On Two Classes of Sublattices of the Subgroup Lattice of a Finite Group

Abstract

Let G denote a finite group; $\sigma =\{{\sigma }_i\mid i\in I\subseteq \left\{0\right\}\cup \mathbb{N}\}$ be some partition of the set of all primes $\mathbb{P}$, where $0\in I$; $\mathfrak{I}$ be a class of finite ${\sigma }_0$-groups that is closed under extensions, epimorphic images, and subgroups and contains all finite soluble ${\sigma }_0$-groups. We say that a subgroup A of G is ${\sigma }_{\mathfrak{I}}$-subnormal in G if there is a subgroup chain $A=A_0\le A_1\le \cdots \le A_t=G$ such that either $A_{i-1}⊴A_i$, $A_i/{\left(A_{i-1}\right)}_{A_i}\in \mathfrak{I}$, or $A_i/{\left(A_{i-1}\right)}_{A_i}$ is a ${\sigma }_j$-group for some $j\ne 0$. In this paper, we prove that the set ${\mathcal{L}}_{{\sigma }_{\mathfrak{I}}}\left(G\right)$ of all ${\sigma }_{\mathfrak{I}}$-subnormal subgroups of G and some proper subsets of ${\mathcal{L}}_{{\sigma }_{\mathfrak{I}}}\left(G\right)$ form sublattices of the lattice of all subgroups of G.

1. Introduction

Throughout this paper, all groups are finite, and G always denotes a finite group; $\mathcal{L}\left(G\right)$ is the lattice of all subgroups of G.

One of the organizing ideas of group theory is the idea of studying a group G depending on the existence of a subgroup system $\mathcal{L}$ with desired properties. Such an approach is the most effective in the case where $\mathcal{L}$ forms a sublattice of $\mathcal{L}\left(G\right)$. This circumstance makes the general problem of finding new sublattices in $\mathcal{L}\left(G\right)$ important and interesting.

The work of Wielandt in [1] stands as one of the first results in this direction, where he proved that the set ${\mathcal{L}}_{sn}\left(G\right)$ of all subnormal subgroups of G forms a sublattice of $\mathcal{L}\left(G\right)$. Later, Kegel [2] found an original generalization of the lattice ${\mathcal{L}}_{sn}\left(G\right)$. Let $\mathfrak{F}$ be a class of groups. A subgroup H of G is called K-$\mathfrak{F}$-subnormal in G [3] (Definition 6.1.4) if there is a subgroup chain

$$H=H_0\le H_1\le \cdots \le H_t=G$$

such that either $H_{i-1}⊴H_i$ or $H_i/{\left(H_{i-1}\right)}_{H_i}$ belongs to $\mathfrak{F}$ for all $i=1,\dots ,t$. Kegel proved [2] that if the class $\mathfrak{F}$ is closed under extensions, epimorphic images, and subgroups, then the set ${\mathcal{L}}_{\mathfrak{F}sn}\left(G\right)$ of all K-$\mathfrak{F}$-subnormal subgroups of G is a sublattice of $\mathcal{L}\left(G\right)$. Subsequently, this result has been generalized on the basis of the methods of formation theory (see, for example, refs. [4,5] and Chapter 6 in [3]).

On the other hand, our focus shifts to the permutability of subgroups. Let A, $B\le G$. We say that A permutes with B if $AB=BA$. A subgroup A is called quasinormal (Ore [6]) (respectively, π-permutable (Kegel [7])) in G if A permutes with all subgroups (respectively, with all Sylow p-subgroups for all $p\in \pi$) of G. The subgroup A is called S-permutable in G [8,9] if A permutes with all Sylow subgroups of G. In general, the set of all quasinormal subgroups of G is not a sublattice of $\mathcal{L}\left(G\right)$ (see [9] (Example 1.2.20)). Nevertheless, Kegel proved [7] that the sets of all subnormal $\pi$-permutable and all S-permutable subgroups form sublattices in $\mathcal{L}\left(G\right)$, respectively.

Let $\sigma =\{{\sigma }_i\mid i\in I\subseteq \left\{0\right\}\cup N\}$ be some partition of $\mathbb{P}$, where $0\in I$. $\mathfrak{I}$ denotes a class of ${\sigma }_0$-groups that is closed under extensions, epimorphic images, and subgroups, and which contains all soluble ${\sigma }_0$-groups. Recall that a σ-property of a group [10,11] is the property that depends on $\sigma$ and which does not imply any restrictions on $\sigma$.

In this paper, we obtain some generalizations of the above-mentioned results in [1,2,7] and some other results on the sublattices of $\mathcal{L}\left(G\right)$ in the theory of $\sigma$-properties of a group.

Before continuing, recall some basic concepts of the theory of $\sigma$-properties of a group. A group G is said to be [11] (i) ${\sigma }_{\mathfrak{I}}$-primary provided G is a ${\sigma }_i$-group for some $i\in I$ and $G\in \mathfrak{I}$ if $i=0$; and (ii) ${\sigma }_{\mathfrak{I}}$-nilpotent if G is the direct product of ${\sigma }_{\mathfrak{I}}$-primary groups. A subgroup A of G is ${\sigma }_{\mathfrak{I}}$-subnormal in G if there is a subgroup chain

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

such that either $A_{i-1}⊴A_i$ or $A_i/{\left(A_{i-1}\right)}_{A_i}$ is ${\sigma }_{\mathfrak{I}}$-primary for all $i=1,\dots ,t$.

In the sequel, we always omit the symbol $\mathfrak{I}$ in all definitions and notations in the case where $\mathfrak{I}$ is the class of all ${\sigma }_0$-groups. For example, G is σ-primary if G is a ${\sigma }_i$-group for some $i\in I$.

To state our results, we also need the following: let $\mathbf{A}$ be a group of operators of G ([12], p. 15) containing $Inn\left(G\right)$. A subgroup H of G is $\mathbf{A}$-normal if $H^a=H$ for all $a\in \mathbf{A}$. We use $H^{\mathbf{A}}$ to denote the $\mathbf{A}$-normal closure of H in G (i.e., the intersection of all $\mathbf{A}$-normal subgroups of G containing H), and $H_{\mathbf{A}}$ is the $\mathbf{A}$-core of H in G (i.e., the subgroup generated by all $\mathbf{A}$-normal subgroups of G contained in H).

Recall that a subgroup H of G is strongly σ-subnormal in G [13] if $H^G/H_G$ is $\sigma$-nilpotent. This leads to the following:

Definition 1.

