A Note on Average Codegrees of All Proper Subgroups of Finite Groups

Wei Xie, Shitian Liu^*

School of Mathematics, Sichuan University of Arts and Science, Dazhou, China.

DOI: 10.4236/am.2025.164018                

Abstract

Let $G$ be a finite group and let $\text{T}\left(G\right)$ be the sum of codegrees of complex irreducible characters of a group $G$ . Say $f\left(G\right)=\frac{\text{T}\left(G\right)}{\left|G\right|}$ . We prove that a finite group $G$ such that for all proper subgroup $H$ , $1.1<f\left(H\right)<1.3$ , is solvable.

Keywords

Simple Group, Character Codegree Sum, Proper Subgroup

1. Introduction

All groups are finite under consideration. Let $G$ be a group and let $\text{Irr}\left(G\right)$ be the set of complex irreducible characters of $G$ . Some authors studied the influence of character degrees on group structure, for instance, see [1]-[4]. Qian [5] gave the definition of the codegree of a finite group as follows: for an irreducible character $\chi \in \text{Irr}\left(G\right)$ , the codegree of a character $\chi$ of $G$ is defined as

$\text{cod}\chi =\frac{\left|G:\ker \chi \right|}{\chi \left(1\right)}$ . Now we have $\text{cod}\chi =1$ when $\chi =1_G$ , the principal character

of $G$ . Let ${\text{cd}}_{\text{cod}}\left(G\right)$ denote the set of irreducible character codegrees of $G$ , i.e. ${\text{cd}}_{\text{cod}}\left(G\right)=\left\{\text{cod}\chi :\chi \in \text{Irr}\left(G\right)\right\}$ . Some good results are gotten between codegree (degree) and group structure; see [6]-[12] for instance.

Wang, Qian, Lv and Chen studied the relation between average codegree and group structure [12]. Define $\text{T}\left(G\right)={\displaystyle \sum }_{\chi \in \text{Irr}\left(G\right)}\text{cod}\chi$ and $f\left(G\right)=\frac{1}{\left|G\right|}{\displaystyle \sum }_{\chi \in \text{Irr}\left(G\right)}\text{cod}\chi$ . Let $k\left(G\right)$ be the number of conjugacy classes of $G$ . Then $k\left(G\right)=\left|\text{Irr}\left(G\right)\right|$ .

Inspired by [12]-[14], we change the condition from “the codegree sum of a finite group” to “the codegree sums of all proper subgroups of a finite group”. In order to talk in short, the following notion is defined.

Definition 1.1. Then call $G$ a CS-group if $1.3>f\left(G\right)>1.1$ .

Let ${\text{PSL}}_2\left(q\right)$ be the projective special linear group of degree 2 over a finite field of order $q$ and $A_n$ be the alternating group on $n$ symbols. Now we give the following result.

Theorem 1.2. If $G$ is a nonabelian simple group with $f\left(G\right)>1.1$ , then $G$ is isomorphic to $A_5$ or ${\text{PSL}}_2\left(7\right)$ .

Now by Theorem 1.2, we have the following corollary.

Corollary 1.3. If all subgroups of $G$ are CS-groups, then $G$ is solvable.

If we only consider the proper subgroups of a group $G$ , then we introduce the definition as follows.

Definition 1.4. Let $prop\left(G\right)$ be the set of proper nonabelian subgroups of $G$ . Then call $G$ an SCS-group if for every $H\in prop\left(G\right)$ , $H$ is a CS-group.

Note that a non-trivial abelian group $G$ has $f\left(G\right)\ge 1.5$ as $f\left(C_3\right)=\frac{1+3+3}{3}\approx 2.33$ where $C_n$ is a cyclic group of order $n$ . Thus in the hypothesis of the Definition, we only consider the non-abelian subgroups of a group.

As an application of Theorem 1.2, we also can prove the following result.

Theorem 1.5. Let $G$ be an SCS-group. Then $G$ is solvable.

Remark 1.6. We know from ([15], p. 2) that $A_4$ , $D_{10}$ and $S_3$ are the maximal

subgroups of $A_5$ and by [16], $f\left(A_4\right)=\frac{11}{12}$ , $f\left(D_{10}\right)=\frac{13}{10}$ and $f\left(S_3\right)=\frac{5}{6}$ . Thus

the condition of Theorem 1.5 “for all $H\in prop\left(G\right)$ , $H$ is a CS-group” is the best possible.

