On σ-Residuals of Subgroups of Finite Soluble Groups

Abstract

Let $\sigma =\{{\sigma }_i:i\in I\}$ be a partition of the set of all prime numbers. A subgroup H of a finite group G is said to be $\sigma$-subnormal in G if H can be joined to G by a chain of subgroups $H=H_0\subseteq H_1\subseteq \cdots \subseteq H_n=G$ where, for every $j=1,\cdots ,n$, $H_{j-1}$ is normal in $H_j$ or $H_j/Cor{e}_{H_j}\left(H_{j-1}\right)$ is a ${\sigma }_i$-group for some $i\in I$. Let B be a subgroup of a soluble group G normalising the ${\mathfrak{N}}_{\sigma }$-residual of every non-$\sigma$-subnormal subgroup of G, where ${\mathfrak{N}}_{\sigma }$ is the saturated formation of all $\sigma$-nilpotent groups. We show that B normalises the ${\mathfrak{N}}_{\sigma }$-residual of every subgroup of G if G does not have a section that is $\sigma$-residually critical.

1. Introduction

All groups considered in this paper are finite.

Recall that a class of groups $\mathfrak{F}$ is called a formation if $\mathfrak{F}$ is closed when taking epimorphic images and every group G has the smallest normal subgroup with quotient in $\mathfrak{F}$. This subgroup is called the $\mathfrak{F}$-residual of G and it is denoted by $G^{\mathfrak{F}}$. It is known that $G^{\mathfrak{F}}$ is epimorphism-invariant (see ([1], Proposition 2.2.8)). A formation $\mathfrak{F}$ is saturated if a group G belongs to $\mathfrak{F}$ provided that the Frattini quotient $G/\Phi \left(G\right)$ is an $\mathfrak{F}$-group.

Gong and Isaacs ([2], Theorem A) showed that if a subgroup A of a group G normalises the nilpotent (respectively, soluble) residual of each non-subnormal subgroup of G, then A normalises the nilpotent (respectively, soluble) residual of every subgroup of G.

It was shown by Ballester-Bolinches et al. [3] that Gong and Isaacs’ results can be obtained owing to a general completeness property of all subgroup-closed saturated formations containing the class of all nilpotent groups.

Theorem 1.

Let $\mathfrak{F}$ be a subgroup-closed saturated formation containing the class $\mathfrak{N}$ of all nilpotent groups. Assume that A is a subgroup of a group G normalising the $\mathfrak{F}$-residual $H^{\mathfrak{F}}$ of every non-subnormal subgroup H of G. Then, A normalises $H^{\mathfrak{F}}$ for all subgroups H of G.

A powerful extension of subnormality within the framework of formation theory is the K-$\mathfrak{F}$-subnormality introduced by Kegel.

Definition 1.

Let $\mathfrak{F}$ be a formation. A subgroup H of a group G is called K-$\mathfrak{F}$-subnormal in G if H can be joined to G by a chain of subgroups

$$H=H_0\subseteq H_1\subseteq \cdots \subseteq H_n=G,$$

with $H_{i-1}$ normal in $H_i$ or $H_i/{Core}_{H_i}\left(H_{i-1}\right)\in \mathfrak{F}$ for every $1\le i\le n$.

The K-$\mathfrak{F}$-subnormal subgroups associated with subgroup-closed saturated formations have been extensively investigated (see ([1], Chapter 6)). Note that if $\mathfrak{F}=\mathfrak{N}$, the K-$\mathfrak{N}$-subnormal subgroups of a group G are the subnormal subgroups of G.

Bearing in mind the crucial role of the K-$\mathfrak{F}$-subnormality associated with subgroup-closed saturated formations in the structural study of groups, we are led naturally to inquire about a possible extension of Theorem 1 to K-$\mathfrak{F}$-subnormal subgroups.

Unfortunately, the solution to this problem seems to be intractable. However, we can show a significant extension of Theorem 1 for the subgroup-closed saturated formation ${\mathfrak{N}}_{\sigma }$ of all $\sigma$-nilpotent groups, where $\sigma =\{{\sigma }_i:i\in I\}$ is a partition of the set of all prime numbers, which was studied by Skiba in his seminal paper [4].