We say that a subgroup H of G is $\mathbf{A}$-strongly ${\sigma }_{\mathfrak{I}}$-subnormal in G if $H^{\mathbf{A}}/H_{\mathbf{A}}$ is ${\sigma }_{\mathfrak{I}}$-nilpotent.

Remark 1.

Since $H_{\mathbf{A}}\le H_G\le H\le H^G\le H^{\mathbf{A}}$, we see that $H^G/H_{\mathbf{A}}\le H^{\mathbf{A}}/H_{\mathbf{A}}$ and $H^G/H_G$ represent a factor group of $H^G/H_{\mathbf{A}}$. By Proposition 2 (ii) below, and in noting that any ${\sigma }_{\mathfrak{I}}$-nilpotent group is σ-nilpotent, we have that every $\mathbf{A}$-strongly ${\sigma }_{\mathfrak{I}}$-subnormal subgroup is also strongly σ-subnormal. The following example shows that the converse is not true. Therefore, the set of all $\mathbf{A}$-strongly ${\sigma }_{\mathfrak{I}}$-subnormal subgroups is a proper subset of the set of all strongly σ-subnormal subgroups.

Indeed, let ${\sigma }_0=\{2,3,5\}$, and let $\mathfrak{I}$ be the class of all ${\sigma }_0$-groups S such that every non-abelian composition factor of S is isomorphic to the alternating group $A_5$. Then the class $\mathfrak{I}$ is closed under extensions, epimorphic images, and subgroups, and it contains all soluble ${\sigma }_0$-groups. Now, let $\sigma =\{{\sigma }_0,{\sigma }_0^{\prime }\}$ and $G=H\times K$, where $H=K=A_6$ is the alternating group of degree 6. Note that $H_G=H=H^G$, and we see that H is strongly σ-subnormal in G. Since G is a ${\sigma }_0$-group and $G\notin \mathfrak{I}$, we have that G is not ${\sigma }_{\mathfrak{I}}$-nilpotent. Take $\mathbf{A}=\mathit{Aut}\left(G\right)$ to be the automorphism group of G. Since H is not characteristic in G, we can deduce that $H_{\mathbf{A}}=1$ and $H^{\mathbf{A}}=G$, and it follows that $H^{\mathbf{A}}/H_{\mathbf{A}}$ is not ${\sigma }_{\mathfrak{I}}$-nilpotent. Therefore, H is not $\mathbf{A}$-strongly ${\sigma }_{\mathfrak{I}}$-subnormal in G.

In the literature, there are several special versions of ${\sigma }_{\mathfrak{I}}$-subnormality.

Remark 2. 

$\left(i\right)$ Let $\mathfrak{F}$ be a class of groups that is closed under extensions, epimorphic images, and subgroups, and that contains all soluble $\pi \left(\mathfrak{F}\right)$-groups, where $\pi \left(\mathfrak{F}\right):={\bigcup }_{G\in \mathfrak{F}}\pi \left(G\right)$. Set ${\sigma }_0=\pi \left(\mathfrak{F}\right)$ and $\mathbb{P}\backslash {\sigma }_0=\{p_1,\dots ,p_n,\dots \}$. In the case where $\sigma =\{{\sigma }_0,\left\{p_1\right\},\dots ,\left\{p_n\right\},\dots \}$ and $\mathfrak{I}=\mathfrak{F}$, we conclude that the concept of K-$\mathfrak{F}$-subnormality is equivalent to that of ${\sigma }_{\mathfrak{I}}$-subnormality.

Let $A\le G$. In fact, if A is K-$\mathfrak{F}$-subnormal in G and $A=A_0\le A_1\le \cdots \le A_t=G$, where either $A_{i-1}⊴A_i$ or $A_i/{\left(A_{i-1}\right)}_{A_i}\in \mathfrak{F}$ for all $i=1,\dots ,t$, then either $A_{i-1}⊴A_i$ or $A_i/{\left(A_{i-1}\right)}_{A_i}$ is ${\sigma }_{\mathfrak{I}}$-primary. Hence, A is ${\sigma }_{\mathfrak{I}}$-subnormal in G.

Conversely, assume that A is ${\sigma }_{\mathfrak{I}}$-subnormal in G and let $A=A_0\le A_1\le \cdots \le A_t=G$, where either $A_{i-1}⊴A_i$, $A_i/{\left(A_{i-1}\right)}_{A_i}\in \mathfrak{I}$, or $A_i/{\left(A_{i-1}\right)}_{A_i}$ is a p-group for some prime p. However, if $A_i/{\left(A_{i-1}\right)}_{A_i}$ is a p-group, then $A_{i-1}$ is subnormal in $A_i$. Consequently, there is a subgroup chain

$$A=B_0\le B_1\le \cdots \le B_n=G$$

such that either $B_{i-1}⊴B_i$ or $B_i/{\left(B_{i-1}\right)}_{B_i}\in \mathfrak{F}$ for all $i=1,\dots ,n$. Therefore, A is K-$\mathfrak{F}$-subnormal in G.

$\left(ii\right)$ Let ${\sigma }_0\in \Pi \subseteq \sigma$ and $\mathfrak{X}$ be a class of ${\sigma }_0$-groups closed under extensions, epimorphic images, and subgroups. Following [14], G is said to be ${\Pi }_{\mathfrak{X}}$-primary, provided either $G\in \mathfrak{X}$ or G is a ${\sigma }_i$-group for some ${\sigma }_i\in \Pi \backslash \left\{{\sigma }_0\right\}$; a subgroup A of G is said to be ${\Pi }_{\mathfrak{X}}$-subnormal in G if there is a subgroup chain

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

such that either $A_{i-1}⊴A_i$ or $A_i/{\left(A_{i-1}\right)}_{A_i}$ is ${\Pi }_{\mathfrak{X}}$-primary for all $i=1,\dots ,t$. Let ${\sigma }_0^{*}=\pi \left(\mathfrak{X}\right)$ and $\{p_1,\dots ,p_n,\dots \}$ be the complement set of ${\sigma }_0^{*}\cup \left({\bigcup }_{{\sigma }_i\in \Pi \backslash \left\{{\sigma }_0\right\}}{\sigma }_i\right)$ in $\mathbb{P}$. Consider the new partition $\sigma =(\Pi \backslash \left\{{\sigma }_0\right\})\cup \left\{{\sigma }_0^{*}\right\}\cup (\left\{p_1\right\},\dots ,\left\{p_n\right\},\dots \})$. By an argument similar to that in Part (i), we obtain that the notions of ${\Pi }_{\mathfrak{X}}$-subnormality and ${\sigma }_{\mathfrak{I}}$-subnormality are equivalent.

