| Izvestiya: Mathematics |
|
|
|
|
|
When is the search of relatively maximal subgroups reduced to quotient groups? Wen Bin Guoab, D. O. Revincde a School of Science, Hainan University, Haikou, Hainan, P. R. China b Department of Mathematics, University of Science and Technology of China, Hefei, P. R. China c Sobolev Institute of Mathematics, Siberian Branch of the Russian Academy of Sciences, Novosibirsk d N.N. Krasovskii Institute of Mathematics and Mechanics, Ural Branch of the Russian Academy of Sciences, Ekaterinburg e Novosibirsk State University
Bibliographic databases:
     § 1. Introduction1.1. The main resultIn what follows, we will consider only finite groups, and a ‘group’ will always mean a ‘finite group’. A group from a class of groups $\mathfrak{X}$ will be simply called an $\mathfrak{X}$-group. The set of (inclusion) maximal $\mathfrak{X}$-subgroups (or $\mathfrak{X}$-maximal subgroups) of a group $G$ will be denoted by $\operatorname{m}_{\mathfrak{X}}(G)$. The group $G$ itself, acting on $\operatorname{m}_{\mathfrak{X}}(G)$ by conjugacies, splits this set into orbits (the conjugacy classes). The number of these classes is denoted by $\mathrm{k}_{\mathfrak{X}}(G)$. The term a ‘relatively maximal subgroup’, which we used in the title of the present paper, was proposed by Wielandt [1] in order to denote $\mathfrak{X}$-maximal subgroups without indication of a concrete class $\mathfrak{X}$ and to distinguish them from maximal subgroups (in the usual sense, that is, from the maximal ones among the proper ones). Following Wielandt [2], [3], we say that the nonempty class $\mathfrak{X}$ of finite groups is complete if it is closed under taking subgroups, homomorphic images, and extensions. The last means that $G\in\mathfrak{X}$, whenever $N\in\mathfrak{X}$ and $G/N\in\mathfrak{X}$ for some normal subgroup $N$ of the group $G$. For a complete class $\mathfrak{X}$, the problem of when is the search of $\mathfrak{X}$-maximal subgroups of a group $G$ reduced to the analogous problem for the quotient group $G/N$, as formulated in the title of the paper, turns out to be equivalent to the problem of when $\mathrm{k}_{\mathfrak{X}}(G)=\mathrm{k}_{\mathfrak{X}}(G/N)$? The main result of the paper is the following Theorem 1. Let $\mathfrak{X}$ be a complete class of groups and $N$ be a normal subgroup of a finite group $G$. Then if $\mathrm{k}_{\mathfrak{X}}(G)=\mathrm{k}_{\mathfrak{X}}(G/N)$, then $\mathrm{k}_{\mathfrak{X}}(N)=1$. The converse result to Theorem 1 was also proved in [4], Theorem 1. The following theorem holds. Theorem 2. Let $\mathfrak{X}$ be a complete class of groups and $N$ be a normal subgroup of a finite group $G$. Then $\mathrm{k}_{\mathfrak{X}}(G)=\mathrm{k}_{\mathfrak{X}}(G/N)$ if and only if $\mathrm{k}_{\mathfrak{X}}(N)=1$. That is, the number of classes of conjugate $\mathfrak{X}$-maximal subgroups remains unchanged when transiting from a group $G$ to the quotient group $G/N$ if and only if in $N$ all $\mathfrak{X}$-maximal subgroups are conjugate. There is also an exhaustive description of the groups $A$ with $\mathrm{k}_{\mathfrak{X}}(A)=1$: this condition is equivalent to saying that each nonabelian composition factor of the group $A$ either lies in $\mathfrak{X}$, or is isomorphic to a simple group indicated in [4], Appendix A. So, in the case of a complete class $\mathfrak{X}$, we give a complete answer to the question raised in the title of the present paper. The assumption in Theorem 2 that the class $\mathfrak{X}$ is complete is essential. Indeed, the conclusion of the theorem fails to hold if $\mathfrak{X}=\mathfrak{A}$ is the class of all abelian groups or $\mathfrak{X}=\mathfrak{N}$ is the class of all nilpotent groups. In both cases, $\mathfrak{X}$ is not closed under extensions and $\mathrm{k}_{\mathfrak{X}}(\operatorname{Sym}_3)=2\ne 1=\mathrm{k}_{\mathfrak{X}}(\operatorname{Sym}_3/\operatorname{Alt}_3)$, even though $\mathrm{k}_{\mathfrak{X}}(\operatorname{Alt}_3)=1$. 1.2. Motivation and historical remarksThe following class of problems appears in the finite group theory and its applications starting from the seminal studies of É. Galois and C. Jordan: given a group $G$ (for example, a symmetric group) and a class $\mathfrak{X}$ of finite groups (for example, the class of solvable groups), find the $\mathfrak{X}$-subgroups of the group $G$. It seems that problems of this kind cannot be successfully attacked in the general setting. If the class $\mathfrak{X}$ is complete (similarly to the class of solvable groups), one may confine oneself with search of $\mathfrak{X}$-maximal subgroups. In what follows, $\mathfrak{X}$ will always denote a fixed complete class. In addition to the class $\mathfrak{S}$ of solvable groups, among typical examples of complete classes we mention the class $\mathfrak{G}_\pi$ of all $\pi$-groups and the class $\mathfrak{S}_\pi$ of all solvable $\pi$-groups for a given subset $\pi$ of the set $\mathbb{P}$ of all primes (recall that a $\pi$-group is a group in which any prime divisor of the order lies in $\pi$). Note that, for the class $\mathfrak{X}$, we have the inclusions:
$$
\begin{equation*}
\mathfrak{S}_\pi\subseteq\mathfrak{X}\subseteq\mathfrak{G}_\pi,
\end{equation*}
\notag
$$
here $\pi$ is the set
$$
\begin{equation*}
\pi(\mathfrak{X})=\{p\in\mathbb{P}\mid \text{ there exists } G\in\mathfrak{X} \text{ such that } p \text{ divides } |G|\}.
\end{equation*}
\notag
$$
The classes of $\pi$-separable and $\pi$-solvable groups1 are also complete. It is natural that the $\mathfrak{X}$-maximal subgroups should be studied up to a conjugacy. By an $\mathfrak{X}$-scheme we will mean a complete system of representatives of its classes of conjugate $\mathfrak{X}$-maximal subgroups. The cardinality of an $\mathfrak{X}$-scheme of a group $G$ is defined as the above number $\mathrm{k}_{\mathfrak{X}}(G)$. The main aim in the problems mentioned above can be looked upon as the search of an $\mathfrak{X}$-scheme and description of the structure of its elements. If $\mathfrak{X}=\mathfrak{G}_p$ is the class of $p$-groups for a prime $p$, then any $\mathfrak{X}$-maximal subgroup is a Sylow $p$-subgroup. Such subgroups in any group are conjugate [5]. It is also worth mentioning that the $\mathfrak{X}$-maximal subgroups of solvable groups are, precisely, the so-called $\pi(\mathfrak{X})$-Hall subgroups, which form a conjugacy class by Hall’s theorem [6]. The search of the Sylow and Hall subgroups is substantially facilitated by their properties that allow one to change from a group to sections of a normal or a subnormal series in inductive arguments. For example, if $H$ is a Sylow $p$-subgroup of the group $G$ and $N\trianglelefteq G$, then $H\cap N$ and $HN/N$ are Sylow $p$-subgroups in $N$ and $G/N$, respectively. In the general case, the problems under considerations are highly noniductive, inasmuch as both the intersection $H\cap N$ of the subgroups $H\in\operatorname{m}_{\mathfrak{X}}(G)$ and $N\trianglelefteq G$, or the image of $HN/N$ in $G/N$ (or, equivalently, the image of $H$ under an arbitrary epimorphism) may fail to be $\mathfrak{X}$-maximal subgroups in $N$ and $G/N$ (see [2], [3]). However, for intersections with normal subgroups, the situation can be partially improved by studying the $\mathfrak{X}$-submaximal subgroups,2 which are generalizations of $\mathfrak{X}$-maximal subgroups (see [3]). The present paper is concerned with the behaviour of $\mathfrak{X}$-maximal subgroups under homomorphisms. It is known (see [2], § 14.2, [3], § 4.3) that if, for a class $\mathfrak{X}$, there exists a group $L$ with nonconjugate $\mathfrak{X}$-maximal subgroups, then any group $G$ is the image of a homomorphism (more precisely, of the natural epimorphism from the regular wreath product $L \wr G$) under which each (not only $\mathfrak{X}$-maximal) $\mathfrak{X}$-subgroup coincides with the image of some $\mathfrak{X}$-maximal subgroup. In other words, an attempt to extend the concept of an $\mathfrak{X}$-maximal subgroup to be in accord with homomorphic images calls for the study of all $\mathfrak{X}$-subgroups. Another challenge associated with the transition to the epimorphic image is that the images of nonconjugate $\mathfrak{X}$-maximal subgroups may happen to be conjugate, and, as a result, information on conjugacy classes may be lost. Consequently, it is important to describe all the cases where a transition from a group $G$ to the quotient group $G/N$ is a reduction for the highlighted type of problems, that is, when $\mathfrak{X}$-maximality of subgroups is preserved and information on their conjugacy is not distorted. In other words, it is important to know when an $\mathfrak{X}$-scheme is carried over to an $\mathfrak{X}$-scheme; in particular,
$$
\begin{equation}
\mathrm{k}_{\mathfrak{X}}(G)=\mathrm{k}_{\mathfrak{X}}(G/N).
