# Closures of permutation groups with restricted nonabelian composition
factors

Ilia Ponomarenko  Steklov Institute of Mathematics at St. Petersburg, Russia
<EMAIL_ADDRESS>,  Saveliy V. Skresanov  Novosibirsk State University,
Novosibirsk 630090, Russia
Sobolev Institute of Mathematics, Novosibirsk 630090, Russia
<EMAIL_ADDRESS>and  Andrey V. Vasil’ev  Novosibirsk State University,
Novosibirsk 630090, Russia
Sobolev Institute of Mathematics, Novosibirsk 630090, Russia
<EMAIL_ADDRESS>
###### Abstract.

Given a permutation group $G$ on a finite set $\Omega$, let $G^{(k)}$ denote
the $k$-closure of $G$, that is, the largest permutation group on $\Omega$
having the same orbits in the induced action on $\Omega^{k}$ as $G$. Recall
that a group is $\operatorname{Alt}(d)$-free if it does not contain a section
isomorphic to the alternating group of degree $d$. Motivated by some problems
in computational group theory, we prove that the $k$-closure of an
$\operatorname{Alt}(d)$-free group is again $\operatorname{Alt}(d)$-free for
$k\geq 4$ and $d\geq 25$.

##  1\. Introduction

Throughout all groups and sets on which they act are assumed to be finite.

One of the motivations for the present paper is a computational problem in
which one needs to find efficiently the automorphism group
$\operatorname{Aut}(\mathfrak{S})$ of a given set $\mathfrak{S}$ of relations
(generally speaking of different arities). In the case when all the relations
of $\mathfrak{S}$ are binary, this problem is equivalent to the famous Graph
Isomorphism Problem and can be solved by the Babai algorithm [2] in
quasipolynomial time in the size of $\mathfrak{S}$. It is currently unknown
whether the automorphism group of a graph can be found in polynomial time. It
is surprising, however, that apparently the problem does not become easier if
a sufficiently large subgroup of $\operatorname{Aut}(\mathfrak{S})$ is known
in advance. In the present paper, we take a step towards the latter problem by
studying the structure of the group $\operatorname{Aut}(\mathfrak{S})$ in the
case when all the relations of $\mathfrak{S}$ are of arity at least $4$. To be
more precise, we recall briefly the method of invariant relations that was
introduced and developed by Wielandt in the late 1960s [32], and used in the
above-mentioned Babai’s paper in the framework of multidimensional coherent
configurations.

The method of invariant relations was considered by Wielandt as one of the
main tools for studying actions of a group on a set [33]. This method is based
on the concept of $k$-equivalence of permutation groups, where $k$ is a
natural number. Namely, two permutation groups $G$ and $H$ on a set $\Omega$
are said to be _$k$ -equivalent_ if for every $k$-ary relation
$s\subseteq\Omega^{k}$, we have $s^{G}=s$ if and only if $s^{H}=s$. In each
class of $k$-equivalent groups there is a unique maximal (with respect to
inclusion) group, which is called the _$k$ -closure_ of each group $G$
belonging to this class and is denoted by $G^{(k)}$. Wielandt proved [32] (see
also Proposition 2.2 below) that

(1) $\operatorname{Sym}(\Omega)\geq G^{(1)}\geq G^{(2)}\geq\cdots\geq
G^{(k)}=G^{(k+1)}=\cdots=G$

for some $k$. Thus, the closures of the group $G$ can be considered as
approximations to $G$. On the other hand, the concept of the closure is a
powerful tool for dealing with the above-mentioned computational problem,
because the $k$-closures are exactly the groups
$\operatorname{Aut}(\mathfrak{S})$ where $\mathfrak{S}$ is the set of
relations of arity at least $k$ left invariant by the action of $G$.

From the computational point of view, the $k$-closure problem consists in
finding the $k$-closure of a given permutation group (it is assumed that
permutation groups are given by generating sets, see [27]). Note that the case
$k=1$ is trivial, because the $1$-closure of any permutation group $G$ is the
direct product of symmetric groups acting on the orbits of $G$. In this
setting, polynomial-time algorithms for computing the $2$-closure were
constructed for the nilpotent groups [21], the groups of odd order [7], and
the supersolvable groups [22]. Very recently it was shown that the $3$-closure
of a solvable group can also be computed in polynomial time [23].

The above computational results are based on some “similarity” of
$k$-equivalent groups. It is clear from (1) that the larger the number $k$ is
the more “similar” the $k$-equivalent groups are. In [32], Wielandt proved
that if groups $G$ and $H$ are $k$-equivalent and $k\geq 1$ (respectively,
$k\geq 2$, $k\geq m$), then $G$ is transitive (respectively, imprimitive,
$m$-transitive) if and only if so is $H$. More interestingly, however, that
$k$-equivalent groups are “similar” not only as permutation groups, but also
as abstract groups. For example, if $G$ is an abelian group (respectively, a
$p$-group, a group of odd order), then, for $k\geq 2$, the group $H$ is an
abelian group (respectively, a $p$-group, a group of odd order), see [32].
Recently, it was proved [19] that a similar statement is true for solvable
groups if $k\geq 3$ (the example of $2$-transitive solvable groups shows that
$k=2$ cannot be taken here). Each of these statements expresses the fact that
the $k$-closure (for the corresponding $k$) of any permutation group of the
corresponding class also lies in it. This fact, a polynomial upper bound on
the orders of solvable primitive groups [20], and the Babai–Luks algorithm [4]
form a foundation of the computational results from the previous paragraph
(except for the case of supersolvable groups, where the algorithm is
constructed on more subtle arguments, because the $2$-closure of a
supersolvable group need not be even solvable).

In fact, the Babai–Luks algorithm [4] can be applied in a much broader
situation, for the class of $\operatorname{Alt}(d)$-free groups. Recall that
$H$ is a section of a group $G$ if $H$ is a homomorphic image of a subgroup of
$G$. Now a group is $\operatorname{Alt}(d)$-free, $d\geq 5$, if it does not
contain a section isomorphic to the alternating group of degree $d$. In
particular, solvable groups are $\operatorname{Alt}(5)$-free. Furthermore, the
classical result by Babai, Cameron, and Pálfy [3] states, up to the language
(see, e.g., a remark before Theorem III in [26]), that the size of an
$\operatorname{Alt}(d)$-free primitive permutation group of degree $n$ is
bounded from above by $n^{c}$, where $c$ is a constant depending only on $d$.
Thus, in the context of the previous paragraph, we arrive at the question when
the $k$-closure of an $\operatorname{Alt}(d)$-free group is also
$\operatorname{Alt}(d)$-free. The goal of the present paper is to answer this
question even for a wider class of groups.

To proceed, we recall another concept introduced by Wielandt [31, Definition
11.3]. A class $\operatorname{\mathfrak{X}}$ of (abstract) groups is said to
be complete if it is closed with respect to taking subgroups, quotients, and
extensions. The examples are the class of all groups, $p$-groups for any fixed
prime $p$, and solvable groups. In fact, given a fixed set $S$ (finite or
infinite) of simple groups, the class of groups whose composition factors are
sections of groups from $S$ is complete.

We are ready to state our main result.

######  Theorem 1.1.

Let $\operatorname{\mathfrak{X}}$ be a complete class including all
$\operatorname{Alt}(25)$-free groups. Then the $k$-closure of every
permutation group from $\operatorname{\mathfrak{X}}$ belongs to
$\operatorname{\mathfrak{X}}$ for each $k\geq 4$.

Since the class of all $\operatorname{Alt}(d)$-free groups is complete, the
statement below is an immediate consequence of Theorem 1.1.

######  Corollary 1.2.

If $G$ is an $\operatorname{Alt}(d)$-free group with $d\geq 25$, then
$G^{(k)}$ is $\operatorname{Alt}(d)$-free group for $k\geq 4$.

Note that the constant $4$ in Theorem 1.1 is the best possible and, if $k=4$,
then the same holds true for the constant $25$. This follows from the two
examples below.

(i) The affine group $G=\mathrm{AGL}_{m}(2)$ is $3$-transitive in its natural
action on a linear space of dimension $m\geq 2$ over the field of order $2$.
It follows that $G^{(3)}=\operatorname{Sym}(2^{m})$, so the theorem does not
hold for $k\leq 3$.

(ii) The Mathieu group $G=M_{24}$ is $\operatorname{Alt}(9)$-free and acts on
24 points 5-transitively, so $G^{(4)}=\operatorname{Sym}(24)$ is not
$\operatorname{Alt}(24)$-free.

It should be also noted that there are a lot of complete classes of groups
distinct from the class of $\operatorname{Alt}(d)$-free groups, where $d\geq
25$, that satisfy the hypothesis of Theorem 1.1. Indeed, consider any
nonabelian simple group $S$ containing a section isomorphic to
$\operatorname{Alt}(d)$. Then the class of all groups whose nonabelian factors
are either $\operatorname{Alt}(25)$-free or sections of $S$ is such an
example.

Let us now briefly discuss the strategy of the proof of the main result.
Suppose that $\operatorname{\mathfrak{X}}$ is a complete class of groups
satisfying the hypothesis of Theorem 1.1 and ${k\geq 4}$. We are interested in
the question of when $G^{(k)}\in\operatorname{\mathfrak{X}}$ for any
permutation group $G\in\operatorname{\mathfrak{X}}$. Note that along with each
pair of permutation groups, the class $\operatorname{\mathfrak{X}}$ contains
their transitive and intransitive direct products, as well as their
imprimitive and primitive wreath products. Explicit formulas for the
$k$-closures of these products are known, see [11, 22], hence our question is
reduced to the case when the group $G$ is primitive and even basic [23].
Recall that a permutation group is _nonbasic_ if it is contained in a wreath
product in the product action; it is _basic_ otherwise, see [5, Section 4.3].

Apparently, the first nontrivial results on the closures of primitive groups
were obtained in [15, 24], where it was proved that in most cases the socle of
the group is preserved when taking the $k$-closure, and all the cases were
described when this is not so. These results allow us to reduce the situation
to the case of the affine groups (see Section 4). When working with affine
groups, it is natural to use Aschbacher’s classification [1], which divides
them structurally into $9$ classes. An important step in the study of
$k$-closures of affine groups was made in [29, 30], where for each Aschbacher
class an explicit value for the number $k$ was found such that the $k$-closure
of each group from this class remains in the same class. Following the
Aschbacher classification, we consider the classes to which group $G$ might
belong. While excluding the Aschbacher classes related with tensor
decompositions of the underlying linear space, we need some new tools to
control the $k$-equivalence and $k$-closure for stabilizers of the tensor
decompositions, which are the subject of Section 3. Finally (Subsections 4.1
and 4.2), we exclude the cases when the group $G$ is of symplectic type or
belongs to the last Aschbacher class (the classes
$\operatorname{\mathcal{C}}_{6}$ and $\operatorname{\mathcal{C}}_{9}$ in the
notation in [30]), thus completing the proof of the theorem.

To make the paper as self-contained as possible, we cite the relevant results
concerning the $k$-equivalence and $k$-closure of permutation groups in
Section 2. The concluding remarks are collected at the end of the paper
(Section 5).

##  2\. Preliminaries

Let $G$ be a permutation group on $\Omega$, and let
$\operatorname{\mathit{k}-rel}(G)$, $k\geq 1$, denote the set of $G$-invariant
$k$-ary relations.

######  Lemma 2.1.

[32, Theorem 4.7] Let $G$ and $H$ be permutation groups on the same set
$\Omega$, and $k\geq 1$. Then
$\operatorname{\mathit{k}-rel}(H)\subseteq\operatorname{\mathit{k}-rel}(G)$ if
and only if for any $x_{1},\dots,x_{k}\in\Omega$ and any $g\in G$ there exists
some $h\in H$ such that $x_{i}^{g}=x_{i}^{h}$ for all $i=1,\dots,k$.

We say that $G$ is _$k$ -equivalent_ to $H$ if
$\operatorname{\mathit{k}-rel}(G)=\operatorname{\mathit{k}-rel}(H)$. Observe
that the $k$-closure of $G$ is the largest permutation group on $\Omega$ which
is $k$-equivalent to $G$. Recall that a _base_ of a permutation group is a
subset whose pointwise stabilizer is trivial. We collect some basic properties
of $k$-closures and $k$-equivalence which will be used all throughout the
text.

######  Proposition 2.2.

Let $G$ and $H$ be permutation groups on $\Omega$.

  1. (i)

For $k\geq 1$, if $G$ and $H$ are $(k+1)$-equivalent, then they are
$k$-equivalent. In particular, $G^{(k+1)}\leq G^{(k)}$.