Our attention now turns to several special partitions of $\mathbb{P}$ that frequently arise in mathematical practice.

Example 1. 

(i) In the case where $\sigma ={\sigma }^1:=\{\left\{2\right\},\left\{3\right\},\left\{5\right\},\dots \}$, the concepts of σ-primality, σ-nilpotence, and σ-subnormality reduce to the familiar concepts of primality, nilpotence, and subnormality, respectively.

(ii) In the case where $\sigma ={\sigma }^{\pi }:=\{\pi ,{\pi }^{\prime }\}$, G is σ-primary if and only if G is either a π-group or a ${\pi }^{\prime }$-group. The concept of σ-nilpotence is equivalent to that of π-decomposability, namely $G=O_{\pi }\left(G\right)\times O_{{\pi }^{\prime }}\left(G\right)$. Moreover, the σ-subnormality of A can be stated as $\pi , {\pi }^{\prime }$-subnormality, that is, G has a subgroup chain

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

satisfying $A_{i-1}⊴A_i$ or $A_i/{\left(A_{i-1}\right)}_{A_i}$ is either a π-group or a ${\pi }^{\prime }$-group for all $i=1,\dots ,n$.

(iii) Let $\sigma ={\sigma }^{1,{\pi }^{\prime }}:=\{\left\{p_1\right\},\dots ,\left\{p_n\right\},{\pi }^{\prime }\}$, where $\pi =\{p_1,\dots ,p_n\}$. Then, G is σ-primary if and only if G is either a ${\pi }^{\prime }$-group or a p-group for some $p\in \pi$. The concept of σ-nilpotence amounts to that of π-specialty [15], namely $G=O_{p_1}\left(G\right)\times \cdots \times O_{p_n}\left(G\right)\times O_{{\pi }^{\prime }}\left(G\right)$. Furthermore, the σ-subnormality of A becomes${\sigma }^{1,{\pi }^{\prime }}$-subnormality, i.e., G has a subgroup chain $A=A_0\le A_1\le \cdots \le A_n=G$ with $A_{i-1}⊴A_i$, or $A_i/{\left(A_{i-1}\right)}_{A_i}$ is either a p-group for some $p\in \pi$ or a ${\pi }^{\prime }$-group for all $i=1,\dots ,n$.

It is worth noting that A is ${\sigma }^{1,{\pi }^{\prime }}$-subnormal in G if and only if A is a K-$\mathfrak{I}$-subnormal subgroup, where $\mathfrak{I}$ is the class of all ${\pi }^{\prime }$-groups.

Let $\sigma \left(n\right)=\{{\sigma }_i\mid {\sigma }_i\cap \pi \left(n\right)\ne \varnothing \}$ and $\sigma \left(G\right)=\sigma \left(\right|G\left|\right)$. If $\sigma \left(H\right)\subseteq \Pi$, then H is called a $\Pi$-subgroup of G. A $\Pi$-subgroup H of G is called (i) a maximal Π-subgroup of G if $H=V$ for every $\Pi$-subgroup V of G such that $H\le V$; or (ii) a Hall Π-subgroup of G if $\sigma \left(\right|G:H\left|\right)\cap \Pi =\varnothing$.

Following [16], a subgroup H of G is (i) a Π-semiprojector of G if $HN/N$ is a maximal $\Pi$-subgroup of $G/N$ for every normal subgroup N of G; or (ii) a Π-projector of G if H is a $\Pi$-semiprojector in every subgroup of G containing H. One can observe that the notion of a $\Pi$-projector generalizes that of a Hall subgroup. Since the class of all $\Pi$-groups is a saturated formation, every group G possesses a $\Pi$-semiprojector by [16] (Theorem 11.15) or [3] (Theorem 2.3.13).

In this paper, we prove the following result:

Theorem 1.

Let H be a subgroup of G.

(i) If H is a Π-semiprojector of G, then the set of all $\mathbf{A}$-strongly ${\sigma }_{\mathfrak{I}}$-subnormal subgroups of G which permute with H forms a sublattice in $\mathcal{L}\left(G\right)$.

(ii) If H is a Π-projector of G, then the set of all ${\sigma }_{\mathfrak{I}}$-subnormal subgroups of G which permute with H forms a sublattice in $\mathcal{L}\left(G\right)$.

This theorem generalizes the aforementioned results of Wielandt and Kegel. Additionally, the following known result is a special case of Theorem 1, where $\mathbf{A}=\mathrm{Aut}\left(G\right)$ and $\mathfrak{I}$ is the class of all ${\sigma }_0$-groups.

Corollary 1

([17] (Theorem B)). Let H be a subgroup of G.

(i) If a subgroup H is a Π-semiprojector of G, then the set of all strongly σ-subnormal subgroups of G which permute with H forms a sublattice in $\mathcal{L}\left(G\right)$.

(ii) If H is a Π-projector of G, then the set of all σ-subnormal subgroups of G which permute with H forms a sublattice in $\mathcal{L}\left(G\right)$.

Furthermore, by taking $H=G$ and $\mathbf{A}=\mathrm{Inn}\left(G\right)$ in Theorem 1, we can see that the set ${\mathcal{L}}_{{\sigma }_{\mathfrak{I}}}\left(G\right)$ of all ${\sigma }_{\mathfrak{I}}$-subnormal subgroups of G forms sublattices of $\mathcal{L}\left(G\right)$, and we obtain the following three known results.

Corollary 2

([13]). The set of all strongly σ-subnormal subgroups of G forms a sublattice in $\mathcal{L}\left(G\right)$.

Corollary 3

([10] (Lemma 2.6(3))). The set of all σ-subnormal subgroups of G forms a sublattice in the lattice $\mathcal{L}\left(G\right)$.

Corollary 4

([14], p. 234). The set of all ${\Pi }_{\mathfrak{I}}$-subnormal subgroups of G forms a sublattice in $\mathcal{L}\left(G\right)$.

Since a Hall $\Pi$-subgroup H of G is evidently a $\Pi$-projector of G, from Theorem 1 (ii), we therefore also get the following results:

Corollary 5.

Let G be a group. If H is a Hall Π-subgroup of G, then the set of all ${\sigma }_{\mathfrak{I}}$-subnormal subgroups of G which permute with H forms a sublattice in $\mathcal{L}\left(G\right)$.

Corollary 6.

Let G be a group. If H is a Hall π-subgroup of G, then the set of all subnormal subgroups of G which permute with H forms a sublattice in $\mathcal{L}\left(G\right)$.

