Transitive characteristically simple subgroups of finite quasiprimitive permutation groups

TL;DR

This paper characterizes nonabelian transitive characteristically simple subgroups within finite quasiprimitive permutation groups, revealing their containment in the socle or base group under specific conditions.

Contribution

It provides a detailed classification of such subgroups, especially when they do not contain nontrivial normal subgroups of the socle.

Findings

Nonabelian transitive characteristically simple subgroups lie in the socle or base group.

Explicit descriptions of subgroup possibilities are given under certain conditions.

The results clarify subgroup structure in finite quasiprimitive permutation groups.

Abstract

The first main result of this paper is that a finite transitive nonabelian characteristically simple subgroup of a wreath product in product action must lie in the base group of the wreath product. This allows us to characterize nonabelian transitive characteristically simple subgroups $H$ of finite quasiprimitive permutation groups $G$. If the socle of $G$, denoted by $\mbox{soc}(G)$, is nonabelian, then $H$ lies in $\mbox{soc}(G)$. An explicit description is given for the possibilities of $H$ under the condition that $H$ does not contain a nontrivial normal subgroup of $\mbox{soc}(G)$.

Tables

Table 1. Table 1. Table for Theorem  1.3 (3)(b)

Table 2. Table 2. Table for Theorem  1.3 (2)-(4)

Equations

$$S_{\alpha }=\{(q,q\alpha )\mid q\in Q_1\}$$

$$D=\{(q,q\alpha )\mid q\in Q_i\}$$

$$\begin{array}[]{ccccc}\pi_{i}&\colon&S&\rightarrow&Q_{i}\\ &&(q_{1},\ldots,q_{r})&\mapsto&q_{i},\end{array}$$

$$\mathrm{supp}(P)=\{Q_i\mid P{\pi }_i\ne 1\}.$$

$$D=\{(q,q{\alpha }_2,\dots ,q{\alpha }_r)\mid q\in Q_1\}.$$

$$Q_1=H{\pi }_1(H{\pi }_2{\alpha }_2^{-1}\cap H{\pi }_3{\alpha }_3^{-1})=H{\pi }_2{\alpha }_2^{-1}(H{\pi }_1\cap H{\pi }_3{\alpha }_3^{-1})=H{\pi }_3{\alpha }_3^{-1}(H{\pi }_1\cap H{\pi }_2{\alpha }_2^{-1}).$$

$$|D||H|=|Q||T|<|Q{|}^2=|S|.$$

$$Q^2=$$

$$|A_1\cap A_2|=|A_3\cap A_4|=|M_{11}{|}^2/|M_{12}|=660.$$

$$X\cap Y=\{(c,c,c\psi ,c\psi )\mid c\in C_1\cap C_2\}\cong C_1\cap C_2.$$

$$|X\cap Y|=\frac{|X||Y|}{|Q{|}^4}=\frac{|M_{11}{|}^3}{|M_{12}{|}^2}=55.$$

$$|C_1{C}_2|=\frac{|C_1||C_2|}{|X\cap Y|}=\frac{660^2}{55}=7920=|A_2|,$$

$$D=\{(q,q)\mid q\in Q\}$$

$$W_j=\mathrm{Sym}(\Gamma )\times (\mathrm{Sym}(\Gamma )\mathrm{wr}{S}_{r-1}),$$

$$(y,y{\beta }_2,\dots ,y{\beta }_r)=m^g=(y^{g_1},(y{\beta }_2{)}^{g_2},\dots ,(y{\beta }_r{)}^{g_r}{)}^h.$$

$$|{\omega }^{H_B}|=\frac{|H_B|}{|(H_B{)}_{\omega }|}=\frac{|H_B^{(1)}|}{|(H_B^{(1)}{)}_{\gamma }|}=|\Gamma |,$$

$$W_1\times W_2=\left(\mathrm{Sym}(\Gamma )\mathrm{wr}{S}_{r_1}\right)\times \left(\mathrm{Sym}(\Gamma )\mathrm{wr}{S}_{r-r_1}\right)$$

$$S_{\alpha }=(Q_1\cap S_{\alpha })\times \cdots \times (Q_r\cap S_{\alpha }).$$

$$\mathrm{Aut}S\cap \left(\bigcap _{i=1}^r{\mathbb{N}}_{\mathrm{Aut}S}(Q_i)\right)=\prod _i(\mathrm{Aut}{Q}_i).$$

$$H/(H\cap S)\cong (HS)/S⩽\prod _i(\mathrm{Aut}{Q}_i)/Q_i.$$

$$H/(H\cap S)\cong (HS)/S⩽\mathrm{Aut}(S)/S=\mathrm{Out}(S).$$

$$H/(H\cap S)\cong (HS)/S⩽G/S⩽\mathrm{Hol}(Q_1)/S\cong \mathrm{Out}(Q_1).$$

$$X=\mathrm{Aut}(Q_1\times Q_{r/2+1})\times \cdots \times \mathrm{Aut}(Q_{r/2}\times Q_r).$$

$$H_0/(H_0\cap S)\cong (H_{0}S)/S⩽\overline{S}/S\cong \mathrm{Out}(Q).$$

$$\frac{|H_1|}{|(H_1{)}_{\alpha }|}=|\Omega |=|Q{|}^{r-1},$$

$$\frac{|Q{|}^{r-1}}{|{\alpha }^{H_0}|}=\frac{|Q{|}^{r-1}}{|T{|}^{k_0}}|(H_0{)}_{\alpha }|.$$

$$Q_i⩾H_0{\pi }_i\cong H_0/(\ker {\pi }_i\cap H_0)\cong T^{s_i},$$

$$H_0=T_1\times \cdots \times T_{k_0},$$

$$\prod _{Q_j\in \mathrm{supp}(T_i)}{Q}_j,$$

$$H_0=T_1\times \cdots \times T_{m_1}\times T_{m_1+1}\times \cdots \times T_{m_1+m_2},$$

Full text

The research presented in this paper formed a part of the first author’s PhD project under the supervision by the second author. Daldegan was supported by a PhD scholarship awarded by the CAPES and CNPq, while Schneider acknowledges the support of the CNPq grants Universal (project no.: 421624/2018-3) and Produtividade em Pesquisa (project no.: 308773/2016-0). We thank the referees for their careful reading of the manuscript.

1. Introduction

The characterization of transitive subgroups of finite permutation groups is an important problem with several applications in algebraic combinatorics and algebraic graph theory. Without the ambition of being exhaustive, we recall some results in this research area. Finite primitive permutation groups with a transitive abelian subgroup were classified by Li [Li03], while finite permutation groups with a transitive cyclic subgroup were described by Li and Praeger [LP12]. Regular subgroups of finite primitive permutation groups were classified by Liebeck, Praeger and Saxl [LPS10] and by Baumeister [Bau07]. A general description of transitive subgroups of finite primitive permutation groups was given by Liebeck, Praeger and Saxl [LPS00].

In this paper we focus on transitive nonabelian characteristically simple subgroups of finite quasiprimitive permutation groups. Transitive simple subgroups of wreath products in product action were classified by the second author in collaboration with Baddeley and Praeger in [BPS04]. The examples of [BPS04] consisted of two isolated groups and of two infinite families. Furthermore, it was proved by Baddeley and Praeger in [BP03, Theorem 1.4], that if $H$ is a nonabelian, nonsimple, characteristically simple subgroup of a finite quasiprimitive almost simple permutation group $G$, then either $H$ is intransitive or $\operatorname{soc}(G)$ is an alternating group in its natural action. The scarcity of such examples gave the hope that nonabelian characteristically simple subgroups of finite quasiprimitive groups could be described. This goal is achieved in this paper.

It is an easy consequence of the divisibility relations of Lemma 5.1 that if $T$ is a transitive nonabelian simple subgroup of a wreath product in product action, then $T$ is always contained in the base group of the wreath product. Our first theorem generalizes this result to transitive characteristically simple groups.

Theorem 1.1**.**

Let $\Gamma$ be a finite set such that $|\Gamma|\geqslant 2$, let $r\geqslant 2$, and let $W=\operatorname{Sym}(\Gamma)\operatorname{wr}S_{r}$ be considered as a permutation group on $\Omega=\Gamma^{r}$ in product action. If $H$ is a transitive nonabelian characteristically simple subgroup of $W$, then $H$ is a subgroup of the base group; that is, $H\leqslant\operatorname{Sym}(\Gamma)^{r}$.

Since finite quasiprimitive permutation groups are often contained in wreath products in product action, Theorem 1.1 leads to the following more general result.

Theorem 1.2**.**

Let $G$ be a finite quasiprimitive permutation group acting on $\Omega$ with nonabelian socle $S$ and let $H$ be a transitive nonabelian characteristically simple subgroup of $G$. Then $H\leqslant\operatorname{soc}(G)$.

Once we know by Theorem 1.2 that $H$ lies in the socle of $G$, a more detailed description of the possibilities of $H$ can be given considering the possible O’Nan–Scott classes of $G$. For the description of the O’Nan–Scott classes of quasiprimitive permutation groups see Section 3, while for the terminology related to strips see Section 2. Given a natural number $r$, the symbol $\underline{r}$ denotes $\{1,\ldots,r\}$.

Theorem 1.3**.**

Let $G$ be a finite quasiprimitive permutation group acting on $\Omega$ and let $S=Q_{1}\times\cdots\times Q_{r}$ be the socle of $G$, where each $Q_{i}$ is isomorphic to a simple group $Q$. Assume that $T$ is a nonabelian finite simple group, and let $H\cong T_{1}\times\cdots\times T_{k}\cong T^{k}$ be a transitive subgroup of $G$ with $k\geqslant 2$. If $H$ does not contain a minimal normal subgroup of $S$, then, after possibly relabeling the $T_{i}$, one of the following holds.

