P$. A finite simple nonabelian group P is considered to be quasirecognizable by the set of element orders if each finite group H with $π e ( H ) = π e ( P )$ 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 $Γ ( H ) = Γ ( S )$, where S is a sporadic simple group [1]. In [2,3], finite groups with the same prime graph as $Γ ( P S L ( 2 , q ) )$, where q is a prime power, are determined. Quasirecognizability by prime graph of groups $G 2 ( 3 2 n + 1 )$ and $2 B 2 ( 2 2 n + 1 )$ has been proved in [4]. In [5,6,7], finite groups with the same prime graphs as $Γ ( L n ( 2 ) )$, $Γ ( U n ( 2 ) )$, $Γ ( D n ( 2 ) )$, $Γ ( 2 D n ( 2 ) )$ and $Γ ( 2 D 2 k ( 3 ) )$ 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 $L n ( p )$ and $U n ( p )$ are quasirecognizable by prime graph. Now as the main result of this paper, we prove the following theorem:
Main Theorem.
The finite simple group $2$$D 2 k ( q )$, where $k ≥ 9$ and $q < 10 5$, 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].

## 2. Preliminary Results

Lemma 1
([10] Theorem 1). Let G be a finite group with $t ( G ) ≥ 3$ and $t ( 2 , G ) ≥ 2$. Then the following hold:
1.
there exists a finite nonabelian simple group S such that
$S ≤ G ¯ = G / K ≤ A u t ( S )$
for the maximal normal soluble subgroup K of G.
2.
for every independent subset ρ of $π ( G )$ with $| ρ | ≥ 3$ at most one prime in ρ divides the product $| K | · | G ¯ / S |$. In particular, $t ( S ) ≥ t ( G ) − 1$.
3.
one of the following holds:
(a)
every prime $r ∈ π ( G )$ nonadjacent to 2 in $Γ ( G )$ does not divide the product $| K | · | G ¯ / S |$; in particular, $t ( 2 , S ) ≥ t ( 2 , G )$;
(b)
there exists a prime $r ∈ π ( K )$ nonadjacent to 2 in $Γ ( G )$; in this case $t ( G ) = 3$, $t ( 2 , G ) = 2$, and $S ≅ A 7$ or $L 2 ( q )$ for some odd q.
Remark 1.
In Lemma 1, for every odd prime $p ∈ π ( S )$, we have $t ( p , S ) ≥ t ( p , G ) − 1$.
If q is a natural number, r is an odd prime and $( q , r ) = 1$, then by $e ( r , q )$ we denote the smallest natural number m such that $q m ≡ 1 ( mod r )$. Given an odd q, put $e ( 2 , q ) = 1$ if $q ≡ 1 ( mod 4 )$ and put $e ( 2 , q ) = 2$ if $q ≡ − 1 ( mod 4 )$. Using Fermat’s little theorem we can see that if r is an odd prime such that $r ∣ ( q n − 1 )$, then $e ( r , q ) ∣ n$.
Lemma 2
([11] Proposition 2.5). Let $G = D n ε ( q )$, where q is power of prime p. Define
Suppose r, s are odd primes and $r , s ∈ π ( D n ε ( q ) ) \ { p }$. Put $k = e ( r , q )$, $l = e ( s , q )$, and $1 ≤ η ( k ) ≤ η ( l )$. Then r and s are non-adjacent if and only if $2 η ( k ) + 2 η ( l ) > 2 n − ( 1 − ε ( − 1 ) k + l )$ and k and l satisfy the following condition:
and if $ε = +$, then the chain of equalities
$n = l = 2 η ( l ) = 2 η ( k ) = 2 k$
is not true.
Lemma 3
([11] Proposition 2.3). Let G be one of the simple groups of Lie type, $B n ( q )$ or $C n ( q )$, over a field of characteristic p. Define
Let r, s be odd primes with $r , s ∈ π ( G ) \ { p }$. Put $k = e ( r , q )$ and $l = e ( s , q )$, and suppose that $1 ≤ η ( k ) ≤ η ( l )$. Then r and s are non-adjacent if and only if $η ( k ) + η ( l ) > n$, and k, l satisfy:
Lemma 4
([12] Proposition 2.1). Let $G = L n ( q )$, where q is a power of prime p. Let r and s be odd primes and $r , s ∈ π ( G ) \ { p }$. Put $k = e ( r , q )$ and $l = e ( s , q )$ and assume that $2 ≤ k ≤ l$. Then r and s are nonadjacent if and only if $k + l > n$ and k does not divide l.
Lemma 5
([12] Proposition 2.2). Let $G = U n ( q )$, where q is a power of prime p. Define
Let r and s be odd primes and r, $s ∈ π ( G ) \ { p }$. Put $k = e ( r , q )$ and $l = e ( s , q )$ and suppose that $2 ≤ ν ( k ) ≤ ν ( l )$. Then r and s are nonadjacent if and only if $ν ( k ) + ν ( l ) > n$ and $ν ( k )$ does not divide $ν ( l )$.
For Lemmas 2 and 5, simultaneously, we define the following function:
which we will use in the proofs. We note that a prime r with $e ( r , q ) = m$ is called a primitive prime divisor of $q m − 1$ (obviously, $q m − 1$ can have more than one primitive prime divisors).
Lemma 6.
(Zsigmondy’s theorem) [13] Let p be a prime and let n be a positive integer. Then one of the following holds:
1.
there is a primitive prime $p ′$ for $p n − 1$, that is , $p ′ ∣ ( p n − 1 )$ but $p ′ ∨ ( p m − 1 )$, for every $1 ≤ m < n$,
2.
$p = 2$, $n = 1$ or 6,
3.
p is a Mersenne prime and $n = 2$.