Recall that a group G is said to be $\sigma$-primary if the primes dividing $\left|G\right|$ belong to the same member of the partition $\sigma$.

Definition 2.

A group G is called σ-nilpotent if it is a direct product of σ-primary groups.

If $\sigma =\left\{\right\{2\},\{3\},\{5\},\cdots \}$, then the class of $\sigma$-nilpotent groups is simply the class of nilpotent groups.

It is known that the class ${\mathfrak{N}}_{\sigma }$ of all $\sigma$-nilpotent groups is a subgroup-closed saturated Fitting formation ([4], Corollary 2.4 and Lemma 2.5). Therefore, every group has ${\mathfrak{N}}_{\sigma }$-projectors ([5], Theorem A.3.10), which are called $\sigma$-projectors here.

The ${\mathfrak{N}}_{\sigma }$-residual of a group G is called the $\sigma$-residual of G and it will be denoted by $G^{*}$. If S is a $\sigma$-subnormal subgroup of a soluble group G, then $S^{*}$ is subnormal in G by ([1], Lemma 6.1.9).

The K-${\mathfrak{N}}_{\sigma }$-subnormal subgroups of every soluble group G are merely the $\sigma$-subnormal subgroups of G introduced by Skiba in [4].

Definition 3.

A subgroup H of a group G is called σ-subnormal in G if H can be joined to G by a chain of subgroups

$$H=H_0\subseteq H_1\subseteq \cdots \subseteq H_n=G,$$

where, for every $i=1,\cdots ,n$, $H_{i-1}$ is normal in $H_i$ or $H_i/{Core}_{H_i}\left(H_{i-1}\right)$ is σ-primary.

It is clear that a subgroup H of a soluble group G is $\sigma$-subnormal in G if and only if there exists $H=H_0\subseteq H_1\subseteq \cdots \subseteq H_n=G$, a chain of subgroups of G such that $H_{i-1}$ is a maximal subgroup of $H_i$ and $H_i/{Core}_{S_i}\left(H_{i-1}\right)$ is $\sigma$-primary for every $i=1,\cdots ,n$.

If we replace “subnormal subgroup” with “$\sigma$-subnormal subgroup” in Theorem 1, the analogous result can fail even in the soluble universe. In fact, there exist exceptions to the generalisation of Theorem 1 to $\sigma$-subnormal subgroups.

Definition 4.

Let $\sigma =\{{\sigma }_i:i\in I\}$ be a partition of the set of all prime numbers. A group S is said to be σ-residually critical, if there exist $p,q,r$ distinct primes with $p\in {\sigma }_i$, $\{q,r\}\subseteq {\sigma }_j$ for some ${\sigma }_i,{\sigma }_j\in \sigma$, $i\ne j$ such that $S=NRQ$, where N and $NR$ are normal in S and

1.

$N\in {Syl}_p\left(S\right)$, $R\in {Syl}_r\left(S\right)$ and $Q\in {Syl}_q\left(S\right)$;

2.

R is cyclic of order r and Q is cyclic of order q and q divides $r-1$;

3.

N is an irreducible and faithful $RQ$-module over the field of p-elements;

4.

N, regarded as a Q-module, is a direct sum of irreducibles and a trivial Q-module.

Example 1.

Let Q be a cyclic group of order 2. Let $p,r$ be odd primes such that r divides $p-1$. Then, Q has a faithful one-dimensional module R over the field of r-elements by ([5], Theorem B. 9.8). Let $B=\left[R\right]Q$ be the corresponding semidirect product. Then, B is a primitive group and Q is a core-free maximal subgroup of B. By ([5], Corollary B.11.7), B has a faithful and irreducible module N over the field of p-elements such that N, regarded as a Q-module, has a quotient isomorphic to the trivial Q-module. Since N is a completely irreducible Q-module, N has a trivial Q-submodule as a direct summand. Furthermore, N, as an R-module, is a direct sum of one-dimensional irreducibles by ([5], Theorem B. 9.8). Since 2 divides $p-1$, we can apply ([6], Lemma 2) to conclude that dim $N=2$. Let $S=\left[N\right]B$ be the corresponding semidirect product. Then, S is a σ-residually critical group.

