On $\sigma$-subnormality criteria in finite $\sigma$-soluble groups

Abstract

Let $\sigma = \{ {\sigma }_{i} : i \in I \}$ be a partition of the set ${\mathbb {P}}$ of all prime numbers. A subgroup X of a finite group G is called $\sigma$-subnormal in G if there is a chain of subgroups

$$\begin{aligned} X=X_{0} \subseteq X_{1} \subseteq \cdots \subseteq X_{n}=G \end{aligned}$$

where for every $j=1, \dots , n$ the subgroup $X_{j-1}$ is normal in $X_{j}$ or $X_{j}/Core_{X_{j}}(X_{j-1})$ is a ${\sigma }_{i}$-group for some $i \in I$. In the special case that $\sigma$ is the partition of ${\mathbb {P}}$ into sets containing exactly one prime each, the $\sigma$-subnormality reduces to the familiar case of subnormality. In this paper some $\sigma$-subnormality criteria for subgroups of $\sigma$-soluble groups, or groups in which every chief factor is a $\sigma _{i}$-group, for some $\sigma _{i} \in \sigma$, are showed.

1 Introduction and statements of results

All groups considered in this paper are finite.

The results of this article are based on a paper of Skiba [13]. There he generalised the concepts of solubility, nilpotency and subnormality introducing $\sigma$-solubility, $\sigma$-nilpotency, and $\sigma$-subnormality in which $\sigma$ is a partition of the set ${\mathbb {P}}$, the set of all primes. Hence ${\mathbb {P}}={\bigcup }_{i \in I}{\sigma }_{i}$, with ${\sigma }_{i} \cap {\sigma }_{j}=\emptyset$ for all $i \ne j$.

We note that in the special case that $\sigma$ is the partition of ${\mathbb {P}}$ containing exactly one prime each, the definitions below reduce to the familiar case of soluble groups, nilpotent groups and subnormal subgroups.

From now on let $\sigma$ denote a partition of ${\mathbb {P}}$. Given a natural number n, we denote by $\sigma (n)$ the set of all elements of $\sigma$ including the primes dividing n. Two natural numbers m and n are called $\sigma$-coprime if $\sigma (m) \cap \sigma (n) = \emptyset$. We say that n is $\sigma$-primary if $|\sigma (n)| = 1$, that is, if its prime factors all belong to the same member of $\sigma$.

A group G is called $\sigma$-primary if |G| is a $\sigma$-primary number.

Definition 1

A group G is said to be $\sigma$-soluble if every chief factor of G is $\sigma$-primary. G is said to be $\sigma$-nilpotent if it is a direct product of $\sigma$-primary groups.

Note that if $\pi$ is a set of primes and $\sigma = \{\pi , \pi '\}$, then a group G is $\sigma$-soluble if and only if G is $\pi$-separable. In this case, G is $\sigma$-nilpotent if and only if G is $\pi$-decomposable. If $\pi = \{p_1, \ldots , p_n\}$, and $\sigma = \{\{p_1\}, \ldots , \{p_n\}, \pi '\}$, then G is $\sigma$-soluble if and only if G is $\pi$-soluble, and G is $\sigma$-nilpotent if and only if G has a normal Hall $\pi '$-subgroup and a normal Sylow $p_i$-subgroup, for all $i=1, \dots , n$.

Many normal and arithmetical properties of soluble groups still hold for $\sigma$-soluble groups (see [13]). In particular, every $\sigma$-soluble group has a conjugacy class of Hall $\sigma _i$-subgroups and a conjugacy class of Hall $\sigma _{i}'$-subgroups, for every $\sigma _i \in \sigma$.

The role of the class ${{{\mathcal {N}}}}_{\sigma }$ of all $\sigma$-nilpotent groups in $\sigma$-soluble groups is analogous to that of nilpotent groups in soluble groups. In particular, ${{{\mathcal {N}}}}_{\sigma }$ is a subgroup-closed saturated Fitting formation [13, Corollary 2.4 and Lemma 2.5] that is closely related to the subgroup embedding property of $\sigma$-subnormality.

Definition 2

Given a partition $\sigma$ of the set of prime numbers, a subgroup X of a group G is called $\sigma$-subnormal in G if there exists a chain of subgroups

$$\begin{aligned} X=X_{0} \le X_{1} \le \cdots \le X_{n}=G, \end{aligned}$$

with $X_{i-1}$ normal in $X_{i}$ or $X_{i}/Core_{X_{i}}(X_{i-1})$$\sigma$-primary for every $1 \le i \le n$.

To know that a non-$\sigma$-nilpotent group possesses a non-trivial proper $\sigma$-subnormal subgroup is equivalent to know that the group is not simple. Therefore criteria for the $\sigma$-subnormality of a subgroup may have some importance in the study of the normal structure of a group. The close relationship between $\sigma$-subnormal subgroups and direct decompositions of a group strongly supports that claim. The significance of the $\sigma$-subnormal subgroups in $\sigma$-soluble groups is apparent since they are precisely the ${{{\mathcal {N}}}}_{\sigma }$-subnormal subgroups, and so they are a sublattice of the subgroup lattice of G. They are also important to analyse the structural impact of some permutability properties (see [14,15,16,17]).

