A New Characterization of Finite $\sigma$-Soluble $P\sigma T$-Groups

Abstract

We prove that $G$ is a finite $\sigma$-soluble group with transitive $\sigma$-permutability if and only if the following hold: (i) $G$ possesses a complete Hall $\sigma$-set ${\mathcal{H}}=\{H_{1},\dots,H_{t}\}$ and a normal subgroup $N$ with $\sigma$-nilpotent quotient $G/N$ such that $H_{i}\cap N\leq Z_{\mathfrak{U}}(H_{i})$ for all $i$; and (ii) every $\sigma_{i}$-subgroup of $G$ is $\tau_{\sigma}$-permutable in $G$ for all $\sigma_{i}\in\sigma(N)$.

1. Introduction

Throughout this paper, all groups are finite and $G$ stands for a finite group. Moreover, $𝕇$ is the set of all primes, $\pi\subseteq 𝕇$ and $\pi^{\prime}=𝕇\setminus\pi$; and $\pi(G)$ is the set of all primes dividing $|G|$. Furthermore, $Z_{\mathfrak{U}_{\pi}}(G)$ is the $\pi$-supersoluble hypercenter of $G$, i.e., the product of all normal subgroups $N$ of $G$ such that every chief factor of $G$ below $N$ is either cyclic or a $\pi^{\prime}$-group, and $Z_{\mathfrak{U}}(G)=Z_{\mathfrak{U}_{𝕇}}(G)$ is the supersoluble hypercenter of $G$.

In what follows, $\sigma$ is some partition of $𝕇$, i.e., $\sigma=\{\sigma_{i}\mid i\in I\}$, where $𝕇=\bigcup_{i\in I}\sigma_{i}$ and $\sigma_{i}\cap\sigma_{j}=\varnothing$ for all $i\neq j$; $\sigma(G)=\{\sigma_{i}\mid\sigma_{i}\cap\pi(G)\neq\varnothing\}$ (see [1]).

A set ${\mathcal{H}}$ of subgroups of $G$ is said to be a complete Hall $\sigma$-set of $G$ (see [1]) if each nonidentity member of ${\mathcal{H}}$ is a Hall $\sigma_{i}$-subgroup of $G$ for some $i\in I$ and $\mathcal{H}$ has exactly one Hall $\sigma_{i}$-subgroup of $G$ for every $i$.

A subgroup $A$ of $G$ is said to be $\sigma$-permutable in $G$ (see [2]) if $G$ possesses a complete Hall $\sigma$-set and $A$ permutes with every Hall $\sigma_{i}$-subgroup $H$ of $G$, i.e., $AH=HA$ for all $i$ and $A$ is $\sigma$-semipermutable in $G$ [3] if $G$ possesses a complete Hall $\sigma$-set ${\mathcal{H}}$ such that $AH^{x}=H^{x}A$ for all $x\in G$ and all $H\in{\mathcal{H}}$ with $\sigma(A)\cap\sigma(H)=\varnothing$.

The theories of $\sigma$-permutable and $\sigma$-semipermutable subgroups are closely related to the theories of $\sigma$-soluble and $\sigma$-nilpotent groups [1,2,3,4,5].

Recall that $G$ is said to be $\sigma$-decomposable (see [6]) or $\sigma$-nilpotent (see [2]) if $G$ is $\sigma_{i}$-closed for all $i$; $\sigma$-soluble (see [2]) if every chief factor $H/K$ of $G$ is a $\sigma_{i}$-group for some $i$; and $G^{\mathfrak{N}_{\sigma}}$ is the $\sigma$-nilpotent residual of $G$, i.e., the smallest normal subgroup of $G$ with $\sigma$-nilpotent quotient.

Let $\tau_{\mathcal{H}}(A)=\{\sigma_{i}\in\sigma(G)\setminus\sigma(A)\mid\sigma(A)\cap\sigma(H^{G})\neq\varnothing$ for a Hall $\sigma_{i}$-subgroup $H\in{\mathcal{H}}$} (see [7]).

Then we say, following Beidleman and Skiba [7], that a subgroup $A$ of $G$ is as follows:

(i) $\tau_{\sigma}$-permutable in $G$ with respect to ${\mathcal{H}}$ if $AH^{x}=H^{x}A$ for all $x\in G$ and all $H\in\mathcal{H}$ such that $\sigma(H)\subseteq\tau_{\mathcal{H}}(A)$;

(ii) $\tau_{\sigma}$-permutable in $G$ if $A$ is $\tau_{\sigma}$-permutable in $G$ with respect to some complete Hall $\sigma$-set $\mathcal{H}$ of $G$.

In the classical case when $\sigma=\sigma{{}^{1}}=\{\{2\},\{3\},\dots\}$ (we use here the notations of [1]), the ${\sigma}$-permutable, $\sigma$-semipermutable, and $\tau_{\sigma}$-permutable subgroups are called respectively $S$-permutable (see [8]), $S$-semipermutable (see [9]), and $\tau$-permutable (see [10]).

Finally, recall that $G$ is said to be a $P\sigma T$-group (see [2]) if $\sigma$-permutability is a transitive relation in $G$; i.e., if $H$ is a $\sigma$-permutable subgroup of $K$ and $K$ is a $\sigma$-permutable subgroup of $G$, then $H$ is $\sigma$-permutable in $G$. In the case when $\sigma=\sigma^{1}$, a $P\sigma T$-group is called a $PST$-group [8].

The theory of $P\sigma T$-groups was developed in [1, 2, 5, 11], and the following theorem is one of the culmination results of the theory.

Theorem A (see Theorem A in [1])

Let $D=G^{\mathfrak{N_{\sigma}}}$. If $G$ is a $\sigma$-soluble $P\sigma T$-group, then the following 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 (i) and (ii) hold for some subgroups $D$ and $M$ of $G$, then $G$ is a $P\sigma T$-group.