Recall that G is said to be a σ-full group of Sylow type [10] if every subgroup E of G is a $D_{{\sigma }_i}$-group for every ${\sigma }_i\in \sigma \left(E\right)$. In view of [10] (Theorem B), if G is a $\sigma$-full group of Sylow type, then every $\sigma$-pemutable subgroup is $\sigma$-subnormal in G. Therefore, since the intersection of any set of sublattices of a lattice is also a sublattice, from Theorem 1(ii), we get the following:

Corollary 7

([10] (Theorem C)). Let G be a σ-full group of Sylow type. Then, the set of all σ-permutable subgroups of G forms a sublattice in $\mathcal{L}\left(G\right)$.

Following [11], if $\mathcal{H}=\{H_1,\dots ,H_t\}$, where $H_j$ is a Hall ${\sigma }_{i_j}$-subgroup of G, and $\{{\sigma }_{i_1},\dots ,{\sigma }_{i_t}\}=\Pi \cap \sigma \left(G\right)$, then we say that $\mathcal{H}$ is a complete Hall Π-set of G; G is Π-full if G has a complete Hall $\Pi$-set. A subgroup A is Π-permutable in G if G has a complete Hall $\Pi$-set $\mathcal{H}$ such that $A{H}^x=H^{x}A$ for all $H\in \mathcal{H}$ and all $x\in G$.

Corollary 8

([18] (Theorem C)). Suppose that G is Π-full. Then, the set of all Π-permutable σ-subnormal subgroups of G forms a sublattice in $\mathcal{L}\left(G\right)$.

2. Proof of Theorem 1

Lemma 1

([12] (Chap. A, Lemma 1.6)). Let $A, B$, and H be subgroups of a group G. If $AH=HA$ and $BH=HB$, then $⟨A,B⟩H=H⟨A,B⟩$.

Lemma 2.

Suppose that H is a Π-semiprojector (respectively, Π-projector) of a group G and $N⊴G$. Then, $HN/N$ is a Π-semiprojector (respectively, Π-projector) of $G/N$.

Proof. 

Let H be a $\Pi$-semiprojector of G and $K/N⊴G/N=\overline{G}$. Then, $HK/K$ is a maximal $\Pi$-subgroup of $G/K$. Since $\overline{H}\overline{K}/\overline{K}\cong HK/K$, and $\overline{G}/\overline{K}\cong G/K$, $\overline{H}\overline{K}/\overline{K}$ is a maximal $\Pi$-subgroup in $\overline{G}/\overline{K}$. Hence, $\overline{H}$ is a $\Pi$-semiprojector of $\overline{G}$, and the remaining claim follows from this. □

Lemma 3.

Let $G\ne 1$ and $\sigma =\{{\sigma }_0,{\sigma }_1,\dots \}$. Then, G is ${\sigma }_{\mathfrak{I}}$-nilpotent if and only if $G=H_0\times H_1\times \cdots \times H_t$, where $H_i$ is a Hall ${\sigma }_i$-subgroup of G for all ${\sigma }_i\in \sigma \left(G\right)$ and $H_0\in \mathfrak{I}$ if $H_0\ne 1$.

Proof. 

Only the necessity needs to be shown. Indeed, assume that $G=A_0\times A_1\times \cdots \times A_n$, where $A_j$ is ${\sigma }_{\mathfrak{I}}$-primary for all $j=0,1,\dots ,n$. Let $H_i$ be the product of all factors $A_k$ such that $A_k$ is a ${\sigma }_i$-group. Then, $H_i$ is a Hall ${\sigma }_i$-subgroup of G for all i and, clearly, $H_0\in \mathfrak{I}$. Therefore, $G=H_0\times H_1\times \cdots \times H_t$. □

We use $O^{\Pi }\left(G\right)$ to denote the subgroup of G generated by all its ${\Pi }^{\prime }$-subgroups, where ${\Pi }^{\prime }=\sigma \backslash \Pi$; $O_{\Pi }\left(G\right)$ is the product of all normal $\Pi$-subgroups of G. We also use $G^{{\sigma }_{\mathfrak{I}}}$ to denote the ${\sigma }_{\mathfrak{I}}$-nilpotent residual of G, i.e., the intersection of all normal subgroups N of G with ${\sigma }_{\mathfrak{I}}$-nilpotent $G/N$.

Proposition 1

(A.N. Skiba [19]). Let A, K, and N be subgroups of G, where A is ${\sigma }_{\mathfrak{I}}$-subnormal and N is normal in G.

(i) $A\cap K$ is ${\sigma }_{\mathfrak{I}}$-subnormal in K;

(ii) $AN/N$ is ${\sigma }_{\mathfrak{I}}$-subnormal in $G/N$;

(iii) If $N\le K$ and $K/N$ is ${\sigma }_{\mathfrak{I}}$-subnormal in $G/N$, then K is ${\sigma }_{\mathfrak{I}}$-subnormal in G;

(iv) If a minimal normal subgroup R of G is not ${\sigma }_{\mathfrak{I}}$-primary, then $R\le N_G\left(A\right)$;

(v) The ${\sigma }_{\mathfrak{I}}$-nilpotent residual $A^{{\mathfrak{N}}_{{\sigma }_{\mathfrak{I}}}}$ of A is subnormal in G;

(vi) If A is ${\sigma }_{\mathfrak{I}}$-nilpotent, then $A^G$ is ${\sigma }_{\mathfrak{I}}$-nilpotent, and $\sigma \left(A\right)=\sigma \left(A^G\right)$;

(vii) If $\sigma \left(\right|G:A\left|\right)\subseteq \Pi$, then $O^{\Pi }\left(A\right)=O^{\Pi }\left(G\right)$.

Recall that a class of groups $\mathfrak{F}$ is called a formation if every homomorphic image of every $\mathfrak{F}$-group is an $\mathfrak{F}$-group, and $G/(M\cap N)\in \mathfrak{F}$ for every two normal subgroups M and N of G such that $G/M,G/N\in \mathfrak{F}$.

A class of groups $\mathfrak{F}$ is said to be saturated if $G\in \mathfrak{F}$ whenever $G/\Phi \left(G\right)\in \mathfrak{F}$; or (normally) hereditary if $H\in \mathfrak{F}$ whenever $G\in \mathfrak{F}$ and H is a (normal) subgroup of G. A class of groups $\mathfrak{F}$ is called a Fitting class if it is normally hereditary and closed under taking products of normal subgroups.

A group G is said to be Π-closed if G has a normal Hall $\Pi$-subgroup.