## 3. Proof of the Main Theorem

Throughout this section, we suppose that $D : = 2 D 2 k ( p α )$ where $k ≥ 9$, $2 < p α < 10 5$ and G is a finite group such that $Γ ( G ) = Γ ( D )$. We denote a primitive prime divisor of $q i − 1$ by $r i$ and a primitive prime divisor of $q ′ i − 1$ by $r i ′$, where $q ′ ≠ q$.
By ([12] Tables 4, 6 and 8), we deduce that $t ( D ) ≥ 14$ and $t ( 2 , D ) ≥ 2$. Therefore, $t ( G ) ≥ 14$ and $t ( 2 , G ) ≥ 2$. Now by Lemma 1, it follows that there exists a finite nonabelian simple group S such that
$S ≤ G ¯ : = G / K ≤ Aut ( S ) ,$
where K is the maximal normal solvable subgroup of G. In addition, $t ( S ) ≥ t ( G ) − 1$ and $t ( 2 , S ) ≥ t ( 2 , G )$ by Lemma 1. Therefore, $t ( S ) ≥ 13$ and $t ( 2 , S ) ≥ 2$. 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 $t ( S ) ≤ 12$. 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 $S ≅ A m$, where $m ≥ 5$. Since $t ( S ) ≥ 13$, Lemma 1, we get that $m ≥ 61$ and so {$47 , 59 } ⊆ π ( S )$.
Case 1. Let ${ 47 , 59 } ⊊ π ( q 2 − 1 )$, where $q = p α < 10 5$. Therefore, we get that $e ( 47 , q ) ≥ 23$ or $e ( 59 , q ) ≥ 29$. ${ 47 , 59 } ⊆ π ( S )$, so G contains an element $a ∈ A m$, such that $e ( a , q ) ≥ 23$, which implies that $n ≥ 24$. Now we have $min { t ( 47 , G ) , t ( 59 , G ) } ≥ 19$. Hence, according to Remark 1, $min { t ( 47 , S ) , t ( 59 , S ) } ≥ 18$ in S. On the other hand, 47 is not connected to the prime numbers in the interval $[ m − 46 , m ]$ in the prime graph of $A m$ and similarly, 59 is not connected to the prime numbers in the interval $[ m − 58 , m ]$, in the prime graph of $A m$. However, these intervals contain at most 16 prime numbers, and this implies that $t ( S ) < t ( G ) − 1$, is a contradiction.
Case 2. Let ${ 47 , 59 } ⊆ π ( q 2 − 1 )$, where $q = p α < 10 5$. Using GAP, we get that:
$q ∈ A : = { 11093 , 21713 , 27259 , 28201 , 38351 , 38821 , 39293 , 44839 , 55931 , 61007 , 66553 , 93811 , 99829 } .$
First let $q ∈ A \ { 21713 }$. Then we have $e ( 23 , q ) ≥ 11$ and so similarly to Case 1, we get that $t ( 23 , G ) − 1 > t ( 23 , S )$, is a contradiction.
Let $q = 21713$. Since $e ( 19 , q ) = 18$, 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 $S ≅ D$.
Proof.
Let S be a nonabelian simple group of Lie type over $GF ( q ′ )$, $q ′ = p β$. By the hypothesis,
$S ≤ G ¯ : = G / N ≤ Aut ( S ) ,$
where N is the maximal normal solvable subgroup of G. In the sequel, we denote by $r i$ a primitive prime divisor of $q i − 1$ and by $r i ′$ a primitive prime divisor of $q ′ i − 1$. We remark that ${ p , r 2 n } ⊆ π ( S )$ and $| ρ ( p , G ) ∩ π ( S ) | ≥ 3$ by Lemma 1.
Now we consider the following cases:
Case 1. Let $r 2 n − 2 ∈ π ( S )$. In addition, let $p 1$ and $p 2$ be two primitive prime divisors of $p ( 2 n − 2 ) α − 1$ and $p 2 n α − 1$, respectively. So we may assume that $p 1$ and $p 2$ are $r 2 n − 2$ and $r 2 n$, respectively. This implies that ${ r 2 n − 2 , r 2 n } ⊆ π ( S )$. Thus $r 2 n − 2$ is a primitive prime divisor of $q ′ s − 1$ and $r 2 n$ is a primitive prime divisor of $q ′ t − 1$, where $s = e ( r 2 n − 2 , p β )$ and $t = e ( r 2 n , p β )$. It follows that $( 2 n − 2 ) α ∣ s β$ and $2 n α ∣ t β$. On the other hand, using Zsigmondy’s theorem, we conclude that $t β ≤ 2 n α$ and so $t β = 2 n α$. Furthermore, since $2 n < 2 ( 2 n − 2 )$, we have $s β = ( 2 n − 2 ) α$ and $s < t$.