In this paper, basing on Theorem A and some results of [7], we obtain the following characterization of $\sigma$-soluble $P\sigma T$-groups:

Theorem B

$G$ is a $\sigma$-soluble $P\sigma T$-group if and only if the following hold:

(i) $G$ possesses a complete Hall $\sigma$-set ${\mathcal{H}}=\{H_{1},\dots,H_{t}\}$ and a normal subgroup $N$ with $\sigma$-nilpotent quotient $G/N$ such that $H_{i}\cap N\leq Z_{\mathfrak{U}_{\pi}}(H_{i})$ for all $i$, where $\pi=\pi(N)$;

(ii) Every $\sigma_{i}$-subgroup of $G$ is $\tau_{\sigma}$-permutable in $G$ for all $\sigma_{i}\in\sigma(N)$.

Since every $\sigma$-semipermutable subgroup is $\tau_{\sigma}$-permutable, we get from Theorem B the following already-known result:

Corollary 1.1 (see Theorem A in [3])

Let $D=G^{{\mathfrak{N}}_{\sigma}}$ and $\pi=\pi(D)$. Suppose that $G$ possesses a complete Hall $\sigma$-set $\mathcal{H}$ all members of which are $\pi$-supersoluble. If every $\sigma_{i}$-subgroup of $G$ is $\sigma$-semipermutable in $G$ for all $\sigma_{i}\in\sigma(D)$, then $G$ is a $\sigma$-soluble $P\sigma T$-group.

Note that Theorem B remains new for each special partition $\sigma$ of $𝕇$. In particular, in the case when $\sigma=\sigma^{1}$ we get from Theorem B the following new characterization of the soluble $PST$-groups.

Corollary 1.2

Let $D=G^{\mathfrak{N}}$ be the nilpotent residual of $G$ and $\pi=\pi(D)$. Then $G$ is a soluble $PST$-group if and only if every $p$-subgroup of $G$ is $\tau$-permutable in $G$ for all $p\in\pi$.

The proof of Theorem B consists of many steps and the following theorem is one of them.

Theorem C

Let $D=G^{\mathfrak{N_{\sigma}}}$ and $\pi=\pi(D)$. Suppose that $G$ possesses a complete Hall $\sigma$-set ${\mathcal{H}}=\{H_{1},\dots,H_{t}\}$ such that $H_{i}\cap D\leq Z_{\mathfrak{U}_{\pi}}(H_{i})$ for all $i$. If all maximal subgroups of every noncyclic Sylow $p$-subgroup of $G$ are $\tau_{\sigma}$-permutable in $G$ for all $p\in\pi$, then

(i) $D$ is a nilpotent Hall subgroup of $G$, $D\leq Z_{\mathfrak{U}}(G)$;

(ii) $(p-1,|G|)\neq 1$ for every prime $p$ dividing $|D|$. Hence, $p\in\pi(G/D)$ for the smallest prime $p$ dividing $|G|$.

Corollary 1.3 (see Theorem 10.3 in [12, VI])

If every Sylow subgroup of $G$ is cyclic, then $G$ is supersoluble.

Corollary 1.4 (see Theorem B in [3])

Let $D=G^{{\mathfrak{N}}_{\sigma}}$ and $\pi=\pi(D)$. Suppose that $G$ possesses a complete Hall $\sigma$-set $\mathcal{H}$ such that every member $H$ of $\mathcal{H}$ with $H\cap D\neq 1$ is $\pi$-supersoluble. If all maximal subgroups of every noncyclic Sylow $p$-subgroup of $G$ are $\sigma$-semipermutable in $G$ for all $p\in\pi$, then $D$ is a nilpotent Hall subgroup of $G$ of odd order and every chief factor of $G$ below $D$ is cyclic.

The unexplained terminology and notation are standard. The reader is referred to [9, 12, 13] if need be.

2. Proof of Theorem C

We use $\mathfrak{N}_{\sigma}$ to denote the class of all $\sigma$-nilpotent groups.

Lemma 2.1 [2, Corollary 2.4 and Lemma 2.5]

The class ${\mathfrak{N}}_{\sigma}$ is closed under direct products, 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.

In view of Proposition 2.2.8 in [14], we get from Lemma 2.1 the following

Lemma 2.2

If $N$ is a normal subgroup of $G$, then $(G/N)^{{\mathfrak{N}}_{\sigma}}=G^{{\mathfrak{N}}_{\sigma}}N/N$.

Lemma 2.3 [15]

Let $H$, $K$, and $N$ be pairwise permutable subgroups of $G$ and let $H$ be a Hall subgroup of $G$. Then $N\cap HK=(N\cap H)(N\cap K)$.

Recall that $G$ is a $D_{\pi}$-group if $G$ possesses a Hall $\pi$-subgroup $E$ and every $\pi$-subgroup of $G$ lies in some conjugate of $E$; a $\sigma$-full group of Sylow type (see [16]), if every subgroup $E$ of $G$ is a $D_{\sigma_{i}}$-group for every $\sigma_{i}\in\sigma(E)$, and $\sigma$-full (see [16]), provided that $G$ possesses a complete Hall $\sigma$-set.

In view of Theorems A and B in [16], the following is true:

Lemma 2.4

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

Lemma 2.5 [2, Lemma 3.1]

Let $H$ be a $\sigma_{i}$-subgroup of a $\sigma$-full group $G$. Then $H$ is $\sigma$-permutable in $G$ if and only if $O^{\sigma_{i}}(G)\leq N_{G}(H)$.

Lemma 2.6 [7, Lemma 2.6]

Suppose that $G$ has a complete Hall $\sigma$-set ${\mathcal{H}}=\{H_{1},\dots,H_{t}\}$ such that the subgroups $H$ and $K$ of $G$ are $\tau_{\sigma}$-permutable in $G$ with respect to ${\mathcal{H}}$. Let $R$ be a normal subgroup of $G$ and $H\leq L\leq G$. Then

(1) ${\mathcal{H}}_{0}=\{H_{1}R/R,\dots,H_{t}R/R\}$ is a complete Hall $\sigma$-set of $G/R$. Moreover, if $\sigma(H)=\sigma(HR/R)$, then $HR/R$ is $\tau_{\sigma}$-permutable in $G/N$ with respect to ${\mathcal{H}}_{0}$.

