Characterizations of some classes of finite $\sigma$-soluble $P\sigma T$-groups

TL;DR

This paper characterizes finite $\sigma$-soluble groups where $\sigma$-permutability is transitive, providing new insights into the structure of these groups based on their subgroup permutability properties.

Contribution

It introduces characterizations of finite $\sigma$-soluble groups with transitive $\sigma$-permutability, a novel property in the context of subgroup permutability.

Findings

Identifies conditions under which $\sigma$-permutability is transitive in finite $\sigma$-soluble groups.

Provides structural descriptions of groups with transitive $\sigma$-permutability.

Enhances understanding of subgroup permutability relations in finite group theory.

Abstract

Let $\sigma =\{\sigma_{i} | i\in I\}$ be some partition of the set of all primes $\Bbb{P}$ and $G$ a finite group. $G$ is said to be $\sigma$-soluble if every chief factor $H/K$ of $G$ is a $\sigma _{i}$-group for some $i=i(H/K)$. A set ${\cal H}$ of subgroups of $G$ is said to be a complete Hall $\sigma $-set of $G$ if every member $\ne 1$ of ${\cal H}$ is a Hall $\sigma _{i}$-subgroup of $G$ for some $\sigma _{i}\in \sigma $ and ${\cal H}$ contains exactly one Hall $\sigma _{i}$-subgroup of $G$ for every $i \in I$ such that $\sigma _{i}\cap \pi (G)\ne \emptyset$. A subgroup $A$ of $G$ is said to be ${\sigma}$-permutable in $G$ if $G$ has a complete Hall $\sigma$-set $\cal H$ such that $AH^{x}=H^{x}A$ for all $x\in G$ and all $H\in \cal H$. We obtain characterizations of finite $\sigma$-soluble groups $G$ in which $\sigma$-permutability is a transitive relation in $G$.

Equations

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

$$(G/N)/(DN/N)\simeq G/DN\simeq (G/D)/(DN/D),$$

$$(H/H_G)(E{H}_G/H_G)=(E{H}_G/H_G)(H/H_G)$$

$$DN/D\simeq N\simeq NL/L=G_p/L=DL/L\simeq D$$

$$H_i\simeq H_{i}N/N=(O_{{\sigma }_i}(D/N))\times (V/N)=(O_{{\sigma }_i}(D)N/N)\times (V/N).$$

$$G=H_i{O}^{{\sigma }_i}(G)=(O_{{\sigma }_i}(D)\times S)O^{{\sigma }_i}(G)=S{O}^{{\sigma }_i}(G)\le N_G(H)$$

$$(HN/N)(EN/N)=(EN/N)(HN/N)=EHN/N$$

$$EHR\cap EHN=E(HR\cap EHN)=EH(R\cap EHN)$$

$$=EH(R\cap H)=EH$$

$$H^G=H^{DM}=H^{O^{{\sigma }_i}(G)M}=H^M\le M$$

$$M=M\cap ⟨A,H_i^x⟩=⟨A,(M\cap H_i^x)⟩.$$

$$H_i^{x}A=H_i^x(M\cap H_i^x)A=H_i^{x}M=M{H}_i^x=H_i^x(M\cap H_i^x)A=H_i^{x}A.$$

$$⟨A,H\cap C⟩=A(H\cap C)=AH\cap C=⟨A,H⟩\cap C,$$

Full text

A set ${\cal H}$ of subgroups of $G$ is said to be a complete Hall $\sigma$-set of $G$ if every member $\neq 1$ of ${\cal H}$ is a Hall $\sigma_{i}$-subgroup of $G$ for some $\sigma_{i}\in\sigma$ and ${\cal H}$ contains exactly one Hall $\sigma_{i}$-subgroup of $G$ for every $i\in I$ such that $\sigma_{i}\cap\pi(G)\neq\emptyset$. A subgroup $A$ of $G$ is said to be ${\sigma}$-permutable in $G$ if $G$ has a complete Hall $\sigma$-set $\cal H$ such that $AH^{x}=H^{x}A$ for all $x\in G$ and all $H\in\cal H$.

We obtain characterizations of finite $\sigma$-soluble groups $G$ in which $\sigma$-permutability is a transitive relation in $G$.

00footnotetext: Keywords: finite group, ${\sigma}$-permutable subgroup, $P\sigma T$-group, $\sigma$-soluble group, $\sigma$-nilpotent group.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, $\mathbb{P}$ is the set of all primes, $\pi=\{p_{1},\ldots,p_{n}\}\subseteq\mathbb{P}$ and $\pi^{\prime}=\mathbb{P}\setminus\pi$. If $n$ is an integer, 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 $G$. $G$ is said to be a $D_{\pi}$-group if $G$ possesses a Hall $\pi$-subgroup $E$ and every $\pi$-subgroup of $G$ is contained in some conjugate of $E$.

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

By the analogy with the notation $\pi(n)$, we write $\sigma(n)$ to denote the set $\{\sigma_{i}|\sigma_{i}\cap\pi(n)\neq\emptyset\}$; $\sigma(G)=\sigma(|G|)$. $G$ is said to be: $\sigma$-primary [1] if $G$ is a $\sigma_{i}$-group for some $i$; $\sigma$-decomposable [2] or $\sigma$-nilpotent [3] if $G=G_{1}\times\dots\times G_{n}$ for some $\sigma$-primary groups $G_{1},\ldots,G_{n}$; $\sigma$-soluble [1] if every chief factor of $G$ is $\sigma$-primary; a $\sigma$-full group of Sylow type [1] if every subgroup $E$ of $G$ is a $D_{\sigma_{i}}$-group for every $\sigma_{i}\in\sigma(E)$. Note in passing, that every $\sigma$-soluble group is a $\sigma$-full group of Sylow type [4].

A set ${\cal H}$ of subgroups of $G$ is a complete Hall $\sigma$-set of $G$ [4, 5] if every member $\neq 1$ of ${\cal H}$ is a Hall $\sigma_{i}$-subgroup of $G$ for some $\sigma_{i}\in\sigma$ and ${\cal H}$ contains exactly one Hall $\sigma_{i}$-subgroup of $G$ for every $\sigma_{i}\in\sigma(G)$.

Recall also that a subgroup $A$ of $G$ is said to be ${\sigma}$-subnormal in $G$ [1] 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 $i=1,\ldots,n$.

Definition 1.1. We say that a subgroup $A$ of $G$ is said to be $\sigma$-quasinormal or $\sigma$-permutable in $G$ if $G$ possesses a complete Hall $\sigma$-set and $A$ permutes with each Hall $\sigma_{i}$-subgroup $H$ of $G$, that is, $AH=HA$ for all $i\in I$.