Now we consider each possibility for S, separately. If $ρ ( p , S ) = { r i ′ ∣ i ∈ I } ∪ { p }$, then using the results in [12], each $r j ′ ∈ π ( S )$, where $j ∉ I$ is adjacent to p in $Γ ( S )$.
Subcase 1.1. Let $S ≅ L m ( q ′ )$. By [[12] Proposition 2.6], we see that each prime divisor of $| L m ( q ′ ) |$ is adjacent to p, except $r m ′$ and $r m − 1 ′$. Hence $ρ ( p , S ) =$${ p , r m − 1 ′ , r m ′ }$. Therefore, $p 1$ and $p 2$ are some primitive prime divisors of $q ′ m − 1$ and $q ′ m − 1 − 1$. Since $s < t$, we conclude that $m = t$ and $m − 1 = s$. Hence $2 n α = m β$ and $( 2 n − 2 ) α = ( m − 1 ) β$. Consequently, we get that $β = 2 α$ and $m = n$, that is $S ≅ L n ( p 2 α )$. Then S has a maximal torus of order $( p 2 n α − 1 ) / ( ( p 2 α − 1 ) ( n , p 2 α − 1 ) )$, say T. Obviously, $r n , r 2 n ∈ π ( T )$. Therefore, $r n ∼ r 2 n$ in $Γ ( L n ( p 2 α ) )$, whereas $r n ≁ r 2 n$ in $Γ ( G )$, by Lemma 2, which is a contradiction.
Subcase 1.2. Let $S ≅ U m ( q ′ )$. If $m = 3$, then $ρ ( p , S ) = { p , r 1 ′ ≠ 2 , r 6 ′ }$ and so $s = 1$ and $t = 6$. Hence $( 2 n − 2 ) α = β$ and $2 n α = 6 β$. Therefore $n = 6 / 5$, is a contradiction.
If $m ≡ 0 ( mod 4 )$, then $ρ ( p , S ) = { p , r 2 m − 2 ′ , r m ′ }$ and so $s = m$ and $t = 2 m − 2$. Hence $( 2 n − 2 ) α = m β$ and $2 n α = ( 2 m − 2 ) β$. Then $n = ( 2 m − 2 ) / ( m − 2 )$ and so $n = 3$, is a contradiction.
If $m ≡ 3 ( mod 4 )$, then Therefore, $s = ( m − 1 ) / 2$ and $t = 2 m$. Hence $( 2 n − 2 ) α = ( m − 1 ) β / 2$ and $2 n α = 2 m β$. Now easy computation shows that it is impossible.
If $m ≡ 1 , 2 ( mod 4 )$, then similarly to the above discussion, we get a contradiction.
Subcase 1.3. Let $S ≅ B m ( q ′ )$ or $C m ( q ′ )$. Since $t ( p , S ) ≥ 3$, using [[12] Table 4], we get that m is odd. In this case, $ρ ( p , S ) = { p , r m ′ , r 2 m ′ }$. Hence $s = m$ and $t = 2 m$ and so $( 2 n − 2 ) α = m β$ and $2 n α = 2 m β$, which implies that $n = 2$, which is a contradiction. Let $S ≅ 2 D m ( q ′ )$, where m is odd. Since $ρ ( p , S ) = { p , r 2 m − 2 ′ , r 2 m ′ }$, we conclude that $( 2 n − 2 ) α = ( 2 m − 2 ) β$ and $2 n α = 2 m β$ and so $m = n$, which is impossible, since n is even.
Similarly, we can prove that $S ≇ 2 D m ( q ′ )$, where m is even and $S ≇ D m ( q ′ )$.
Case 2. Let $r 2 n − 2 ∉ π ( S )$. Hence $r n − 1 ∈ π ( S )$. Let $p 1$ and $p 2$ be as $r n − 1$ and $r 2 n$, respectively. Therefore $r n − 1$ and $r 2 n$ are primitive prime divisors of $q ′ s − 1$ and $q ′ t − 1$, respectively, where $s = e ( r n − 1 , p β )$ and $t = e ( r 2 n , p β )$. Now we conclude that $( n − 1 ) α ∣ s β$ and $2 n α ∣ t β$. On the other hand, using Zsigmondy’s theorem, we conclude that $t β ≤ 2 n α$ and so $t β = 2 n α$. If $s β > ( n − 1 ) α$, then using Zsigmondy’s theorem, we conclude that $s β = ( 2 n − 2 ) α$, which implies that $r 2 n − 2 ∈ π ( S )$, which is a contradiction. Hence we suppose that $s β = ( n − 1 ) α$.