In this paper, which is a natural continuation of [3], extensions of some well-known subnormality criteria are presented. For instance, according to a result of Wielandt (see [9, Theorem 7.3.3]), a subgroup X of a group G is subnormal in G if and only if X is subnormal in $\langle X, X^{g}\rangle$ for all $g \in G$.

In [8, Question 19.84] (see also [16]), Skiba asked whether it is enough to know that X is $\sigma$-subnormal in $\langle X, X^{g}\rangle$ for all $g \in G$ to deduce that X is $\sigma$-subnormal in G. It is certainly true in the soluble universe by virtue of [2, Proposition 6.1.10 and Theorem 6.2.17] (see [3, Lemma 2]). Our first main result shows that the answer is also affirmative for $\sigma$-soluble groups.

Theorem A

Suppose that G is a $\sigma$-soluble group and X is a subgroup of G that is $\sigma$-subnormal in $\langle X, X^{g} \rangle$ for all $g \in G$. Then X is $\sigma$-subnormal in G.

Theorem A is not true for arbitrary groups. Therefore Question 19.84 in [8] is answered.

Example 1

Let $\pi =\{2,3 \}$ and $\sigma =\{ \pi , {\pi }' \}$. The simple group $G = {\text {PSL}}_2(7)$ of order $168 = 2^3\cdot 3\cdot 7$ has a unique conjugacy class of elements of order 2. Let x be an element of this class. Given $g \in G$, the group $\langle x, x^g \rangle$ is isomorphic to $C_2$, to $C_2\times C_2$, to $\Sigma _3$ or to $D_8$. Therefore $X = \langle x \rangle$ is $\sigma$-subnormal in $\langle X, X^{g} \rangle$ for all $g \in G$ but X is not $\sigma$-subnormal in G.

Another important subnormality criterion asserts that if $G = AB$ is a group which is the product of the subgroups A and B and X is a subgroup of G contained in $A \cap B$ that is subnormal in A and B, then X is subnormal in G. This result was proved by Maier in [10] for soluble groups and then for arbitrary groups by Wielandt [17]. Applying Theorem A, we show that Maier–Wielandt’s result also holds for $\sigma$-subnormal subgroups not only in the soluble universe, but also in the $\sigma$-soluble one.

Theorem B

Let the $\sigma$-soluble group G be the product of two subgroups A and B. If X is a subgroup of $A \cap B$ which is $\sigma$-subnormal in both A and B, then X is $\sigma$-subnormal in G.

Theorem B does not hold in general as the following example shows (see [7]).

Example 2

Let $\pi =\{2,5 \}$ and $\sigma =\{ \pi , {\pi }' \}$. The alternating group of degree five $A_{5}$ is the product of the subgroups A and B, where A is the alternating group of degree 4 and B is a dihedral group of order 10. Then $A \cap B$ is $\sigma$-subnormal in both A and B, but $A \cap B$ is not $\sigma$-subnormal in $A_{5}$.

On the other hand, Wielandt [17] conjectured that if X is a subgroup of G such that X is subnormal in $\langle X,X^{g} \rangle$ for all $g \in A \cup B$, then X is subnormal in G.

Wielandt’s conjecture was proved to be true in the soluble universe by Maier and Sidki [11] for subgroups X of prime power order and then for every subgroup X of a soluble group by Casolo in [4].

In [3, Theorem A], we show that the following $\sigma$-version of the aforementioned result holds.

Theorem 1

Assume that G is a soluble group factorised as a product of the subgroups A and B. Let X be a subgroup of G such that X is $\sigma$-subnormal in $\langle X,X^{g} \rangle$ for all $g \in A \cup B$. Then X is $\sigma$-subnormal in G.

A natural question to ask is now whether Theorem 1 holds for $\sigma$-soluble groups. Unfortunately we have been unable to answer this question; however, our third main result could be regarded as a significant step to solve it.

Theorem C

Assume that G is a $\sigma$-soluble group factorised as a product of the subgroups A and B. Let X be a subgroup of G such that X is $\sigma$-subnormal in $\langle X,X^{g} \rangle$ for all $g \in A \cup B$. Then X is $\sigma$-subnormal in G if one of the following conditions is true:

• (i) |G : A| and |G : B| are $\sigma$-primary.

• (ii) |G : A| is $\sigma$-primary and |G : A| and |G : B| are $\sigma$-coprime.

The proof of Theorem C strongly depends on the following extension of [6, Theorem 3].

Theorem D

Let G be a $\sigma$-soluble group, and A and X two subgroups of G such that X is $\sigma$-subnormal in $\langle X, X^{a}\rangle$ for all $a \in A$. If |G : A| is $\sigma$-primary, then X is $\sigma$-subnormal in $\langle X,A \rangle$.

