Huaiyu He
School of Economics and Management, Shanghai University of Political Science and Law, Shanghai, China.
DOI: 10.4236/jamp.2026.1410208   PDF    HTML   XML   0 Downloads   28 Views   Citations

Abstract

Let $L$ be finite simple linear groups $A_p\left(5\right)$ with prime $p$ , $\omega \left(G\right)$ denote the spectrum of a finite group $G$ , i.e., the set of orders of its elements. We show that if $p>3$ is prime, then any finite group $G$ with $\omega \left(G\right)=\omega \left(L\right)$ is isomorphic to $L$ , where $L=A_p\left(5\right)$ has disconnected Gruenberg-Kegel graph. As an immediate consequence of the known results, the recognition problem for all simple groups $A_p\left(5\right)$ by spectrum is completely resolved.

Keywords

Finite Simple Groups, Recognition, Spectrum, Gruenberg-Kegel Graph

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1. Introduction

Within group theory, quantity is an interesting subject and also occupies an important position. Examples include the order of a group, the order of an element, the element of maximum order in a group, and the set of orders of its elements, and so on. Early results show that we can determine the type of a group using quantities alone. A celebrated result is the Feit-Thompson theorem: every finite group of odd order is solvable. Since the classification of finite simple groups was completed, the quantitative properties within simple groups have become highly interesting. In a recent paper [1], finite simple groups without elements of order 10 are obtained. And then, finite non-solvable groups without elements of order 10 and finite groups without elements of order 10 are obtained successively ([2] and [3]).

All groups in this paper are finite, and all simple groups are nonabelian. For a finite group $G$ , $\pi \left(G\right)$ stands for the set of prime divisors of the order of $G$ , and $\omega \left(G\right)$ denotes its spectrum, namely the set of orders of elements of $G$ . Since $\omega \left(G\right)$ is closed under divisibility, it is uniquely determined by the subset $\mu \left(G\right)$ of maximal elements of $\omega \left(G\right)$ under divisibility.

The spectrum $\omega \left(G\right)$ of a finite group $G$ defines its Gruenberg-Kegel graph $GK\left(G\right)$ . The vertex set of $GK\left(G\right)$ is $\pi \left(G\right)$ , and two primes $p,q\in \pi \left(G\right)$ are adjacent whenever G contains an element of order pq. Clearly, $\omega \left(G\right)$ uniquely determines $GK\left(G\right)$ . Hence, $\omega \left(G\right)=\omega \left(L\right)$ implies $GK\left(G\right)=GK\left(L\right)$ .

A finite group $G$ is called quasi-recognizable by spectrum if every finite group $H$ satisfying $\omega \left(H\right)=\omega \left(G\right)$ possesses a unique nonabelian composition factor isomorphic to $G$ . A finite group $G$ is recognizable by spectrum provided that any finite group $H$ with $\omega \left(H\right)=\omega \left(G\right)$ is isomorphic to $G$ . Equivalently, $G$ is recognizable if $h\left(G\right)=1$ , where $h\left(G\right)$ denotes the number of pairwise non-isomorphic finite groups $H$ with $\omega \left(H\right)=\omega \left(G\right)$ . By a well-known observation of W. J. Shi, a finite group admitting a nontrivial normal soluble subgroup cannot be recognizable. Therefore, the recognizability problem is of interest mainly for groups with trivial soluble radical, especially for simple and almost simple groups.

Let S be a finite simple group of Lie type of characteristic $r$ , and let A be an abelian $r$ -group equipped with an S-action. An element $s\in S$ is said to be unisingular on A if $s$ has a nonzero fixed point on A. The group S is called unisingular if every element $s\in S$ acts unisingularly on every finite abelian $r$ -group A admitting an S -action.

In [4], finite simple linear groups over fields of characteristic 2 are recognized by spectrum. Applying the adjacency criterions of finite simple groups ([5] and [6]), we solve the recognizability problem for simple groups of type $A_p\left(5\right)$ , which is the goal of this paper. This work enriches the existing body of research on the recognition of finite simple groups by spectrum.

For prime $p$ , if the Gruenberg-Kegel graph of simple group $A_p\left(5\right)$ is disconnected ([7] Table 2a-2c), then $4|p+1$ . After treating with them separately, we draw the following conclusions:

Theorem 1. Let $L=A_p\left(5\right)$ be a finite simple group with disconnected Gruenberg-Kegel graph. Then $L$ is recognizable by spectrum whenever $p>3$ .

