On the lattice of the $\sigma$-permutable subgroups of a finite group

TL;DR

This paper characterizes finite groups whose lattice of ${\sigma}$-permutable subgroups is distributive, based on the structure of Hall ${\sigma}$-subgroups and their permutability properties.

Contribution

It provides a new characterization of finite groups with a distributive lattice of ${\sigma}$-permutable subgroups, extending previous subgroup lattice theories.

Findings

Identifies conditions for the distributivity of the subgroup lattice

Links subgroup permutability with lattice properties

Provides structural insights into ${\sigma}$-permutable subgroups

Abstract

Let $\sigma =\{\sigma_{i} | i\in I\}$ be some partition of the set of all primes $\Bbb{P}$, $G$ a finite group and $\sigma (G) =\{\sigma_{i} |\sigma_{i}\cap \pi (G)\ne \emptyset \}$. 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 $\sigma_{i}\in \sigma (G)$. A subgroup $A$ of $G$ is said to be ${\sigma}$-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 $i \in I$. We characterize finite groups with distributive lattice of the ${\sigma}$-permutable subgroups.

Equations

$$(A/V)(HV/V)=AH/V=HA/V=(HV/V)(A/V)$$

$$(A\cap B/V)(HV/V)=((A/V)\cap (B/V))(HV/V)=$$

$$=(HV/V)((A/V)\cap (B/V))=(HV/V)(A\cap B/V).$$

$$(A\cap B)H=(A\cap B)HV=HV(A\cap B)=H(A\cap B).$$

$$VH=((A_1\cap V)\times \cdots \times (A_t\cap V))H=H((A_1\cap V)\times \cdots \times (A_t\cap V))=HV.$$

$$AH=HA=(A_1\times V\times A_i)H=(A_1\times V)H=(A_1\times V)H,$$

$$A{L}^x/R=(A/R)(LR/R{)}^{xR}=(LR/R{)}^{xR}(A/R)=L^{x}A/R$$

$$AH/R=(A/R)(HR/R)=(HR/R)(A/R)=HA/R$$

$$VH/R=(V/R)(HR/R)=(HR/R)(V/R)=HV/R$$

$$(HR/R)(LR/R{)}^{xR}=H{L}^{x}R/R=L^{x}HR/R=(LR/R{)}^{xR}(HR/R)$$

$$G=⟨A,B⟩O^{{\sigma }_i}(G)=⟨A,B⟩\times O^{{\sigma }_i}(G),$$

$$(l_{0}K{)}^g=((hK{)}^f{)}^g=((hK{)}^g{)}^f=(h^{g}K{)}^f=(h_{0}K{)}^f,$$

$$(H_0/K)(L_0/K)=(H_0/K)\times (L_0/K).$$

$$(H_0/K)\times (L_0/K)=(H_0/K)\times (E_0/K)=(L_0/K)\times (E_0/K).$$

$$(hK(hK{)}^f{)}^g=(hK{)}^g((hK{)}^f{)}^g=(h^{g}K)(h^{g}K{)}^f\in E_0/K$$

$$C\cap ⟨A,B⟩=C\cap AB=A(C\cap B)=⟨A,C\cap B⟩$$

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

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

$$D\cap ⟨K,H⟩=D\cap ⟨D\cap K,D\cap H⟩=(D\cap K)(D\cap H)$$

$$(G/R{)}^{{\mathfrak{N}}_{\sigma }}\cap ⟨(K/R),(H/R)⟩=(DR/R)\cap (K/R)(H/R)=(DR\cap KH)/R=R(D\cap KH)/R=$$

$$=R(D\cap K)(D\cap H)/R=((D\cap K)R/R)((D\cap H)R/R)=$$

$$=((DR/R)\cap (K/R))((DR/R)\cap (H/R))=⟨(G/R{)}^{{\mathfrak{N}}_{\sigma }}\cap (K/R)),(G/R{)}^{{\mathfrak{N}}_{\sigma }}\cap (H/R)⟩.$$

$$(A_G{B}_G\cap A_G{C}_G)\cap B_G{C}_G=A_G(B_G\cap A_G{C}_G)\cap B_G{C}_G=$$

$$=A_G(B_G\cap A_G)(B_G\cap C_G)\cap B_G{C}_G=$$

$$=A_G\cap B_G{C}_G=(A_G\cap B_G)(A_G\cap C_G)=1$$

$$T/A_G{B}_G=AB/A_G{B}_G=(A{A}_G{B}_G/A_G{B}_G)(B{A}_G{B}_G/A_G{B}_G),$$

$$A{A}_G{B}_G/A_G{B}_G\simeq A/A\cap A_G{B}_G=$$

$$=A/A_G(A\cap B_G)\simeq (A/A_G)/(A_G(A\cap B_G)/A_G)$$

$$B{A}_G{B}_G/A_G{B}_G\simeq (B/B_G)/(B_G(B\cap A_G)/B_G)$$

$$K_0=A_0\cap B_0=A_i\cap B_i\cap D=A_i\cap C_i\cap D=A_0\cap C_0=B_i\cap C_i\cap D=B_0\cap C_0.$$

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 $\sigma_{i}\in\sigma(G)$. A subgroup $A$ of $G$ is said to be ${\sigma}$-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 $i\in I$.

We characterize finite groups with distributive lattice of the ${\sigma}$-permutable subgroups.

00footnotetext: Keywords: finite group, $\sigma$-permutable subgroup, subgroup lattice, modular lattice, distributive lattice00footnotetext: Mathematics Subject Classification (2010): 20D10, 20D30, 20E15

1 Introduction

Throughout this paper, $G$ always denotes a finite group. Moreover, we use ${\cal L}(G)$ to denote the lattice of all subgroups of $G$. $\mathbb{P}$ is the set of all primes, $\pi\subseteq\mathbb{P}$ and $\pi^{\prime}=\mathbb{P}\setminus\pi$. As usual, $\pi(G)$ is the set of all primes dividing the order $|G|$ of $G$. The subgroups $A$ and $B$ of $G$ are said to be permutable if $AB=BA$. In this case they also say that $A$ *permutes * with $B$. If $A$ permutes with all Sylow subgroups of $G$, then $A$ is called $S$-permutable in $G$ [1]. Recall also that an element $a$ of the lattice $\cal L$ is called meet-distributive [2, p. 136] if $a\wedge(b\vee c)=(a\wedge b)\vee(a\wedge c)$ for all $b,c\in\cal L$.

