On $H_{\sigma}$-permutably embedded and $H_{\sigma}$-subnormaly embedded subgroups of finite groups

TL;DR

This paper investigates the structure of finite groups with specific embedded subgroups called $H_{\sigma}$-permutably embedded subgroups, generalizing known results in subgroup embedding theory.

Contribution

It introduces the concept of $H_{\sigma}$-permutably embedded subgroups and extends existing theorems to this broader context.

Findings

Characterization of finite groups with $H_{\sigma}$-permutably embedded subgroups

Generalization of known subgroup embedding results

Conditions for the existence of such subgroups in finite groups

Abstract

Let $G$ be a finite group. Let $\sigma =\{\sigma_{i} | i\in I\}$ be a partition of the set of all primes $\Bbb{P}$ and $n$ an integer. We write $\sigma (n) =\{\sigma_{i} |\sigma_{i}\cap \pi (n)\ne \emptyset \}$, $\sigma (G) =\sigma (|G|)$. A set $ {\cal H}$ of subgroups of $G$ is said to be a complete Hall $\sigma $-set of $G$ if every member of ${\cal H}\setminus \{1\}$ is a Hall $\sigma_{i}$-subgroup of $G$ for some $\sigma_{i}$ and ${\cal H}$ contains exact one Hall $\sigma_{i}$-subgroup of $G$ for every $\sigma_{i}\in \sigma (G)$. A subgroup $A$ of $G$ is called: (i) a $\sigma$-Hall subgroup of $G$ if $\sigma (|A|) \cap \sigma (|G:A|)=\emptyset$; (ii) ${\sigma}$-permutable in $G$ if $G$ possesses a complete Hall $\sigma$-set ${\cal H}$ such that $AH^{x}=H^{x}A$ for all $H\in {\cal H}$ and all $x\in G$. We say that a subgroup $A$ of $G$ is $H_{\sigma}$-permutably embedded in $G$ if $A$ is a ${\sigma}$-Hall subgroup of some ${\sigma}$-permutable subgroup of $G$. We study finite groups $G$ having an $H_{\sigma}$-permutably embedded subgroup of order $|A|$ for each subgroup $A$ of $G$. Some known results are generalized.

Equations

$$A=A_0\le A_1\le \cdots \le A_t=G$$

$$B=(\dots ((L_1\text{Yup}{L}_2)\text{Yup}{L}_3)\text{Yup}\cdots )\text{Yup}{L}_n$$

$$D=P_1{P}_2\cdots P_n=G^{{\mathfrak{N}}_{\sigma }}$$

$$H=(H\cap D)⋊(A\times (H\cap A_5))$$

$$G=H{N}_G(E)=(E(H\cap \Phi (G)))N_G(E)=N_G(E).$$

$$D/D\cap R\simeq DR/R=(G/R{)}^{{\mathfrak{N}}_{\sigma }}$$

$$G=M_G{N}_G(B)=M{N}_G(B)=H{N}_M(B)N_G(B)=H{N}_G(B).$$

$$G/V_G\simeq (H/K)⋊(G/C_G(H/K)).$$

$$|A_2|=|H_2|=|K_2|.$$

$$|W:V\cap K_i|=|V{K}_2:V\cap K_i|=|V||K_2|:|V\cap K_2||V\cap K_i|$$

$$|W|=p|H_2||V\cap H_3|\dots |V\cap H_t|$$

$$|W|=|A_1|\cdots |A_t|$$

$$(E/D)\times (A_{i}D/D)=E{A}_i/D$$

$$G/D=(E/D)\times (H_{i}D/D).$$

$$AE/D=(DA/D)\times (ED/D)$$

$$(|G:E_i|,|G:E_j|)=1$$

$$R\le D\le S_G\le S$$

$$E=E_0\le E_1\le \cdots \le E_r=G$$

$$ER/R=(ER/R{)}^{{\mathfrak{N}}_{\sigma }}(RM/R).$$

$$(ER/R{)}^{{\mathfrak{N}}_{\sigma }}=E^{{\mathfrak{N}}_{\sigma }}R/R.$$

$$DE/D\simeq E/D\cap E$$

$$E^{{\mathfrak{N}}_{\sigma }}\cap RM=(E^{{\mathfrak{N}}_{\sigma }}\cap R)(E^{{\mathfrak{N}}_{\sigma }}\cap M)=1.$$

$$E/E^{{\mathfrak{N}}_{\sigma }}=E^{{\mathfrak{N}}_{\sigma }}MR/E^{{\mathfrak{N}}_{\sigma }}\simeq MR$$

$$G/V_G\simeq (H/K)⋊(G/C_G(H/K))$$

Full text

We study finite groups $G$ having an $H_{\sigma}$-permutably embedded subgroup of order $|A|$ for each subgroup $A$ of $G$. Some known results are generalized.

00footnotetext: Keywords: finite group, $\sigma$-Hall subgroup, $\sigma$-subnormal subgroup, $\sigma$-nilpotent group, $H_{\sigma}$-permutably embedded subgroup.00footnotetext: Mathematics Subject Classification (2010): 20D10, 20D15, 20D30

1 Introduction

Throughout this paper, all groups are finite and $G$ always denotes a finite group. Moreover, $n$ is an integer, $\mathbb{P}$ is the set of all primes, and if $\pi\subseteq\mathbb{P}$, then $\pi^{\prime}=\mathbb{P}\setminus\pi$. The symbol $\pi(n)$ denotes the set of all primes dividing $n$; as usual, $\pi(G)=\pi(|G|)$, the set of all primes dividing the order of $G$. We use $n_{\pi}$ to denote the ${\pi}$-part of $n$, that is, the largest ${\pi}$-number dividing $n$; $n_{p}$ denotes the largest degree of $p$ dividing $n$.

In what follows, $\sigma=\{\sigma_{i}|i\in I\}$ is some partition of $\mathbb{P}$, that is, $\mathbb{P}=\cup_{i\in I}\sigma_{i}$ and $\sigma_{i}\cap\sigma_{j}=\emptyset$ for all $i\neq j$; $\Pi$ is a subset of $\sigma$ and $\Pi^{\prime}=\sigma\setminus\Pi$.

Let $\sigma(n)=\{\sigma_{i}|\sigma_{i}\cap\pi(n)\neq\emptyset\}$ and $\sigma(G)=\sigma(|G|)$. Then we say that $G$ is $\sigma$-primary [1] if $G$ is a $\sigma_{i}$-group for some $\sigma_{i}\in\sigma$.

A set ${\cal H}$ of subgroups of $G$ is said to be a * complete Hall $\sigma$-set* of $G$ [2, 3] if every member of ${\cal H}\setminus\{1\}$ is a Hall $\sigma_{i}$-subgroup of $G$ for some $\sigma_{i}$ and ${\cal H}$ contains exact one Hall $\sigma_{i}$-subgroup of $G$ for every $\sigma_{i}\in\sigma(G)$. We say that $G$ is $\sigma$-full if $G$ possesses a complete Hall $\sigma$-set. Throughout this paper, $G$ is always supposed to be a $\sigma$-full group.