\end{equation}
\tag{1.1}
$$
The situation $N\in\mathfrak{X}$ is an example. Less straightforward examples were given by Wielandt in [2], who, following Chunikhin [7]–[10], proposed a programme for finding all ‘good’ cases. The present paper concludes this program. Equality (1.1) is not only necessary but also a sufficient condition that a canonical (or any other) epimorphism $\overline{\phantom{G}}\colon G\to G/N$ would send an $\mathfrak{X}$-scheme of a group $G$ into the $\mathfrak{X}$-scheme $G/N$. Indeed, it is known that each $\mathfrak{X}$-subgroup of $\overline{G}$ is the image of an $\mathfrak{X}$-subgroup from $G$ (see Lemma 1). Hence $\operatorname{m}_{\mathfrak{X}}(\overline{G})\subseteq\{\overline{H}\mid H\in \operatorname{m}_{\mathfrak{X}}(G)\}$ and $\mathrm{k}_{\mathfrak{X}}(\overline{G})\leqslant \mathrm{k}_{\mathfrak{X}}(G)$. So, (1.1) implies
$$
\begin{equation*}
\operatorname{m}_{\mathfrak{X}}(\overline{G})=\{\overline{H}\mid H\in \operatorname{m}_{\mathfrak{X}}(G)\}.
\end{equation*}
\notag
$$
Therefore, the existence of some one-one correspondence between classes of conjugate $\mathfrak{X}$-maximal subgroups in the groups $G$ and $\overline{G}=G/N$, as implied by (1.1), gives evidence about the presence of a natural one-one correspondence between these classes induced by the mapping $H\mapsto \overline{H}$. We will say that the reduction $\mathfrak{X}$-theorem – holds for a pair $(G,N)$, $N\trianglelefteq G$, if (1.1) holds; – holds for a group $A$ if it holds for any pair $(G,N)$ with $N\cong A$. Setting $G=A$, we see that the reduction $\mathfrak{X}$-theorem for a group $A$ implies the conjugacy of the $\mathfrak{X}$-maximal subgroups: $\mathrm{k}_{\mathfrak{X}}(A)=\mathrm{k}_{\mathfrak{X}}(A/A)=1$. Wielandt [2], § 15.4, noted that the reduction $\mathfrak{X}$-theorem itself for $A$, would, in turn, follow from the conjugacy of the $\mathfrak{X}$-submaximal subgroups, and raised the conjecture (see [2], Offene Frage zu 15.4, which was proved later in [4], Theorem 1), to the effect that the conjugacy of the $\mathfrak{X}$-maximal and $\mathfrak{X}$-submaximal subgroups are equivalent. Therefore, the reduction $\mathfrak{X}$-theorem for the group $A$ is equivalent to the equality $\mathrm{k}_{\mathfrak{X}}(A)=1$. Next, the condition $\mathrm{k}_{\mathfrak{X}}(A)=1$ is equivalent to saying that $\mathrm{k}_{\mathfrak{X}}(S)=1$ for any composition factor $S$ for the group $A$. If $S$ is a simple group, then necessary and sufficient arithmetic conditions on natural parameters3 of the group $S$ for the equality $\mathrm{k}_{\mathfrak{X}}(S)=1$ to hold are known (see [4], Theorem 1, Appendix A). So, the results of [4] can be interpreted as a description of all such pairs $(G,N)$ for which equality (1.1) is controlled only by the isomorphism type of the group $N$. The isomorphism type of a group $G$ and of its normal subgroup $N$ do not define uniquely the number $\mathrm{k}_{\mathfrak{X}}(G/N)$. For example, the group $G=\operatorname{PSL}_2(7)\times \operatorname{PGL}_2(7)$ has two normal subgroups $N_1$ and $N_2$ such that $N_1\cong N_2\cong\operatorname{PSL}_2(7)$, $G/N_1\cong\mathbb{Z}_2\times \operatorname{PSL}_2(7)$, and $G/N_2\cong\operatorname{PGL}_2(7)$. However, for the class $\mathfrak{X}=\mathfrak{S}$ of solvable groups it can be easily shown (see, for example, [11]) that $\mathrm{k}_{\mathfrak{X}}(G/N_1)=3$, while $\mathrm{k}_{\mathfrak{X}}(G/N_2)=4$. Nevertheless, in view of Theorem 1, the equality $\mathrm{k}_{\mathfrak{X}}(G)=\mathrm{k}_{\mathfrak{X}}(G/N)$ for a group $G$ and its normal subgroup $N$ is an intrinsic property of the group $N$, which does not depend only on the particularities of the embedding of $N$ into $G$, but also on the group $G$ itself, and which implies the reduction $\mathfrak{X}$-theorem for $N$. This is the fact which advances Theorem 1 in comparison with Theorem 1 in [4]. It is worth pointing out that whereas Theorem 1 in [4] was proved by reducing the general situation to the known particular case $ \mathfrak{X}=\mathfrak{G}_\pi$ (see [13], [12]), our result in Theorem 1 was unknown even for this case. From the description of all groups for which the reduction $\mathfrak{X}$-theorem (see [4], Theorem 1) holds, we get a description of all pairs for which it holds. Corollary 1. Let $\mathfrak{X}$ be a complete class. Then a necessary and sufficient condition that the reduction $\mathfrak{X}$-theorem hold for a pair $(G,N)$ is that, for any composition factor $S$ of the group $N$, either $S\in\mathfrak{X}$, or one of conditions I–VII in [4], Appendix A, is met for the pair $(S,\mathfrak{X})$. 1.3. Some corollariesSince $\mathrm{k}_{\mathfrak{X}}(G/N)\leqslant \mathrm{k}_{\mathfrak{X}}(G)$ for any normal subgroup $N$ of a group $G$, the following result is a direct consequence of Theorem 1. Corollary 2. Let $\mathfrak{X}$ be a complete class of groups and $N$ be a normal subgroup of a finite group $G$ such that $\mathrm{k}_{\mathfrak{X}}(N)>1$. Then $\mathrm{k}_{\mathfrak{X}}(G)>\mathrm{k}_{\mathfrak{X}}(G/N)$. Moreover, as the above example of the group $G=\operatorname{PSL}_2(7)\times \operatorname{PGL}_2(7)$ shows, the precise value $\mathrm{k}_{\mathfrak{X}}(G/N)$ does not only depend on the numbers $\mathrm{k}_{\mathfrak{X}}(G)$ and $\mathrm{k}_{\mathfrak{X}}(N)$ themselves, but is not even controlled by the isomorphism type of the groups $G$ and $N$. From Theorem 2 it follows that any group $G$ contains the largest normal subgroup $R$ such that $\mathrm{k}_{\mathfrak{X}}(G)=\mathrm{k}_{\mathfrak{X}}(G/R)$. Corollary 3. Let $\mathfrak{X}$ be a complete class. For an arbitrary finite group $G$, consider the subgroup $R=\langle {N\mid N\trianglelefteq G}\text{ and } \mathrm{k}_{\mathfrak{X}}(G)=\mathrm{k}_{\mathfrak{X}}(G/N)\rangle$. This subgroup has the following properties: (i) $\mathrm{k}_{\mathfrak{X}}(G)=\mathrm{k}_{\mathfrak{X}}(G/R)$; (ii) if $N\trianglelefteq G$ and $N\leqslant R$, then $\mathrm{k}_{\mathfrak{X}}(G)=\mathrm{k}_{\mathfrak{X}}(G/N)$; (iii) if $\overline{G}=G/R$, then $\mathrm{k}_{\mathfrak{X}}(\overline{G})=\mathrm{k}_{\mathfrak{X}}(\overline{G}/\overline{N})$ implies $\overline{N}=1$ for any $\overline{N}\trianglelefteq \overline{G}$. In view of Theorem 2, the subgroup $R\leqslant G$ from Corollary 3 coincides with the $\mathscr{D}_{\mathfrak{X}}$-radical of the group $G$, where, as in [14], [4], by $\mathscr{D}_{\mathfrak{X}}$ we denote the class of finite groups in which all $\mathfrak{X}$-maximal subgroups are conjugate. The class $\mathscr{D}_{\mathfrak{X}}$ is closed under taking normal subgroups of homomorphic images and extensions,4 and, in particular, is a Fitting class (see the definition in [15]), and hence in any group there exits the $\mathscr{D}_{\mathfrak{X}}$-radical. Note that, in general, $\mathscr{D}_{\mathfrak{X}}$ is not a complete class, because it may fail to be closed under taking subgroups (see [16], Theorem 1.7). The quotient group $G/R$ will be called a complete reduction over $\mathfrak{X}$ of a group $G$, and the subgroup $R=\langle N\mid {N\trianglelefteq G}\text{ and } \mathrm{k}_{\mathfrak{X}}(G)=\mathrm{k}_{\mathfrak{X}}(G/N)\rangle$ itself will be called the kernel of reduction. A group $G$ will be called completely reduced over $\mathfrak{X}$ if the kernel of its reduction is trivial. The search problem of an $\mathfrak{X}$-scheme can be reduced to consideration of completely reduced groups. Given a group $G$, let
$$
\begin{equation*}
\mathrm{om}_{\mathfrak{X}}(G)=\{K\leqslant G\mid \operatorname{m}_{\mathfrak{X}}(K)\cap\operatorname{m}_{\mathfrak{X}}(G)\ne\varnothing\}
\end{equation*}
\notag
$$