(2) If $HK=KH$ and $\sigma(H\cap K)=\sigma(H)=\sigma(K)$, then $H\cap K$ is $\tau_{\sigma}$-permutable in $G$ with respect to ${\mathcal{H}}$.

(3) If $H\leq O_{\sigma_{i}}(G)$ for some $i$, then $H$ is ${\sigma}$-permutable in $G$.

(4) If $G$ is a $\sigma$-full group of Sylow type, then $H$ is $\tau_{\sigma}$-permutable in $L$.

Lemma 2.7

Let $Z=Z_{{\mathfrak{U}}_{\pi}}(G)$. Then

(1) each chief factor of $G$ below $Z$ is either cyclic or a $\pi^{\prime}$-group;

(2) $Z\cap E\leq Z_{{\mathfrak{U}}_{\pi}}(E)$ for every subgroup $E$ of $G$;

(3) $NZ/N\leq Z_{{\mathfrak{U}}_{\pi}}(G/N)$ for every normal subgroup $N$ of $G$.

Proof

(1): In fact, it suffices to prove that if $A$ and $B$ are normal subgroups of $G$ such that each chief factor of $G$ below $A$ is either cyclic or a $\pi^{\prime}$-group and each chief factor of $G$ below $B$ is either cyclic or a $\pi^{\prime}$-group, then each chief factor $H/K$ of $G$ below $AB$ is either cyclic or a $\pi^{\prime}$-group. Moreover, in view of the Jordan–Hölder Theorem for chief series, it suffices to show that if $A\leq K<H\leq AB$, then $H/K$ is either cyclic or a $\pi^{\prime}$-group. But this follows from $H=A(H\cap B)=K(H\cap B)$ and the $G$-isomorphism $K(H\cap B)/K\simeq(H\cap B)/(K\cap B)$. Therefore, each chief factor of $G$ below $Z$ is either cyclic or a $\pi^{\prime}$-group.

(2): Let $1=Z_{0}<Z_{1}<\cdots<Z_{t-1}<Z_{t}=Z$ be a chief series of $G$ below $Z$. Then each factor $Z_{i}/Z_{i-1}$ of the series is either cyclic or a $\pi^{\prime}$-group by (1).

Consider the normal series

$$ 1=Z_{0}\cap E\leq Z_{1}\cap E\leq\cdots\leq Z_{t-1}\cap E\leq Z_{t}\cap E=Z\cap E $$

in $E$. Assume that $(Z_{i}\cap E)/(Z_{i-1}\cap E)$ is not a $\pi^{\prime}$-group. Then, in view of the isomorphism,

$$ (Z_{i}\cap E)/(Z_{i-1}\cap E)\simeq(Z_{i}\cap E)Z_{i-1}/Z_{i-1}\leq Z_{i}/Z_{i-1} $$

we get that $Z_{i}/Z_{i-1}$ is cyclic, and so $(Z_{i}\cap E)/(Z_{i-1}\cap E)$ is cyclic. Therefore, in view of the Jordan–Hölder Theorem, each chief factor of $E$ below $Z\cap E$ is either cyclic or a $\pi^{\prime}$-group. Hence $Z\cap E\leq Z_{{\mathfrak{U}}_{\pi}}(E)$.

(3): Let $(H/N)/(K/N)$ be a chief factor of $G/N$ such that $H/N\leq NZ/N$. Then, in view of the isomorphism $(H\cap Z)K/K\simeq(H\cap Z)/(K\cap Z)$, we have that $H/K=(H\cap Z)K/K$ is a chief factor of $G$ such that $H/K$ is either cyclic or a $\pi^{\prime}$-group by (1). Hence $NZ/N\leq Z_{{\mathfrak{U}}_{\pi}}(G/N)$. The lemma is proved.

The following lemma is a corollary of Theorem 6.7 in [13, IV].

Lemma 2.8

Let $N\leq E$ be normal subgroups of $G$ such that $N\leq\Phi(E)$ and every chief factor of $G$ between $E$ and $N$ is cyclic. Then each chief factor of $G$ below $E$ is cyclic.

A group $G$ is said to be $\sigma$-primary (see [2]) if $G$ is a $\sigma_{i}$-group for some $i$.

Lemma 2.9

Let $D=G^{\mathfrak{N_{\sigma}}}$ and $p\in\pi=\pi(D)$, where $p$ is the smallest prime dividing $|D|$. If all maximal subgroups of every Sylow $p$-subgroup of $G$ are $\tau_{\sigma}$-permutable in $G$, then $D$ is $p$-soluble.

Proof

Suppose that this lemma is false and let $G$ be a counterexample of minimal order. Then $D\neq 1$. Assume that $p\in\sigma_{k}$.

We show first that $DR/R\simeq D/(D\cap R)$ is $p$-soluble for every minimal normal subgroup $R$ of $G$. Indeed, in case $p$ does not divide $|DR/R|$, it is clear. Suppose that $p\in\pi(DR/R)$. Then $p$ is the smallest prime dividing $|DR/R|$, where $DR/R=(G/R)^{{\mathfrak{N}}_{\sigma}}$ by Lemma 2.2.

Let $V/R$ be a maximal subgroup of a Sylow $p$-subgroup $P/R$ of $G/R$. Then $P/R=G_{p}R/R$ and $V=R(V\cap G_{p})$ for some Sylow $p$-subgroup $G_{p}$ of $G$. Hence

$$ p=|(P/R):(V/R)|=|G_{p}R:R(V\cap G_{p})|=|G_{p}|:|V\cap G_{p}|=|G_{p}:(V\cap G_{p})|, $$

and so $V\cap G_{p}$ is a maximal subgroup of $G_{p}$. Therefore, $V\cap G_{p}$ is $\tau_{\sigma}$-permutable in $G$ by hypothesis, and so $V/R=R(V\cap G_{p})/R$ is $\tau_{\sigma}$-permutable in $G/R$ by Lemma 2.6(1). The choice of $G$ implies that $(G/R)^{{\mathfrak{N}}_{\sigma}}=DR/R\simeq D/(D\cap R)$ is $p$-soluble.