We shall adhere to the notation and terminology of [2, 5].

2 Preliminaries

Our first lemma collects some basic properties of $\sigma$-subnormal subgroups which are very useful in induction arguments.

Lemma 1

[13] Let H, K and N be subgroups of a group G. Suppose that H is $\sigma$-subnormal in G and N is normal in G. Then the following statements hold:

• 1. $H \cap K$ is $\sigma$-subnormal in K.

• 2. If K is a $\sigma$-subnormal subgroup of H, then K is $\sigma$-subnormal in G.

• 3. If K is $\sigma$-subnormal in G, then $H \cap K$ is $\sigma$-subnormal in G.

• 4. HN/N is $\sigma$-subnormal in G/N.

• 5. If $N \subseteq K$ and K/N is $\sigma$-subnormal in G/N, then K is $\sigma$-subnormal in G.

• 6. If $L \le K$ and K is $\sigma$-nilpotent, then L is $\sigma$-subnormal in K.

• 7. If |G : H| is a $\sigma _i$-number, then $O^{\sigma _i}(H)=O^{\sigma _i}(G)$.

• 8. If N is a $\sigma _i$-subgroup of G, then $N \le N_{G}(O^{\sigma _i}(H))$.

A standard induction argument using Lemma 1 allows us to prove the following result.

Lemma 2

Let X be a subgroup of a $\sigma$-soluble group G. Then X is $\sigma$-subnormal in G if and only if X is $\mathcal{N}_{\sigma }$-subnormal in G, that is, there exists a chain of subgroups

$$\begin{aligned} X=X_{0} \le X_{1} \le \cdots \le X_{n}=G, \end{aligned}$$

such that $X_{i-1}$ is a maximal subgroup of $X_{i}$ and $X_{i}/Core_{X_{i}}(X_{i-1}) \in {{{\mathcal {N}}}}_{\sigma }$, for $1 \le i \le n$.

The fact that $\sigma$-subnormal subgroups are $\mathcal{N}_{\sigma }$-subnormal in the $\sigma$-soluble universe allows us to prove some relevant properties of these subgroups which are crucial in the proof of our main results.

Lemma 3

Let X be a subgroup of a group G.

• 1. [2, Lemma 6.1.9 and Proposition 6.1.10] If X is $\sigma$-subnormal in G, then the ${{{\mathcal {N}}}_{\sigma }}$-residual $X^{{{\mathcal {N}}}_{\sigma }}$ of X is subnormal in G.

• 2. [2, Lemma 6.1.9] If X is subnormal in G, then X is $\sigma$-subnormal in G.

• 3. [2, Lemmas 6.3.11 and 6.3.12 and Example 6.3.13] ${{{\mathcal {N}}}}_{\sigma }$ is a lattice formation, that is, the set of all $\sigma$-subnormal subgroups of a $\sigma$- soluble group G forms a sublattice of the subgroup lattice of G.

• 4. [2, Theorem 6.3.3] If X is a $\sigma$-subnormal $\sigma$-nilpotent subgroup of a $\sigma$-soluble group G, then X is contained in $F_{\sigma }(G)$, the ${{{\mathcal {N}}}}_{\sigma }$-radical of G. In particular, if X is $\sigma _{i}$-group, then $X \le {{\,\mathrm{O}\,}}_{\sigma _{i}}(G)$.

• 5. [2, Theorem 6.5.46] If $G=\langle A,B \rangle$ is a a $\sigma$-soluble group generated by two $\sigma$-subnormal subgroups A and B, then $G^{{{{\mathcal {N}}}}_{\sigma }}=\langle A^{\mathcal{N}_{\sigma }},B^{{{{\mathcal {N}}}}_{\sigma }} \rangle$.

Note that by Lemmas 1 (2) and 3 (2), subnormal subgroups of $\sigma$-subnormal subgroups of a group G are $\sigma$-subnormal in G. This fact will be applied in the sequel without further reference.

Our third lemma shows that the residual associated with the class of all $\sigma _i$-groups (also called $\sigma _i$-residual) respects the $\sigma$-subnormal generation of $\sigma$-soluble groups.

Lemma 4

Let $\sigma _i \in \sigma$. If A and B are $\sigma$-subnormal subgroups of a $\sigma$-soluble group $G=\langle A,B \rangle$, then $O^{\sigma _i}(G)=\langle O^{\sigma _i}(A), O^{\sigma _i}(B) \rangle$.

Proof

Assume the result is false and let G be a counterexample of least order. Denote $H=\langle O^{\sigma _i}(A),O^{\sigma _i}(B) \rangle$ and $X=O^{\sigma _i}(G)$. Clearly $1 \ne X$. Let N be a minimal normal subgroup of G such that $N \le X$. Since G is $\sigma$-soluble, it follows that N is $\sigma _j$-group for some $\sigma _j \in \sigma$. The minimality of G yields $X=HN$ and $Core_{G}(H)=1$.