A subgroup $A$ of $G$ is called [1]: (i) a $\sigma$-Hall subgroup of $G$ if $\sigma(|A|)\cap\sigma(|G:A|)=\emptyset$; (ii) ${\sigma}$-subnormal in $G$ if there is a subgroup chain

[TABLE]

such that either $A_{i-1}\trianglelefteq A_{i}$ or $A_{i}/(A_{i-1})_{A_{i}}$ is ${\sigma}$-primary for all $i=1,\ldots,t$; (iii) ${\sigma}$-quasinormal or ${\sigma}$-permutable in $G$ if $G$ possesses a complete Hall $\sigma$-set ${\cal H}$ such that $AH^{x}=H^{x}A$ for all $H\in{\cal H}$ and all $x\in G$. In particular, $A$ is called $S$-quasinormal or $S$-permutable in $G$ [4, 5] provided $AP=PA$ for all Sylow subgroups $P$ of $G$.

Definition 1.1. We say that a subgroup $A$ of $G$ is * $H_{\sigma}$-subnormally* (respectively $H_{\sigma}$-permutably, $H_{\sigma}$-normally) embedded in $G$ if $A$ is a ${\sigma}$-Hall subgroup of some ${\sigma}$-subnormal (respectively ${\sigma}$-permutable, normal) subgroup of $G$.

In the special case, when $\sigma=\{\{2\},\{3\},\ldots\}$, the definition of $H_{\sigma}$-normally embedded subgroups is equivalent to the concept of Hall normally embedded subgroups in [6], the definition of $H_{\sigma}$-permutably embedded subgroups is equivalent to the concept of Hall $S$-quasinormally embedded subgroups in [7] and the definition of $H_{\sigma}$-subnormally embedded subgroups is equivalent to the concept of Hall subnormally embedded subgroups in [8].

Example 1.2. (i) For any $\sigma$, all $\sigma$-Hall subgroups and all ${\sigma}$-subnormal subgroups of any group $S$ are $H_{\sigma}$-subnormally embedded in $S$. Now, let $G=(C_{7}\rtimes C_{3})\times A_{5}$, where $C_{7}\rtimes C_{3}$ is a non-abelian group of order 21 and $A_{5}$ is the alternating group of degree 5, and let $H=(C_{7}\rtimes C_{3})A$, where $A$ is a Sylow 2-subgroup of $A_{5}$. Let $\sigma=\{\sigma_{1},\sigma_{2}\}$, where $\sigma_{1}=\{7\}$ and $\sigma_{2}=\{7\}^{\prime}$. Then $H$ is $\sigma$-subnormal in $G$ and $C_{3}A_{5}$ is a $\sigma$-Hall subgroup of $G$. In view of Lemma 2.1(1)(5) below, the subgroup $C_{3}A$ is neither $\sigma$-subnormal in $G$ nor $H_{\sigma}$-normally embedded in $G$.

(ii) For any $\sigma$, all $\sigma$-Hall subgroups and all $\sigma$-permutable subgroups of any group $S$ are $H_{\sigma}$-permutably embedded in $S$. Now, let $p>q>r$ be primes, where $r^{2}$ divides $q-1$. Let $\sigma=\{\sigma_{1},\sigma_{2}\}$, where $\sigma_{1}=\{q,r\}$ and $\sigma_{2}=\{q,r\}^{\prime}$. Let $H=Q\rtimes R$ be a group of order $qr^{2}$, where $C_{H}(Q)=Q$. Let $P$ be a simple ${\mathbb{F}}_{p}H$-module which is faithful for $H$ and $G=P\rtimes H$. Let $R_{1}$ be a subgroup of $R$ of order $r$. Then the subgroup $V=PR_{1}$ is $\sigma$-permutable in $G$ and $R_{1}$ is a $\sigma$-Hall subgroup of $V$. Hence $R_{1}$ is $H_{\sigma}$-permutably embedded in $G$. It is also clear that $G$ has no an $S$-permutable subgroup $W$ such that $R_{1}$ is a Hall subgroup of $W$, so $R_{1}$ is neither $H_{\sigma}$-normally embedded nor $S$-permutably embedded in $G$.

(iii) For any $\sigma$, all $\sigma$-Hall subgroups and all normal subgroups of any group $S$ are $H_{\sigma}$-normally embedded in $S$. Now, let $P$ be a simple ${\mathbb{F}}_{11}(C_{7}\rtimes C_{3})$-module which is faithful for $C_{7}\rtimes C_{3}$. Let $G=(P\rtimes(C_{7}\rtimes C_{3}))\times A_{5}$. Let $\sigma=\{\sigma_{1},\sigma_{2}\}$, where $\sigma_{1}=\{5,7,11\}$ and $\sigma_{2}=\{5,7,11\}^{\prime}$. Then the subgroup $M=(P\rtimes C_{7})\times A_{5}$ is normal in $G$ and a subgroup $B$ of $A_{5}$ of order 12 is ${\sigma}$-Hall subgroup of $M$, so $B$ is $H_{{\sigma}}$-normally embedded in $G$. Finally, if $\sigma=\{\sigma_{1},\sigma_{2}\}$, where $\sigma_{1}=\{7\}$ and $\sigma_{2}=\{7\}^{\prime}$, then $B$ is not $H_{{\sigma}}$-normally embedded in $G$.

Recall that $G$ is $\sigma$-nilpotent [9] if $G=H_{1}\times\cdots\times H_{t}$ for some $\sigma$-primary groups $H_{1},\ldots,H_{t}$. The $\sigma$-nilpotent residual $G^{{\mathfrak{N}}_{\sigma}}$ of $G$ is the intersection of all normal subgroups $N$ of $G$ with $\sigma$-nilpotent quotient $G/N$, $G^{\mathfrak{N}}$ denotes the nilpotent residual of $G$. It is clear that every subgroup of a $\sigma$-nilpotent group $G$ is $\sigma$-permutable and $\sigma$-subnormal in $G$.

Theorem 1.3. Let ${\cal H}=\{1,H_{1},\ldots H_{t}\}$ be a complete Hall $\sigma$-set of $G$ and $D=G^{{\mathfrak{N}}_{\sigma}}$. Then any two of the following conditions are equivalent:

(i) $G$ has an $H_{\sigma}$-permutably embedded subgroup of order $|A|$ for each subgroup $A$ of $G$.

(ii) $D$ is a complemented cyclic of square-free order subgroup of $G$ and $|\sigma_{i}\cap\pi(G)|=1$ for all $i$ such that $\sigma_{i}\cap\pi(D)\neq\emptyset$.

(iii) For each set $\{A_{1},\ldots,A_{t}\}$, where $A_{i}$ is a subgroup (respectively normal subgroup) of $H_{i}$ for all $i=1,\ldots,t$, $G$ has an $H_{\sigma}$-permutably embedded (respectively $H_{\sigma}$-normally embedded) subgroup of order $|A_{1}|\cdots|A_{t}|$.