To avoid sections that are $\sigma$-residually critical is necessary to obtain a $\sigma$-subnormal extension of Theorem 1 for the saturated formation of all $\sigma$-nilpotent groups.

Lemma 1.

If S is σ-residually critical, then $NR$ normalises the σ-residual of every non-σ-subnormal subgroup of S, but $NR$ does not normalise the σ-residual of the σ-subnormal subgroup $NQ$.

Proof. 

Assume that $S=NRQ$ is a $\sigma$-residually critical group. Then, S is a primitive group, $C_S\left(N\right)=N$ is the unique minimal normal subgroup of S, $S^{*}=N$ and $NQ$ is a $\sigma$-subnormal maximal subgroup of S.

Note that $NQ=(N_1\times N_2)Q$, where $N_1\ne N$ is a faithful irreducible Q-module and $N_2\subseteq Z\left(NQ\right)$. Therefore, ${\left(NQ\right)}^{*}=N_1$ and $N_1$ cannot be normalised by R since otherwise $N_1$ would be normal in S.

Let U be a non-$\sigma$-subnormal subgroup of S. Then, U is contained in a maximal subgroup of G. We may assume that U is contained in either $RQ$ or $NQ$ or $NR$. If $U\subseteq RQ$, then $U^{*}=1$. Suppose that U is contained in $NQ$. Then, either $U^{*}=1$ or $U^{*}=N_1$. Assume that $U^{*}=N_1$. Then, U is $\sigma$-subnormal in $NQ$, which is $\sigma$-subnormal in G, and this is not possible. Hence, $U^{*}=1$. In both cases, $NR$ normalises $U^{*}=1$.

Assume that $U\subseteq NR$. Since U is not $\sigma$-subnormal in S and $NR$ is normal in S, it follows that U is not $\sigma$-subnormal in $NR$. In particular, R is contained in U and $U=N_{0}R$ for some normal subgroup $N_0$ of U contained in N. Then, $U^{*}\subseteq N_0$ and so $U^{*}$ is normalised by $NR$. □

We prove the following.

Theorem 2.

Let $\sigma =\{{\sigma }_i:i\in I\}$ be a partition of the set of all prime numbers. Assume that a subgroup A of a soluble group G normalises the σ-residual of every non-σ-subnormal subgroup of G. If G has no σ-residually critical sections, then A normalises the σ-residual of every subgroup of G.

The notation and terminology agree with the books [1,5].

2. Preparatory Lemmas

Our first lemma collects some basic properties of $\sigma$-subnormal subgroups.

Lemma 2

([4]). Let H, K and N be subgroups of a group G. Suppose that H is σ-subnormal in G and N is normal in G. Then,

1.

$H\cap K$ is σ-subnormal in K.

2.

If K is a σ-subnormal subgroup of H, then K is σ-subnormal in G.

3.

If K is σ-subnormal in G, then $H\cap K$ is σ-subnormal in G.

4.

$HN/N$ is σ-subnormal in $G/N$.

5.

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

6.

If $K\subseteq H$ and H is σ-nilpotent, then K is σ-subnormal in G.

7.

If H is a ${\sigma }_i$-group, where ${\sigma }_i\in \sigma$, then $H\subseteq O_{{\sigma }_i}\left(G\right)$.

8.

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

9.

If N is a ${\sigma }_i$-subgroup of G, then $N\subseteq N_G\left(O^{{\sigma }_i}\left(H\right)\right)$.

Lemma 3

([1], Theorem 6.5.46). If $G=\langle A,B\rangle$ is a soluble group generated by σ-subnormal subgroups A and B, then $G^{*}=\langle A^{*},B^{*}\rangle$.

3. Proof of Theorem 2

Proof of Theorem 2.

Assume, arguing by contradiction, that the theorem is false. Then, there exists a group G with a subgroup A normalising the $\sigma$-residual of every non-$\sigma$-subnormal subgroup of G and a $\sigma$-subnormal subgroup S such that A does not normalise $S^{*}$. We choose $(G,A,S)$ such that $\left|G\right|+|G:A|+\left|S\right|$ is minimal. Then, S is a proper subgroup of G. Let B be the intersection of the normalisers of the $\sigma$-residuals of all non-$\sigma$-subnormal subgroups of G. Then, B is a normal subgroup of G containing A. The choice of $(G,A,S)$ yields $A=B$. We reach a contradiction after the following steps.