1. (0)

$G$* has type HA, $T\cong\mbox{SL}(3,2)$, $|\Omega|=8^{k}$.* 2. (1)

$G$* has type As, $S=A_{n}$ and $G=A_{n}$ or $G=S_{n}$ acting naturally on $n$ points with $n\geqslant 10$.* 3. (2)

$G$* has type Sd or HS, $k=r=2$ and $T_{i}<Q_{i}$, for $i=1,2$. Moreover, the groups $Q$ and $T$ are described in one of the rows of Table 2. Furthermore, if*

[TABLE]

for some isomorphism $\alpha:Q_{1}\rightarrow Q_{2}$, then $Q_{1}=T_{1}(T_{2}\alpha^{-1})$. 4. (3)

$G$* has type Pa and one of the following is valid.*

1. (a)

$k=r$* and $T_{i}<Q_{i}$ for all $i\in\underline{r}$.* 2. (b)

The $T_{i}$ are pairwise disjoint strips in $S$ with $|\operatorname{supp}(T_{i})|\in\{1,2\}$ for all $i$, with $|\operatorname{supp}(T_{i})|=2$ for at least one $i$, and the groups $T$ and $Q$ are as in one of the rows of Table 1. 3. (c)

$S\cong(A_{n})^{r}$, where $|\Omega|=n^{r}$ and $n\geqslant 10$. 5. (4)

$G$* has type Cd or HC, $k=r$ and $T_{i}<Q_{i}$ for all $i$; the groups $Q$ and $T$ are described in one of the rows of Table 2. Further, a point stabilizer $S_{\alpha}$ is the direct product of pairwise disjoint strips $D$ with $|\operatorname{supp}(D)|=2$. If*

[TABLE]

is a strip in $S_{\alpha}$ with some isomorphism $\alpha:Q_{i}\rightarrow Q_{j}$, then $Q_{i}=T_{i}(T_{j}\alpha^{-1})$.

In part (1) of Theorem 1.3, we paraphrase the aforementioned result [BP03, Theorem 1.4] by Baddeley and Praeger.

Note that we stated Theorem 1.3 under the condition that $H$ contains no minimal normal subgroup of $\operatorname{soc}(G)$, which is the same condition as the one applied in [LPS00]. If we allow that $H$ can contain a minimal normal subgroup of $\operatorname{soc}(G)$, then the simple components of $H$ and $\operatorname{soc}(G)$ must be isomorphic and $H$ is the direct product of full strips of $\operatorname{soc}(G)$ (see Definition 2.2). Such examples do arise and their classification can be achieved by considering the factorization $\operatorname{soc}(G)=H(\operatorname{soc}(G)_{\alpha})$ where $\operatorname{soc}(G)_{\alpha}$ is a point stabilizer in $\operatorname{soc}(G)$. We also note that the group $G$ in Theorem 1.3 is actually primitive, except perhaps in part (3). This shows that a finite imprimitive quasiprimitive group $G$ very rarely contains a transitive characteristically simple subgroup that does not contain a minimal normal subgroup of $\operatorname{soc}(G)$.

In the proofs of Theorems 1.2 and 1.3 we could have relied more on the descriptions of transitive subgroups of primitive permutations groups presented in [LPS00]. However, since we are interested only in transitive characteristically simple subgroups, our arguments are simpler that those in [LPS00], and so we preferred a more direct approach.

Using Theorem 1.3, we obtain a characterization of regular characteristically simple subgroups of quasiprimitive permutation groups.

Corollary 1.4**.**

Suppose that $G$ and $H$ are as in Theorem 1.3. If $H$ is regular then the O’Nan–Scott type of $G$ is either As or Pa.

If we allow that $H$ may contain a minimal normal subgroup of $\operatorname{soc}(G)$, then we obtain several examples of regular characteristically simple subgroups of $G$ with $G$ having other O’Nan–Scott types. For example, if $G$ has type Sd and $H$ is a maximal normal subgroup of $\operatorname{soc}(G)$, then $H$ is regular and characteristically simple.

Regular subgroups occur in the automorphisms groups of Cayley graphs. Quasiprimitive subgroups of automorphisms of Cayley graphs of finite simple groups were studied in [FPW02]. Our results allow us to state the following corollary concerning quasiprimitive subgroups of the automorphism groups of Cayley graphs of characteristically simple groups.

Corollary 1.5**.**

Let $H$ be a finite nonabelian characteristically simple group and let $\Gamma$ be a noncomplete Cayley graph of $H$. Suppose that $G$ is a quasiprimitive subgroup of $\operatorname{Aut}(\Gamma)$ with nonabelian socle such that $H\leqslant G$. Then either $H$ contains a minimal normal subgroup of $\operatorname{soc}(G)$ or the O’Nan–Scott type of $H$ is Pa.

Theorems 1.1 and 1.2 are proved in Sections 5 and 6, respectively. The proofs of Theorem 1.3, and Corollaries 1.4 and 1.5 can be found in Section 7. Examples are described in Section 4 to show that all possibilities described in Theorem 1.3 arise.

2. Subgroups of characteristically simple groups and factorizations

We say that a group $G$ is characteristically simple if it has no proper and nontrivial characteristic subgroups; that is, its only subgroups invariant under $\operatorname{Aut}(G)$ are $1$ and $G$ itself. In particular, when $G$ is finite, this is equivalent to saying that $G$ is a direct product of isomorphic simple groups [Rob96, 3.3.15].

If $A$ and $B$ are proper subgroups of a group $G$ such that $G=AB$, then we call this expression a factorization of $G$. In a transitive permutation group $G$, a subgroup $H\leqslant G$ is transitive if and only if $G=HG_{\alpha}$ where $G_{\alpha}$ is the stabilizer of a point. Hence factorizations play a natural role in the study of transitive subgroups of permutation groups and in this section we collect some auxiliary results concerning subgroups and factorizations of finite characteristically simple groups.

Lemma 2.1**.**

Let $Q$ be a finite simple group, and suppose that $Q=AB$ is a factorization of $Q$ where $A\cong T^{s_{1}}$ and $B\cong T^{s_{2}}$ for some finite nonabelian simple group $T$ and positive integers $s_{1}$ and $s_{2}$. Then $s_{1}=s_{2}=1$ and $Q$ and $T$ are as in Table 2.

Proof.

For a group $G$, let $p(G)$ denote the set of primes that divide $|G|$. We claim that $p(Q)=p(A)=p(B)$. Clearly, $p(A)\subseteq p(Q)$ and $p(B)\subseteq p(Q)$. On the other hand, given a prime number $p\in p(Q)$, since $|Q|=|A||B|/|A\cap B|$ and $|A|$ and $|B|$ are powers of $|T|$, the prime $p$ divides $|T|$, and so $p$ divides $|A|$ and $|B|$. Thus $p(Q)=p(A)=p(B)$, as claimed. In the terminology of [BP98], the expression $Q=AB$ is a full factorization of the simple group $Q$ and such factorizations are described in [BP98, Theorem 1.1]. Keeping in mind that $\mbox{Sp}_{6}(2)\cong\Omega_{7}(2)$, inspection of [BP98, Table I] shows that the only options where $|A|$ and $|B|$ are powers of the same finite simple group occur when $A\cong B\cong T$ and $Q$ and $T$ are as in Table 2. ∎

Next we introduce some terminology to describe diagonal subgroups in wreath products.

Definition 2.2**.**

Let $Q_{1},\ldots,Q_{r}$ be groups and set $S=Q_{1}\times\cdots\times Q_{r}$. Consider, for $i\in\underline{r}$, the projections

[TABLE]

and assume that $P\leqslant S$.

1. (i)

$P$ is a strip of $S$ if $P\neq 1$ and, for each $i\in\underline{r}$, either the restriction of $\pi_{i}$ to $P$ is injective or $P\pi_{i}=1$. We define the support of $P$ as

[TABLE] 2. (ii)

A strip $P$ of $S$ is said to be nontrivial if $|\operatorname{supp}(P)|>1$. 3. (iii)

A strip $P$ of $S$ is said to be a full strip if $P\pi_{i}=Q_{i}$ for all $Q_{i}\in\operatorname{supp}(P)$. 4. (iv)

Two strips $P$ and $Q$ are said to be disjoint if $\operatorname{supp}(P)\cap\operatorname{supp}(Q)=\varnothing$. 5. (v)

$P$ is a subdirect subgroup of $S$ if $P\pi_{i}=Q_{i}$ for each $i\in\underline{r}$. 6. (vi)

$P$ is a diagonal subgroup of $S$ if the restriction of $\pi_{i}$ to $P$ is injective for each $i\in\underline{r}$. 7. (vii)

$P$ is a full diagonal subgroup of $S$ if $P$ is both a subdirect and a diagonal subgroup of $S$.

Note that the definitions above depend on the given direct decomposition of the group $S$. However, we will usually assume that $S$ is the direct product of pairwise isomorphic nonabelian simple groups and, unless explicitly stated otherwise, the conditions in Definition 2.2 will be interpreted with respect to the unique finest direct decomposition of $S$. The concepts introduced in Definition 2.2 are studied in more depth in [PS18b, Section 4.4]. If $P$ is a strip of $S$, as above, and, for some $i\in\underline{r}$, the restriction $\pi_{i}|_{P}$ is injective, then $P\pi_{i}\cong P$.

The first part of the following lemma appeared in Scott’s paper [Sco80, Lemma p. 328], and is known as Scott’s Lemma (see also [PS18b, Section 4.6] for several generalizations). It describes the structure of the subdirect subgroups of a direct product of nonabelian simple groups. The second part can be found, for example, in [KL90, Proposition 5.2.5(i)]. Note that Lemma 2.3 does not assume that the groups should be finite.

Lemma 2.3**.**

Consider $S=Q_{1}\times\cdots\times Q_{r}$, where each $Q_{i}$ is a nonabelian simple group, and let $P$ be a nontrivial subgroup of $S$.

1. (1)

If $P$ is a subdirect subgroup of $S$, then $P$ is the direct product of full strips whose supports form a partition of $\{Q_{1},\ldots,Q_{r}\}$. 2. (2)

If $P$ is a normal subgroup of $S$, then $P=\prod_{j\in J}Q_{j}$, where $J\subseteq\underline{r}$.

The next result characterizes the factorizations $S=HD$ of finite nonabelian characteristically simple groups $S$, in which $H$ is a nonabelian characteristically simple subgroup and $D$ is a full diagonal subgroup of $S$.

Lemma 2.4**.**

Let $Q$ and $T$ be nonabelian finite simple groups, let $S=Q_{1}\times\cdots\times Q_{r}$ where $r\geqslant 2$ and $Q_{i}\cong Q$ for all $i$, and let $H\cong T^{k}$ be a nonabelian characteristically simple subgroup of $S$. Consider, for $i\in\underline{r}$, the projections $\pi_{i}\colon S\rightarrow Q_{i}$ and suppose that $1<H\pi_{i}<Q_{i}$ for all $i\in\underline{r}$. Assume, for $i=2,\ldots,r$, that $\alpha_{i}:Q_{1}\rightarrow Q_{i}$ is an isomorphism and define

[TABLE]

If $DH=S$, then $r=k=2$ and $H=T_{1}\times T_{2}$ where $T_{1}=H\cap Q_{1}$, $T_{2}=H\cap Q_{2}$ and $T_{1}\cong T_{2}\cong T$. Further, in this case, $Q_{1}=T_{1}(T_{2}\alpha_{2}^{-1})$ and the groups $Q$ and $T$ are as in one of the rows of Table 2.