Remark 1.2. Using Theorem B in [1], it is not difficult to show that 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 $x\in G$, then $A$ is $\sigma$-permutable in $G$.

Remark 1.3. (i) In the classical case when $\sigma=\sigma^{0}=\{\{2\},\{3\},\ldots\}$: $G$ is $\sigma^{0}$-soluble (respectively $\sigma^{0}$-nilpotent) if and only if $G$ is soluble (respectively nilpotent); ${\sigma}^{0}$-permutable subgroups are also called $S$-quasinormal or $S$-permutable [6, 7]. A subgroup $A$ of $G$ is subnormal in $G$ if and only if it is $\sigma^{0}$-subnormal in $G$.

(ii) In the other classical case when $\sigma=\sigma^{\pi}=\{\pi,\pi^{\prime}\}$: $G$ is $\sigma^{\pi}$-soluble (respectively $\sigma^{\pi}$-nilpotent) if and only if $G$ is $\pi$-separable (respectively $\pi$-decomposable, that is, $G=O_{\pi}(G)\times O_{\pi^{\prime}}(G)$); a subgroup $A$ of a $\pi$-separable group $G$ is $\sigma^{\pi}$-permutable in $G$ if and only if $A$ permutes with all Hall $\pi$-subgroups and with all Hall $\pi^{\prime}$-subgroups of $G$. A subgroup $A$ of a $\pi$-separable group $G$ is $\sigma^{\pi}$-subnormal in $G$ if and only if there is a subgroup chain

[TABLE]

such that $A_{i}/(A_{i-1})_{A_{i}}$ is either a $\pi$-group or a $\pi^{\prime}$-group for all $i=1,\ldots,n$.

(iii) In fact, in the theory of $\pi$-soluble groups ($\pi=\{p_{1},\ldots,p_{n}\}$) we deal with the partition $\sigma=\sigma^{0\pi}=\{\{p_{1}\},\ldots,\{p_{n}\},\pi^{\prime}\}$ of $\mathbb{P}$. Note that $G$ is $\sigma^{0\pi}$-soluble (respectively $\sigma^{0\pi}$-nilpotent) if and only if $G$ is $\pi$-soluble (respectively $G=O_{p_{1}}(G)\times\cdots\times O_{p_{n}}(G)\times O_{\pi^{\prime}}(G)$). A subgroup $A$ of a $\pi$-soluble group $G$ is $\sigma^{0\pi}$-permutable in $G$ if and only if $A$ permutes with all Hall $\pi^{\prime}$-subgroups and with all Sylow $p$-subgroups of $G$ for all $p\in\pi$. Note also that a subgroup $A$ of $G$ is $\sigma^{0\pi}$-subnormal in $G$ if and only if it is $\mathfrak{F}$-subnormal in $G$ in the sense of Kegel [8], where $\mathfrak{F}$ is the class of all $\pi^{\prime}$-groups.

We say that $G$ is a $P\sigma T$-group [1] if ${\sigma}$-permutability is a transitive relation in $G$, that is, if $K$ is a ${\sigma}$-permutable subgroup of $H$ and $H$ is a ${\sigma}$-permutable subgroup of $G$, then $K$ is a ${\sigma}$-permutable subgroup of $G$. In the case when $\sigma=\sigma^{0}$, a $P\sigma T$-group is also called a $PST$-group [6]. Note that if $G=(Q_{8}\rtimes C_{3})\curlywedge(C_{7}\rtimes C_{3})$ (see [9, p. 50]), where $Q_{8}\rtimes C_{3}=SL(2,3)$ and $C_{7}\rtimes C_{3}$ is a non-abelian group of order 21, then $G$ is not a $PST$-group but $G$ is a $P\sigma T$-group, where $\sigma=\{\{2,3\},\{2,3\}^{\prime}\}$.

The description of $PST$-groups was first obtained by Agrawal [10], for the soluble case, and by Robinson in [11], for the general case. In the further publications, authors (see, for example, the recent papers [12]–[22] and Chapter 2 in [6]) have found out and described many other interesting characterizations of soluble $PST$-groups.

The purpose of this paper is to study $\sigma$-soluble $P\sigma T$-groups in the most general case (i.e., without any restrictions on $\sigma$). In view of Theorem B in [1], $G$ is a $P\sigma T$-group if and only if every $\sigma$-subnormal subgroup of $G$ is $\sigma$-permutable. Being based on this result, here we prove the following revised version of Theorem A in [1].

**Theorem A. ** Let $D=G^{\mathfrak{N_{\sigma}}}$. If $G$ is a $\sigma$-soluble $P\sigma T$-group, then the following conditions hold:

(i) $G=D\rtimes M$, where $D$ is an abelian Hall subgroup of $G$ of odd order, $M$ is $\sigma$-nilpotent and every element of $G$ induces a power automorphism in $D$;

(ii) $O_{\sigma_{i}}(D)$ has a normal complement in a Hall $\sigma_{i}$-subgroup of $G$ for all $i$.

Conversely, if Conditions (i) and (ii) hold for some subgroups $D$ and $M$ of $G$, then $G$ is a $P\sigma T$-group.

In this theorem, $G^{\mathfrak{N_{\sigma}}}$ denotes the $\sigma$-nilpotent residual of $G$, that is, the intersection of all normal subgroups $N$ of $G$ with $\sigma$-nilpotent quotient $G/N$.

Corollary 1.4. If $G$ is a $\sigma$-soluble $P\sigma T$-group, then every quotient and every subgroup of $G$ are $P\sigma T$-groups.

In the case when $\sigma=\sigma^{0}$, we get from Theorem A the following

Corollary 1.5 (Agrawal [10, Theorem 2.3]). Let $D=G^{\mathfrak{N}}$ be the nilpotent residual of $G$. If $G$ is a soluble $PST$-group, then $D$ is an abelian Hall subgroup of $G$ of odd order and every element of $G$ induces a power automorphism in $D$.

In the case when $\sigma=\sigma^{\pi}$, we get from Theorem A the following

Corollary 1.6. $G$ is a $\pi$-separable $P\sigma^{\pi}T$-group if and only if the following conditions hold:

(i) $G=D\rtimes M$, where $D$ is an abelian Hall subgroup of $G$ of odd order, $M$ is $\pi$-decomposable and every element of $G$ induces a power automorphism in $D$;

(ii) $O_{\pi}(D)$ has a normal complement in a Hall $\pi$-subgroup of $G$;

(iii) $O_{\pi^{\prime}}(D)$ has a normal complement in a Hall $\pi^{\prime}$-subgroup of $G$.