  2. (ii)

For $k\geq 2$, if $G$ and $H$ are $k$-equivalent, then their orbits and
systems of imprimitivity are the same.

  3. (iii)

For $k\geq 2$ and $\alpha\in\Omega$, if $G$ and $H$ are $k$-equivalent, then
$G_{\alpha}$ is $(k-1)$-equivalent to $H_{\alpha}$.

  4. (iv)

For $k\geq 1$, if $G\leq H$, then $G^{(k)}\leq H^{(k)}$.

  5. (v)

For $k\geq 1$, we have $(G^{(k)})^{(k)}=G^{(k)}$.

  6. (vi)

Suppose $G$ has a base of size $k\geq 1$, then $G^{(k+1)}=G$. In particular,
if $G$ has a faithful regular orbit, then $G^{(2)}=G$.

###### Proof.

Parts (i)-(vi) follow from Theorems 4.3, 4.11, 4.12, 5.7, 5.9, and 5.12 of
[32], respectively. ∎

For $\Delta\subseteq\Omega$ let $G^{\Delta}$ denote the group induced on
$\Delta$ by the setwise stabilizer of $\Delta$ in $G$. We may view
$G^{\Delta}$ as a subgroup of $\operatorname{Sym}(\Delta)$.

Let $\Delta$ be some block of imprimitivity for a transitive permutation group
$G$. Translates of $\Delta$ give a system of imprimitivity of $G$ which we
will denote by $\Omega/\Delta$. Let $G^{\Omega/\Delta}$ denote the permutation
group induced by $G$ on $\Omega/\Delta$ by permuting the blocks. Clearly,
$G^{\Delta}$ and $G^{\Omega/\Delta}$ are sections of $G$, so if $G$ lies in
some complete class $\operatorname{\mathfrak{X}}$ then
$G^{\Delta}\in\operatorname{\mathfrak{X}}$ and
$G^{\Omega/\Delta}\in\operatorname{\mathfrak{X}}$. Moreover, $G$ can be
embedded into the _wreath product in imprimitive action_ $G^{\Delta}\wr
G^{\Omega/\Delta}$.

Given two permutation groups $L\leq\operatorname{Sym}(\Delta)$ and
$K\leq\operatorname{Sym}(\Gamma)$, let $L\uparrow
K\leq\operatorname{Sym}(\Delta^{\Gamma})$ denote the _wreath product in
product action_. As an abstract group it is isomorphic to $L^{|\Gamma|}\rtimes
K$, where $L^{|\Gamma|}$ acts on $\Delta^{\Gamma}$ coordinatewise, while $K$
permutes the coordinates. A primitive group $G$ is called _basic_ if it cannot
be embedded into the wreath product in product action for $|\Delta|>1$,
$|\Gamma|>1$.

Let $G\leq\operatorname{Sym}(\Omega)$ be a primitive permutation group
preserving a nontrivial product decomposition $\Omega=\Delta^{\Gamma}$.
Without loss of generality we may assume that $\Gamma=\\{1,\dots,m\\}$ for
some $m>1$. For $\gamma\in\Gamma$ define a partition of $\Omega$ into
$|\Delta|$ parts

$\Delta_{\gamma}=\\{\Delta\times\dots\times\Delta\times\\{x\\}\times\Delta\times\dots\times\Delta\mid
x\in\Delta\\},$

i.e. each part of $\Delta_{\gamma}$ consists of those tuples from
$\Delta^{\Gamma}$ which have its $\gamma$-th coordinate constant. There is a
natural bijective correspondence between the partitions $\Delta_{\gamma}$,
$\gamma\in\Gamma$, and the points of $\Gamma$. The group $G$ permutes the
partitions $\Delta_{\gamma}$, $\gamma\in\Gamma$, so let $G^{\Gamma}$ denote
the permutation group on $\Gamma$ obtained through this correspondence. With
some abuse of notation, let $G^{\Delta_{\gamma}}$ denote the permutation group
on $\Delta$ induced by the stabilizer of $\Delta_{\gamma}$ in $G$. In the
terminology of [25, Section 5.3], the group $G^{\Delta_{\gamma}}$ is the
_$\gamma$ -component_ of $G$, and there is an explicit formula for it [25,
(5.11)].

Clearly, $G^{\Gamma}$ and $G^{\Delta_{\gamma}}$ are sections of $G$, and [25,
Theorem 5.14 (ii)] implies that $G$ can be embedded into the wreath product
$G^{\Delta_{\gamma}}\uparrow G^{\Gamma}$ in product action.

To compute the $k$-closure of a group in product action we need another
closure operator. Observe that a permutation group
$K\leq\operatorname{Sym}(\Gamma)$ acts on the set of ordered partitions of
$\Gamma$ in at most $r\geq 1$ classes. Let $K^{[r]}$ denote the largest
permutation group on $\Gamma$ having the same orbits on the set of ordered
partitions into at most $r$ parts as $K$. This new operator, which we call the
_$r$ -closure with respect to partitions_, is indeed a closure operator and
enjoys some properties similar to the $k$-closure, see [28, Section 3]. In
particular, $K^{[r+1]}\geq K^{[r]}$ and $K^{[m]}=K$ for $m=|\Gamma|$.

There is a direct connection between these two types of closures.

######  Proposition 2.3 ([28, Lemma 3.2]).

Let $K\leq\operatorname{Sym}(\Gamma)$ be a permutation group. For $k\geq 1$ we
have $K^{[k+1]}\leq K^{(k)}$.

Let $\mathrm{Orb}_{k}(K)$ denote the set of orbits of
$K\leq\operatorname{Sym}(\Gamma)$ on $\Gamma^{k}$.

######  Proposition 2.4.

Let $G$ and $H$ be permutation groups on $\Omega$.

  1. (i)

Let $\Delta\subset\Omega$ be an orbit of $G$. For any $k\geq 1$ we have
$(G^{(k)})^{\Delta}\leq(G^{\Delta})^{(k)}$. Moreover, if $H$ is $k$-equivalent
to $G$ then $G^{\Delta}$ is $k$-equivalent to $H^{\Delta}$.

  2. (ii)

Let $G$ be a transitive imprimitive group with a nontrivial block
$\Delta\subset\Omega$. For any $k\geq 2$,

$(G^{\Delta}\wr
G^{\Omega/\Delta})^{(k)}=(G^{\Delta})^{(k)}\wr(G^{\Omega/\Delta})^{(k)}.$

Moreover, if $H$ is $k$-equivalent to $G$, then $H$ is also imprimitive, and
$G^{\Delta}$ is $k$-equivalent to $H^{\Delta}$ and $G^{\Omega/\Delta}$ is
$k$-equivalent to $H^{\Omega/\Delta}$.

  3. (iii)

Let $G$ be a primitive nonbasic group preserving a nontrivial product
decomposition $\Omega=\Delta^{\Gamma}$, and let $\gamma\in\Gamma$. For any
$k\geq 2$ we have

$(G^{\Delta_{\gamma}}\uparrow
G^{\Gamma})^{(k)}=(G^{\Delta_{\gamma}})^{(k)}\uparrow(G^{\Gamma})^{[r]},$

where $r=\min\\{|\mathrm{Orb}_{k}(G^{\Delta_{\gamma}})|,\,|\Gamma|\\}$.
Moreover, if $H$ is $k$-equivalent to $G$, then $H$ is also nonbasic, and
$G^{\Delta_{\gamma}}$ is $k$-equivalent to $H^{\Delta_{\gamma}}$ and
$(G^{\Gamma})^{[r]}=(H^{\Gamma})^{[r]}$.

###### Proof.

(i) Since
$\operatorname{\mathit{k}-rel}(G^{\Delta})=\operatorname{\mathit{k}-rel}(G)\cap\Delta^{k}$,
groups $G^{\Delta}$ and $H^{\Delta}$ are $k$-equivalent and the second part of
the claim follows. Now take $H=G^{(k)}$ and observe that $(G^{(k)})^{\Delta}$
is $k$-equivalent to $G^{\Delta}$, hence
$(G^{(k)})^{\Delta}\leq(G^{\Delta})^{(k)}$.

(ii) Follows from [11, Theorem 2.4 and Lemma 2.5].

(iii) The closure formula is the main result of [28]. It follows from the
closure formula that
$H\leq(G^{\Delta_{\gamma}})^{(k)}\uparrow(G^{\Gamma})^{[r]}$ so $H$ is
nonbasic. It also follows that
$H^{\Delta_{\gamma}}\leq(G^{\Delta_{\gamma}})^{(k)}$ and
$H^{\Gamma}\leq(G^{\Gamma})^{[r]}$. Therefore
$(H^{\Delta_{\gamma}})^{(k)}\leq(G^{\Delta_{\gamma}})^{(k)}$ and
$(H^{\Gamma})^{[r]}\leq(G^{\Gamma})^{[r]}$. By exchanging the roles of $H$ and
$G$ we derive $(G^{\Delta_{\gamma}})^{(k)}\leq(H^{\Delta_{\gamma}})^{(k)}$ and
$(G^{\Gamma})^{[r]}\leq(H^{\Gamma})^{[r]}$. Hence
$(G^{\Delta_{\gamma}})^{(k)}=(H^{\Delta_{\gamma}})^{(k)}$ which implies that
$G^{\Delta_{\gamma}}$ is $k$-equivalent to $H^{\Delta_{\gamma}}$, and
$(G^{\Gamma})^{[r]}=(H^{\Gamma})^{[r]}$ as claimed. ∎

Note that in Proposition 2.4 (iii) when $k\geq 3$ we have
$|\mathrm{Orb}_{k}(G^{\Delta_{\gamma}})|\geq k+1$ since $\Delta\geq 2$, so
Proposition 2.3 implies that $(G^{\Delta_{\gamma}}\uparrow
G^{\Gamma})^{(k)}\leq(G^{\Delta_{\gamma}})^{(k)}\uparrow(G^{\Gamma})^{(k)}$,
cf. [28, Theorem 1.2].

Recall that a permutation group $G$ on $\Omega$ is called _affine_ if it
contains a normal elementary abelian subgroup $V$ acting regularly on
$\Omega$. In that case $\Omega$ can be identified with $V$ and $G$ decomposes
into a semidirect product $G=V\rtimes G_{0}$, where $G_{0}$ is the zero
stabilizer. The group $G_{0}$ acting on $V$ can be viewed as a subgroup of
$\operatorname{GL}(V)$, and $G$ is primitive if and only if $G_{0}$ acts
irreducibly on $V$. In this case, $V=\operatorname{Soc}(G)$.

######  Lemma 2.5.

Let $V$ be a vector space over a prime field, and let $G=V\rtimes G_{0}$,
$G_{0}\leq\operatorname{GL}(V)$, be an affine group with socle $V$. Suppose
that $G^{(k)}$, $k\geq 2$, is also an affine group with socle $V$, i.e.
$G^{(k)}=V\rtimes H_{0}$ for some $H_{0}\leq\operatorname{GL}(V)$. Then
$H_{0}=G_{0}^{(k-1)}\cap\operatorname{GL}(V)$.

###### Proof.

By Proposition 2.2 (iii), $G_{0}$ is $(k-1)$-equivalent to $H_{0}$, hence
$H_{0}\leq G_{0}^{(k-1)}\cap\operatorname{GL}(V)$. To prove the converse
inclusion, let $h\in G_{0}^{(k-1)}\cap\operatorname{GL}(V)$ be arbitrary. Let
$v_{1},\dots,v_{k}\in V$ be arbitrary vectors. Since $h\in G_{0}^{(k-1)}$, by
Lemma 2.1 there exists $g\in G_{0}$ such that
$(v_{i}-v_{1})^{g}=(v_{i}-v_{1})^{h}$ for $i=2,\dots,k$. As elements $h$ and
$g$ lie in $\operatorname{GL}(V)$, we have
$v_{i}^{h}=v_{i}^{g}+v_{1}^{h}-v_{1}^{g}$ for all $i=2,\dots,k$. There exists
some element $t\in V\leq G$ acting on $V$ by translation by the vector
$v_{1}^{h}-v_{1}^{g}$, that is, $v^{t}=v+v_{1}^{h}-v_{1}^{g}$ for all $v\in
V$. Therefore $v_{i}^{h}=v_{i}^{gt}$ for all $i=2,\dots,k$. Notice that this
equality holds for $i=1$ as well, indeed,
$v_{1}^{gt}=v_{1}^{g}+v_{1}^{h}-v_{1}^{g}=v_{1}^{h}$. Finally, $gt$ lies in
$G$, so $h$ lies in $G^{(k)}$ by Lemma 2.1, as the choice of
$v_{1},\dots,v_{k}\in V$ was arbitrary. Hence
$G_{0}^{(k-1)}\cap\operatorname{GL}(V)\leq G^{(k)}$ and since
$G_{0}^{(k-1)}\cap\operatorname{GL}(V)$ stabilizes the zero vector, we derive
$G_{0}^{(k-1)}\cap\operatorname{GL}(V)\leq H_{0}$. ∎

The $k$-closure of a primitive affine group is also an affine group with the
same socle for $k\geq 4$ according to the following result.