Remark 1.7. We do not know whether in Theorem 1.5, the condition “$f\left(G\right)>1.1$ ” can be changed into “$f\left(G\right)>1$ ”.

Here we introduce the structure of this paper. In Section 2, we compute the $\text{T}\left({\text{PSL}}_2\left(q\right)\right)$ . In Section 3, some results about CS-groups are given and then prove the solvability of an SCS-group. All other symbols are standard, see [15] and [17] for instance.

2. Results for Section 3

Lemma 2.1. Let $H<G$ .

(1) Let $H$ be subnormal in $G$ and let $\psi$ be an irreducible constituent of ${\chi }_M$ . Then $\text{cod}\left(\psi \right)|\text{cod}\left(\chi \right)$ .

(2) For $\chi \in \text{Irr}\left(G\right)\backslash \left\{1_G\right\}$ , $\text{cod}\chi \ge \chi \left(1\right)+\frac{1}{\chi \left(1\right)}>\chi \left(1\right)$ .

Proof. (1) It follows from Lemma 5.11 of [17].

(2) $\text{cod}\left(\chi \right)=\frac{\left|G:\ker \chi \right|}{\chi \left(1\right)}\ge \frac{1+\chi {\left(1\right)}^2}{\chi \left(1\right)}>\frac{\chi {\left(1\right)}^2}{\chi \left(1\right)}=\chi \left(1\right)$ .

Lemma 2.2. Let $q$ be a prime power and let $G={\text{PSL}}_2\left(q\right)$ . Then $\left|G\right|=\frac{1}{\left(2,q-1\right)}q\left(q^2-1\right)$ . Furthermore,

$\text{T}\left(G\right)=\{\begin{array}{ll}q\left(q^2+1\right),\hfill & q\equiv 0\left(\mathrm{mod}2\right),\hfill \\ \frac{q^3+7q^2+8q+2}{4},\hfill & q\equiv -1\left(\mathrm{mod}4\right),\hfill \\ \frac{q^3+7q^2-6q+2}{4},\hfill & q\equiv 1\left(\mathrm{mod}4\right).\hfill \end{array}$

Proof. We know that for a nonabelian simple group $G$ , $G$ has faithful irreducible characters except for $1_G$ , the principal character. By ([18], pp. 402-403), we have that, if $q\equiv 0\left(\mathrm{mod}2\right)$ , then

$\begin{array}{c}\text{T}\left(G\right)=1\cdot 1+\frac{1}{2}q\cdot \frac{q\left(q^2-1\right)}{q-1}+1\cdot \frac{q\left(q^2-1\right)}{q}+\left(\frac{1}{2}q-1\right)\cdot \frac{q\left(q^2-1\right)}{q+1}\\ =q^3+q\\ =q\left(q^2+1\right).\end{array}$

$\left|G\right|=q\left(q^2-1\right);$

if $q=-1\left(\mathrm{mod}4\right)$ , then

$\begin{array}{c}\text{T}\left(G\right)=1+1\cdot \frac{q\left(q^2-1\right)/2}{q}+2\cdot \frac{q\left(q^2-1\right)/2}{\left(q-1\right)/2}+\frac{q-3}{4}\cdot \frac{q\left(q^2-1\right)/2}{q+1}\\ +\frac{q-3}{4}\cdot \frac{q\left(q^2-1\right)/2}{q-1}\\ =1+\frac{q^2-1}{2}+2\left(q^2+q\right)+\frac{1}{8}\left(q-3\right)\left(q^2-q\right)+\frac{1}{8}\left(q-3\right)\left(q^2+q\right)\\ =\frac{q^3+7q^2+8q+2}{4}.\end{array}$

$\left|G\right|=\frac{1}{2}q\left(q^2-1\right);$

if $q=1\left(\mathrm{mod}4\right)$ , then

$\begin{array}{c}\text{T}\left(G\right)=1+1\cdot \frac{q\left(q^2-1\right)/2}{q}+2\cdot \frac{q\left(q^2-1\right)/2}{\left(q+1\right)/2}+\frac{q-5}{4}\cdot \frac{q\left(q^2-1\right)/2}{q+1}\\ +\frac{q-1}{4}\cdot \frac{q\left(q^2-1\right)/2}{q-1}\\ =\frac{q^3+7q^2-6q+2}{4}.\end{array}$

$\left|G\right|=\frac{1}{2}q\left(q^2-1\right).$

This completes the proof.