In what follows, $\sigma=\{\sigma_{i}|i\in I\}$ is some partition of $\mathbb{P}$, that is, $\mathbb{P}=\bigcup_{i\in I}\sigma_{i}$ and $\sigma_{i}\cap\sigma_{j}=\emptyset$ for all $i\neq j$.

A set ${\cal H}$ of subgroups of $G$ is a complete Hall $\sigma$-set of $G$ [3] 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 $i$ such that $\sigma_{i}\cap\pi(G)\neq\emptyset$. $G$ is said to be * $\sigma$-full* if it possesses a complete Hall $\sigma$-set.

Definition 1.1. We say that a subgroup $A$ of $G$ is said to be $\sigma$-quasinormal or $\sigma$-permutable in $G$ [4] if $G$ is $\sigma$-full and $A$ permutes with each Hall $\sigma_{i}$-subgroup $H$ of $G$ for all $i\in I$.

**Remark 1.2. ** (i) 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 $G$ (see Proposition 3.1 below).

(ii) $G$ is called $\sigma$-decomposable (Shemetkov [5]) or $\sigma$-nilpotent (Guo and Skiba [6]) if $G=H_{1}\times\cdots\times H_{t}$, where $\{H_{1},\ldots,H_{t}\}$ is a complete Hall $\sigma$-set of $G$. It is not difficult to show that $G$ is $\sigma$-nilpotent if and only if every subgroup of $G$ is $\sigma$-permutable in $G$.

(iii) In the classical case when $\sigma=\sigma^{0}=\{\{2\},\{3\},\ldots\}$: $G$ is $\sigma^{0}$-nilpotent if and only if $G$ is nilpotent; a subgroup $A$ of $G$ is $\sigma^{0}$-permutable in $G$ if and only if it is $S$-permutable in $G$.

(iv) In the other classical case when $\sigma=\sigma^{\pi}=\{\pi,\pi^{\prime}\}$: $G$ is $\sigma^{\pi}$-nilpotent if and only if it is $\pi$-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 $G$.

(v) If $G$ is $\sigma$-full, the set ${\cal L}_{\sigma per}(G)$ of all $\sigma$-permutable subgroups of $G$ is partially ordered with respect to set inclusion. Moreover, ${\cal L}_{\sigma per}(G)$ is a lattice since $1\in{\cal L}_{\sigma per}(G)$ and, by Lemma 2.1 below, for any $A_{1},\ldots,A_{n}\in{\cal L}_{\sigma per}(G)$ the subgroup $\langle A_{1},\ldots,A_{n}\rangle$ is the least upper bound for $\{A_{1},\ldots,A_{n}\}$ in ${\cal L}_{\sigma per}(G)$.

The conditions under which the lattice ${\cal L}_{sn}(G)$ of all subnormal subgroups of $G$ is modular or distributive are known (see [2, Theorems 9.2.3, 9.2.4]). It is well-known also that the lattice ${\cal L}_{n}(G)$ of all normal subgroups of $G$ is modular and this lattice is distributive if and only if in every factor group $G/R$, any two $G/R$-isomorphic normal subgroups coincide (see [7] and [2, Theorem 9.1.6]). Kegel proved [8] that the set ${\cal L}_{S}(G)$ of all $S$-permutable subgroups of $G$ forms a sublattice of the lattice ${\cal L}_{sn}(G)$. Since ${\cal L}_{n}(G)\subseteq{\cal L}_{S}(G)\subseteq{\cal L}_{sn}(G)$, where both inclusions in general are strict, it seems natural to ask: Under what conditions the lattice ${\cal L}_{S}(G)$ is modular or distributive? Moreover, in view of Remark 1.2(v), it makes sense to consider the following general

Question 1.3 (See Questions 6.10 and 6.11 in [3]). Under what conditions the lattice ${\cal L}_{\sigma per}(G)$ is modular or distributive?

Note that if $K\trianglelefteq H$ and $K,H\in{\cal L}_{\sigma_{i}}(G)$, where ${\cal L}_{\sigma_{i}}(G)$ is the set of all $\sigma$-permutable $\sigma_{i}$-subgroups of $G$, then $O^{\sigma_{i}}(G)$ normalizes both subgroups $K$ and $H$ (see Lemma 2.4(1) below) and hence we can consider $O^{\sigma_{i}}(G)$ as a group of operators for $H/K$ (assuming, as usual, that $(hK)^{a}=h^{a}K$ for all $hK\in H/K$ and $a\in O^{\sigma_{i}}(G)$).

We do not know under what conditions on $G$ the lattice ${\cal L}_{\sigma}(G)$ is modular. Nevertheless, we give a full answer to the second part of Question 1.3.

Theorem A. Suppose that $G$ is $\sigma$-full. Let $D=G^{{\mathfrak{N}}_{\sigma}}$ and ${\cal L}={\cal L}_{\sigma}(G)$. Then the lattice ${\cal L}$ is distributive if and only if the following conditions hold:

(i) Every two members of ${\cal L}$ are permutable.

(ii) The lattice ${\cal L}_{n}(G)$ of all normal subgroups of $G$ is distributive.

(iii) $G/D$ is cyclic and $D$ is a meet-distributive element of ${\cal L}$.

(iv) In every factor group $\bar{G}=G/R$, any two $O^{\sigma_{i}}(\bar{G})$-isomorphic sections $\bar{H}/\bar{K}$ and $\bar{L}/\bar{K}$, where $\bar{K},\bar{H},\bar{L}\in{\cal L}_{\sigma_{i}}(\bar{G})$ for some $i$, coincide.

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 $G/N$.

Theorem A remains to be new for each special partition of $\mathbb{P}$. In particular, in the case when $\sigma=\sigma^{0}$ we get from Theorem A the following

Corollary 1.4. Let $D=G^{\mathfrak{N}}$ be the nilpotent residual of $G$ and ${\cal L}={\cal aL}_{S}(G)$. Then the lattice ${\cal L}$ is distributive if and only if the following conditions hold:

(1) Conditions (i), (ii) and (iii) in Theorem A fold for $G$.