On the other hand, by Lemma 3 (5), we have that $G^{\mathcal{N}_{\sigma }}=\langle A^{{{{\mathcal {N}}}}_{\sigma }}, B^{{{{\mathcal {N}}}}_{\sigma }} \rangle \le \langle O^{\sigma _i}(A),O^{\sigma _i}(B) \rangle =H.$ Since $G^{{{{\mathcal {N}}}}_{\sigma }}$ is normal in G and $Core_{G}(H)=1$, it follows that G is $\sigma$-nilpotent.

Then $G=X \times Y$ with $Y=O_{\sigma _i}(G)$. If $Y \ne 1$, then by the minimal choice of G, we have that $G=X \times Y=H \times Y$, and therefore $X=H$. Thus $Y=1$ and so $G = O^{\sigma _i}(G)$, $A =O^{\sigma _i}(A)$ and $B=O^{\sigma _i}(B)$. This contradiction proves the lemma. $\square$

Lemma 5

Let $H^{*}$ denote either the ${{{\mathcal {N}}}}_{\sigma }$-residual or the $\sigma _i$-residual of a subgroup H of a $\sigma$-soluble group G, for $\sigma _i \in \sigma$. Let A be a subgroup of G. If H is a $\sigma$-subnormal subgroup of $\langle H,H^{a} \rangle$ for all $a \in A$, then H normalises $(H^{*})^{A}$.

Proof

Let $a \in A$. Since H is a $\sigma$-subnormal subgroup of $\langle H, H^{a^{-1}} \rangle$, it follows that $H^{a}$ is $\sigma$-subnormal in $\langle H^{a},H \rangle =\langle H, H^{a} \rangle$. By Lemmas 3 (5) and 4, we have $\langle H,H^a \rangle ^{*}=\langle H^{*}, (H^{a})^{*} \rangle =\langle H^{*}, (H^{*})^a \rangle$, thus

$$\begin{aligned} {[}H,(H^{*})^a] \le [H, \langle H,H^{a}\rangle ^{*}] \le \langle H,H^{a} \rangle ^{*} \le (H^{*})^{A}. \end{aligned}$$

$\square$

Lemma 6

Let G be a $\sigma$-soluble group, X a $\sigma _i$-subgroup of G and H a Hall $\sigma _i$-subgroup of G. If X is $\sigma$-subnormal in $\langle X, X^{h} \rangle$ for all $h \in H$, then $X \le H$.

Proof

Suppose that the result is false. Let G be a counterexample of the smallest possible order. Clearly the hypotheses of the lemma hold in $G/O_{\sigma _i}(G)$. Therefore, if $O_{\sigma _i}(G) \ne 1$, we have that $XO_{\sigma _i}(G)/O_{\sigma _i}(G) \le H/O_{\sigma _i}(G)$ by minimality of G. Hence $X \le H$, contrary to supposition. Thus $O_{\sigma _i}(G)=1$. Let N be a minimal normal subgroup of G. Then N is a $\sigma _j$-group for some $j \ne i$. Since $X \le HN$ by the minimal choice of G, there exists $n \in N$ with $X^{n} \le H$. Let $x \in X$ and $h = x^{-n} \in H$. Then $[x,h]=[x,n][x^{-1},n] \in N$ and $[x,h]=x^{-1}x^{h} \in \langle x,x^{h} \rangle$. Hence $[x,h] \in N \cap \langle x,x^{h}\rangle$. Then X is $\sigma$-subnormal in $\langle X,X^{h} \rangle$ by hypothesis. Since X is a $\sigma _i$-subgroup, we have that $X \le O_{\sigma _i}(\langle X,X^{h} \rangle )$ by Lemma 3 (4). Therefore, $\langle X,X^{h} \rangle = O_{\sigma _i}(\langle X,X^{h} \rangle )X^{h}$ is a $\sigma _i$-subgroup of HN. Thus $[x,h] \in N \cap \langle X,X^{h} \rangle =1$ and $[x,h]=1$. In particular, $[x,n]=[x^{-1},n]$ is a $\sigma _i$-element. Since N is a $(\sigma _i)'$-group and $[x,n] \in N$, it follows that $[x,n]=1$ and $X^{n}=X \le H$. $\square$

Lemma 7

Let H be a subgroup of a $\sigma$-soluble group G such that $O^{\sigma _i}(H)= H$ for some $\sigma _i \in \sigma$. Assume K is a normal $\sigma _i$-subgroup of G and $k \in K$ such that H is a $\sigma$-subnormal subgroup of $\langle H,H^{k} \rangle$. Then k normalises H.

Proof

Denote $L= \langle H,H^{k} \rangle$. Let Z denote the normal closure of H in L. By Lemma 4, $O^{\sigma _i}(Z)= Z$. Since $O^{\sigma _i}(L/Z) = L/Z$, it follows that $L = O^{\sigma _i}(L)Z$. By [2, Proposition 6.5.5], it follows that $O^{\sigma _i}(L)=O^{\sigma _i}(L)O^{\sigma _i}(Z) = O^{\sigma _i}(L)Z = L$.