3. Results

In this section, we first show the solvability of a CS-group, and then give that an SCS-group is solvable, and finally some other common-use results for character degree sums are given.

3.1. CS-Groups

In this subsection, we will prove Theorem 1.2 and Corollary 1.3. By Lemma 2.2,

$f\left(A_5\right)=\frac{68}{60}=\frac{17}{15}$ and $f\left({\text{PSL}}_2\left(7\right)\right)=\frac{186}{168}=\frac{31}{28}$ . We will first determine the structures of CS-groups.

Theorem 3.1. Let $G$ be a nonabelian simple group with $f\left(G\right)>1.1$ . Then $G$ is isomorphic to $A_5$ or ${\text{PSL}}_2\left(7\right)$ .

Proof. As $G$ is a nonabelian simple group, we have that for all $\chi \in \text{Irr}\left(G\right)\backslash \left\{1_G\right\}$ , $\ker \chi =1$ . By the definition of $f\left(G\right)$ and Lemma 2.1, we have that

$\begin{array}{c}f\left(G\right)=\frac{1}{\left|G\right|}{\displaystyle \sum }_{\chi \in \text{Irr}\left(G\right)}\frac{\left|G:\ker \chi \right|}{\chi \left(1\right)}\\ ={\displaystyle \sum }_{\chi \in \text{Irr}\left(G\right)}\frac{1}{\left|\ker \chi \right|\chi \left(1\right)}={\displaystyle \sum }_{\chi \in \text{Irr}\left(G\right)}\frac{1}{\chi \left(1\right)}\\ =1+{\displaystyle \sum }_{\chi \in \text{Irr}\left(G\right)\backslash \left\{1_G\right\}}\frac{1}{\chi \left(1\right)}\ge 1+\frac{\left(k-1\right)\sqrt{k-1}}{\sqrt{\left|G\right|-1}}.\end{array}$

where $k=k\left(G\right)$ and the last inequality is gotten by using the well-known inequality $\frac{n}{{{\displaystyle \sum }}_{i=1}^n\frac{1}{x_i}}\le \sqrt{\frac{{{\displaystyle \sum }}_{i=1}^n{x}_i^2}{n}}$ . Note that $k\left(A_5\right)=5$ , so

$f\left(G\right)\ge 1+\frac{\left(5-1\right)\sqrt{5-1}}{\sqrt{\left|G\right|-1}}=1+\frac{8}{\sqrt{\left|G\right|-1}}$ .

By hypothesis, we can assume that $1+\frac{8}{\sqrt{\left|G\right|-1}}>1.1$ . It follows that

$60\le \left|G\right|\le 6401$ as a nonabelian simple group $G$ is of order $\ge 60$ . Now by ([15], p. 239), the possibilities for $G$ are $A_5\cong {\text{PSL}}_2\left(5\right)$ , ${\text{PSL}}_2\left(7\right)$ , $A_6\cong {\text{PSL}}_2\left(9\right)$ , ${\text{PSL}}_2\left(11\right)$ , ${\text{PSL}}_2\left(13\right)$ , ${\text{PSL}}_2\left(17\right)$ , $A_7$ , ${\text{PSL}}_2\left(19\right)$ , ${\text{PSL}}_2\left(16\right)$ , ${\text{PSL}}_3\left(3\right)$ , ${\text{PSU}}_3\left(3\right)$ and ${\text{PSL}}_2\left(23\right)$ .