In the case when $\sigma=\sigma^{0\pi}$, we get from Theorem A the following

Corollary 1.7. $G$ is a $\pi$-soluble $P\sigma^{0\pi}T$-group if and only if the following conditions hold:

(i) $G=D\rtimes M$, where $D$ is an abelian Hall subgroup of $G$ of odd order, $M=O_{p_{1}}(M)\times\cdots\times O_{p_{n}}(M)\times O_{\pi^{\prime}}(M)$ and every element of $G$ induces a power automorphism in $D$;

(ii) $O_{\pi^{\prime}}(D)$ has a normal complement in a Hall $\pi^{\prime}$-subgroup of $G$.

A natural number $n$ is said to be a $\Pi$-number if $\sigma(n)\subseteq\Pi$. A subgroup $A$ of $G$ is said to be: a Hall $\Pi$-subgroup of $G$ [1] if $|A|$ is a $\Pi$-number and $|G:A|$ is a $\Pi^{\prime}$-number; a $\sigma$-Hall subgroup of $G$ if $A$ is a Hall $\Pi$-subgroup of $G$ for some $\Pi\subseteq\sigma$.

The proof of Theorem A is based on many results and observations of the paper [1]. We use also the remarkable result of the paper by Alejandre, Ballester-Bolinches and Pedraza-Aguilera [14] (see also Theorem 2.1.8 in [6]) that in a soluble $PST$-group $G$ any two isomorphic chief factors are $G$-isomorphic. Finally, in the proof of Theorem A, the following fact is useful, which is possibly independently interesting.

Theorem B. Let $G$ have a normal $\sigma$-Hall subgroup $D$ such that: (i) $G/D$ is a $P\sigma T$-group, and (ii) every $\sigma$-subnormal subgroup of $D$ is normal in $G$. If $G$ is a $\sigma$-full group of Sylow type, then $G$ is a $P\sigma T$-group.

Corollary 1.8 (See Theorem A in [1]). Let $G$ have a normal $\sigma$-Hall subgroup $D$ such that: (i) $G/D$ is $\sigma$-nilpotent, and (ii) every subgroup of $D$ is normal in $G$. Then $G$ is a $P\sigma T$-group.

In the case when $\sigma=\sigma^{0}$, we get from Theorem B the following

Corollary 1.9 (Agrawal [10, Theorem 2.4]). Let $G$ have a normal Hall subgroup $D$ such that: (i) $G/D$ is a $PST$-group, and (ii) every subnormal subgroup of $D$ is normal in $G$. Then $G$ is a $PST$-group.

Some other applications of Theorems A and B and some other characterizations of $\sigma$-soluble $P\sigma T$-groups we discuss in Section 4.

2 Some preliminary results

In view of Theorem B in [4], the following fact is true.

Lemma 2.1. If $G$ is $\sigma$-soluble, then $G$ is a $\sigma$-full group of Sylow type.

**Lemma 2.2 ** (See Corollary 2.4 and Lemma 2.5 in [1]). The class of all $\sigma$-nilpotent groups ${\mathfrak{N}}_{\sigma}$ is closed under taking products of normal subgroups, homomorphic images and subgroups. Moreover, if $E$ is a normal subgroup of $G$ and $E/E\cap\Phi(G)$ is $\sigma$-nilpotent, then $E$ is $\sigma$-nilpotent.

Lemma 2.3 (See Proposition 2.2.8 in [23]). If $N$ is a normal subgroup of $G$, then $(G/N)^{{\mathfrak{N}}_{\sigma}}=G^{{\mathfrak{N}}_{\sigma}}N/N.$

Lemma 2.4 (See Knyagina and Monakhov [24]). Let $H$, $K$ and $N$ be pairwise permutable subgroups of $G$ and $H$ be a Hall subgroup of $G$. Then $N\cap HK=(N\cap H)(N\cap K).$

Lemma 2.5 (See Lemma 2.8 in [1]). Let $A$, $K$ and $N$ be subgroups of $G$. Suppose that $A$ is $\sigma$-subnormal in $G$ and $N$ is normal in $G$.

(1) If $N\leq K$ and $K/N$ is $\sigma$-subnormal in $G/N$, then $K$ is $\sigma$-subnormal in $G$.

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

(3) If $A$ is a $\sigma$-Hall subgroup of $G$, then $A$ is normal in $G$.

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

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

Lemma 2.6 (See Lemmas 2.8, 3.1 and Theorem B in [1]). Let $H$, $K$ and $R$ be subgroups of $G$. Suppose that $H$ is $\sigma$-permutable in $G$ and $R$ is normal in $G$. Then:

(1) $H$ is $\sigma$-subnormal in $G$.

(2) The subgroup $HR/R$ is $\sigma$-permutable in $G/R$.

(3) If $K$ is a $\sigma_{i}$-group, then $K$ is $\sigma$-permutable in $G$ if and only if $O^{\sigma_{i}}(G)\leq N_{G}(K)$.

(4) If $G$ is a $\sigma$-full group of Sylow type and $H\leq K$, then $H$ is $\sigma$-permutable in $K$.

(5) If $G$ is a $\sigma$-full group of Sylow type, $R\leq K$ and $K/R$ is $\sigma$-permutable in $G/R$, then $K$ is $\sigma$-permutable in $G$.

(6) $H/H_{G}$ is $\sigma$-nilpotent.

Lemma 2.7. The following statements hold:

(i) $G$ is a $P\sigma T$-group if and only if every $\sigma$-subnormal subgroup of $G$ is $\sigma$-permutable in $G$.

(ii) If $G$ is a $P\sigma T$-group, then every quotient $G/N$ of $G$ is also a $P\sigma T$-group.

Proof. (i) This follows from Lemmas 2.5(6) and 2.6(1).

(ii) Let $H/N$ be a $\sigma$-subnormal subgroup of $G/N$. Then $H$ is a $\sigma$-subnormal subgroup of $G$ by Lemma 2.5(1), so $H$ is $\sigma$-permutable in $G$ by hypothesis and Part (i). Hence $H/N$ is $\sigma$-permutable in $G/N$ by Lemma 2.6(2). Hence $G/N$ is a $P\sigma T$-group by Part (i).

The lemma is proved.

3 Proofs of Theorems A and B

Proof of Theorem B. Since $G$ is a $\sigma$-full group of Sylow type by hypothesis, it possesses a complete Hall $\sigma$-set ${\cal H}=\{H_{1},\ldots,H_{t}\}$ and a subgroup $H$ of $G$ is $\sigma$-permutable in $G$ if and only if $HH_{i}^{x}=H_{i}^{x}H$ for all $H_{i}\in{\cal H}$ and $x\in G$. We can assume without loss of generality that $H_{i}$ is a $\sigma_{i}$-group for all $i=1,\ldots,t$.

