Characterizations of $\sigma$-Solvable Finite Groups

Abstract

All the groups considered in this paper are finite, and $G$ always denotes a finite group; $\sigma$ is a partition of the set $\mathbb{P}$ of all primes, i.e., $\sigma=\{\sigma_{i} \mid i \in I\}$, where $\mathbb{P}=\bigcup_{i \in I} \sigma_{i}$ and $\sigma_{i} \cap \sigma_{j}=\varnothing$ for all $i \ne j$. A group $G$ is said to be $\sigma$-primary if $G$ is a $\sigma_{i}$-group for some $i=i(G)$, and $\sigma$-solvable if every chief factor of $G$ is $\sigma$-primary. A set of subgroups $\mathcal{H}$ of a group $G$ is called a complete Hall $\sigma$-set of $G$ if every element $\ne 1$ of the set $\mathcal{H}$ is a Hall $\sigma_{i}$-subgroup $G$ for some $i$, and $\mathcal{H}$ contains exactly one Hall $\sigma_{i}$-subgroup of the group $G$ for all $i$ such that $\sigma_{i}\cap \pi(G)\ne \varnothing$. A subgroup $A$ of a group $G$ is said to be $K$-$\mathfrak{S}_{\sigma}$-subnormal in $G$ if $G$ contains a series of subgroups $A=A_{0} \le A_{1} \le\cdots\le A_{t}=G$ such that either $A_{i-1} \trianglelefteq A_{i}$ or the group $A_{i}/(A_{i-1})_{A_{i}}$ is $\sigma$-solvable for all $i=1,\dots,t$.

We say that a subgroup $A$ of a group $G$ is weakly $K$- $\mathfrak{S}_{\sigma}$-subnormal in $G$ if $G$ contains $K$-$\mathfrak{S}_{\sigma}$-subnormal subgroups $S$ and $T$ such that $G=AT$ and $A \cap T \le S \le A$. In the present paper, we study conditions under which a group is $\sigma$-solvable. In particular, we prove that a group $G$ is $\sigma$-solvable if and only if at least one of the following two conditions is satisfied: (i) $G$ has a complete Hall $\sigma$-set $\mathcal H$ all of whose elements are weakly $K$-$\mathfrak{S}_{\sigma}$-subnormal in $G$; (ii) in every maximal chain of subgroups $\cdots < M_{3} < M_{2} < M_{1} < M_{0}=G$ of the groups $G$, at least one of the subgroups $M_{3}$, $M_{2}$, or $M_{1}$ is weakly $K$-$\mathfrak{S}_{\sigma}$-subnormal in $G$.

1. Introduction

All the groups considered in the paper are finite, and $G$ always denotes a finite group. Moreover, $\mathbb{P}$ is the set of all primes, and $\sigma$ is some partition of $\mathbb{P}$, i.e.,

$$\sigma=\{\sigma_{i} \mid i \in I\},$$

where $\mathbb{P}=\bigcup_{i \in I} \sigma_{i}$ and $\sigma_{i} \cap \sigma_{j}=\varnothing$ for all $i \ne j$. The symbol $\pi(G)$ denotes the set of all prime divisors of $|G|$, and $\sigma(G)=\{\sigma_{i}\mid\sigma_{i}\cap \pi(G)\ne\varnothing\}$. We say that a chain of subgroups

$$\cdots<M_{3}<M_{2}<M_{1}<M_{0}=G$$

of $G$ is a maximal chain in $G$ if $M_{i}$ is a maximal subgroup in $M_{i-1}$ for all $i$. The groups $A$ and $B$ are called groups of equal order if $|A|=|B|$. If $A$ is a subgroup of a group $G$, then $A_{G}$ is the largest normal subgroup of $G$ contained in $A$.

Let $\mathfrak{F}$ be a class of groups. Then a subgroup $A$ of a group $G$ is said to be $\mathfrak{F}$-subnormal in the sense of Kegel [1] or $K$-$\mathfrak{F}$-subnormal in $G$ [2] if $G$ contains a series of subgroups

$$A=A_{0} \le A_{1} \le \cdots \le A_{t}=G$$

such that either $A_{i-1} \trianglelefteq A_{i}$ or $A_{i}/(A_{i-1})_{A_{i}} \in \mathfrak{F}$ for all $i=1,\dots,t$.

Recall some notions of the theory of $\sigma$-properties of a group [3]–[6].

A group $G$ is said to be $\sigma$-primary if $G$ is a $\sigma_{i}$-group for some $i=i(G)$; $\sigma$-solvable if every chief factor of $G$ is $\sigma$-primary. We use the symbols $\mathfrak{S}_{\sigma}$ and $\mathfrak{S}$ to denote the classes of all $\sigma$-solvable and all solvable groups, respectively; $G^{\mathfrak{S}_{\sigma}}$ stands for the intersection of all normal subgroups $N$ of a group $G$ with $G/N\in \mathfrak{S}_{\sigma}$.

A set of subgroups $\mathcal{H}$ of a group $G$ is called a complete Hall $\sigma$-set of $G$ if every element $\ne 1$ of the set $\mathcal{H}$ is a Hall $\sigma_{i}$-subgroup of $G$ for some $i$ and $\mathcal{H}$ contains exactly one Hall $\sigma_{i}$-subgroup of $G$ for every $\sigma_{i}\in \sigma(G)$.

Definition 1.

We say that a subgroup $A$ of a group $G$ is weakly $K$-$\mathfrak{S}_{\sigma}$-subnormal in $G$ if there are $K$-$\mathfrak{S}_{\sigma}$-subnormal subgroups $T$ and $S$ such that $G=AT$ and $A \cap T \le S \le A$.

Remark 1.

(i) Each $K$-$\mathfrak{S}_{\sigma}$-subnormal subgroup $S=S \cap G$ is weakly $K$-$\mathfrak{S}_{\sigma}$-subnormal in the group $G$, since $G$ is a $K$-$\mathfrak{S}_{\sigma}$-subnormal subgroup of $G$.