Let $\mathfrak{F}$ be a class of groups. A subgroup $H$ of $G$ is said to be an $\mathfrak{F}$-covering subgroup of $G$ [10, VI, Definition 7.8] if $H\in\mathfrak{F}$ and for every subgroup $E$ of $G$ such that $H\leq E$ and $E/N\in\mathfrak{F}$ it follows that $E=NH$. We say that a subgroup $H$ of $G$ is a $\sigma$-Carter subgroup of $G$ if $H$ is an $\mathfrak{N}_{\sigma}$-covering subgroup of $G$, where $\mathfrak{N}_{\sigma}$ is the class of all $\sigma$-nilpotent groups.

A group $G$ is said to have a Sylow tower if $G$ has a normal series $1=G_{0}<G_{1}<\cdots<G_{t-1}<G_{t}=G$, where $|G_{i}/G_{i-1}|$ is the order of some Sylow subgroup of $G$ for each $i\in\{1,\ldots,t\}$. A chief factor of $G$ is said to be $\sigma$-central (in $G$) [1] if the semidirect product $(H/K)\rtimes(G/C_{G}(H/K))$ is $\sigma$-primary. Otherwise, $H/K$ is called $\sigma$-eccentric (in $G$).

We say that $G$ is a $H\sigma E$-group if the following conditions are hold: (i) $G=D\rtimes M$, where $D=G^{{\mathfrak{N}}_{\sigma}}$ is a $\sigma$-Hall subgroup of $G$ and $|\sigma(D)|=|\pi(D)|$; (ii) $D$ has a Sylow tower and every chief factor of $G$ below $D$ is $\sigma$-eccentric; (iii) $M$ acts irreducibly on every $M$-invariant Sylow subgroup of $D$.

We do not still know the structure of a group $G$ having a $H_{\sigma}$-subnormally embedded subgroup of order $|A|$ for each subgroup $A$ of $G$. Nevertheless, the following fact is true.

Theorem 1.4. Any two of the following conditions are equivalent:

(i) Every subgroup of $G$ is $H_{\sigma}$-subnormally embedded in $G$.

(ii) Every ${\sigma}$-subnormal subgroup $H$ of $G$ is an $H\sigma E$-group of the form $H=D\rtimes M$, where $D=H^{{\mathfrak{N}}_{\sigma}}$ and $M$ is a $\sigma$-Carter subgroup of $H$.

(iii) Every ${\sigma}$-subnormal subgroup of $G$ is an $H\sigma E$-group.

Now, let us consider some corollaries of Theorems 1.3 and 1.4. First note that since a nilpotent group $G$ possesses a normal subgroup of order $n$ for each integer $n$ dividing $|G|$, in the case when $\sigma=\{\{2\},\{3\},\ldots\}$, Theorem 1.3 covers Theorem 11 in [6], Theorem 2.7 in [8] and Theorems 3.1 and 3.2 in [7].

From Theorem 1.3 we also get the following result.

Corollary 1.5. Suppose that $G$ possesses a complete Hall $\sigma$-set ${\cal H}=\{1,H_{1},\ldots,H_{t}\}$ such that $H_{i}$ is nilpotent for all $i=1,\ldots,t$. Then $G$ has an $H_{\sigma}$-normally embedded subgroup of order $|H|$ for each subgroup $H$ of $G$ if and only if the nilpotent residual $D=G^{\mathfrak{N}}$ of $G$ is cyclic of square-free order and $|\sigma_{i}\cap\pi(G)|=1$ for all $i$ such that $\sigma_{i}\cap\pi(D)\neq\emptyset$.

In the case when $\sigma=\{\{2\},\{3\},\ldots\}$ we get from Corollary 1.5 the following known result.

Corollary 1.6 (Ballester-Bolinches, Qiao [11]). $G$ has a Hall normally embedded subgroup of order $|H|$ for each subgroup $H$ of $G$ if and only if the nilpotent residual $G^{\mathfrak{N}}$ of $G$ is cyclic of square-free order.

On the basis of Theorems 1.3 and 1.4 we prove also the next two theorems.

Theorem 1.7. Any two of the following conditions are equivalent:

(i) Every subgroup of $G$ is $H_{\sigma}$-normally embedded in $G$.

(ii) $G=D\rtimes M$ is a $H\sigma E$-group, where $D$ is a cyclic group of square-free order and $M$ is a Dedekind group.

(iii) $G=D\rtimes M$, where $D$ is a $\sigma$-Hall cyclic subgroup of $G$ of square-free order with $|\sigma(D)|=|\pi(D)|$ and $M$ is a Dedekind group.

In the case when $\sigma=\{\{2\},\{3\},\ldots\}$ we get from Theorem 1.7 the following known result.

Corollary 1.8 (Li, Liu [8]). Every subgroup of $G$ is a Hall normally embedded subgroup of $G$ if and only if $G=D\rtimes M$, where $D=G^{\mathfrak{N}}$ is a cyclic Hall subgroup of $G$ of square-free order and $M$ is a Degekind group.

Theorem 1.9. Any two of the following conditions are equivalent:

(i) Every subgroup of $G$ is $H_{\sigma}$-permutably embedded in $G$.

(ii) $G=D\rtimes M$ is a $H\sigma E$-group, where $D=G^{{\mathfrak{N}}_{\sigma}}$ is a cyclic group of square-free order.

(iii) $G=D\rtimes M$, where $D$ is a $\sigma$-Hall cyclic subgroup of $G$ of square-free order with $|\sigma(D)|=|\pi(D)|$ and $M$ is ${\sigma}$-nilpotent

Corollary 1.10. Every subgroup of $G$ is a Hall $S$-quasinormally embedded subgroup of $G$ if and only if $G=D\rtimes M$, where $D=G^{{\mathfrak{N}}}$ is a cyclic Hall subgroup of $G$ of square-free order and $M$ is a Carter subgroup of $G$.

In conclusion of this section, consider the following example.

Example 1.11. Let $5<p_{1}<p_{2}<\cdots<p_{n}$ be a set of primes and $p$ a prime such that either $p>p_{n}$ or $p$ divides $p_{i}-1$ for all $i=1,\ldots,n$. Let $A$ be a group of order $p$ and $P_{i}$ a simple ${\mathbb{F}}_{p_{i}}A$-module which is faithful for $A$. Let $L_{i}=P_{i}\rtimes A$ and

[TABLE]

(see [10, p. 50]). We can assume without loss of generality that $L_{i}\leq B$ for all $i=1,\ldots,n$. Let $G=B\times A_{5}$, where $A_{5}$ is the alternating group of degree 5, and let $\sigma$ be a partition of $\mathbb{P}$ such that for some different indices $i,j,i_{1},\ldots,i_{n}\in I$ we have $p\in\sigma_{i}$, $\{2,3,5\}\subseteq\sigma_{j}$ and $p_{k}\in\sigma_{i_{k}}$ for all $k=1,\ldots,n$. Then

[TABLE]

is a $\sigma$-Hall subgroup of $G$ and $G=D\rtimes(A\times A_{5})$.