######  Lemma 2.6.

[24, Lemma 4.1] Let $G$ be a primitive affine permutation group with socle
$V$, and assume $k\geq 4$. Then $G^{(k)}$ is also an affine group with socle
$V$.

The following proposition implies that the groups of Lie type of bounded rank
are $\operatorname{Alt}(d)$-free for $d$ bounded in terms of the rank. A less
precise inequality but with a more elementary proof can also be found in [6,
Theorem 5.7A].

######  Proposition 2.7.

Suppose that $\operatorname{Alt}(d)$ is a section of
$\operatorname{GL}_{a}(q)$ for some $a\geq 1$, $q\geq 2$. Then $a\geq d-2$ for
$d\geq 9$.

###### Proof.

Assume that $a$ is the minimal integer such that
$\operatorname{GL}_{a}(\operatorname{\mathbb{F}})$ contains a section
isomorphic to $\operatorname{Alt}(d)$, $d\geq 9$, for some field
$\operatorname{\mathbb{F}}$ of positive characteristic. By [6, Lemma 5.7D],
for some algebraically closed field $\mathbb{K}$ of characteristic $p>0$ there
exists a subgroup $G$ of $\operatorname{GL}_{a}(\mathbb{K})$ such that
$G/Z(G)\simeq\operatorname{Alt}(d)$, and $Z(G)$ is a group of scalars. In
particular, $\operatorname{Alt}(d)$ has a faithful projective $p$-modular
representation of degree $a$. By [12, Proposition 5.3.7 (i)] we have $a\geq
d-2$. ∎

When dealing with $k$-closures of affine permutation groups, we will use
Aschbacher’s classification of subgroups of linear groups [1]. Xu, Giudici, Li
and Praeger [30] proved that Aschbacher’s classes in
$\operatorname{{\rm\Gamma}L}_{a}(q)$ are preserved by $k$-closures for
suitably chosen $k$, depending on the class; here we view a subgroup of
$\operatorname{{\rm\Gamma}L}_{a}(q)$ as a permutation group on
$\operatorname{\mathbb{F}}_{q}^{a}$.

We will follow [30] in our definitions and notations of Aschbacher classes.
Recall that a subgroup of $\operatorname{{\rm\Gamma}L}_{a}(q)$ not containing
$\mathrm{SL}_{a}(q)$ lies in one of the nine classes
$\operatorname{\mathcal{C}}_{1},\dots,\operatorname{\mathcal{C}}_{9}$, defined
mostly in terms of some geometric structure preserved by the groups lying in
the class. We briefly outline the classes and relevant structures:

$\operatorname{\mathcal{C}}_{1}$: Groups preserving a nontrivial proper
subspace.

$\operatorname{\mathcal{C}}_{2}$: Groups acting imprimitively on the vector
space.

$\operatorname{\mathcal{C}}_{3}$: Groups preserving the structure of an
extension field.

$\operatorname{\mathcal{C}}_{4}$: Groups preserving a nontrivial decomposition
of the vector space into the tensor product of two spaces of unequal
dimensions.

$\operatorname{\mathcal{C}}_{5}$: Groups preserving the structure of a proper
subfield.

$\operatorname{\mathcal{C}}_{6}$: Groups normalizing a subgroup of symplectic
type.

$\operatorname{\mathcal{C}}_{7}$: Groups preserving a nontrivial decomposition
of the vector space into the tensor product of several spaces of equal
dimensions.

$\operatorname{\mathcal{C}}_{8}$: Groups preserving a nondegenerate
alternating, hermitian or quadratic form. We will use notation
$\operatorname{\mathcal{C}}_{\mathrm{Sp}}$,
$\operatorname{\mathcal{C}}_{\mathrm{U}}$ and
$\operatorname{\mathcal{C}}_{\mathrm{O}}$, respectively, depending on the form
preserved.

$\operatorname{\mathcal{C}}_{9}$: Groups which are not contained in any of
$\operatorname{\mathcal{C}}_{1},\dots,\operatorname{\mathcal{C}}_{8}$. These
groups are almost simple modulo center.

Strictly speaking, each class $\operatorname{\mathcal{C}}_{i}$, $i=1,\dots,8$,
is defined by its maximal members. For example,
$G\leq\operatorname{{\rm\Gamma}L}_{a}(q)$ lies in
$\operatorname{\mathcal{C}}_{4}$ if $V=\operatorname{\mathbb{F}}_{q}^{a}$ can
be decomposed as $V=U\otimes W$, $\dim U>\dim W>1$, and $G$ can be conjugated
inside the central product
$(\operatorname{GL}(U)\otimes\operatorname{GL}(W))\rtimes\operatorname{Aut}(\operatorname{\mathbb{F}}_{q})$
acting naturally on $V$. By [30, Proposition 4.4.1], this is equivalent to
saying that $G$ preserves the set of simple tensors, i.e. a unary relation of
the form $\\{u\otimes w\mid u\in U,\,w\in W\\}$. This implies that
$G^{(1)}\cap\operatorname{{\rm\Gamma}L}_{a}(q)$ also lies in the class
$\operatorname{\mathcal{C}}_{4}$. Essentially [30, Theorem 1.1] says that one
can find similar relations for all of Aschbacher classes. Here we give a
summary of the main results of [30].

######  Proposition 2.8.

Let $G\leq\operatorname{{\rm\Gamma}L}_{a}(q)$, $a\geq 2$, act on
$\operatorname{\mathbb{F}}_{q}^{a}$ naturally.

  1. (i)

If $G\in\operatorname{\mathcal{C}}_{i}$ for
$i\in\\{1,\dots,7,\mathrm{Sp},\mathrm{U},\mathrm{O}\\}$, then
$G^{(k)}\cap\operatorname{{\rm\Gamma}L}_{a}(q)\in\operatorname{\mathcal{C}}_{i}$
for $k=2$ if $i\in\\{3,6,\mathrm{Sp}\\}$ and $k=1$ otherwise.

  2. (ii)

If $G\in\operatorname{\mathcal{C}}_{9}$, then
$G^{(2)}\cap\operatorname{{\rm\Gamma}L}_{a}(q)\in\operatorname{\mathcal{C}}_{9}$,
unless $a=4$, $q=2$ and $G=\operatorname{Alt}(7)$.

###### Proof.

Part (i) follows from [30, Theorem 1.1], and part (ii) from [30, Proposition
3.3.1]. ∎

Note that Aschbacher’s theorem [1] applies not only to
$\operatorname{{\rm\Gamma}L}_{a}(q)$ but to the automorphism groups of the
classical groups in general. Classes
$\operatorname{\mathcal{C}}_{1},\dots,\operatorname{\mathcal{C}}_{9}$ are
defined similarly for the classical groups, and if a group $G$ belongs to some
$\operatorname{\mathcal{C}}_{i}$, $i=1,\dots,8$, defined for a classical
group, then $G$ lies in the same class but defined for
$\operatorname{{\rm\Gamma}L}_{a}(q)$, if one forgets about the classical
geometry preserved by $G$.

##  3\. Closures of tensor products

In this section $\operatorname{\mathfrak{X}}$ is an arbitrary complete class
which contains all solvable groups. In our descriptions of subgroups
stabilizing tensor decompositions we follow [30, Sections 4.4 and 4.5].

###  3.1. Stabilizers of $X\otimes Y$

Let $V=X\otimes Y$ be a tensor product of vector spaces over a finite field
$\operatorname{\mathbb{F}}$. We will also use the symbol $\otimes$ to denote
the central product of two linear groups acting naturally on the corresponding
tensor product of vector spaces. Recall that the central product
$\operatorname{GL}(X)\otimes\operatorname{GL}(Y)$ acts on $V$ by the following
rule, if $g_{X}\in\operatorname{GL}(X)$, $g_{Y}\in\operatorname{GL}(Y)$ and
$x\in X$, $y\in Y$, then $(x\otimes y)^{g_{X}\otimes g_{Y}}=x^{g_{X}}\otimes
y^{g_{Y}}$.

The group of field automorphisms
$\operatorname{Aut}(\operatorname{\mathbb{F}})$ acts on $V$ as follows. First,
choose an $\operatorname{\mathbb{F}}$-basis $x_{1},\dots,x_{r}$ of $X$. Every
$x\in X$ can be decomposed uniquely into a sum
$x=\alpha_{1}x_{1}+\dots+\alpha_{r}x_{r}$, and one can define the action of
$\sigma\in\operatorname{Aut}(\operatorname{\mathbb{F}})$ on $X$ by
$x^{\sigma}=\alpha_{1}^{\sigma}x_{1}+\dots+\alpha_{r}^{\sigma}x_{r}$. By
choosing some basis of $Y$ one defines the action of
$\operatorname{Aut}(\operatorname{\mathbb{F}})$ in the similar way. Finally,
the action of $\sigma\in\operatorname{Aut}(\operatorname{\mathbb{F}})$ on
$V=X\otimes Y$ is defined by $(x\otimes y)^{\sigma}=x^{\sigma}\otimes
y^{\sigma}$, where $x\in X$, $y\in Y$.

It follows that the group
$(\operatorname{GL}(X)\otimes\operatorname{GL}(Y))\rtimes\operatorname{Aut}(\operatorname{\mathbb{F}})$
preserves the tensor decomposition $V=X\otimes Y$, and when $\dim X\neq\dim Y$
it is the largest group preserving that decomposition, see [30, Formula
(4.4.1)]. Every element $g$ of that group can be written in the form
$g=(g_{X}\otimes g_{Y})\sigma$, where $g_{X}\in\operatorname{GL}(X)$,
$g_{Y}\in\operatorname{GL}(Y)$ and
$\sigma\in\operatorname{Aut}(\operatorname{\mathbb{F}})$. If $g$ admits a
second decomposition $g=(g_{X}^{\prime}\otimes
g_{Y}^{\prime})\sigma^{\prime}$, then $\sigma^{\prime}=\sigma$ and there
exists some nonzero $\alpha\in\operatorname{\mathbb{F}}$ such that
$g_{X}^{\prime}=\alpha\cdot g_{X}$ and $g_{Y}^{\prime}=\frac{1}{\alpha}\cdot
g_{Y}$.

Take
$G\leq(\operatorname{GL}(X)\otimes\operatorname{GL}(Y))\rtimes\operatorname{Aut}(\operatorname{\mathbb{F}})$
where $X$ and $Y$ are vector spaces over $\operatorname{\mathbb{F}}$. Recall
that we can identify $\operatorname{\mathbb{F}}^{\times}$ with the center of
$\operatorname{GL}(X)$ and $\operatorname{GL}(Y)$. For $g=(g_{X}\otimes
g_{Y})\sigma\in G$ define the cosets
$\pi_{X}(g)=\operatorname{\mathbb{F}}^{\times}\cdot g_{X}\sigma$ and
$\pi_{Y}(g)=\operatorname{\mathbb{F}}^{\times}\cdot g_{Y}\sigma$. Note that
these elements are defined correctly. Indeed, if $g=(g_{X}^{\prime}\otimes
g_{Y}^{\prime})\sigma^{\prime}$, then $\sigma^{\prime}=\sigma$ and
$g_{X}^{\prime}=\alpha\cdot g_{X}$ for some
$\alpha\in\operatorname{\mathbb{F}}^{\times}$. Hence
$\operatorname{\mathbb{F}}^{\times}\cdot
g_{X}^{\prime}\sigma^{\prime}=\operatorname{\mathbb{F}}^{\times}\cdot\alpha
g_{X}\sigma=\operatorname{\mathbb{F}}^{\times}\cdot g_{X}\sigma$, so
$\pi_{X}(g)$ does not depend on the choice of the decomposition of $g$.
Similar reasoning holds for $\pi_{Y}$, so $\pi_{X}:G\to\operatorname{P\Gamma
L}(X)$ and $\pi_{Y}:G\to\operatorname{P\Gamma L}(Y)$ are correctly defined
maps, and moreover, $\pi_{X}$ and $\pi_{Y}$ are homomorphisms.