Hence $R\leq D$ and $R$ is nonabelian. It is easy to see that $R$ is the unique minimal normal subgroup of $G$ and $C_{G}(R)=1$. By [12, IV, Theorem 2.8], a Sylow $p$-subgroup $Q$ of $R$ is not cyclic. Hence $|Q|>p$.

Let $P$ be a Sylow $p$-subgroup of $G$ such that $Q=P\cap R$. Then by the Tate Theorem [12, IV, Theorem 4.7] there exists some maximal subgroup $V$ of $P$ such that $Q\nleq V$, which implies that $P=QV$ and so $V\cap R<P\cap R=Q$. If $V\cap R=1$, then $V\cap R=P\cap V\cap R=Q\cap V=1$ and so $|Q|=p$; a contradiction. Hence $V\cap R\neq 1$. Since $R=R_{1}\times\cdots\times R_{n}$, where $R_{1}\simeq\cdots\simeq R_{n}$ are nonabelian simple groups, $Q=(P\cap R_{1})\times\cdots\times(P\cap R_{n})$ and so $V\cap R_{i}<P\cap R_{i}$ for some $i$. Note also that $V\cap R_{i}\neq 1$. Otherwise from the isomorphism

$$ V(P\cap R_{i})/V\simeq(P\cap R_{i})/(V\cap(P\cap R_{i}))=(P\cap R_{i})/1 $$

we get that the order of a Sylow $p$-subgroup of $P\cap R_{i}$ divides $p$ and so $P\cap R_{i}$ is $p$-nilpotent by [12, IV, Theorem 2.8], which implies that $R$ is $p$-nilpotent.

We show first that $R$ is $\sigma$-primary. Suppose the contrary. We can assume without loss of generality that $V$ is $\tau_{\sigma}$-permutable in $G$ with respect to ${\mathcal{H}}$. Then there exists some $j\neq k$, and for $H=H_{j}$ we have $H\cap R_{i}\neq 1$ because $R$ is not $\sigma$-primary. Note also that $\sigma_{k}\in\sigma(H^{G})$. If not, then $R\cap H^{G}=1$, which implies that $1<H^{G}\leq C_{G}(R)=1$. Therefore $\sigma_{k}\in\tau_{\mathcal{H}}(V)$, and so $VH^{x}=H^{x}V$ for all $x\in G$. By [13, Chapter A, Lemma 14.1(a)], $L=VH^{x}\cap R_{i}$ is a subnormal subgroup of $VH^{x}$, where $V$ is a Hall $\sigma_{k}$-subgroup of $VH^{x}$ and $H^{x}$ is a Hall $\sigma_{j}$-subgroup of $VH^{x}$. Therefore, $L=(L\cap V)(L\cap H^{x})$ by [13, I, Lemma 3.2]. Hence,

$$ \begin{gathered}\displaystyle L=(L\cap V)(L\cap H^{x})=(VH^{x}\cap R_{i}\cap V)(VH^{x}\cap R_{i}\cap H^{x})\\ \displaystyle=(R_{i}\cap V)(R_{i}\cap H^{x})=(V\cap R_{i})(H\cap R_{i})^{x}=(H\cap R_{i})^{x}(V\cap R_{i})\end{gathered} $$

for all $x\in R_{i}$, where $(H\cap R_{i})(V\cap R_{i})\neq R_{i}$ because $V\cap R_{i}<P\cap R_{i}$. Therefore, $R_{i}$ is not simple by [8, Lemma 1.1.9(1)] because $H\cap R_{i}\neq 1$ and $V\cap R_{i}\neq 1$. This contradiction shows that $R$ is $\sigma$-primary.

Then $H\cap R_{i}\neq 1$ for some $j\neq k$ and $H=H_{j}$. Therefore, $V\cap R$ is $\tau_{\sigma}$-permutable in $G$ by Lemma 2.6(2). But $V\cap R\leq R\leq O_{\sigma_{k}}(G)$ and so $V\cap R$ is $\sigma$-permutable in $G$ by Lemma 2.6(3). Because $R\leq D\leq O^{\sigma_{i}}(G)$ and so $R\leq N_{G}(V\cap R)$ by Lemma 2.5, it follows that $V\cap R\leq O_{p}(R)=1$; a contradiction. Thus $R$ is abelian, and so $D$ is $p$-soluble. The lemma is proved.

Lemma 2.10

Let $D=G^{\mathfrak{N_{\sigma}}}$ and $\pi=\pi(D)$. If ${\mathcal{H}}=\{H_{1},\dots,H_{t}\}$ is a complete Hall $\sigma$-set of $G$ such that $H_{i}\cap D\leq Z_{\mathfrak{U}_{\pi}}(H_{i})$ for all $i$, then ${\mathcal{H}}_{0}=\{H_{1}N/N,\dots,H_{t}N/N\}$ is a complete Hall $\sigma$-set of $G/N$ such that $(H_{i}N/N)\cap(G/N)^{\mathfrak{N_{\sigma}}}\leq Z_{\mathfrak{U}_{\pi_{0}}}(H_{i}N/N)$ for all $i$, where $\pi_{0}=\pi((G/N)^{\mathfrak{N_{\sigma}}})$.