We show that every subnormal subgroup $H$ of $G$ satisfies Condition (ii) in Theorem 1.4. If $H^{{\mathfrak{N}}_{\sigma}}=1$, it is evident. Hence we can assume without loss of generality $A\leq H$ since every $p^{\prime}$-subgroup of $G$ is $\sigma$-nilpotent. But then

[TABLE]

by Lemma 2.1(4), where $H\cap D$ is a normal $\sigma$-Hall subgroup of $H$ and $M=A\times(H\cap A_{5})$ is a $\sigma$-nilpotent subgroup of $H$. Moreover, $H\cap A_{5}$ induces on every non-identity Sylow subgroup of $H\cap D$ a non-trivial irreducible group of automorphisms. Therefore $H^{{\mathfrak{N}}_{\sigma}}=H\cap D$ and $|\sigma(H^{{\mathfrak{N}}_{\sigma}})|=|\pi(H^{{\mathfrak{N}}_{\sigma}})|$. It is also clear that $M$ is a $\sigma$-Carter subgroup of $H$ and every chief factor of $H$ below $H^{{\mathfrak{N}}_{\sigma}}$ is $\sigma$-eccentric in $H$. Thus $G$ satisfies Condition (ii) in Theorem 1.4, and so every subgroup $H$ of $G$ is $H_{\sigma}$-subnormally embedded in $G$. On the other hand, the subgroup $DAC_{2}$, where $C_{2}$ is a subgroup of order 2 of $G$, is not Hall subnormally embedded in $G$ since $C_{2}$ is not a Sylow subgroup of any subnormal subgroup of $G$.

Finally, if $p$ divides $p_{i}-1$ for all $i=1,\ldots,n$, then $|P_{i}|=p_{i}$ for all $i=1,\ldots,n$, so $G$ satisfies Condition (ii) in Theorem 1.9 and hence satisfies Condition (ii) in Theorem 1.3.

2 Basic lemmas

An integer $n$ is called a $\Pi$-number if $\sigma(n)\subseteq\Pi$. A subgroup $H$ of $G$ is called a Hall $\Pi$-subgroup of $G$ [1] if $|H|$ is a $\Pi$-number and $|G:H|$ is a $\Pi^{\prime}$-number.

Lemma 2.1 (See Lemma 2.6 in [1]). Let $A$, $K$ and $N$ be subgroups of $G$, where $A$ is $\sigma$-subnormal in $G$ and $N$ is normal in $G$.

(1) $A\cap K$ is $\sigma$-subnormal in $K$.

(2) If $K$ is $\sigma$-subnormal in $G$, then $A\cap K$ and $\langle A,K\rangle$ are $\sigma$-subnormal in $G$.

(3) $AN/N$ is $\sigma$-subnormal in $G/N$.

(4) If $H\neq 1$ is a Hall $\Pi$-subgroup of $G$ and $A$ is not a $\Pi^{\prime}$-group, then $A\cap H\neq 1$ is a Hall $\Pi$-subgroup of $A$.

(5) If $|G:A|$ is a $\sigma_{i}$-number, then $O^{{\sigma_{i}}}(A)=O^{{\sigma_{i}}}(G)$.

(6) If $V/N$ is a $\sigma$-subnormal subgroup of $G/N$, then $V$ is $\sigma$-subnormal in $G$.

(7) If $K$ is a $\sigma$-subnormal subgroup of $A$, then $K$ is $\sigma$-subnormal in $G$.

A group $G$ is said to be *$\sigma$-soluble *[1] if every chief factor of $G$ is $\sigma$-primary.

Lemma 2.2 (See Lemmas 2.8 and 3.2 and Theorems B and C in [1]). Let $A$, $K$ and $N$ be subgroups of $G$, where $A$ is $\sigma$-permutable in $G$ and $N$ is normal in $G$.

(1) $AN/N$ is $\sigma$-permutable in $G/N$.

(2) If $G$ is $\sigma$-soluble, then $A\cap K$ is $\sigma$-permutable in $K$.

(3) If $N\leq K$, $K/N$ is $\sigma$-permutable in $G/N$ and $G$ is $\sigma$-soluble, then $K$ is $\sigma$-permutable in $G$.

(4) $A$ is $\sigma$-subnormal in $G$.

(5) If $G$ is $\sigma$-soluble and $K$ is $\sigma$-permutable in $G$, then $K\cap A$ is $\sigma$-permutable in $G$.

Lemma 2.3. Let $H$ be a normal subgroup of $G$. If $H/H\cap\Phi(G)$ is a $\Pi$-group, then $H$ has a a Hall $\Pi$-subgroup, say $E$, and $E$ is normal in $G$. Hence, if $H/H\cap\Phi(G)$ is $\sigma$-nilpotent, then $H$ is $\sigma$-nilpotent.

Proof. Let $D=O_{\Pi^{\prime}}(H)$. Then, since $H\cap\Phi(G)$ is nilpotent, $D$ is a Hall $\Pi^{\prime}$-subgroup of $H$. Hence by the Schur-Zassenhaus theorem, $H$ has a Hall $\Pi$-subgroup, say $E$. It is clear that $H$ is $\pi^{\prime}$-soluble where $\pi^{\prime}=\cup_{\sigma_{i}\in\Pi^{\prime}}\sigma_{i}$, so any two Hall $\Pi$-subgroups of $H$ are conjugate. By the Frattini argument,

[TABLE]

Therefore $E$ is normal in $G$. The lemma is proved.

Lemma 2.4. If every chief factor of $G$ below $D=G^{{\mathfrak{N}}_{\sigma}}$ is cyclic, then $D$ is nilpotent.

Proof. Assume that this is false and let $G$ be a counterexample of minimal order. Let $R$ be a minimal normal subgroup of $G$. Then from the $G$-isomorphism

[TABLE]

we know that every chief factor of $G/R$ below $DR/R$ is cyclic, so the choice of $G$ implies that $D/D\cap R\simeq DR/R$ is nilpotent. Hence $R\leq D$ and $R$ is the unique minimal normal subgroup of $G$. In view of Lemma 2.3, $R\nleq\Phi(G)$ and so $R=C_{R}(R)$ by [13, A, 15.2]. But by hypothesis, $|R|$ is a prime, hence $G/R=G/C_{G}(R)$ is cyclic, so $G$ is supersoluble and so $G^{{\mathfrak{N}}_{\sigma}}$ is nilpotent since $G^{{\mathfrak{N}}_{\sigma}}\leq G^{{\mathfrak{N}}}$. The lemma is proved.

The following lemma is evident.

Lemma 2.5. The class of all $\sigma$-soluble groups is closed under taking direct products, homomorphic images and subgroups. Moreover, any extension of the $\sigma$-soluble group by a $\sigma$-soluble group is a $\sigma$-soluble group as well.

Let $A$, $B$ and $R$ be subgroups of $G$. Then $A$ is said to $R$-permute with $B$ [12] if for some $x\in R$ we have $AB^{x}=B^{x}A$.