(2) In every factor group $\bar{G}=G/R$, any two $O^{p}(\bar{G})$-isomorphic sections $\bar{H}/\bar{K}$ and $\bar{L}/\bar{K}$, where $\bar{K},\bar{H},\bar{L}\in{\cal L}_{pS}(\bar{G})$ and $p$ is a prime, coincide.

In this corollary ${\cal L}_{pS}(\bar{G})$ denotes the set of all $S$-permutable $p$-subgroups of $\bar{G}$.

In the case when $\sigma=\sigma^{\pi}$ (see Remark 1.2(iv)) we get from Theorem A the following fact.

Corollary 1.5. Let $D$ be the $\pi$-decomposable residual of $G$, that is, the smallest normal subgroup of $G$ with $\pi$-decomposable quotient $G/D$. Suppose that $G$ is $\pi$-separable and let ${\cal L}={\cal L}_{\sigma^{\pi}}(G)$. Then the lattice ${\cal L}$ is distributive if and only if the following conditions hold:

(1) Conditions (i), (ii) and (iii) in Theorem A hold for $G$.

(2) In every factor group $\bar{G}=G/R$, any two $O^{\pi}(\bar{G})$-isomorphic sections $\bar{H}/\bar{K}$ and $\bar{L}/\bar{K}$, where $\bar{K},\bar{H}$ and $\bar{L}$ are $\sigma^{\pi}$-permutable $\pi$-subgroups of $\bar{G}$, coincide.

(3) In every factor group $\bar{G}=G/R$, any two $O^{\pi^{\prime}}(\bar{G})$-isomorphic sections $\bar{H}/\bar{K}$ and $\bar{L}/\bar{K}$, where $\bar{K},\bar{H}$ and $\bar{L}$ are $\sigma^{\pi}$-permutable $\pi^{\prime}$-subgroups of $\bar{G}$, coincide.

The proof of Theorem A consists of many steps and the next theorems are two of them.

Theorem B. Suppose that $G$ is $\sigma$-full. Then ${\cal L}_{\sigma per}(G)$ is a sublattice of the lattice ${\cal L}(G)$.

Corollary 1.6 (Kegel [8]). The set ${\cal L}_{S}(G)$ of all $S$-permutable subgroups of $G$ forms a sublattice of the lattice ${\cal L}(G)$.

There are at least three different proofs of Corollary 1.6 (see, for example, [8, 9, 4]). One more, the shortest one, gives the proof of Theorem B.

Theorem C. A $\sigma$-nilpotent subgroup $A$ of $G$ is $\sigma$-permutable in $G$ if and only if each characteristic subgroup of $A$ is $\sigma$-permutable in $G$.

Corollary 1.7 (See [9] or [1, Theorem 1.2.17]). Let $A$ be a nilpotent subgroup of $G$. Then the following statements are equivalent:

(i) $A$ is $S$-permutable in $G$.

(ii) Each Sylow subgroup of $A$ is $S$-permutable in $G$.

(iii) Each characteristic subgroup of $A$ is $S$-permutable in $G$.

2 Proof of Theorems B and C

Lemma 2.1 (See [10, A, Lemma 1.6]). Let $A$, $B$ and $H$ be subgroups of $G$. If $AH=HA$ and $BH=HB$, then $\langle A,B\rangle H=H\langle A,B\rangle$.

A subgroup $A$ of $G$ is called ${\sigma}$-subnormal in $G$ [4] if there is a subgroup chain $A=A_{0}\leq A_{1}\leq\cdots\leq A_{t}=G$ such that either $A_{i-1}\trianglelefteq A_{i}$ or $A_{i}/(A_{i-1})_{A_{i}}$ is ${\sigma}$-primary for all $i=1,\ldots,t$.

The importance of this concept is related to the following result.

Lemma 2.2 (See [4, Theorem B]). Let $A$ be a subgroup of $G$. 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 $x\in G$, then $A$ is $\sigma$-subnormal in $G$.

Lemma 2.3 (See Lemma 2.6 in [4]). Let $A$, $K$ and $N$ be subgroups of $G$. Suppose that $A$ is $\sigma$-subnormal in $G$ and $N$ is normal in $G$.

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

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

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

(4) If $A$ is a $\sigma_{i}$-group, then $A\leq O_{\sigma_{i}}(G)$.

(5) If $H\neq 1$ is a Hall $\sigma_{i}$-subgroup of $G$, then $A\cap H\neq 1$ is a Hall $\sigma_{i}$-subgroup of $A$.

(6) If $K$ is $\sigma$-subnormal in $G$ and the subgroups $A$ and $K$ are $\sigma$-nilpotent, then $\langle A,K\rangle$ is $\sigma$-nilpotent.

Proof of Theorem B. In fact, in view of Lemmas 2.1 and 2.2, it is enough to show that if $A$ and $B$ are $\sigma$-subnormal subgroups of $G$ such that for a Hall $\sigma_{i}$-subgroup $H$ of $G$ we have $AH=HA$ and $BH=HB$, then $(A\cap B)H=H(A\cap B)$. Assume that this is false and let $G$ be a counterexample of minimal order. Then $G$ is not a $\sigma_{i}$-group, since otherwise we have $H=G$ and so $G=(A\cap B)H=H(A\cap B)$.

Let $E=AH\cap BH$. Then $A\cap E$ and $B\cap E$ are $\sigma$-subnormal subgroups in $E$ by Lemma 2.3(1). Moreover, $AH\cap E=H(A\cap E)=(A\cap E)H$. Similarly, $(B\cap E)H=H(B\cap E)$. Hence the hypothesis holds for $(A\cap E,B\cap E,H,E)$. Assume that $E<G$. Then the choice of $G$ implies that $A\cap B=(A\cap E)\cap(B\cap E)$ is permutable with $H$. Hence $E=G$, so $G=AH=BH$. Thus $|G:A|$ and $|G:B|$ are $\sigma_{i}$-numbers. Hence by Lemma 2.3(2) we have $O^{\sigma_{i}}(A)=O^{\sigma_{i}}(G)=O^{\sigma_{i}}(B).$ Therefore, since $G$ is not a $\sigma_{i}$-group, it follows that $V=A_{G}\cap B_{G}\neq 1$. Moreover, $A/V$ and $B/V$ are $\sigma$-subnormal subgroups of $G/V$ by Lemma 2.3(3). Also we have

[TABLE]

and $(B/V)(HV/V)=(HV/V)(B/V)$, where $HV/V$ is a Hall $\sigma_{i}$-subgroup of $G/V$. Hence the choice of $G$ implies that

[TABLE]

[TABLE]

But then

[TABLE]

This contradiction completes the proof of the result.