be the set of all overgroups of $\mathfrak{X}$-maximal subgroups. The following result, which follows from Theorem 2 and the main result of [16], is not so obvious. Corollary 4. Let $\mathfrak{X}$ be a complete class of groups and let $N$ be a normal subgroup of a finite group $G$. Then the following conditions are equivalent: (i) $\mathrm{k}_{\mathfrak{X}}(G)=\mathrm{k}_{\mathfrak{X}}(G/N)$; (ii) $\mathrm{k}_{\mathfrak{X}}(K)=\mathrm{k}_{\mathfrak{X}}(K/(K\cap N))$ for all $K\in \mathrm{om}_{\mathfrak{X}}(G)$. 1.4. The category of groups and $\mathfrak{X}$-isoschematismsLet us formulate Theorem 2 in the language of homomorphisms. An epimorphism $\phi\colon G \to G^*$ will be said to be an isoschematism over $\mathfrak{X}$ (or, simply, an $\mathfrak{X}$-isoschematism) if it maps an $\mathfrak{X}$-scheme of the group $G$ (each or some) into an $\mathfrak{X}$-scheme of the group $G^*$. Theorem 2 is equivalent to saying that an epimorphism $\phi$ is an $\mathfrak{X}$-isoschematism if and only if $\mathrm{k}_{\mathfrak{X}}(\ker\phi)=1$. According to the above, $\mathfrak{X}$-isoschematicity of an epimorphism $\phi\colon G \to G^*$ is completely determined only by the groups $G$ and $G^*$, and is independent of a concrete mapping. In other words, the following result holds. Proposition 1. Let $G$ be a finite group, and let $G^*$ be its epimorphic image. Given a complete class $\mathfrak{X}$, the following assertions are equivalent: (i) there exists an $\mathfrak{X}$-isoschematism $\phi\colon G \to G^*$; (ii) any epimorphism $\phi\colon G \to G^*$ is an $\mathfrak{X}$-isoschematism; (iii) $\mathrm{k}_{\mathfrak{X}}(G)=\mathrm{k}_{\mathfrak{X}}(G^*)$. The kernels of two $\mathfrak{X}$-isoschematisms from $G$ onto $G^*$ may fail to be isomorphic, even though they have the same set of composition factors by the Jordan–Hölder theorem. Existence of an $\mathfrak{X}$-isoschematism from $G$ onto $G^*$ is written as The same symbol will also be used in the notation
$$
\begin{equation*}
\phi\colon G\underset{\mathfrak{X}}{\twoheadrightarrow} G^*,
\end{equation*}
\notag
$$
which means that the mapping $\phi$ is an $\mathfrak{X}$-isoschematism from $G$ onto $G^*$. The relation $\underset{\mathfrak{X}}{\twoheadrightarrow}$ can be considered as a relation between groups. Clearly, this relation is reflexive and transitive, but not symmetric. Let us symmetrize it. We say that two groups $G_1$ and $G_2$ are isoschemic over $\mathfrak{X}$ (or $\mathfrak{X}$-isoschemic), written if there exist $\mathfrak{X}$-isoschematisms from $G_1$ and $G_2$ onto the same group: The relation $\underset{\mathfrak{X}}{\equiv}$ is, clearly, reflexive and symmetric. In the actual fact, this relation defines an equivalence on groups; its transitivity follows from Theorem 2. Using this relation, one can describe the category of groups and $\mathfrak{X}$-isoschematisms.Corollary 5. For finite groups $G_1$ and $G_2$, $G_1\underset{\mathfrak{X}}{\equiv} G_2$ if and only if the complete reductions of these groups over $\mathfrak{X}$ are isomorphic. The relation $\underset{\mathfrak{X}}{\equiv}$ is an equivalence on finite groups. Each equivalence class is a subcategory in the category of all groups and $\mathfrak{X}$-isoschematisms and contains a unique (up to isomorphism) completely reduced over $\mathfrak{X}$ group which is an universally attracting object5 in this subcategory. In the language of homomorphisms, Corollary 4 can be stated as follows. Let $\phi$ be an $\mathfrak{X}$-isoschematism defined on a group $G$, and let $K$ be an overgroup of an $\mathfrak{X}$-maximal subgroup from $G$. Then the restriction of $\phi$ to $K$ is an $\mathfrak{X}$-isoschematism $K \underset{\mathfrak{X}}{\twoheadrightarrow} K^\phi$. § 2. Notation and preliminary lemmasWe will use the following standard notation from the group theory (see, for example, [11], [15], [18]–[20]). Given a natural number $n$, by $\pi(n)$ we denote the set of its prime divisors; for a group $G$, we set $\pi(G)=\pi(|G|)$. For a fixed set $\pi\subseteq\mathbb{P}$ of primes and a complete class $\mathfrak{X}$ of finite groups, we will use the following not quite standard notation: $\Omega/G$ is the set of orbits of an action of the group $G$ on a set $\Omega$; $|\Omega:G|$ is the number of orbits of an action of $G$ on $\Omega$, i.e. $|\Omega:G|=|\Omega/G|$ $\operatorname{Hall}_{\mathfrak{X}}(G)$ is the set of $\mathfrak{X}$-Hall subgroups of the group $G$ — these being the $\mathfrak{X}$-subgroups whose index is not divisible by any number from $\pi(\mathfrak{X})$; $\operatorname{Hall}_\pi(G)$ is the set of $\pi$-Hall subgroups in $G$, that is, $\operatorname{Hall}_{\mathfrak{X}}(G)$ for $\mathfrak{X} = \mathfrak{G}_\pi$; $\operatorname{m}_{\mathfrak{X}}(G)$ is the set of $\mathfrak{X}$-maximal subgroups of the group $G$; $\mathrm{k}_{\mathfrak{X}}(G)$ is the number of the conjugacy classes of $\mathfrak{X}$-maximal subgroups of the group $G$, that is, $\mathrm{k}_{\mathfrak{X}}(G)=|{\operatorname{m}_{\mathfrak{X}}(G):G}|$ for the action of the group $G$ by conjugations on the set $\operatorname{m}_{\mathfrak{X}}(G)$; $\operatorname{h}_{\mathfrak{X}}(G)$ is the number of conjugacy classes of $\mathfrak{X}$-Hall subgroups of the group $G$, that is, $\operatorname{h}_{\mathfrak{X}}(G)=|{\operatorname{Hall}_{\mathfrak{X}}(G):G}|$ for the action of the group $G$ by conjugations on the set $\operatorname{Hall}_{\mathfrak{X}}(G)$; $\mathscr{E}_{\mathfrak{X}}$ is the class of all finite groups $G$ such that $\operatorname{h}_{\mathfrak{X}}(G)\geqslant 1$ (or, equivalently, $\operatorname{Hall}_{\mathfrak{X}}(G)\ne\varnothing$); $\mathscr{C}_{\mathfrak{X}}$ is the class of all finite groups $G$ such that $\operatorname{h}_{\mathfrak{X}}(G)=1$; $\mathscr{D}_{\mathfrak{X}}$ is the class of all finite groups $G$ such that $\mathrm{k}_{\mathfrak{X}}(G)=1$; $\mathscr{M}_{\mathfrak{X}}$ is the class of all finite groups $G$ such that $\mathrm{k}_{\mathfrak{X}}(G)=\operatorname{h}_{\mathfrak{X}}(G)$ (or, equivalently, $\operatorname{m}_{\mathfrak{X}}(G)=\operatorname{Hall}_{\mathfrak{X}}(G)$). The notation $\mathscr{E}_{\mathfrak{X}}$, $\mathscr{C}_{\mathfrak{X}}$ and $\mathscr{D}_{\mathfrak{X}}$, which generalizes P. Hall’s notation $\mathscr{E}_\pi$, $\mathscr{C}_\pi$ and $\mathscr{D}_\pi$ (see [21], and also [15], Ch. I, § 3, [20], Ch. 5, § 3), is equivalent to that used by P. Hall if $\mathfrak{X}=\mathfrak{G}_\pi$ is the class of all $\pi$-groups. By definition, $\mathscr{D}_{\mathfrak{X}}=\mathscr{C}_{\mathfrak{X}}\cap \mathscr{M}_{\mathfrak{X}}$. The inclusions between the classes $\mathscr{E}_{\mathfrak{X}}$, $\mathscr{C}_{\mathfrak{X}}$, $\mathscr{M}_{\mathfrak{X}}$ and $\mathscr{D}_{\mathfrak{X}}$ are shown in the diagram In the case $\mathfrak{X}=\mathfrak{G}_\pi$, we will use the natural notation $\mathrm{k}_\pi(G)$ and $\operatorname{h}_\pi(G)$, respectively, for the number of conjugacy classes of $\pi$-maximal and $\pi$-Hall subgroups of a group $G$. We will say that $n\in\mathbb{N}$ is a $\pi$-number if $\pi(n)\subseteq\pi$.Lemma 1 (see [22], Ch. III, Theorem 3.9). Let $\mathfrak{X}$ be a complete class. Given a group homomorphism $\phi\colon G\to G_0$, suppose that $K\in\mathfrak{X}$ for some subgroup $K\leqslant G^\phi$. Then $K=H^\phi$ for some $\mathfrak{X}$-subgroup $H\leqslant G$. In particular, $\operatorname{m}_{\mathfrak{X}}(G^\phi)\subseteq\operatorname{m}_{\mathfrak{X}}(G)^\phi$. By $\mathfrak{X}'$ we denote the class of all groups $G$ such that $\operatorname{m}_{\mathfrak{X}}(G)=\{1\}$. A group is called $\mathfrak{X}$-separable if it admits a (sub)normal series in which each factor is either an $\mathfrak{X}$- or an $\mathfrak{X}'$-group. The following lemma summarizes some known results on the behaviour of $\mathfrak{X}$-maximal and $\mathfrak{X}$-Hall subgroups. Lemma 2. Let $N$ be a normal subgroup of the group $G$. Then the following assertions hold. (i) If $H\in\operatorname{Hall}_{\mathfrak{X}}(G)$, then $H\cap N\in \operatorname{Hall}_{\mathfrak{X}}(N)$ and $HN/N\in \operatorname{Hall}_{\mathfrak{X}}(G/N)$ (see [20], Ch. IV, § 5.11). (ii) Let $G/N\in\mathfrak{X}$. Then a necessary and sufficient condition that, for $H\in\operatorname{Hall}_{\mathfrak{X}}(N)$, there exist a subgroup $K\in\operatorname{Hall}_{\mathfrak{X}}(G)$ such that $H=K\cap N$ is that $H^N=H^G$ (that is, when the class $H^N\in\operatorname{Hall}_{\mathfrak{X}}(N)/N$ is invariant under the action of the group $G$ on the set $\operatorname{Hall}_{\mathfrak{X}}(N)/N$; see [23], Lemma 2.1, (e)). (iii) Let $N$ be an $\mathfrak{X}$-separable group. Then $\mathrm{k}_{\mathfrak{X}}(G)=\mathrm{k}_{\mathfrak{X}}(G/N)$. In particular, $G\in \mathscr{D}_{\mathfrak{X}}$ if and only if $G/N\in \mathscr{D}_{\mathfrak{X}}$ (see [2], § 12.9). Lemma 3 (see [4], Theorem 1). Let $N$ be a normal subgroup of a group $G$ and $\mathrm{k}_{\mathfrak{X}}(N)=1$. Then $\mathrm{k}_{\mathfrak{X}}(G)=\mathrm{k}_{\mathfrak{X}}(G/N)$. Given subgroups $S, H$ of the group $G$, we denote by $\operatorname{Aut}_H(S)$ the $H$-induced automorphism group of the group $S$, that is, the image in $\operatorname{Aut}(S)$ of the homomorphism
$$
\begin{equation*}
\alpha_H\colon \mathrm{N}_H(S)\to \operatorname{Aut}(S),
\end{equation*}
\notag
$$
which associates with each $x\in \mathrm{N}_H(S)$ the automorphism of the group $S$ defined by $s\mapsto s^x=x^{-1}sx$ for all $s\in S$. The kernel of this homomorphism is $\mathrm{C}_H(S)$, and hence
$$
\begin{equation*}
\operatorname{Aut}_H(S)\cong \mathrm{N}_H(S)/\mathrm{C}_H(S).