Proof

It is clear that ${\mathcal{H}}_{0}$ is a complete Hall $\sigma$-set of $G/N$. Put $D_{0}=(G/N)^{{\mathfrak{N}}_{\sigma}}$. Then $D_{0}=DN/N$ by Lemma 2.2, and so

$$ \pi_{0}=\pi(D_{0})=\pi(DN/N)=\pi(D/(D\cap N))\subseteq\pi(D)=\pi. $$

Hence, $Z_{\mathfrak{U}_{\pi}}(H_{i}N/N)\leq Z_{\mathfrak{U}_{\pi_{0}}}(H_{i}N/N)$. On the other hand, $D\cap H_{i}N=(D\cap H_{i})(D\cap N)$ by Lemma 2.3. Thus,

$$ D_{0}\cap(H_{i}N/N)=(D\cap H_{i})N/N. $$

Note that, in view of Lemma 2.7(3),

$$ (D\cap H_{i})(N\cap H_{i})/(N\cap H_{i})\leq Z_{\mathfrak{U}_{\pi}}(H_{i}/(N\cap H_{i})) $$

since $D\cap H_{i}\leq Z_{\mathfrak{U}_{\pi}}(H_{i})$. Hence

$$ f((D\cap H_{i})(N\cap H_{i})/(N\cap H_{i}))=(D\cap H_{i})N/N\leq Z_{\mathfrak{U}_{\pi}}(H_{i}N/N), $$

where $f:H_{i}/(N\cap H_{i})\to H_{i}N/N$ is the canonical isomorphism, since

$$ f(Z_{\mathfrak{U}_{\pi}}(H_{i}/(N\cap H_{i})))=Z_{\mathfrak{U}_{\pi}}(H_{i}N/N). $$

Therefore, $(D\cap H_{i})N/N\leq Z_{\mathfrak{U}_{\pi_{0}}}(H_{i}N/N)$ for all $i$. The lemma is proved.

Lemma 2.11

Let $D=G^{\mathfrak{N_{\sigma}}}$ and $\pi=\pi(D)$. Suppose that $G$ is $\sigma$-soluble and all maximal subgroups of every noncyclic Sylow $p$-subgroup of $G$ are $\tau_{\sigma}$-permutable in $G$ for all $p\in\pi$. Then

(1) the hypothesis holds for $G/L$ for every minimal normal subgroup $L$ of $G$;

(2) if $D$ is nilpotent, then $D$ is a Hall subgroup of $G$.

Proof

(1): See the proof of Lemma 2.9.

(2): Suppose that this assertion is false. Let $P$ be a Sylow $p$-subgroup of $D$ and let $G_{p}$ be a Sylow $p$-subgroup of $G$ such that $1<P<G_{p}$. We can assume without loss of generality that $G_{p}\leq H_{1}$.

(a) $D=P$ is a minimal normal subgroup of $G$. Hence $D\leq G_{p}=H_{1}\trianglelefteq G$.

Let $R$ be a minimal normal subgroup of $G$ lying in $D$. Since $D$ is nilpotent by hypothesis, $R$ is a $q$-group for some prime $q$. Moreover, by (1) and the choice of $G$ we have that $D/R=(G/R)^{\mathfrak{N}_{\sigma}}$ is a Hall subgroup of $G/R$. Suppose now that $PR/R\neq 1$. Then $PR/R$ is a Sylow $p$-subgroup of $G/R$. If $q\neq p$, then $P$ is a Sylow $p$-subgroup of $G$. This contradicts the fact that $P<G_{p}$. Hence $q=p$ and so $R\leq P$. It implies that $P/R$ is a Sylow $p$-subgroup of $G/R$, and so $P$ is a Sylow $p$-subgroup of $G$. This contradiction shows that $PR/R=1$, which implies that $R=P$ is the unique minimal normal subgroup of $G$ lying in $D$. Since $D$ is nilpotent, a $p^{\prime}$-complement $E$ of $D$ is characteristic in $D$ and so $E$ is normal in $G$. Hence $E=1$. This implies that $R=D=P$. Finally, $G/D$ is $\sigma$-nilpotent by Lemma 2.1 and so $H_{1}/D$ is normal in $G/D$. Hence (a) holds.

(b) $D\nleq\Phi(G)$. Hence there exists a maximal subgroup $M$ of $G$ such that $G=D\rtimes M$. (This follows from (2) and Lemma 2.1 because $G$ is not $\sigma$-nilpotent.)

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

By Lemma 2.2, $(G/L)^{\mathfrak{N}_{\sigma}}=LD/L$. Therefore, by (1), (a), and the choice of $G$ we have that $LD/L\simeq D$ is a Hall subgroup of $G/L$. Hence $G_{p}L/L=DL/L$, and so $G_{p}=D\times(L\cap G_{p})$. Since $D<G_{p}$ by (a), $O_{p^{\prime}}(G)=1$.

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

In view of (a) and (b), $C_{G}(D)=D\times V$, where $V=C_{G}(D)\cap M$ is a normal subgroup of $G$. By (a), $V\cap D=1$ and so $V\simeq DV/D$ is $\sigma$-nilpotent by Lemma 2.1. Let $W$ be a $\sigma_{1}$-complement of $V$. Then $W$ is characteristic in $V$ and so it is normal in $G$. Therefore, (d) holds in view of (c).

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

Assume that $G_{p}=H_{1}$. Then $D<G_{p}\leq C_{G}(D)$ by (a) and [13, Chapter A, Theorem 10.6(b)]. It follows from (d) that $L\leq C_{G}(D)\cap M\leq G_{p}$ for some minimal normal subgroup $L$ of $G$. Hence $G_{p}=D\times L$ is a normal elementary abelian $p$-subgroup of $G$ by (c). This ensues from Lemmas 2.6(3) and 2.5 that every maximal subgroup of $G_{p}$ is normal in $G$. It follows that every subgroup of $G_{p}$ is normal in $G$.

Hence $|D|=|L|=p$. Let $D=\langle a\rangle$, $L=\langle b\rangle$, and $N=\langle ab\rangle$. Then $N\nleq D$ and so, in view of the $G$-isomorphisms

$$ DN/D\simeq N\simeq NL/L=G_{p}/L=DL/L\simeq D, $$

we get that $G/C_{G}(D)=G/C_{G}(N)$ is a $p$-group since $G_{p}=H_{1}$ and $G/D$ is $\sigma$-nilpotent by Lemma 2.1. It follows from (d) that $G$ is a $p$-group. This contradiction shows that we have (e).