The following conclusion is an immediate consequence of the known results.

Theorem 2. Let $L=A_p\left(5\right)$ be a finite simple group with disconnected Gruenberg-Kegel graph. Then $L$ is recognizable by spectrum for $p>3$ , quasi-recognizable by spectrum for $p=3$ , and satisfies $h\left(G\right)=2$ for $p=2$ .

2. Preliminary Results

This section contains several auxiliary results required in the sequel.

Gruenberg and Kegel obtained the following description of finite groups with disconnected graph $GK\left(G\right)$ , which was summarized in ([8] Lemma 1).

Lemma 2.1 (Gruenberg-Kegel Theorem). If G is a finite group with disconnected graph $GK\left(G\right)$ , then one of the following holds:

(a) $s\left(G\right)=2$ , $G$ is a Frobenius group;

(b) $s\left(G\right)=2$ , $G$ is a 2-Frobenius group;

(c) Let K denote a nilpotent normal subgroup of $G$ . Then there exists a non-abelian simple group S satisfying $S\le \overline{G}=G/K\le Aut\left(S\right)$ ; moreover, $K$ and $\overline{G}/S$ are ${\pi }_1\left(G\right)$ -groups, the graph $GK\left(S\right)$ is disconnected, $s\left(S\right)\ge s\left(G\right)$ , and for every $i$ , $2\le i\le s\left(G\right)$ , there exists $j$ satisfying $2\le j\le s\left(S\right)$ such that ${\omega }_i\left(G\right)={\omega }_j\left(S\right)$ .

A more constructive result, which generalizes the Gruenberg-Kegel theorem in a certain sense, was obtained in [5].

Lemma 2.2 ([5] Corollary 7.2). Let L be a finite non-abelian simple group other than $A_2\left(3\right),{}^{2}A{}_3\left(3\right),C_2\left(3\right),Al{t}_{10}$ and $Al{t}_n$ with $n$ satisfying $\left\{r|n-3\le r\le n,risprime\right\}=\varnothing$ . Let $G$ be a finite group with $\omega \left(G\right)=\omega \left(L\right)$ , and K denote the maximal normal soluble subgroup of G. Then there exists a finite nonabelian simple group S satisfying $S\le \overline{G}=G/K\le Aut\left(S\right)$ . Moreover, $t\left(S\right)\ge t\left(G\right)-1$ , and one of the following statements holds:

(1) $S\cong Al{t}_7$ or $A_1\left(q\right)$ for some odd $q$ , and $t\left(S\right)=t\left(2,S\right)=3$ .

(2) Each prime $p\in \pi \left(G\right)$ non-adjacent to 2 in $GK\left(G\right)$ satisfies that the Sylow p-subgroup of G is isomorphic to the Sylow p-subgroup of S. In particular, $t\left(2,S\right)\ge t\left(2,G\right)$ .

For a finite group, we write $\rho \left(G\right)$ $\left( \rho \left(r,G\right)\right)$ for some independent set in $GK\left(G\right)$ (containing $r$ ) with a maximal number of vertices, and put $t\left(G\right)=\left|\rho \left(G\right)\right|$ and $t\left(r,G\right)=\left|\rho \left(r,G\right)\right|$ . Let $s=s\left(G\right)$ denote the number of connected components of $GK\left(G\right)$ , and let ${\pi }_i={\pi }_i\left(G\right)$ be its $i$ th connected component for $i=1,\cdots ,s$ . For any even order group G, we adopt the convention that $2\in {\pi }_1$ . For each $i$ , let ${\mu }_i={\mu }_i\left(G\right)$ (resp. ${\omega }_i={\omega }_i\left(G\right)$ ) be the collection of all $n\in \mu \left(G\right)$ (resp. all $n\in \omega \left(G\right)$ ) such that every prime divisor of $n$ lies in ${\pi }_i$ . The symbol $\left[x\right]$ stands for the integral part of $x$ .

Lemma 2.3 ([7], Lemma 1.1). Let G be a finite simple group whose graph $GK\left(G\right)$ is disconnected. Then ${\mu }_i\left(G\right)$ is a singleton for $2\le i\le s\left(G\right)$ . Write $n_i$ for the unique element in ${\mu }_i\left(G\right)$ whenever $i>1$ . The group $G$ , the value $t\left(G\right)$ , and $n_i\left(G\right)$ with $2\le i\le s\left(G\right)$ are listed in Tables 1-3, where p stands for an odd prime.