\end{equation*}
\notag
$$
If $H\leqslant K\leqslant G$, then the homomorphism $\alpha_H$ is the restriction of the homomorphism $\alpha_K\colon \mathrm{N}_K(S)\to\operatorname{Aut}(S)$ to $\mathrm{N}_H(S)$. Therefore, $\operatorname{Aut}_H(S)\leqslant \operatorname{Aut}_K(S)$. Lemma 4. Let $S$ be a simple nonabelian subnormal subgroup of a group $G$, let $H\in\operatorname{Hall}_{\mathfrak{X}}(S)$, and let the group $\operatorname{Aut}_G(S)$ stabilize the conjugacy class of the subgroup $H$ in $S$ (that is, $H^{\operatorname{Aut}_G(S)}=H^S$). Consider an arbitrary right transversal $g_1,\dots, g_n$ of the subgroup $\operatorname{N}_G(S)$ in $G$. Let Then $M=\langle S^G \rangle$ is a minimal normal subgroup of the group $G$, and (i) $V^G=V^M$; (ii) $V\in\operatorname{Hall}_{\mathfrak{X}}(M)$; (iii) if $G/M\in {\mathfrak{X}}$, then $V=K\cap M$ and $H=K\cap S$ for some $K\in\operatorname{Hall}_{\mathfrak{X}}(G)$. Proof. For any $g\in G$ we have $S^g\in\{S^{g_i}\mid i=1,\dots, n\}$, and hence $M=\langle S^G \rangle\trianglelefteq G$. We also note that $[S^{g_i},S^{g_j}]=1$ for $i\ne j$ since the subgroup $S$ is simple and subnormal.
Let $g\in G$. There exist a permutation $\sigma\in\operatorname{Sym}_n$ and elements $x_1,\dots,x_n\in \mathrm{N}_G(S)$ such that $g_ig=x_ig_{i\sigma}$. Consider the automorphisms $\gamma_i\in\operatorname{Aut}_G(S)$ defined by $\gamma_i\colon s\mapsto s^{x_i}$. By the assumption, $H^{x_i}=H^{\gamma_i}=H^{s_i}$ for some $s_i\in S$. We set $a_i=s_{i\sigma^{-1}}^{g_i}$ and $a=a_1\cdots a_n$. It is clear that $a\in M$. The equality $V^G=V^M$ will be verified if show that $V^g=V^a$. By definition $a_i\in S^{g_i}$ and $H^{g_ia}=H^{g_ia_i}$. We have
$$
\begin{equation*}
\begin{aligned} \, V^g &= \langle H^{g_ig}\mid i=1,\dots, n\rangle= \langle H^{x_ig_{i\sigma}}\mid i=1,\dots, n\rangle \\ &=\langle H^{s_ig_{i\sigma}}\mid i=1,\dots, n\rangle= \langle H^{s_{i\sigma^{-1}}g_{i}}\mid i=1,\dots, n\rangle \\ &=\langle H^{g_is_{i\sigma^{-1}}^{g_i}}\mid i=1,\dots, n\rangle=\langle H^{g_ia_i}\mid i=1,\dots, n\rangle= \langle H^{g_ia}\mid i=1,\dots, n\rangle = V^a\!. \end{aligned}
\end{equation*}
\notag
$$
This proves assertion (i). Next, $V$ is a direct product of the $\mathfrak{X}$-groups $H^{g_i}$, $i=1,\dots,n$. Hence $V\in\mathfrak{X}$. Further, since $H\in\operatorname{Hall}_{\mathfrak{X}}(S)$, the number is not divisible by any number from $\pi(\mathfrak{X})$. Hence $V\in\operatorname{Hall}_{\mathfrak{X}}(M)$, which proves (ii). Finally, (iii) is secured by (i) and Lemma 2, (ii). Lemma 4 is proved. Lemma 5. Let a normal subgroup $N$ of a group $G$ be a direct product of nonabelian simple groups, and $S$ be one of these groups. Suppose that $G=KN$ for some subgroup $K$. Then (i) $\mathrm{N}_G(S)=N\mathrm{N}_K(S)$; (ii) $\operatorname{Aut}_G(S)=\operatorname{Inn}(S)\operatorname{Aut}_K(S)$. Proof. Let $N=S_1\times S_2\times\dots\times S_n$ and $S_1=S$. Then
$$
\begin{equation*}
N\leqslant \mathrm{N}_G(S)\text{ and }S_2\times\dots\times S_n=\mathrm{C}_N(S)\leqslant \mathrm{C}_G(S).
\end{equation*}
\notag
$$
Hence $\mathrm{N}_G(S)=N\mathrm{N}_K(S)$, as claimed in (i). Let denote the natural homomorphism induced by conjugations. Its kernel is $\mathrm{C}_G(S)$. We have $S^\alpha=\operatorname{Aut}_S(S)=\operatorname{Inn}(S)$, $N=S\mathrm{C}_N(S)$. Hence $N^\alpha= \operatorname{Inn}(S)$ and This proves assertion (ii), and, therefore, the lemma. The key role in the proof of Theorem 1 is played by the theorem on the number of classes of conjugate $\pi$-Hall subgroups in simple groups (see [23]). We will use the following refined version of this result. Lemma 6 (see [23], Theorem 1.1). Let $S$ be a simple finite group possessing a $\pi$-Hall subgroup for some set $\pi$ of primes. Then one of the following assertions holds: (i) $2\notin\pi$ and $\mathrm{h}_\pi(S)=1$; (ii) $3\notin\pi$ and $\mathrm{h}_\pi(S)\in\{1,2\}$; (iii) $2,3\in\pi$ and $\mathrm{h}_\pi(S)\in\{1,2,3,4,9\}$. Lemma 7 (see [14], Lemma 12). Let $\mathfrak{X}$ be a complete class. We set $\pi=\pi(\mathfrak{X})$. Suppose also that $\mathrm{h}_\pi(S)=9$. Then $\mathrm{h}_{\mathfrak{X}}(S)$ is one of the numbers $0$, $1$ or $9$. The following result is a consequence of Lemmas 6 and 7. Lemma 8. Let $S$ be a simple finite group. Then one of the following assertions holds: (i) $2\notin\pi(\mathfrak{X})$ and $\mathrm{h}_{\mathfrak{X}}(S)\in\{0,1\}$; (ii) $3\notin\pi(\mathfrak{X})$ and $\mathrm{h}_{\mathfrak{X}}(S)\in\{0,1,2\}$; (iii) $2,3\in\pi(\mathfrak{X})$ and $\mathrm{h}_{\mathfrak{X}}(S)\in\{0,1,2,3,4, 9\}$. Assume that a simple group $S$ satisfies $\mathrm{h}_{\mathfrak{X}}(S) = 9$. Since $\mathrm{h}_{\mathfrak{X}}(S)\leqslant\mathrm{h}_\pi(S)$ for $\pi=\pi(\mathfrak{X})$, by Lemma 6 we have $\mathrm{h}_{\mathfrak{X}}(S)=\mathrm{h}_\pi(S)$, and, therefore, $\operatorname{Hall}_{\mathfrak{X}}(S)=\operatorname{Hall}_\pi(S)$. Now, in view of Lemmas 2.3, 3.1, 4.4, 8.1 in [23], we get the following result on the structure of $\mathfrak{X}$-Hall subgroups of a group $S$. Lemma 9. Let $S$ be a simple finite group and $\mathrm{h}_{\mathfrak{X}}(S)=9$ for some complete class $\mathfrak{X}$. Then the following assertions hold. (i) $S\cong \operatorname{PSp}_{2n}(q)\cong \operatorname{PSp}(V)$, where $q$ is a power of a prime $p\notin \pi(\mathfrak{X})$, and $V$ is the vector space of dimension $2n$ over $\mathbb{F}_q$ with nondegenerate skew-symmetric form associated with $\operatorname{PSp}_{2n}(q)$. (ii) $\pi(\mathfrak{X})\cap\pi(S)\subseteq\pi(q^2-1)$ and – either $\pi(\mathfrak{X})\cap\pi(S)=\{2,3\}$ and $n\in\{5,7\}$, – or $\pi(\mathfrak{X})\cap\pi(S)=\{2,3,5\}$ and $n=7$. (iii) Any $\pi(\mathfrak{X})$-Hall subgroup of the group $\operatorname{PSp}_{2n}(q)$ is contained in the stabilizer $M$ of a decomposition of the space $V$ in the orthogonal sum of nondegenerate isometric subspaces of dimension $2$. There exists a subgroup $A\trianglelefteq M$ such that $A=L_1\dots L_n$, where $L_i\cong \operatorname{Sp}(V_i)\cong \operatorname{Sp}_2(q)\cong \operatorname{SL}_2(q)$, $[L_i,L_j]= 1$, $i,j=1,\dots,n$, $i\ne j$, and $M/A\cong \operatorname{Sym}_n$.(iv) $\mathrm{h}_{\mathfrak{X}}(\operatorname{Sym}_n)=1$. In addition,
(v) $\mathrm{h}_{\mathfrak{X}}(\operatorname{Sp}_2(q))=3$. In addition,
(vi) The number of fixed points of any subgroup $G\leqslant\operatorname{Aut}(S)$ under its action on the set $\operatorname{Hall}_{\mathfrak{X}}(S)/S$ is either $1$ or $9$. § 3. Frattini argument for $\mathfrak{X}$-Hall subgroupsOur main purpose in this section is to prove the following result. Proposition 2. Let a group $G$ have a normal subgroup $A$ such that $A=KN$ for some subgroup $N$ normal in $G$, where $N$ is a direct product of nonabelian simple groups and some $K\in\operatorname{Hall}_{\mathfrak{X}}(G)$. Then there exists $L\in\operatorname{Hall}_{\mathfrak{X}}(A)$ such that $G=A\mathrm{N}_G(L)$. Proof. Let $\pi=\pi(\mathfrak{X})$. Since $A$ contains $H\in\operatorname{Hall}_{\mathfrak{X}}(G)$, the index $|{G:A}|$ is a $\pi'$-number. Since $A\trianglelefteq G$, we have $\operatorname{Hall}_{\mathfrak{X}}(G)=\operatorname{Hall}_{\mathfrak{X}}(A)$, and $G/A$ is a $\pi'$-group.