Proposition 2.

(i) The class $\mathfrak{F}$ of all Π-closed groups is both a hereditary formation and a Fitting class.

(ii) The class ${\mathfrak{N}}_{{\sigma }_{\mathfrak{I}}}$ of all ${\sigma }_{\mathfrak{I}}$-nilpotent groups is both a hereditary formation and a Fitting class.

(iii) If a group G is ${\sigma }_{\mathfrak{I}}$-nilpotent, then every subgroup of G is ${\sigma }_{\mathfrak{I}}$-subnormal in G.

Proof. 

(i) Let $G\in \mathfrak{F}$, where H is the normal Hall $\Pi$-subgroup of G. Suppose that $N⊴G$ and $K\le G$; then, $HN/N$ is the normal Hall $\Pi$-subgroup of $G/N$, and $H\cap K$ is the normal Hall $\Pi$-subgroup of K, so $\mathfrak{F}$ is closed under taking homomorphic images and subgroups.

Now, let $N,M⊴G$, and $G/N,G/M$ are $\Pi$-closed. Then, $(G/N)\times (G/M)$ is $\Pi$-closed, so every subgroup of $(G/N)\times (G/M)$ is also $\Pi$-closed. Therefore, $G/(N\cap M)$ is $\Pi$-closed according to [20] (Chap. I, Lemma 9.6). Hence, $\mathfrak{F}$ is a hereditary formation.

Next, we show that if $G=AB$, where A and B are normal $\Pi$-closed subgroups of G, then G is $\Pi$-closed. Let $A_{\Pi }$ and $B_{\Pi }$ be the normal Hall $\Pi$-subgroups of A and B, respectively. Then, $A_{\Pi }$ is characteristic in A, so $A_{\Pi }$ is normal in G. Similarly, $B_{\Pi }$ is normal in G. Hence, $H=A_{\Pi }{B}_{\Pi }$ is a normal $\Pi$-subgroup of G. Since

$$|G:H|=|AB:A_{\Pi }{B}_{\Pi }|=\left({\displaystyle \frac{\left|A\right|}{|A_{\Pi }|}}\right)\left({\displaystyle \frac{\left|B\right|}{|B_{\Pi }|}}\right)\left({\displaystyle \frac{|A_{\Pi }\cap B_{\Pi }|}{|A\cap B|}}\right)$$

is a ${\Pi }^{\prime }$-number, $A_{\Pi }{B}_{\Pi }$ is a normal Hall $\Pi$-subgroup of G. Hence, $\mathfrak{F}$ is a Fitting class.

(ii) In view of Lemma 2.2, and in noting that $\mathfrak{I}$ is closed under extensions, epimorphic images, and subgroups, this assertion is a corollary of Part (i).

(iii) Assume that a subgroup A of G is not ${\sigma }_{\mathfrak{I}}$-subnormal in G. Then, G is not ${\sigma }_{\mathfrak{I}}$-primary. Moreover, $1<A<G$, and so for some maximal subgroup M of G, we have $A\le M$. In view of Part (ii), M is ${\sigma }_{\mathfrak{I}}$-nilpotent, so A is ${\sigma }_{\mathfrak{I}}$-subnormal in M by induction on $\left|M\right|$.

Now, we show that M is ${\sigma }_{\mathfrak{I}}$-subnormal in G. Assume that $M_G\ne 1$. The hypothesis holds for $G/M_G$ by Part (ii), so $M/M_G$ is ${\sigma }_{\mathfrak{I}}$-subnormal in $G/M_G$ by induction on $|G/M_G|$. Hence, M is ${\sigma }_{\mathfrak{I}}$-subnormal in G by Proposition 1 (iii). Then, A is ${\sigma }_{\mathfrak{I}}$-subnormal in G; this contradicts the initial assumption.

Hence, we may assume that $M_G=1$. Then, G is primitive, and by applying [12] (Chap. A, Theorem 15.2), either G has a unique minimal normal subgroup R, $G=RM$, and $C_G\left(R\right)\le R$ or G has exactly two minimal normal subgroups, R and N, and the following holds: R and N are isomorphic non-abelian groups, $C_G\left(R\right)=N$, and $C_G\left(N\right)=R$. We can conclude that $Z\left(G\right)=1$ in both cases, and so $C_G\left(R\right)<G$.

By Lemma 2.2, $G=H_0\times H_1\times \cdots \times H_t$, where $H_i$ is a Hall ${\sigma }_i$-subgroup of G for all ${\sigma }_i\in \sigma \left(G\right)$. Assume that $R\le H_j$. Then, $K=H_0\cdots H_{j-1}{H}_{j+1}\cdots H_t\le C_G\left(R\right)$. In the former case, since $C_G\left(R\right)\le R$, $K\le H_j$. Thus, $K=1$, and so $G=H_j$ is ${\sigma }_{\mathfrak{I}}$-primary, a contradiction. In the latter case, since $C_G\left(R\right)=N$ and $C_G\left(N\right)=R$, we deduce that $K=N$ and $R=H_j$. However, this is impossible since $R\cong N$. Hence, $\left(iii\right)$ holds. □

Proof of Theorem 1.

$\left(\mathbf{i}\right)$ Let $\mathcal{L}$ be the set of all $\mathbf{A}$-strongly ${\sigma }_{\mathfrak{I}}$-subnormal subgroups of G which permute with H. For any $A,B\in \mathcal{L}$, let $K=⟨A,B⟩$ and $V=A\cap B$. By the hypothesis, we have that $A^{\mathbf{A}}/A_{\mathbf{A}}$ and $B^{\mathbf{A}}/B_{\mathbf{A}}$ are ${\sigma }_{\mathfrak{I}}$-nilpotent.