(ii) A subgroup $A$ of a group $G$ is said to be $\sigma$-subnormal in $G$ [3] if there is a series of subgroups $A=A_{0} \le A_{1} \le \cdots A_{n}=G$ such that either $A_{i-1} \trianglelefteq A_{i}$ or the quotient $A_{i}/(A_{i-1})_{A_{i}}$ is $\sigma$-primary for all $i=1,\dots,n$. Every subnormal subgroup is $\sigma$-subnormal, and every $\sigma$-subnormal subgroup is $K$-$\mathfrak{S}_{\sigma}$-subnormal in the group.

(iii) A subgroup $S$ of a group $G$ is said to be $\sigma$-permutable in $G$ [3] if $G$ has a complete Hall $\sigma$-set $\mathcal H$ such that $SH^{x}=H^{x}S$ for all $H\in \mathcal{H}$ and all $x\in G$.

By Theorem B of [3], every $\sigma$-permutable subgroup is $\sigma$-subnormal and, therefore, $K$-$\mathfrak{S}_{\sigma}$-subnormal in the group.

Now consider the following example.

Example 1.

(i) A subgroup $A$ of a group $G$ is said to be weakly $\sigma$-permutable in $G$ ([7], [8]) if $G$ contains a $\sigma$-permutable subgroup $S$ and a $\sigma$-subnormal subgroup $T$ such that $G=AT$ and $A \cap T \le S \le A$. By Remark 1 (ii, iii), every weakly $\sigma$-permutable subgroup is weakly $K$-$\mathfrak{S}_{\sigma}$-subnormal in the group.

(ii) In the general case, the class of all weakly $K$-$\mathfrak{S}_{\sigma}$-subnormal subgroups is wider than the class of all weakly $\sigma$-permutable subgroups. Let $\sigma=\{\{2\},\{3\},\dots\}$. In this case, every $\sigma$-subnormal subgroup is subnormal, and any $\sigma$-permutable subgroup $H$ is $S$-permutable in the group ([9], [10]), i.e., $HP=PH$ for all Sylow subgroups $P$ of the group.

Now let $A$ be a non-Abelian group of order $p^{3}$ of a simple odd exponent $p$. Let $G=A \wr C_{2}=B \rtimes C_{2}$, where $B$ is the base of the regular wreath product $G$. Let $L=\Phi(A_{1})$, where $A_{1}$ is the first copy of the group $A$ in $B$. Then $|L|=p$, $L \le \Phi(B)$, and $L$ is subnormal, and hence weakly $K$-$\mathfrak{S}$-subnormal in $G$, considering Remark 1.2 (ii). It is also clear that $L$ is a nonnormal subgroup of $G$. Now let us show that $L$ is not weakly $S$-permutable in $G$. Suppose that $G$ contains a subnormal subgroup $T$ and an $S$-permutable subgroup $S$ such that $G=LT$ and $L \cap T \le S \le L$. Then $B=L(T \cap B)$, and thus $T\cap B=B$, since $L \le \Phi(B)$. Hence $L \cap T=L=S$ is $S$-permutable in $G$. Then $L$ is normal in $G$ by [9, Lemma 1.2.16]. This contradiction shows that $L$ is not weakly $S$-permutable in $G$.

(iii) A subgroup $A$ of a group $G$ is said to be $c$-normal in $G$ [11] if $AT=G$ and $A \cap T \le A_{G}$ for some normal subgroup $T$ of $G$. Consequently, any $c$-normal subgroup is weakly $K$-$\mathfrak{S}_{\sigma}$-subnormal in the group for every partition $\sigma$ of $\mathbb{P}$.

(iv) A subgroup $M$ of a group $G$ is said to be

• (a) modular in $G$ [12] if $M$ is a modular element (in the sense of Kurosh [12, p. 43]) of the lattice of all subgroups $\mathcal{L}(G)$ of $G$, i.e., the following conditions hold:

  • (1) $\langle X,M \cap Z \rangle=\langle X,M \rangle \cap Z$ for all $X \le G$ and $Z \le G$ such that $X \le Z$;

  • (2) $\langle M,Y \cap Z \rangle=\langle M,Y \rangle \cap Z$ for all $Y \le G$ and $Z \le G$ such that $M \le Z$;

• (b) submodular in $G$ ([13], [14]) if $G$ contains a series of subgroups $A=A_{0} \le A_{1} \le\cdots\le A_{t}=G$ such that $A_{i-1}$ is a modular subgroup of $A_{i}$ for all $i=1,\dots,t$.

It follows from the main result of the theory of modular subgroups [12, Theorem 5.1.14] that every submodular subgroup is $K$-$\mathfrak{S}$-subnormal and, therefore, weakly $K$-$\mathfrak{S}$-subnormal in the group.

Although the concept of $\sigma$-solvable group was first introduced in a recent paper [3], this concept proved to be very useful in the analysis of many open questions (see, for example, [3]–[8], [15]–[26]), and thus the problem of finding conditions under which a group is $\sigma$-solvable is very interesting and relevant. In this paper, we prove the following result.

Theorem 1.

The following conditions are equivalent:

• (i) a group $G$ is $\sigma$ -solvable;

• (ii) every subgroup of $G$ is $K$ - $\mathfrak{S}_{\sigma}$ -subnormal;

• (iii) $G$ has a complete Hall $\sigma$ -set $\mathcal H$ all of whose elements are weakly $K$ - $\mathfrak{S}_{\sigma}$ -subnormal in $G$ ;

• (iv) in every maximal chain $\cdots < M_{3} < M_{2} < M_{1} < M_{0}=G$ of $G$ , at least one of the subgroups $M_{3}$ , $M_{2}$ or $M_{1}$ is weakly $K$ - $\mathfrak{S}_{\sigma}$ -subnormal in $G$ ;

• (v) $i_{K,\mathfrak{S}_{\sigma}}(G)\le 2|\sigma(G)|$ , where $i_{K,\mathfrak{S}_{\sigma}}(G)$ is the number of classes of non- $K$ - $\mathfrak{S}_{\sigma}$ -subnormal subgroups of $G$ of equal order.