First we consider ${\text{PSL}}_2\left(q\right)$ with $q\ge 5$ . By Lemma 2.2, we have that

$f\left({\text{PSL}}_2\left(q\right)\right)=\{\begin{array}{ll}\frac{q^2+1}{q^2-1}\ge 1.1,\hfill & \text{if}q\equiv 0\left(\mathrm{mod}2\right),\hfill \\ \frac{q^3+7q^2+8q+2}{2q\left(q^2-1\right)}\ge 1.1,\hfill & \text{if}q\equiv -1\left(\mathrm{mod}4\right),\hfill \\ \frac{q^3+7q^2-6q+2}{2q\left(q^2-1\right)}\ge 1.1,\hfill & \text{if}q\equiv 1\left(\mathrm{mod}4\right),\hfill \end{array}$

i.e.

$0.1q^2-2.1\le 0,\text{if}q\equiv 0\left(\mathrm{mod}2\right),$

$1.2q^3-7q^2-10.2q-2\le 0,\text{if}q\equiv -1\left(\mathrm{mod}4\right),$

$1.2q^3-7q^2+1.5q-2\le 0,\text{if}q\equiv 1\left(\mathrm{mod}4\right).$

It follows from $q\ge 5$ that $q$ is equal to 5 or 7. Now by ([15], pp. 2-3), we obtain that $f\left(A_5\right)=\frac{68}{60}\approx 1.13>1.1$ and $f\left({\text{PSL}}_2\left(7\right)\right)=\frac{186}{168}\approx 1.107>1.1$ .

Now we consider these groups $A_7$ , ${\text{PSL}}_3\left(3\right)$ and ${\text{PSU}}_3\left(3\right)$ . By ([15], p. 10, 13-14), we obtain that

$f\left(A_7\right)=\frac{1645}{2520}\approx 0.652\cancel{\ge }1.1,$

$f\left({\text{PSL}}_3\left(3\right)\right)=\frac{3305}{5616}\approx 0.588\cancel{\ge }1.1,$

$f\left({\text{PSU}}_3\left(3\right)\right)=\frac{5931}{6048}\approx 0.98\cancel{\ge }\mathrm{1.1.}$

Therefore $G\cong {\text{PSL}}_2\left(q\right)$ with $q=5$ or 7.

Here we rewrite Corollary 1.3 here.

Corollary 3.2. If all subgroups of $G$ are CS-groups, then $G$ is solvable.

Proof. Assume the result is not true with minimal order $\left|G\right|$ , then $G$ is nonsolvable but its proper subgroups are solvable. Thus, we can assume that $G$ is a nonabelian simple group. Now hypothesis shows that $G$ is a CS-group and by Lemma 3.1, $G$ is isomorphic to $A_5$ or ${\text{PSL}}_2\left(7\right)$ . Thus, two cases are done within the following.

Case 1: $A_5$ .

We know from ([15], p. 2) that $A_4$ is a maximal subgroup of $A_5$ . Now by [16], $f\left(A_4\right)=\frac{11}{12}\cancel{\ge }1.1$ , a contradiction.

Case 2: ${\text{PSL}}_2\left(7\right)$ .

Then by ([15], p. 3), ${\text{PSL}}_2\left(7\right)$ has $S_4$ as a maximal subgroup. Now $f\left(S_4\right)=\frac{22}{24}\cancel{\ge }1.1$ arrives at a contradiction.

The two contradictions show that $G$ is solvable.

We also can get the following results.

Theorem 3.3. A finite group with $f\left(G\right)>\frac{17}{15}$ is solvable.

Proof. If the result is wrong, then $G$ is nonsolvable with minimal order $\left|G\right|$ . Thus we can assume that $G$ is a nonabelian simple group. By Theorem 3.1, we see that for a nonabelian simple group $G$ , $G$ has order at most 6401 with

$f\left(G\right)>\frac{17}{15}>1.1$ . We can check these groups by [15] and get that $f\left(G\right)\le \frac{17}{15}$ , a contradiction. Thus $G$ is solvable.

Corollary 3.4. Let $G$ be a nonabelian simple group. Then $f\left(G\right)\le \frac{17}{15}$ .

Proof. If the theorem is not true, then $f\left(G\right)>\frac{17}{15}$ , a contradiction to Theorem 3.3.

3.2. SCS-Groups

In this subsection, we give the proof of Theorem 1.5. We first need the following result which is due to Thompson [19].