$\left(a\right)$ First, we show that $K\in \mathcal{L}$. In view of Lemma 1, we have that $KH=HK$. Thus, it is enough to show that K is $\mathbf{A}$-strongly ${\sigma }_{\mathfrak{I}}$-subnormal in G. Since $A^{\mathbf{A}}\le ⟨A$, ${B⟩}^{\mathbf{A}}$, $B^{\mathbf{A}}\le {⟨A,B⟩}^{\mathbf{A}}$, we have $A^{\mathbf{A}}{B}^{\mathbf{A}}\le {⟨A,B⟩}^{\mathbf{A}}$. The reverse containment is clear since $A^{\mathbf{A}}{B}^{\mathbf{A}}$ is $\mathbf{A}$-normal in G. Hence, ${⟨A,B⟩}^{\mathbf{A}}=A^{\mathbf{A}}{B}^{\mathbf{A}}$ and $A_{\mathbf{A}}{B}_{\mathbf{A}}\le {⟨A,B⟩}_{\mathbf{A}}$ in a similar manner. Therefore, ${⟨A,B⟩}^{\mathbf{A}}/{⟨A,B⟩}_{\mathbf{A}}=A^{\mathbf{A}}{B}^{\mathbf{A}}/{⟨A,B⟩}_{\mathbf{A}}$ is a factor group of $A^{\mathbf{A}}{B}^{\mathbf{A}}/A_{\mathbf{A}}{B}_{\mathbf{A}}$

Since

$$A^{\mathbf{A}}\left(A_{\mathbf{A}}{B}_{\mathbf{A}}\right)/A_{\mathbf{A}}{B}_{\mathbf{A}}\cong A^{\mathbf{A}}/(A^{\mathbf{A}}\cap A_{\mathbf{A}}{B}_{\mathbf{A}})=A^{\mathbf{A}}/A_{\mathbf{A}}(A^{\mathbf{A}}\cap B_{\mathbf{A}})$$

is a factor group of $A^{\mathbf{A}}/A_{\mathbf{A}}$, and the class ${\mathfrak{N}}_{{\sigma }_{\mathfrak{I}}}$ is closed under taking homomorphic images by Proposition 2 (ii), we obtain that $A^{\mathbf{A}}\left(A_{\mathbf{A}}{B}_{\mathbf{A}}\right)/A_{\mathbf{A}}{B}_{\mathbf{A}}\in {\mathfrak{N}}_{{\sigma }_{\mathfrak{I}}}$. Similarly, we can also conclude that $B^{\mathbf{A}}\left(A_{\mathbf{A}}{B}_{\mathbf{A}}\right)/A_{\mathbf{A}}{B}_{\mathbf{A}}\in {\mathfrak{N}}_{{\sigma }_{\mathfrak{I}}}$.

Note that

$$A^{\mathbf{A}}{B}^{\mathbf{A}}/A_{\mathbf{A}}{B}_{\mathbf{A}}=(A^{\mathbf{A}}\left(A_{\mathbf{A}}{B}_{\mathbf{A}}\right)/A_{\mathbf{A}}{B}_{\mathbf{A}})(B^{\mathbf{A}}\left(A_{\mathbf{A}}{B}_{\mathbf{A}}\right)/A_{\mathbf{A}}{B}_{\mathbf{A}}),$$

and the class ${\mathfrak{N}}_{{\sigma }_{\mathfrak{I}}}$ is a Fitting class by Proposition 2 (ii); we see that $A^{\mathbf{A}}{B}^{\mathbf{A}}/A_{\mathbf{A}}{B}_{\mathbf{A}}\in {\mathfrak{N}}_{{\sigma }_{\mathfrak{I}}}$. It follows that

$${⟨A,B⟩}^{\mathbf{A}}/{⟨A,B⟩}_{\mathbf{A}}\in {\mathfrak{N}}_{{\sigma }_{\mathfrak{I}}}$$

since the class ${\mathfrak{N}}_{{\sigma }_{\mathfrak{I}}}$ is closed under taking homomorphic images by Proposition 2 (ii). Hence, $⟨A,B⟩$ is $\mathbf{A}$-strongly ${\sigma }_{\mathfrak{I}}$-subnormal in G. Therefore, $K\in \mathcal{L}$.

$\left(b\right)$ Now, we show that $V\in \mathcal{L}$.

(1) First, we prove that V is $\mathbf{A}$-strongly ${\sigma }_{\mathfrak{I}}$-subnormal in G. By an argument similar to the one above, we see that ${(A\cap B)}_{\mathbf{A}}=A_{\mathbf{A}}\cap B_{\mathbf{A}}$ and ${(A\cap B)}^{\mathbf{A}}\le A^{\mathbf{A}}\cap B^{\mathbf{A}}$. Then, ${(A\cap B)}^{\mathbf{A}}/{(A\cap B)}_{\mathbf{A}}\le (A^{\mathbf{A}}\cap B^{\mathbf{A}})/(A_{\mathbf{A}}\cap B_{\mathbf{A}})$. Since

$$(A^{\mathbf{A}}\cap B^{\mathbf{A}})/(A_{\mathbf{A}}\cap B^{\mathbf{A}})=(A^{\mathbf{A}}\cap B^{\mathbf{A}})/(A_{\mathbf{A}}\cap B^{\mathbf{A}}\cap A^{\mathbf{A}})$$

$$\cong A_{\mathbf{A}}(B^{\mathbf{A}}\cap A^{\mathbf{A}})/A_{\mathbf{A}}\le A^{\mathbf{A}}/A_{\mathbf{A}},$$

and the class ${\mathfrak{N}}_{{\sigma }_{\mathfrak{I}}}$ is hereditary by Proposition 2 (ii), we see that $(A^{\mathbf{A}}\cap B^{\mathbf{A}})/(A_{\mathbf{A}}\cap B^{\mathbf{A}})\in {\mathfrak{N}}_{{\sigma }_{\mathfrak{I}}}$. Similarly, we obtain that $(A^{\mathbf{A}}\cap B^{\mathbf{A}})/(A^{\mathbf{A}}\cap B_{\mathbf{A}})\in {\mathfrak{N}}_{{\sigma }_{\mathfrak{I}}}$. Since the class ${\mathfrak{N}}_{{\sigma }_{\mathfrak{I}}}$ is a formation by Proposition 2 (ii), it follows that

$$(A^{\mathbf{A}}\cap B^{\mathbf{A}})/(A_{\mathbf{A}}\cap B_{\mathbf{A}})=(A^{\mathbf{A}}\cap B^{\mathbf{A}})/(A_{\mathbf{A}}\cap B^{\mathbf{A}}\cap A^{\mathbf{A}}\cap B_{\mathbf{A}})\in {\mathfrak{N}}_{{\sigma }_{\mathfrak{I}}}.$$

Note that ${(A\cap B)}^{\mathbf{A}}/{(A\cap B)}_{\mathbf{A}}$ is a subgroup of $(A^{\mathbf{A}}\cap B^{\mathbf{A}})/(A_{\mathbf{A}}\cap B_{\mathbf{A}})$. Therefore, we obtain that

$${(A\cap B)}^{\mathbf{A}}/{(A\cap B)}_{\mathbf{A}}\in {\mathfrak{N}}_{{\sigma }_{\mathfrak{I}}}.$$

Hence, $V=A\cap B$ is $\mathbf{A}$-strongly ${\sigma }_{\mathfrak{I}}$-subnormal in G.