If $G$ has a complete Hall $\sigma$-set ${\cal H}=\{1,H_{1},\ldots,H_{t}\}$ such that $H_{i}H_{j}=H_{j}H_{i}$ for all $i,j,$ then we say that $\{H_{1},\ldots,H_{t}\}$ is a $\sigma$-basis of $G$.

Lemma 2.6 (See Theorems A and B in [2]). Assume that $G$ is $\sigma$-soluble.

(i) $G$ has a $\sigma$-basis $\{H_{1},\ldots,H_{t}\}$ such that for each $i\neq j$ every Sylow subgroup of $H_{i}$ $G$-permutes with every Sylow subgroup of $H_{j}$.

(ii) For any $\Pi$, the following hold: $G$ has a Hall $\Pi$-subgroup $E$, every $\Pi$-subgroup of $G$ is contained in some conjugate of $E$ and $E$ $G$-permutes with every Sylow subgroup of $G$.

Lemma 2.7. Let $H$, $E$ and $R$ be subgroups of $G$. Suppose that $H$ is $H_{\sigma}$-subnormally embedded in $G$ and $R$ is normal in $G$.

(1) If $H\leq E$, then $H$ is $H_{\sigma}$-subnormally embedded in $E$.

(2) $HR/R$ is $H_{\sigma}$-subnormally embedded in $G/R$.

(3) If $S$ is a ${\sigma}$-subnormal subgroup of $G$, then $H\cap S$ is $H_{\sigma}$-subnormally embedded in $G$.

(4) If $|G:H|$ is ${\sigma}$-primary, then $H$ is either a ${\sigma}$-Hall subgroup of $G$ or ${\sigma}$-subnormal in $G$.

Proof. Let $V$ be a ${\sigma}$-subnormal subgroup of $G$ such that $H$ is a $\sigma$-Hall subgroup of $V$.

(1) This assertion is a corollary of Lemma 2.1(1).

(2) In view of Lemma 2.1(3), $VR/R$ is ${\sigma}$-subnormal subgroup of $G/R$. It is also clear that $HR/R$ is a $\sigma$-Hall subgroup of $VR/R$. Hence $HR/R$ is $H_{\sigma}$-subnormally embedded in $G/R$.

(3) By Lemma 2.1(1)(2), $V\cap S$ is ${\sigma}$-subnormal both in $V$ and in $G$ and so $H\cap(V\cap S)=H\cap S$ is a ${\sigma}$-Hall subgroup of $V\cap S$ by Lemma 2.1(4), as required.

(4) Assume that $H$ is not ${\sigma}$-subnormal in $G$. Then $H<V$. By hypothesis, $|G:H|$ is ${\sigma}$-primary, say $|G:H|$ is a ${\sigma_{i}}$-number. Then $|V:H|$ is a ${\sigma_{i}}$-number. But $H$ is a ${\sigma}$-Hall subgroup of $V$. Hence $H$ is a ${\sigma}$-Hall subgroup of $G$.

The lemma is proved.

Lemma 2.8. Let $H$ be a ${\sigma}$-subnormal subgroup of a ${\sigma}$-soluble group $G$. If $|G:H|$ is a $\sigma_{i}$-number and $B$ is a $\sigma_{i}$-complement of $H$, then $G=HN_{G}(B)$.

Proof. Assume that this lemma is false and let $G$ be a counterexample of minimal order. Then $H<G$, so $G$ has a proper subgroup $M$ such that $H\leq M$, $|G:M_{G}|$ is a $\sigma_{i}$-number and $H$ is ${\sigma}$-subnormal in $M$. The choice of $G$ implies that $M=HN_{M}(B)$. On the other hand, clearly that $B$ is a $\sigma_{i}$-complement of $M_{G}$. Since $G$ is ${\sigma}$-soluble, Lemma 2.6 and the Frattini argument imply that

[TABLE]

The lemma is proved.

The following lemma is well-known (see for example [14, Lemma 3.29] or [15, 1.10.10]).

Lemma 2.9. Let $H/K$ be an abelian chief factor of $G$ and $V$ a maximal subgroup of $G$ such that $K\leq V$ and $HV=G$. Then

[TABLE]

3 Proofs of the results

Proof of Theorem 1.3. Without loss of generality we may assume that $H_{i}$ is a $\sigma_{i}$-group for all $i=1,\ldots,t$.

(i), (iii) $\Rightarrow$ (ii) Assume that this is false. Then $D\neq 1$ and so $t>1$.

First we prove the following claim.

(*) If $p\in\sigma_{i}\cap\pi(G)$, then $G$ has a $\sigma$-permutable subgroup $E$ with $|E|=|G|_{\sigma_{i}^{\prime}}p$.

We can assume without loss of generality that $i=1$. In fact, to prove Claim (*), we consistently build the $\sigma$-permutable subgroups $E_{2},\ldots,E_{t}$ such that $|H_{2}|\cdots|H_{j}|$ divides $|E_{j}|$ and $|E_{j}|_{\sigma_{1}}=p$ for all $j=2,\ldots,t$.

By hypothesis, $G$ has an $H_{\sigma}$-permutably embedded subgroup $X$ of order $p$. Let $V$ be a $\sigma$-permutable subgroup of $G$ such that $X$ is a $\sigma$-Hall subgroup of $V$. Then $|V|_{\sigma_{1}}=p$ and $G$ has a complete Hall $\sigma$-set $\{1,K_{1},\ldots,K_{t}\}$, where $K_{i}$ is a $\sigma_{i}$-group for all $i=1,\ldots,t$, such that $VK_{i}=K_{i}V$ for all $i=1,\ldots,t$. Let $W=VK_{2}$. Then $|W|_{\sigma_{1}}=p$.

Next we show that there is an $H_{\sigma}$-permutably embedded subgroup $Y$ of $G$ such that $|Y|=|W|$. It is enough to consider the case when Condition (iii) holds. Let $A_{1}$ be a subgroup of $H_{1}$ of order $p$, $A_{2}=H_{2}$ and $A_{i}=H_{i}\cap V$ for all $i>2$. Then

[TABLE]

On the other hand, $V\cap K_{i}$ and $V\cap H_{i}$ are Hall $\sigma_{i}$-subgroups of $V$ by Lemmas 2.1(4) and 2.2(4) and so $|V\cap K_{i}|=|V\cap H_{i}|$. Also, for every $i>2$ we have

[TABLE]

is a $\sigma_{i}^{\prime}$-number and hence $V\cap K_{i}=W\cap K_{i}$ is a Hall $\sigma_{i}$-subgroup of $W$. Therefore,

[TABLE]

and so $G$ has an $H_{\sigma}$-permutably embedded subgroup $Y$ of order

[TABLE]

by hypothesis.

Let $E_{2}$ be a $\sigma$-permutable subgroup of $G$ such that $Y$ is a $\sigma$-Hall subgroup of $E_{2}$. Then $|H_{2}|=|K_{2}|$ divides $|E_{2}|$ and $|E_{2}|_{\sigma_{1}}=p$. Now, arguing by induction, assume that $G$ has a $\sigma$-permutable subgroup $E_{t-1}$ such that $|H_{2}|\cdots|H_{t-1}|$ divides $|E_{t-1}|$ and $|E_{t-1}|_{\sigma_{1}}=p$. Then for some Hall $\sigma_{t}$-group $L$ we have $E_{t-1}L=LE_{t-1}$, and if $E_{t}=E_{t-1}L$, then $|E_{t}|=|G|_{\sigma_{1}^{\prime}}p$ and $E_{t}$ clearly is $\sigma$-permutable in $G$, as required.