Assume that $\operatorname{\mathbb{F}}^{\times}=Z(\operatorname{GL}(V))\leq
G$, so $\ker\pi_{X}\cap\ker\pi_{Y}=\operatorname{\mathbb{F}}^{\times}$. Since
$\pi_{X}(G)\leq\operatorname{P\Gamma L}(X)$, we can consider the full preimage
$G_{X}$ of $\pi_{X}(G)$ in $\operatorname{GL}(X)$; define $G_{Y}$ accordingly.
By construction, $G_{X}$ and $G_{Y}$ contain
$\operatorname{\mathbb{F}}^{\times}$. Now, by Remak’s theorem,
$G/\operatorname{\mathbb{F}}^{\times}$ embeds into a direct product
$\pi_{X}(G)\times\pi_{Y}(G)\leq\operatorname{P\Gamma
L}(X)\times\operatorname{P\Gamma L}(Y)$, so
$G/\operatorname{\mathbb{F}}^{\times}$ embeds into
$G_{X}/\operatorname{\mathbb{F}}^{\times}\times
G_{Y}/\operatorname{\mathbb{F}}^{\times}$.

######  Proposition 3.1.

Let $X$ and $Y$ be vector spaces over a finite field
$\operatorname{\mathbb{F}}$, and let $V=X\otimes Y$. Let $G$ and $H$ be
subgroups of
$(\operatorname{GL}(X)\otimes\operatorname{GL}(Y))\rtimes\operatorname{Aut}(\operatorname{\mathbb{F}})\leq\operatorname{{\rm\Gamma}L}(V)$.
Assume that $\operatorname{\mathbb{F}}^{\times}\leq G$ and
$\operatorname{\mathbb{F}}^{\times}\leq H$, and $G$ and $H$ are $k$-equivalent
for some $k\geq 1$. Then $G_{X}$ is $k$-equivalent to $H_{X}$, and $G_{Y}$ is
$k$-equivalent to $H_{Y}$.

###### Proof.

We will show that
$\operatorname{\mathit{k}-rel}(H_{X})\subseteq\operatorname{\mathit{k}-rel}(G_{X})$.
By Lemma 2.1, it is sufficient to show that for all $x_{1},\dots,x_{k}\in X$
and $\gamma_{X}\in G_{X}$ there exists $\beta_{X}\in H_{X}$ such that
$x_{i}^{\gamma_{X}}=x_{i}^{\beta_{X}}$, $i=1,\dots,k$. Notice that we may
assume that the vector $x_{i}$ is nonzero, $i=1,\dots,k$.

Let $y\in Y$ be an arbitrary nonzero vector. Since $\gamma_{X}$ lies in
$G_{X}$, there exists $g\in G$ such that $g=(g_{X}\otimes g_{Y})\sigma$, where
$g_{X}\in\operatorname{GL}(X)$, $g_{Y}\in\operatorname{GL}(Y)$,
$\sigma\in\operatorname{Aut}(\operatorname{\mathbb{F}})$ and
$\gamma_{X}=g_{X}\sigma$. Let $\gamma_{Y}=g_{Y}\sigma$. As $G$ is
$k$-equivalent to $H$, there exists an element $h\in H$ such that
$(x_{i}\otimes y)^{g}=(x_{i}\otimes y)^{h}$, $i=1,\dots,k$.

Write $h=(h_{X}\otimes h_{Y})\sigma^{\prime}$, where
$h_{X}\in\operatorname{GL}(X)$, $h_{Y}\in\operatorname{GL}(Y)$,
$\sigma^{\prime}\in\operatorname{Aut}(\operatorname{\mathbb{F}})$. Let
$\gamma_{X}^{\prime}=h_{X}\sigma^{\prime}$ and
$\gamma_{Y}^{\prime}=h_{Y}\sigma^{\prime}$. We have

$x_{i}^{\gamma_{X}}\otimes y^{\gamma_{Y}}=(x_{i}\otimes y)^{g}=(x_{i}\otimes
y)^{h}=x_{i}^{\gamma_{X}^{\prime}}\otimes y^{\gamma_{Y}^{\prime}},$

for $i=1,\dots,k$. It follows that there exist
$\alpha_{i}\in\operatorname{\mathbb{F}}^{\times}$ such that
$x_{i}^{\gamma_{X}}=\alpha_{i}\cdot x_{i}^{\gamma_{X}^{\prime}}$ and
$y^{\gamma_{Y}}=\frac{1}{\alpha_{i}}\cdot y^{\gamma_{Y}^{\prime}}$ for
$i=1,\dots,k$. As $y$ is nonzero, $\alpha_{i}$ does not depend on $i$, i.e.
$\alpha_{i}=\alpha$ for $i=1,\dots,k$.

Since $\operatorname{\mathbb{F}}^{\times}\leq H_{X}$, there exists $z\in
H_{X}$ such that $x^{z}=\alpha\cdot x$ for all $x\in X$. Therefore
$x_{i}^{\gamma_{X}}=x_{i}^{\gamma_{X}^{\prime}z}$ for $i=1,\dots,k$. Now
$\beta_{X}=\gamma_{X}^{\prime}\cdot z\in H_{X}$ is the required element.

By switching the roles of $G$ and $H$, we obtain
$\operatorname{\mathit{k}-rel}(G_{X})\subseteq\operatorname{\mathit{k}-rel}(H_{X})$.
Thus
$\operatorname{\mathit{k}-rel}(H_{X})=\operatorname{\mathit{k}-rel}(G_{X})$
and $G_{X}$ is $k$-equivalent to $H_{X}$, as claimed. Similarly,
$\operatorname{\mathit{k}-rel}(H_{Y})=\operatorname{\mathit{k}-rel}(G_{Y})$
and $G_{Y}$ is $k$-equivalent to $H_{Y}$. ∎

######  Corollary 3.2.

Let $V$ be a vector space over a finite field $\operatorname{\mathbb{F}}$, and
let $G$ be a primitive affine group with socle $V$. Set $G=V\rtimes G_{0}$,
where $G_{0}$ is the zero stabilizer, and assume that
$\operatorname{\mathbb{F}}^{\times}\leq G_{0}$. Suppose that $G_{0}$
stabilizes a nontrivial tensor decomposition $V=X\otimes Y$ over
$\operatorname{\mathbb{F}}$, where $\dim X\neq\dim Y$. Finally, assume that
$k\geq 4$ and the $k$-closures of $X\rtimes(G_{0})_{X}$ and
$Y\rtimes(G_{0})_{Y}$ lie in the class $\operatorname{\mathfrak{X}}$. Then
$G^{(k)}$ lies in the class $\operatorname{\mathfrak{X}}$.

###### Proof.

By Lemma 2.6, the $k$-closures of $G$, $X\rtimes(G_{0})_{X}$ and
$Y\rtimes(G_{0})_{Y}$ are affine groups with socles $V$, $X$ and $Y$,
respectively. Set $H=G^{(k)}$ and $H=V\rtimes H_{0}$, so $G_{0}$ is
$(k-1)$-equivalent to $H_{0}$ by Proposition 2.2 (iii). By Proposition 2.8
(i), $H_{0}$ preserves some tensor decomposition of $V$, and by [30, Lemma
4.4.5 (1)] it preserves the decomposition $V=X\otimes Y$. By Proposition 3.1,
$(G_{0})_{X}$ is $(k-1)$-equivalent to $(H_{0})_{X}$, and $(G_{0})_{Y}$ is
$(k-1)$-equivalent to $(H_{0})_{Y}$. Therefore
$(H_{0})_{X}\leq(G_{0})_{X}^{(k-1)}$ and $(H_{0})_{Y}\leq(G_{0})_{Y}^{(k-1)}$,
hence

$X\rtimes(H_{0})_{X}\leq
X\rtimes((G_{0})_{X}^{(k-1)}\cap\operatorname{GL}(X))=(X\rtimes(G_{0})_{X})^{(k)},$

where the last equality follows from Lemma 2.5, and $\operatorname{GL}(X)$ is
the general linear group over the prime subfield of
$\operatorname{\mathbb{F}}$. Similarly,
$Y\rtimes(H_{0})_{Y}\leq(Y\rtimes(G_{0})_{Y})^{(k)}$. Therefore $(H_{0})_{X}$
and $(H_{0})_{Y}$ lie in $\operatorname{\mathfrak{X}}$, and thus $H$ lies in
$\operatorname{\mathfrak{X}}$ as claimed. ∎

###  3.2. Subfield stabilizers

Let $V$ be a vector space over a finite field $\operatorname{\mathbb{F}}_{q}$
of order $q$, and let $v_{1},\dots,v_{d}$ be the
$\operatorname{\mathbb{F}}_{q}$-basis of $V$. Let
$\operatorname{\mathbb{F}}_{q_{0}}$ be a proper subfield of
$\operatorname{\mathbb{F}}_{q}$, and let $V_{0}$ be the
$\operatorname{\mathbb{F}}_{q_{0}}$-span of $v_{1},\dots,v_{d}$. We say that
$G\leq\operatorname{{\rm\Gamma}L}(V)$ is a subfield subgroup, if it preserves
the set $\operatorname{\mathbb{F}}_{q}V_{0}=\\{\lambda
v\mid\lambda\in\operatorname{\mathbb{F}}_{q},\,v\in V_{0}\\}$ of lines passing
through $V_{0}$. The space $V$ can be identified with a tensor product
$V_{0}\otimes\operatorname{\mathbb{F}}_{q}$ over
$\operatorname{\mathbb{F}}_{q_{0}}$, and under such an identification the
group $G$ preserves this tensor product, see [30, Formula (4.5.2)]. As in the
previous section, let $G_{V_{0}}\leq\operatorname{GL}(V_{0})$ and
$G_{\operatorname{\mathbb{F}}_{q}}\leq\operatorname{GL}_{1}(\operatorname{\mathbb{F}}_{q})$
denote the groups induced by $G$ on $V_{0}$ and
$\operatorname{\mathbb{F}}_{q}$ respectively.

######  Proposition 3.3.

Let $V$ be a vector space over a finite field $\operatorname{\mathbb{F}}_{q}$
of order $q$, and let $G$ be a primitive affine group with socle $V$. Set
$G=V\rtimes G_{0}$, where $G_{0}\leq\operatorname{{\rm\Gamma}L}(V)$, and
assume that $\operatorname{\mathbb{F}}_{q}^{\times}\leq G_{0}$. Suppose that
$G_{0}$ is a subfield subgroup, in particular, $G_{0}$ preserves the tensor
decomposition $V=V_{0}\otimes\operatorname{\mathbb{F}}_{q}$ over the field
$\operatorname{\mathbb{F}}_{q_{0}}$. Assume that $k\geq 4$ and the $k$-closure
of $V_{0}\rtimes(G_{0})_{V_{0}}$ lies in the class
$\operatorname{\mathfrak{X}}$. Then $G^{(k)}$ lies in the class
$\operatorname{\mathfrak{X}}$.

###### Proof.

By Lemma 2.6, the $k$-closures of $G$ and $V_{0}\rtimes(G_{0})_{V_{0}}$ are
affine groups with socles $V$ and $V_{0}$, respectively. By Proposition 2.8
(i), $G^{(k)}$ is also a subfield subgroup preserving
$V=V_{0}\otimes\operatorname{\mathbb{F}}_{q}$ over
$\operatorname{\mathbb{F}}_{q_{0}}$. Set $H=G^{(k)}$ and $H=V\rtimes H_{0}$,
$H_{0}\leq\operatorname{{\rm\Gamma}L}(V)$. Since $G_{0}$ is $(k-1)$-equivalent
to $H_{0}$, Proposition 3.1 implies that $(H_{0})_{V_{0}}$ is
$(k-1)$-equivalent to $(G_{0})_{V_{0}}$. By Lemma 2.5 and our assumptions,
$(H_{0})_{V_{0}}$ lies in $\operatorname{\mathfrak{X}}$. Since
$(H_{0})_{\operatorname{\mathbb{F}}_{q}}\leq\operatorname{GL}_{1}(q)$, it
follows that $H_{0}$ lies in class $\operatorname{\mathfrak{X}}$, as required.
∎

###  3.3. Stabilizers of $X\otimes\dots\otimes X$

Let $X$ be a vector space over a finite field $\operatorname{\mathbb{F}}$, and
let $V=X\otimes\dots\otimes X$ be the tensor product of $m\geq 2$ copies of
$X$. An element $g_{1}\otimes\dots\otimes g_{m}$ of the central product
$\operatorname{GL}(X)\otimes\dots\otimes\operatorname{GL}(X)$ acts on $V$
naturally, indeed, if $x_{1}\otimes\dots\otimes x_{m}\in V$, where $x_{i}\in
X$, $i=1,\dots,m$, then $(x_{1}\otimes\dots\otimes
x_{m})^{g_{1}\otimes\dots\otimes g_{m}}=x_{1}^{g_{1}}\otimes\dots\otimes
x_{m}^{g_{m}}$.