$\left(1\right)$ $G=SB$.

Assume that $SB$ is a proper subgroup of G. Then, $(SB,B,S)$ satisfies the hypotheses of the theorem by Lemma 2. Minimality of $(G,B,S)$ yields that B normalises $S^{*}$, and this is not the case.

$\left(2\right)$ Let N be a minimal normal subgroup of G. Then, ${\langle S^{*}\rangle }^G=S^{*}N$ is the normal closure of $S^{*}$ in G. In particular, ${Core}_G\left(S^{*}\right)=1$ and $S^{*}$ is σ-nilpotent.

By Lemma 2, $(G/N,BN/N,SN/N)$ satisfies the hypotheses of the theorem. Consequently, $BN/N$ normalises ${(SN/N)}^{*}=S^{*}N/N$ by the minimal choice of $(G,B,S)$. In particular, B normalises $S^{*}N$ and so $S^{*}N$ is a normal subgroup of G. This yields $1\ne D={\langle S^{*}\rangle }^G\subseteq S^{*}N$.

Assume that ${Core}_G\left(S^{*}\right)\ne 1$. Then, we can assume that $N\subseteq {Core}_G\left(S^{*}\right)$ and $S^{*}$ is normal in G, a contradiction. Therefore, ${Core}_G\left(S^{*}\right)=1$.

Assume that N is not contained in D. Then, $D\cap N=1$ and so $D=S^{*}(D\cap N)=S^{*}$ is normal in G, and this is not the case. Thus, N is contained in D and $D=S^{*}N$.

Note that $S^{*}$ is subnormal in G. Then, the $\sigma$-residual T of $S^{*}$ is subnormal in G. In particular, T is $\sigma$-subnormal in G. Since $S^{*}$ is soluble, we have that T is a proper subgroup of $S^{*}$. The choice of $(G,B,S)$ implies that T is normalised by B and so T is normal in G. Hence, $T\subseteq {Core}_G\left(S^{*}\right)=1$, and $S^{*}$ is $\sigma$-nilpotent.

$\left(3\right)$ $S^{*}$ is a ${\sigma }_i$-group for some ${\sigma }_i\in \sigma$, and every minimal normal subgroup of N is contained in $O_{{\sigma }_i}\left(G\right)$.

Assume that $O_{{\sigma }_i}\left(G\right)\ne 1$ and $O_{{\sigma }_j}\left(G\right)\ne 1$ for some $i\ne j\in I$. Let N be a minimal normal subgroup of G contained in $O_{{\sigma }_i}\left(G\right)\ne 1$ and let W be a minimal normal subgroup of G contained in $O_{{\sigma }_j}\left(G\right)\ne 1$. By statement (2), ${\langle S^{*}\rangle }^G=S^{*}N=S^{*}W$. Hence, ${\langle S^{*}\rangle }^G=S^{*}$ by order considerations. This contradicts our assumption. Furthermore, every minimal normal subgroup of G is abelian. Consequently, $O_{{\sigma }_i}\left(G\right)\ne 1$ for some ${\sigma }_i\in \sigma$ and $O_{{\sigma }_i^{\prime }}\left(G\right)=1$. In particular, every minimal normal subgroup of N is contained in $O_{{\sigma }_i}\left(G\right)$.

Note that $S^{*}$ is a direct product of Hall ${\sigma }_j$-subgroups by statement (2), which are subnormal in G. Since $O_{{\sigma }_i^{\prime }}\left(G\right)=1$, we conclude that $S^{*}$ is a ${\sigma }_i$-group.

$\left(4\right)$ S is non-nilpotent and every proper subgroup of S is nilpotent. $V=S^{*}$ is a minimal normal subgroup of S, which is an elementary abelian p-group for some prime $p\in {\sigma }_i$, and it is complemented by a cyclic group Q of prime order $q\in {\sigma }_j$ for some $j\ne i$.