Assume that this theorem is false and let $G$ be a counterexample of minimal order. Then $D\neq 1$ and for some $\sigma$-subnormal subgroup $H$ of $G$ and for some $x\in G$ and $k\in I$ we have $HH_{k}^{x}\neq H_{k}^{x}H$ by Lemma 2.7(i). Let $E=H_{k}^{x}$.

(1) The hypothesis holds for every quotient $G/N$ of $G$.

It is clear that $G/N$ is a $\sigma$-full group of Sylow type and $DN/N$ is a normal $\sigma$-Hall subgroup of $G/N$. On the other hand,

[TABLE]

so $(G/N)/(DN/N)$ is a $P\sigma T$-group by Lemma 2.7(ii). Finally, let $H/N$ be a $\sigma$-subnormal subgroup of $DN/N$. Then $H=N(H\cap D)$ and, by Lemma 2.5(1), $H$ is $\sigma$-subnormal in $G$. Hence $H\cap D$ is $\sigma$-subnormal in $D$ by Lemma 2.5(2), so $H\cap D$ is normal in $G$ by hypothesis. Thus $H/N=N(H\cap D)/N$ is normal in $G/N$. Therefore the hypothesis holds for $G/N$.

(2) $H_{G}=1$.

Assume that $H_{G}\neq 1$. Clearly, $H/H_{G}$ is $\sigma$-subnormal in $G/H_{G}$. Claim (1) implies that the hypothesis holds for $G/H_{G}$, so the choice of $G$ implies that $G/H_{G}$ is a $P\sigma T$-group. Hence

[TABLE]

by Lemma 2.7(i). Therefore $HE=EH$, a contradiction. Hence $H_{G}=1$.

(3) $DH=D\times H$.

By Lemma 2.5(2), $H\cap D$ is $\sigma$-subnormal in $D$. Hence $H\cap D$ is normal in $G$ by hypothesis, which implies that $H\cap D=1$ by Claim (2). Lemma 2.5(2) implies also that $H$ is $\sigma$-subnormal in $DH$. But $H$ is a $\sigma$-Hall subgroup of $DH$ since $D$ is a $\sigma$-Hall subgroup of $G$ and $H\cap D=1$. Therefore $H$ is normal in $DH$ by Lemma 2.5(3), so $DH=D\times H$.

Final contradiction. Since $D$ is a $\sigma$-Hall subgroup of $G$, then either $E\leq D$ or $E\cap D=1$. But the former case is impossible by Claim (3) since $HE\neq EH$, so $E\cap D=1$. Therefore $E$ is a $\Pi^{\prime}$-subgroup of $G$, where $\Pi=\sigma(D)$. By the Schur-Zassenhaus theorem, $D$ has a complement $M$ in $G$. Then $M$ is a Hall $\Pi^{\prime}$-subgroup of $G$ and so for some $x\in G$ we have $E\leq M^{x}$ since $G$ is a $\sigma$-full group of Sylow type. On the other hand, $H\cap M^{x}$ is a Hall $\Pi^{\prime}$-subgroup of $H$ by Lemma 2.5(4) and hence $H\cap M^{x}=H\leq M^{x}$. Lemma 2.5(2) implies that $H$ is $\sigma$-subnormal in $M^{x}$. But $M^{x}\simeq G/D$ is a $P\sigma T$-group by hypothesis, so $HE=EH$ by Lemma 2.7(i). This contradiction completes the proof of the theorem.

Sketch of the proof of Theorem A. Since $G$ is $\sigma$-soluble by hypothesis, $G$ is a $\sigma$-full group of Sylow type by Lemma 2.1. Let ${\cal H}=\{H_{1},\ldots,H_{n}\}$ be a complete Hall $\sigma$-set of $G$. We can assume without loss of generality that $H_{i}$ is a $\sigma_{i}$-group for all $i=1,\ldots,n$.

First suppose that $G$ is a $P\sigma T$-group. We show that Conditions (i) and (ii) hold for $G$ in this case. Assume that this is false and let $G$ be a counterexample of minimal order. Then $D\neq 1.$

(1) If $R$ is a non-identity normal subgroup of $G$, then Conditions (i) and (ii) hold for $G/R$ (Since the hypothesis holds for $G/R$ by Lemma 2.7(ii), this follows from the choice of $G$).

(2) If $E$ is a proper $\sigma$-subnormal subgroup of $G$, then $E^{\mathfrak{N_{\sigma}}}\leq D$ and Conditions (i) and (ii) hold for $E$.

Every $\sigma$-subnormal subgroup $H$ of $E$ is $\sigma$-subnormal in $G$ by Lemma 2.5(6), so $H$ is $\sigma$-permutable in $G$ by Lemma 2.7(i). Thus $H$ is $\sigma$-permutable in $E$ by Lemma 2.6(4). Therefore $E$ is a $\sigma$-soluble $P\sigma T$-group by Lemma 2.7(i), so Conditions (i) and (ii) hold for $E$ by the choice of $G$. Moreover, since $G/D\in{\mathfrak{N_{\sigma}}}$ and ${\mathfrak{N_{\sigma}}}$ is a hereditary class by Lemma 2.2, $E/E\cap D\simeq ED/D\in{\mathfrak{N_{\sigma}}}$ and so $E^{\mathfrak{N_{\sigma}}}\leq E\cap D\leq D$.

(3) $D$ is nilpotent.

(4) $D$ is a Hall subgroup of $G$. Hence $D$ has a $\sigma$-nilpotent 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 $G_{p}\leq H_{1}$.

($a^{0}$) $D=P$ is a minimal normal subgroup of $G$.

Let $R$ be a minimal normal subgroup of $G$ contained in $D$. Since $D$ is nilpotent by Claim (3), $R$ 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 Lemma 2.3. 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 therefore $P/R\in\text{Syl}_{p}(G/R)$ and we again get that $P\in\text{Syl}_{p}(G)$. This contradiction shows that $PR/R=1$, which implies that $R=P$ is the unique minimal normal subgroup of $G$ contained in $D$. Since $D$ is nilpotent, a $p$-complement $E$ of $D$ is characteristic in $D$ and so it is normal in $G$. Hence $E=1$, which implies that $R=D=P$.

($b^{0}$) $D\nleq\Phi(G)$. Hence for some maximal subgroup $M$ of $G$ we have $G=D\rtimes M$ (This follows from ($a^{0}$) and Lemma 2.2 since $G$ is not $\sigma$-nilpotent).