Let q be a natural number and r an odd prime satisfying $\left(r,q\right)=1$ . We write $e\left(r,q\right)$ for the minimal positive integer number $n$ such that $q^n\equiv 1\left(\mathrm{mod}r\right)$ . If q is odd, we set $e\left(2,q\right)=1$ when $q\equiv 1\left(\mathrm{mod}4\right)$ and $e\left(2,q\right)=2$ when $q\equiv -1\left(\mathrm{mod}4\right)$ . An odd prime $r$ with $e\left(r,q\right)=n$ is called a primitive prime divisor of $q^n-1$ , denoted by $r_n$ . Zsigmondy proved the existence of such a prime in [9].

Lemma 2.4 ([10], Lemma 1). Suppose G is a finite group and $N⊲G$ . Assume $G/N$ is a Frobenius group with kernel F and a cyclic complement C. If $\left(\left|F\right|,\left|N\right|=1\right)$ and $F⊈N{C}_G\left(N\right)/N$ , then there exists a prime divisor $p$ of $\left|N\right|$ such that $p\left|C\right|\in \omega \left(G\right)$ .

3. Proof of Theorem 1

In this section, we investigate the spectrum recognizability of the simple group $A_p\left(5\right)$ whose Gruenberg-Kegel graph is disconnected, which is stated in Theorem 1.

At first we assume the following hypothesis:

Let $L=A_p\left(5\right)$ where $p>3$ , and suppose $G$ is a finite group satisfying $\omega \left(G\right)=\omega \left(L\right)$ . Then $GK\left(G\right)=GK\left(L\right)$ , $t\left(L\right)=\left[\frac{p+2}{2}\right]$ , $\rho \left(2,L\right)=\left\{2,r_p\right\}$ , $\rho \left(5,L\right)=\left\{5,r_p,r_{p+1}\right\}$ , $s\left(L\right)=2$ and $n_2\left(G\right)=n_2\left(L\right)=\frac{5^p-1}{4}>1093$ .

By Lemma 2.1 and Lemma 2.2, if $K$ denotes the maximal normal soluble subgroup of $G$ , then there exists a finite non-abelian simple group $S$ such that $S\le \overline{G}=G/K\le Aut\left(S\right)$ . Furthermore, $s\left(S\right)\ge s\left(G\right)=2$ , and $n_i\left(S\right)=n_2\left(G\right)=\frac{5^p-1}{4}$ for some $i$ with $2\le i\le s\left(S\right)$ .

By Lemma 2.3, the group $G$ , the value $t\left(G\right)$ and $n_i\left(G\right)$ with $2\le i\le s\left(G\right)$ are listed in the following Table 1.

Table 1. Finite simple groups G satiffying $s\left(G\right)\ge 2$ , $t\left(G\right)\ge 4$ and $n_i\ge 1093$ .

All $n_i\left(G\right)$ ($2\le i\le s\left(G\right)$ ) are quoted from the three tables above and will not be reproduced unless required.

Suppose $K\ne 1$ . For every prime $r\in \pi \left(K\right)$ , we have $O^r\left(K\right)\ne K$ ([6] Lemma 2.2). Let $\tilde{G}$ = $G/O^r\left(K\right)$ and $\tilde{K}$ =$K/O^r\left(K\right)$ denote these factor groups. Then $\tilde{K}$ is a nontrivial r-group. Write $\varphi \left(\tilde{K}\right)$ for the Frattini subgroup of $\tilde{K}$ , and define the factor groups $\widehat{G}=\tilde{G}/\varphi \left(\tilde{K}\right)$ and $\widehat{K}=\tilde{K}/\varphi \left(\tilde{K}\right)$ . As $G/K\cong \widehat{G}/\widehat{K}$ , the relation $\omega \left(\widehat{G}\right)⊈\omega \left(L\right)$ implies $\omega \left(G\right)⊈\omega \left(L\right)$ . It follows that we are allowed to replace G and K by $\widehat{G}$ and $\widehat{K}$ , respectively. After this reduction, we may assume that K is a nontrivial elementary abelian $r$ -group ([4], Proposition 1).