Let T be a proper subgroup of S. If T is not $\sigma$-subnormal in G, then B normalises $T^{*}$, and if T is $\sigma$-subnormal in S, then T is $\sigma$-subnormal in G by Lemma 2 and so B normalises $T^{*}$ by the minimal choice of S. In any case, $T^{*}$ is normalised by B and so ${\langle T^{*}\rangle }^G={\langle T^{*}\rangle }^S$ is a normal subgroup of G contained in $S^{*}$. Hence, ${\langle T^{*}\rangle }^G\subseteq {Core}_G\left(S^{*}\right)=1$ and T is $\sigma$-nilpotent. Consequently, S is an ${\mathfrak{N}}_{\sigma }$-critical group. By [1], Corollary 6.4.5, S is an $\mathfrak{N}$-critical group, i.e., S is non-nilpotent and every proper subgroup of S is nilpotent.

By [7], $\pi \left(S\right)=\{p,q\}$, $S=\left[V\right]\langle a\rangle$, where V is a normal Sylow p-subgroup of S, and $Q=\langle a\rangle$ is a non-normal Sylow q-subgroup of S. Moreover, $V/\Phi \left(V\right)$ is a non-central chief factor of S. Note that $V=S^{*}$ and so $p\in {\sigma }_i$ by statement (3). Assume that $\Phi \left(V\right)\ne 1$. Since $\Phi \left(V\right)\le \Phi ({\langle S^{*}\rangle }^G)$, there exists a minimal normal subgroup N of G contained in $\Phi ({\langle S^{*}\rangle }^G)$. By statement (2), ${\langle S^{*}\rangle }^G=S^{*}N=S^{*}$. Since this is not the case, we conclude that $V=S^{*}$ is a minimal normal subgroup of S. Because $\Phi \left(Q\right)$ is a normal subgroup of S, it follows that $\Phi \left(Q\right)$ is $\sigma$-subnormal in G by Lemma 2, and since S is not $\sigma$-nilpotent, we conclude that $q\in {\sigma }_j$ for some $j\ne i$. Then, $\Phi \left(Q\right)\le O_{{\sigma }_i^{\prime }}\left(G\right)=1$. Thus, $\left|Q\right|=q$, as claimed.

$\left(5\right)$ $Y=N_G\left(S^{*}\right)$ is the unique maximal subgroup of G containing S. Y is σ-subnormal in G and $Soc\left(G\right)$ is contained in Y. Furthermore, if $S\subseteq H\subseteq Y$, then $S^{*}$ is a minimal normal subgroup of H.

Let M be a maximal subgroup of G containing S. Then, $M=S(M\cap B)$ and $(M,M\cap B,S)$ satisfies the hypotheses of the theorem by Lemma 2. The choice of $(G,B,S)$ guarantees that $M\cap B$ normalises $S^{*}$. Thus, M normalises $S^{*}$ and so $M=N_G\left(S^{*}\right)$. Consequently, $Y=N_G\left(S^{*}\right)$ is the unique maximal subgroup of G containing S. Note that S is contained in a $\sigma$-subnormal maximal subgroup of G because S is $\sigma$-subnormal in G. Therefore, Y is $\sigma$-subnormal in G. Since $S^{*}$ is subnormal in G, it follows that $Soc\left(G\right)$ is contained in Y by ([5], Lemma A.14.3). Assume that L is a non-trivial normal subgroup of H contained in $S^{*}$. Since $S^{*}$ is a minimal normal subgroup of S, we have that $L=S^{*}$.

$\left(6\right)$ $Y^{*}=S^{*}$. In particular, Y is not normal in G and $G^{*}=S^{*}N$ for each minimal normal subgroup N of G.