Let $N=S_1\times\dots\times S_n$, where $S_i$, $i=1,\dots,n$, are nonabelian simple groups. Let us establish some facts on $\mathfrak{X}$-Hall subgroups of $S_i$, on conjugacy classes of such subgroups, and on the action on these classes of the groups of $G$-induced automorphisms. We fix $S\in\{S_1,\dots,S_n\}$. Since $K\cap S\in\operatorname{Hall}_{\mathfrak{X}}(S)$, we have $S\in\mathscr{E}_{\mathfrak{X}}$, and hence, by Lemma 8, Let $\Omega$ be the set of all fixed points of the group $\operatorname{Aut}_A(S)$ acting on the set $\operatorname{Hall}_{\mathfrak{X}}(S)/S$ of conjugacy classes of $\mathfrak{X}$-Hall subgroups of the group $S$, that is,
$$
\begin{equation*}
\Omega =\{H^S\mid H\in\operatorname{Hall}_{\mathfrak{X}}(S),\ \forall\, a\in \mathrm{N}_A(S)\ \exists\, x\in S\colon H^a=H^x\}.
\end{equation*}
\notag
$$
Note that $\Omega\ne\varnothing$, because $(K\cap S)^S\in\Omega$. Indeed, $N\leqslant \mathrm{N}_A(S)$, and hence, $\mathrm{N}_A(S)=\mathrm{N}_{KN}(S)=\mathrm{N}_K(S)N$. In addition, the conjugacy class $(K\cap S)^S$ is invariant under both groups $\mathrm{N}_K(S)$ and $N$. Hence it is invariant under both $\mathrm{N}_A(S)$ and $\operatorname{Aut}_A(S)$, and, therefore, lies in $\Omega$.
Since $A\trianglelefteq G$, we have $\mathrm{N}_A(S)\trianglelefteq \mathrm{N}_G(S)$. Hence $\operatorname{Aut}_A(S)\trianglelefteq \operatorname{Aut}_G(S)$ and, therefore, the group $\operatorname{Aut}_G(S)$ acts on $\Omega$. We assert that Indeed, $|\Omega|\leqslant \operatorname{h}_{\mathfrak{X}}(S)$, and the length of any orbit in $\operatorname{Aut}_G(S)$ on $\Omega$ is at most $|\Omega|$. If $2\notin\pi$ or $3\notin\pi$, then from Lemma 8 it follows that the length of any orbit of the group $\operatorname{Aut}_G(S)$ on $\Omega$ is a $\pi$-number. So, we can assume that $2,3\in\pi$. Now if $\operatorname{h}_{\mathfrak{X}}(S)\leqslant 4$, then the length of any orbit of the group $\operatorname{Aut}_G(S)$ on $\Omega$ is again a $\pi$-number. So, we can assume that $\operatorname{h}_{\mathfrak{X}}(S)=9$. From Lemma 9 it follows that $|\Omega|\in\{1,9\}$. The case $|\Omega|=1$ is clear. The only non-$\{2,3\}$-numbers majorized by $9$ are $5$ and $7$, and so, as is easily checked, any partition of $9$ into a sum of natural numbers involves a $\{2,3\}$-number and, therefore, a $\pi$-number. Hence the length of one of the orbits into which $\Omega$ splits relative to the action of$\operatorname{Aut}_G(S)$ is a $\pi$-number. Assertion $1^\circ)$ can be refined as follows: The hypotheses of the theorem imply that $G/A$ is a $\pi'$-group. So, both the group $ \mathrm{N}_G(S)A/A\cong \mathrm{N}_G(S)/\mathrm{N}_A(S)$ and its homomorphic image
$$
\begin{equation*}
\begin{aligned} \, \mathrm{N}_G(S)/\mathrm{N}_A(S)\mathrm{C}_G(S) &\cong (\mathrm{N}_G(S)/\mathrm{C}_G(S))/(\mathrm{N}_A(S)\mathrm{C}_G(S)/\mathrm{C}_G(S)) \\ &\cong \operatorname{Aut}_G(S)/\operatorname{Aut}_A(S) \end{aligned}
\end{equation*}
\notag
$$
are also $\pi'$-groups. By definition of the set $\Omega$, $\operatorname{Aut}_A(S)$ stabilizes any element from $\Omega$. Hence the length of any orbit on $\Omega$ of the group $\operatorname{Aut}_G(S)$ divides a $\pi'$-number $|{\operatorname{Aut}_G(S):\operatorname{Aut}_A(S)}|$ and, therefore, is itself a $\pi'$-number. If a number is simultaneously a $\pi$- and a $\pi'$-number, then it is equal to $1$.
From $1^\circ)$ and $2^\circ)$ we conclude that We now assert the following: It can be assumed that $M=\langle S^G\rangle$. Assertion $4^\circ)$ follows from Lemma 4. We also assert that By $\Lambda$ we denote the set of all minimal normal subgroups of the group $G$ lying in $N$. By the assumption, where the product is direct. For each $M\in \Lambda$, we choose, according to $4^\circ)$, a subgroup $V_M\in\operatorname{Hall}_{\mathfrak{X}}(M)$ so that $V^M_M=V^G_M$. Let Then $V\in\operatorname{Hall}_{\mathfrak{X}}(N)$. Consider any $g\in G$. Any $M\in\Lambda$ contains an element $x_M$ such that $V_M^{g}=V_M^{x_M}$. We set It is clear that $x\in N$ and $V_M^{x}=V_M^{x_M}=V_M^{g}$ for any $M\in\Lambda$. Therefore,
$$
\begin{equation*}
V^g=\langle V_M^g\mid M\in \Lambda \rangle=\langle V_M^x\mid M\in \Lambda \rangle=V^x.
\end{equation*}
\notag
$$
This proves assertion $5^\circ)$.
Consider the normal series
$$
\begin{equation*}
\mathrm{N}_G(V)\trianglerighteq \mathrm{N}_A(V)\trianglerighteq \mathrm{N}_N(V)\trianglerighteq V\trianglerighteq 1
\end{equation*}
\notag
$$
and the sections of this series. The section $\mathrm{N}_G(V)/\mathrm{N}_A(V)$ is isomorphic to the subgroup $\mathrm{N}_G(V)A/A$ in the $\mathfrak{X}'$-group $G/A$ and, therefore, is also an $\mathfrak{X}'$-group. Similarly
$$
\begin{equation*}
\mathrm{N}_A(V)/\mathrm{N}_N(V)\cong \mathrm{N}_A(V)N/N\leqslant A/N=KN/N\cong K/(K\cap N),
\end{equation*}
\notag
$$
which gives $\mathrm{N}_A(V)/\mathrm{N}_N(V)\in\mathfrak{X}$. Since $V\in\operatorname{Hall}_{\mathfrak{X}}(N)$, we have $\mathrm{N}_N(V)/V\in \mathfrak{X}'$. Finally, $V\in\mathfrak{X}$. This proves $6^\circ)$.
Now the required proposition follows from $5^\circ)$ and $6^\circ)$. Using $5^\circ)$, we see that $V^A=V^N$. By Lemma 2, there exists $L\in \operatorname{Hall}_{\mathfrak{X}}(A)$ such that $V=L\cap N$. Let us show that $L$ is as claimed in the proposition. It suffices to prove the inclusion $G\leqslant A\mathrm{N}_G(L)$. It is clear that $L\leqslant \mathrm{N}_A(V)$, that is, $L$ is an $\mathfrak{X}$-Hall subgroup of the $\mathfrak{X}$-separable normal subgroup $\mathrm{N}_A(V)$ of the group $\mathrm{N}_G(V)$. From conjugacy of the $\mathfrak{X}$-Hall subgroups in $\mathfrak{X}$-separable groups, we have
$$
\begin{equation*}
L^{\mathrm{N}_G(V)}=L^{\mathrm{N}_A(V)},\quad\text{which implies}\quad \mathrm{N}_G(V)\leqslant \mathrm{N}_A(V)\mathrm{N}_G(L).
\end{equation*}
\notag
$$
Now an appeal to $5^\circ)$ shows that
$$
\begin{equation*}
G=N\mathrm{N}_G(V)\leqslant N\mathrm{N}_A(V)\mathrm{N}_G(L)\leqslant A\mathrm{N}_G(L),
\end{equation*}
\notag
$$
which completes the proof of Proposition 2. It seems that by using [24], Theorem 3.1 (see also [25], Theorem 2), Proposition 2 might be strengthened to the hollowing hypothetical result: if $G\in \mathscr{E}_\mathfrak{X}$ and $A\trianglelefteq G$, then $G=A\mathrm{N}_G(H)$ for some $H\in\operatorname{Hall}_{\mathfrak{X}}(A)$. If $\mathfrak{X}$ is the class of $\pi$-groups, this result is known (see [12], Corollary 3.7). § 4. On simple groups with nine conjugacy classes of $\mathfrak{X}$-Hall subgroupsProposition 3. Let $\mathfrak{X}$ be a complete class of finite groups, $S$ be a nonabelian simple group and $\operatorname{h}_{\mathfrak{X}}(S)=9$. Then $S\notin\mathscr{M}_{\mathfrak{X}}$. Proof. Assume that $S\in\mathscr{M}_{\mathfrak{X}}$. Let $\pi=\pi(\mathfrak{X})$. In view of Lemma 9, we can assume that Any $\mathfrak{X}$-Hall subgroup of the group $\operatorname{PSp}_{2n}(q)$ is contained in the stabilizer $M$ of a decomposition of the associated space $V$ in the orthogonal sum of nondegenerate isometric subspaces of dimension $2$. There exists a subgroup $A\trianglelefteq M$ such that $A=L_1\dots L_n$, where $L_i\cong \operatorname{Sp}(V_i)\cong \operatorname{Sp}_2(q)\cong \operatorname{SL}_2(q)$, $[L_i,L_j]= 1$, $i,j=1,\dots,n$, $i\ne j$, and $M/A\cong \operatorname{Sym}_n$.