Final contradiction for (2). By Theorem A in [16], $G$ has a $\sigma_{1}$-complement $E$ such that $W=EG_{p}=G_{p}E$. Then $D\leq G_{p}\leq W$ by (a). Moreover, since $W/D\leq G/D\in{\mathfrak{N_{\sigma}}}$ and ${\mathfrak{N_{\sigma}}}$ is a hereditary class by Lemma 2.1, $W/D\in{\mathfrak{N_{\sigma}}}$, and thereby $V=W^{\mathfrak{N_{\sigma}}}\leq D$. Therefore, in view of Lemmas 2.4 and 2.6(4), the hypothesis holds for $W$. From (e) we derive that $W\neq G$. Hence the conclusion of the lemma holds for $W$ by the choice of $G$, which implies that $V$ is a Hall subgroup of $W$. Moreover, $V\leq D$ and so $|V_{p}|\leq|P|<|G_{p}|$ for a Sylow $p$-subgroup $V_{p}$ of $V$. Hence $V$ is a $p^{\prime}$-group. It implies from (d) that $V\leq C_{G}(D)\leq H_{1}\cap W$. Therefore $V=1$, which shows that $W=EG_{p}=E\times G_{p}$ is $\sigma$-nilpotent and so $E\leq C_{G}(D)\leq H_{1}$. Hence $E=1$. It follows that $D=1$, which is a contradiction. Thus $D$ is a Hall subgroup of $G$. The lemma is proved.

Proof of Theorem C

Suppose that this theorem is false and let $G$ be a counterexample of minimal order. Then $D\neq 1$. Let ${\mathcal{H}}=\{H_{1},\dots,H_{t}\}$. We can assume without loss of generality that $H_{i}$ is a $\sigma_{i}$-group for all $i=1,\dots,t$. Let $R$ be a minimal normal subgroup of $G$.

(1) The hypothesis holds for $G/R$ (see the proof of Lemma 2.9 and use Lemma 2.10).

(2) $D$ is soluble, and so $G$ is $\sigma$-soluble. Hence $G$ is a $\sigma$-full group of Sylow type (in view of Theorem 2.8 in [12, IV], this follows from Lemmas 2.4, 2.9, and the Feit–Thompson Theorem).

(3) $D$ is nilpotent.

Assume that this is false. Note that $(G/R)^{{\mathfrak{N_{\sigma}}}}=RD/R$ is nilpotent by (1) and the choice of $G$. Therefore $R\leq D$, while $R$ is the unique minimal normal subgroup of $G$ and $R\nleq\Phi(G)$ by Lemma 2.1. It implies from (2) that $R$ is a $p$-group for some prime $p$. Therefore, by [13, Chapter A, Theorem 15.2] $R=C_{G}(R)$, $G=R\rtimes M$ for some maximal subgroup $M$ of $G$ and $|R|>p$, if not, then $G/C_{G}(R)=G/R$ is a cyclic group and so $D$ is nilpotent, contrary to our assumption on $D$.

Clearly, $R\leq H_{i}\cap D$ for some $i$. Then $H_{i}=R\rtimes(H_{i}\cap M)$ and $R\leq Z_{\mathfrak{U}_{\pi}}(H_{i})$ by hypothesis. It shows that there exists a maximal subgroup $V$ of $R$ such that $V$ is normal in $H_{i}$ because $p\in\pi$. Let $P$ be a Sylow $p$-subgroup of $H_{i}\cap M$. Then $RP$ is a Sylow $p$-subgroup of $G$, and $VP$ is a maximal subgroup of $RP$. Hence, by the hypothesis of the theorem $VP$ is $\tau_{\sigma}$-permutable in $G$. It follows from Lemma 2.6(2)(3) that $V=V(R\cap P)=R\cap VP$ is $\sigma$-permutable in $G$. Therefore $O^{\sigma_{i}}(G)\leq N_{G}(V)$ by Lemma 2.5, and thereby $G=H_{i}O^{\sigma_{i}}(G)\leq N_{G}(V)$. The minimality of $R$ implies that $V=1$ and so $|R|=p$; a contradiction. Hence, we have (3).

(4) $D$ is a Hall subgroup of $G$. (This is straightforward from (2), (3), and Lemma 2.11.)

(5) If $p$ is a prime such that $(p-1,|G|)=1$, then $p$ does not divide $|D|$. In particular, the smallest prime divisor of $|G|$ divides $|G:D|$.

Assume the contrary and let $P$ be the Sylow $p$-subgroup of $D$. Then, arguing as in the proof of (3), we can show that some maximal subgroup $E$ of $P$ is normal in $G$. Hence $C_{G}(D/E)=G$ because $(p-1,|G|)=1$ by hypothesis. Since $D$ is a Hall subgroup of $G$ by (4), $D$ has a complement $M$ in $G$. Therefore $G/E=(D/E)\times(ME/E)$, where $ME/E\simeq M\simeq G/D$ is $\sigma$-nilpotent. Thus, $G/E$ is $\sigma$-nilpotent. It follows that $D\leq E$; a contradiction. Hence $p$ does not divide $|D|$. In particular, the smallest prime divisor of $|G|$ divides $|G:D|$.

(6) Every chief factor of $G$ below $D$ is cyclic.

Suppose the contrary. Assume that $\Phi(D)\neq 1$ and let $R\leq\Phi(D)$. Then the choice of $G$ and (1) imply that every chief factor of $G/R$ below $(G/R)^{{\mathfrak{N}}_{\sigma}}=D/R$ is cyclic, and so every chief factor of $G$ below $D$ is cyclic by Lemma 2.8. Hence $\Phi(D)=1$, and so every Sylow subgroup of $D$ is elementary. Moreover, there is $p\in\pi(D)$ such that the Sylow $p$-subgroup $P$ of $D$ has a minimal normal subgroup $N$ of $G$ such that $|N|>p$. Let $V$ be a maximal subgroup of $P$ such that $P=NV$. Then $N\cap V\neq 1$. Since $D$ is a Hall subgroup of $G$, $P$ is the Sylow $p$-subgroup of $G$. Therefore $V$ is $\tau_{\sigma}$-permutable in $G$, and so $N\cap V$ is $\sigma$-permutable in $G$ by Lemma 2.6(2)(3). Arguing as in the proof of (3), we can show that $N\cap V$ is normal in $G$. The minimality of $N$ implies that $N\cap V=1$, and so $|N|=p$. This contradiction completes the proof of (6).