Subcase 2.1. Let $S ≅ L m ( q ′ )$, where $q ′ = p β$. We know that $ρ ( p , S ) =$ ${ p , r m − 1 ′ , r m ′ }$. Hence $t = m$, $s = m − 1$, $2 n α = m β$, and $( n − 1 ) α = ( m − 1 ) β$. These equations imply that $m = 2 − 2 / ( n + 1 )$, which is impossible.
Subcase 2.2. Let $S ≅ 2 D m ( q ′ )$, where m is odd. We note that $ρ ( p , S ) = { p , r 2 m − 2 ′ , r 2 m ′ }$ and so $( n − 1 ) α = ( 2 m − 2 ) β$ and $2 n α = 2 m β$ and so $m = 2 − 2 / ( n + 1 )$, which is impossible.
Subcase 2.3. Let $S ≅ 2 D m ( q ′ )$, where m is even. Since $ρ ( p , S ) =$ ${ p , r m − 1 ′ , r 2 m − 2 ′ , r 2 m ′ }$, we get that $2 n α = 2 m β$ and $( n − 1 ) α = ( 2 m − 2 ) β$ or $( n − 1 ) α = ( m − 1 ) β$. If $( n − 1 ) α = ( 2 m − 2 ) β$, then we get that $m = 2 − 2 / ( n + 1 )$, which is impossible. Hence $( n − 1 ) α = ( m − 1 ) β$, which implies that $m = n$ and $α = β$, and so $S ≅ D$, which is a contradiction, since $r 2 n − 2 ∉ π ( S )$.
We can use a similar proof for groups $U m ( q ′ )$, $B m ( q ′ )$, $C m ( q ′ )$ and $D m ( q ′ )$ 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 $p ′ ≠ p$, then $S ≇ D$.
Proof.
Let S be isomorphic to a classical simple group of Lie type over a field with $q ′$ elements, where $q ′ = p ′ β$. Using [[12] Table 4], $t ( p ′ , S ) ≤ 4$ and so Lemma 1 implies that $t ( p ′ , G ) ≤ 5$. On the other hand, by Lemma 2, we deduce that if $r ∈ π ( G ) \ { r 1 , r 2 , r 3 , r 4 , r 6 }$, then $t ( r , G ) > 5$. Hence $p ′ ∈ { r 1 , r 2 ,$$r 3 ,$$r 4 , r 6 }$ and so $p ′ ∣ ( q 2 + 1 ) ( q 6 − 1 )$.
Consider $r 3 ′ ∈ π ( S ) ⊆ π ( G )$ and $3 = e ( r 3 ′ , q ′ ) ≤ e ( r 3 ′ , p ′ )$. By Lemmas 2, 3, 4 and 5 we get that for each classical simple group of Lie type S, we have $t ( r 3 ′ , S ) ≤ 6$. On the other hand, using Remark 1, we have $t ( r 3 ′ , G ) ≤ t ( r 3 ′ , S ) + 1$ and so $t ( r 3 ′ , G ) ≤ 7$. We note that $r 3 ′ ∈ π ( S ) ⊆ π ( G ) = π ( 2 D 2 k ( q ) )$. Hence, by Lemma 2, it follows that $e ( r 3 ′ , q ) ≤ 10$. Since $t ( S ) ≥ 13$, we conclude that $r i ′ ∈ π ( S )$, where $2 ≤ i ≤ 10$ and we have a similar argument for $r i ′$, $2 ≤ i ≤ 10$ and $e ( r i ′ , q ) ≤ 2 i + 4$.
Hence, according to the above discussion, if $p ′ ∈ π ( ( q 2 + 1 ) ( q 6 − 1 ) )$, then the following condition holds:
Using GAP, we get that the above condition holds only for $q = 54251$, where $p ′ = 2$. Since $t ( S ) ≥ 13$, we conclude that $r 13 ′ ∈ π ( S )$. If $p ′ = 2$ and $q = 54251$, then $r 13 ′ = 8191$ and so $e ( 8191 , q ) = 1365$, 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 $p ′ ≠ p$. ☐
Using the Classification Theorem of finite simple groups and Lemmas 7–9, we get that the finite simple group $2 D 2 k ( q )$, where $k ≥ 9$ and $q < 10 5$ is quasirecognizable by prime graph.