One of the following two cases holds. (1) $\pi(\mathfrak{X})\cap\pi(S)=\pi(\mathfrak{X})\cap\pi(\operatorname{SL}_2(q))=\{2,3\}$ and $n\in\{5,7\}$. In addition, the $\mathfrak{X}$-Hall subgroups of any group $L_i\cong\operatorname{SL}_2(q)$ are, precisely, the generalized quaternion groups of order $48$ and groups of the form $2.\operatorname{Sym}_4$. (2) $\pi(\mathfrak{X})\cap\pi(S)=\pi(\mathfrak{X})\cap\pi(\operatorname{SL}_2(q))=\{2,3,5\}$ and $n=7$. In addition, the $\mathfrak{X}$-Hall subgroups in any $L_i\cong\operatorname{SL}_2(q)$ are, precisely, the generalized quaternion groups of order $120$ and groups of the form $2.\operatorname{Alt}_5$; any $\mathfrak{X}$-Hall subgroup in $M/A\cong \operatorname{Sym}_7$ is isomorphic to $\operatorname{Sym}_6$. In each of these cases, we choose in $S$ a subgroup $U$ as follows. Consider case (1). The group $S$ contains a subgroup of the form
$$
\begin{equation*}
\operatorname{Sp}_6(q)\circ \operatorname{Sp}_{2(n-3)}(q)
\end{equation*}
\notag
$$
which stabilizes in $S$ a nondegenerate subspace of dimension $6$ and its orthogonal complement, and hence, contains a subgroup isomorphic to $\operatorname{Sp}_6(q)$. For any $\varepsilon\in\{+,-\}$, this subgroup contains a subgroup7 $\operatorname{GL}^\varepsilon_3(q).2$ (see [18], Table 8.28); here $\varepsilon$ can be chosen so that the number $q-\varepsilon1$ would be divisible by $3$. With this choice of $\varepsilon$, in view of [18], Tables 8.3, 8.5, we can choose a $\{2,3\}$-subgroup in the subgroup $\operatorname{SL}^\varepsilon_3(q)\leqslant \operatorname{GL}^\varepsilon_3(q).2$. By solvability, $U\in\mathfrak{X}$. Since $S\in\mathscr{M}_{\mathfrak{X}}$, we have $U\leqslant H$ for some $H\in\operatorname{Hall}_{\mathfrak{X}}(S)$. Proceeding as above, we choose a subgroup $M$ and a normal subgroup $A$ in $M$ such that $H\leqslant M$. Consider the canonical epimorphism We have $\overline{U}\leqslant \overline{H}\leqslant \overline{M}\cong \operatorname{Sym}_n$, where $n\in\{5,7\}$. On the other hand, $\overline{H}\cong H/(H\cap A)$ and $\overline{U}\cong U(H\cap A)/(H\cap A)$. We choose in $H\cap A$ the characteristic subgroups $B$, $C$, and $D$ defined by
$$
\begin{equation*}
B:=\mathrm{O}_2(H\cap A),\quad C/B:=\mathrm{O}_3((H\cap A)/B)\quad\text{and}\quad D/C:=\mathrm{O}_2((H\cap A)/C).
\end{equation*}
\notag
$$
By the choice, $ B\leqslant C\leqslant D$. From Lemma 2 it follows that the subgroup $H\cap A$ is generated by pairwise permutational $\mathfrak{X}$-Hall subgroups of factors $L_i$, each of which is either a generalized quaternion $\{2,3\}$-group, or is isomorphic to $2.\operatorname{Sym}_4$. Now it is clear that $D=H\cap A$, and, therefore, $\overline{U}\cong U/(U\cap D)$. Since $\mathrm{O}_2(U)=1$, we have Since, in each factor, the Sylow $3$-subgroups forming the group $H\cap A$ are cyclic groups of order $3$, any Sylow 3-subgroup of the group $H\cap A$ which is isomorphic to $C/B$ is abelian, and its section is a normal abelian 3-subgroup of the group $UB/B\cong U$. It follows that it is contained in $\mathrm{Z}(\mathrm{O}_3(UB/B))$, inasmuch as and $Q_8$ acts irreducibly on the quotient group of the group $3_+^{1+2}$ to its centre. Therefore,
$$
\begin{equation*}
\text{either}\quad UC/C\cong U/(U\cap C)\cong 3_+^{1+2}:Q_8,\quad\text{or}\quad UC/C\cong 3^{2}:Q_8.
\end{equation*}
\notag
$$
Now, since $\mathrm{O}_2(3_+^{1+2}:Q_8)=1$ and $\mathrm{O}_2(3^{2}:Q_8)=1$, we find that
$$
\begin{equation*}
(UC/C)\cap D/C=1\quad\text{and}\quad\overline{U}=UD/D\cong UC/C.
\end{equation*}
\notag
$$
But $\overline{U}$ (and, therefore, its subgroup $Q_8$) is isomorphic to a subgroup of the group $\operatorname{Sym}_n$ for $n\in\{5,7\}$. At the same time, it is quite clear that the group $Q_8$ has not faithful permutation representations of degree $<8$. A contradiction.
Consider case (2). Proceeding with the group $S=\operatorname{PSp}_{14}(q)$ as in case (1), we find a subgroup isomorphic to $\operatorname{Sp}_{10}(q)$. Since 5 divides $q^2-1$, we choose $\varepsilon\in\{+,-\}$ so that $5$ would divide $q-\varepsilon1$. The group $\operatorname{Sp}_{10}(q)$, and hence, the group $S$, contains a subgroup $\operatorname{GL}_5^\varepsilon(q).2$ (see [18], Table 8.64), which, in turn, contains $\operatorname{SL}_5^\varepsilon(q)$ as a subgroup. Further, $\operatorname{SL}_5^\varepsilon(q)$ contains a subgroup see [18], Tables 8.18 and 8.20. Moreover, since $\operatorname{Sp}_2(5)\cong 2.\operatorname{Alt}_5$, we have $U\in\mathfrak{X}$. Next, since $S\in\mathscr{M}_{\mathfrak{X}}$, we see that $U\leqslant H$ for some $H\in\operatorname{Hall}_{\mathfrak{X}}(S)$. Proceeding as above, we choose a subgroup $M$ and in it a normal subgroup $A$ so as to have $H\leqslant M$. Let be the canonical epimorphism. Then $\overline{U}\leqslant \overline{H}\leqslant \overline{M}\cong \operatorname{Sym}_7$. Therefore, $|\overline{U}|_5\leqslant 5$. From the structure of the group $U$ any homomorphism image of the group $U$ whose order is not divisible by $5^2$ is an image of the group $\operatorname{Sp}_2(5)\cong \operatorname{SL}_2(5)$. Hence the extra special subgroup $5_+^{1+2}$ of the group $U$ should lie in the kernel of the homomorphism $\overline{\phantom{x}}$, and hence, in $U\cap A$. But the Sylow $5$-subgroups of the group $A$ are abelian (in each factor $L_i$ the order of the Sylow $5$-subgroup is 5). This contradiction proves Proposition 3.§ 5. Proof of Theorem 1 and its corollaries Proof of Theorem 1. Assume on the contrary that there exists a group $G$ with the following properties:
(a) $G$ has a normal subgroup $N$ such that $\mathrm{k}_{\mathfrak{X}}(N)>1$, but the reduction $\mathfrak{X}$-theorem holds for the pair $(G,N)$, that is, $\mathrm{k}_{\mathfrak{X}}(G)=\mathrm{k}_{\mathfrak{X}}(\overline{G})$, where bar denotes the canonical epimorphism;(b) the order of $G$ is smallest among the groups with property (a). Recall (see § 1.2) that the reduction $\mathfrak{X}$-theorem for the pair $(G,N)$ implies the following properties: $1^\circ)$ if $K\in\operatorname{m}_{\mathfrak{X}}(G)$, then $\overline{K}\in\operatorname{m}_{\mathfrak{X}}(\overline{G})$; $2^\circ)$ if for $K,L\in\operatorname{m}_{\mathfrak{X}}(G)$, the subgroups $\overline{K}$ and $\overline{L}$ are conjugate in $\overline{G}$ (for example, are equal), then $K$ and $L$ are conjugate in $G$. Note that if $M\ne 1$ is a normal subgroup of $G$, $M\leqslant N$, then, since $G/N$ is a homomorphic image of the group $G/M$, we have
$$
\begin{equation*}
\mathrm{k}_{\mathfrak{X}}(G)=\mathrm{k}_{\mathfrak{X}}(G/N)\leqslant\mathrm{k}_{\mathfrak{X}}(G/M) \leqslant\mathrm{k}_{\mathfrak{X}}(G).
\end{equation*}
\notag
$$
As a result, we get the reduction $\mathfrak{X}$-theorem for the pairs $(G/M,N/M)$ and $(G,M)$. By (b) and since $|G/M|<|G|$, we have $\mathrm{k}_{\mathfrak{X}}(N/M)=1$. Hence if $\mathrm{k}_{\mathfrak{X}}(M)=1$, then by Lemma 3
$$
\begin{equation*}
\mathrm{k}_{\mathfrak{X}}(N)=\mathrm{k}_{\mathfrak{X}}(N/M)=1.