Theorem 1 covers many known results. In particular, taking into account Remark 1 and Example 1, we see that the following well-known results are special cases of this theorem.

Corollary 1 (Guo, Skiba [19]).

If, in every maximal chain $\cdots<M_{3}<M_{2}<M_{1}<M_{0}=G$ of a group $G$ , at least one of the subgroups $M_{3}$ , $M_{2}$ , or $M_{1}$ is $\sigma$ -subnormal in $G$ , then $G$ is $\sigma$ -solvable.

Corollary 2 (Zhang, Wu, Guo [7]).

If $G$ has a complete Hall $\sigma$ -set $\mathcal H$ all of whose elements are weakly $\sigma$ -permutable in $G$ , then $G$ is $\sigma$ -solvable.

Corollary 3 (Zimmermann [14]).

If, in every maximal chain $\cdots<M_{3}<M_{2}<M_{1}<M_{0}=G$ of a group $G$ , at least one of the subgroups $M_{3}$ , $M_{2}$ , or $M_{1}$ is submodular in $G$ , then $G$ is solvable.

Corollary 4 (Spencer [27]).

If, in every maximal chain $\cdots<M_{3}<M_{2}<M_{1}<M_{0}=G$ of a group $G$ , at least one of the subgroups $M_{3}$ , $M_{2}$ , or $M_{1}$ is subnormal in $G$ , then $G$ is solvable.

Corollary 5 (Schmid [28]).

A group $G$ is solvable if each of its 3-maximal subgroups is modular.

Corollary 6 (Kovaleva [15]).

If $i_{\sigma}(G) \le 2|\sigma(G)|$ , where $i_{\sigma}(G)$ stands for the number of classes of non- ${\sigma}$ -subnormal subgroups of equal order of a group $G$ , then $G$ is $\sigma$ -solvable.

Corollary 7 (Lu, Meng [29]).

If the number of conjugacy classes of nonsubnormal subgroups of a group $G$ does not exceed $2|\pi(G)|$ , then $G$ is solvable.

2. Proof of Theorem 1

Lemma 1.

(1) The class $\mathfrak{S}_{\sigma}$ is closed with respect to taking products of normal $\mathfrak{S}_{\sigma}$-subgroups, homomorphic images, and subgroups. Moreover, any extension of a $\sigma$-solvable group with the help of a $\sigma$-solvable group is $\sigma$-solvable.

(2) A group is $\sigma$-solvable if and only if all its maximal subgroups are $K$-$\mathfrak{S}_{\sigma}$-subnormal.

Proof.

(1) This assertion is obvious.

(2) It is clear that, in every $\sigma$-solvable group, all its maximal subgroups are $K$-$\mathfrak{S}_{\sigma}$-subnormal.

Suppose now that all maximal subgroups of a group $G$ are $K$-$\mathfrak{S}_{\sigma}$-subnormal in $G$. Then $G/M_{G}\in \mathfrak{S}_{\sigma}$ for every maximal subgroup $M$ of $G$. Therefore, $G/\Phi(G) \in \mathfrak{S}_{\sigma}$ and, consequently, $G \in \mathfrak{S}_{\sigma}$, by part (1).

This completes the proof of the lemma.

From Lemma 1 and from the main result of [1], we obtain the following assertion.

Lemma 2.

The set of all $K$ - $\mathfrak{S}_{\sigma}$ -subnormal subgroups of a group $G$ forms a sublattice of the lattice of all subgroups of $G$ .

Lemma 3.

Let $A$ , $H$ , and $N$ be subgroups of $G$ , where $A$ is $K$ - $\mathfrak{S}_{\sigma}$ -subnormal and $N$ is normal in $G$ . Then

• (1) $A\cap H$ is $K$ - $\mathfrak{S}_{\sigma}$ -subnormal in $H$ ;

• (2) $AN/N$ is $K$ - $\mathfrak{S}_{\sigma}$ -subnormal in $G/N$ ;

• (3) if $N\le H$ and $H/N$ is $K$ - $\mathfrak{S}_{\sigma}$ -subnormal in $G/N$ , then $H$ is $K$ - $\mathfrak{S}_{\sigma}$ -subnormal in $G$ ;

• (4) the subgroup $A^{\mathfrak{S}_{\sigma}}$ is subnormal in $G$ ;

• (5) if $A$ is $\sigma$ -solvable and $N$ is a non- $\sigma$ -primary minimal normal subgroup of $G$ , then $A\le C_{G}(N)$ .

Proof.

(1)–(4). These assertions are corollaries of Lemmas 6.1.6, 6.1.7, and 6.1.9 in [2].

(5) Assume that this assertion fails to hold; let $G$ be a counterexample of minimal order. Then $A\ne G$. By assumption, there is a series of subgroups

$$A=A_{0} \le A_{1} \le \cdots \le A_{r}=G$$

such that either $A_{i-1} \trianglelefteq A_{i}$ or $A_{i}/(A_{i-1})_{A_{i}}$ is $\sigma$-solvable for all $i=1,\dots,r$. Let $M=A_{r-1}$. Without loss of generality, we can assume that $M\ne G$.

Let $E=NA$. Let us first assume that $E < G$. It is clear that $N=N_{1}\times \cdots \times N_{n}$ for some minimal normal subgroups $N_{1},\dots,N_{n}$ of $E$ and $N_{i}$ is not $\sigma$-primary for all $i$. By Lemma 3 (1), $A$ is $K$-$\mathfrak{S}_{\sigma}$-subnormal in $E$ and, therefore, due to the choice of $G$, this means that $A \le C_{E}(N_{i})$ for all $i$ and, therefore, $A \le C_{E}(N)$. Hence $NA=E=G$. Then $N\nleq M$, and hence $G/M_{G}$ is not $\sigma$-solvable, since $N\simeq NM_{G}/M_{G}$ is not $\sigma$-primary. This implies that $M$ is normal in $G$, and hence $N\cap M=1$. This implies that $[N,M]=1$, and thus $A\le C_{G}(N)$.

This completes the proof of the lemma.

Lemma 4.