Claims (3)–(6) show that the conclusion of the theorem holds for $G$, which contradicts the choice of $G$. The theorem is proved.

3. Proof of Theorem B

Lemma 3.1

Suppose that $D=G^{{\mathfrak{N}}_{\sigma}}$ is a nilpotent Hall subgroup of $G$. If every $\sigma_{i}$-subgroup of $G$ is $\tau_{\sigma}$-permutable in $G$ for all $\sigma_{i}\in\sigma(D)$, then $D$ is an abelian group of odd order and each element of $G$ induces a power automorphism in $D$.

Proof

Suppose that this lemma is false and let $G$ be a counterexample of minimal order. Let ${\mathcal{H}}=\{H_{1},\dots,H_{t}\}$. We can assume without loss of generality that $H_{i}$ is a $\sigma_{i}$-group for all $i=1,\dots,t$.

Note first that

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

is a nilpotent Hall subgroup of $G/N$ for every minimal normal subgroup $N$ of $G$ by Lemma 2.2. Let $V/N$ be a nonidentity $\sigma_{i}$-subgroup of $G/N$ for some

$$ \sigma_{i}\in\sigma((G/N)^{{\mathfrak{N}}_{\sigma}})=\sigma(DN/N)=\sigma(D/(D\cap N))\subseteq\sigma(D). $$

Let $U$ be a minimal supplement to $N$ in $V$. Then $U\cap N\leq\Phi(U)$, and so $U$ is a $\sigma_{i}$-subgroup of $G$ since $V/N=UN/N\simeq U/(U\cap N)$. Thus, $U$ is $\tau_{\sigma}$-permutable in $G$ by hypothesis and $\sigma(U)=\sigma(UN/N)=\{\sigma_{i}\}$, which implies that $V/N=UN/N$ is $\tau_{\sigma}$-permutable in $G/N$ by Lemma 2.6(1). Hence the hypothesis holds for $G/N$.

Let $H$ be a subgroup of the Sylow $p$-subgroup $P$ of $D$ for some prime $p\in\pi$. We show that $H$ is normal in $G$. For some $i$ we have $P\leq O_{\sigma_{i}}(D)=H_{i}\cap D$. On the other hand, $D=O_{\sigma_{i}}(D)\times O^{\sigma_{i}}(D)$ since $D$ is nilpotent. Assume that $O^{\sigma_{i}}(D)\neq 1$ and let $N$ be a minimal normal subgroup of $G$ lying in $O^{\sigma_{i}}(D)$. Then $HN/N\leq DN/N=(G/N)^{{\mathfrak{N}}_{\sigma}}$, and so the choice of $G$ implies that $HN/N$ is normal in $G/N$. Hence, $H=H(N\cap O_{\sigma_{i}}(D))=HN\cap O_{\sigma_{i}}(D)$ is normal in $G$.

Assume now that $O^{\sigma_{i}}(D)=1$. Then $D$ is a $\sigma_{i}$-group. Since $G/D$ is $\sigma$-nilpotent by Lemma 2.1, $H_{i}/D$ is normal in $G/D$ and so $H_{i}$ is normal in $G$. It follows from Lemma 2.6(3) and the hypothesis of the theorem that all subgroups of $H_{i}$ are $\sigma$-permutable in $G$. Since $D$ is a normal Hall subgroup of $H_{i}$; therefore, $D$ has a complement $S$ in $H_{i}$ by the Schur–Zassenhaus Theorem. It implies from Lemma 2.5 that $D\leq O^{\sigma_{i}}(G)\leq N_{G}(S)$. Hence $H_{i}=D\times S$, and so

$$ G=H_{i}O^{\sigma_{i}}(G)=SO^{\sigma_{i}}(G)\leq N_{G}(H). $$

This implies that $H$ is normal in $G$. Hence $D$ is a Dedekind group, and so $|D|$ is odd by Theorem C. Hence, $D$ is abelian and each element of $G$ induces a power automorphism in $D$. The lemma is proved.

The following lemma is a corollary of Theorem A of this paper and Theorem B in [2].

A subgroup $A$ of $G$ is said to be ${\sigma}$-subnormal in $G$ [2] if 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 $i=1,\dots,n$.

Lemma 3.2

The following hold:

(i) $G$ is a $P\sigma T$-group if and only if every $\sigma$-subnormal subgroup of $G$ is $\sigma$-quasinormal 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 of Theorem B

Sufficiency: Assume the contrary and let $G$ be a counterexample with $|G|+|N|$ minimal. By Lemma 2.1, $D:=G^{\mathfrak{N}_{\sigma}}$ is the smallest normal subgroup of $G$ with $\sigma$-nilpotent quotient. Therefore $D\leq N$ and so the hypothesis holds for $(G,D)$. Hence the choice of $G$ shows that $D=N$. We can assume without loss of generality that $H_{i}$ is a $\sigma_{i}$-group for all $i=1,\dots,t$.

(1) $G=D\rtimes M$, 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 $D$. (This is straightforward from Lemma 3.1 and Theorem C.)

(2) If $R$ is a nonidentity normal subgroup of $G$, then the hypothesis holds for $G/R$, and so $G/R$ is a $\sigma$-soluble $P\sigma T$-group (see the proof of Lemma 3.1 and use Lemma 2.10).

(3) $H_{i}=O_{\sigma_{i}}(D)\times S$ for some subgroup $S$ of $H_{i}$ for each $\sigma_{i}\in\sigma(D)$.