A field automorphism $\sigma\in\operatorname{Aut}(\operatorname{\mathbb{F}})$
acts on $V$ componentwise:

$(x_{1}\otimes\dots\otimes x_{m})^{\sigma}=x_{1}^{\sigma}\otimes\dots\otimes
x_{m}^{\sigma}.$

Finally, a permutation $\tau\in\operatorname{Sym}(m)$ acts on $V$ by permuting
components of simple tensors:

$(x_{1}\otimes\dots\otimes x_{m})^{\tau}=x_{\tau^{-1}(1)}\otimes\dots\otimes
x_{\tau^{-1}(m)},$

and this action extends to the rest of $V$ by linearity. Since the action of
field automorphisms commutes with the action of $\operatorname{Sym}(m)$, we
obtain the action of
$(\operatorname{GL}(X)\otimes\dots\otimes\operatorname{GL}(X))\rtimes(\operatorname{Aut}(\operatorname{\mathbb{F}})\times\operatorname{Sym}(m))$
on $V$. This is the stabilizer of the tensor decomposition
$V=X\otimes\dots\otimes X$, see [30, Formula (4.4.3)].

Set
$L=(\operatorname{GL}(X)\otimes\dots\otimes\operatorname{GL}(X))\rtimes(\operatorname{Aut}(\operatorname{\mathbb{F}})\times\operatorname{Sym}(m))$
and as usual, identify $\operatorname{\mathbb{F}}^{\times}$ with the center of
$\operatorname{GL}(X)$. For $g\in L$ we can write $g=(g_{1}\otimes\dots\otimes
g_{m})\sigma\tau$, where $g_{i}\in\operatorname{GL}(X)$, $i=1,\dots,m$,
$\sigma\in\operatorname{Aut}(\operatorname{\mathbb{F}})$ and
$\tau\in\operatorname{Sym}(m)$. For $i\in\\{1,\dots,m\\}$ set
$\pi_{i}(g)=\operatorname{\mathbb{F}}^{\times}\cdot g_{i}\sigma$, and notice
that it gives a correctly defined map $\pi_{i}:L\to\operatorname{P\Gamma
L}(X)$. The restriction of $\pi_{i}$ to
$\operatorname{GL}(X)\otimes\dots\otimes\operatorname{GL}(X)$ is a
homomorphism into $\operatorname{P\Gamma L}(X)$ with kernel
$K_{i}=\operatorname{GL}(X)\otimes\dots\otimes\operatorname{\mathbb{F}}^{\times}\otimes\dots\otimes\operatorname{GL}(X)$,
where $\operatorname{\mathbb{F}}^{\times}$ is on the $i$-th position in the
central product. We can also define a homomorphism
$\pi:L\to\operatorname{Sym}(m)$ by the rule $\pi(g)=\tau$. The kernel of $\pi$
is equal to
$(\operatorname{GL}(X)\otimes\dots\otimes\operatorname{GL}(X))\rtimes\operatorname{Aut}(\operatorname{\mathbb{F}})$.

Now define a map $\Pi:L\to\operatorname{P\Gamma L}(X)\wr\operatorname{Sym}(m)$
by

$\Pi(g)=(\pi_{1}(g),\dots,\pi_{m}(g))\pi(g),$

where $g\in L$. Clearly $\Pi$ is a homomorphism with kernel
$\bigcap_{i}K_{i}=\operatorname{\mathbb{F}}^{\times}$.

We can define a faithful action of the abstract wreath product
$\operatorname{P\Gamma L}(X)\wr\operatorname{Sym}(m)$ on

$\mathcal{L}=\\{\operatorname{\mathbb{F}}\cdot(v_{1}\otimes\dots\otimes
v_{m})\mid v_{i}\in X,\,i=1,\dots,m\\}$

which is the set of lines spanned by simple tensors. To do this, let
$\operatorname{P\Gamma L}(X)$ act on the lines of $X$, and let
$\operatorname{Sym}(m)$ permute the coordinates. The corresponding action is a
product action, so we may identify $\operatorname{P\Gamma
L}(X)\wr\operatorname{Sym}(m)$ with a wreath product in the product action
$\operatorname{P\Gamma
L}(X)\uparrow\operatorname{Sym}(m)\leq\operatorname{Sym}(\mathcal{L})$, The
group $L$ also acts on $\mathcal{L}$ naturally, and one can easily see that
$g\in L$ and $\Pi(g)$ act on $\mathcal{L}$ in the same way, i.e. they
represent the same permutation from $\operatorname{Sym}(\mathcal{L})$.

Let $G\leq L$ be a subgroup containing $\operatorname{\mathbb{F}}^{\times}$,
and assume that $\pi(G)$ is a transitive subgroup of $\operatorname{Sym}(m)$.
Let $G_{X}\leq\operatorname{{\rm\Gamma}L}(X)$ be the projection of $G$ into
the first component of the tensor product, i.e.

$G_{X}=\\{g_{1}\sigma\mid(g_{1}\otimes\dots\otimes g_{m})\sigma\tau\in
G,\text{ where
}\sigma\in\operatorname{Aut}(\operatorname{\mathbb{F}}),\,\tau\in\operatorname{Sym}(m),\,\tau(1)=1\\}.$

Since $\pi(G)$ is transitive, the projection $G_{X}$ does not depend on the
labelling of points $\\{1,\dots,m\\}$ up to conjugation in
$\operatorname{{\rm\Gamma}L}(X)$. One can similarly define the projection of
$\Pi(G)\leq\operatorname{P\Gamma L}(X)\uparrow\operatorname{Sym}(m)$ into
$\operatorname{P\Gamma L}(X)$; notice that this projection is
$G_{X}/\operatorname{\mathbb{F}}^{\times}$.

By [25, Section 5.3] the group $\Pi(G)\leq\operatorname{P\Gamma
L}(X)\uparrow\operatorname{Sym}(m)$ can be embedded into a wreath product
$(G_{X}/\operatorname{\mathbb{F}}^{\times})\uparrow\pi(G)$, where
$G_{X}/\operatorname{\mathbb{F}}^{\times}$ is the 1-component of $\Pi(G)$. In
particular, $G$ lies in the class $\operatorname{\mathfrak{X}}$ if and only if
$G_{X}$ and $\pi(G)$ lie in the class $\operatorname{\mathfrak{X}}$.

######  Proposition 3.4.

Let $X$ be a vector space over a finite field $\operatorname{\mathbb{F}}$ with
$\dim X\geq 2$. Let $G$ and $H$ be subgroups of
$(\operatorname{GL}(X)\otimes\dots\otimes\operatorname{GL}(X))\rtimes(\operatorname{Aut}(\operatorname{\mathbb{F}})\times\operatorname{Sym}(m))$,
$m\geq 2$, preserving the $m$-fold tensor power $X\otimes\dots\otimes X$, and
assume that $G$, $H$ contain $\operatorname{\mathbb{F}}^{\times}$ and
$\pi(G)$, $\pi(H)$ are transitive subgroups of $\operatorname{Sym}(m)$.
Suppose that $G$ and $H$ are $k$-equivalent for some $k\geq 3$. Then $G_{X}$
and $H_{X}$ are $k$-equivalent, and $\pi(G)^{[r]}=\pi(H)^{[r]}$ for
$r=\min\\{|\mathrm{Orb}_{k}(G_{X}/\operatorname{\mathbb{F}}^{\times})|,\,m\\}$,
where $G_{X}/\operatorname{\mathbb{F}}^{\times}$ acts on the set of lines of
$X$ and $\mathrm{Orb}_{k}(G_{X}/\operatorname{\mathbb{F}}^{\times})$ is the
set of orbits on $k$-tuples in this action.

###### Proof.

We will use the following notation throughout the proof. Given $s$ simple
tensors $v_{i}=v_{i1}\otimes\dots\otimes v_{im}$, $i=1,\dots,s$, let
$T(v_{1},\dots,v_{s})$ denote the $s\times m$ matrix:

$T(v_{1},\dots,v_{s})=\begin{pmatrix}\operatorname{\mathbb{F}}\cdot
v_{11}&\dots&\operatorname{\mathbb{F}}\cdot v_{1m}\\\ \vdots&&\vdots\\\
\operatorname{\mathbb{F}}\cdot v_{s1}&\dots&\operatorname{\mathbb{F}}\cdot
v_{sm}\\\ \end{pmatrix}$

Since each entry of this matrix is a line in $X$, the matrix
$T(v_{1},\dots,v_{s})$ is defined correctly and does not depend on the choice
of $v_{i1},\dots,v_{im}$ in the decomposition of $v_{i}$, $i=1,\dots,s$.

Claim: $G_{X}$ is $k$-equivalent to $H_{X}$. We will prove that
$\operatorname{\mathit{k}-rel}(H_{X})\subseteq\operatorname{\mathit{k}-rel}(G_{X})$.
Let $x_{1},\dots,x_{k}\in X$ be arbitrary nonzero vectors, and let $g_{X}\in
G_{X}$ be an arbitrary element. Choose $g\in G$ such that
$g=(g_{1}\otimes\cdots\otimes g_{m})\sigma\tau$, where
$\sigma\in\operatorname{Aut}(\operatorname{\mathbb{F}})$,
$\tau\in\operatorname{Sym}(m)$, $\tau(1)=1$ and $g_{X}=g_{1}\sigma$. Assume
first that $x_{1},\dots,x_{k}$ lie on the same line, i.e. for some nonzero
$u\in X$ we have $x_{i}\in\operatorname{\mathbb{F}}^{\times}\cdot u$ for all
$i=1,\dots,k$.

Let $\omega$ be a cyclic generator of $\operatorname{\mathbb{F}}^{\times}$.
Since $\dim X\geq 2$ we can find a vector $w\in X$ which is linearly
independent from $u$. Setting $v=u^{g_{X}}$ we have $(\omega\cdot
u)^{g_{X}}=\omega^{\sigma}\cdot v$. Now consider three simple tensors:
$a=u\otimes w\otimes\dots\otimes w$, $b=\omega\cdot u\otimes
w\otimes\dots\otimes w$ and $c=w\otimes w\otimes\dots\otimes w$, which differ
only in the first component. By the definition of $G_{X}$, the element $g_{X}$
stabilizes the first component, therefore

$\displaystyle a^{g}=v\otimes$ $\displaystyle
w^{g_{\tau^{-1}(2)}\sigma}\otimes\dots\otimes w^{g_{\tau^{-1}(m)}\sigma},$
$\displaystyle b^{g}=\omega^{\sigma}\cdot v\otimes$ $\displaystyle
w^{g_{\tau^{-1}(2)}\sigma}\otimes\dots\otimes w^{g_{\tau^{-1}(m)}\sigma},$
$\displaystyle c^{g}=w^{g_{X}}\otimes$ $\displaystyle
w^{g_{\tau^{-1}(2)}\sigma}\otimes\dots\otimes w^{g_{\tau^{-1}(m)}\sigma}.$

It is readily seen that each column of the matrix $T(a^{g},b^{g},c^{g})$ has
all coordinates equal to each other with the exception of the first column.
The first column does not have all coordinates equal to each other, as vectors
$v$, $\omega^{\sigma}\cdot v$ and $w^{g_{X}}$ do not lie on the same line.

Now, groups $G$ and $H$ are $k$-equivalent, in particular, they are
$3$-equivalent, so there exists an element $h\in H$ such that $a^{g}=a^{h}$,
$b^{g}=b^{h}$ and $c^{g}=c^{h}$. Write $h=(h_{1}\otimes\dots\otimes
h_{m})\phi\mu$, where $\phi\in\operatorname{Aut}(\operatorname{\mathbb{F}})$
and $\mu\in\operatorname{Sym}(m)$. It follows that

$\displaystyle a^{h}$
$\displaystyle=w^{h_{\mu^{-1}(1)}\phi}\otimes\dots\otimes
u^{h_{1}\phi}\otimes\dots\otimes w^{h_{\mu^{-1}(m)}\phi},$ $\displaystyle
b^{h}$
$\displaystyle=w^{h_{\mu^{-1}(1)}\phi}\otimes\dots\otimes\omega^{\phi}\cdot
u^{h_{1}\phi}\otimes\dots\otimes w^{h_{\mu^{-1}(m)}\phi},$ $\displaystyle
c^{h}$ $\displaystyle=w^{h_{\mu^{-1}(1)}\phi}\otimes\dots\otimes
w^{h_{1}\phi}\otimes\dots\otimes w^{h_{\mu^{-1}(m)}\phi},$

where the “central” component of each tensor is on position $\mu(1)$. Since
$u^{h_{1}\phi}$, $\omega^{\phi}\cdot u^{h_{1}\phi}$ and $w^{h_{1}\phi}$ do not
lie on the same line, each column of the matrix $T(a^{h},b^{h},c^{h})$ has
equal coordinates with the exception of column number $\mu(1)$. Since
$T(a^{g},b^{g},c^{g})=T(a^{h},b^{h},c^{h})$, that must be the first column and
hence $\mu(1)=1$.