Let $A$ , $H$ , and $N$ be subgroups of a group $G$ , where $A$ is weakly $K$ - $\mathfrak{S}_{\sigma}$ -subnormal and $N$ is normal in $G$ .

• (1) If either $N\le A$ or $(|N|,|A|)=1$ , then $AN/N$ is weakly $K$ - $\mathfrak{S}_{\sigma}$ -subnormal in $G/N$ .

• (2) If $N \le H$ and $H/N$ is weakly $K$ - $\mathfrak{S}_{\sigma}$ -subnormal in $G/N$ , then $H$ is weakly $K$ - $\mathfrak{S}_{\sigma}$ -subnormal in $G$ .

• (3) If $A \le E \le G$ , then $A$ is weakly $K$ - $\mathfrak{S}_{\sigma}$ -subnormal in $E$ .

Proof.

Let $T$ and $S$ be $K$-$\mathfrak{S}_{\sigma}$-subnormal subgroups of $G$ such that $G=AT$ and $A \cap T \le S \le A$.

(1) First, note that

$$(AN/N)(TN/N)=ATN/N=G/N,\qquad SN/N\le AN/N,$$

where the subgroups $TN/N$ and $SN/N$ are $K$-$\mathfrak{S}_{\sigma}$-subnormal in $G/N$ by Lemma 3 (2). Thus, it remains only to show that

$$(AN/N) \cap (TN/N) \le SN/N.$$

If $N \le A$, then

$$(AN/N) \cap (TN/N)=(A \cap TN)/N=N(A \cap T)/N \le SN/N.$$

Suppose now that $(|N|,|A|)=1$. Since $G=AT$, it follows that $|G:T|$ divides $|A|$. Hence $|NT:T|$ divides $|A|$. However, $|NT:T|=|N: N \cap T|$ divides $|N|$. Thus, $N \le T$, and hence

$$(AN/N) \cap (TN/N)=(AN/N) \cap (T/N)=(AN \cap T)/N= N(A \cap T)/N \le SN/N.$$

Therefore, the subgroup $AN/N$ is weakly $K$-$\mathfrak{S}_{\sigma}$-subnormal in $G/N$.

(2) By assumption, $G/N$ contains $K$-$\mathfrak{S}_{\sigma}$-subnormal subgroups $Z/N$ and $D/N$ such that

$$G/N=(H/N)(Z/N),\qquad (H\cap Z)/N=(H/N) \cap (Z/N) \le D/N \le H/N.$$

Then $G=HZ$ and $H \cap Z \le D \le H$, where $Z$ and $D$ are $K$-$\mathfrak{S}_{\sigma}$-subnormal subgroups of $G$ by Lemma 3 (3) and, therefore, $H$ is weakly $K$-$\mathfrak{S}_{\sigma}$-subnormal in $G$.

(3) Note that

$$E=E \cap AT=A(E \cap T),\qquad A \cap (E \cap T)=A \cap T \le S \le A,$$

where $E \cap T$ and $S$ are weakly $K$-$\mathfrak{S}_{\sigma}$-subnormal in $E$ by Lemma 3 (1). Hence $A$ is weakly $K$-$\mathfrak{S}_{\sigma}$-subnormal in $E$.

This completes the proof of the lemma.

Proof of Theorem 1.

(i) $\Rightarrow$ (ii) Let $H\le M < G$, where $M$ is a maximal subgroup of $G$. Then $M$ is $K$-$\mathfrak{S}_{\sigma}$-subnormal in $G$ by Lemma 1 (2). On the other hand, $H$ is $K$-$\mathfrak{S}_{\sigma}$-subnormal in $M$ by induction. Hence $H$ is $K$-$\mathfrak{S}_{\sigma}$-subnormal in $G$.

The implication (ii) $\Rightarrow$ (i) follows from Lemma 1 (2).

Since a $\sigma$-solvable group is $\sigma_{i}$-separable, and hence has a Hall $\sigma_{i}$-subgroup for all $i$, it follows from the implication (ii) $\Rightarrow$ (i) and from Remark 1 (i) that (ii) $\Rightarrow$ (iii), (iv), (v).

(iii) $\Rightarrow$ (i) Suppose that this assertion fails to hold; let $G$ be a counterexample of minimal order. Let $\mathcal{H}=\{H_{1},\dots,H_{t}\}$. Then $t > 1$. Without loss of generality, we can assume that $H_{i}$ is a nonidentity $\sigma_{i}$-group for all $i=1,\dots,t$. According to the condition, for any $i$, the group $G$ has $K$-$\mathfrak{S}_{\sigma}$-subnormal subgroups $T_{i}$ and $S_{i}$ such that $G=H_{i} T_{i}$ and $H_{i} \cap T_{i} \le S_{i} \le H_{i}$.

Let us first show that $G/L$ is $\sigma$-solvable for every nontrivial $\sigma$-primary normal subgroup $L$ of $G$ and, therefore, there is no $\sigma$-primary normal subgroup in $G$. Indeed, suppose that $G$ contains a minimal normal subgroup $L$ which is a $\sigma_{i}$-group for some $i$. It can readily be seen that $\{H_{1}L/L,\dots,H_{t}L/L\}$ is a complete Hall $\sigma$-set of $G/L$. Moreover, $L \le H_{i}$ and $(|L|,|H_{j}|)=1$ for all $j \ne i$. However, then $H_{k}L/L$ is weakly $K$-$\mathfrak{S}_{\sigma}$-subnormal in $G/L$ for all $k=1,\dots,t$ by Lemma 4 (1). Then condition (iii) holds for $G/L$, and hence $G/L$ is $\sigma$-solvable due to the choice of $G$. Thus, $G$ is $\sigma$-solvable, which contradicts the choice of $G$. Hence every minimal normal subgroup $G$ is not $\sigma$-primary.