Since $D$ is an abelian Hall subgroup of $G$ by (1), $D=L\times N$, where $L=O_{\sigma_{i}}(D)$ and $N=O^{\sigma_{i}}(D)=O_{\sigma^{\prime}_{i}}(D)$ are Hall subgroups of $G$. Assume first that $N\neq 1$. Then

$$ O_{\sigma_{i}}((G/N)^{\mathfrak{N}_{\sigma}})=O_{\sigma_{i}}(D/N)=LN/N $$

has a normal complement $V/N$ in $H_{i}N/N\simeq H_{i}$ by (2) and Theorem A. On the other hand, $N$ has a complement $S$ in $V$ by the Schur–Zassenhaus Theorem. Hence $H_{i}=H_{i}\cap LSN=LS$ and $L\cap S=1$ since

$$ (L\cap S)N/N\leq(LN/N)\cap(V/N)=(LN/N)\cap(SN/N)=1. $$

It is clear that $V/N$ is a Hall subgroup of $H_{i}N/N$, and so $V/N$ is characteristic in $H_{i}N/N$. On the other hand, $H_{i}N/N$ is normal in $G/N$ by Lemma 2.2 since $D/N\leq H_{i}N/N$. Hence $V/N$ is normal in $G/N$. Thus $H_{i}\cap V=H_{i}\cap NS=S(H_{i}\cap N)=S$ is normal in $H_{i}$, and so $H_{i}=O_{\sigma_{i}}(D)\times S$.

Assume that $D=O_{\sigma_{i}}(D)$. Then $H_{i}$ is normal in $G$, and so all subgroups of $H_{i}$ are $\sigma$-permutable in $G$ by Lemma 2.6(3). Since $D$ is a normal Hall subgroup of $H_{i}$, $D$ has a complement $S$ in $H_{i}$. Using Lemma 2.5, we imply that $D\leq O^{\sigma_{i}}(G)\leq N_{G}(S)$. Hence, $H_{i}=D\times S=O_{\sigma_{i}}(D)\times S$.

It follows from Theorem A, (2), and (3) that $G$ is a $\sigma$-soluble $P\sigma T$-group, contrary to our assumption on $G$. This completes the proof of sufficiency.

Assume now that $G$ is a $\sigma$-soluble $P\sigma T$-group and let $D=G^{\mathfrak{N}_{\sigma}}$. Then $G$ possesses a complete $\sigma$-set ${\mathcal{H}}=\{H_{1},\dots,H_{t}\}$ by Lemma 2.4. Moreover, $G/D$ is $\sigma$-nilpotent by Lemma 2.1 and every subgroup of $D$ is normal in $G$ by Theorem A. Then $H_{i}\cap N\leq Z_{\mathfrak{U}}(H_{i})\leq Z_{\mathfrak{U}_{\pi}}(H_{i})$, where $\pi=\pi(N)$ for all $i$. Therefore, (i) holds for $G$.

We show now that every $\sigma_{i}$-subgroup of $G$ is $\tau_{\sigma}$-permutable in $G$ for each $\sigma_{i}\in\sigma(D)$. It suffices to show that if $H$ is a $\sigma_{i}$-subgroup of $G$, and so $H$ permutes with every Hall $\sigma_{j}$-subgroup of $G$ for all $j\neq i$. Assume the contrary and let $G$ be a counterexample of minimal order. Then $D\neq 1$ and there are $\sigma_{i}$ and $\sigma_{j}$ ($i\neq j$) such that $\sigma_{i}\in\sigma(D)$ and $HE\neq EH$ for some $\sigma_{i}$-subgroup $H$ and some Hall $\sigma_{j}$-subgroup $E$ of $G$. Then $H$ is not $\sigma$-subnormal in $G$ by Lemma 3.2. Hence a Hall $\sigma_{i}$-subgroup $H_{i}$ of $G$ is not normal in $G$ since otherwise $H\leq H_{i}$ and so $H$ is $\sigma$-subnormal in $G$ by Lemma 2.6. Note that $|\sigma(D)|>1$. Indeed, if $|\sigma(D)|=1$, then $\sigma(D)=\{\sigma_{i}\}$ and so $D\leq H_{i}$, which implies that $H_{i}/D$ is normal in $G/D$ because $G/D$ is $\sigma$-nilpotent. Hence $H_{i}$ is normal in $G$; a contradiction.

We show now that $EHN$ is a subgroup of $G$ for every minimal normal subgroup $N$ of $G$. Note first that the hypothesis holds for $G/N$ by Lemma 3.2. Moreover, $HN/N\simeq H/(H\cap N)$ is a $\sigma_{i}$-subgroup of $G/N$. Therefore, if $\sigma_{i}\in\sigma(DN/N)=\sigma((G/N)^{{\mathfrak{N}}_{\sigma}})$, then the choice of $G$ implies that

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

is a subgroup of $G/N$. Hence $EHN$ is a subgroup of $G$. Assume now that $\sigma_{i}\not\in\sigma(DN/N)$. Then a Hall $\sigma_{i}$-subgroup $H_{i}$ of $G$ lies in $N$. Clearly, $H_{i}=N$ because $N$ is $\sigma$-primary. It follows that $H\leq N$ and so $H$ is $\sigma$-subnormal in $G$; a contradiction. Hence $EHN$ is a subgroup of $G$. Since $|\sigma(D)|>1$ and $D$ is abelian by Theorem A, $G$ has at least two $\sigma$-primary 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 $N$, say $R$, is a $\sigma_{k}$-group for some $k\neq j$. Moreover,

$$ R\cap E(HN)=(R\cap E)(R\cap HN)=R\cap HN $$

by Lemma 2.3 and $R\cap HN\leq O_{\sigma_{k}}(HN)\leq V$, where $V$ is a Hall $\sigma_{k}$-subgroup of $H$, because $N$ is a $\sigma_{k}^{\prime}$-group and $G$ is a $\sigma$-full group of Sylow type by Lemma 2.4. Hence

$$ EHR\cap EHN=EH(R\cap E(HN))=EH(R\cap HN)=EH(R\cap H)=EH $$

is a subgroup of $G$, i.e., $HE=EH$. This contradicts $HE\neq EH$. Therefore, (ii) holds for $G$. Hence the necessity of the condition of the theorem holds for $G$. The theorem is proved.