\end{equation*}
\notag
$$
Therefore, it can be assumed that
$3^\circ)$ $N$ is a minimal normal subgroup of the group $G$, and $N$ is nonabelian, because $\mathrm{k}_{\mathfrak{X}}(N)\,{>}\, 1$. Therefore, according to [19], Ch. 2, Corollary 3 to Theorem 4.14, for some nonabelian simple subgroups $S_1,\dots ,S_n$ conjugate in $G$. Let $S$ be one of the $S_i$’s.We will obtain a contradiction by examining the action of the group $\operatorname{Aut}_G(S)$ on the set of the conjugacy classes of $\mathfrak{X}$-Hall subgroups of the group $S$. By Lemma 8,$4^\circ)$ $|\Delta|= \operatorname{h}_{\mathfrak{X}}(S)\in\{0,1,2,3,4,9\}$. Let us exclude all six possibilities. We will first verify that $5^\circ)$ $\operatorname{h}_{\mathfrak{X}}(S)\ne 0$. To this end, we will show that $6^\circ)$ if $K\in\operatorname{m}_{\mathfrak{X}}(G)$, then $K\cap N\in\operatorname{Hall}_{\mathfrak{X}}(N)$ and $K\in\operatorname{Hall}_{\mathfrak{X}}(KN)$. As a result, $\operatorname{Hall}_{\mathfrak{X}}(N)\ne\varnothing$ and $\operatorname{h}_{\mathfrak{X}}(S)\ne 0$, because
$$
\begin{equation*}
\varnothing\ne\{H\cap S\mid H\in\operatorname{Hall}_{\mathfrak{X}}(N)\}\subseteq\operatorname{Hall}_{\mathfrak{X}}(S).
\end{equation*}
\notag
$$
In addition, from the inclusion $K\cap N\in\operatorname{Hall}_{\mathfrak{X}}(N)$ it will also follow that $K\in\operatorname{Hall}_{\mathfrak{X}}(KN)$, inasmuch as
We choose an arbitrary $K\in\operatorname{m}_{\mathfrak{X}}(G)$, $p\in\pi(\mathfrak{X})$ and $P\in\operatorname{Syl}_p(N)$. We have $P\in\mathfrak{X}$. It suffices to prove that $P$ is conjugate to a subgroup from $K$. We set From Frattini’s argument (see [15], Ch. A, (6.3)) it follows that $A=\mathrm{N}_{A}(P)N$. Hence
$$
\begin{equation*}
\overline{K}=\overline{A}=\overline{\mathrm{N}_{A}(P)}
\end{equation*}
\notag
$$
and now, by Lemma 1, we have $\overline{\mathrm{N}_{A}(P)}=\overline{U}$ for some $U\in\operatorname{m}_{\mathfrak{X}}({\mathrm{N}_{A}(P)})$. Since $U$ normalizes the $\mathfrak{X}$-subgroup $P$, we have $P\leqslant U$. We embed $U$ into a maximal $\mathfrak{X}$-subgroup $L$ of the group $G$. By $1^\circ)$, we have $\overline{L}\in\operatorname{m}_{\mathfrak{X}}(\overline{G})$. In addition,
$$
\begin{equation*}
\overline{K}=\overline{\mathrm{N}_{A}(P)}=\overline{U}\leqslant \overline{L}.
\end{equation*}
\notag
$$
Similarly by $1^\circ)$ and since $K\in\operatorname{m}_{\mathfrak{X}}(G)$, we have $\overline{K}\in\operatorname{m}_{\mathfrak{X}}(\overline{G})$. Now we have $\overline{K}=\overline{L}$, and hence, using $2^\circ)$, we conclude that $L^g=K$ for some $g\in G$. Hence as claimed.
We note some other corollaries to $6^\circ)$. We assert that $7^\circ)$ any $\mathfrak{X}$-subgroup in $G$ normalizes some member of $\operatorname{Hall}_{\mathfrak{X}}(N)$; in particular $8^\circ)$ $N\in\mathscr{M}_{\mathfrak{X}}$ and $S\in\mathscr{M}_{\mathfrak{X}}$; $9^\circ)$ any $\mathfrak{X}$-subgroup in $\operatorname{Aut}_G(S)$ stabilizes some element from $\Delta$. Let us verify $7^\circ)$ and $8^\circ)$. If $U$ is an $\mathfrak{X}$-subgroup of the group $G$, then $U\leqslant K$ for some $K\in\operatorname{m}_{\mathfrak{X}}(G)$ and $U$ normalizes $K\cap N\in\operatorname{Hall}_{\mathfrak{X}}(N)$ in view of $6^\circ)$. This proves $7^\circ)$. If here we take $U\in\operatorname{m}_{\mathfrak{X}}(N)$, then $U(K\cap N)$ is an $\mathfrak{X}$-subgroup in $N$ containing $U\in \operatorname{m}_{\mathfrak{X}}(N)$ and $K\cap N\in\operatorname{Hall}_{\mathfrak{X}}(N)\subseteq \operatorname{m}_{\mathfrak{X}}(N)$, and hence
$$
\begin{equation*}
U=U(K\cap N)=K\cap N\in\operatorname{Hall}_{\mathfrak{X}}(N).
\end{equation*}
\notag
$$
This shows that $\operatorname{m}_{\mathfrak{X}}(N)=\operatorname{Hall}_{\mathfrak{X}}(N)$. Therefore, $N\in\mathscr{M}_{\mathfrak{X}}$. From Lemma 2 it follows that any normal subgroup of an $\mathscr{M}_{\mathfrak{X}}$-group is an $\mathscr{M}_{\mathfrak{X}}$-group. Hence $S\in\mathscr{M}_{\mathfrak{X}}$, as claimed in $8^\circ)$.
Let us verify $9^\circ)$. We denote by
$$
\begin{equation*}
\alpha\colon \mathrm{N}_G(S)\to \operatorname{Aut}_G(S)
\end{equation*}
\notag
$$
the epimorphism which associates with each $g\in \mathrm{N}_G(S)$ the automorphism of the group $S$ acting by the rule $x\mapsto x^g$ for all $x\in S$. By Lemma 1, an arbitrary $\mathfrak{X}$-subgroup $T$ of the group $\operatorname{Aut}_G(S)$ has the form $U^\alpha$, where $U$ is some $\mathfrak{X}$-subgroup of the group $\mathrm{N}_G(S)$. Let $t\in T$, and let that $t=g^\alpha$ for $g\in U$. Then $x^t=x^{g^\alpha}=x^g$ for all $x\in S$. By $7^\circ)$, the subgroup $U$ normalizes some $H\in\operatorname{Hall}_{\mathfrak{X}}(N)$. Since $U$ also normalizes $S$, we have Therefore, $T$ leaves invariant the conjugacy class of $\mathfrak{X}$-Hall subgroups of the group $S$ that contains $H\cap S\in\operatorname{Hall}_{\mathfrak{X}}(S)$.
$10^\circ)$ $\operatorname{h}_{\mathfrak{X}}(S)\ne 1$. Indeed, if $\operatorname{h}_{\mathfrak{X}}(S)=1$, then $S\in\mathscr{C}_{\mathfrak{X}}$, and hence, by $8^\circ)$, we have
$$
\begin{equation*}
S\in\mathscr{C}_{\mathfrak{X}}\cap \mathscr{M}_{\mathfrak{X}}=\mathscr{D}_{\mathfrak{X}}.
\end{equation*}
\notag
$$
But then $\mathrm{k}_{\mathfrak{X}}(S_1)=\dots=\mathrm{k}_{\mathfrak{X}}(S_n)=1$ and $\mathrm{k}_{\mathfrak{X}}(N)=1$ by Lemma 3, in contrast to (a).
$11^\circ)$ $\operatorname{h}_{\mathfrak{X}}(S)\ne 9$. We have $S\in\mathscr{M}_{\mathfrak{X}}$. But if $\operatorname{h}_{\mathfrak{X}}(S)=9$, then $S\notin\mathscr{M}_{\mathfrak{X}}$ by Proposition 3. $12^\circ)$ $\mathrm{N}_G({KN})\in\mathscr{C}_{\mathfrak{X}}$ and $KN\in\mathscr{C}_{\mathfrak{X}}$ for any $K\in\operatorname{m}_{\mathfrak{X}}(G)$. Let $K\in\operatorname{m}_{\mathfrak{X}}(G)$. We set Since $\overline{A}=\overline{K}\in \operatorname{m}_{\mathfrak{X}}(\overline{G})$, we conclude that $B/A\cong \mathrm{N}_{\overline{G}}(\overline{K})/\overline{K}$ is a $\pi'$-group. Therefore,
$$
\begin{equation*}
\operatorname{Hall}_{\mathfrak{X}}(A)=\operatorname{Hall}_{\mathfrak{X}}(B).
\end{equation*}
\notag
$$
According to $6^\circ)$ we have $K\in\operatorname{Hall}_{\mathfrak{X}}(A)$. In particular, $\operatorname{Hall}_{\mathfrak{X}}(A)\ne\varnothing$. Therefore,
$$
\begin{equation*}
A,B\in\mathscr{E}_{\mathfrak{X}}\quad\text{and}\quad\operatorname{h}_{\mathfrak{X}}(A) \geqslant\operatorname{h}_{\mathfrak{X}}(B)\geqslant 1.
\end{equation*}
\notag
$$
Given an arbitrary $L\in\operatorname{Hall}_{\mathfrak{X}}(A)$, we will show that $L$ and $K$ are conjugate in $B$. First, it is clear that $A=KN=LN$, and hence, $\overline{K}=\overline{L}$. Therefore,
$$
\begin{equation*}
\overline{L}\in\operatorname{m}_{\mathfrak{X}}(\overline{G})\quad\text{and}\quad L\cap N\in\operatorname{Hall}_{\mathfrak{X}}(N)\subseteq\operatorname{m}_{\mathfrak{X}}(N).