This implies that $S^{\mathfrak{S}_{\sigma}}\ne 1$ for any nontrivial $K$-$\mathfrak{S}_{\sigma}$-subnormal subgroup $S$ of $G$ by Lemma 3 (5) and, therefore, $S_{i}=1$ for all $i$, and $T_{1}$ is not a $\sigma_{2}$-group. Hence $t > 2$, and $T_{i}$ is a supplement to $H_{i}$ to $G$. Therefore, $T_{i}$ is a Hall $\sigma_{i}'$-subgroup in $G$. Then $T_{2}\cap \cdots \cap T_{t}$ is a nonidentity Hall $\sigma_{1}$-subgroup of $G$ according to [31, A, 1.6 (b)], and this subgroup is $K$-$\mathfrak{S}_{\sigma}$-subnormal in $G$ by Lemma 2, a contradiction. Therefore, the implication (iii) $\Rightarrow$ (i) holds.

(iv) $\Rightarrow$ (i) Assume that this assertion fails to hold; let $G$ be a counterexample of minimal order.

(1) If $N$ is a minimal normal subgroup of $G$, then $G/N$ is a $\sigma$-solvable group. Thus, $N$ is not a $\sigma$-primary group. Moreover, $N$ is a unique minimal normal subgroup of $G$, $C_{G}(N)=1$, and $L\le N$ for any minimal subnormal subgroup $L$ of $G$.

By Lemma 4 (1), the hypothesis is true for $G/N$, which means that $G/N$ is $\sigma$-solvable by the choice of $G$. Consequently, $N$ is not $\sigma$-primary.

If $G$ contains a minimal normal subgroup $R \ne N$, then $G/N$ and $G/R$ are $\sigma$-solvable groups, and hence it follows from the isomorphisms $R \simeq R/1=R/(R \cap N) \simeq RN/N$ that $R$ is a $\sigma$-primary group, which contradicts the fact proved above. Hence $N$ is a unique minimal normal subgroup of $G$ and $C_{G}(N)=1$, since the subgroup $N$ is non-Abelian.

It follows from [31, A, 13.4] that $N\le N_{G}(L)$. Moreover, $L \nleq C_{G}(N)=1$ and, therefore, $L\le N$. Hence assertion (1) holds.

(2) If $p$ is an arbitrary odd prime dividing $|N|$ and $N_{p}$ is the Sylow $p$-subgroup of $N$, then $N_{p}=N \cap G_{p}$, $G_{p} \le N_{G}(N_{p}) \le M$, and $G=NM$ for some Sylow $p$-subgroup $G_{p}$ and some maximal subgroup $M$ of $G$. Hence $p$ does not divide $|G: M|$, and $M_{G}=1$.

By Frattini’s argument, $G=NN_{G}(N_{p})$. Since $N$ is not an Abelian group, it follows that $N_{G}(N_{p})\ne G$. Then $N \nleq M$ for a maximal subgroup $M$ of $G$ containing $N_{G}(N_{p})$. Hence $G=NM$ and $M_{G}=1$. Moreover, if $N_{p} \le G_{p}$, where $G_{p}$ is a Sylow $p$-subgroup of $G$, then $N_{p}=N\cap G_{p}$, and hence $G_{p} \le N_{G}(N_{p}) \le M$.

(3) The intersection $D:=M \cap N$ is not nilpotent. In particular, $D \nleq \Phi (M)$.

Assume that $D$ is nilpotent; let $N_{p}$ be the Sylow $p$-subgroup of $D$. Then $N_p$ is normal in $M$, because $N_p$ is a characteristic subgroup of $D$, and $D$ is normal in $M$. Therefore, the subgroup $Z(J (N_{p}))$ is normal in $M$. Since $M_{G}=1$, it follows that $N_{G}(Z(J(N_{p})))=M$. Then $N_{N}(Z(J(N_{p})))=D$ is a nilpotent group. This implies that $N$ has a normal $p$-complement by the Glauberman–Thompson theorem, since $p$ is odd. However, then $N$ is an $p$-group, which contradicts assertion (1). Therefore, (3) holds.

(4) $V^{\mathfrak {S}_\sigma} \ne 1$ for every nontrivial $K$-$\mathfrak{S}_{\sigma}$-subnormal subgroup $V$ of $G$.

Indeed, $V^{\mathfrak {S}_{\sigma}}=1$ implies that $V$ is $\sigma$-solvable and, therefore, $1 < V\le C_{G}(N)=1$ by Lemma 3 (5). Hence (4) holds.

(5) If $H$ is $K$-$\mathfrak{S}_{\sigma}$-subnormal in $G$ and is contained in $M$, then $H=1$. (5)

Let $W=H^{\mathfrak{S}_{\sigma}}$ be the $\sigma$-solvable residual of $H$. Then $W$ is subnormal in $G$ by Lemma 3 (4) and, therefore,

$$W^{G}=W^{NM}=W^{M} \le M_{G}=1$$

by [31, A, 14.3]. Hence $W=1$. Thus, $H$ is $\sigma$-solvable and, therefore, $H=1$ by part (4).

(6) If $H$ is a weakly $K$-$\mathfrak {S}_{\sigma}$-subnormal subgroup of $G$ and is contained in $M$, then $G=HT$ and $H \cap T=1$ for some $K$-$\mathfrak {S}_{\sigma}$-subnormal subgroup $T$ of $G$ (this follows from part (5) and from the definition of weakly $K$-$\mathfrak {S}_{\sigma}$-subnormal subgroup).

(7) $M$ is not weakly $K$-$\mathfrak{S}_{\sigma}$-subnormal in $G$. In particular, $M$ is $\sigma$-solvable.

Suppose that $M$ is weakly $K$-$\mathfrak{S}_{\sigma}$-subnormal in $G$. Then $G=MT$ and $M \cap T=1$ for some $K$-$\mathfrak{S}_{\sigma}$-subnormal subgroup $T$ of $G$ by assertion (6). Moreover, $W:=T^{\mathfrak {S}_{\sigma}} \ne 1$ according to part (4), and $W$ is a subnormal subgroup of $G$ according to Lemma 3 (4).