Since S is $\sigma$-subnormal in Y, there exists $S=S_0\subseteq S_1\subseteq \cdots \subseteq S_n=Y$, a chain of subgroups of Y such that $S_{i-1}$ is a maximal subgroup of $S_i$ and $S_i/{Core}_{S_i}\left(S_{i-1}\right)$ is $\sigma$-primary for every $i=1,\cdots ,n$. We seek to show that $Y^{*}=S^{*}$ by induction on n. Assume that $n=1$. Then, S is a maximal subgroup of Y and $Y/{Core}_Y\left(S\right)$ is $\sigma$-nilpotent. In particular, $S^{*}\subseteq Y^{*}\subseteq {Core}_Y\left(S\right)\subseteq S$. Since $S^{*}$ is maximal in S, it follows that either $S^{*}=Y^{*}$ or $Y^{*}={Core}_Y\left(S\right)=S$. Assume that $Y^{*}=S$. Then, S is a normal subgroup of Y. By the Frattini Argument, $Y=S{N}_Y\left(Q\right)=S^{*}{N}_Y\left(Q\right)$. Note that $L=N_Y\left(Q\right)$ is a non-$\sigma$-subnormal subgroup of G. Therefore, B normalises $L^{*}$ and so ${\langle L^{*}\rangle }^G={\langle L^{*}\rangle }^S\subseteq S$. If L were $\sigma$-nilpotent, then $Y^{*}$ would be contained in $S^{*}$, and this is a contradiction. Hence, ${\langle L^{*}\rangle }^G$ is a non-trivial normal subgroup of G contained in S and thus either $S^{*}={\langle L^{*}\rangle }^G$ or $S={\langle L^{*}\rangle }^G$ is normal in G. Since this is not case, we deduce that $S^{*}=Y^{*}$. Assume that $W=S_{n-1}$ and $S^{*}=W^{*}$. We argue next that $S^{*}=Y^{*}$. Suppose that W is not normal in Y. Then, $Y=\langle W,W^x\rangle$ for some $x\in Y$ since W is maximal in Y. Applying Lemma 3, we conclude that $Y^{*}=\langle S^{*},{\left(S^{*}\right)}^x\rangle =S^{*}$ because $S^{*}$ is a normal subgroup of Y. Assume that W is a normal subgroup of Y. Let F be a $\sigma$-projector of W. Then, $W=S^{*}F$ and $S^{*}\cap F=1$ by ([5], Theorem IV.5.18). Therefore, F is a maximal subgroup of W since $S^{*}$ is a minimal normal subgroup of W by statement (5). Since the $\sigma$-projectors of W are conjugate by ([1], Theorem 4.2.1), we have that $Y=W{N}_Y\left(F\right)=S^{*}{N}_Y\left(F\right)$. Set $F=E\times C$, where E is the Hall ${\sigma }_i$-subgroup of F and C is the Hall ${\sigma }_i^{\prime }$-subgroup of F. Observe that $N_W\left(F\right)=F$ and $C\ne 1$ because W is not $\sigma$-nilpotent. Since $N_Y\left(F\right)$ is a proper subgroup of Y and $S^{*}$ is a minimal normal subgroup of Y by statement (5), it follows that $N_Y\left(F\right)$ is a maximal subgroup of Y. Since $N_Y\left(F\right)\subseteq N_Y\left(C\right)$, we have that either $N_Y\left(F\right)=N_Y\left(C\right)$ or $N_Y\left(C\right)=Y$. In the latter case, C would be $\sigma$-subnormal in G and so $C\subseteq O_{{\sigma }_i^{\prime }}\left(G\right)=1$ by Lemma 2, and this would be a contradiction. Thus, $N_Y\left(F\right)=N_Y\left(C\right)$ and there exists a prime q and a q-element $x\in N_Y\left(F\right)\backslash W$. Then, $Y=W\langle x\rangle =W\left(S^{*}\langle x\rangle \right)$ and $\langle x\rangle$ normalises $S^{*}C$. Suppose that $S^{*}\langle x\rangle$ were $\sigma$-subnormal in G. Then, $S^{*}\langle x\rangle$ would be $\sigma$-subnormal in Y by Lemma 2 and so $Y^{*}=W^{*}{\left(S^{*}\langle x\rangle \right)}^{*}$ by Lemma 3. Since ${\left(S^{*}\langle x\rangle \right)}^{*}$ is contained in $S^{*}$, we conclude that $Y^{*}=S^{*}$. Hence, we may assume that $S^{*}\langle x\rangle$ is not $\sigma$-subnormal in G. By hypothesis, B normalises ${\left(S^{*}\langle x\rangle \right)}^{*}$ and so ${\langle {\left(S^{*}\langle x\rangle \right)}^{*}\rangle }^G={\langle {\left(S^{*}\langle x\rangle \right)}^{*}\rangle }^S\subseteq {\langle S^{*}\rangle }^S=S^{*}$. Consequently, ${\langle {\left(S^{*}\langle x\rangle \right)}^{*}\rangle }^G=1$ by statement (2). In particular, $S^{*}\langle x\rangle$ is $\sigma$-nilpotent. Observe that $C\langle x\rangle$ is $\sigma$-nilpotent. Moreover, $Y=\left(S^{*}E\right)\left(C\langle x\rangle \right)$ and $S^{*}E$ is normal in Y. Since $Y/S^{*}E$ is $\sigma$-nilpotent, it follows that $Y^{*}$ is contained in $S^{*}E$. In particular, $Y^{*}$ is a ${\sigma }_i$-group.