($c^{0}$) If $G$ has a minimal normal subgroup $L\neq D$, then $G_{p}=D\times(L\cap G_{p})$. Hence $O_{p^{\prime}}(G)=1$.

Indeed, $DL/L\simeq D$ is a Hall subgroup of $G/L$ by Claim (1) and lemma 2.3. Hence $G_{p}L/L=DL/L$, so $G_{p}=D\times(L\cap G_{p})$. Thus $O_{p^{\prime}}(G)=1$ since $D<G_{p}$ by Claim ($a^{0}$).

($d^{0}$) $V=C_{G}(D)\cap M$ is a normal subgroup of $G$ and $C_{G}(D)=D\times V\leq H_{1}$.

In view of Claims ($a^{0}$) and ($b^{0}$), $C_{G}(D)=D\times V$, where $V=C_{G}(D)\cap M$ is a normal subgroup of $G$. Moreover, $V\simeq DV/D$ is $\sigma$-nilpotent by Lemma 2.2. Let $W$ be a $\sigma_{1}$-complement of $V$. Then $W$ is characteristic in $V$ and so it is normal in $G$. Therefore we have ($d^{0}$) by Claim ($c^{0}$).

($e^{0}$) $G_{p}\neq H_{1}$.

Assume that $G_{p}=H_{1}$. Let $Z$ be a subgroup of order $p$ in $Z(G_{p})\cap D$. Then, since $O^{\sigma_{1}}(G)=O^{p}(G)$, $Z$ is normal in $G$ by Lemmas 2.6(3) and 2.7(i). Hence $D=Z<G_{p}$ by Claim ($a^{0}$) and so $D<C_{G}(D)$. Then $V=C_{G}(D)\cap M\neq 1$ is a normal subgroup of $G$ and $V\leq H_{1}=G_{p}$ by Claim ($d^{0}$). Let $L$ be a minimal normal subgroup of $G$ contained in $V$. Then $G_{p}=D\times L$ is a normal elementary abelian subgroup of $G$ by Claim ($c^{0}$). Therefore every subgroup of $G_{p}$ is normal in $G$ by Lemma 2.6(3). Hence $|D|=|L|=p$. Let $D=\langle a\rangle$, $L=\langle b\rangle$ and $N=\langle ab\rangle$. Then $N\nleq D$, so in view of the $G$-isomorphisms

[TABLE]

we get that $G/C_{G}(D)=G/C_{G}(N)$ is a $p$-group since $G/D$ is $\sigma$-nilpotent by Lemma 2.2. But then Claim ($d^{0}$) implies that $G$ is a $p$-group. This contradiction shows that we have ($e^{0}$).

Final contradiction for (4). In view of Theorem A in [4], $G$ has a $\sigma_{1}$-complement $E$ such that $EG_{p}=G_{p}E$. Let $V=(EG_{p})^{{\mathfrak{N}}_{\sigma}}$. By Claim ($e^{0}$), $EG_{p}\neq G$. On the other hand, since $D\leq EG_{p}$ by Claim ($a^{0}$), $EG_{p}$ is $\sigma$-subnormal in $G$ by Lemma 2.5(1). Therefore Claim (2) implies that $V$ is a Hall subgroup of $EG_{p}$ and $V\leq D$, so for a Sylow $p$-subgroup $V_{p}$ of $V$ we have $|V_{p}|\leq|P|<|G_{p}|$. Hence $V$ is a $p^{\prime}$-group and so $V\leq C_{G}(D)\leq H_{1}=G_{p}$ by Claim ($d^{0}$). Thus $V=1$. Therefore $EG_{p}=E\times G_{p}$ is $\sigma$-nilpotent and so $E\leq C_{G}(D)\leq H_{1}$. Hence $E=1$ and so $D=1$, a contradiction. Thus, $D$ is a Hall subgroup of $G$. Hence $D$ has a complement $M$ in $G$ by the Schur-Zassenhaus theorem and $M\simeq G/D$ is $\sigma$-nilpotent by Lemma 2.2.

(5) $H_{i}=O_{\sigma_{i}}(D)\times S$ for each $\sigma_{i}\in\sigma(D)$.

First assume that $N=O^{\sigma_{i}}(D)\neq 1$. Since $D$ is nilpotent by Claim (3), $N$ is a $\sigma_{i}^{\prime}$-group. Moreover, $G/N$ is a $P\sigma T$-group by Lemma 2.7(ii) and so the choice of $G$ implies that

[TABLE]

Since $D$ is a Hall subgroup of $H_{i}$ by Claim (4), $DN/N$ is a Hall subgroup of $H_{i}N/N$ and so $V/N$ is a Hall subgroup of $H_{i}N/N$. Hence $V/N$ is characteristic in $H_{i}N/N$. On the other hand, since $D/N=(G/N)^{{\mathfrak{N}}_{\sigma}}$ is $\sigma$-nilpotent by Lemma 2.2, $H_{i}N/N$ is normal in $G/N$ and so $V/N$ is normal in $G/N$. The subgroup $N$ has a complement $S$ in $V$ by the Schur-Zassenhaus theorem. Thus $H_{i}\cap V=H_{i}\cap NS=S(H_{i}\cap N)=S$ is normal in $H_{i}$.

Now assume that $O^{\sigma_{i}}(D)=1$, that is, $D$ is a $\sigma_{i}$-group. Then $H_{i}$ is normal in $G$, so all subgroups of $H_{i}$ are $\sigma$-permutable in $G$ by Lemmas 2.5(6), 2.7(i) and hypothesis. Since $D$ is a normal Hall subgroup of $H_{i}$, it has a complement $S$ in $H_{i}$. Lemma 2.6(3) implies that $D\leq O^{\sigma_{i}}(G)\leq N_{G}(S)$. Hence $H_{i}=D\times S$.

(6) Every subgroup $H$ of $D$ is normal in $G$. Hence every element of $G$ induces a power automorphism in $D$.

Since $D$ is nilpotent by Claim (3), it is enough to consider the case when $H\leq O_{\sigma_{i}}(D)=H_{i}\cap D$ for some $\sigma_{i}\in\sigma(D)$. Claim (5) implies that $H_{i}=O_{\sigma_{i}}(D)\times S$. It is clear that $H$ is subnormal in $G$, so $H$ is $\sigma$-permutable in $G$. Therefore

[TABLE]

by Lemma 2.6(3).

(7) If $p$ is a prime such that $(p-1,|G|)=1$, then $p$ does not divide $|D|$. Hence the smallest prime in $\pi(G)$ belongs to $\pi(|G:D|)$. In particular, $|D|$ is odd.