Let now $L$ be a minimal subnormal subgroup of $G$ contained in $W$. Then $L$ is a minimal normal subgroup of $N$ by part (1). Hence $p$ divides $|L|$, where $|L|$ divides $|T|=|G: M|$. However, $p$ does not divide $|G: M|$ by part (2). This contradiction completes the proof of the first part of (7). Then, in any maximal chain

$$\cdots < M_{3} < M_{2} < M_{1}=M < G$$

of $G$, one of the subgroups $M_{3}$ and $M_{2}$ is weakly $K$-$\mathfrak{S}_{\sigma}$-subnormal in $G$ by assumption. Hence this subgroup is weakly $K$-$\mathfrak{S}_{\sigma}$-subnormal in $M$ by Lemma 4 (3). Thus, the conjecture holds for $M$, and thus $M$ is $\sigma$-solvable by the choice of $G$.

(8) The relation $N < G$ holds.

Suppose that $N=G$ is a non-Abelian simple group; let $H$ be an arbitrary proper $K$-$\mathfrak{S}_{\sigma}$-subnormal subgroup of $G$. Suppose that $H\ne 1$.

Then $G$ has a proper nonidentity subgroup $V$ such what $H \le V$ and either $V$ is normal in $G$ or $G/V_{G}$ is a $\sigma$-solvable group. However, the first condition is impossible, since the group $G$ is simple. Hence $V_{G}=1$, and then $G/V_{G}=G/1\simeq G$ is a $\sigma$-solvable group, which contradicts the choice of $G$. Thus, any proper $K$-$\mathfrak{S}_{\sigma}$-subnormal subgroup of $G$ is trivial.

Let $Q$ be a Sylow $q$-subgroup of $G$, where $q$ is the least prime dividing $|G|$, and let $L$ be a maximal subgroup of $G$ containing $Q$. Then, taking into account [32, IV, 2.8], we obtain $|Q| >q$. Let $V$ be a maximal subgroup of $Q$ and $S$ be a maximal subgroup of $V$. Then there is a 3-maximal subgroup $1 < W$ of $G$ such that $W\le Q$. Indeed, if $S\ne 1$, then this is obvious. On the other hand, if $S=1$, then $Q$ is an Abelian group and, therefore, $Q < L$ by [32, IV, 7.4]. Therefore, in this case, there exists a 3-maximal subgroup $W$ of $G$ such that $V \le W \le Q$.

Due to condition (iv), $G$ has a weakly $K$-$\mathfrak{S}_{\sigma}$-subnormal subgroup $U$ such what $W\le U < G$. Hence $G$ contains $K$-$\mathfrak{S}_{\sigma}$-subnormal subgroups $T$ and $R$ such that $G=UT$ and $U \cap T \le R \le U$. Then $T=G$, and thus $U=U \cap T \le R \le U$.

Hence $U=R$ is $K$-$\mathfrak{S}_{\sigma}$-subnormal in $G$. Then $U=1$, a contradiction. Thus, assertion (8) holds.

(9) If $G_{p} \le V \le M$, then $V$ is not weakly $K$-$\mathfrak{S}_{\sigma}$-subnormal in $G$.

Assume that $V$ is weakly $K$-$\mathfrak{S}_{\sigma}$-subnormal in $G$. Then $G=VT$ and $V \cap T=1$ for some $K$-$\mathfrak{S}_{\sigma}$-subnormal subgroup $T$ of $G$ by assertion (6). Then $1 < T < G$, and the subgroup $T^{\mathfrak {S}_{\sigma}}$ is nontrivial by part (4). Moreover, this subgroup is subnormal in $G$ by Lemma 3 (4). Hence $Z \le T$ for a minimal subnormal subgroup $Z$ of $G$ contained in $T^{\mathfrak {S}_{\sigma}}$. Then $Z$ is a $p'$-group, because $V \cap T=1$ and $G_{p} \le V$. On the other hand, $Z$ is a minimal normal subgroup of $N$ by part (1) and, therefore, $p$ divides $|Z|$. This contradiction completes the proof of (9).

(10) The group $M=D \rtimes L$ is solvable, where $L$ is a group of prime order.

By [32, IV, 7.4], $G_{p} \le L$ for some maximal subgroup $L$ of $M$, since $p > 2$ and $G$ is unsolvable. Let us first show that $G_{p}=L$. Indeed, suppose that $G_{p} \le V$ for some maximal subgroup $V$ of $L$. Then, in every maximal chain

$$\cdots < V=M_{i} < \cdots <M_{3} < M_{2} < M_{1}=M < M_{0}=G$$

of $G$ that contains $V$ below $M_{3}$, all subgroups $M_{3}$, $M_{2}$, and $M_{1}$ are not weakly $K$-$\mathfrak{S}_{\sigma}$-subnormal in $G$ by part (9), which contradicts the condition. Hence $G_{p}=L$ is a maximal subgroup of $M$ and, therefore, $M$ is solvable according to [32, IV, 7.4]. If $G_{p} \le D$, then $D=M \cap N=M$, since $G_{p} \ne D$ by part (2), whence it follows that $N=G$, which contradicts part (8). Hence $G_{p} \nleq D$, and thus $DL=M$. Therefore, $D \cap L=1$ in the case of $|G_{p}|=p$, whence it follows that $M=D \rtimes L$, and thus assertion (10) holds because $M$ is solvable.

Finally, suppose that $|G_{p}| >p$. First of all, note what $L$ is not weakly $\sigma$-subnormal in $G$ by part (9). However, on the other hand, $M$ is also not weakly $K$-$\mathfrak{S}_{\sigma}$-subnormal in $G$ by (7). Therefore, every maximal subgroup of $L=G_{p}$ is weakly $K$-$\mathfrak {S}_{\sigma}$-subnormal in $G$.

It follows from $G_{p} \nleq D$ that $N_{p}=N \cap G_{p} \le V$ for some maximal subgroups $V$ of $G_{p}$. Since $V$ is weakly $K$-$\mathfrak{S}_{\sigma}$-subnormal in $G$, it follows that $V$ has a $K$-$\mathfrak{S}_{\sigma}$-subnormal complement $T$ in $G$ by part (7). Hence $T \cap N_{p}= 1$. Lemma 3 (4) and part (4) imply that $Z\le T^{\mathfrak{S}_{\sigma}}\le T$ for some minimal subnormal subgroup $Z$ of $G$. Then $Z \cap N_{p}=1$ and, therefore, $Z \nleq N$, which contradicts part (1). Therefore, assertion (10) holds.