Proof.

By [PS18b, Lemma 8.16], $r\leqslant 3$. Further, if $r=3$, then the same lemma implies that

[TABLE]

Hence, in the terminology of [BP98], the set $\{H\pi_{1},H\pi_{2}\alpha_{2}^{-1},H\pi_{3}\alpha_{3}^{-1}\}$ is a strong multiple factorization of $Q_{1}$. Noting that $H\pi_{i}\cong T^{s_{i}}$ with $s_{i}\geqslant 1$, and considering the classification of strong multiple factorizations of finite simple groups in [BP98, Table V], we obtain that $r=2$. Furthermore, $Q_{1}=(H\pi_{1})(H\pi_{2}\alpha_{2}^{-1})$ where $H\pi_{1}\cong T^{s_{1}}$ and $H\pi_{2}\alpha_{2}^{-1}\cong H\pi_{2}\cong T^{s_{2}}$ with some $s_{1},s_{2}\geqslant 1$. Now Lemma 2.1 implies that $H\pi_{1}\cong H\pi_{2}\cong T$ and that the groups $Q$ and $T$ are as in Table 2. Finally, note that $H\cap Q_{i}\unlhd H\pi_{i}$ holds for $i=1,2$. Hence either $H\cap Q_{i}=H\pi_{i}$ or $H\cap Q_{i}=1$. Thus, if $H\neq(H\cap Q_{1})\times(H\cap Q_{2})$, then $H\cong H\pi_{1}\cong H\pi_{2}\cong T$, and hence the factorization $S=DH$ is impossible, since

[TABLE]

Thus $H=(H\cap Q_{1})\times(H\cap Q_{2})$ must hold. Setting $T_{1}=H\cap Q_{1}$ and $T_{2}=H\cap Q_{2}$, the rest of the lemma follows. ∎

The following result was originally proved in [BP03, Lemma 2.2]; see also [PS18a, Theorem 1.2] and [PS18b, Section 4.8] for generalizations.

Lemma 2.5**.**

Let $T$ be a nonabelian finite simple group and let $X$ and $Y$ be direct products of pairwise disjoint nontrivial full strips in $T^{r}$ with $r\geqslant 2$. Then $XY\neq T^{r}$.

The following technical lemma, which characterizes certain factorizations related to the Mathieu group $M_{12}$, will be used in the last section.

Lemma 2.6**.**

Let $S=Q^{4}$ and $A_{1},A_{2},A_{3},A_{4}\leqslant Q$ such that $Q=M_{12}$ and each $A_{i}\cong M_{11}$. Let $X=\{(p,p,q,q)\mid p,q\in Q\}$ and $Y=\{(a_{1},a_{2},a_{2}\psi,a_{4})\mid a_{i}\in A_{i}\}$ be subgroups of $S$, where $\psi\colon A_{2}\rightarrow A_{3}$ is an isomorphism. Then $S\neq XY$.

Proof.

Let $D_{1}=\{(p,p,1,1)\mid p\in Q\}$ and $D_{2}=\{(1,1,q,q)\mid q\in Q\}$. We have that $X=D_{1}\times D_{2}\cong Q^{2}$ is the direct product of two full strips of $S$ and $Y\cong(M_{11})^{3}$ is the direct product of three strips of $S$, where the second strip is a diagonal subgroup of $Q^{2}$.

Assume that $Q^{4}=XY$ and consider the projections $\pi_{12}\colon S\rightarrow Q^{2}$ and $\pi_{34}\colon S\rightarrow Q^{2}$, where $\pi_{12}$ projects onto the first two coordinates and $\pi_{34}$ projects onto the last two coordinates. Applying these projections to $Q^{4}=XY$, we obtain that

[TABLE]

It follows from Lemma 2.4 that $Q=A_{1}A_{2}=A_{3}A_{4}$. Thus,

[TABLE]

Set $C_{1}=A_{1}\cap A_{2}$ and $C_{2}=(A_{3}\cap A_{4})\psi^{-1}$. Note that

[TABLE]

Since $Q^{4}=XY$,

[TABLE]

As $X\cap Y\cong C_{1}\cap C_{2}\leqslant A_{2}$ and

[TABLE]

we find that $A_{2}=C_{1}C_{2}$. Since $M_{11}$ has a unique conjugacy class of subgroups of order 660 [CCN*+*85, p.18], $C_{1}$ and $C_{2}$ are conjugate in $A_{2}$, which contradicts to $A_{2}=C_{1}C_{2}$. Then $Q^{4}\neq XY$ and the result is proved. ∎

3. Quasiprimitive permutation groups

A permutation group is quasiprimitive if all its nontrivial normal subgroups are transitive. For example, primitive permutation groups are quasiprimitive. Since a transitive simple group is always quasiprimitive, but not necessarily primitive, the class of quasiprimitive permutation groups is strictly larger than the class of primitive permutation groups.

Finite primitive and quasiprimitive groups were classified by the respective versions of the O’Nan–Scott Theorem; see [PS18b, Chapter 7]. In this classification, we distinguish between 8 classes of finite primitive groups, namely HA, HS, HC, SD, CD, PA, AS, TW, and 8 classes of finite quasiprimitive groups, namely HA, HS, HC, Sd, Cd, Pa, As, Tw. As the notation suggests, quasiprimitive groups in classes HA, HS and HC are always primitive. Here we only give a brief summary of these classes; the interested reader can find a more detailed treatment in [PS18b, Chapter 7]. The type of a primitive or quasiprimitive group $G$ can be recognized from the structure and the permutation action of its socle, denoted $\operatorname{soc}(G)$. Let $G\leqslant\operatorname{Sym}(\Omega)$ be a quasiprimitive permutation group, let $M$ be a minimal normal subgroup of $G$, and let $\omega\in\Omega$. Note that $M$ is a characteristically simple group. If $M$ is nonabelian, then by a subdirect subgroup of $M$ we mean one that is subdirect with respect to the unique finest direct decomposition of $M$ (see Definition 2.2). For a group $G$, the holomorph $\operatorname{Hol}G$ is the semidirect product $G\rtimes\operatorname{Aut}G$ viewed as a permutation group acting on the set $G$; see [PS18b, Section 3.3] for more details. The main characteristics of $G$ and $M$ in each primitive and quasiprimitive type are as follows.

HA: $M$ is abelian, ${\mathbb{C}}_{G}(M)=M$ and $G\leqslant\operatorname{Hol}M$. A quasiprimitive permutation group in this class is always primitive.

HS: $M$ is nonabelian, simple, and regular; $\operatorname{soc}(G)=M\times{\mathbb{C}}_{G}(M)\cong M\times M$ and $G\leqslant\operatorname{Hol}M$. Such a quasiprimitive permutation group is always primitive.

HC: $M$ is nonabelian, nonsimple, and regular; $\operatorname{soc}(G)=M\times{\mathbb{C}}_{G}(M)\cong M\times M$ and $G\leqslant\operatorname{Hol}M$. A quasiprimitive permutation group of this type is always primitive.

Sd: $M$ is nonabelian and nonsimple; $M_{\omega}$ is a simple subdirect subgroup of $M$ and ${\mathbb{C}}_{G}(M)=1$. If, in addition, $G$ is primitive, then the type of $G$ is SD.

Cd: $M$ is nonabelian and nonsimple; $M_{\omega}$ is a nonsimple subdirect subgroup of $M$ and ${\mathbb{C}}_{G}(M)=1$. If, in addition, $G$ is primitive, then the type of $G$ is CD.

Pa: $M$ is nonabelian and nonsimple; $M_{\omega}$ is not a subdirect subgroup of $M$ and $M_{\omega}\neq 1$; ${\mathbb{C}}_{G}(M)=1$. If, in addition, $G$ is primitive, then the type of $G$ is PA.

As: $M$ is nonabelian and simple; ${\mathbb{C}}_{G}(M)=1$. If, in addition, $G$ is primitive, then the type of $G$ is AS.

Tw: $M$ is nonabelian and nonsimple; $M_{\omega}=1$; ${\mathbb{C}}_{G}(M)=1$. If, in addition, $G$ is primitive, then the type of $G$ is TW.

Note that if $G$ is a primitive permutation group of type HS or HC, then there exists a primitive group $\overline{G}$ of type SD or CD, respectively, that contains $G$ as a subgroup of index 2 (see [PS18b, Corollary 3.11]). Hence in the proof of the inclusion $H\leqslant G$ in Theorem 1.2, the cases when $G$ has type HS or HC can be reduced to the cases when $G$ has type SD or CD, respectively.

4. Examples of transitive characteristically simple subgroups

In this section we describe some examples to show that all the possibilities described in Theorem 1.3 arise. In Examples 4.1–4.5, $T$ is a nonabelian finite simple group.

Example 4.1** ($G$ has type HA).**

The following example goes back to at least D. G. Higman [Hig62, Lemma 4]. Suppose that $R=\mbox{SL}(3,2)$. Then $R$ acts irreducibly on $V=\mathbb{F}_{2}^{3}$ and set $G=V\rtimes R$. It is well-known (see for example [Pra90, Proposition 5.2]) that $G$ has a transitive simple subgroup $T$ isomorphic to $R$. Letting $W=G\operatorname{wr}S_{r}$ with $r\geqslant 2$, the direct product $T^{r}$ is a transitive characteristically simple subgroup in the primitive permutation group $W$ of type HA acting on $\Omega=\mathbb{F}_{2}^{3r}$. Hence $T^{r}$ is contained in any primitive subgroup of $\mathbb{F}_{2}^{3r}\rtimes\mbox{GL}(2,3r)$ that contains $W$. These inclusions are as in Theorem 1.3(0).

Example 4.2** ($G$ has type As).**

Set $H=T^{k}$ with $k\geqslant 2$. Suppose that $X$ is a corefree subgroup of $H$ and consider the right coset space $\Omega=[H:X]$. Then $H$ can be viewed as a transitive subgroup of $Q=\mbox{Alt}(\Omega)$ which is a primitive permutation group of type AS. These examples satisfy the conditions of Theorem 1.3(1).