Write $J=S^{*}C\langle x\rangle$. Then, $Y=WJ$, and $J^{*}\subseteq S^{*}$. Assume that J is $\sigma$-subnormal in Y. Then, $Y^{*}=W^{*}{J}^{*}=S^{*}$ by Lemma 3. Assume that J is not $\sigma$-subnormal in Y. Then, J cannot be $\sigma$-subnormal in G by Lemma 2, and so B normalises $J^{*}$. Arguing as above, we conclude that $J^{*}=1$ and J is $\sigma$-nilpotent. In particular, $S^{*}\subseteq N_Y\left(C\right)=N_Y\left(F\right)$, and this is a contradiction because $Y\ne N_Y\left(F\right)$. We conclude that $Y^{*}=S^{*}$, as claimed.

Since $S^{*}$ is not normal in G, it follows that Y cannot be normal in G and so there exists $z\in G$ such that $G=\langle Y,Y^z\rangle$. By Lemma 3 and statement (2), $G^{*}=\langle Y^{*},{\left(Y^{*}\right)}^z\rangle =\langle S^{*},{\left(S^{*}\right)}^z\rangle ={\langle S^{*}\rangle }^G=S^{*}N$ for each minimal normal subgroup N of G.

$\left(7\right)$ G is a σ-residually critical group, the final contradiction.

Let $G_i$ be a Hall ${\sigma }_i$-subgroup of G. Then, $G^{*}\subseteq G_i$ and so $G_i$ is a normal subgroup of G. By Lemma 2, $G_i$ normalises $O^{{\sigma }_i}\left(S\right)=S$. Hence, $G_i$ is contained in Y. Since G is soluble, there exists a prime r and a Sylow r-subgroup R of G such that $G=YR$. Then, $r\in {\sigma }_k$ for some $k\ne i$. Suppose that $r=q$. We may assume that $Q\subseteq R$ and so $R=Q(B\cap R)$. If $G^{*}R$ were a proper subgroup of G, then $B\cap R$ would normalise $S^{*}$ by the choice of $(G,B,S)$ (note that $S\subseteq G^{*}R$). Then, R would be contained in Y, and this would be a contradiction. Therefore, $G=G^{*}R$, Y is a normal subgroup of G and $S^{*}$ is also normal in G, and this is a contradiction. Consequently, $r\ne q$ and thus R is contained in B by statement (1). Let $Q_1$ be a Sylow q-subgroup of G contained in Y and containing Q. We may assume that $Q_{1}R$ is a subgroup of G. Suppose that $G^{*}\left(Q_{1}R\right)$ is a proper subgroup of G. Then, S is contained in $G^{*}\left(Q_{1}R\right)$. The choice of $(G,B,S)$ guarantees that R normalises $S^{*}$. In particular, R is contained in Y, a contradiction. Therefore, $G=G^{*}\left(Q_{1}R\right)$.