A concluding contradiction for (iv) $\Rightarrow$ (i). It follows from part (10) that $|D|$ is a power of a prime, which contradicts part (2). Therefore, the implication (iv) $\Rightarrow$ (i) holds.

(v) $\Rightarrow$ (i) Suppose that this assertion fails to hold; let $G$ be a counterexample of minimal order.

First, note that, if $H\le A\le G$, where $H$ is not $K$-$\mathfrak{S}_{\sigma}$-subnormal in $A$, then $H$ is not $K$-$\mathfrak{S}_{\sigma}$-subnormal in $G$ by Lemma 3 (1). Hence $i_{K,\mathfrak{S}_{\sigma}}(A)\le i_{K,\mathfrak{S}_{\sigma}}(G)$ for every subgroup $A$ of $G$. Moreover, if $H/N\le G/N$, where $H/N$ is not $K$-$\mathfrak{S}_{\sigma}$-subnormal in $G/N$, then $H$ is not $K$-$\mathfrak{S}_{\sigma}$-subnormal in $G$ by Lemma 3 (2). Hence $i_{K,\mathfrak{S}_{\sigma}}(G/N)\le i_{K,\mathfrak{S}_{\sigma}}(G)$.

(1) If $N$ is a nonidentity normal subgroup of $G$ such that $|\sigma(G/N)|=|\sigma(G)|$, then $G/N$ is $\sigma$-solvable. Moreover, if $L$ is a proper subgroup of $G$ such that $|\sigma(L)|=|\sigma(G)|$, then $L$ is $\sigma$-solvable.

Since

$$i_{K,\mathfrak{S}_{\sigma}}(G/N)\le i_{K,\mathfrak{S}_{\sigma}}(G)\le 2|\sigma(G)|=2|\sigma(G/N)|,$$

it follows that the conjecture holds for $G/N$ and, therefore, $G/N$ is $\sigma$-solvable by the choice of $G$. Similarly, it follows from $i_{K,\mathfrak{S}_{\sigma}}(L) \le i_{K,\mathfrak{S}_{\sigma}}(G)$ that $L$ is $\sigma$-solvable.

(2) The equality $O_{\sigma_{i}}(G)=1$ holds for every $\sigma_i\in \sigma(G)$.

Suppose that $O:= O_{\sigma_i} (G)\ne 1$ for some $i$. Let us first show that $O\nleq \Phi(G)$. Indeed, if $O \le \Phi(G)$, then $\pi(G/O)=\pi(G)$ and, therefore, $\sigma(G/O)=\sigma(G)$, and thus $G/O$ is $\sigma$-solvable by part (1). Consequently, $G$ is $\sigma$-solvable; a contradiction. Hence $O\nleq \Phi(G)$, and thus there is a maximal subgroup $M$ of $G$ such that $G=OM$. Then $M$ is not a $\sigma$-solvable group and, therefore, $|\sigma(M)|\ne| \sigma(G)|$ by part (1). On the other hand, $|G: M|=|O: Q\cap M|$ is a $\sigma_{i}$-number and, therefore, $|\sigma(M)|=|\sigma(G)|-1$. This implies that $M\cap O=1$ and, therefore, $O$ is a Hall $\sigma_i$-subgroup of $G$.

The inequality $i_{\sigma}(M)\ge 2|\sigma(M)|+1$ holds by the choice of $G$. Let $i_{K,\mathfrak{S}_{\sigma}}(M)=k$. If $k=0$, then all subgroups of $M$ are $K$-$\mathfrak{S}_{\sigma}$-subnormal in $M$ and, therefore, $M$ is $\sigma$-solvable by implication (ii) $\Rightarrow$ (i). Thus, $k\ne 0$. Let $T_1,\dots,T_k$ be representatives of the classes of non-$K$-$\mathfrak{S}_{\sigma}$-subnormal subgroups of $M$ of equal order. Since $T_j=T_j(O\cap M)=OT_j\cap M$ is not a $K$-$\mathfrak{S}_{\sigma}$-subnormal subgroup of $M$, it follows that $OT_j$ is not $K$-$\mathfrak{S}_{\sigma}$-subnormal in $G$ by Lemma 3 (1). Moreover, since $M\cap O=1$ and $|T_{j_1}|\ne |T_{j_2}|$ for $j_1\ne j_2$, it follows that $|OT_{j_1}|\ne |OT_{j_2}|$ for all $j_1\ne j_2$.

Thus, taking into account the inequality $i_{K,\mathfrak{S}_{\sigma}}(G)\ge i_{K,\mathfrak{S}_{\sigma}}(M)$ and counting the subgroups $OT_i$, we obtain

$$i_{K,\mathfrak{S}_{\sigma}}(G)\ge 2i_{K,\mathfrak{S}_{\sigma}}(M)\ge 2(2|\sigma (M)|+1).$$

However, since $2\le |\sigma (M)|=|\sigma (G)|-1$, it follows that

$$i_{K,\mathfrak{S}_{\sigma}}(G)\ge 2(|\sigma(G)|-1)+ 2|\sigma(M)|+2\ge 2|\sigma (G)|+4.$$

This contradiction completes the proof of assertion (2).

(3) If $H$ is a nonidentity $\sigma$-solvable subgroup of $G$, then $H$ is not $K$-$\mathfrak {S}_{\sigma}$-subnormal in $G$ (this follows from part (2) and Lemma 3 (5)).

(4) The equality $O^{\sigma_i}(G)=G$ holds for every $\sigma_i\in\sigma(G)$.

Suppose that $O=O^{\sigma_i}(G)< G$ for some $i$. Since $G/O$ is a $\sigma_i$-group, it follows that $G/O$ is $\sigma$-solvable. Hence

$$i_{K,\mathfrak{S}_{\sigma}}(O)\ge 2|\sigma(O)|+1$$

by the choice of $G$.