Example 4.3** ($G$ has type Sd or HS).**

Suppose that $T$ and $Q$ are as in one of the rows of Table 2. Set $D$ to be the diagonal subgroup

[TABLE]

of $Q^{2}$ and let $\Omega$ denote the right coset space $[Q^{2}:D]$. The normalizer ${\mathbb{N}}_{\operatorname{Sym}(\Omega)}(Q^{2})$ is a primitive group of type SD whose abstract group structure is $(Q^{2}\cdot\mbox{Out}(Q))\rtimes C_{2}$. Suppose that $G\leqslant{\mathbb{N}}_{\operatorname{Sym}(\Omega)}(Q^{2})$ such that $G$ contains $Q^{2}$. Then $G$ is a primitive permutation group of type SD or HS depending on the projection of $G$ on $C_{2}$. Assume that $T_{1}$ and $T_{2}$ are subgroups of $Q$ such that $T_{1}\cong T_{2}\cong T$ and $T_{1}T_{2}=Q$ and $H=T_{1}\times T_{2}$. Viewing $H$ as a subgroup of $Q^{2}$, easy calculation shows that $DH=Q^{2}$ and so $H$ is a transitive characteristically simple subgroup of $G$. These examples are as in Theorem 1.3(2).

Example 4.4** ($G$ has type Pa).**

Examples for Theorem 1.3(3) can be constructed by wreathing smaller examples. Suppose that $T$ is a simple transitive subgroup of a nonabelian simple group $Q$ acting on a set $\Gamma$. Then, for $r\geqslant 2$, the wreath product $W=Q\operatorname{wr}S_{r}$ contains the transitive characteristically simple subgroup $T^{r}$. Furthermore, the inclusion $T^{r}\leqslant W$ is as in Theorem 1.3(3)(a).

Suppose that $H$ and $Q$ are as in Example 4.2. Then, for $r\geqslant 2$, $H^{r}$ is contained in $Q\operatorname{wr}S_{r}$ and this inclusion is as in Theorem 1.3(3)(c).

Finally, let $Q$ and $T$ be as in one of the rows of Table 1. Then $T$ can be embedded as a transitive subgroup of $G=Q\operatorname{wr}S_{2}$, and so, for $r\geqslant 2$, $T^{r}$ is a transitive characteristically simple subgroup of $W=G\operatorname{wr}S_{k}$. The inclusion $T^{r}\leqslant W$ is as in Theorem 1.3(3)(b).

Example 4.5** ($G$ has type Cd or HC).**

Suppose that $G$ and $H$ are as in Example 4.3. If $r\geqslant 2$, then the wreath product $W=G\operatorname{wr}S_{r}$ is a primitive group of type CD or HC (depending on the type of $G$) that contains the transitive characteristically simple subgroup $H^{r}=T^{2r}$. The inclusion $H\leqslant W$ is as in Theorem 1.3(4).

5. The proof of Theorem 1.1

This section contains the proof of Theorem 1.1. We start by stating a number theoretic result which is a corollary of Legendre’s Formula for the largest prime-power that divides $n!$.

Lemma 5.1**.**

Given natural numbers $p,n\geqslant 2$, the following are valid.

1. (1)

$p^{n}\nmid n!$. 2. (2)

If $p^{n-1}\mid n!$, then $p=2$ and $n$ is a power of $2$. 3. (3)

$4^{n-1}\nmid n!$.

Next we review some concepts related to subgroups of wreath products in product action. Let $\Gamma$ be a finite set such that $|\Gamma|\geqslant 2$, let $r\geqslant 2$, and let $W=\operatorname{Sym}(\Gamma)\operatorname{wr}S_{r}$ be considered as a permutation group on $\Omega=\Gamma^{r}$ in product action. An element of $W$ is written as $(a_{1},\ldots,a_{r})b$ where $a_{i}\in\operatorname{Sym}(\Gamma)$ for all $i\in\underline{r}$, and $b\in S_{r}$. Suppose that $X$ is a subgroup of $W$. For $j\in\underline{r}$, we define the $j$-th component $X^{(j)}$ of $X$ as follows. Suppose that $W_{j}$ is the stabilizer in $W$ of $j$ under the permutation representation $\pi\colon W\rightarrow S_{r}$. Then

[TABLE]

where the first factor of the direct product acts on the $j$-th coordinate, while the second factor acts on the other coordinates. In particular, ‘$S_{r-1}$’ is taken to be the stabilizer of $j$ in $S_{r}$. We define $X^{(j)}$ as the projection of $X_{j}=X\cap W_{j}$ onto the first factor of $W_{j}$. We view $X^{(j)}$ as a subgroup of $\operatorname{Sym}(\Gamma)$. The following theorem was stated in [PS12, Theorem 1.2]; see also [PS18b, Corollary 5.17].

Theorem 5.2**.**

If $X$ is a transitive subgroup of $W$, then each component of $X$ is transitive on $\Gamma$. Moreover, if $X$ acts transitively on $\underline{r}$, then each component of the intersection $X\cap(\operatorname{Sym}(\Gamma))^{r}$ is transitive on $\Gamma$.

We turn to the proof of Theorem 1.1.

Proof of Theorem 1.1.

Suppose that $H=T_{1}\times\cdots\times T_{k}=T^{k}$ for some nonabelian finite simple group $T$. Suppose, as above, that $\pi:W\rightarrow S_{r}$ is the natural projection. Let $B$ be the base group $\operatorname{Sym}(\Gamma)^{r}$ of $W$. Then $B=\ker\pi$. Assume for contradiction that $H\not\leqslant B$; that is $H\pi\neq 1$.

First we assume that $H\pi$ is transitive on $\underline{r}$. The case when $H\pi$ is intransitive will be treated afterwards. Set $H_{B}=H\cap B$. Then $H_{B}$ is a normal subgroup of $H$ and, by Lemma 2.3(2), it is of the form $T^{s}$, with some $s$. Further, $H=H_{B}\times\overline{H}_{B}$ where similarly $\overline{H}_{B}=T^{k-s}$, and $\overline{H}_{B}$ acts transitively and faithfully by $\pi$ on $\underline{r}$. For $j\in\underline{r}$, consider the component $H_{B}^{(j)}$ as a permutation group on $\Gamma$. By Theorem 5.2, $H_{B}^{(j)}$ is transitive on $\Gamma$ for all $j$.

Claim 1. $H_{B}^{(j)}\cong H_{B}$ for all $j$.

Proof of Claim 1. Suppose, for $j\in\underline{r}$, that $\sigma_{j}:\operatorname{Sym}(\Gamma)^{r}\rightarrow\operatorname{Sym}(\Gamma)$ denotes the $j$-th coordinate projection. Then $H_{B}^{(j)}\cong H_{B}/(\ker\sigma_{j}\cap H_{B})$. Let $m$ be an element of $\ker\sigma_{1}\cap H_{B}$. Thus $m=(1,m_{2},\ldots,m_{r})$ with $m_{j}\in H_{B}^{(j)}$. Let $j\in\underline{r}$. Since $\overline{H}_{B}$ is transitive on $\underline{r}$, there is some element $g=(g_{1},\ldots,g_{r})h$ of $\overline{H}_{B}$ such that $1(g\pi)=1h=j$. Then $m^{g}=(1,m_{2},\ldots,m_{r})^{g}=(1,m_{2}^{g_{2}},\ldots,m_{r}^{g_{r}})^{h},$ and so the $j$-th coordinate of $m^{g}$ is 1. Hence $m^{g}\in\ker\sigma_{j}\cap H_{B}$, and then $(\ker\sigma_{1}\cap H_{B})^{g}\leqslant\ker\sigma_{j}\cap H_{B}.$ Analogously, the same argument above shows that $\ker\sigma_{j}\cap H_{B}\leqslant(\ker\sigma_{1}\cap H_{B})^{g}$, and so $\ker\sigma_{j}\cap H_{B}=(\ker\sigma_{1}\cap H_{B})^{g}$. On the other hand, $\ker\sigma_{1}\cap H_{B}$ is a subgroup of $H_{B}$ and $\overline{H}_{B}$ centralizes $H_{B}$, and so $\ker\sigma_{1}\cap H_{B}=\ker\sigma_{j}\cap H_{B}$ for all $j$. Thus $\ker\sigma_{1}\cap H_{B}$ acts trivially on $\Omega$, and so $\ker\sigma_{1}\cap H_{B}=1$, which gives $\ker\sigma_{j}\cap H_{B}=1$ for all $j$. Therefore, $H_{B}^{(j)}\cong H_{B}$ for all $j$, which proves Claim 1.

Thus the restrictions to $H_{B}$ of the projection maps $\sigma_{j}$ are monomorphisms. Then $\beta_{j}=\sigma_{1}^{-1}\sigma_{j}:H_{B}^{(1)}\rightarrow H_{B}^{(j)}$ is an isomorphism for all $j$. As a consequence, every element $m\in H_{B}$ can be expressed uniquely as $m=(y,y\beta_{2},\ldots,y\beta_{r})$, for some $y\in H_{B}^{(1)}$.

Claim 2. For all $j\in\underline{r}$, there is some element $x_{j}\in\operatorname{Sym}(\Gamma)$ such that $y\beta_{j}=y^{x_{j}}$ for all $y\in H_{B}^{(1)}$.

Proof of Claim 2. Suppose that $y\in H_{B}^{(1)}$. Then $m=(y,y\beta_{2},\ldots,y\beta_{r})\in H_{B}$. Let $j\in\underline{r}$ and, using the transitivity of $\overline{H}_{B}$ on $\underline{r}$, suppose that $g=(g_{1},\ldots,g_{r})h\in\overline{H}_{B}$ is such that $1(g\pi)=1h=j$. Then $g$ centralizes $m$ and hence

[TABLE]

Comparing the $j$-th coordinates in the two sides of the last equation, we find that $y\beta_{j}=y^{g_{1}}$. Taking $x_{j}=g_{1}$, thus we have $y\beta_{j}=y^{x_{j}}$, which proves Claim 2.

Claim 3. If $\Sigma$ is a $H_{B}$-orbit in $\Omega$, then $|\Sigma|=|\Gamma|$.