Lemma 2.4 (See Lemmas 2.8, 3.1 and Theorem B in [4]). Let $A$ and $B$ be subgroups of $G$. Suppose that $G$ 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:

(1) If $A$ is a $\sigma_{i}$-group, then $O^{\sigma_{i}}(G)\leq N_{G}(A)$.

(2) $A/A_{G}$ is $\sigma$-nilpotent.

(3) If $B$ is a $\sigma_{i}$-group and $O^{\sigma_{i}}(G)\leq N_{G}(B)$, then $G$ possesses a complete Hall $\sigma$-set ${\cal L}$ such that $BL^{x}=L^{x}B$ for all $L\in{\cal L}$ and all $x\in G$.

Proposition 2.5. Let $A$ be a $\sigma$-nilpotent $\sigma$-subnormal subgroup of $G$ and $V$ a characteristic subgroup of $A$. Let $H$ be a Hall $\sigma_{i}$-subgroup of $G$. If $AH=HA$, then $VH=HV$.

**Proof. ** Assume that this proposition is false and let $G$ be a counterexample with $|G|+|V|+|A|$ minimal.

By hypothesis, $A=A_{1}\times\cdots\times A_{t}$, where $\{A_{1},\ldots,A_{t}\}$ is a complete Hall $\sigma$-set of $A$. Hence $V=(A_{1}\cap V)\times\cdots\times(A_{t}\cap V)$, where $\{A_{1}\cap V,\ldots,A_{t}\cap V\}$ is a complete Hall $\sigma$-set of $V$. We can assume without loss of generality that $A_{k}$ is a $\sigma_{k}$-subgroup of $A$ for all $k=1,\ldots,t$.

It is clear that $A_{i}\cap V$ is characteristic in $A$ for all $i=1,\ldots,t$. Therefore, if $A_{i}\cap V<V$, then $(A_{i}\cap V)H=H(A_{i}\cap V)$ by the choice of $G$ and so for some $j$, $j=1$ say, we have $A_{1}\cap V=V$ since otherwise we have

[TABLE]

Thus $V\leq A_{1}$. It is clear that $A_{1}$ is a $\sigma$-subnormal subgroup of $G$, so in the case when $i=1$ we have $V\leq A_{1}\leq H$ by Lemma 2.3(5). But then $VH=H=HV$, a contradiction. Thus $i>1.$

Now we show that $A_{1}H=HA_{1}$. Indeed, it is clear that $A=A_{1}\times V\times A_{i}$ and $A_{i}$ is $\sigma$-subnormal in $G$. Thus $A_{i}\leq H$ by Lemma 2.3(5). Therefore

[TABLE]

where $A_{1}\times V$ is a $\sigma$-subnormal $\sigma_{i}^{\prime}$-subgroup of $G$. Then $A_{1}\times V$ is $\sigma$-subnormal in $(A_{1}\times V)H$ by Lemma 2.3(1). Hence $H\leq N_{G}(A_{1}\times V)$ by Lemma 2.3(2). Since $A_{1}$ is a characteristic subgroup of $A_{1}\times V$, we have $H\leq N_{G}(A_{1})$ and so $A_{1}H=HA_{1}$. Therefore $H\leq N_{G}(A_{1})$ by Lemma 2.3(2). But $V$ is a characteristic subgroup of $A_{1}$ since $V$ is characteristic in $A$ by hypothesis and $A=A_{1}\times\cdots\times A_{t}$. Therefore $H\leq N_{G}(V)$ and so $VH=HV$, a contradiction. The proposition is proved.

Corollary 2.6. Let $A$ be a $\sigma$-nilpotent subgroup of a $\sigma$-full group $G$. Then the following statements are equivalent:

(i) $A$ is $\sigma$-permutable in $G$.

(ii) Each Hall $\sigma_{i}$-subgroup of $A$ is $\sigma$-permutable in $G$ for all $i$.

(iii) Each characteristic subgroup of $A$ is $\sigma$-permutable in $G$.

**Proof. ** By hypothesis, $A=A_{1}\times\cdots\times A_{t}$, where $\{A_{1},\ldots,A_{t}\}$ is a complete Hall $\sigma$-set of $A$. Then $A_{i}$ is characteristic in $A$ for all $i=1,\ldots,t$. Therefore (ii), (iii) $\Rightarrow$ (i).

(i) $\Rightarrow$ (ii), (iii) This follows from Proposition 2.5.

The corollary is proved.

**Proof of Theorem C. ** This directly follows from Corollary 2.6.

3 Proof of Theorem A

Proposition 3.1. Let $A$ be a subgroup of $G$. If $G$ possesses a complete Hall $\sigma$-set ${\cal H}$ such that $AL^{x}=L^{x}A$ for all $L\in{\cal H}$ and all $x\in G$, then $A$ is $\sigma$-permutable in $G$.

**Proof. ** Assume that this proposition is false and let $G$ be a counterexample with $|G|+|A|$ minimal. Then for some $i$ and some Hall $\sigma_{i}$-subgroup $H$ of $G$ we have $AH\neq HA$. Let ${\cal H}=\{H_{1},\ldots,H_{t}\}$. We can assume without loss of generality that $H_{k}$ is a $\sigma_{k}$-group for all $k=1,\ldots,t$. Let $V=H_{i}$.

First we show that $A_{G}=1$. Indeed, assume that $R=A_{G}\neq 1$. Then ${\cal H}_{0}=\{H_{1}R/R,\ldots,H_{t}R/R\}$ is a complete Hall $\sigma$-set of $G/R$ such that

[TABLE]

for all $LR/R\in{\cal H}_{0}$ and all $xR\in G/R$. On the other hand, $HR/R$ is Hall $\sigma_{i}$-subgroup of $G/R$. Hence the choice of $G$ implies that

[TABLE]

and so $AH=HA$, a contradiction. Therefore $A_{G}=1$, hence $A=A_{1}\times\cdots\times A_{t}$, where $\{A_{1},\ldots,A_{t}\}$ is a complete Hall $\sigma$-set of $A$ by Lemma 2.4(2). Moreover, Lemma 2.2 implies that $A$ is $\sigma$-subnormal in $G$.