Lemma 3.5 (Corollary 1 of [19]). Every minimal simple group is isomorphic to one of the following minimal simple groups:

(1) ${\text{PSL}}_2\left(2^p\right)$ for $p$ a prime;

(2) ${\text{PSL}}_2\left(3^p\right)$ for $p$ an odd prime;

(3) ${\text{PSL}}_2\left(p\right)$ , for $p$ any prime exceeding 3 such that $p^2+1\equiv 0\left(\mathrm{mod}5\right)$ ;

(4) $Sz\left(2^p\right)$ for $p$ an odd prime;

(5) ${\text{PSL}}_3\left(3\right)$ .

In order to argue in brief, we introduce the notation $\left[x\right]$ which denotes the maximal integer part of a rational number $x$ , for example $\left[-\pi \right]=-4$ and $\left[\pi \right]=3$ . Let

${\ker }_m\left(G\right)=\left\{\left[1_G\left(1\right),\left|G\right|,1\right],\left[{\chi }_1\left(1\right),\left|\ker {\chi }_1\right|,m_1\right],\cdots ,\left[{\chi }_s\left(1\right),\left|\ker {\chi }_s\right|,m_s\right]\right\}$

where $m_i$ denotes the number of ${\chi }_i\in \text{Irr}\left(G\right)$ with the same $\left|\ker {\chi }_i\right|$ .

To control the structure of a group, we also need the following result.

Lemma 3.6. Let $D_{2n}$ be a dihedral group of order $2n$ .

(1) Let $n$ be odd. Then

${\text{ker}}_m\left(D_{2n}\right)=\left\{\left[1,2n,1\right],\left[1,n,1\right],\left[2,1,\frac{n-1}{2}\right]\right\}$ .

(2) Let $n$ be even. Then

${\text{ker}}_m\left(D_{2n}\right)=\left\{\left[1,2n,1\right],\left[1,n,1\right],\left[2,\left(3+{\left(-1\right)}^j\right)/2,\frac{n}{2}-1\right]\right\}$ .

(3) $\text{T}\left(D_{2n}\right)=1+2+2\cdot 2+{\displaystyle \sum }_{j=1}^{\left[\frac{n}{2}\right]-1}\frac{2n}{3+{\left(-1\right)}^j}$ .

Table 1. Character table of $D_{2n}$ with $n$ odd ([20], p. 182).

$g_i$	1	$a^r\left(1\le r\le \left(n-1\right)/2\right)$	$b$
$\left|C_{D_{2n}}\left(g_i\right)\right|$	$2n$	$n$	2
${\chi }_1$	1	1	1
${\chi }_2$	1	1	−1
${\psi }_j$	2	${\epsilon }^{jr}+{\epsilon }^{-jr}$	0
$\left(1\le j\le \left(n-1\right)/2\right)$

In Table 1, $\epsilon ={\text{e}}^{2\pi \text{i}/n}$ .

Proof. (1) By Table 1, we have that ${\chi }_1\left(g\right)={\chi }_1\left(1\right)$ for $g\in G$ , so $\ker {\chi }_1=G$ ; ${\chi }_2\left(a^r\right)={\chi }_2\left(1\right)$ , so $\left|\ker {\chi }_2\right|=1+\frac{2n}{n}\cdot \frac{n-1}{2}=n$ ; if ${\psi }_j\left(g\right)={\psi }_i\left(1\right)$ , then $g=1$ , so $\ker {\psi }_j=1$ . Thus, we have

$\text{T}\left(D_{2n}\right)=1+\frac{\left(2n/n\right)}{{\chi }_2\left(1\right)}+{\displaystyle \sum }_{1\le j\le \left(n-1\right)/2}\frac{\left(2n/1\right)}{{\psi }_j\left(1\right)}=3+n\cdot \frac{n-1}{2}$ .

(2) Obviously $\ker {\chi }_1=G$ , so $\text{cod}{\chi }_1=1$ .

We see $\left|\ker {\chi }_2\right|=1+1+\frac{2n}{n}\cdot \left(m-1\right)=2m=n$ and $\text{cod}{\chi }_2=\frac{\left(2n/n\right)}{1}=2$ .

Now $\left|\ker {\psi }_j\right|$ equals to 1 + 1 when $j$ is even or 1 when $j$ is odd, so

$\left|\ker {\psi }_j\right|=1+\frac{2n}{2n}\cdot \left(1+{\left(-1\right)}^j\right)/2$ .