Equality $a^{g}=a^{h}$ of simple tensors implies that there exist scalars
$\alpha_{i}\in\operatorname{\mathbb{F}}^{\times}$, $i=1,\dots,m$, such that
$v=\alpha_{1}\cdot u^{h_{1}\phi}$ and
$w^{g_{\tau^{-1}(i)}\sigma}=\alpha_{i}\cdot w^{h_{\mu^{-1}(i)}\phi}$ for
$i=2,\dots,m$, and $\alpha_{1}\cdot\alpha_{2}\cdot\dots\cdot\alpha_{m}=1$.
Similarly, $b^{g}=b^{h}$ implies that there exist scalars
$\beta_{i}\in\operatorname{\mathbb{F}}^{\times}$, $i=1,\dots,m$, with
$\omega^{\sigma}\cdot v=\beta_{1}\omega^{\phi}\cdot u^{h_{1}\phi}$ and
$w^{g_{\tau^{-1}(i)}\sigma}=\beta_{i}\cdot w^{h_{\mu^{-1}(i)}\phi}$ for
$i=2,\dots,m$, and $\beta_{1}\cdot\dots\beta_{m}=1$. Since the vector $w$ is
nonzero, equalities

$\displaystyle w^{g_{\tau^{-1}(i)}\sigma}$ $\displaystyle=\alpha_{i}\cdot
w^{h_{\mu^{-1}(i)}\phi},$ $\displaystyle w^{g_{\tau^{-1}(i)}\sigma}$
$\displaystyle=\beta_{i}\cdot w^{h_{\mu^{-1}(i)}\phi}$

imply $\alpha_{i}=\beta_{i}$ for $i=2,\dots,m$. It follows that
$\alpha_{1}=\beta_{1}=\alpha$ and hence $v=\alpha\cdot u^{h_{1}\phi}$ and
$\omega^{\sigma}\cdot v=\alpha\omega^{\phi}\cdot u^{h_{1}\phi}$. We can choose
an element $z\in\operatorname{\mathbb{F}}^{\times}$ such that
$v=u^{zh_{1}\phi}$ and $\omega^{\sigma}\cdot v=\omega^{\phi}\cdot
u^{zh_{1}\phi}$. We derive $\omega^{\sigma}=\omega^{\phi}$, and since $\omega$
is a cyclic generator of $\operatorname{\mathbb{F}}^{\times}$, we have
$\sigma=\phi$. Finally, as $\operatorname{\mathbb{F}}^{\times}\leq H_{X}$ and
$\mu(1)=1$, the element $h^{\prime}_{X}=zh_{1}\sigma$ lies in $H_{X}$. To sum
up, $u^{g_{X}}=v=u^{h^{\prime}_{X}}$ and $(\alpha\cdot
u)^{g_{X}}=\alpha^{\sigma}\cdot v=(\alpha\cdot u)^{h^{\prime}_{X}}$ for any
$\alpha\in\operatorname{\mathbb{F}}^{\times}$.

Recall that $x_{i}\in\operatorname{\mathbb{F}}^{\times}\cdot u$,
$i=1,\dots,k$. There exist scalars
$\alpha_{i}\in\operatorname{\mathbb{F}}^{\times}$ such that
$x_{i}=\alpha_{i}\cdot u$, $i=1,\dots,k$. It follows that for any
$i\in\\{1,\dots,m\\}$

$x_{i}^{g_{X}}=(\alpha_{i}\cdot u)^{g_{X}}=(\alpha_{i}\cdot
u)^{h^{\prime}_{X}}=x_{i}^{h^{\prime}_{X}},$

and this finishes the proof of the case when all $x_{i}$ lie on the same line.

Now assume $x_{1},\dots,x_{k}\in X$ do not lie on the same line. Choose some
nonzero vector $w\in X$ and define simple tensors $v_{i}=x_{i}\otimes
w\otimes\dots\otimes w$, $i=1,\dots,k$, which differ only in the first
component. Arguing as before, we see that each column of the matrix
$T(v_{1}^{g},\dots,v_{k}^{g})$ has equal coordinates, with the exception of
the first column. The first column does not have all coordinates equal to each
other as $x_{i}^{g_{X}}$, $i=1,\dots,k$, do not lie on the same line.

Since $G$ and $H$ are $k$-equivalent, there exists an element $h\in H$ such
that $v_{i}^{g}=v_{i}^{h}$ for all $i=1,\dots,k$. Write
$h=(h_{1}\otimes\dots\otimes h_{m})\phi\mu$, where
$\phi\in\operatorname{Aut}(\operatorname{\mathbb{F}})$,
$\mu\in\operatorname{Sym}(m)$. All columns of the matrix
$T(v_{1}^{h},\dots,v_{k}^{h})$ have equal coordinates with the exception of
column number $\mu(1)$, which has coordinates
$x_{1}^{h_{1}\phi},\dots,x_{k}^{h_{1}\phi}$. Since
$T(v_{1}^{g},\dots,v_{k}^{g})=T(v_{1}^{h},\dots,v_{k}^{h})$, we derive
$\mu(1)=1$.

Equality $v_{i}^{g}=v_{i}^{h}$, $i=1,\dots,k$, of simple tensors implies that
there exist scalars $\alpha_{ij}\in\operatorname{\mathbb{F}}^{\times}$ such
that for every $i\in\\{1,\dots,k\\}$ we have $x_{i}^{g_{X}}=\alpha_{i1}\cdot
x_{i}^{h_{1}\phi}$ and $w^{g_{j}\sigma}=\alpha_{ij}\cdot w^{h_{j}\phi}$ for
$j=2,\dots,m$, and $\prod_{j=1}^{m}\alpha_{ij}=1$. Since the vector $w$ is
nonzero, scalars $\alpha_{ij}$, $j=2,\dots,m$, do not depend on $i$. It
follows that $\alpha_{i1}$ does not depend on $i$ and we can set
$\alpha=\alpha_{i1}$ for all $i=1,\dots,k$.

We have $x_{i}^{g_{X}}=\alpha\cdot x_{i}^{h_{1}\phi}$ for $i=1,\dots,k$.
Choose an element $z\in\operatorname{\mathbb{F}}^{\times}$ such that
$x_{i}^{g_{X}}=x_{i}^{zh_{1}\phi}$ for $i=1,\dots,k$. Since
$\operatorname{\mathbb{F}}^{\times}\leq H_{X}$ and $\mu(1)=1$, the element
$h^{\prime}_{X}=zh_{1}\phi$ lies in $H_{X}$ and we have
$x_{i}^{g_{X}}=x_{i}^{h^{\prime}_{X}}$, $i=1,\dots,k$. This finishes the proof
of the case when not all $x_{i}$, $i=1,\dots,k$, lie on the same line.

It follows that
$\operatorname{\mathit{k}-rel}(H_{X})\subseteq\operatorname{\mathit{k}-rel}(G_{X})$,
and by repeating the same argument for $G$ and $H$ with interchanged roles, we
derive
$\operatorname{\mathit{k}-rel}(H_{X})=\operatorname{\mathit{k}-rel}(G_{X})$,
i.e. $G_{X}$ and $H_{X}$ are $k$-equivalent. The first claim is proved.

Claim: $\pi(G)^{[r]}=\pi(H)^{[r]}$. By Proposition 2.4 (iii) it is sufficient
to prove that $\Pi(G)$ is $k$-equivalent to $\Pi(H)$. Let
$(\operatorname{\mathbb{F}}v_{1},\dots,\operatorname{\mathbb{F}}v_{k})$ be a
$k$-tuple of lines for some $v_{1},\dots,v_{k}\in X\otimes\dots\otimes X$.
Take some $g\in\Pi(G)$ and let $g^{\prime}\in G$ be such that
$g=\Pi(g^{\prime})$. Define $u_{i}=v_{i}^{g^{\prime}}$ for $i=1,\dots,k$. It
follows that

$(\operatorname{\mathbb{F}}v_{1},\dots,\operatorname{\mathbb{F}}v_{k})^{g}=(\operatorname{\mathbb{F}}v_{1},\dots,\operatorname{\mathbb{F}}v_{k})^{g^{\prime}}=(\operatorname{\mathbb{F}}u_{1},\dots,\operatorname{\mathbb{F}}u_{k}).$

Since $G$ and $H$ are $k$-equivalent, there exists $h^{\prime}\in H$ such that
$u_{i}=v_{i}^{h^{\prime}}$, $i=1,\dots,k$. Setting $h=\Pi(h)\in\Pi(H)$ we have

$(\operatorname{\mathbb{F}}v_{1},\dots,\operatorname{\mathbb{F}}v_{k})^{h}=(\operatorname{\mathbb{F}}v_{1},\dots,\operatorname{\mathbb{F}}v_{k})^{h^{\prime}}=(\operatorname{\mathbb{F}}u_{1},\dots,\operatorname{\mathbb{F}}u_{k})=(\operatorname{\mathbb{F}}v_{1},\dots,\operatorname{\mathbb{F}}v_{k})^{g}.$

Hence $\Pi(G)$ and $\Pi(H)$ are $k$-equivalent by Lemma 2.1 and the claim is
proved. ∎

######  Corollary 3.5.

Let $V$ be a vector space over a finite field $\operatorname{\mathbb{F}}$, and
let $G$ be a primitive affine permutation group with socle $V$. Set
$G=V\rtimes G_{0}$, where $G_{0}\leq\operatorname{{\rm\Gamma}L}(V)$, and
assume that $\operatorname{\mathbb{F}}^{\times}\leq G_{0}$. Suppose that
$G_{0}$ preserves a tensor decomposition $V=X\otimes\dots\otimes X$, where $X$
is a vector space over $\operatorname{\mathbb{F}}$ and there are $m\geq 2$
tensor factors in the decomposition of $V$. Assume that $k\geq 4$ and the
$k$-closures of $X\rtimes(G_{0})_{X}$ and $\pi(G_{0})$ lie in the class
$\operatorname{\mathfrak{X}}$. Then $G^{(k)}$ lies in the class
$\operatorname{\mathfrak{X}}$.

###### Proof.

By Lemma 2.6, the $k$-closures of $G$ and $X\rtimes(G_{0})_{X}$ are affine
groups with socles $V$ and $X$, respectively. Set $H=G^{(k)}$ and $H=V\rtimes
H_{0}$, so $G_{0}$ is $(k-1)$-equivalent to $H_{0}$. By Proposition 2.8 (i)
and [30, Lemma 4.4.5 (2)], $H_{0}$ preserves the tensor decomposition
$V=X\otimes\dots\otimes X$, and by Proposition 3.4, the group $(G_{0})_{X}$ is
$(k-1)$-equivalent to $(H_{0})_{X}$, and $\pi(G_{0})^{[r]}=\pi(H_{0})^{[r]}$,
where
$r=\min\\{|\mathrm{Orb}_{k-1}((G_{0})_{X}/\operatorname{\mathbb{F}}^{\times})|,\,m\\}$.
By Lemma 2.5 and our assumptions, $(H_{0})_{X}$ lies in the class
$\operatorname{\mathfrak{X}}$.

Now, as $k-1\geq 3$, we have
$|\mathrm{Orb}_{k-1}((G_{0})_{X}/\operatorname{\mathbb{F}}^{\times})|\geq k+1$
since $X$ contains at least 3 lines. If $m<k+1$, then $r=m$ and
$\pi(G_{0})^{[r]}=\pi(G_{0})^{[m]}=\pi(G_{0})$, where the last equality
follows from the definition of the closure with respect to partitions. If
$m\geq k+1$, then $r\geq k+1$ and
$\pi(G_{0})^{[r]}\leq\pi(G_{0})^{[k+1]}\leq\pi(G_{0})^{(k)}$, where the last
inclusion follows from Proposition 2.3. In either case
$\pi(H_{0})\leq\pi(G_{0})^{[r]}$ lies in the class
$\operatorname{\mathfrak{X}}$, so the claim is proved. ∎

##  4\. Proof of Theorem 1.1

Set $H=G^{(4)}$. By Proposition 2.2 (i) it suffices to prove that
$H\in\operatorname{\mathfrak{X}}$. Suppose that $G$ is a counterexample to
Theorem 1.1 of minimal degree $n$. Note that $n>24$, because otherwise $H$ is
$\operatorname{Alt}(25)$-free and hence lies in $\operatorname{\mathfrak{X}}$.
By [23, Theorem 3.1 (i)] (see also Proposition 2.4), we may assume that $G$ is
a primitive basic permutation group.