First assume that $A=A_{1}$ is a $\sigma_{j}$-group. If $j=i$, then $A\cap H=A$ by Lemma 2.3(5) and so $AH=H=HA$. Hence $j\neq i$. By hypothesis, $AV^{x}=V^{x}A$ for each $x\in G$. Then $V^{x}\leq N_{G}(A)$ for all $x\in G$ by Lemma 2.3(1)(2). Hence $V^{G}\leq N_{G}(A)$. But then $H\leq V^{G}\leq N_{G}(A)$, which implies that $AH=HA$. This contradiction shows that $t>1$.

The subgroups $A_{1},\ldots,A_{t}$ are characteristic in $A$, so $A_{i}L^{x}=L^{x}A_{i}$ for all $L\in{\cal H}$ and all $x\in G$ by Proposition 2.5. Therefore the minimality of $|G|+|A|$ implies that $A_{i}H=HA_{i}$ for all $i=1,\ldots,t$, so $AH=HA$. This contradiction completes the proof of the result.

Lemma 3.2. Let $R\leq V$ and $H$ be subgroups of a $\sigma$-full group $G$. Suppose that $H$ is $\sigma$-permutable in $G$ and $R$ is normal in $G$. Then:

(1) If $V/R$ is a $\sigma$-permutable subgroup of $G/R$, then $V$ is a $\sigma$-permutable subgroup of $G$.

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

**Proof. ** (1) Let $i\in I$ and $H$ be a Hall $\sigma_{i}$-subgroup of $G$. Then $HR/R$ is a Hall $\sigma_{i}$-subgroup of $G/R$, so

[TABLE]

by hypothesis and hence $VH=HV$.

(2) By hypothesis, $G$ possesses a complete Hall $\sigma$-set ${\cal H}=\{H_{1},\ldots,H_{t}\}$ and $HL^{x}=L^{x}H$ for all $L\in{\cal H}$ and all $x\in G$. Then ${\cal H}_{0}=\{H_{1}R/R,\ldots,H_{t}R/R\}$ is a complete Hall $\sigma$-set in $G/R$ and

[TABLE]

for all $LR/R\in{\cal H}_{0}$ and $xR\in G/R$. Therefore $HR/R$ is $\sigma$-permutable in $G/R$ by Proposition 3.1.

The lemma is proved.

Lemma 3.3 (See Lemma 5.2 in [11]). Let $\cal L$ be a modular sublattice of the lattice ${\cal L}(G)$, and $U,V,N\in{\cal L}$ with $N\trianglelefteq\langle U,V\rangle$. If $U$ permutes both with $V\cap UN$ and $VN$, then $U$ permutes with $V$.

Proposition 3.4. Let $G$ be $\sigma$-full and ${\cal L}={\cal L}_{\sigma_{i}}(G)$. Then: (i) ${\cal L}$ is a sublattice of ${\cal L}_{\sigma per}(G)$, and (ii) If ${\cal L}$ is distributive, then $AB=BA$ for all $A,B\in{\cal L}$.

Proof. (i) Let $A,B\in{\cal L}$. The subgroups $A$ and $B$ are $\sigma$-subnormal in $G$ by Lemma 2.2, so $A,B\leq O_{\sigma_{i}}(G)$ by Lemma 2.3(4). Thus $\langle A,B\rangle$ is a $\sigma_{i}$-subgroup of $G$ and this subgroup is $\sigma$-permutable in $G$ by Lemma 2.1. Finally, $A\cap B$ is also a $\sigma_{i}$-subgroup of $G$ and this subgroup is $\sigma$-permutable in $G$ by Theorem B. Thus we have (i).

(ii) Suppose that this assertion is false and let $G$ be a counterexample with $|G|+|A|+|B|$ minimal. Thus $AB\neq BA$ but $A_{1}B_{1}=B_{1}A_{1}$ for all $A_{1},B_{1}\in{\cal L}$ such that $A_{1}\leq A$, $B_{1}\leq B$ and either $A_{1}\neq A$ or $B_{1}\neq B$. Let $V=\langle A,B\rangle O^{\sigma_{i}}(G)$ and $R=\langle A,B\rangle\cap O^{\sigma_{i}}(G)$. Then $V$ is $\sigma$-subnormal in $G$.

(1) ${\cal L}_{\sigma_{i}}(V)$ is a sublattice of ${\cal L}$.

Indeed, let $H\in{\cal L}_{\sigma_{i}}(V)$. Then $H$ is $\sigma$-subnormal in $V$ by Lemma 2.2 and so, because of Lemma 2.3(4), $H\leq O_{\sigma_{i}}(V)\leq O_{\sigma_{i}}(G)$. Therefore $H$ permutes with each Hall $\sigma_{i}$-subgroup of $G$. On the other hand, each Hall $\sigma_{j}$-subgroup $W$ of $G$, where $j\neq i$, is contained in $V$ by Lemma 2.3(5) since $|W|$ divides $|V|$, so $HW=WH$. Hence $H\in{\cal L}_{\sigma}(G)$, which implies (1).

(2) $V=G,$ so $\langle A,B\rangle\trianglelefteq G$.

Claim (1) implies that the hypothesis holds for ${\cal L}_{\sigma_{i}}(V)$ and so in the case when $V\neq G$ the choice of $G$ implies that $AB=BA$. Thus $G=\langle A,B\rangle O^{\sigma_{i}}(G)$. Therefore, since $O^{\sigma_{i}}(G)\leq N_{G}(\langle A,B\rangle)$ by Lemma 2.4(1), $\langle A,B\rangle$ is normal in $G$.

(3) $R=1$.