Thus $\text{cod}{\psi }_j=\frac{2n}{3+{\left(-1\right)}^j}$ .

To compute the kernels of the remaining irreducible characters of a group, by Table 2, we need to consider when ${\left(-1\right)}^k=1$ . As the computation methods of $\ker {\chi }_3$ and $\ker {\chi }_4$ are similar, we only consider to compute $\ker {\chi }_3$ . We divide the computation into two cases in light of $m$ .

Table 2. Character table of $D_{2n}$ , $n=2m$ ([20], p. 183).

$g_i$	1	$a^m$	$a^r\left(1\le r\le m-1\right)$	$b$	$ab$
$\left|C_{D_{2n}}\left(g_i\right)\right|$	$2n$	$2n$	$n$	4	4
${\chi }_1$	1	1	1	1	1
${\chi }_2$	1	1	1	−1	−1
${\chi }_3$	1	${\left(-1\right)}^m$	${\left(-1\right)}^r$	1	−1
${\chi }_4$	1	${\left(-1\right)}^m$	${\left(-1\right)}^r$	−1	1
${\psi }_j$	2	$2{\left(-1\right)}^j$	${\epsilon }^{jr}+{\epsilon }^{-jr}$	0	0
$\left(1\le j\le m-1\right)$

In Table 2, $\epsilon ={\text{e}}^{2\pi \text{i}/n}$ .

Case 1: $m$ is even.

Then in the interval $\left[1,m-1\right]$ , there are $\left[\frac{m-1}{2}\right]$ even numbers $r$ such that ${\left(-1\right)}^r=1$ . Thus

$\left|\ker {\chi }_3\right|=1+\frac{2n}{2n}+\frac{2n}{n}\cdot \left[\frac{\frac{n}{2}-1}{2}\right]+\frac{2n}{4}=2+2\left(\frac{n-4}{4}\right)+\frac{n}{2}=n.$

so $\text{cod}{\chi }_3=2=\text{cod}{\chi }_4$ .

Therefore $\text{T}\left(D_{2n}\right)=1+2+2\cdot 2+{\displaystyle \sum }_{j=1}^{\frac{n}{2}-1}\frac{2n}{3+{\left(-1\right)}^j}$ .

Case 2: $m$ is odd.

Then there are $\left[\frac{m-1}{2}\right]$ even integers $r$ in the interval $\left[1,m-1\right]$ such that ${\left(-1\right)}^r=1$ , so

$\begin{array}{c}\left|\ker {\chi }_3\right|=1+\frac{2n}{n}\cdot \left[\frac{m-1}{2}\right]+\frac{2n}{4}\\ =1+2\cdot \left(\frac{m-1}{2}\right)+\frac{n}{2}=n.\end{array}$

It follows that $\text{cod}{\chi }_3=2=\text{cod}{\chi }_4$ .

Now we have $\text{T}\left(D_{2n}\right)=1+2+2\cdot 2+{\displaystyle \sum }_{j=1}^{\frac{n}{2}-1}\frac{2n}{3+{\left(-1\right)}^j}$ .

The lemma is complete.

Let $\max G$ or $\max \left(G\right)$ denote the set of maximal proper subgroups with regard to subgroup-order divisibility. Let $F:Q$ or $\left|F\right|:\left|Q\right|$ be the Frobenius group with kernel $F$ and complement $Q$ respectively. In order to read easily, we write Theorem 1.5 here.

Theorem 3.7. Let $G$ be an SCS-group. Then $G$ is solvable.

Proof. Assume the theorem is wrong with minimal order $\left|G\right|$ . Let $H\in prop\left(G\right)$ be a nontrivial minimal normal subgroup in $G$ . Then $H$ is a CS-group, and by Theorem 1.3, $H$ is solvable. Thus $H$ is abelian. If there is a normal subgroup $K$ with $H<K<G$ such that $K/H$ is non-abelian simple, then $K$ is solvable, a contradiction. It means that $K=G$ . If $H>1$ , then $G/H$ is an SCS-group. It follows from the minimal choice of $G$ that $G/H$ solvable and so is $G$ , a contradiction. It follows that $H=1$ and that $G$ is a minimal simple group such that its proper subgroups are solvable. Thus $G$ is a minimal nonabelian simple group. So in the following, three cases are considered.