Now, let $p\in\sigma_{i}\cap\pi(D)$ and let $P$ be a Sylow $p$-subgroup of $D$. Then, by Claim (*), $G$ possesses a $\sigma$-permutable subgroup $E$ such that $|E|=|G|_{\sigma_{i}^{\prime}}p$. Lemma 2.2(4) implies that $E$ is $\sigma$-subnormal in $G$. Let $j\neq i$. Then $H_{j}^{x}\cap E$ is a Hall $\sigma_{j}$-subgroup of $E$ by Lemma 2.1(4), so $|E:H_{j}^{x}\cap E|$ is a ${\sigma_{j}}^{\prime}$-number. But $|H_{j}^{x}|$ divides $|E|$ and hence $|H_{j}^{x}|$ divides $|H_{j}\cap E|$. Thererore $H_{j}^{x}\leq E$ for all $x\in G$. Thus $H_{j}^{G}\leq E$ and so $G/E_{G}$ is a $\sigma_{i}$-group. Hence $D\leq E_{G}\leq E$, so $|P|=p$. Therefore $G$ is supersoluble by [10, IV, 2.9] and so every chief factor of $G$ below $D$ is cyclic. Hence $D$ is nilpotent by Lemma 2.4, so $D$ is cyclic of square-free order. Hence $D$ is complemented in $G$ by Theorem 11.8 in [17].

Finally, assume that $|\sigma_{i}\cap\pi(G)|>1$ and let $q\in\sigma_{i}\cap\pi(G)\setminus\{p\}$. Then $G$ possesses a $\sigma$-permutable subgroup $F$ such that $|F|=|G|_{\sigma_{i}^{\prime}}q$. Then $D\leq F_{G}\leq F.$ Therefore $D\leq E\cap F$ and so $p$ does not divide $|D|$. This contradiction completes the proof of the implications (i) $\Rightarrow$ (ii) and (iii) $\Rightarrow$ (ii).

(ii) $\Rightarrow$ (iii) First we show that for every $i$ and for every subgroup (respectively normal subgroup) $A_{i}$ of $H_{i}$, there is an $H_{\sigma}$-permutably embedded (respectively $H_{\sigma}$-normally embedded) subgroup $E_{i}$ of $G$ such that $|E_{i}|=|A_{i}||G|_{\sigma_{i}^{\prime}}$. Since $G$ evidently is $\sigma$-soluble, it has a $\sigma_{i}$-complement $E$ by Lemma 2.6. Therefore, it is enough to consider the case when $A_{i}\neq 1$ since every $\sigma$-Hall subgroup of $G$ is an $H_{\sigma}$-normally embedded in $G$.

First suppose that $D\leq E$. Then $E/D$ is normal in $G$ since $G/D$ is $\sigma$-nilpotent. Therefore

[TABLE]

is $\sigma$-permutable (respectively normal) in

[TABLE]

Hence $E_{i}=EA_{i}$ is $\sigma$-permutable (respectively normal) in $G$ by Lemma 2.2(3) and $|E_{i}|=|A_{i}||G|_{\sigma_{i}^{\prime}}$.

Now suppose that $D\nleq E$. Then $D\cap H_{i}\neq 1$, so $H_{i}$ is a $p$-group for some prime $p$ since for each $\sigma_{i}\in\sigma(D)$ we have $|\sigma_{i}\cap\pi(G)|=1$ by hypothesis. Hence $H_{2}$ has a normal subgroup $A$ such that $D_{p}\leq A$ and $|A|=|A_{i}|$, where $D_{p}$ is a Sylow $p$-subgroup of $D$. Then $D\leq AE$. Moreover,

[TABLE]

since $ED/D$ is a Hall $\sigma_{i}^{\prime}$-subgroup of $G/D$. Therefore $E_{i}=AE$ is ${\sigma}$-permutable (respectively normal) in $G$ by Lemma 2.2(3) and $|E_{i}|=|A_{i}||G|_{\sigma_{i}^{\prime}}$.

Let $E=E_{1}\cap\cdots\cap E_{t}$. Then $|E|=|A_{1}|\ldots|A_{t}|$ since

[TABLE]

for all $i\neq j$. Note that $E_{i}$ is either a $\sigma$-Hall subgroup of $G$ or $\sigma$-permutable (respectively normal) in $G$. Indeed, let $V$ be a $\sigma$-permutable (respectively normal) subgroup of $G$ such that $E_{i}$ is a $\sigma$-Hall subgroup of $V$. Assume that $E_{i}$ is not $\sigma$-permutable (respectively not normal) in $G$. Then $E_{i}<V$. Since $|G:E_{i}|$ is ${\sigma_{i}}$-number, $|V:E_{i}|$ is a ${\sigma_{i}}$-number. But $E_{i}$ is a ${\sigma}$-Hall subgroup of $V$. Hence $E_{i}=V$ is a ${\sigma}$-Hall subgroup of $G$.

Assume that $E_{1},\ldots,E_{r}$ are $\sigma$-permutable (respectively normal) in $G$ and $E_{i}$ is a $\sigma$-Hall subgroup of $G$ for all $i>r$. Then $E^{0}=E_{1}\cap\cdots\cap E_{r}$ is $\sigma$-permutable (respectively normal) in $G$ by Lemma 2.2(5) and $E^{*}=E_{r+1}\cap\cdots\cap E_{t}$ is a $\sigma$-Hall subgroup of $G$. Now, $E=E^{0}\cap E^{*}$ is a $\sigma$-Hall subgroup of $E^{0}$ by Lemmas 2.1(4) and 2.2(4), so $E$ is $H_{\sigma}$-permutably (respectively $H_{\sigma}$-normally) embedded in $G$. Hence (ii) $\Rightarrow$ (iii).

(ii) $\Rightarrow$ (i) Since $G$ is $\sigma$-soluble, $H$ is $\sigma$-soluble. Hence $H$ has a $\sigma$-basis $\{L_{1},\ldots,L_{r}\}$ such that $L_{i}\leq H_{i}$ for all $i=1,\ldots,r$ by Lemma 2.6. Therefore from the implication (ii) $\Rightarrow$ (iii) we get that $G$ has an $H_{\sigma}$-permutably embedded subgroup of order $|L_{1}|\cdots|L_{r}|=|H|$.

The theorem is proved.

Proof of Theorem 1.4. (i) $\Rightarrow$ (ii) Assume that this is false and let $G$ be a counterexample of minimal order. Then some $\sigma$-subnormal subgroup $V$ of $G$ is not an $H\sigma E$-group. Moreover, $D=G^{{\mathfrak{N}}_{\sigma}}\neq 1$, so $|\sigma(G)|>1$.