On the other hand, $L=L \cap HK=H(L \cap K)$. By Lemma 4, $L=O^{\sigma _i}(L)=O^{\sigma _i}(H)O^{\sigma _i}(L \cap K)=H$. Thus $L=H$ and $H^{k}=H$. $\square$

3 Proofs of the main theorems

Proof of Theorem A

Suppose the result is not true and let G be a counterexample with $|G| + |X|$ minimal. Then $G^{{{\mathcal {N}}}_{\sigma }} \ne 1$. Let N be a minimal normal subgroup of G contained in $G^{{{\mathcal {N}}}_{\sigma }}$. Then N is a $\sigma _i$-group for some $\sigma _i \in \sigma$. Note that XN/N is $\sigma$-subnormal in G/N by the minimality of the pair (G, X). If XN were a proper subgroup of G, then X would be $\sigma$-subnormal in XN. By Lemma 1, X would be $\sigma$-subnormal in G, contrary to our assumption. Hence $G=XN$. Assume that X is a $\sigma _i$-group. Then G is a $\sigma _i$-group, and X is $\sigma$-subnormal in G. This contradiction implies that X is not a $\sigma _i$-group, and so $O^{\sigma _i}(X) \ne 1$.

Assume that $O^{\sigma _i}(X) < X$. By minimality of (G, X), it follows that $O^{\sigma _i}(X)$ is $\sigma$-subnormal in G. By Lemma 1 (8), N normalises $O^{\sigma _i}(O^{\sigma _i}(X))=O^{\sigma _i}(X)$. Hence $O^{\sigma _i}(X)$ is a normal subgroup of G. The minimal choice of G implies that $X/O^{\sigma _i}(X)$ is $\sigma$-subnormal in $G/O^{\sigma _i}(X)$ and then X is $\sigma$-subnormal in G by Lemma 1 (5). This is not possible. Thus $X=O^{\sigma _i}(X)$.

If $n \in N$ then X is $\sigma$-subnormal in $U_{n}=\langle X,X^{n} \rangle =(U_{n} \cap N)X$ by hypothesis. By Lemma 1 (7), we have that

$$\begin{aligned} O^{\sigma _i}(U_{n})=O^{\sigma _i}((U_{n} \cap N)X)=O^{\sigma _i}(X)=X. \end{aligned}$$

In particular, X is normal in $U_{n}$. Consequently, X is normal in $V=\langle X^{n}: n \in N \rangle$. Since V is normal in G, we have X is subnormal in G, and we have reached the desired contradiction. $\square$

Proof of Theorem B

Assume the result is false and let G be a counterexample such that $|G:A| + |X|$ is minimal. Suppose that M is a maximal subgroup of G containing A. Then $M=A(M \cap B)$ and X is $\sigma$-subnormal in both A and $M \cap B$ by Lemma 1 (1). By minimality of G, X is $\sigma$-subnormal in M. On the other hand, $G=MB$. If $|G:M| < |G:A|$, we have X is $\sigma$-subnormal in G, which is a contradiction. Therefore $A=M$ is a maximal subgroup of G. Let $K={{\,\mathrm{Core}\,}}_{G}(A)$. If $K \ne 1$, then XK/K is $\sigma$-subnormal in G/K by the minimal choice of G. By Lemma 1 (5), XK is $\sigma$-subnormal in G. Moreover $X \le XK \le A$. Thus X is $\sigma$-subnormal in XK by Lemma 1 (1). Thus X is $\sigma$-subnormal in G. This contradiction yields $K=1$ and G is a primitive group. By Lemma 3 (1), $X^{{{\mathcal {N}}}_{\sigma }}$ is a subnormal subgroup of A and B. Applying the result of Maier–Wielandt, we have that $X^{{{\mathcal {N}}}_{\sigma }}$ is a subnormal subgroup of G. By [9, Lemma 7.3.16], $X^{{{\mathcal {N}}}_{\sigma }} \le {{\,\mathrm{Core}\,}}_{G}(A)=1$. Hence X is $\sigma$-nilpotent. By Lemma 1 (6), every subgroup of X is $\sigma$-subnormal in X. Therefore every proper subgroup of X is $\sigma$-subnormal in A and B by Lemma 1 (2). The minimal choice of X implies that every proper subgroup of X is $\sigma$-subnormal in G. By Lemma 3 (3), X is cyclic of prime power order. Assume X is a $\sigma _i$-group. Since X is $\sigma$-subnormal in A, by Lemma 3 (4), X is contained in $O_{\sigma _i}(A)$. Then $X^{A}$, the normal closure of X in A, is a $\sigma _i$-group. Analogously, $X^{B}$ is a $\sigma _i$-group. According to [1, Lemma 1.3.2], there exist Hall $\sigma _i$-subgroups $A_{\sigma _i}$ of A and $B_{\sigma _i}$ of B such that $A_{\sigma _i}B_{\sigma _i}$ is a Hall $\sigma _i$-subgroup of G. Then $\langle X^{A}, X^{B} \rangle$ is a $\sigma _i$-group because it is contained in $A_{\sigma _i}B_{\sigma _i}$. Let $g=ab \in G$ with $a \in A$ and $b \in B$. Then

$$\begin{aligned} \langle X, X^{g} \rangle =\langle X^{b^{-1}},X^{a} \rangle ^{b} \le \langle X^{B},X^{A} \rangle ^b. \end{aligned}$$