## Author Contributions

Hossein Morasdi is Phd student and Professor Mohammad Reza Darafsheh is supervisor and Professor Ali Iranmanesh is advisor. Select subject and edit of all article is by Mohammad Reza Darafsheh and calculation and software working done by Hossein Morasdi. Ali Iranmanesh has advised to use type of software and where use it and check details of calculations. Main Theorem was solved by Mohammad Reza Darafsheh.

## Conflicts of Interest

The authors declare no conflict of interest.

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MDPI and ACS Style

Moradi, H.; Darafsheh, M.R.; Iranmanesh, A. Quasirecognition by Prime Graph of the Groups 2D2n(q) Where q < 105. Mathematics 2018, 6, 57. https://doi.org/10.3390/math6040057

AMA Style

Moradi H, Darafsheh MR, Iranmanesh A. Quasirecognition by Prime Graph of the Groups 2D2n(q) Where q < 105. Mathematics. 2018; 6(4):57. https://doi.org/10.3390/math6040057

Chicago/Turabian Style

Moradi, Hossein, Mohammad Reza Darafsheh, and Ali Iranmanesh. 2018. "Quasirecognition by Prime Graph of the Groups 2D2n(q) Where q < 105" Mathematics 6, no. 4: 57. https://doi.org/10.3390/math6040057

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