Assume that this is false. Then, by Claim (6), $D$ has a maximal subgroup $E$ such that $|D:E|=p$ and $E$ is normal in $G$. It follows that $C_{G}(D/E)=G$ since $(p-1,|G|)=1$. Hence $G/E=(D/E)\times(ME/E)$, where $ME/E\simeq M\simeq G/D$ is $\sigma$-nilpotent. Therefore $G/E$ is $\sigma$-nilpotent. But then $D\leq E$, a contradiction. Hence we have (7).

(8) $D$ is abelian.

In view of Claim (6), $D$ is a Dedekind group. Hence $D$ is abelian since $|D|$ is odd by Claim (7).

From Claims (4)–(8) we get that Conditions (i) and (ii) hold for $G$.

Now we show that if Conditions (i) and (ii) hold for $G$, then $G$ is a $P\sigma T$-group. Assume that this is false and let $G$ be a counterexample of minimal order. Then $D\neq 1$ and, by Lemma 2.7(i), for some $\sigma$-subnormal subgroup $H$ of $G$ and for some $x\in G$ and $k\in I$ we have $HH_{k}^{x}\neq H_{k}^{x}H$. Let $E=H_{k}^{x}$.

($1^{0}$) If $N$ is a minimal normal subgroup of $G$, then $G/N$ is a $P\sigma T$-group (Since the hypothesis holds for $G/N$, this follows from the choice of $G$).

$(2^{0})$ If $N$ is a minimal normal subgroup of $G$, then $EHN$ is a subgroup of $G$. Hence $E\cap N=1$.

Claim $(1^{0})$ implies that $G/N$ is a $P\sigma T$-group. On the other hand, $EN/N$ is a Hall $\sigma_{k}$-subgroup of $G/N$ and, by Lemma 2.5(5), $HN/N$ is a $\sigma$-subnormal subgroup of $G/N$. Note also that $G/N$ is $\sigma$-soluble, so every two Hall $\sigma_{k}$-subgroups of $G/N$ are conjugate by Lemma 2.1. Thus,

[TABLE]

by Lemma 2.7(i). Hence $EHN$ is a subgroup of $G$. Since $G$ is $\sigma$-soluble, $N$ is a $\sigma_{j}$-group for some $j$. Hence in the case $E\cap N\neq 1$ we have $j=k$, so $N\leq E$. But then $EHN=EH=HE$, a contradiction. Thus $E\cap N=1$.

$(3^{0})$ $|\sigma(D)|>1$.

Indeed, suppose that $\sigma(D)=\{\sigma_{i}\}$. Then $H_{i}/D$ is normal in $G/D$ since $G/D\simeq M$ is $\sigma$-nilpotent by hypothesis, so $H_{i}=D\times S$ is normal in $G$. The subgroup $S$ is also normal in $G$ since it is characteristic in $H_{i}$. On the other hand, Theorem B and the choice of $G$ imply that $S\neq 1$.

Let $R$ and $N$ be minimal normal subgroups of $G$ such that $R\leq D$ and $N\leq S$. Then $R$ is a group of order $p$ for some prime $p$ and $N$ is a $p^{\prime}$-group since $D$ is a Hall subgroup of $H_{i}$. Hence $R\cap HN\leq O_{p}(HN)\leq P$, where $P$ is a Sylow $p$-subgroup of $H$, so $R\cap HN=R\cap H$. Claim $(2^{0})$ implies that $EHR$ and $EHN$ are subgroups of $G$. Therefore from Lemma 2.4 and Claim ($2^{0}$) we get that $R\cap EHN=R\cap E(HN)=(R\cap E)(R\cap HN)=R\cap H$. Hence

[TABLE]

[TABLE]

is a subgroup of $G$. Thus $HE=EH$, a contradiction. Hence we have $(3^{0})$.

Final contradiction. Since $|\sigma(D)|>1$ by Claim ($3^{0}$) and $D$ is nilpotent, $G$ has at least two minimal normal subgroups $R$ and $N$ such that $R,N\leq D$ and $\sigma(R)\neq\sigma(N)$. Then at least one of the subgroups $R$ or $N$, $R$ say, is a $\sigma_{i}$-group for some $i\neq k$. Hence $R\cap HN\leq O_{\sigma_{i}}(HN)\leq V$, where $V$ is a Hall $\sigma_{i}$-subgroup of $H$, since $N$ is a $\sigma_{i}^{\prime}$-group and $G$ is a $\sigma$-full group of Sylow type. Hence $R\cap HN=R\cap H$. Claim ($2^{0}$) implies that $EHR$ and $EHN$ are subgroups of $G$. Now, arguing similarly as in the proof of $(3^{0})$, one can show that $EHR\cap EHN=EH=HE$. This contradiction completes the proof of the fact that $G$ is a $P\sigma T$-group.

The theorem is proved.

4 Some other characterizations of $\sigma$-soluble $P\sigma T$-groups

Theorem A and Theorem B in [1] are basic in the sense that many other characterizations of $\sigma$-soluble $P\sigma T$-groups can be obtained by using these two results. As a partial illustration to this, we give in this section our next three characterizations of $\sigma$-soluble $P\sigma T$-groups.

1. Recall that $Z_{\sigma}(G)$ denotes the *$\sigma$-hypercentre * of $G$ [26], that is, the largest normal subgroup of $G$ such that for every chief factor $H/K$ of $G$ below $Z_{\sigma}(G)$ the semidirect product $(H/K)\rtimes(G/C_{G}(H/K))$ is $\sigma$-primary.

We say, following [6, p. 20], that a subgroup $H$ of $G$ is $\sigma$-hypercentrally embedded in $G$, if $H/H_{G}\leq Z_{\sigma}(G/H_{G})$.

Theorem 4.1. Let $G$ be $\sigma$-soluble. Then $G$ is a $P\sigma T$-group if and only if every $\sigma$-subnormal subgroup of $G$ is $\sigma$-hypercentrally embedded in $G$.

**Proof. ** Let $D=G^{{\mathfrak{N}}_{\sigma}}$. First we show that if $G$ is a $P\sigma T$-group, then every $\sigma$-subnormal subgroup $H$ of $G$ is $\sigma$-hypercentrally embedded in $G$. Assume that this is false and let $G$ be a counterexample with $|G|+|H|$ minimal. Then $G/H_{G}$ is a $\sigma$-soluble $P\sigma T$-group by Lemma 2.7(ii) and $H/H_{G}$ is $\sigma$-subnormal in $G/H_{G}$ by Lemma 2.5(5). Hence the choice of $G$ implies that $H_{G}=1$, so $H$ is $\sigma$-nilpotent by Lemma 2.6(6). Therefore every subgroup of $H$ is $\sigma$-subnormal in $G$ by Proposition 2.3 in [1] and Lemma 2.5(6). Assume that $H$ possesses two distinct maximal subgroups $V$ and $W$. Then $V,W\leq Z_{\sigma}(G)$ by minimality of $|G|+|H|$ since $V_{G}=1=W_{G}$, which implies that $H\leq Z_{\sigma}(G)$. Hence $H$ is a cyclic $p$-group for some $p\in\sigma_{i}$.