Consequently $\langle X,X^g \rangle$ is a $\sigma _i$-group and then X is $\sigma$-subnormal in $\langle X, X^g \rangle$ for every $g \in G$ by Lemma 1 (6). Applying Theorem A, X is $\sigma$-subnormal in G, a contradiction. $\square$

Proof of Theorem D

Suppose that the result is false. We choose a counterexample G with $|G| + |X|$ minimal and proceed to derive a contradiction. The minimal choice of G and Theorem A show that $G=\langle X,A \rangle$ and X is not contained in A. Suppose that |G : A| is a $\sigma _i$-number for some $\sigma _i \in \sigma$. Then A contains a Hall $\sigma _{i}'$-subgroup of G.

If $C = {{\,\mathrm{Core}\,}}_{G}(A) \ne 1$, then XC is a $\sigma$-subnormal subgroup of G by minimality of G. Moreover, by Theorem A, X is $\sigma$-subnormal in XC. Thus X is $\sigma$-subnormal in G by Lemma 1 (2). This contradiction shows that ${{\,\mathrm{Core}\,}}_{G}(A) = 1$.

Let N be a minimal normal subgroup of G. Then N is a $\sigma _j$-group for some $\sigma _j \in \sigma$. If $i \ne j$, then N is contained in every Hall $\sigma _{i}'$-subgroup of G. In particular, N is contained in A, a contradiction. Therefore N is a $\sigma _i$-group, $O_{\sigma _i}(G) \ne 1$, and $O_{{\sigma _i}^{'}}(G)=1$.

Suppose that X is not $\sigma$-nilpotent. Then $1 \ne X^{\mathcal{N}_{\sigma }}$ is a proper subgroup of X which is $\sigma$-subnormal in $\langle X,X^{a} \rangle$ for all $a \in A$. The choice of the pair (G, X) yields that $X^{{{{\mathcal {N}}}}_{\sigma }}$ is $\sigma$-subnormal in $\langle X^{{{{\mathcal {N}}}}_{\sigma }},A \rangle$. Hence $X^{{{{\mathcal {N}}}}_{\sigma }}$ is $\sigma$-subnormal in $(X^{\mathcal{N}_{\sigma }})^A$. By Lemma 5, X normalises $(X^{\mathcal{N}_{\sigma }})^A$. Therefore $(X^{{{{\mathcal {N}}}}_{\sigma }})^A$ is a normal subgroup of G and $X^{{{{\mathcal {N}}}}_{\sigma }}$ is a $\sigma$-subnormal subgroup of G. Since X is not a $\sigma _i$-group, it follows that $1 \ne O^{\sigma _i}(X)$. Moreover, since $1 \ne X^{\mathcal{N}_{\sigma }}$ is a $\sigma$-soluble group, it follows that $F_{\sigma }(X^{{{{\mathcal {N}}}}_{\sigma }}) \ne 1$. Thus $F_{\sigma }(X^{{{{\mathcal {N}}}}_{\sigma }}) \ne 1$ is a $\sigma$-nilpotent $\sigma$-subnormal subgroup of G. By Lemma 3 (4), $F_{\sigma }(X^{{{{\mathcal {N}}}}_{\sigma }}) \le F_{\sigma }(G)=O_{\sigma _i}(G)$ and then $1 \ne O_{\sigma _i}(X^{{{{\mathcal {N}}}}_{\sigma }}) \le O_{\sigma _i}(G)$. Hence $Z = X \cap O_{\sigma _i}(G) \ne 1$ and $Z^{A}$ is a $\sigma$-subnormal $\sigma _i$-subgroup of G. Let $a \in A$. Then X is $\sigma$-subnormal in $\langle X, Z^{a} \rangle$ and so $O_{\sigma _i}(\langle X,Z^{a} \rangle )$ normalises $O^{\sigma _i}(X)$ by Lemma 1 (8). Since $Z^{a} \le O_{\sigma _i}(\langle X,Z^{a} \rangle )$, it follows that $Z^{a}$ normalises $O^{\sigma _i}(X)$. Therefore $Z^A$ normalises $O^{\sigma _i}(X)$.