(1) Condition (ii) is true on every proper section $H/K$ of $G$, that is, $K\neq 1$ or $H\neq G$. Hence $V=G$ (This directly follows from Lemma 2.7(1)(2) and the choice of $G$).

(2) $G$ is ${\sigma}$-soluble.

In view of Claim (1) and Lemma 2.5, it is enough to show that $G$ is not simple. Assume that this is false. Then 1 is the only proper $\sigma$-subnormal subgroup of $G$ since $|\sigma(G)|>1$. Hence every subgroup of $G$ is a $\sigma$-Hall subgroup of $G$. Therefore for a Sylow $p$-subgroup $P$ of $G$, where $p$ is the smallest prime divisor of $|G|$, we have $|P|=p$ and so $|G|=p$ by [10, IV, 2.8]. This contradiction shows that we have (2).

(3) If $|G:H|$ is a $\sigma_{i}$-number and $H$ is not a $\sigma$-Hall subgroup of $G$, then $H$ is $\sigma$-subnormal in $G$ and a $\sigma_{i}$-complement $E$ of $H$ is normal in $G$ (This follows from Lemmas 2.7(4) and 2.8).

(4) $D$ is a Hall subgroup of $G$. Hence $D$ has a complement $M$ in $G$.

Suppose that this is false and let $P$ be a Sylow $p$-subgroup of $D$ such that $1<P<G_{p}$, where $G_{p}\in\text{Syl}_{p}(G)$. We can assume without loss of generality that $G_{p}\leq H_{1}$. Let $R$ be a minimal normal subgroup of $G$ contained in $D$.

Since $D$ is soluble by Claim (2), $R$ is a $q$-group for some prime $q$. Moreover, $D/R=(G/R)^{\mathfrak{N}_{\sigma}}$ is a Hall subgroup of $G/R$ by Claim (1) and Proposition 2.2.8 in [16]. Suppose that $PR/R\neq 1$. Then $PR/R\in\text{Syl}_{p}(G/R)$. If $q\neq p$, then $P\in\text{Syl}_{p}(G)$. This contradicts the fact that $P<G_{p}$. Hence $q=p$, so $R\leq P$ and therefiore $P/R\in\text{Syl}_{p}(G/R)$. It follows that $P\in\text{Syl}_{p}(G)$. This contradiction shows that $PR/R=1$, which implies that $R=P$ is a Sylow $p$-subgroup of $D$. Therefore $R$ is a unique minimal normal subgroup of $G$ contained in $D$. It is also clear that a $p$-complement of $D$ is a Hall subgroup of $G$.

Now we show that $R\nleq\Phi(G)$. Indeed, assume that $R\leq\Phi(G)$. Then $D\neq R$ by Lemma 2.3 since $D=G^{{\mathfrak{N}}_{\sigma}}$. On the other hand, $D/R$ is a $p^{\prime}$-group. Hence $O_{p^{\prime}}(D)\neq 1$ by Lemma 2.3. But $O_{p^{\prime}}(D)$ is characteristic in $D$ and so it is normal $G$. Therefore $G$ has a minimal normal subgroup $L$ such that $L\neq R$ and $L\leq D$. This contradiction shows that $R\nleq\Phi(G)$.

Let $S$ be a maximal subgroup of $G$ such that $RS=G$. Then $|G:S|$ is a $p$-number. Hence, since $R$ is not a Sylow $p$-subgroup of $G$, $p$ divides $|S|$. Then $S$ is not a Hall subgroup of $G$ and so $S$ is not a $\sigma$-Hall subgroup of $G$. Therefore $S$ is $\sigma$-subnormal in $G$ by Claim (3) and so $G/S_{G}$ is a $\sigma_{i}$-group, which implies that

[TABLE]

and so $G=RS=S$. This contradiction completes the proof of (4).

(5) If $M\leq E<G$, then $E$ is not $\sigma$-permutable in $G$ and so $E$ a $\sigma$-Hall subgroup of $G$.

Assume that $E$ is $\sigma$-permutable in $G$. Then $E$ is $\sigma$-subnormal in $G$ by Lemma 2.2(4). Then there is a subgroup chain

[TABLE]

such that either $E_{i-1}$ is normal in $E_{i}$ or $E_{i}/(E_{i-1})_{E_{i}}$ is $\sigma$-primary for all $i=1,\ldots,r$. Let $V=E_{r-1}$. We can assume without loss of generality that $V\neq G$. Therefore, since $G$ is $\sigma$-soluble by Claim (2), for some $\sigma$-primary chief factor $G/W$ of $G$ we have $E\leq V\leq W$. Also we have $D\leq W$ and so $G=DE\leq W$, a contradiction. Hence $E$ is not $\sigma$-permutable in $G$.

By hypothesis, $G$ has a $\sigma$-permutable subgroup $S$ such that $E$ is a $\sigma$-Hall subgroup of $S$. But then $S=G$, by the above argument, so $E$ is a $\sigma$-Hall subgroup of $G$. In particular, $M$ is a $\sigma$-Hall subgroup of $G$ and so $D$ is a $\sigma$-Hall subgroup of $G$.

(6) $D$ is soluble, $|\sigma(D)|=|\pi(D)|$ and $M$ acts irreducibly on every $M$-invariant Sylow subgroup of $D$.

Let $p\in\sigma_{i}\in\sigma(D)$. Lemma 2.6 and Claims (2) and (4) imply that for some Sylow $p$-subgroup $P$ of $G$ we have $PM=MP$. Moreover, $MP$ is a $\sigma$-Hall subgroup of $G$ by Claim (5). Hence $|\sigma_{i}\cap\pi(G)|=1$ for all $i$ such that $\sigma_{i}\cap\pi(D)\neq\emptyset$ and so $|\sigma(D)|=|\pi(D)|$. Therefore, since $D$ is soluble by Claim (2), $M$ acts irreducibly on every $M$-invariant Sylow subgroup of $D$ by Claim (5).

(7) $M$ is a $\sigma$-Carter subgroup of $G$.