Since $G$ is primitive, [24, Theorem 2] implies that

(2) $\operatorname{Soc}(G)=\operatorname{Soc}(H)$

unless $G$ is $4$-transitive. By [5, Tables 7.3 and 7.4]), if $G$ is
$4$-transitive and $n>24$, then $G\geq\operatorname{Alt}(n)$, so
$\operatorname{Alt}(n)\in\operatorname{\mathfrak{X}}$ and hence
$H=\operatorname{Sym}(n)\in\operatorname{\mathfrak{X}}$, a contradiction. The
case $n\leq 24$ is not possible, so we may assume that equality (2) holds.

Since $G$ is basic, applying the O’Nan-Scott theorem [16], we arrive to the
following three cases:

  1. (i)

$G$ is almost simple,

  2. (ii)

$G$ is in a diagonal action,

  3. (iii)

$G$ is an affine group.

In the case (i), equality (2) implies that the group $H/\operatorname{Soc}(G)$
is solvable and hence $H\in\operatorname{\mathfrak{X}}$, a contradiction.

Suppose that $G$ is in a diagonal action as in the case (ii). Then
$S=\operatorname{Soc}(G)=T^{m}$, where $T$ is a nonabelian simple group, and
$G/S$ is a subgroup of $\operatorname{Out}(T)\times L$, where $L$ is the
symmetric group of degree $m$ acting faithfully on the set simple factors of
the socle by conjugation. By (2), $\operatorname{Soc}(H)=S$, so $H/S$ is also
a subgroup of $\operatorname{Out}(T)\times L$. If $m\leq 4$ or $G/S$ includes
the alternating subgroup of $L$, then the nonabelian composition factors of
$G$ and $H$ are the same, and we are done. Otherwise, $G$ has a base of size 2
in view of [8, Theorem 1.1], and $G^{(3)}=G$ by Proposition 2.2 (vi). Hence
$H=G\in\operatorname{\mathfrak{X}}$, a contradiction.

Thus, we may assume that $G$ is an affine group. Observe that in this case $H$
is also an affine group by (2), and a primitive basic group by Proposition 2.2
(ii) and Proposition 2.4 (iii).

By [19], we may assume that $G$ is not solvable. Let $V$ be the socle of $G$
and let $G_{0}\leq\operatorname{GL}(V)$ be the zero stabilizer, so $G=V\rtimes
G_{0}$. We may write $H=V\rtimes H_{0}$ for the zero stabilizer
$H_{0}\leq\operatorname{GL}(V)$. Moreover,
$H_{0}=G_{0}^{(3)}\cap\operatorname{GL}(V)$ by Lemma 2.5.

We wish to apply Aschbacher’s theorem [1] to $G_{0}$ and $H_{0}$. By
Proposition 2.8, $G_{0}$ and $H_{0}$ lie in the same Aschbacher classes. Set
$|V|=p^{s}$, where $p$ is a prime. Choose the minimal $a\geq 1$ dividing $s$
such that $G_{0}\leq\operatorname{{\rm\Gamma}L}_{a}(q)$ where $q=p^{s/a}$.
Note that $a\geq 2$ since $G$ is not solvable. By Proposition 2.2 (iv), we may
also assume that $\operatorname{\mathbb{F}}_{q}^{\times}\leq G_{0}$, since
$\operatorname{\mathbb{F}}_{q}^{\times}\cdot G_{0}$ still lies in the class
$\operatorname{\mathfrak{X}}$. Note that the corresponding affine group
$V\rtimes(\operatorname{\mathbb{F}}_{q}^{\times}\cdot G_{0})$ is a permutation
group of degree $n=|V|$, so it is still a counterexample of minimal degree.

Since $G_{0}$ preserves the structure of the extension field
$\operatorname{\mathbb{F}}_{q}$ of $\operatorname{\mathbb{F}}_{p}$ (class
$\operatorname{\mathcal{C}}_{3}$ for Aschbacher’s theorem in
$\operatorname{GL}(V)$), the group $H_{0}$ also preserves this field and hence
$H_{0}\leq\operatorname{{\rm\Gamma}L}_{a}(q)$. Now, we use Aschbacher’s
classification inside $\operatorname{{\rm\Gamma}L}_{a}(q)$. The group $G_{0}$
is irreducible (since $G$ is primitive) and primitive as a linear group (since
$G$ is basic), hence the same is true for $H_{0}$. As $a$ is minimal, $G_{0}$
does not preserve any extension field of $\operatorname{\mathbb{F}}_{q}$,
hence $G_{0}$ and thus $H_{0}$ do not lie in class
$\operatorname{\mathcal{C}}_{3}$. If $G_{0}$ preserves a tensor product
decomposition (class $\operatorname{\mathcal{C}}_{4}$), a subfield (class
$\operatorname{\mathcal{C}}_{5}$), or a tensor power (class
$\operatorname{\mathcal{C}}_{7}$), then from the minimality of the
counterexample $G$ and Corollary 3.2, Proposition 3.3 or Corollary 3.5
respectively, it follows that $H\in\operatorname{\mathfrak{X}}$, a
contradiction.

If $\mathrm{SL}_{a}(q)\leq G_{0}$, then $\mathrm{SL}_{a}(q)\leq
H_{0}\leq\operatorname{{\rm\Gamma}L}_{a}(q)$. Hence $G_{0}$ and $H_{0}$ have
the same nonabelian composition factors, and
$H\in\operatorname{\mathfrak{X}}$, a contradiction. Therefore, Aschbacher’s
theorem in $\operatorname{{\rm\Gamma}L}_{a}(q)$ applies, and we deduce that
either $G_{0}$ and $H_{0}$ normalize a subgroup of symplectic type (class
$\operatorname{\mathcal{C}}_{6}$), or they stabilize a symplectic, unitary or
orthogonal form (class $\operatorname{\mathcal{C}}_{8}$), or they lie in class
$\operatorname{\mathcal{C}}_{9}$. If $G_{0}$ stabilizes a form, then by
Proposition 2.8 (i), $H_{0}$ preserves the same form and hence we can apply
Aschbacher’s theorem inside a symplectic, unitary or orthogonal group. Note
that if a group preserving some form also stabilizes a tensor decomposition or
a subfield, then we have classified that group as belonging to
$\operatorname{\mathcal{C}}_{4}$, $\operatorname{\mathcal{C}}_{7}$ or
$\operatorname{\mathcal{C}}_{5}$ on the previous step. Thus, either $G_{0}$
and $H_{0}$ lie in $\operatorname{\mathcal{C}}_{6}$ or
$\operatorname{\mathcal{C}}_{9}$, or $G_{0}$ contains $\mathrm{Sp}_{a}(q)$,
$\mathrm{SU}_{a}(q^{1/2})$ or $\Omega_{a}^{\pm}(q)$, depending on the type of
the form $G_{0}$ stabilizes. In the latter case $H_{0}$ stabilizes the same
form and contains the same classical group as $G_{0}$. Hence $H_{0}$ and
$G_{0}$ have the same nonabelian composition factors, which implies
$H_{0}\in\operatorname{\mathfrak{X}}$, a contradiction.

Therefore, to finish the proof we are left to consider the cases when $G_{0}$
and $H_{0}$ lie in class $\operatorname{\mathcal{C}}_{6}$ or
$\operatorname{\mathcal{C}}_{9}$.

###  4.1. The class $\operatorname{\mathcal{C}}_{6}$.

The following lemma is probably well-known.

######  Lemma 4.1.

Let $R$ be a nilpotent group of class at most two acting faithfully and
irreducibly on a finite vector space $V$. If $|R|\leq|V|^{1/2}$, then $R$ has
a regular orbit on $V$.

###### Proof.

Let $x\in R$ be a nontrivial element. If $x$ is an element of the center
$Z(R)$, then $C_{V}(x)$ is an $R$-invariant subspace of $V$, and since $R$ is
faithful and irreducible, $C_{V}(x)=0$. If $x\not\in Z(R)$, then there exists
some $g\in R$ such that $[x,g]=x^{-1}x^{g}$ is nontrivial. It follows that
$C_{V}([x,g])\supseteq C_{V}(x^{-1})\cap C_{V}(x^{g})$, and since
$[x,g]\in[R,R]\leq Z(R)$, we have $C_{V}(x^{-1})\cap C_{V}(x^{g})=0$. Observe
that $|C_{V}(x^{-1})|=|C_{V}(x)|=|C_{V}(x^{g})|$, hence
$|C_{V}(x)|\leq|V|^{1/2}$.

Now, suppose that $R$ does not have a regular orbit on $V$. Then

$|V|\leq\sum_{x\in R\setminus 1}|C_{V}(x)|<|R|\cdot|V|^{1/2},$

hence $|V|^{1/2}<|R|$, a contradiction. ∎

If $G_{0}$ lies in class $\operatorname{\mathcal{C}}_{6}$, then it contains a
normal $r$-subgroup $R$ for some prime $r\neq p$ with the following properties
(see [12, Section 4.6] for details). We have $|R/Z(R)|=r^{2m}$, $m\geq 1$, and
if $r$ is odd, then $|Z(R)|=r$ and $R$ is an extraspecial group, while if
$r=2$, then $|Z(R)|\leq 4$ and $R$ is either an extraspecial group or a
central product of a cyclic group of order 4 and an extraspecial group. The
dimension of $V$ over $\operatorname{\mathbb{F}}_{q}$ is $a=r^{m}$, and $R$
acts absolutely irreducibly on $V$. Moreover, $G_{0}$ is contained in a
subgroup $N\leq\operatorname{{\rm\Gamma}L}_{a}(q)$, such that $R$ is normal in
$N$ and $N/R$ is isomorphic to $\mathrm{Sp}_{2m}(r)$ or to $O_{2m}^{\pm}(2)$
if $r=2$. By Proposition 2.8 (i), the group $H_{0}$ also lies in the class
$\operatorname{\mathcal{C}}_{6}$. Note that the order of the subgroup of
symplectic type $R$ can be determined from $a$, so it is the same for both
$G_{0}$ and $H_{0}$.

######  Proposition 4.2.

If $m<12$, then $G$ and $H$ are $\operatorname{Alt}(25)$-free. If $m\geq 12$,
then $G$ has a base of size at most $3$, in particular, $G=H$.

###### Proof.

Recall that $G_{0}/R$ embeds into $\operatorname{GL}_{2m}(r)$. If
$\operatorname{Alt}(25)$ is a section of $G$, then it is a section of
$\operatorname{GL}_{2m}(r)$ as well. By Proposition 2.7, we have $2m\geq 23$.
This implies that if $m<12$ then $G$ (and similarly $H$) is
$\operatorname{Alt}(25)$-free. So we may assume $m\geq 12$.

Suppose that $|R|>|V|^{1/2}$. Recall that $|V|=q^{r^{m}}$, $r\leq q-1$ and
$|R|\leq r^{2m+2}$, therefore

$q^{r^{m}/2}=|V|^{1/2}<|R|\leq r^{2m+2}\leq q^{2m+2}.$

This is impossible for $m\geq 12$, so $|R|\leq|V|^{1/2}$ and $R$ has a regular
orbit on $V$ by Lemma 4.1.

Let $v\in V$ be such that the stabilizer $R_{v}$ is trivial and set
$J=(G_{0})_{v}$. We will now prove that $J$ has a regular orbit on $V$, our
argument following the argument in [17, Lemma 3.6] (see also [10, Proposition
5.6]).

Since $R\cap J=1$, the group $J$ is isomorphic to a subgroup of
$\mathrm{Sp}_{2m}(r)$ or $O^{\pm}_{2m}(2)$. Suppose that $J$ has no regular
orbit and therefore

(3) $|V|\leq\sum_{h\in J\setminus 1}|C_{V}(h)|.$

For $h\in J\setminus 1$, [9, Theorem 4.1] implies that there are $2m+2$
conjugates of $h$ in $N$ generating $N$. Hence

$\dim C_{V}(h)\leq\left(1-\frac{1}{2m+2}\right)\dim V$

and $|C_{V}(h)|\leq|V|^{1-1/(2m+2)}$. The obtained inequality and (3) imply

$|V|\leq|J|\cdot|V|^{1-1/(2m+2)}.$

We derive that

(4)
$q^{\frac{r^{m}}{2m+2}}=|V|^{1/(2m+2)}\leq|J|\leq|N/R|\leq|\operatorname{GL}_{2m}(r)|<r^{4m^{2}}.$

Since $r\leq q-1$, we obtain $r^{m}<4m^{2}(2m+2)$. The only integer solutions
for that inequality for $m\geq 12$ and $r\geq 2$ are $r=2$ and $m=12,13,14$.