Applying Lemma 5, it follows that X normalises $(O^{\sigma _i}(X))^{A}$. Hence $(O^{\sigma _i}(X))^{A}$ is a normal subgroup of G. Assume that $O^{\sigma _i}(X)$ is a proper subgroup of X. By minimality of the pair (G, X), we have that $O^{\sigma _i}(X)$ is a $\sigma$-subnormal subgroup of $\langle O^{\sigma _i}(X), A \rangle$. Therefore $O^{\sigma _i}(X)$ is a $\sigma$-subnormal subgroup of $(O^{\sigma _i}(X))^{A}$, and so $O^{\sigma _i}(X)$ is $\sigma$-subnormal in G. By Lemma 1 (8), $O_{\sigma _i}(G)$ normalises $O^{\sigma _i}(O^{\sigma _i}(X))=O^{\sigma _i}(X)$ and hence $XO_{\sigma _i}(G)$ normalises $O^{\sigma _i}(X)$. Then $X/O^{\sigma _i}(X)$ is $\sigma$-subnormal in $XO_{\sigma _i}(G)/O^{\sigma _i}(X)$. Thus X is $\sigma$-subnormal in $XO_{\sigma _i}(G)$ which is $\sigma$-subnormal in G by minimality of G and Lemma 1 (5). Lemma 1 (2) yields that X is $\sigma$-subnormal in G, contrary to assumption. Hence $O^{\sigma _i}(X)=X$ and so $Z^A$ normalises X. In addition, $[Z^A, X] \le [{{\,\mathrm{N}\,}}_G(X) \cap O_{\sigma _i}(G), X] \le X \cap O_{\sigma _i}(G) = Z \le Z^A$. Hence $Z^A$ is normalised by X and so it is a normal subgroup of G. Again the minimality of G and Lemma 1 (5) imply that $XZ^A$ is $\sigma$-subnormal in G. Since X is normal in $XZ^A$, we have that X is $\sigma$-subnormal in G. This contradiction shows that X is $\sigma$-nilpotent.

Suppose that $O^{\sigma _i}(X) \ne 1$. Since X is $\sigma$-nilpotent, it follows that either X is a $\sigma _i^{'}$-group or $O^{\sigma _i}(X)$ is a proper subgroup of X. Assume that X is a $\sigma _i^{'}$-group. Then, by Lemma 6, X is contained in A. Hence $G = A$ and X is $\sigma$-subnormal in G by Theorem A, which is not possible. Suppose that $O^{\sigma _i}(X)$ is a proper subgroup of X. By minimality of (G, X), $O^{\sigma _i}(X)$ is $\sigma$-subnormal in $\langle O^{\sigma _i}(X), A \rangle$, and, by Lemma 5, X normalises $(O^{\sigma _i}(X))^{A}$. Therefore $O^{\sigma _i}(X)$ is a $\sigma$-subnormal subgroup of $O^{\sigma _i}(X)^{A}$ which is a normal subgroup of G. Consequently $O^{\sigma _i}(X)$ is a $\sigma$-subnormal $\sigma$-nilpotent subgroup of G. By Lemma 3 (4), $O^{\sigma _i}(X)$ is contained in $F_{\sigma }(G) = O_{\sigma _i}(G)$. Hence X is a $\sigma _i$-group, contrary to supposition.

Consequently, $O^{\sigma _i}(X) = 1$ and X is a $\sigma _i$-group. Since every minimal normal subgroup N of G is a $\sigma _i$-group, and XN is $\sigma$-subnormal in G, it follows that X is $\sigma$-subnormal in G. This final contradiction proves the theorem. $\square$

Proof of Theorem C

Suppose that the theorem is false and let G be a counterexample for which $|G|+ |X| + |G : A| + |G : B|$ is minimal. Note that every proper $\sigma$-subnormal subgroup Z of X satifies the hypotheses of the theorem. Therefore Z is a $\sigma$-subnormal subgroup of G by the choice of (G, X).

We proceed in a number of steps.

Step 1. IfXis not contained inA, then$G = \langle A,X \rangle$and |G : A| is not$\sigma$-primary.

Let $A_{0}=\langle A,X \rangle$. We have that $A_{0}=A_{0} \cap AB=A(A_{0} \cap B)$ and $G=A_{0}B$. If $A_{0} \ne G$, then $A_{0}$ is not a counterexample to the theorem. Then X is $\sigma$-subnormal in $A_{0}$, and the 4-tuple $(G, X, A_0, B)$ satisfies the hypotheses of the theorem. The minimal choice of (G, X, A, B) implies that X is $\sigma$-subnormal in G. Consequently, $G=\langle A,X \rangle$. If |G : A| were $\sigma$-primary, then we would have X is $\sigma$-subnormal in G by Theorem D. This is not the case. Thus |G : A| is not $\sigma$-primary.

Step 2. Assume thatXis contained inAand |G : A| is$\sigma$-primary. IfXis not contained inB, then |G : A| and |G : B| are not$\sigma$-coprime.

Assume that X is not contained in B and |G : A| and |G : B| are $\sigma$-coprime and derive a contradiction. Let $B_{0}=\langle X,B \rangle =B(B_{0} \cap A)$. Then B is a proper subgroup of $B_0$ and $G = AB_0$. Then $(B_0, X, B_{0} \cap A, B)$ satisfies the hypotheses of the theorem. Suppose that $B_{0}$ is a proper subgroup of G. Then the theorem holds in $B_0$, and hence X is $\sigma$-subnormal in $B_0$. Applying Theorems A and B, we conclude that X is $\sigma$-subnormal in G. This contradicts the choice of G, however, and we conclude that $G = \langle X,B \rangle$.