(2) Finally, we show that V is permutable with H. Assume that this is false, and let G be a counterexample of minimal order. Then, G is not a $\Pi$-group, since otherwise we have $H=G$ and so $G=VH=HV$, contrary to our assumption on V.

Assume that $R:={(A\cap B)}_{\mathbf{A}}\ne 1$. Consider the factor group $G/R$. Then,

$${(A/R)}^{\mathbf{A}}/{(A/R)}_{\mathbf{A}}=(A^{\mathbf{A}}/R)/(A_{\mathbf{A}}/R)\cong A^{\mathbf{A}}/A_{\mathbf{A}}$$

is ${\sigma }_{\mathfrak{I}}$-nilpotent, so $A/R$ is $\mathbf{A}$-strongly ${\sigma }_{\mathfrak{I}}$-subnormal in $G/R$ by Proposition 2.3(ii). Similarly, $B/R$ is also $\mathbf{A}$-strongly ${\sigma }_{\mathfrak{I}}$-subnormal in $G/R$. Moreover, we see that $A/R$ and $B/R$ permute with $HR/R$. By Lemma 2, $HR/R$ is a $\Pi$-semiprojector of $G/R$. So the minimal choice of G implies that

$$\left(\right(A/R)\cap (B/R\left)\right)(HR/R)=(HR/R)\left(\right(A/R)\cap (B/R\left)\right).$$

It follows that $(A\cap B)H=H(A\cap B)$, contrary to our assumption on the subgroups $A\cap B$ and H.

Therefore, we may suppose that $R=1$; then, ${(A\cap B)}^{\mathbf{A}}$ is ${\sigma }_{\mathfrak{I}}$-nilpotent by Claim (1). Hence, ${(A\cap B)}^{\mathbf{A}}=V\times W$, where W is a Hall $\Pi$-subgroup of ${(A\cap B)}^{\mathbf{A}}$. Since W is normal in G, we have $W\le H$. By Proposition 2 (ii), $A\cap B$ is also ${\sigma }_{\mathfrak{I}}$-nilpotent. Again, $A\cap B=L\times T$, where T is a Hall $\Pi$-subgroup of $A\cap B$, and so $T\le W\le H$. Moreover, $L=O_{{\Pi }^{\prime }}(A\cap B)$.

We claim that $H\le N_G\left(L\right)$. In fact, since $\sigma \left(\right|HA:A\left|\right)\subseteq \Pi$ and $\sigma \left(\right|HB:B\left|\right)\subseteq \Pi$, we deduce that $H\le N_G\left(O^{\Pi }\left(A\right)\right)$ and $H\le N_G\left(O^{\Pi }\left(B\right)\right)$ by Proposition 1 (vii), and so $H\le N_G(O^{\Pi }\left(A\right)\cap O^{\Pi }\left(B\right))$. Observe that $O^{\Pi }\left(A\right)\cap O^{\Pi }\left(B\right)$ is normal in $A\cap B$. Thus,

$$(A\cap B)/(O^{\Pi }\left(A\right)\cap O^{\Pi }\left(B\right))=(A\cap B)/((A\cap B\cap O^{\Pi }\left(A\right))\cap (A\cap B\cap O^{\Pi }\left(B\right)))$$

$$\lesssim (A\cap B)/(A\cap B\cap O^{\Pi }\left(A\right))\times (A\cap B)/(A\cap B\cap O^{\Pi }\left(B\right)).$$

Since $(A\cap B)/(A\cap B\cap O^{\Pi }\left(A\right))\cong (A\cap B)O^{\Pi }\left(A\right)/O^{\Pi }\left(A\right)$ and $(A\cap B)/(A\cap B\cap O^{\Pi }\left(B\right))\cong (A\cap B)O^{\Pi }\left(B\right)/O^{\Pi }\left(B\right)$ are both $\Pi$-groups, we have that $(A\cap B)/(O^{\Pi }\left(A\right)\cap O^{\Pi }\left(B\right))$ is also a $\Pi$-group. Hence,

$$L=O_{{\Pi }^{\prime }}(A\cap B)=O_{{\Pi }^{\prime }}(O^{\Pi }\left(A\right)\cap O^{\Pi }\left(B\right)).$$

Then, L is characteristic in $O^{\Pi }\left(A\right)\cap O^{\Pi }\left(B\right)$, and so $H\le N_G\left(L\right)$, as claimed.

Note that $VH=(A\cap B)H=(L\times T)H=LH$ since $T\le H$. It follows that $VH=HV$; this again contradicts the assumption made about V.

Therefore, $V\in \mathcal{L}$, and so $\mathcal{L}$ is a sublattice of the lattice $\mathcal{L}\left(G\right)$.

$\left(\mathbf{ii}\right)$ Let $\mathcal{L}$ denote the set of all ${\sigma }_{\mathfrak{I}}$-subnormal subgroups of G which permute with H. Let $A,B\in \mathcal{L}$ and $K=⟨A,B⟩$, and $V=A\cap B$.

$\left(a\right)$ First, we show that $K\in \mathcal{L}$. By Lemma 1 again, it is enough to show that K is ${\sigma }_{\mathfrak{I}}$-subnormal in G. Assume that this is false, and consider a counterexample G of minimal order. Then, $A\ne 1\ne B$ and $K\ne G$. Let R be a minimal normal subgroup of G and $L=KR$. Then, we have the following:

$\left(1\right)$ $L=G$, and thus $K_G=1$.

Suppose that $L\ne G$. By Proposition 1 (i), we have that A and B are ${\sigma }_{\mathfrak{I}}$-subnormal in L. Hence, the minimal choice of G implies that K is ${\sigma }_{\mathfrak{I}}$-subnormal in L. By Proposition 1 (ii), $AR/R$ and $BR/R$ are ${\sigma }_{\mathfrak{I}}$-subnormal in $G/R$. Note that

$$L/R=⟨A,B⟩R/R=⟨AR/R,BR/R⟩.$$

Thus, the choice of G implies that $L/R$ is ${\sigma }_{\mathfrak{I}}$-subnormal in $G/R$. Hence, L is ${\sigma }_{\mathfrak{I}}$-subnormal in G by Proposition 1 (iii). It follows that K is ${\sigma }_{\mathfrak{I}}$-subnormal in G. This contradiction shows that $L=G$. If $K_G\ne 1$, then we can choose an R such that $R\le K$, and thus, $K=L=G$. This contradicts the assumption that $L\ne G$. Therefore, $\left(1\right)$ holds.