Assume that $R=\langle A,B\rangle\cap O^{\sigma_{i}}(G)\neq 1$. First we show that $BRA=\langle A,B\rangle R$. Indeed, let $H/R$ be a $\sigma_{i}$-subgroup of $G/R$. Then $H$ is a $\sigma_{i}$-group since $\langle A,B\rangle\leq O_{\sigma_{i}}(G)$. Moreover, Lemma 3.2(1)(2) implies that $H/R$ is $\sigma$-permutable in $G/R$ if and only if $H$ is $\sigma$-permutable in $G$. Therefore the lattice ${\cal L}_{\sigma_{i}}(G/R)$ is isomorphic to the interval $[G/R]$ in the distributive lattice ${\cal L}$. Therefore, by the minimality of $G$, $(AR/R)(BR/R)=(BR/R)(AR/R)$ by Lemma 3.2(2) and so $BRA=\langle A,B\rangle R$.

Now we show that $BRA=BR$. Assume that this is false. Then $A\cap BR<A$. But Theorem B implies that $A\cap BR$ is $\sigma$-permutable in $G$, so the minimality of $|G|+|A|+|B|$ implies that $B$ permutes with $A\cap BR$. Also, $B$ permutes with $RA$ since $B(RA)=\langle A,B\rangle R$, so $AB=BA$ by Lemma 3.3, Part (i) and Theorem B. This contradiction shows that $A\leq BR$, so $BRA=BR$. But $R\leq O^{\sigma_{i}}(G)\leq N_{G}(B)$ by Lemma 2.4(1), hence $B$ is normal in $BR$ and since $A\leq BR$ it follows that $AB=BA$. This contradiction shows that we have (3).

Final contradiction. Claims (2) and (3) imply that

[TABLE]

so every subgroup $H$ of $\langle A,B\rangle$ is $O^{\sigma_{i}}(G)$-invariant. It follows that every subgroup of $\langle A,B\rangle$ is $\sigma$-permutable in $G$ by Lemma 2.4(3) and Proposition 3.1. Hence ${\cal L}(\langle A,B\rangle)$ is a sublattice of the distributive lattice ${\cal L}.$ Thus $\langle A,B\rangle$ is cyclic by the Ore theorem by [2, Theorem 1.2.3], so $AB=BA$, a contradiction. The proposition is proved.

Corollary 3.5. If the lattice ${\cal L}={\cal L}_{\sigma}(G)$ is distributive, then every two members $A$ and $B$ of ${\cal L}$ are permutable.

Proof. Suppose that this corollary is false and let $G$ be a counterexample with $|G|+|A|+|B|$ minimal.

Let $R$ be a minimal normal subgroup of $G$. Lemma 3.2 implies that ${\cal L}_{\sigma}(G/R)$ is isomorphic to the interval $[G/R]$ in the modular lattice ${\cal L}_{\sigma per}(G)$. Therefore, Lemma 3.2(2) and the minimality of $G$ imply that $(AR/R)(BR/R)=(BR/R)(AR/R)$. It follows that $RAB=\langle A,B\rangle R$ is a subgroup of $G$, so $A_{G}=1$ and $B_{G}=1$. Hence, because of Lemma 2.4(2), $A$ and $B$ are $\sigma$-nilpotent. The minimality of $|G|+|A|+|B|$ implies that for some $i$ we have $A,B\leq O_{\sigma_{i}}(G)$ and so $A,B\in{\cal L}_{\sigma_{i}}(G)$. But ${\cal L}_{\sigma_{i}}(G)$ is a sublattice of the distributive lattice ${\cal L}_{\sigma}(G)$ by Proposition 3.4(i). Therefore $AB=BA$ by Proposition 3.4(ii), a contradiction. The corollary is proved.

Lemma 3.6 (See [12, p. 59]). A modular lattice $\cal L$ is distributive if and only if $\cal L$ has no distinct elements $a,b$ and $c$ such that $a\vee b=a\vee c=b\vee c$ and $a\wedge b=a\wedge c=b\wedge c$.

Lemma 3.7 (See [2, Theorem 1.6.2]). Let $G=A\times B$, $f:A\to B$ is an isomorphism and $C=\{aa^{f}\ |\ \ a\in A\}$. Then $G=AC=BC$ and $A\cap C=1=B\cap C$.

**Lemma 3.8 ** (See Corollary 2.4 and Lemma 2.5 in [4]). The class of all $\sigma$-nilpotent groups ${\mathfrak{N}}_{\sigma}$ is closed under taking products of normal subgroups, homomorphic images and subgroups.

In view of Proposition 2.2.8 in [13], we get from Lemma 3.8 the following

Lemma 3.9. If $N$ is a normal subgroup of $G$, then $(G/N)^{{\mathfrak{N}}_{\sigma}}=G^{{\mathfrak{N}}_{\sigma}}N/N.$

**Proof of Theorem A. ** Necessity. First note that every two members of ${\cal L}$ are permutable by Theorem B. Moreover, since the lattice ${\cal L}_{n}(G)$ is a sublattice of the lattice $\cal L$, it is distributive. Since $G/D=G/G^{{\mathfrak{N}}_{\sigma}}$ is $\sigma$-nilpotent by Lemmas 3.8 and 3.9, every subgroup $E$ of $G$ satisfying $D\leq E\leq G$ is $\sigma$-permutable in $G$ by Lemma 3.2(1) and Remark 1.2(ii). Hence ${\cal L}(G/D)={\cal L}_{\sigma per}(G/D)$ is distributive and so $G/D$ is cyclic by the Ore theorem [2, Theorem 1.2.3]. It is clear also that $D$ is a meet-distributive element of ${\cal L}$. Thus Conditions (i)-(iii) hold on $G$.

Now we show that Condition (iv) holds on $G$. First note that since, in view of Lemma 3.2, the lattice ${\cal L}_{\sigma per}(G/R)$ is isomorphic to the interval $[G/R]$ in the distributive lattice ${\cal L}$, it is enough to consider the case when $\bar{G}=G$ and $\bar{K}=K,\bar{H}=H,\bar{L}=L\in{\cal L}$.

Suppose that $H\neq L$. Then $H\neq K$. Let $K<H_{0}\leq H$, where $H_{0}$ covers $K$ in $\cal L$, and let $L_{0}/K=(H_{0}/K)^{f}$, where $f:H/K\to L/K$ is a $O^{\sigma_{i}}(G)$-isomorphism. For $g\in O^{\sigma_{i}}(G)$ and $l_{0}K=(hK)^{f}\in L_{0}/K$, where $h\in H$, we have

[TABLE]

where $h_{0}\in H_{0}$ since $H_{0}$ is $O^{\sigma_{i}}(G)$-invariant by Lemma 2.4(1). Hence $(l_{0}K)^{g}\in L_{0}/K$. It follows that $L_{0}$ is $O^{\sigma_{i}}(G)$-invariant and so $L_{0}$ covers $K$ in $\cal L$ since the inverse map $f^{-1}:L/K\to H/K$ is a $O^{\sigma_{i}}(G)$-isomorphism too.