Proof of Claim 3. Since $H$ is transitive on $\Omega$ and $H_{B}\unlhd H$, all the $H_{B}$-orbits have the same size. Hence it suffices to show the claim for just one $H_{B}$-orbit. Choose the elements $1,x_{2},\ldots,x_{r}$ as in the previous claim, let $\gamma\in\Gamma$ and consider the element $\omega=(\gamma,\gamma x_{2},\ldots,\gamma x_{r})$. Suppose that $m\in H_{B}$. By the previous claim, $m$ has the form $m=(y,y^{x_{2}},\ldots,y^{x_{r}})$ for some $y\in H_{B}^{(1)}$. Hence $\omega^{m}=(\gamma y,\gamma yx_{2},\ldots,\gamma yx_{r})$. Thus $m$ stabilizes $\omega$ if, and only if, $y\in\operatorname{Sym}(\Gamma)$ stabilizes $\gamma$. Thus $(H_{B})_{\omega}=(H_{B}^{(1)})_{\gamma}$. So by applying the Orbit-Stabilizer Theorem twice, and using that $|H_{B}|=|H_{B}^{(1)}|$ and that $H_{B}^{(1)}$ is transitive on $\Gamma$, we have

[TABLE]

and so Claim 3 is proved.

Claim 4. The case when $H\pi$ is transitive is impossible.

Proof of Claim 4. $H_{B}$ is a normal subgroup of $H$ and every $H_{B}$-orbit has size $|\Gamma|$. Hence the number of $H_{B}$-orbits is $|\Gamma|^{r-1}.$ Since $H$ is transitive on $\Omega$, $\overline{H}_{B}$ is transitive on the set of $H_{B}$-orbits and hence $|\Gamma|^{r-1}\mid|\overline{H}_{B}|$. Since $\overline{H}_{B}$ has a faithful action on $\underline{r}$, this leads to $|\Gamma|^{r-1}\mid r!$. Now, since $\Gamma$ is an orbit for the characteristically simple group $H_{B}^{(1)}$, we find that $|\Gamma|\geqslant 5$. Hence $|\Gamma|$ is divisible by $p$, where $p$ is either an odd prime or $p=4$, which is a contradiction by Lemma 5.1. Then Claim 4 is proved.

This completes the proof for the case when $H\pi$ is a transitive subgroup of $S_{r}$. Let us now turn to the case when $H\pi$ is intransitive. Recall that $B$ is the base group of $W$. Assuming that $H\not\leqslant B$ implies that there exists an $H\pi$-orbit $\Delta$ in $\underline{r}$ with size at least 2. Set $\overline{\Delta}=\underline{r}\setminus\Delta$ and $r_{1}=|\Delta|$. Then $H$ can be embedded into the direct product

[TABLE]

such that the projection $H_{1}$ of $H$ into $W_{1}$ acts transitively on $\underline{r_{1}}$. Now, since $H$ is transitive on $\Gamma^{r}$, $H_{1}$ is also transitive on $\Gamma^{r_{1}}$. Further, as $H$ is characteristically simple, so is $H_{1}$. Hence using the theorem in the case when $H\pi$ is transitive gives a contradiction. Therefore, $H\leqslant B$. ∎

Since in several classes of quasiprimitive permutation groups, the individual groups are subgroups in wreath products in product action, Theorem 1.1 leads to the following corollary.

Corollary 5.3**.**

Let $G\leqslant\operatorname{Sym}(\Omega)$ be a finite quasiprimitive permutation group with nonabelian socle $S$ and let $H$ be a transitive nonabelian characteristically simple subgroup of $G$. Let $\alpha\in\Omega$ and assume that $S=Q_{1}\times\cdots\times Q_{r}$, where $r\geqslant 2$ and the $Q_{i}$ are pairwise $G$-conjugate normal subgroups of $S$ such that

[TABLE]

Then $H\leqslant{\mathbb{N}}_{G}(Q_{i})$ for all $i\in\underline{r}$. Further, if the $Q_{i}$ are simple groups, then $H\leqslant S$.

Proof.

Set $\Gamma$ to be the right coset space $[Q_{1}\colon Q_{1}\cap S_{\alpha}]$. By [PS18b, Theorem 4.24], we may assume without loss of generality that $G$ is a subgroup of $W=\operatorname{Sym}(\Gamma)\operatorname{wr}S_{r}$ acting in product action on $\Gamma^{r}$, and so $H$ can also be viewed as a transitive subgroup of $W$. By Theorem 1.1, $H$ is a subgroup of the base group $\operatorname{Sym}(\Gamma)^{r}$ of $W$. Now [PS18b, Theorem 4.24] also implies that the conjugation action of $G$ on the set $\{Q_{1},\ldots,Q_{r}\}$ is equivalent to its action on $\underline{r}$ induced by the natural projection $\pi:W\rightarrow S_{r}$. Since $\ker\pi=\operatorname{Sym}(\Gamma)^{r}$, we have that $H$ acts trivially on $\{Q_{1},\ldots,Q_{r}\}$ by conjugation, and so $H\leqslant{\mathbb{N}}_{G}(Q_{i})$ holds for all $i$.

Suppose now that the $Q_{i}$ are simple. Since ${\mathbb{C}}_{G}(\operatorname{soc}(G))=1$ and since $H\leqslant{\mathbb{N}}_{G}(Q_{i})$ for all $i$, the group $H$ can be viewed as a subgroup of

[TABLE]

Therefore

[TABLE]

Since Schreier’s Conjecture holds, the group on the right-hand side of the last display is soluble. On the other hand, if $H\cap S\neq H$, then $H/(H\cap S)$ is nonabelian characteristically simple, which is impossible. Hence $H\cap S=H$, which is equivalent to $H\leqslant S$. ∎

6. The proof of Theorem 1.2

In this section we prove Theorem 1.2. Suppose that $G\leqslant\operatorname{Sym}(\Omega)$ is a finite quasiprimitive permutation group with nonabelian socle $S$ and let $H$ be a transitive nonabelian characteristically simple subgroup of $G$. We are required to show that $H\leqslant S$. Our strategy is to analyze each of the possible O’Nan–Scott classes of $G$.

$\boldsymbol{G}$** has type AS:** $S$ is a simple group and $S\leqslant G\leqslant\operatorname{Aut}(S)$. Then

[TABLE]

As $S$ is simple, it follows from Schreier’s Conjecture that $\operatorname{Out}(S)$ is soluble. Since $H$ is nonabelian and characteristically simple, $H\cap S=H$. Therefore $H\leqslant S=\operatorname{soc}(G)$, as desired.

$\boldsymbol{G}$** has type HS:** $S=Q_{1}\times Q_{2}$ where $Q_{1}$ and $Q_{2}$ are simple minimal normal subgroups of $G$. Furthermore, $S\leqslant G\leqslant\operatorname{Hol}(Q_{1})$ and

[TABLE]

Therefore, arguing as we did for the type AS, it follows from Schreier’s Conjecture that $H\cap S=H$, and so $H\leqslant S$.

$\boldsymbol{G}$** has type TW:** $S$ is regular, and so $S$ clearly satisfies equation (4) with respect to its finest direct decomposition into the direct product of simple groups. Thus, by Corollary 5.3, $H\leqslant\operatorname{soc}(G)$.

For the next four O’Nan–Scott types, namely for HC, PA, SD, and Cd, we will assume that $S=Q_{1}\times\cdots\times Q_{r}$ where $r\geqslant 2$ and the $Q_{i}$ are pairwise isomorphic nonabelian simple groups. We let $Q$ denote the common isomorphism type of the $Q_{i}$. We define $\pi_{i}$ as the coordinate projection $\pi_{i}\colon S\rightarrow Q_{i}$.

$\boldsymbol{G}$** has type HC:** $S=S_{1}\times S_{2}$ where $S_{1}$ and $S_{2}$ are nonsimple, regular minimal normal subgroups of $G$. Then $r$ is even and, after possibly reordering the $Q_{i}$, a point stabilizer in $S$ is a direct product $D_{1}\times\cdots\times D_{r/2}$ where each $D_{i}$ is a full diagonal subgroup in $Q_{i}\times Q_{r/2+i}$. Since $r\geqslant 4$, Corollary 5.3 applies to the direct decomposition $S=(Q_{1}\times Q_{r/2+1})\times\cdots\times(Q_{r/2}\times Q_{r})$ and this gives that $H$ normalizes $Q_{i}\times Q_{r/2+i}$ for all $i\in\underline{r/2}$. Since ${\mathbb{C}}_{G}(S)=1$, the conjugation action of $H$ on $S$ embeds $H$ into

[TABLE]

Now $X/S$ is a subgroup of $((\operatorname{Out}Q)\operatorname{wr}C_{2})^{r}$, which is a soluble group by Schreier’s Conjecture. Now the usual argument implies that $H\leqslant\operatorname{soc}(G)$.

$\boldsymbol{G}$** has type PA:** for all $\alpha\in\Omega$, $S_{\alpha}$ is not a subdirect subgroup of $S$ and $S$ is not regular. In general, the groups of this class do not satisfy equation (4) in Corollary 5.3, but it is well-known, for $\alpha\in\Omega$, that $G$ has a faithful quotient action on the right coset space $\overline{\Omega}=[S:P]$ where $P=(S_{\alpha}\pi_{1})\times\cdots\times(S_{\alpha}\pi_{r})$. Since $H$ also acts transitively on $\overline{\Omega}$ and $P$ is a point stabilizer in $S$ under this action, equation (4) holds for $P$ with the finest direct product decomposition of $S$, and so $H\leqslant S$ must hold.

$\boldsymbol{G}$** has type SD:** this is the most difficult O’Nan–Scott type to deal with. In this case, for $\alpha\in\Omega$, $S_{\alpha}$ is a simple subdirect subgroup of $S$. Since $S$ is transitive on $\Omega$ and $S_{\alpha}\cong Q$, $|\Omega|=|Q|^{r-1}$. In this case, $G$ can be considered as a subgroup of $\overline{G}=(S\cdot\operatorname{Out}(Q))\rtimes S_{r}$, where $S_{r}$ permutes by conjugation the factors of $S$ and $\operatorname{Out}(Q)$ acts on $S\cong Q^{r}$ diagonally; see [PS18b, Section 7.4].