Assume that $k\ne j$. Then, $G^{*}{Q}_1$ is normal in G and so $Y/G^{*}{Q}_1$ is normal in $G/G^{*}{Q}_1$. This means that Y is normal in G and so is $S^{*}$, a contradiction. Consequently, $\{q,r\}\subseteq {\sigma }_j$. Let L be a maximal subgroup of $Q_{1}R$ containing $Q_1$. Then, $G^{*}L$ is a maximal subgroup of G normalising $S^{*}$ by the choice of $(G,B,S)$. Hence, $Y=G^{*}L$. Suppose that $S^{*}\cap J=1$ for all minimal normal subgroups J of G. By statements (2) and (4), $Soc\left(G\right)\subseteq S^{*}N$ for some minimal normal subgroup N of G. Then, N is ${\mathfrak{N}}_{\sigma }$-central in Y and so it is contained in every $\sigma$-projector $Y_0$ of Y by ([1], Theorem 4.1.18 and Proposition 4.1.22). Since $S^{*}\cap Y_0=1$, we have that $Soc\left(G\right)=N{S}^{*}\cap Y_0=N(S^{*}\cap Y_0)=N$. Thus, G has a unique minimal normal subgroup N. Observe that $G^{*}$ is a completely reducible $Q_{1}R$-module over the field of p-elements by Maschke’s theorem ([5], Theorem A.11.5). Hence, $G^{*}$ is a direct product of distinct minimal normal subgroups of G, and this is a contradiction. Thus, $S^{*}\cap N\ne 1$, $S^{*}\subseteq N$ and $N=G^{*}$ by statement (6). Therefore, N is the unique minimal normal subgroup of G and it is complemented in G by every $\sigma$-projector of G by ([5], Theorem IV.5.18). Hence, G is a primitive group and $N=C_G\left(N\right)$, and $R{Q}_1$ is a core-free maximal subgroup of G.

We have that ${Core}_G\left(Y\right)$ contains N, and since ${Core}_G{\left(Y\right)}^{*}$ is a normal subgroup of G contained in $S^{*}$, it follows that ${Core}_G{\left(Y\right)}^{*}=1$ and ${Core}_G\left(Y\right)$ is $\sigma$-nilpotent. Hence, $N={Core}_G\left(Y\right)$. Then, $G/N=(RN/N)(Y/N)$ is a primitive group and so $RN/N=Soc(G/N)$ is an elementary abelian r-group and $R\cap Y=1$. In particular, $L=Q_1$ is a maximal subgroup of $Q_{1}R$ and R is an elementary abelian r-subgroup of G. However, $RN$ is normal in G. Hence, $R=RN\cap R{Q}_1$ is normal in $R{Q}_1$. Note that N is an irreducible and faithful $R{Q}_1$-module over the field of p-elements. Then, by ([5], Theorem A.11.5), either N is an irreducible R-module or N is a direct product of irreducible R-modules that are faithful for R. In any case, R is cyclic by ([5], Corollary B.9.4). Thus, $\left|R\right|=r$.

Since $S^{*}$ is a minimal normal subgroup of Y and N is a completely reducible Y-module, it follows that $N=S^{*}\times X$ for some normal subgroup X of Y. Now, $Y^{*}=S^{*}$ by statement (6). Hence, X is contained in the ${\mathfrak{N}}_{\sigma }$-hypercentre of Y and so it is contained in every $\sigma$-projector $Y_0$ of Y by ([1], Theorem 4.1.18 and Proposition 4.1.22). In particular, X is centralised by $Q_1$ and so $X\subseteq Z\left(Y\right)$. Then, $C_Y\left(S^{*}\right)=C_Y\left(N\right)=N$. Let A be a maximal subgroup of $Q_1$ and assume that $A\ne 1$. Then, $N\left(RA\right)$ is a proper subgroup of G containing S. The choice of $(G,B,S)$ guarantees that $R\subseteq N\left(RA\right)\cap B$ normalises $S^{*}$. Hence, R is contained in Y, and since this is not the case, we conclude that $A=1$ and $Q=Q_1$ is a cyclic group of order q. Note that $N=S^{*}\times X$ and $X\subseteq Z\left(Y\right)$. Hence, X is a trivial Q-module. Assume that $RQ=R\times Q$. Then, $U^{\mathfrak{N}}=U^{*}$ for all subgroups U of G. This contradicts ([2], Theorem A). Hence, $C_{RQ}\left(R\right)=R$ and q divides $r-1$. Consequently, G is a $\sigma$-residually critical group, and this is the final contradiction. □