First assume that $H_{0}\neq L_{0}$ and let $E_{0}/K=\{hK(hK)^{f}|hK\in H_{0}/K\}$. Then

[TABLE]

Indeed, if $H_{0}^{x}\not=H_{0}$ for some $x\in L_{0}$, then (i) and the fact that $H_{0}$ and $L_{0}$ cover $K$ in ${\cal L}$ would imply that $\{K;H_{0};H_{0}^{x};L_{0};H_{0}L_{0}\}$ would be a diamond in the distributive lattice $\cal L$, contradicting Lemma 3.6. Hence, by Lemma 3.7, $E_{0}/K$ is a subgroup of $(H_{0}/K)\times(L_{0}/K)$ and we have

[TABLE]

Note that if $g\in O^{\sigma_{i}}(G)$ and $hK(hK)^{f}\in E_{0}/K$, then

[TABLE]

since $f_{H_{0}/K}$ is a $O^{\sigma_{i}}(G)$-isomorphism from $H_{0}/K$ onto $L_{0}/K=(H_{0}/K)^{f}$. Hence $E_{0}/K$ is $O^{\sigma_{i}}(G)$-invariant, so $O^{\sigma_{i}}(G)\leq N_{G}(E_{0})$. Therefore $H_{0}$, $L_{0}$ and $E_{0}$ are distinct elements of $\cal L$ such that $H_{0}\cap L_{0}=H_{0}\cap E_{0}=L_{0}\cap E_{0}=K$ and $H_{0}L_{0}=H_{0}E_{0}=L_{0}E_{0}$, which is impossible by Lemma 3.6 since $H_{0}L_{0}$ is a $\sigma$-permutable subgroup of $G$. Therefore $H_{0}=L_{0}$. Now $f$ induces a $O^{\sigma_{i}}(G)$-isomorphism $f^{\prime}:H/H_{0}\to L/H_{0}$ and an obvious induction yields that $H=L$. Hence we have (iv).

Sufficiency. This follows from the following

Proposition 3.10. Let $D=G^{{\mathfrak{N}}_{\sigma}}$ and ${\cal L}={\cal L}_{\sigma per}(G)$. Suppose that the following conditions hold:

(i) Every two members of ${\cal L}$ are permutable.

(ii) The lattice ${\cal L}_{n}(G)$ of all normal subgroups of $G$ is distributive.

(iii) $G/D$ is cyclic and $D$ is a meet-distributive element of ${\cal L}$.

(iv) In every factor group $\bar{G}=G/R$, any two $O^{\sigma_{i}}(\bar{G})$-isomorphic sections $\bar{H}/\bar{K}$ and $\bar{L}/\bar{K}$, where $\bar{K},\bar{H},\bar{L}\in{\cal L}_{\sigma_{i}}(\bar{G})$ (for some $i$) and the subgroups $\bar{H}$ and $\bar{L}$ cover $\bar{K}$ in ${\cal L}_{\sigma per}(\bar{G})$, coincide.

Then $\cal L$ is distributive.

Proof. Suppose that this is false and let $G$ be a counterexample of minimal order.

First note that if $A,B,C\in{\cal L}_{\sigma per}(G)$ and $A\leq C$, then

[TABLE]

by Condition (i), so the lattice ${\cal L}_{\sigma}(G)$ is modular. Hence, by Lemma 3.6, there are distinct $\sigma$-permutable subgroups $A$, $B$ and $C$ of $G$ such that for some $\sigma$-permutable subgroups $E$ and $T$ of $G$ we have $E=A\cap B=A\cap C=B\cap C$ and $T=AB=AC=BC$.

(1) The lattice ${\cal L}_{\sigma per}(G/R)$ is distributive for each non-identity normal subgroup $R$ of $G$.

In view of the choice of $G$, it is enough to show that Conditions (i), (ii), (iii) and (iv) hold for $G/R$.

Let $K/R,H/R\in{\cal L}_{\sigma per}(G/R)$. Then $K,H\in{\cal L}(G)$ by Lemma 3.2(1) and so $KH=HK$ by Condition (i), which implies that $(K/R)(H/R)=(H/R)(K/R)$. It is clear also that the lattice ${\cal L}_{n}(G/R)$ is isomorphic to some sublattice of the lattice ${\cal L}_{n}(G)$, so ${\cal L}_{n}(G/R)$ is distributive. Thus Conditions (i) and (ii) hold on $G/R$.

By Lemma 3.9 we have

[TABLE]

Thus

[TABLE]

is cyclic by Condition (iii). Conditions (i) and (iii) imply that

[TABLE]

since $D\cap K$ and $D\cap H$ are $\sigma$-permutable in $G$ by Theorem B, so

[TABLE]

[TABLE]

[TABLE]

Hence $(G/R)^{{\mathfrak{N}}_{\sigma}}$ is a meet-distributive element of ${\cal L}_{\sigma per}(G/R)$. Thus Condition (iii) hold on $G/R$. Finally, Condition (iv), evidently, hold on $G/R$. Thus we have (1).

(2) $E_{G}=1$ (In view of Lemma 3.2, this follows from Claim (1), Lemma 3.6 and the choice of $G$).

(3) $A_{G}B_{G}\cap A_{G}C_{G}\cap B_{G}C_{G}=1$.

Since $A\cap B=E$, we have $B_{G}\cap A_{G}\leq E_{G}=1$ by Claim (2). Similarly, $B_{G}\cap C_{G}=1$ and $A_{G}\cap C_{G}=1$. Therefore

[TABLE]

[TABLE]

[TABLE]

by Claim (ii).

(4) The subgroup $T$ is $\sigma$-nilpotent.