Lemma 3.1. $S$ is not isomorphic to any sporadic groups.

Proof. When $S$ is a sporadic group, we have $n_i\left(S\right)\le 71<\frac{5^p-1}{4}$ , then $S$ is not isomorphic to any sporadic groups by Table 1. $\square$

Since $t\left(L_2\left(q\right)\right)=3<t\left(L\right)-1$ , we have $S\ncong A_1\left(q\right)$ . Thus the second statement of Lemma 2.2 holds, then $r_p\notin \pi \left(K\right)$ and $\rho \left(2,G\right)=\left\{2,r_p\right\}\subseteq \rho \left(2,S\right)$ . We now examine all possibilities for the non-abelian simple group $S$ listed in Table 1.

Lemma 3.2. $S\ncong Al{t}_n\left(n\ge 5\right)$ .

Proof. Let $S\cong Al{t}_n\left(n\ge 5\right)$ , then at least one prime among $n,n-1,n-2,n-3$ equals $r_p$ . Since $n_i\left(S\right)=n_2\left(L\right)$ , we have $n\ge 11$ and $r_p=\frac{5^p-1}{4}$ . On the other hand, $r_{2p}|5^p+1$ and $5^p+1\equiv 0\left(\mathrm{mod}4\right)$ , then $r_{2p}\le \frac{5^p+1}{4}<\frac{5^p-1}{4}-6\le r_p-6$ . So $r_{2p}\in \pi \left(S\right)$ and $r_{2p}<n-3$ , then $r_{2p}$ is adjacent to 5 in $GK\left(S\right)$ ([5], Proposition 1.1). However they are non-adjacent in $GK\left(L\right)$ , we obtain a contradiction. $\square$ .

Lemma 3.3. $S$ is not isomorphic to any simple exceptional groups of Lie type.

Proof. Let $S\cong { }^3{D}_4\left(q\right)$ , then $t\left(L\right)\le t\left(S\right)+1\le 4$ , therefore $p\le 5$ . However in this case where $n_2\left(L\right)=n_2\left(S\right)<1093$ , a contradiction.

Let $S\cong E_8\left(q\right)$ , then $t\left(L\right)\le t\left(S\right)+1=13$ , therefore $p\le 17$ . And since $31\in \pi \left(S\right)$ , then $p=17$ .But in this case, $GK\left(L\right)$ is not disconnected since $q-1$ does not divisible $p+1$ , a contradiction by Table 1.

Let $S\cong E_7\left(q\right)$ , then $t\left(L\right)\le t\left(S\right)+1=8$ , therefore $p\le 7$ . If $q\ge 4$ , then $s\left(S\right)=1<s\left(G\right)$ , a contradiction. If $q<4$ , then $n_i\left(S\right)<1093$ , we obtain a contradiction by $n_i\left(S\right)=n_2\left(L\right)$ .

Let $S\cong G_2\left(q\right)$ , then $t\left(L\right)\le t\left(S\right)+1=4$ , therefore $p=3$ or 5. By $n_2\left(L\right)=n_i\left(S\right)$ , $p=3$ and $q=4$ or 3, then $21\notin \omega \left(L\right)$ and $30\notin \omega \left(L\right)$ . If $q=4$ , then $21\in \omega \left(S\right)$ , a contradiction. Thus $q=3$ , then $5\in \pi \left(K\right)$ for $G/K\le Aut\left(S\right)$ . And since $G_2\left(3\right)$ includes a Frobenius subgroup with kernel of order 13 and cyclic complement of order 6, then $30\in \omega \left(G\right)$ , a contradiction.

Let $S\cong F_4\left(q\right)$ or $E_6^{\epsilon }\left(q\right)$ , then $t\left(L\right)\le t\left(S\right)+1\le 6$ , therefore $p\le 7$ . If p = 3, then $21\notin \omega \left(L\right)$ , however $21\in \omega \left(S\right)$ , a contradiction. If p = 5, then $33\notin \omega \left(L\right)$ . Since $11=r_p\in \pi \left(S\right)$ , we have $33\in \omega \left(S\right)$ , a contradiction. Thus p = 7, and $n_2\left(L\right)=1093$ . However the equation $n_i\left(S\right)=n_2\left(L\right)$ admits no solutions ([11] Theorem 3), a contradiction.