Consider the extension $\overline{S}=S\cdot\operatorname{Out}(Q)$. We have that $\overline{G}$ permutes the elements in $\Sigma=\{Q_{1},\ldots,Q_{r}\}$ and the kernel of this action is precisely $\overline{S}$. If we denote by $H_{0}$ the kernel of $H$ acting on $\Sigma$, we obtain that $H_{0}=H\cap\overline{S}$. Since $H$ is characteristically simple, we have by Lemma 2.3(2) that $H_{0}\cong T^{k_{0}}$ for some integer $k_{0}$, and there exists a normal subgroup $H_{1}$ of $H$ such that $H=H_{0}\times H_{1}$. It follows from the Isomorphism Theorem and from the definition of $\overline{S}$ that

[TABLE]

Since $\operatorname{Out}(Q)$ is soluble by Schreier’s Conjecture, and $H_{0}$ is nonabelian and characteristically simple, we conclude that $H_{0}=H_{0}\cap S$, which means that $H_{0}\leqslant S$.

If $H_{1}=1$, then, since $H_{0}\leqslant S$, we obtain at once that $H\leqslant S$. Hence it suffices to prove that $H_{1}=1$. Suppose that $H_{1}\neq 1$. Since $H_{1}\cap\overline{S}=1$, $H_{1}$ permutes the elements in $\Sigma$ faithfully, and so $|H_{1}|\mid r!$. In particular, since the size of the smallest nonabelian simple group is 60, we have $r\geqslant 5$.

We claim that $H_{0}\neq 1$. In fact, if that is not the case, then $H=H_{1}$, so $H_{1}$ is transitive on $\Omega$. Then applying the Orbit–Stabilizer Theorem and the transitivity of $S$, we obtain

[TABLE]

so $|Q|^{r-1}\mid|H_{1}|$. Since $|H_{1}|\mid r!$, we obtain that $|Q|^{r-1}\mid r!$. Since the order of every finite nonabelian simple group is divisible by four, this implies that $4^{r-1}\mid r!$, which is not possible by Lemma 5.1. Therefore, $H_{0}\neq 1$.

Let us analyze the action of $H_{0}$ on $\Omega$. Since $H_{0}\trianglelefteq H$ and $H$ is transitive, the orbits of $H_{0}$ form a block system for $H$. In particular, the $H_{0}$-orbits have the same size. By the Orbit–Stabilizer Theorem, it follows that $|(H_{0})_{\alpha}|$ is independent of $\alpha$. Therefore, the number of $H_{0}$-orbits on $\Omega$ is equal to

[TABLE]

Since $H=H_{0}\times H_{1}$ and $H$ is transitive on $\Omega$, we have that $H_{1}$ is transitive on the set of $H_{0}$-orbits. Then the Orbit–Stabilizer Theorem gives that the number of $H_{0}$-orbits divides $|H_{1}|$. Therefore, the number of $H_{0}$-orbits divides $r!$. From the Isomorphism Theorem we have that

[TABLE]

where $s_{i}\geqslant 0$ for all $i\in\underline{r}$.

We claim that $s_{i}\leqslant 1$ for all $i$. Suppose, on the contrary, that there exists $i\in\underline{r}$ such that $s_{i}\geqslant 2$. In particular, $Q$ contains a subgroup isomorphic to $T^{2}$. As every finite simple group has a cyclic Sylow subgroup (see [KLST90, Theorem 4.9]), we can choose a prime $p$ such that the Sylow $p$-subgroups of $Q$ are cyclic. Since a Sylow $p$-subgroup of $T^{2}$ is contained in a Sylow $p$-subgroup of $Q$, we have that $p\nmid|T|$. In particular, $p\neq 2$. Considering the number of $H_{0}$-orbits in (5), we obtain that $p^{r-1}$ divides the number of $H_{0}$-orbits, and so $p^{r-1}$ divides $r!$. However, this contradicts Lemma 5.1. Therefore, $s_{i}\leqslant 1$ for all $i$, as claimed.

Since $H_{0}\neq 1$, $s_{i}=1$ for some $i\in\underline{r}$. Thus (6) gives that $Q$ has a subgroup isomorphic to $T$. Since each $T_{i}\leqslant H_{0}$ is simple, each $T_{i}\leqslant H_{0}$ is a strip of $S$. We assert that if $i\neq j$, then $\operatorname{supp}(T_{i})\cap\operatorname{supp}(T_{j})=\varnothing$. In fact, if there is $m\in\underline{r}$ such that $T_{i}\pi_{m}\cong T$ and $T_{j}\pi_{m}\cong T$, then $(T_{i}\times T_{j})\pi_{m}\cong T^{2}$. However, this is impossible, since $s_{m}\leqslant 1$. Therefore

[TABLE]

where each $T_{i}\cong T$ is a diagonal subgroup of

[TABLE]

in such a way that $\operatorname{supp}(T_{i})\cap\operatorname{supp}(T_{j})=\varnothing$ for all $i\neq j$. As a consequence, we obtain that $k_{0}\leqslant r$. Assume first that $k_{0}<r$; we will treat the case $k_{0}=r$ separately. Since $|T|\mid|Q|$, we have by (5) that $|Q|^{r-k_{0}-1}$ divides the number of $H_{0}$-orbits in $\Omega$. As $H_{1}$ is transitive on the set of $H_{0}$-orbits, this implies that $|Q|^{r-k_{0}-1}$ divides $|H_{1}|$.

Let $d_{i}=|\operatorname{supp}(T_{i})|$. Moreover, let $m_{1}$ be the number of factors $T_{i}$ for which $d_{i}\geqslant 5$, and let $m_{2}$ be the number of factors $T_{i}$ such that $d_{i}<5$. So $m_{1}+m_{2}=k_{0}$. Relabeling if necessary, we can write

[TABLE]

such that $d_{i}\geqslant 5$ if and only if $i\leqslant m_{1}$. Set $m_{3}=r-\sum^{k_{0}}_{i=1}d_{i}$; that is, $m_{3}$ is the number of factors $Q_{i}$ such that $H_{0}\pi_{i}=1$.

Since $H_{1}$ centralizes $H_{0}$, we have that $H_{1}$ centralizes each $T_{i}$. for each $i\leqslant k_{0}$. As, for $h_{1}\in H_{1}$ and $i\in\underline{k_{0}}$, we have

[TABLE]

we conclude that each $\operatorname{supp}(T_{i})$ is $H_{1}$-invariant. In particular, $H_{1}$ acts by conjugation on $\operatorname{supp}(T_{i})$ for all $i\in\underline{k_{0}}$, and, since $H_{1}$ is a nonabelian characteristically simple group, this action is trivial whenever $|\operatorname{supp}(T_{i})|<5$. Hence $H_{1}$ acts trivially on

[TABLE]

Since the action of $H_{1}$ on $\Sigma$ is faithful, we obtain that

[TABLE]

As we assumed $H_{1}\neq 1$, we have from (8) that either $m_{1}=0$ and $m_{3}\geqslant 5$, or $m_{1}\neq 0$. In both cases we conclude that

[TABLE]

Recall that $|Q|^{r-k_{0}-1}$ divides $|H_{1}|$. So (8) implies that

[TABLE]

On the other hand, we have that $r\geqslant d_{1}+d_{2}+\cdots+d_{m_{1}}+m_{2}+m_{3}$. So

[TABLE]

Since (9) is valid and $k_{0}=m_{1}+m_{2}$, we obtain

[TABLE]

Therefore, using equation (10), we obtain that

[TABLE]

Since $4\mid|Q|$, the previous line gives that

[TABLE]

which is a contradiction by Lemma 5.1. This implies that $H_{1}=1$, which means that if $k_{0}<r$, then $H\leqslant S$.

Now consider the case where $k_{0}=r$. Then (7) implies that $d_{i}=1$ for all $i\in\underline{r}$. Since each $\operatorname{supp}(T_{i})$ is $H_{1}$-invariant, we have that $H_{1}$ acts faithfully and trivially on $\Sigma$, thus $H_{1}=1$. Therefore if $k_{0}=r$, we also obtain $H=H_{0}\leqslant S$.

Therefore, if $G$ has type Sd, then $H\leqslant S$.

$\boldsymbol{G}$** has type CD:** for $\alpha\in\Omega$, $S_{\alpha}$ is a subdirect subgroup of $S$, but it is not simple. So Lemma 2.3 implies that there exist sets $\overline{\Sigma}=\{S_{1},\ldots,S_{q}\}$ and $\{D_{1},\ldots,D_{q}\}$, where $q\geqslant 2$, each $D_{i}$ is a full diagonal subgroup of $S_{i}$ and $S_{i}=\prod_{Q_{j}\in\operatorname{supp}(D_{i})}Q_{j}$, such that $S=S_{1}\times\cdots\times S_{q}$, $G_{\alpha}$ acts transitively by conjugation on $\overline{\Sigma}$ and, considering the projections $\overline{\pi}_{i}\colon S\rightarrow S_{i}$, we have that

[TABLE]

where each $S_{\alpha}\overline{\pi}_{i}=D_{i}$. By [PS18b, Theorem 11.13], $G$ is permutationally isomorphic to a subgroup of a wreath product of the form $W=G_{0}\operatorname{wr}S_{q}$ acting on $\Gamma^{q}$ in product action where $G_{0}$ is a quasiprimitive permutation group on $\Gamma$ of type Sd. Further, setting $S=S_{1}$, $\operatorname{soc}(G_{0})=S$ and $\operatorname{soc}(G)=\operatorname{soc}(W)=S^{q}=S_{1}\times\cdots\times S_{q}$. By Corollary 5.3, $H\leqslant{\mathbb{N}}_{G}(S_{i})$ for all $i$, and hence $H$ lies in the base group $(G_{0})^{q}$ of $W$.

For $i\in\underline{q}$, let $\sigma_{i}:(G_{0})^{q}\rightarrow G_{0}$ denote the $i$-th projection map. Then $H\sigma_{i}$ is a transitive characteristically simple subgroup of $G_{0}$ acting on $\Gamma$. Since $G_{0}$ is quasiprimitive of Sd type, we find that $H\sigma_{i}\leqslant\operatorname{soc}(G_{0})=S$. Since this is true for all $i$, we obtain that $H\leqslant S^{q}=\operatorname{soc}(G)$.

This concludes the proof of Theorem 1.2. We note that the cases when $G$ has type HS or HC could have been reduced to the types Sd and Cd, respectively, as explained at the end of Section 3. However, we chose not to do this, since these arguments are significantly easier than the one given for the type Sd.

7. The proofs of Theorem 1.3

and Corollaries 1.4 and 1.5

In this section we prove Theorem 1.3 and Corollaries 1.4 and 1.5.

The proof of Theorem 1.3.