Note that $q\geq 3$. Inequalities (4) imply

$q^{\frac{2^{m}}{2m+2}}\leq|N/R|\leq\max\\{|\mathrm{Sp}_{2m}(2)|,\,|O^{\pm}_{2m}(2)|\\}.$

A direct computation shows that this is possible only for $q=3$ and $m=12$.

We are left with the case $r=2$, $q=3$, $m=12$. By [12, Table 4.6.B], we have
$N/R\simeq O^{\varepsilon}_{24}(2)$, where $\varepsilon\in\\{+,-\\}$. Recall
that $J$ is isomorphic to a subgroup of $O^{\varepsilon}_{24}(2)$, and
consider an arbitrary element $h\in J\setminus 1$. By [9, Theorem 4.4], there
are $m+3=15$ conjugates of $h$ which generate $O^{\varepsilon}_{24}(2)$ unless
$h$ is a transvection, in which case there are $2m=24$ conjugates. There are
at most $2^{24}$ transvections in $O^{\varepsilon}_{24}(2)$, so at most
$2^{24}$ elements $h\in J\setminus 1$ corresponding to a transvection; denote
this set of elements by $T$.

Now, if $h\in T$, there are $25$ conjugates of $h$ generating $N$, hence

$\dim C_{V}(h)\leq\left(1-\frac{1}{25}\right)\dim V$

and $|C_{V}(h)|\leq|V|^{1-1/25}$. If $h\not\in T$, there are $16$ conjugates
of $h$ generating $N$, so

$\dim C_{V}(h)\leq\left(1-\frac{1}{16}\right)\dim V$

and $|C_{V}(h)|\leq|V|^{1-1/16}$. We apply our estimates to inequality (3) and
obtain

$|V|\leq|T|\cdot|V|^{1-1/25}+(|J|-|T|)\cdot|V|^{1-1/16}.$

Recall that $|T|\leq 2^{24}$, $|J|\leq|O^{-}_{24}(2)|$ and $|V|=3^{2^{12}}$.
Thus

$1\leq 2^{24}\cdot 3^{-\frac{2^{12}}{25}}+|O^{-}_{24}(2)|\cdot
3^{-\frac{2^{12}}{16}},$

which is a contradiction. Hence for any $m\geq 12$ the group $J$ has a regular
orbit and so $G$ has a base of size at most 3. Now $G=H$ by Proposition 2.2
(vi). ∎

The proposition above contradicts $G$ being a counterexample, so we can move
to the next class.

###  4.2. The class $\operatorname{\mathcal{C}}_{9}$.

We need the following general statement about permutation groups.

######  Proposition 4.3.

Let $P$ and $Q$ be permutation groups on the same set $V$, and suppose that
they are $3$-equivalent. If $P/Z(P)$ and $Q/Z(Q)$ are almost simple groups,
then either $\operatorname{Soc}(P/Z(P))=\operatorname{Soc}(Q/Z(Q))$ or $P$ and
$Q$ are $\operatorname{Alt}(25)$-free.

###### Proof.

Recall that since $P$ and $Q$ are 3-equivalent, they have the same orbits and
systems of imprimitivity by Proposition 2.2 (ii). There exists an orbit
$\Delta\subset V$ such that $P^{\Delta}$ is nonsolvable. Groups $P^{\Delta}$
and $Q^{\Delta}$ are 3-equivalent by Proposition 2.4 (i), hence $Q^{\Delta}$
is also nonsolvable (otherwise $P^{\Delta}\leq(Q^{\Delta})^{(3)}$ would be
solvable by [19]). The group $P^{\Delta}/Z(P^{\Delta})$ is almost simple with
the same socle as $P$, and similar properties hold for $Q^{\Delta}$. We may
thus replace $P$ and $Q$ by $P^{\Delta}$ and $Q^{\Delta}$, and assume that $P$
is transitive on $\Delta$.

If $P$ is imprimitive on $\Delta$, then let $\Pi\subset\Delta$ be the minimal
nontrivial block of imprimitivity. If $P^{\Pi}$ is nonsolvable, then
$P^{\Pi}/Z(P^{\Pi})$ is almost simple with the same socle as $P$, and hence
again, we can replace $P$ and $Q$ by $P^{\Pi}$ and $Q^{\Pi}$ by Proposition
2.4 (ii). If $P^{\Pi}$ is solvable, then the action of $P$ on the
imprimitivity system $\Delta/\Pi$ defined by $\Pi$ is nonsolvable and almost
simple, therefore we can replace our groups by $P^{\Delta/\Pi}$ and
$Q^{\Delta/\Pi}$ by Proposition 2.4 (ii). We can thus assume that $P$ acts
primitively on $\Delta$.

Since $P$ is a nonsolvable primitive group, $Z(P)=1$. Hence $P$ is almost
simple and by [15, Theorem 2], either
$\operatorname{Soc}(P)=\operatorname{Soc}(P^{(3)})$ or $P$ is 3-transitive.
The same is true for $Q$, and as $P$ and $Q$ are 3-equivalent, $P$ is
3-transitive if and only if $Q$ is 3-transitive. Since $P^{(3)}=Q^{(3)}$, we
either have
$\operatorname{Soc}(P)=\operatorname{Soc}(P^{(3)})=\operatorname{Soc}(Q^{(3)})=\operatorname{Soc}(Q)$,
or $P$ and $Q$ are 3-transitive. In the latter case, [5, Table 7.4] implies
that $|\Delta|\leq 24$ or $\operatorname{Soc}(Q)=\mathrm{PSL}_{2}(u)$ for some
prime power $u$, which means that $P$ and $Q$ are
$\operatorname{Alt}(25)$-free. ∎

Let $G_{0}$ lie in $\operatorname{\mathcal{C}}_{9}$. Since $n>16$, the group
$H_{0}$ also lies in $\operatorname{\mathcal{C}}_{9}$ by Proposition 2.8 (ii).
Both $G_{0}/Z(G_{0})$ and $H_{0}/Z(H_{0})$ are almost simple groups. By
applying Proposition 4.3 to $P=G_{0}$ and $Q=H_{0}$ we derive that
$H_{0}\in\operatorname{\mathfrak{X}}$, which is the final contradiction.

##  5\. Concluding remarks

### 5.1.

As mentioned in the introduction, it is important from the computational point
of view to have an upper bound on the size of primitive sections of a
permutation group from the class under consideration. Combining the main
result of [3] with our Corollary 1.2, we obtain the following.

######  Corollary 5.1.

Suppose that $G$ is an $\operatorname{Alt}(d)$-free primitive permutation
group with $d\geq 25$. Then the order of $G^{(k)}$ is bounded from above by
$n^{c}$ for every $k\geq 4$, where $c$ is a constant depending only on $d$.

### 5.2.

Example (ii) from the introduction shows that for $k=4$ (and even $k=5$), the
best possible value of $d$ providing that the $k$-closure of every
$\operatorname{Alt}(d)$-free group is again $\operatorname{Alt}(d)$-free is
$d=25$. It is clear that for greater value of $k$, the value of $d$ can be
reduced. It would be interesting to find the minimal possible $d$ as a
function on $k$, or at least to find the minimal possible $k$ such that the
$k$-closure of $\operatorname{Alt}(d)$-free group is
$\operatorname{Alt}(d)$-free for all $d\geq 5$.

### 5.3.

The concept of the $\operatorname{Alt}(d)$-free groups provides a natural way
to restrict the composition factors of a group. Namely, only nonabelian
composition factors that do not have linear representations of bounded
dimension must be excluded. For example, if $\operatorname{\mathfrak{X}}$ is
the class of $\operatorname{Alt}(25)$-free groups, then the list of simple
groups of $\operatorname{\mathfrak{X}}$ includes the groups of order $p$ for
all primes $p$, all the sporadic groups and exceptional groups of Lie type, as
well as the classical groups of bounded degree (cf. Proposition 2.7), and the
alternating groups of degree less than $25$.

It is worth noting that historically, there were other ways to restrict
composition factors of a group, see the classical papers by Luks [18] and
Babai–Luks [4]. Namely, let $\Gamma_{d}$ with $d\geq 4$ denote the class of
finite groups every nonabelian composition factor of which can be embedded in
$\operatorname{Alt}(d)$. In fact, the restriction in the Luks paper was even
stronger: it requires that every (not necessarily nonabelian) composition
factor can be embedded in $\operatorname{Alt}(d)$. In our definition of
$\Gamma_{d}$ we follow the Babai–Luks paper, since it provides a wider class
of groups.

Obviously, every group from $\Gamma_{d}$ is $\operatorname{Alt}(d+1)$-free.
Therefore, if $k\geq 4$ and $d\geq 24$, then $G^{(k)}$ is an
$\operatorname{Alt}(d+1)$-free group for every $G\in\Gamma_{d}$ by virtue of
Corollary 1.2. The question is whether $G^{(k)}\in\Gamma_{d}$?

Inspecting the proof of Theorem 1.1, one can see that applying our arguments
to a group $G\in\Gamma_{d}$ for $d\geq 24$, it follows that for $k\geq 4$,
either $G^{(k)}$ belongs to $\Gamma_{d}$, or $G$ is a basic affine group such
that the zero stabilizer $G_{0}$ lies in one of the Aschbacher classes
$\operatorname{\mathcal{C}}_{6}$ and $\operatorname{\mathcal{C}}_{9}$ (as in
Subsections 4.1 and 4.2).

For $G_{0}\in\operatorname{\mathcal{C}}_{9}$, it is not hard to show that
$G^{(4)}\in\Gamma_{d}$. Here the only problem is (cf. the proof of Proposition
4.3) the case when $G_{0}$ is $3$-transitive. Since $d\geq 24$, we may assume
that $\operatorname{Soc}(G_{0})=PSL_{2}(u)$. Applying the results from [13,
14] that describe the irreducible quasisimple linear groups acting without
regular orbits, we conclude that for a finite number of exceptions
$b(G_{0})\leq 2$, and we are done in view of Proposition 2.2(vi).

For $G_{0}\in\operatorname{\mathcal{C}}_{6}$, the situation is more
complicated. Let us keep notation of Proposition 4.2. If $m\geq 12$, then
$G^{(4)}=H^{(4)}$ and everything is fine. However, the inequality $m<12$ on
dimension of $G_{0}$ can only guarantee that $m<12$ for $H_{0}$, that is the
dimension (but not the size) of $H_{0}$ is bounded. We do not know how to
overcome this problem for $k=4$. Nevertheless, it can be done for $k\geq 15$.
Indeed, according to [10, Proposition 5.6], in our situation, for each $m$
(even $m<12$), the base number of $G_{0}$ is at most $13$, so the base number
of $G$ is at most $14$. Now Proposition 2.2(vi) implies that
$G^{(15)}=H^{(15)}$, as required. Thus, the following is true.

######  Corollary 5.2.

If $G\in\Gamma_{d}$ with $d\geq 24$, then $G^{(k)}\in\Gamma_{d}$ for $k\geq
15$.

It would certainly be interesting to prove the similar statement for $k<15$ if
it is possible.

### 5.4.

Let us finish by returning to the computational problem from which we started
the introduction. In the context of the present paper, it can be formulated as
follows: find the $k$-closure of a given $\operatorname{Alt}(d)$-free
permutation group in polynomial time in its degree for fixed $k$ and $d$.
Using [23, Theorem 3.1], this problem can be reduced in polynomial time to
finding (a) the $k$-closure of every primitive basic
$\operatorname{Alt}(d)$-free group, and (b) the intersection of the
$k$-closure of an $\operatorname{Alt}(d)$-free group with any other
$\operatorname{Alt}(d)$-free group. Problem (b) can be solved by the
Babai–Luks algorithm for all $k$ and $d$. Problem (a) is open for now. We
believe that the results of this paper could be used to solve problem (a).

##  6\. Acknowledgements

The authors thank A. Maróti for helpful suggestions, improving the treatment
of class $\operatorname{\mathcal{C}}_{6}$ in Subsection 4.1.

The work is supported by the Russian Science Foundation, project no.
24-11-00127, https://rscf.ru/en/project/24-11-00127/.

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