Let $S$ be isomorphic to a finite simple Suzuki or Ree group, then $t\left(L\right)\le t\left(S\right)+1\le 6$ , therefore $p\le 7$ . However in this case the equation $n_i\left(S\right)=$ $n_2\left(L\right)$ admits no solutions, a contradiction ([11], Theorem 7). $\square$

Lemma 3.4. $S$ is not isomorphic any group of $B_n\left(q\right)$ , $C_n\left(q\right)$ or $D_n^{\epsilon }\left(q\right)$ .

Proof. If $S\cong B_n\left(q\right)$ or $C_n\left(q\right)$ , then $S\cong C_m\left(2\right)$ by the condition $t\left(2,S\right)\ge 3$ , with $m$ a prime. Then $n_2\left(S\right)=2^m-1=\frac{5^p-1}{4}$ , and this equation admits no solutions whenever $p>3$ ([11], Theorem 7).

If $S\cong {}^{2}D{}_n\left(q\right)$ , then $n_i\left(S\right)\in \left\{q^n+1,2^{n-1}+1,\frac{3^{n-1}+1}{2},\frac{3^n+1}{4}\right\}$ by Table 1 and $t\left(2,S\right)$ , where $q$ is even. Obviously, when $n_i\left(S\right)\in \left\{q^n+1,2^{n-1}+1,\frac{3^{n-1}+1}{2}\right\}$ , the equation $n_i\left(S\right)=n_2\left(G\right)=\frac{5^p-1}{4}$ has no solutions. And for $n_i\left(S\right)=\frac{3^n+1}{4}$ , the equation has also no solutions ([11] Theorem 1).

If $S\cong D_n\left(q\right)$ , then $S\cong D_n\left(5\right)$ or $D_n\left(2\right)$ by Table 1 and $t\left(2,S\right)$ . Obviously, since $n_i\left(S\right)=n_2\left(G\right)$ , we have $S\ncong D_n\left(2\right)$ ; otherwise, $2^n-1=n_2\left(D_n\left(2\right)\right)=\frac{5^p-1}{4}$ , which has no solutions whenever p > 3 ([11] Theorem 7). When $S\cong D_n\left(5\right)$ , by $n_i\left(S\right)=n_2\left(G\right)$ , we infer that $n=p$ , i.e., $S\cong D_p\left(5\right)$ . Noticing that $S<G/K\le Aut\left(S\right)$ implies $s\left(G/K\right)=1$ ([10] Theorem 5), we have $S=G/K$ . Suppose $K\ne 1$ and $l\in \pi \left(K\right)$ , then K can taken as an element abelian $l$ -group. Suppose that $C=C_G\left(K\right)\ne K$ . Since $C$ is normal in $G$ and $G/K$ is simple, we have $C/K=G/K=S$ . Therefore $r\cdot r_p\in \omega \left(G\right)=\omega \left(L\right)$ . However $r_p$ is non-adjacent to any other primes in $GK\left(L\right)$ , a contradiction. Thus $C=K$ and S acts faithfully on $K$ . If $l=5$ , then S acts unisingularly on K ([12] Theorem 1.3). Denote the natural semidirect product $K⋋S$ by $M$ , then $\omega \left(M\right)\subseteq \omega \left(G\right)$ . Denote by $o\left(x\right)$ the order of an element $x$ . Let $x\in K$ and $y\in S$ such that $o\left(x\right)=5$ and $o\left(y\right)=r_p$ , then $\left(x,y\right)\in M$ and $o\left(x,y\right)=5\cdot r_p$ , therefore $5r_p\in \omega \left(G\right)$ , a contradiction. If $l\ne 5$ , then S includes a Frobenius subgroup with kernel of order $q^{2p}$ and cyclic complement of order $r_p$ . Therefore $l\cdot r_p\in \omega \left(G\right)$ , a contradiction. Thus $K=1$ , i.e., S = G. However $t\left(S\right)=\left[\frac{3p+1}{4}\right]$ and $t\left(G\right)=\frac{p+1}{2}$ , we obtain a contradiction. Then $S\ncong D_p\left(5\right)$ .

$\square$

Lemma 3.5. $S\cong A_p\left(5\right)$ .