Let $R$ be a minimal normal subgroup of $G$ contained in $D$ and $E$ a subgroup of $G$ containing $M$. We need to show that $E=E^{{\mathfrak{N}}_{\sigma}}M$. Claim (1) implies that $RM/R$ is a $\sigma$-Carter subgroup of $G/R$, so

[TABLE]

Hence $ER=E^{{\mathfrak{N}}_{\sigma}}MR$ since

[TABLE]

Claim (6) implies that $R$ is a $p$-group for some prime $p$. Claims (4), (5) and (6) imply that $R$, $E$ and $E^{{\mathfrak{N}}_{\sigma}}M$ are $\sigma$-Hall subgroups of $G$. Therefore, if $R\nleq E$, then $E$ and $E^{{\mathfrak{N}}_{\sigma}}M$ are Hall $p^{\prime}$-subgroups of $ER=E^{{\mathfrak{N}}_{\sigma}}MR$, so $E=E^{{\mathfrak{N}}_{\sigma}}M$. Finally, assume that $R\leq E$ but $R\nleq E^{{\mathfrak{N}}_{\sigma}}M$. Then $R\cap E^{{\mathfrak{N}}_{\sigma}}=1$. On the other hand, since

[TABLE]

is $\sigma$-nilpotent, $E^{{\mathfrak{N}}_{\sigma}}\leq D$ and so $M\cap E^{{\mathfrak{N}}_{\sigma}}=1$. Therefore

[TABLE]

Then

[TABLE]

is $\sigma$-nilpotent. Hence $M\leq C_{G}(R)$. Suppose that $C_{G}(R)<G$ and let $C_{G}(R)\leq W<G$, where $G/W$ is a chief factor of $G$. Claim (2) implies that $G/W$ is $\sigma$-primary, so $D\leq W$. But then $G=DM\leq W<G$, a contradiction. Therefore $C_{G}(R)=G$, that is, $R\leq Z(G)$. Let $V$ be a complement to $R$ in $D$. Then $V$ is a Hall normal subgroup of $D$, so it is characteristic in $D$. Hence $V$ is normal in $G$ and $G/V\simeq RM$ is $\sigma$-nilpotent, so $D\leq V<D$. This contradiction completes the proof of (7).

(8) $D$ possesses a Sylow tower.

Let $R$ be a minimal normal subgroup of $G$ contained in $D$. Then $R$ is a $p$-group for some prime $p$ by Claim (6). Moreover, the Frattini argument implies that for some Sylow $p$-subgroup $P$ of $D$ we have $M\leq N_{G}(P)$ and so $R=P$ since $M$ acts irreducible on $P$ by Claim (6). On the other hand, by Claim (1), $D/R$ possesses a Sylow tower. Hence we have (8).

(9) Every chief factor of $G$ below $D$ is $\sigma$-eccentric.

Let $H/K$ be a chief factor of $G$ below $D$. Then $H/K$ is a $p$-group for some prime $p$ since $D$ is soluble by Claim (6). By the Fratiini argument, there exist a Sylow $p$-subgroup $P$ and a $p$-complement $E$ of $D$ such that $M\leq N_{G}(P)$ and $M\leq N_{G}(E)$. Then $M\leq N_{G}(P\cap K)$ and $M\leq N_{G}(P\cap H)$. Hence $P\cap K=1$ and $P\cap H=P$ by Claim (6), so $H=K\rtimes P$. Let $V=EM$. Then $K\leq V$ and $HV=G$, so $V$ is a maximal subgroup of $G$. Hence

[TABLE]

by Lemma 2.9. Therefore, if $H/K$ is $\sigma$-central in $G$, then $D\leq V_{G}$, which is impossible since evidently $p$ does not divide $|V|$. Thus we have (9).

From Claims (4)–(9) it follows that $G$ is a $H\sigma E$-group, contrary to our assumption on $G=V$. Hence (i) $\Rightarrow$ (ii).

(ii) $\Rightarrow$ (iii) This implication is evident.

(iii) $\Rightarrow$ (i) By hypothesis, $G=D\rtimes M$, where $D=G^{{\mathfrak{N}}_{\sigma}}$ is a $\sigma$-Hall subgroup of $G$, $|\sigma(D)|=|\pi(D)|$ and $M$ acts irreducibly on every $M$-invariant Sylow subgroup of $D$.

(*) Every subgroup $A$ of $G$ containing $M$ is a $\sigma$-Hall subgroup of $G$.

Let $D_{0}=D\cap A$. Then $A=D_{0}\rtimes M$ and $D_{0}\neq 1$. Let $p\in\pi(D_{0})$. The Frattini argument and Lemma 2.6 imply that for some Sylow $p$-subgroup $P_{0}$ of $D_{0}$ and some Sylow $p$-subgroup $P$ of $D$ we have $M\leq N_{G}(P_{0})$, $M\leq N_{G}(P)$ and $P_{0}M\leq PM$. Hence, since $M$ acts irreducibly on every $M$-invariant Sylow subgroup of $D$, $P_{0}=P$. Therefore every Sylow subgroup of $A$ is a Sylow subgroup of $G$. Hence $A$ is a $\sigma$-Hall subgroup of $G$ since $|\sigma(D)|=|\pi(D)|$ and $M$ is a $\sigma$-Hall subgroup of $G$.

Now, let $A$ be a subgroup of $G$. First assume that $DA<G$. By Lemma 2.1(6), $DA$ is $\sigma$-subnormal in $G$. Therefore every $\sigma$-subnormal subgroup of $DA$ is also $\sigma$-subnormal in $G$. Hence Condition (iii) holds for $DA$, so $A$ is $H_{\sigma}$-subnormally embedded in $DA$ by induction. But then $A$ is $H_{\sigma}$-subnormally embedded in $G$ by Lemma 2.1(7).

Finally, suppose that $DA=G$. Then, since $G$ is $\sigma$-soluble, for some $x$ we have $M\leq A^{x}$ by Lemma 2.6. Hence $A^{x}$ is a ${\sigma}$-Hall subgroup of $G$ by Claim (*), so $A^{x}$ is an $H_{\sigma}$-subnormally embedded subgroup of $G$. But then $A$ is an $H_{\sigma}$-subnormally embedded subgroup of $G$. Therefore the implication (iii) $\Rightarrow$ (i) is true.

The theorem is proved.

Proof of Theorem 1.9. (i) $\Rightarrow$ (ii) This follows from Lemma 2.2(4) and Theorems 1.3 and 1.4.

(ii) $\Rightarrow$ (iii) This implication is evident.

(iii) $\Rightarrow$ (i) Let $A$ be any subgroup of $G$. Then $DA$ is $\sigma$-permutable in $G$ by Lemma 2.2(3) since $G$ is $\sigma$-soluble. On the other hand, since $|\sigma(D)|=|\pi(D)|$ and $D$ is a cyclic $\sigma$-Hall subgroup of $G$ of square-free order, $A$ is a $\sigma$-Hall subgroup of $DA$. Hence $A$ is $H_{\sigma}$-permutably embedded in $G$. Therefore the implication (iii) $\Rightarrow$ (i) is true.

The theorem is proved.

**Proof of Theorem 1.7. ** (i) $\Rightarrow$ (ii) In view of Theorem 1.9, it is enough to show that if $D\leq L\leq G$ and $L$ is a $\sigma$-Hall subgroup of some normal subgroup $V$ of $G$, then $L$ is normal in $G$. But since $G/D$ is $\sigma$-nilpotent, $L/D$ is $\sigma$-subnormal in $G/D$, so $L$ is $\sigma$-subnormal in $G$ by Lemma 2.1(6). Hence $L$ is $\sigma$-subnormal in $V$ by Lemma 2.1(1). But then $L$ is a normal in $V$ by Lemma 2.1(4) and so $L$ is a characteristic subgroup of $V$. It follows that $L$ is normal in $G$.

(ii) $\Rightarrow$ (iii) This implication is evident.

(iii) $\Rightarrow$ (i) See the proof of the implication (iii) $\Rightarrow$ (i) in Theorem 1.9.

The theorem is proved.