Case 1: ${\text{PSL}}_2\left(q\right)$ for certain $q$ .

If $q\in \left\{5,7,9,11\right\}$ , then by ([15], p. 2-3, 5-7) and [16], we respectively have

$f\left(D_{10}\right)=\frac{13}{10}=1.3\cancel{<}1.3,D_{10}\in \max {\text{PSL}}_2\left(5\right),$

$f\left(7:3\right)=1\cancel{\ge }1.1,7:3\in \max {\text{PSL}}_2\left(7\right),$

$f\left(3^2:4\right)=\frac{31}{36}\cancel{\ge }1.1,3^2:4\in \max {\text{PSL}}_2\left(9\right),$

$f\left(11:5\right)=\frac{43}{55}\cancel{\ge }1.1,11:5\in \max {\text{PSL}}_2\left(11\right).$

Thus $q\ne 5,7,9,11$ and we can assume that $q\ge 13$ when $q$ is odd and $q\ge 8$ when $q$ is even, so let $n=\frac{q\pm 1}{k}$ , $n\ge 6$ when $q$ is odd and $n\ge 7$

when $q$ is even. By Lemma 3.6 and Table 3, we have that $D_{2\left(q\pm 1\right)/k}\in \max {\text{PSL}}_2\left(q\right)$ and either

$1.1<\frac{n^2-n+6}{4n}<1.3$ (1)

or

$1.1<\frac{7+{{\displaystyle \sum }}_{j=1}^{\frac{n}{2}-1}\frac{2n}{3+{\left(-1\right)}^j}}{2n}<\mathrm{1.3.}$ (2)

For the inequality (1), we have $n=4\cancel{\ge }6$ , a contradiction. But the inquality (2) has no solution in N since $n\ge 6$ .

Table 3. ${\text{PSL}}_2\left(q\right)$ , $q\ge 5$ ([21], Chap II, Theorem 8.27).

	$\max {\text{PSL}}_2\left(q\right)$	Condition
${\mathcal{C}}_1$	$E_q:C_{\left(q-1\right)/k}$	$k=\text{gcd}\left(q-1,2\right)$
${\mathcal{C}}_2$	$D_{2\left(q-1\right)/k}$	$q\notin \left\{5,7,9,11\right\}$
$C_3$	$D_{2\left(q+1\right)/k}$	$q\notin \left\{7,9\right\}$
${\mathcal{C}}_5$	${\text{PSL}}_2\left(q_0\right).\left(k,b\right)$	$q=q_0^b$ , $b$ a prime, $q_0\ne 2$
${\mathcal{C}}_6$	$S_4$	$q=p\equiv \pm 1\left(\mathrm{mod}8\right)$
	$A_4$	$q=p\equiv 3,5,13,27,37\left(\mathrm{mod}40\right)$
$\mathcal{S}$	$A_5$	$q\equiv \pm 1\left(\mathrm{mod}10\right),F_q=F_p\left[\sqrt{5}\right]$

Case 2: $Sz\left(2^p\right)$ for $p$ an odd prime.

Let $q=2^p$ . Then by ([22], p. 385), $D_{2\left(q-1\right)}\in \max Sz\left(q\right)$ , and so by Lemma 3.6, we have $1.1<\frac{{(q-1)}^2-\left(q-1\right)+6}{4\left(q-1\right)}<1.3$ , so $q=5$ is non-even, a contradiction.

Case 3: ${\text{PSL}}_3\left(3\right)$ .

By ([15], p. 13), $13:3\in \max {\text{PSL}}_3\left(3\right)$ and by [16], $f\left(13:3\right)=\frac{59}{39}\approx 1.5\cancel{\le }1.3$ , a contradiction.

From the above three cases, $G$ is solvable, the wanted result.

Acknowledgements

This work was financially supported by the National Natural Science Foundation of China (Grant No: 11871360) and by the Project of High-Level Talent of Sichuan University of Arts and Science (Grant No: 2021RC001Z).

Conflicts of Interest

The authors declare no conflicts of interest regarding the publication of this paper.

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