By assertion (1), $|\sigma(O)|=|\sigma(G)|-1$ and, therefore, $O$ is a Hall $\sigma_i'$-subgroup of $G$. Therefore, by the Schur–Zassenhaus theorem, there is a complement $H$ for $O$ in $G$, $H$ is a Hall $\sigma_i$-subgroup of $G$, and every Hall $\sigma_i$-subgroup of $G$ is conjugate to $H$.

Note that $H$ is simple. Indeed, suppose what $N\ne 1$ is a proper normal subgroup of $H$. Then $ON$ is a proper normal subgroup of $G$, and $|\sigma(ON)|=|\sigma(G)|$. Hence $ON$ is $\sigma$-solvable by assertion (1), which implies that $G$ is $\sigma$-solvable, since $G/ON$ is a $\sigma_i$-group, a contradiction. Thus, $H$ is a simple group.

Since $O$ and $H$ are Hall subgroups of $G$, it follows that either $O$ or $H$ is solvable by the Feit–Thompson theorem on the solvability of groups of odd order. However, if $O$ is solvable, then $O$ is $\sigma$-solvable and, therefore, $G$ is $\sigma$-solvable. Hence $H$ is solvable, and thus $|H|=q$ for some $q\in \pi(G)$. Consequently, $H$ is not $K$-$\mathfrak{S}_{\sigma}$-subnormal in $G$ by part (3). Moreover, every nonidentity Sylow subgroup of $O$ is not $K$-$\mathfrak{S}_{\sigma}$-subnormal in $O$. Let $P$ be some Sylow $p$-subgroup of $O$, $p\in \pi(O)$. According to Frattini’s argument, $G=ON_G(P)$. Since $q$ does not divide $|O|$, it follows that $q$ divides $|N_G(P)|$. Hence there is an $x\in G$ such that $H^x\le N_G(P)$ and, therefore, $H^{x}P$ is a subgroup of $G$. Since

$$P=P(H^{x}\cap O)=H^{x}P\cap O$$

is not $K$-$\mathfrak{S}_{\sigma}$-subnormal in $O$, it follows that $H^{x}P$ is not $K$-$\mathfrak{S}_{\sigma}$-subnormal in $G$ by Lemma 3 (1). We also note that $|\pi (O)|\ge 2$ by the $p^aq^b$-Burnside theorem. Therefore, the number of subgroups of the form $H^{x}P$ is not less than 2. Thus, taking into account that

$$i_{K,\mathfrak{S}_{\sigma}}(G)\ge i_{K,\mathfrak{S}_{\sigma}}(O)\ge 2|\sigma (O)|+1=2(|\sigma (G)|-1)+1=2(|\sigma (G)|-1$$

and taking into account the subgroups $H^{x}P$ and $H$, we obtain

$$i_{K, \mathfrak{S}_{\sigma}}(G)\ge i_{K,\mathfrak{S}_{\sigma}}(O)+2+1\ge 2|\sigma (G)|+2.$$

This contradiction completes the proof of assertion (4).

A concluding contradiction for (v) $\Rightarrow$ (i). Since $G$ is unsolvable, it follows that $G$ has a noncyclic Sylow subgroup $P$ by [32, VI, 10.3]. Then a maximal subgroup $V$ of $P$ is not identity and, therefore, $V$ is not $K$-$\mathfrak{S}_{\sigma}$-subnormal in $G$ by part (3). Moreover, every nonidentity Sylow subgroup of $G$ is not $K$-$\mathfrak{S}_{\sigma}$-subnormal in $G$. Therefore, $m \ge |\pi(G)|+1$ for the number $m$ of all classes of non-$K$-$\mathfrak{S}_{\sigma}$-subnormal primary subgroups of equal order.

By assertion (4), $G$ is not $p$-nilpotent for any $p\in\pi(G)$. Therefore, taking into account [32, III, 5.2], we see that there is a $p$-closed Schmidt subgroup $E$ of $G$ with $p\in \pi(E)$ for every $p\in \pi(G)$.

Now let $p> q$ be distinct prime divisors of $|G|$. Let $V$ be a $p$-closed Schmidt subgroup with $p\in \pi(V)$, and let $W$ be a $q$-closed Schmidt subgroup with $q\in \pi(W)$. By [32, IV, 5.4], $V=V_p\rtimes V_r$ and $W=W_q\rtimes W_t$, where $V_p$ is a Sylow $p$-subgroup and $V_r$ is a cyclic Sylow $r$-subgroup of $V$ ($r\ne p$); $W_q$ is a Sylow $q$-subgroup and $W_t$ is a cyclic Sylow $t$-subgroup of $W$ ($t\ne q$). By part (4), $V$ and $W$ are not $K$-$\mathfrak{S}_{\sigma}$-subnormal in $G$. If $|V|\ne |W|$, then $V$ and $W$ are representatives of two distinct classes of non-$K$-$\mathfrak{S}_{\sigma}$-subnormal subgroups in $G$ of equal order.

Now suppose that $|V|=|W|$. In this case, $r=q$ and $t=p$. Let $T$ be a maximal subgroup of $V$ such that $V_p\le T$. If $V_p=T$, then $|V_q|=q$, and thus, taking into account that $|V|=|W|$, we see that $|W_q|=q$. Hence $W$ is supersolvable. As is well known, every supersolvable group is Ore dispersive. Therefore, $W$ is Ore dispersive, which implies that $q>p$, a contradiction. Thus, $V_p <T$. Moreover, $T$ is not $K$-$\mathfrak{S}_{\sigma}$-subnormal in $G$ by part (4), and $T$ is not a primary group. Thus, for every $p\in \pi(G)$, there is a class of non-$K$-$\mathfrak{S}_{\sigma}$-subnormal subgroups in $G$ of equal order, and each representative of any of these classes is not a primary group. It is also clear that $r \ge\pi(G)|$ for the number $r$ of such classes. Hence

$$i_{K,\mathfrak{S}_{\sigma}}(G)\ge |\pi(G)|+1+|\pi(G)| \ge 2|\sigma (G)|+1.$$

This contradiction completes the proof of the implication (v) $\Rightarrow$ (i).

This completes the proof of the theorem.