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

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

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.

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

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

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

(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 $m$.

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

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

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 $n\ge 6$.

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.