By hypothesis, |G : A| is a $\sigma _i$-number, for some $\sigma _i \in \sigma$. Since |G : A| and |G : B| are $\sigma$-coprime, it follows that |G : B| is a $\sigma _i^{'}$-number. Therefore B contains a Hall $\sigma _i$-subgroup of G.

Let N be a minimal normal subgroup of G. Then N is $\sigma$-primary. Assume that N is a $\sigma _j$-group, where $j \ne i$. Since |G : A| is $\sigma _i$-number, then $N \le A$. By the choice of G, XN is a $\sigma$-subnormal subgroup of G. Moreover, $XN \le A$. Therefore X is $\sigma$-subnormal in XN and then in G, a contradiction. Consequently, every minimal normal subgroup of G is a $\sigma _i$-group and $F_{\sigma }(G)=O_{\sigma _i}(G)$. Moreover, $R = O_{\sigma _i}(G)$ is contained in B.

Suppose that X is not $\sigma$-nilpotent. Then $O^{\sigma _i}(X) \ne 1$. Suppose that $O^{\sigma _i}(X)$ is a proper subgroup of X. Then it is $\sigma$-subnormal in G. By Lemma 1 (8), $O_{\sigma _i}(G)$ normalises $O^{\sigma _i}(O^{\sigma _i}(X))=O^{\sigma _i}(X)$ and hence $XO_{\sigma _i}(G)$ normalises $O^{\sigma _i}(X)$. Then $X/O^{\sigma _i}(X)$ is $\sigma$-subnormal in $XO_{\sigma _i}(G)/O^{\sigma _i}(X)$. Thus X is $\sigma$-subnormal in $XO_{\sigma _i}(G)$ which is $\sigma$-subnormal in G by minimality of G and Lemma 1 (5). Lemma 1 (2) yields that X is $\sigma$-subnormal in G, contrary to supposition. Thus $O^{\sigma _i}(X)=X$.

On the other hand, since X is not $\sigma$-nilpotent, $1 \ne X^{{{{\mathcal {N}}}}_{\sigma }}$ is $\sigma$-subnormal in G. Therefore $1 \ne F_{\sigma }(X^{{{{\mathcal {N}}}}_{\sigma }})$ is a $\sigma$-nilpotent $\sigma$-subnormal subgroup of G contained in $F_{\sigma }(G)=O_{\sigma _i}(G)$ by Lemma 3 (4). In particular, $O_{\sigma _i}(X) \ne 1$. Applying Lemma 5, we conclude that X normalises $(O^{\sigma _i}(X))^{B}$. Hence $(O^{\sigma _i}(X))^{B}$ is a normal subgroup of G. Write $Z = X \cap O_{\sigma _i}(G)$. Then $1 \ne Z$ is a $\sigma$-subnormal $\sigma _i$-subgroup of G. Let $b \in B$. Then X is $\sigma$-subnormal in $\langle X, Z^{b} \rangle$ and so $O_{\sigma _i}(\langle X,Z^{b} \rangle )$ normalises $O^{\sigma _i}(X) = X$ by Lemma 1 (8). Since $Z^{b} \le O_{\sigma _i}(\langle X,Z^{b} \rangle )$, it follows that $Z^{b}$ normalises X. Therefore $Z^B$ normalises X. Then $[Z^{B},X] \le X \cap O_{\sigma _i}(G)=Z \le Z^{B}$ and $Z^{B}$ is normal in G. By the choice of G, it follows that $XZ^{B}$ is a $\sigma$-subnormal subgroup of G and then X is $\sigma$-subnormal in G, a contradiction.

Thus X is $\sigma$-nilpotent. By assumption every proper subgroup of X is $\sigma$-subnormal in G. Applying Lemma 3 (3), X is a cyclic p-group for some prime $p \in \sigma _j$, for some $\sigma _j \in \sigma$. Assume that $i = j$. Then XN is a $\sigma$-subnormal $\sigma _i$-subgroup of G. Consequently, X is $\sigma$-subnormal in G, which contradicts our assumption that G is a counterexample. Thus $i \ne j$ and $O^{\sigma _i}(X)=X$. By Lemma 7, $R=O_{\sigma _i}(G)$ normalises X, and so X is normal in XR. Since XR is $\sigma$-subnormal in G by minimality of G and Lemma 1 (5), we conclude that X is $\sigma$-subnormal in G, which is not the case.

Step 3. We have a contradiction

Assume that either |G : A| and |G : B| are $\sigma$-primary or |G : A| is $\sigma$-primary and |G : A| and |G : B| are $\sigma$-coprime. Then, by Steps 1 and 2, $X \subseteq A \cap B$. Then, by Theorem A, X is $\sigma$-subnormal in A and B. Therefore X is $\sigma$-subnormal in G by Theorem B. $\square$