By Theorem A, $G=D\rtimes M$, where $D$ is a Hall subgroup of $G$, $M$ is $\sigma$-nilpotent and every subgroup of $D$ is normal in $G$. Then $H\cap D=1$ and so, in view of Lemma 2.1, we can assume without loss of generality that $H\leq M$. Lemma 2.7(i) implies that $H$ is $\sigma$-permutable in $G$, so

[TABLE]

by Lemma 2.6(3). Hence $H^{G}\cap D=1$ and then, from the $G$-isomorphism $H^{G}D/D\simeq H^{G}$, we deduce that $H\leq H^{G}\leq Z_{\sigma}(G)$. Therefore $H$ is $\sigma$-hypercentrally embedded in $G$. This contradiction completes the proof of the necessity of the condition of the theorem.

Sufficiency. It is enough to show that if a $\sigma$-subnormal subgroup $H$ of a $\sigma$-soluble group $G$ is $\sigma$-hypercentrally embedded in $G$, then $H$ is $\sigma$-permutable in $G$. Assume that this is false and let $G$ be a counterexample with $|G|+|H|$ minimal. Since $G$ is $\sigma$-soluble, it is a $\sigma$-full group of Sylow type by Lemma 2.1. Therefore, in view of Lemma 2.6(5), $H_{G}=1$ and so $H\leq Z_{\sigma}(G)$. It is clear that $Z_{\sigma}(G)$ is $\sigma$-nilpotent, so $H=H_{1}\times\cdots\times H_{t}$ for some $\sigma$-primary groups $H_{1},\ldots,H_{t}$. Moreover, the minimality of $|G|+|H|$ implies that $H=H_{1}$ is a $\sigma_{i}$-group for some $i$. Hence $H\leq N$, where $N$ is a Hall $\sigma_{i}$-subgroup of $Z_{\sigma}(G)$. Since $Z_{\sigma}(G)$ is $\sigma$-nilpotent, $N$ is characteristic in $Z_{\sigma}(G)$ and so $N$ is normal in $G$.

Let $1=Z_{0}<Z_{1}<\cdots<Z_{t}=N$ be a chief series of $G$ below $N$ and $C_{i}=C_{G}(Z_{i}/Z_{i-1})$. Let $C=C_{1}\cap\cdots\cap C_{t}$. Then $G/C$ is a ${\sigma_{i}}$-group. On the other hand, $C/C_{G}(N)\simeq A\leq\text{Aut}(N)$ stabilizes the series $1=Z_{0}<Z_{1}<\cdots<Z_{t}=N$, so $C/C_{G}(N)$ is a $\pi(N)$-group by [25, Ch. A, 12.4(a)]. Hence $G/C_{G}(N)$ is a ${\sigma_{i}}$-group and so $O^{\sigma_{i}}(G)\leq C_{G}(N)$. But then $O^{\sigma_{i}}(G)\leq C_{G}(H)$, so $H$ is $\sigma$-permutable in $G$ by Lemma 2.6(3). This contradiction completes the proof of the sufficiency of the condition of the theorem.

The theorem is proved.

In the case when $\sigma=\sigma^{0}$, we have $Z_{\sigma}(G)=Z_{\infty}(G)$. Hence from Theorem 4.1 we get

Corollary 4.2 (See Theorem 2.4.4 in [6]). Let $G$ be soluble. Then $G$ is a $PST$-group if and only if every subnormal subgroup $H$ of $G$ is hypercentrally embedded in $G$ (that is, $H/H_{G}\leq Z_{\infty}(G/H_{G}))$.

2. We say, following [6, p. 68], that $G$ satisfies property ${\cal Y}_{\sigma_{i}}$ if whenever $H\leq K$ are two $\sigma_{i}$-subgroups of $G$, $H$ is $\sigma$-permutable in $N_{G}(K)$.

The idea of the next theorem goes back to the paper [15].

Theorem 4.3. Let $G$ be $\sigma$-soluble. Then $G$ is a $P\sigma T$-group if and only if $G$ satisfies ${\cal Y}_{\sigma_{i}}$ for all primes $i$.

Lemma 4.4. Let $K\leq H$ and $N$ be subgroups of $G$. Suppose that $K$ is $\sigma$-permutable in $H$ and $N$ is normal in $G$. Then $KN/N$ is $\sigma$-permutable in $HN/N$.

Proof. Let $f:H/H\cap N\to HN/N$ be the canonical isomorphism from $H/H\cap N$ onto $HN/N$. Then $f(K(H\cap N)/(H\cap N))=KN/N,$ so $KN/N$ is $\sigma$-permutable in $HN/N$ by Lemma 2.6(2).

The lemma is proved.

**Sketch of the proof of Theorem 4.3. ** Necessity. Let $H\leq K$ be two $\sigma_{i}$-subgroups of $G$ and $N=N_{G}(K)$. Then $H$ is $\sigma$-subnormal in $N$ by Lemma 2.5(6). On the other hand, Corollary 1.2 implies that $N$ is a $\sigma$-soluble $P\sigma T$-group. Therefore $H$ is $\sigma$-permutable in $N$ by Lemma 2.7(i).

Sufficiency. It is enough to show that Conditions (i) and (ii) of Theorem A hold for $G$. Assume that this is false and let $G$ be a counterexample of minimal order. Since $G$ is $\sigma$-soluble, it is a $\sigma$-full group of Sylow type by Lemma 2.1. Let $D=G^{{\mathfrak{N}}_{\sigma}}$.

(1) Every proper subgroup $E$ of $G$ is a $\sigma$-soluble $P\sigma T$-group and $E^{\mathfrak{N_{\sigma}}}\leq D$ (This follows from Lemmas 2.2, 2.3, 2.6(4) and the choice of $G$).

(2) $G/N$ is a $\sigma$-soluble $P\sigma T$-group for every minimal normal subgroup $N$ of $G$.