Note that

[TABLE]

where

[TABLE]

[TABLE]

and

[TABLE]

are $\sigma$-nilpotent by Lemma 2.4(2). The subgroups $AA_{G}B_{G}/A_{G}B_{G}$ and $BA_{G}B_{G}/A_{G}B_{G}$ are $\sigma$-subnormal in $G/A_{G}B_{G}$ by Lemmas 2.2 and 3.2(2). Hence $T/A_{G}B_{G}$ is $\sigma$-nilpotent by Lemma 2.3(6). Similarly, $T/A_{G}C_{G}$ and $T/C_{G}B_{G}$ are $\sigma$-nilpotent. Hence from Claim (3) it follows that $T\simeq T/(A_{G}B_{G}\cap A_{G}C_{G}\cap B_{G}C_{G})$ is $\sigma$-nilpotent by Lemma 3.8.

(5) For some prime $i$, there are distinct $\sigma_{i}$-subgroups $A_{i},B_{i},C_{i}\in{\cal L}$ such that $H_{i}=A_{i}B_{i}=A_{i}C_{i}=B_{i}C_{i}$ and $K_{i}=A_{i}\cap B_{i}=A_{i}\cap C_{i}=B_{i}\cap C_{i}$ are $\sigma$-permutable subgroups of $G$.

Let $\sigma_{i}\in\sigma(T)$, that is, $\sigma_{i}\cap\pi(T)\neq\emptyset$. Then, by Claim (4), $H_{i}=O_{\sigma_{i}}(T)$ is the Hall $\sigma_{i}$-subgroup of $T$ and $A_{i}=O_{\sigma_{i}}(A)$, $B_{i}=O_{\sigma_{i}}(B)$ and $C_{i}=O_{\sigma_{i}}(C)$ are the Hall $\sigma_{i}$-subgroups of $A$, $B$ and $C$, respectively. Hence $H_{i}=A_{i}B_{i}=A_{i}C_{i}=B_{i}C_{i}$. Moreover, $A_{i}$, $B_{i}$ and $C_{i}$ are $\sigma$-permutable in $G$ by Theorem C. It is clear also that $K_{i}=A_{i}\cap B_{i}=A_{i}\cap C_{i}=B_{i}\cap C_{i}=O_{\sigma_{i}}(E)$.

Suppose that $A_{i}=B_{i}$. Then $H_{i}=A_{i}B_{i}=A_{i}=B_{i}=K_{i}\leq C_{i}\leq H_{i}$. Hence $A_{i}=B_{i}=C_{i}$. Therefore, since $A\neq B\neq C$ and $A\neq C$, there is $\sigma_{i}\in\sigma(T)$ such that $A_{i}\neq B_{i}\neq C_{i}$ and $A_{i}\neq C_{i}$. Finally, $H_{i}$ and $K_{i}$ are $\sigma$-permutable subgroups of $G$ by Condition (i), so we have (5).

(6) There are distinct $\sigma_{i}$-subgroups $A_{0},B_{0},C_{0}\in{\cal L}$ such that $H_{0}=A_{0}B_{0}=A_{0}C_{0}=B_{0}C_{0}$ and $K_{0}=A_{0}\cap B_{0}=A_{0}\cap C_{0}=B_{0}\cap C_{0}$ are $\sigma$-permutable subgroups of $G$ and $A_{0},B_{0},C_{0}$ are normal subgroups of $O^{\sigma_{i}}(G)$.

Let $A_{0}=A_{i}\cap D$, $B_{0}=B_{i}\cap D$ and $C_{0}=C_{i}\cap D$. Then $A_{0}$, $B_{0}$ and $C_{0}$ are $\sigma$-permutable $\sigma_{i}$-subgroups of $G$ by Claim (5) and Theorem B. Moreover, Claim (5) implies that

[TABLE]

Since $D$ is a meet-distributive element of ${\cal L}$ by Condition (iii),

[TABLE]

Now we show that $A_{0},B_{0},C_{0}$ are distinct elements of ${\cal L}$. First note that

[TABLE]

so $|A_{i}:K_{i}|=|B_{i}:K_{i}|=|C_{i}:K_{i}|$. Hence $|A_{i}|=|B_{i}|=|C_{i}|.$ Suppose that $A_{0}=B_{0}$. Then

[TABLE]

Hence $K_{i}D\cap H_{i}=K_{i}(D\cap H_{i})=K_{i}$ is normal in $H_{i}$ and

[TABLE]

is cyclic since $G/D$ is cyclic by Condition (iii). On the other hand, $H_{i}/K_{i}=(A_{i}/K_{i})(B_{i}/K_{i})$, where $|A_{i}/K_{i}|=|B_{i}/K_{i}|$, so $A_{i}/K_{i}=B_{i}/K_{i}=1$, which implies that $A_{i}=B_{i}$. This contradiction shows that $A_{0}\neq B_{0}$. Similarly, one can show that $A_{0}\neq C_{0}$ and $B_{0}\neq C_{0}$. Finally, $A_{0},B_{0},C_{0}$ are normal subgroups of $O^{\sigma_{i}}(G)$ by Lemma 2.4(1), and Claim (5) and Theorem B imply that $K_{0}$ and $H_{0}$ are $\sigma$-permutable in $G$.

(7) $A_{0}/K_{0}$ and $B_{0}/K_{0}$ are $O^{\sigma_{i}}(G)$-isomorphic.

From Claim (6) we get that

[TABLE]

Therefore

[TABLE]

and

[TABLE]

are $O^{\sigma_{i}}(G)$-isomorphisms by Lemma 2.4(1). Hence we have (7).

Final contradiction. Let $f:A_{0}/K_{0}\to B_{0}/K_{0}$ be a $O^{\sigma_{i}}(G)$-isomorphism. Let $K_{0}<X\leq A_{0}$, where $X$ covers $K_{0}$ in ${\cal L}$. Then $X/K_{0}$ is a chief factor of $O^{\sigma_{i}}(G)$ by Lemma 2.4(3) and Proposition 3.1, so $L/K_{0}=f(A_{0}/K_{0})$ is also a chief factor of $O^{p}(G)$. Hence $L$ covers $K_{0}$ in ${\cal L}$ by Lemma 2.4(1). Now $f$ induces a $O^{\sigma_{i}}(G)$-isomorphism from $X/K_{0}$ onto $L/K_{0}$ and so $L=T$ by Condition (iv). Hence $K_{0}<A_{0}\cap B_{0}$, contrary to (6).

The proposition is proved.