(2) R is a ${\sigma }_{\mathfrak{I}}$-primary ${\sigma }_i$-group for some i.

Suppose that R is not ${\sigma }_{\mathfrak{I}}$-primary. By Proposition 1 (iv), $R\le N_G\left(A\right)$. Then,

$$A^G=A^{RK}=A^K\le K_G=1$$

by Claim (1). Thus, $A=1$; this contradicts the initial assumption that $A\ne 1$. Hence, we have $\left(2\right)$.

$\left(3\right)$ A and B are ${\sigma }_{\mathfrak{I}}$-nilpotent.

Assume, for example, that A is not ${\sigma }_{\mathfrak{I}}$-nilpotent. By Proposition 1 (v), the ${\sigma }_{\mathfrak{I}}$-nilpotent residual $D=A^{{\mathfrak{N}}_{{\sigma }_{\mathfrak{I}}}}$ of A is subnormal in G. Applying [12] (Chap. A, Lemma 14.3), we have $R\le N_G\left(D\right)$. Again, $D^G=D^{RK}=D^K\le K_G=1$, and so $D=1$. This contradicts our assumption that A is not ${\sigma }_{\mathfrak{I}}$-nilpotent. Claim (3) is therefore established.

The final contradiction. We have demonstrated that R is ${\sigma }_{\mathfrak{I}}$-nilpotent, and A and B are ${\sigma }_{\mathfrak{I}}$-nilpotent ${\sigma }_{\mathfrak{I}}$-subnormal subgroups of G. By Proposition 1 (vi), we obtain that $A^G$ and $B^G$ are both ${\sigma }_{\mathfrak{I}}$-nilpotent. Note that ${\mathfrak{N}}_{{\sigma }_{\mathfrak{I}}}$ is a Fitting class. Then,

$$G=L=A^G{B}^{G}R$$

is also ${\sigma }_{\mathfrak{I}}$-nilpotent. From Proposition 2(iii), it follows that K is ${\sigma }_{\mathfrak{I}}$-subnormal in G, and this is the final contradiction, which contradicts our assumption on K.

Therefore, K is ${\sigma }_{\mathfrak{I}}$-subnormal in G, and thus, $K\in \mathcal{L}$.

$\left(b\right)$ Now, we show that $V\in \mathcal{L}$. By applying Proposition 1 (i) and noting that ${\sigma }_{\mathfrak{I}}$-subnormality is transitive, V is ${\sigma }_{\mathfrak{I}}$-subnormal in G, and so it is enough to show that $VH=HV$. We proceed by induction on $\left|G\right|$. If $H=G$, there is nothing to prove; thus, we can assume that G is not a $\Pi$-group.

Let $E=AH\cap BH$. Then, $H\le E$ and $A\cap B\le E$. By Proposition 1 (i), again, we see that $A\cap E$ and $B\cap E$ are ${\sigma }_{\mathfrak{I}}$-subnormal in E. Moreover, $AH\cap E=H(A\cap E)$, and so $H(A\cap E)=(A\cap E)H$. Similarly, $(B\cap E)H=H(B\cap E)$. Observe that H is a Π-projector of E. Hence, the hypothesis holds for $(E,A\cap E,B\cap E,H)$. Assume that $E<G$. Then, our induction hypothesis implies that $A\cap B=(A\cap E)\cap (B\cap E)$ is permutable with H, i.e., $VH=HV$.

Now, turn to the case $E=G$ and, thus, $G=AH=BH$. Then, $|G:A|$ and $|G:B|$ are $\Pi$-numbers. By Proposition 1 (vii), we have $O^{\Pi }\left(A\right)=O^{\Pi }\left(G\right)=O^{\Pi }\left(B\right)$. Since G is not a $\Pi$-group, $O^{\Pi }\left(G\right)\ne 1$, and it follows that $W:=A_G\cap B_G\ne 1$. Consider the factor group $G/W$. By Proposition 1 (ii), $A/W$ and $B/W$ are ${\sigma }_{\mathfrak{I}}$-subnormal in $G/W$. Note that $HW/W$ is a $\Pi$-projector of $G/W$ by Lemma 2. Hence, $(A/W)\cap (B/W)$ permutes with $HW/W$ by the induction hypothesis. Again, we conclude that $(A\cap B)H=H(A\cap B)$. This completes the proof of (ii). □

3. Conclusions

In this work, we aim to explore new structures in the theory of subgroup lattices of finite groups by introducing two important classes of sublattices. The theoretical framework is built on the theory of $\sigma$-properties, formation theory, and operator groups. Our main results provide a unified framework that generalizes and extends numerous classical results in the literature, including the fundamental theorems of Wielandt [1] on the lattice of the subnormal subgroups and Kegel’s results on the lattice of K-$\mathfrak{F}$-subnormal subgroups [2] and on S-permutable subgroups [7].

Since these three classes of lattices have already found numerous applications [2,9], we are confident that their generalizations, considered in this paper, will also be useful in studying various classes of finite groups.

It will be interesting to find natural examples or applications where these sublattices arise. Another promising direction is to investigate the conditions under which they possess other lattice-theoretic properties beyond being sublattices. More concretely, we propose the following specific problems, motivated by the results in Sections $9.1$ and $9.2$ of the book [21]:

Problem 1.

Let $\mathcal{L}$ and $\mathcal{M}$ be the lattices of all $\mathbf{A}$-strongly ${\sigma }_{\mathfrak{I}}$-subnormal subgroups and all ${\sigma }_{\mathfrak{I}}$-subnormal subgroups of G, respectively. What are the conditions on G under which $\mathcal{L}$ (respectively, $\mathcal{M}$) is modular, distributive, or complemented?

Problem 2.

Let H be a Hall Π-subgroup of G. Denote by ${\mathcal{L}}_H$ and ${\mathcal{M}}_H$ the lattices of all ${\sigma }_{\mathfrak{I}}$-subnormal subgroups and all $\mathbf{A}$-strongly ${\sigma }_{\mathfrak{I}}$-subnormal subgroups of G that permute with H, respectively. What are the conditions on G under which ${\mathcal{L}}_H$ (respectively, ${\mathcal{M}}_H$) is modular, distributive, or complemented?

We note that solutions exist for restricted versions of these problems. Problem 1 is solved when $\mathcal{L}$ is the lattice of all subnormal subgroups [21] (Section 9.2). For Problem 2, a solution is known in the restricted case, where ${\mathcal{L}}_H$ is the lattice of all S-permutable subgroups of G [22]. The resolution of the general cases will therefore mark a notable development.