Let $H/N\leq K/N$ be two $\sigma_{i}$-subgroups of $G/N$. Since $G$ is $\sigma$-soluble, $N$ is a $\sigma_{j}$-subgroup for some $j$. Assume that $j\neq i$. Then there are a Hall $\sigma_{i}$-subgroup $V$ of $H$ and a Hall $\sigma_{i}$-subgroup $W$ of $K$ such that $V\leq W$ since $G$ is a $\sigma$-full group of Sylow type. Then $V$ is $\sigma$-permutable in $N_{G}(W)$ by hypothesis, so $H/N=VN/N$ is $\sigma$-permutable in $N_{G}(W)N/N=N_{G/N}(WN/N)=N_{G/N}(K/N)$ by Lemma 4.4. Similarly we get that $H/N$ is $\sigma$-permutable in $N_{G/N}(K/N)$ in the case when $j=i$.

(3) $D$ is nilpotent.

(4) $D$ is a Hall subgroup of $G$ and $H_{i}=O_{\sigma_{i}}(D)\times S$ for each $\sigma_{i}\in\sigma(D)$ (See Claims (4) and (5) in the proof of Theorem A and use Claims (1), (2) and (3)).

(5) Every subgroup $H$ of $D$ is normal in $G$. Hence every element of $G$ induces a power automorphism in $D$.

Since $D$ is nilpotent by Claim (3), it is enough to consider the case when $H\leq O_{\sigma_{i}}(D)=H_{i}\cap D$ for some $\sigma_{i}\in\sigma(D)$. Hence $H$ is $\sigma$-permutable in $G$ by hypothesis. Claim (4) implies that $H_{i}=O_{\sigma_{i}}(D)\times S$. Therefore $G=H_{i}O^{\sigma_{i}}(G)=SO^{\sigma_{i}}(G)\leq N_{G}(H)$ by Lemma 2.6(3).

(6) $D$ is abelian of odd order (See Claims (7) and (8) in the proof of Theorem A and use Claim (5)).

The theorem is proved.

Corollary 4.5 (Ballester-Bolinches and Esteban-Romero [15], see also Theorem 2.2.9 in [6]). $G$ is a soluble $PST$-group if and only if $G$ satisfies ${\cal Y}_{p}$ for all primes $p$.

Proof. It is enough to note that, as it was remarked at the beginning of the proof of Theorem 2.2.9 in [6], every group which satisfies ${\cal Y}_{p}$ for all primes $p$ is soluble.

3. We say that a subgroup $A$ of $G$ is $\sigma$-modular ($S$-modular in the case $\sigma=\sigma^{0}$) provided $G$ possesses a complete Hall $\sigma$-set and $\langle A,H\cap C\rangle=\langle A,H\rangle\cap C$ for every Hall $\sigma_{i}$-subgroup $H$ of $G$ and all $i\in I$ and $A\leq C\leq G$.

Theorem 4.6. Let $G$ be $\sigma$-soluble. Then $G$ is a $P\sigma T$-group if and only if every $\sigma$-subnormal subgroup $A$ of $G$ is $\sigma$-modular in every subgroup of $G$ containing $A$.

Proof. Since $G$ is $\sigma$-soluble, $G$ is a $\sigma$-full group of Sylow type by Lemma 2.1. Hence $G$ possesses a complete Hall $\sigma$-set ${\cal H}=\{H_{1},\ldots,H_{t}\}$ and, for each $\sigma_{i}\in\sigma(G)$, a subgroup $H$ of $G$ is a Hall $\sigma_{i}$-subgroup of $G$ if and only if $H=H_{k}^{x}$ for some $x\in G$ and $H_{k}\in{\cal H}$.

Sufficiency. Assume that this is false and let $G$ be a counterexample of minimal order. Then, in view of Lemma 2.7(i), $G$ has a $\sigma$-subnormal subgroup $A$ which is not $\sigma$-permutable in $G$. Hence, for some $H_{i}\in{\cal H}$ and $x\in G$, we have $AH_{i}^{x}\neq H_{i}^{x}A$. Note also that every proper $\sigma$-subnormal subgroup $E$ of $G$ is a $P\sigma T$-group. Indeed, $E$ is clearly $\sigma$-soluble and if $H$ is a $\sigma$-subnormal subgroup of $E$, then $H$ is $\sigma$-subnormal in $G$ by Lemma 2.5(6). Hence $H$ is $\sigma$-modular in every subgroup of $E$ containing $H$ by hypothesis. Thus the hypothesis holds for $E$ and so $E$ is a $P\sigma T$-group by the choice of $G$.

By definition, there is a subgroup chain $A=A_{0}\leq A_{1}\leq\cdots\leq A_{n}=G$ such that either $A_{i-1}\trianglelefteq A_{i}$ or $A_{i}/(A_{i-1})_{A_{i}}$ is ${\sigma}$-primary for all $i=1,\ldots,n$. We can assume without loss of generality that $M=A_{n-1}<G$. Then $M$ is a $P\sigma T$-group since $M$ is clearly $\sigma$-subnormal in $G$, so $A$ is $\sigma$-permutable in $M$ by Lemma 2.7(i). Moreover, the $\sigma$-modularity of $A$ in $G$ implies that

[TABLE]

On the other hand, by Lemma 2.5(4), $M\cap H_{i}^{x}$ is a Hall $\sigma_{i}$-subgroup of $M$, where $\{\sigma_{i}\}=\sigma(H_{i})$. Hence $M=A(M\cap H_{i}^{x})=(M\cap H_{i}^{x})A$. If $H_{i}^{x}\leq M_{G}$, then $A(M\cap H_{i}^{x})=AH_{i}^{x}=H_{i}^{x}A$ and so $H_{i}^{x}\nleq M_{G}$.

Now note that $H_{i}^{x}M=MH_{i}^{x}$. Indeed, if $M$ is normal in $G$, it is clear. Otherwise, $G/M_{G}$ is $\sigma$-primary and so $G=MH_{i}^{x}=H_{i}^{x}M$ since $H_{i}^{x}\nleq M_{G}$ and $H_{i}\in{\cal H}$. Therefore

[TABLE]

This contradiction completes the proof of the sufficiency of the condition of the theorem.

Necessity. In view of Lemma 2.6(4), it is enough to show that if $A$ is a $\sigma$-subnormal subgroup of $G$, then $A$ is $\sigma$-modular in $G$. First note that $A$ is $\sigma$-permutable in $G$ by Lemma 2.7(i). Therefore for every subgroup $C$ of $G$ containing $A$, for every $i\in I$, and for all Hall $\sigma_{i}$-subgroup $H$ of $G$ we have

[TABLE]

so $A$ is $\sigma$-modular in $G$.

The theorem is proved.

From Theorem 4.6 we get the following characterization of soluble $PST$-groups.

Corollary 4.7. Let $G$ be soluble. Then $G$ is a $PST$-group if and only if every subnormal subgroup $A$ of $G$ is $S$-modular in every subgroup of $G$ containing $A$.