\end{equation*}
\notag
$$
Hence $L\in\operatorname{m}_{\mathfrak{X}}(G)$. According to $2^\circ)$, the equality $\overline{K}=\overline{L}$ implies the conjugacy of the subgroups $K$ and $L$ in $G$ and, therefore, in $B=\mathrm{N}_G(KN)=\mathrm{N}_G(LN)$. So, we have shown that the group $B$ acts transitively by conjugations on the nonempty set $\operatorname{Hall}_{\mathfrak{X}}(A)=\operatorname{Hall}_{\mathfrak{X}}(B)$. Therefore, $\mathrm{N}_G(KN)=B\in \mathscr{C}_{\mathfrak{X}}$.
From Proposition 2 it follows that there exists an $L\in\operatorname{Hall}_{\mathfrak{X}}(A)$ satisfying $B=\mathrm{N}_B(L)A$. Next, since $\operatorname{Hall}_{\mathfrak{X}}(A)=\operatorname{Hall}_{\mathfrak{X}}(B)$ and $B\in\mathscr{C}_{\mathfrak{X}}$, this equality holds for any $L\in\operatorname{Hall}_{\mathfrak{X}}(A)$. From $B\in \mathscr{C}_{\mathfrak{X}}$ it also follows that, for any $L\in\operatorname{Hall}_{\mathfrak{X}}(A)$, there exist an element $b\in B$ such that $K=L^b$. In addition, we have $b=ga$ for some $g\in \mathrm{N}_B(L)$ and $a\in A$. Therefore, and which shows that $KN=A\in\mathscr{C}_{\mathfrak{X}}$, as claimed.From $12^\circ)$ we can establish the following fact, which will be crucial in the proof of the remaining cases: $13^\circ)$ if $K\in\operatorname{m}_{\mathfrak{X}}(G)$, then $\operatorname{Aut}_K(S)$ stabilizes precisely one element from $\Delta$. Let $K\in\operatorname{m}_{\mathfrak{X}}(G)$. That $\Delta$ has a fixed point for $\operatorname{Aut}_K(S)$ follows from $9^\circ)$. Let $H\in\operatorname{Hall}_{\mathfrak{X}}(S)$. We will prove $13^\circ)$ if we establish that the invariancy of the class $H^S\in\operatorname{Hall}_{\mathfrak{X}}(S)/S$ with respect to $\operatorname{Aut}_K(S)$ implies that $H\in(K\cap S)^S$. Let, as before, $A=KN$. By Lemma 5, we have
$$
\begin{equation*}
H^{\operatorname{Aut}_A(S)}=(H^S)^{\operatorname{Aut}_K(S)}=H^S.
\end{equation*}
\notag
$$
Now from Lemma 4 it follows that, for there exist $L\in\operatorname{Hall}_{\mathfrak{X}}(KM)$ such that $H=L\cap S$. In addition, $|L|=|K|$, and, therefore, $L\in\operatorname{Hall}_{\mathfrak{X}}(A)$. But $A=KN\in\mathscr{C}_{\mathfrak{X}}$ by $12^\circ)$, and hence, there exist $u\in K$ and $v\in N$ such that $L=K^{uv}=K^v$. It is clear that if $w\in S$ is the projection of $v$ to $S$ (recall that, $S$ is one of the direct factors $S_1,\dots,S_n$ whose product makes up the group $N$), then
$$
\begin{equation*}
H=L\cap S=K^v\cap S=(K\cap S)^v=(K\cap S)^{w}\in (K\cap S)^{S}.
\end{equation*}
\notag
$$
This proves $13^\circ)$.
From $4^\circ)$, $5^\circ)$, $10^\circ)$ and $11^\circ)$ it follows that $\operatorname{h}_{\mathfrak{X}}(S)\in\{2,3,4\}$. We claim that $14^\circ)$ $\operatorname{h}_{\mathfrak{X}}(S)\ne 2$. Indeed, we fix some $K\in\operatorname{m}_{\mathfrak{X}}(G)$. According to $13^\circ)$, the group $\operatorname{Aut}_K(S)$ stabilizes precisely one element from $\Delta$. But for $\operatorname{h}_{\mathfrak{X}}(S)=2$ the group $\operatorname{Aut}_K(S)$ also stabilizes the remaining element. A contradiction. By the above and in view of Lemma 8, it can be assumed that $15^\circ)$ $\operatorname{h}_{\mathfrak{X}}(S)\in\{3,4\}$ and $2,3\in\pi(\mathfrak{X})$. Now by Burnside’s theorem [15], § I.2, and since each solvable $\pi(\mathfrak{X})$-group is an $\mathfrak{X}$-group, we have $16^\circ)$ each $\{2,3\}$-group is an $\mathfrak{X}$-group. The theorem will be proved once we obtain a contradiction to $9^\circ)$ by showing that $17^\circ)$ some $\mathfrak{X}$-subgroup in $\operatorname{Aut}_G(S)$ acts transitively on $\Delta$. The action of the group $\operatorname{Aut}_G(S)$ on the set $\Delta$ induces the homomorphism
$$
\begin{equation*}
{}^*\colon \operatorname{Aut}_G(S)\to \operatorname{Sym}(\Delta),\quad \text{where}\quad \operatorname{Sym}(\Delta)\cong \begin{cases} \operatorname{Sym}_3 &\text{if }\operatorname{h}_{\mathfrak{X}}(S)=3, \\ \operatorname{Sym}_4 & \text{if }\operatorname{h}_{\mathfrak{X}}(S)=4. \end{cases}
\end{equation*}
\notag
$$
Consider an arbitrary $H\in\operatorname{Hall}_{\mathfrak{X}}(S)$ and consider the class $H^S=\{H^x\mid x\in S\} \in \Delta$. Let $H\leqslant K$ for some $K\in\operatorname{m}_{\mathfrak{X}}(G)$. Since the subgroup $H=K\cap S$ is invariant under $\mathrm{N}_K(S)$, the class $H^S$ is stabilized by the group $\operatorname{Aut}_K(S)$, and according to $11^\circ)$, this group acts without fixed points on the set of the remaining two or three classes. Since
$$
\begin{equation*}
|\Gamma|=\operatorname{h}_{\mathfrak{X}}(S)-1=\begin{cases} 2 & \text{if }\operatorname{h}_{\mathfrak{X}}(S)=3, \\ 3 & \text{if }\operatorname{h}_{\mathfrak{X}}(S)=4, \end{cases}
\end{equation*}
\notag
$$
it follows that the action of $\operatorname{Aut}_K(S)$ on $\Gamma$ should be transitive. But then so is the action of the stabilizer in $\operatorname{Aut}_G(S)$ of the point $H^S$ on $\Gamma$, because this stabilizer contains the subgroup $\operatorname{Aut}_K(S)$. Therefore, $\operatorname{Aut}_G(S)^*$ is a transitive (and even 2-transitive) subgroup in $\operatorname{Sym}(\Delta)$, that is,
$$
\begin{equation*}
\operatorname{Aut}_G(S)^*\cong \begin{cases} \operatorname{Sym}_3 &\text{if }\operatorname{h}_{\mathfrak{X}}(S)=3, \\ \operatorname{Alt}_4\text{ or }\operatorname{Sym}_4 &\text{if } \operatorname{h}_{\mathfrak{X}}(S)=4. \end{cases}
\end{equation*}
\notag
$$
In any case, $\operatorname{Aut}_G(S)^*$ is a $\{2,3\}$-group and, therefore, is an $\mathfrak{X}$-group. By Lemma 1, there exists an $\mathfrak{X}$-subgroup $U$ of the group $\operatorname{Aut}_G(S)$ such that $U^*=\operatorname{Aut}_G(S)^*$. This subgroup $U$ acts transitively on $\Delta$, in contrast to $9^\circ)$.
This completes the proof of Theorem 1. For proofs of Corollaries 1–3, see §§ 1.2, 1.3. Proof of Corollary 4. It suffices to show that (i) $\Rightarrow$ (ii). Let $H\in \operatorname{m}_{\mathfrak{X}}(G)$ and $H\leqslant K\leqslant G$. Let a subgroup $N\trianglelefteq G$ be such that $\mathrm{k}_{\mathfrak{X}}(G/N)=\mathrm{k}_{\mathfrak{X}}(G)$. Then $N\in\mathscr{D}_{\mathfrak{X}}$ by Theorem 1. We claim that $\mathrm{k}_{\mathfrak{X}}(K/(K\cap N))=\mathrm{k}_{\mathfrak{X}}(K)$.
Assume that $N$ is minimal in $G$ as a normal subgroup. Then $N$ is a direct product of its simple subgroups conjugate in $G$. From Theorem 2 and Lemma 2.28 in [4] we conclude that either $N\in\mathfrak{X}$, or, for $\pi=\pi(\mathfrak{X})$, any $\pi$-Hall subgroup of the group $N$ is solvable (in particular, lies in $\mathfrak{X}$) and $N\in \mathscr{D}_\pi$. In the first case, the required result is clear. In the second case, from Lemma 2 it follows that $H\cap N\in\operatorname{Hall}_{\mathfrak{X}}(N)\subseteq\operatorname{Hall}_\pi(N)$ and $H\cap N\leqslant K\cap N\leqslant N$. According to Theorem 1.4, [16], $K\cap N\in\mathscr{D}_\pi$. Since any $\pi$-Hall subgroup from $K\cap N$ lies in $\mathfrak{X}$, we have $K\cap N\in \mathscr{D}_{\mathfrak{X}}$. As a result, $\mathrm{k}_{\mathfrak{X}}(K/(K\cap N))=\mathrm{k}_{\mathfrak{X}}(K)$ by Theorem 2. The general case can be derived from that considered above by induction on $|N|$ after passing to the quotient group relative to the minimal normal subgroup of the group $G$ contained in $N$, and then applying Theorem 2. This proves the corollary. Proof of Corollary 5. By definition of the relation $\underset{\mathfrak{X}}{\equiv}$, $G_1\underset{\mathfrak{X}}{\equiv} G_2$ if and only if the complete $\mathfrak{X}$-reductions of the groups $G_1$ and $G_2$ are isomorphic. Now all the assertions in Corollary 5 are clear. This proves the corollary.
Citation in format AMSBIB
\Bibitem{GuoRev22} |
|
||||||||||||||||||||||||||||||
|
Contact us: |
Terms of Use
|
Registration to the website |
Logotypes |
|