(0) Suppose that $G\leqslant\operatorname{Sym}(\Omega)$ has type HA. Then $G$ has a unique minimal normal subgroup $V$ which is elementary abelian of order $p^{d}$. Suppose that $H=T^{k}$ is a nonabelian transitive characteristically simple subgroup of $G$. Then $H\cap V$ is a normal subgroup of $H$ which is elementary abelian, and so, by Lemma 2.3(2), $H\cap V=1$. Hence one may consider the group $X=V\rtimes H$. Since both $V$ and $H$ are transitive on $\Omega$, so is $X$.

Though the group $X$ is transitive on $\Omega$, it may be imprimitive. Suppose that $\Delta$ is a maximal proper $X$-block in $\Omega$ (if $X$ is primitive, then $\Delta$ is a singleton) and let $\mathcal{B}$ denote the corresponding $X$-invariant block system. For a subgroup $Y$ of $\operatorname{Sym}(\Omega)$ preserving $\mathcal{B}$, let $Y^{\mathcal{B}}$ denote the permutation group of $\operatorname{Sym}(\mathcal{B})$ induced by $Y$. By the maximality of $\Delta$, $X^{\mathcal{B}}$ is a primitive group in which $V^{\mathcal{B}}$ is a transitive abelian normal subgroup and $H^{\mathcal{B}}$ is a transitive nonabelian characteristically simple subgroup. In particular, $V^{\mathcal{B}}$ and $H^{\mathcal{B}}$ are nontrivial and $V^{\mathcal{B}}$ is regular. Since $X=VH$, we obtain that $X^{\mathcal{B}}=V^{\mathcal{B}}H^{\mathcal{B}}$. Furthermore, $V^{\mathcal{B}}\cap H^{\mathcal{B}}$ is an elementary abelian normal subgroup in the nonabelian characteristically simple group $H^{\mathcal{B}}$, which gives that $X^{\mathcal{B}}=V^{\mathcal{B}}\rtimes H^{\mathcal{B}}$. As $X^{\mathcal{B}}$ is primitive of HA type, the conjugation action of $H^{\mathcal{B}}$ induces an irreducible linear group on $V^{\mathcal{B}}$. If $|V^{\mathcal{B}}|=p^{d_{0}}$, then the group $H^{\mathcal{B}}$ has a transitive permutation representation of degree $p^{d_{0}}$ and also a faithful irreducible representation of degree $d_{0}$ over the field of $p$ elements. Now [Bau00, Theorem 1.1] implies that $H^{\mathcal{B}}\cong T\cong\mbox{SL}(3,2)$, $p=2$, $d_{0}=3$ and the stabilizer $K$ in $H^{\mathcal{B}}=T$ is a subgroup of index eight111The cited theorem of Baumeister contains a misprint and $\mbox{SL}(3,3)$ is written instead of $\mbox{SL}(3,2)$.. Since, up to conjugacy, $K$ is the unique subgroup of $T=\mbox{SL}(3,2)$ with $2$-power index, we find that the stabilizer of a point $\alpha\in\Omega$ in $H$ must be of the form $K^{k}$, and so $|\Omega|=8^{k}$.

In the rest of this proof, let $G$ be a finite quasiprimitive permutation group on $\Omega$ of type HS, HC, AS, TW, PA, SD or CD, and let $H\cong T^{k}$ be a transitive nonabelian characteristically simple subgroup of $G$ where $k\geqslant 2$ and $T$ is a nonabelian finite simple group. Assume that $S=Q_{1}\times\cdots\times Q_{r}$ is the socle of $G$, where each $Q_{i}\cong Q$ for a nonabelian simple group $Q$. Consider the projections $\pi_{i}\colon\operatorname{soc}(G)\rightarrow Q_{i}$ of $\operatorname{soc}(G)$ onto its direct factors. According to Theorem 1.2, $H\leqslant S$ and we assume that $H$ does not contain a nontrivial normal subgroup of $\operatorname{soc}(G)$. If $G$ had type Tw, then $S$ would be regular, and no proper subgroup of $S$ would be transitive. Therefore, under these conditions, the type of $G$ cannot be Tw. The rest of the proof of Theorem 1.3 is by considering each possible O’Nan–Scott type for $G$.

(1) In this case, $S\cong Q$ is a nonabelian simple group. For $\alpha\in\Omega$, the factorization $S=HS_{\alpha}$ holds. Now [BP03, Theorem 1.4] implies that $S=A_{n}$ with $n\geqslant 10$ acting naturally on $n$ points and hence $G=A_{n}$ or $G=S_{n}$ must follow.

(2) $G$ has type Sd or HS. Since $S$ is transitive on $\Omega$ and $S_{\alpha}\cong Q$, $|\Omega|=|Q|^{r-1}$. For a fixed $j\in\underline{r}$, denote

[TABLE]

Given ${i_{0}}\in\underline{r}$, a priori, we have three options: $H\pi_{i_{0}}=1$, $H\pi_{i_{0}}=Q_{i_{0}}$ or $1<H\pi_{i_{0}}<Q_{i_{0}}$.

We claim, for all $i_{0}\in\underline{r}$, that $1<H\pi_{i_{0}}<Q_{i_{0}}$. First we assume that $H\pi_{i_{0}}=1$ for some ${i_{0}}\in\underline{r}$. Without loss of generality, assume that ${i_{0}}=1$. Then $H$ is a transitive subgroup of $\overline{Q}_{1}$, which is a regular subgroup. Since no proper subgroup of $\overline{Q}_{1}$ is transitive, $H$ would have to be equal to $\overline{Q}_{1}$, which is impossible, as we assume that $H$ contains no nontrivial normal subgroup of $S$. Hence $H\pi_{i_{0}}\neq 1$ holds for all $i_{0}\in\underline{r}$.

Consider now the case in which $H\pi_{i_{0}}=Q_{i_{0}}$ for some ${i_{0}}\in\underline{r}$. In this case $Q$ is a composition factor of $H$, which means by the Jordan-Hölder Theorem that $Q\cong T$, and so $k\leqslant r$. Further, the transitivity of $H$ implies that $k=r-1$ or $k=r$. If $k=r$, then $H=S$, which is impossible in our conditions. Assume now that $k=r-1$. By the previous paragraph, $H\pi_{i_{0}}\neq 1$ for all $i_{0}$, and so $H\pi_{i_{0}}=Q_{i_{0}}$ must hold for all $i_{0}$. Therefore $H$ is a subdirect subgroup of $S$. By Lemma 2.3, $H$ must be the direct product of full strips with pairwise disjoint supports. Since $k=r-1$, either $r=2$ and $k=1$ or one of these strips must be equal to a direct factor of $S$. The latter possibility cannot hold by our conditions. In the former case, $Q_{1}\times Q_{2}=S_{\alpha}H$ is a factorization with two full diagonal subgroups, which is not possible, by Lemma 2.5.

Therefore $1<H\pi_{j}<Q_{j}$ must hold for all $j\in\underline{r}$ as claimed. We have that $S_{\alpha}$ is a full diagonal subgroup of $S$. Since $H$ is transitive, $S=HS_{\alpha}$. Now Lemma 2.4 implies that $k=r=2$, $H=T^{2}=T_{1}\times T_{2}$ where $T_{1}<Q_{1}$ and $T_{2}<Q_{2}$ and $Q$ and $T$ are described in one of the rows of Table 2. Assuming, as we may, that

[TABLE]

for some isomorphism $\alpha:Q_{1}\rightarrow Q_{2}$, the same lemma implies that $Q_{1}=T_{1}(T_{2}\alpha^{-1})$.

(3) $G$ has type Pa. As $S_{\alpha}$ is nontrivial and is not a subdirect subgroup of $S$, $1<S_{\alpha}\pi_{i}<Q_{i}$ holds for all $i$ and the projections $S_{\alpha}\pi_{i}$ are permuted by $G_{\alpha}$ transitively. Furthermore, since $H\pi_{i}$ is a homomorphic image of $H$, we have that $H\pi_{i}=T^{s_{i}}$ with some $s_{i}\geqslant 0$. Since $H$ is transitive, we have that $S_{\alpha}H=S$, which implies that $s_{i}\geqslant 1$ for all $i\in\underline{r}$. Further, each minimal normal subgroup $T_{i}$ of $H$ is a strip in $S$. Thus we may consider the support $\operatorname{supp}(T_{i})$. The transitivity of $H$ implies that

[TABLE]

Case 1: Suppose that $s_{i_{0}}\geqslant 2$ for some $i_{0}\in\underline{r}$. In this case, the factorization in (12) and [BP03, Theorem 1.4] implies $Q\cong A_{n}$ and $S_{\alpha}\pi_{i_{0}}\cong A_{n-1}$ where $n\geqslant 10$. Since the $S_{\alpha}\pi_{i}$ are pairwise isomorphic, $S_{\alpha}\pi_{i}\cong A_{n-1}$ holds for all $i\in\underline{r}$. Set $P=S_{\alpha}\pi_{1}\times\cdots\times S_{\alpha}\pi_{r}$. In particular, $P\cong(A_{n-1})^{r}$.

We claim that $S_{\alpha}=P$. Assume that $S_{\alpha}\neq P$. Since $P$ is a nonabelian characteristically simple group and $S_{\alpha}$ is a subdirect subgroup of $P$, by Scott’s Lemma (Lemma 2.3), $S_{\alpha}$ is the direct product of diagonal subgroups $S_{\alpha}=D_{1}\times\cdots\times D_{l}$ for some $l\leqslant r$. Renumbering if necessary and using that $S_{\alpha}\neq P$, we can assume that $\operatorname{supp}(D_{1})=\{Q_{1},\ldots,Q_{m}\}$ with $2\leqslant m\leqslant r$. Consider the projection $\overline{\pi}\colon S\rightarrow Q_{1}\times Q_{2}$. Since $S=S_{\alpha}H$, we have $Q_{1}\times Q_{2}=S\overline{\pi}=(S_{\alpha}\overline{\pi})(H\overline{\pi})$, where $S_{\alpha}\overline{\pi}=\{(x,x\alpha)\mid x\in S_{\alpha}\pi_{1}\}$ for an isomorphism $\alpha:S_{\alpha}\pi_{1}\rightarrow S_{\alpha}\pi_{2}$. Since $n\geqslant 10$, the automorphisms of $A_{n}$ are induced by conjugations by elements of $S_{n}$ [Wil09, 2.4.1]. Thus we can extend the isomorphism $\alpha$ to an isomorphism $\overline{\alpha}$ between $Q_{1}$ and $Q_{2}$. Now

[TABLE]

where $\overline{D}=\{(x,x\overline{\alpha})\mid x\in Q_{1}\}$ is a full diagonal subgroup in $Q_{1}\times Q_{2}$. By Lemma 2.4, the possibilities for $Q$ and $T$ are in Table 2. Since $Q\cong A_{n}$ with $n\geqslant 10$, we obtain a contradiction. Therefore $S_{\alpha}=P$, as desired.