Proof. Suppose false. Let $S\cong A_n^{\epsilon }\left(q\right)$ where $q=r^k$ and $q\ne 5$ , then the group $S$ is one of the groups in the following list by Table 1:

$A_1\left(q\right)$ , where $q>2$ and $q$ is even; $A_{n-1}\left(q\right)$ , where $n$ is an odd prime, $\left(n,q\right)\ne \left(3,2\right),\left(3,4\right)$ ; $A_n\left(q\right)$ , where n is an odd prime, $\left(q-1\right)|\left(n+1\right)$ ; ${}^2{A}_{n-1}\left(q\right)$ , where n is an odd prime; ${}^2{A}_n\left(q\right)$ , where n is an odd prime, $\left(q+1\right)|\left(n+1\right)$ , $\left(n,q\right)\ne \left(3,3\right),\left(5,2\right)$ .

In this case $n-2\ge p$ for $p\ge 11$ . We have $n_2\left(S\right)\ge \frac{q^{n-2}}{2}$ , therefore $q<5$ by $n_2\left(S\right)=n_2\left(G\right)$ . When $q=2$ and $p\ge 11$ , we have $n_2\left(S\right)=2^n-1$ or $\frac{2^n+1}{3}$ . If $n_2\left(S\right)=2^n-1$ , then $2^{n+2}-5^p=3$ , which is impossible. If $n_2\left(S\right)=\frac{2^n+1}{3}$ , we have $2^{n+2}=3\cdot 5^p-7$ , which is also impossible. Similarly, we infer that $q\ne 3$ or 4 for $p\ge 11$ . Thus we have p < 11. Let p = 5, then $s\cong A_4\left(5\right)$ or $A_5\left(5\right)$ by $n_2\left(S\right)=n_2\left(G\right)$ . However in this case, $t\left(2,S\right)=2<t\left(2,G\right)$ , a contradiction. Similarly we can exclude $p=7$ . Similarly $S$ isn’t isomorphic to any other case.

When $S\cong A_n\left(5\right)$ , by $n_i\left(S\right)=n_2\left(G\right)$ , we infer that $n=p$ , i.e., $S\cong A_p\left(5\right)$ .

Thus the Lemma is proved. $\square$

Lemma 3.6. $S=G$

Proof. By Lemma 2.2, we obtain $S\le \overline{G}=G/K\le Aut\left(S\right)$ .

Suppose $K\ne 1$ , then $K$ may be assumed to be an elementary abelian $r$ -group with $r\in \pi \left(K\right)$ . Let $r=5$ , then the group $A_p\left(5\right)$ acts unisingular on $K$ , so $5\cdot r_p\in \omega \left(G\right)$ , which is a contradiction. Hence $r\ne 5$ , and $r\cdot r_p\in \omega \left(G\right)$ . However $r_p$ is non-adjacent to any other prime in $GK\left(L\right)$ , which gives another contradiction. Therefore $K=1$ , and $S\le G\le Aut\left(S\right)$ .

If $S<G\le Aut\left(S\right)$ , then there exists an order-2 graph automorphism β such that $G/S\cong \langle \beta \rangle$ ([13] Theorem 5). Consider $C_S\left(\beta \right)$ , the centralizer of β in S. Since $\pi \left(C_S\left(\beta \right)\right)=\pi \left(B_{p-1}\left(5\right)\right)$ ([5] Table 4.5.2), we deduce that $5\cdot r_{2p-2}\in \omega \left(G\right)$ , which yields a contradiction. Consequently, $G=S\cong A_p\left(5\right)$ . $\square$

By virtue of Lemmas 3.1-3.6 above, the proof of Theorem 1 is now complete.

Grechkoseeva and Zvezdina investigated the spectrum recognition problem for simple classical groups $L_4\left(q\right)$ and $U_4\left(q\right)$ over finite fields of odd characteristic (see [14]). They proved that every finite group isospectral to $L_4\left(q\right)$ is almost simple with socle $L_4\left(q\right)$ , so $L_4\left(q\right)$ is quasi-recognizable. In particular, $L_4\left(5\right)\cong A_3\left(5\right)$ is only known to be quasi-recognizable, and the pure spectrum-recognition remains open for this group. Mazurov showed in [10] that $h\left(A_2\left(5\right)\right)=2$ . Hence, the spectrum-recognition problem for all simple groups $A_p\left(5\right)$ is fully settled, and Theorem 2 is proved.

Conflicts of Interest

The author declares no conflicts of interest regarding the publication of this paper.

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