Applying the Orbit-Stabilizer Theorem, we see that

[TABLE]

So if $s_{i}\geqslant 2$ for some $i\in\underline{r}$, then $S\cong(A_{n})^{r}$, $S_{\alpha}\cong(A_{n-1})^{r}$ and $|\Omega|=n^{r}$, where $n\geqslant 10$. Hence in this case we obtain Theorem 1.3(3)(c).

Case 2: Suppose that $s_{i}=1$ for all $i\in\underline{r}$. Since $T_{i}$ is simple, we have that $T_{i}$ is a strip of $S$ for all $i\in\underline{r}$. Moreover, the supports of the $T_{i}$ are pairwise disjoint. In fact, if for some $l$ we have $T_{i}\pi_{l}\cong T\cong T_{j}\pi_{l}$ for distinct $i,j\in\underline{r}$, then $T^{2}\cong(T_{i}\times T_{j})\pi_{l}\leqslant H\pi_{l}\cong T$, which is absurd. Then the supports $\operatorname{supp}(T_{i})$ are pairwise disjoint and we can write

[TABLE]

First suppose that $l_{i}\geqslant 2$ for some $i\in\underline{r}$. Renumbering, if necessary, assume that $l_{1}\geqslant 2$. Write $l=l_{1}$ and consider the projection map $\overline{\pi}\colon S\rightarrow Q_{1}\times\cdots\times Q_{l}$. As $S=HS_{\alpha}$, it follows that $(H\overline{\pi})(S_{\alpha}\overline{\pi})=Q_{1}\times\cdots\times Q_{l}$. Write $L=S_{\alpha}\pi_{1}\times\cdots\times S_{\alpha}\pi_{l}$. Since $H\overline{\pi}=T_{1}$ and $S_{\alpha}\overline{\pi}\leqslant L$, $T_{1}L=Q_{1}\times\cdots\times Q_{l}$. Therefore, $T_{1}$ is a transitive subgroup of $Q_{1}\times\cdots\times Q_{l}$ under its faithful action by right multiplication on the right coset space $[Q_{1}\times\cdots\times Q_{l}\colon L]$. On the other, hand $Q_{1}\times\cdots\times Q_{l}$ can be embedded into the quasiprimitive permutation group $W=Q_{1}\operatorname{wr}S_{l}$ acting in product action on $[Q_{1}:S_{\alpha}\pi_{1}]^{l}$. Hence $T_{1}$ is a transitive simple subgroup of a wreath product in product action. According to [BPS04, Theorem 1.1] and [BPS06, Theorem 1.1(b)], $l=2$ and $T$ and $Q$ are as in one of the rows of Table 1. Thus, in this case, Theorem 1.3(3)(b) is valid.

Now suppose that $l_{i}=1$ for all $i\in\underline{r}$. Then $k=r$ and $T_{i}<Q_{i}$ for all $i\in\underline{r}$. In this case, Theorem 1.3(3)(a) holds.

(4) $G$ has type Cd or HC. Lemma 2.3 gives two sets

[TABLE]

where $l\geqslant 2$, each $D_{i}$ is a full diagonal subgroup of $S_{i}$ and $S_{i}=\prod_{Q_{j}\in\operatorname{supp}(D_{i})}Q_{j}$, such that $S=S_{1}\times\cdots\times S_{l}$ and $S_{\alpha}=D_{1}\times\cdots\times D_{l}\cong Q^{l}$.

We claim that $1<H\pi_{j}<Q_{j}$ for all $j\in\underline{r}$. Suppose that this is not the case and assume first that $H\pi_{i_{0}}=Q_{i_{0}}$ for some $i_{0}\in\underline{r}$. In this case $Q$ is a composition factor of $H$, which means by the Jordan-Hölder Theorem that $Q\cong T$, and so $k\leqslant r$. Therefore $H$ is the direct product of disjoint full strips in $S$. Since $H$ is transitive, $S=S_{\alpha}H$. Now Lemma 2.5 implies that $H$ must contain a direct factor of $S$. In our case, this is impossible.

Suppose now that $H\pi_{i_{0}}=1$ for some $i_{0}\in\underline{r}$. Let, for $i\in\underline{l}$, $\bar{\pi}_{i}$ denote the coordinate projection $\bar{\pi}_{i}:S\rightarrow S_{i}$. We may assume that $Q_{i_{0}}\in\operatorname{supp}(D_{1})$. Applying $\bar{\pi}_{1}$ to the factorization $S=S_{\alpha}H$, we obtain that $S_{1}=D_{1}(H\bar{\pi}_{1})$ and $H\bar{\pi}_{1}$ is contained in $X=\prod_{Q_{i}\in\operatorname{supp}(D_{1})\setminus\{Q_{i_{0}}\}}Q_{i}$. Now $S_{1}=D_{1}X$ is a factorization with $D_{1}\cap X=1$, which gives that $H\bar{\pi}_{1}=X$. Therefore $H\pi_{j}=Q_{j}$ must hold for some $j\neq i_{0}$. Hence by the analysis in the previous paragraph, this is also impossible. Therefore $1<H\pi_{j}<Q_{j}$ holds for all $j\in\underline{r}$, as claimed.

Considering the factorization

[TABLE]

we obtain, as in part (2), that $|\operatorname{supp}(D_{i})|=2$ for all $i\in\underline{l}$. That is, $r$ is even and $S_{i}\cong Q^{2}$, $H\pi_{i}\cong T$ where and $Q$ and $T$ are as in one of the rows of Table 2.

Renumbering, if necessary, assume that

[TABLE]

We claim that $H=H\pi_{1}\times\cdots\times H\pi_{r}$. It suffices to prove that $H\pi_{i}\leqslant H$ for all $i\in\underline{r}$. So assume the opposite, that is, $H\pi_{i}\not\leqslant H$ for some $i\in\underline{r}$. Then $H$ has a nontrivial strip $X\cong T$ such that $|\operatorname{supp}(X)|\geqslant 2$.

We claim that $\operatorname{supp}(D_{i})\nsubseteq\operatorname{supp}(X)$ for each $i\in\underline{l}$. To prove the claim, assume that $Q_{1},Q_{2}\in\operatorname{supp}(X)$ and consider the projection $\pi_{1}\colon S\rightarrow S_{1}$. As $S=S_{\alpha}H$,

[TABLE]

However, we obtain, analyzing the orders, that

[TABLE]

which is a contradiction. Thus, $\operatorname{supp}(D_{i})\nsubseteq\operatorname{supp}(X)$ for all $i\in\underline{l}$. Again, renumbering if necessary, we may assume that $Q_{2},Q_{3}\in\operatorname{supp}(X)$. Considering the projection $\overline{\pi}\colon S\rightarrow S_{1}\times S_{2}$, we have

[TABLE]

where $H\overline{\pi}$ is contained in a subgroup $\overline{H}$ of $S_{1}\times S_{2}$ that is isomorphic to $T^{3}$. Set $P=(D_{1}\times D_{2})\cap\overline{H}$. Then $|Q|^{4}=|Q|^{2}|T|^{3}/|P|$, thus

[TABLE]

Suppose that $Q\cong A_{6}$ and $T\cong A_{5}$. Then, by (14), $|P|=5/3$, which is impossible. If $Q\cong M_{12}$ and $T\cong M_{11}$, then the claim follows from Lemma 2.6.

Assume that $Q\cong\operatorname{P\Omega}_{8}^{+}(q)$ and $T\cong\Omega_{7}(q)$. Then by (14)

[TABLE]

We prove that there exists an odd prime that divides $q^{2}+1$ but does not divide $q^{3}\cdot(q^{6}-1)$. If $q$ is even, then $q^{2}+1$ is odd, and so there exists an odd prime $p$ that divides $q^{2}+1$. On the other hand, if $q$ is odd, then q^{2}+1\equiv 2\mbox{ (mod 4)}. Thus $q^{2}+1$ is even but it is not a 2-power, so there exists an odd prime $p$ that divides $q^{2}+1$. Therefore, in both cases, there exists an odd prime $p$ that divides $q^{2}+1$. We claim that $p$ does not divide $q^{3}\cdot(q^{6}-1)$. Since $p$ is odd, $p$ does not divide $q^{2}-1$. As $q^{6}-1=(q^{2}-1)(q^{4}+q^{2}+1)$, and $p$ does not divide $q^{2}-1$ but divides $q^{2}+1$, we find that $p$ does not divide $q^{6}-1$. Hence $p$ is a prime we are looking for. This means that also $|P|$ is not an integer, which is impossible.

Therefore, $H=H\pi_{1}\times\cdots\times H\pi_{r}$ as asserted.

Thus $1<H\pi_{i}<Q_{i}$ for all $i\in\underline{r}$, $k=r$ and we may assume after possibly reordering the $T_{i}$ that $T_{i}<Q_{i}$ for all $i$. The proof of Theorem 1.3 is now complete. ∎

Now it only remains to prove Corollaries 1.4 and 1.5.

The proof of Corollary 1.4.

In part (0) of Theorem 1.3, $H$ is not regular. In part (2), $H=T_{1}\times T_{2}\leqslant Q^{2}$ such that $T_{1}T_{2}=Q$. The stabilizer of $H$ is isomorphic to $T_{1}\cap T_{2}$ which, as can be verified in each of the lines of Table 2, is nontrivial. Hence $H$ is nonregular. Since, in case (4), the group $G$ is a blow-up of a group that appears in case (2), $H$ is nonregular in case (4) also. Hence $G$ and $H$ must be as either in Theorem 1.3(1) or in Theorem 1.3(3) and the O’Nan–Scott type of $G$ is either As or Pa. ∎

The proof of Corollary 1.5.

Let $H$, $G$ and $\Gamma$ be as in Corollary 1.5. Then $H$ is a regular nonabelian and characteristically simple subgroup of the quasiprimitive group $G$ acting on the vertex set of $\Gamma$. Noticing that if $G$ is as in Theorem 1.3(1), then $G$ is either an alternating or a symmetric group, and so $\Gamma$, in this case, would be a complete graph, the result follows from Corollary 1.4. ∎