# Finite groups satisfying
the independence property

Saul D. Freedman  Saul D. Freedman
School of Mathematics and Statistics
University of St Andrews
St Andrews, KY16 9SS, UK
email<EMAIL_ADDRESS>,  Andrea Lucchini  Andrea Lucchini
Università di Padova
Dipartimento di Matematica “Tullio Levi-Civita”
Via Trieste 63, 35121 Padova, Italy
email<EMAIL_ADDRESS>,  Daniele Nemmi  Daniele Nemmi
Università di Padova
Dipartimento di Matematica “Tullio Levi-Civita”
Via Trieste 63, 35121 Padova, Italy
email<EMAIL_ADDRESS>and  Colva M. Roney-Dougal  Colva M. Roney-Dougal
School of Mathematics and Statistics
University of St Andrews
St Andrews, KY16 9SS, UK
email<EMAIL_ADDRESS>
###### Abstract.

We say that a finite group $G$ satisfies the _independence property_ if, for
every pair of distinct elements $x$ and $y$ of $G$, either $\\{x,y\\}$ is
contained in a minimal generating set for $G$ or one of $x$ and $y$ is a power
of the other. We give a complete classification of the finite groups with this
property, and in particular prove that every such group is supersoluble. A key
ingredient of our proof is a theorem showing that all but three finite almost
simple groups $H$ contain an element $s$ such that the maximal subgroups of
$H$ containing $s$, but not containing the socle of $H$, are pairwise non-
conjugate.

###### Key words and phrases:

Generating sets, supersoluble groups, simple groups

###### 2010 Mathematics Subject Classification:

20D60 (primary); 20F16, 20E32 (secondary)

##  1\. Introduction

Let $G$ be a finite group. A generating set $X$ for $G$ is said to be
_minimal_ if no proper subset of $X$ generates $G.$ Let $d(G)$ and $m(G)$
denote, respectively, the smallest and largest cardinality of a minimal
generating set for $G$. A nice result in universal algebra, due to Tarski and
known as the Tarski Irredundant Basis Theorem (see, for example, [10, Theorem
4.4]), implies that, for every positive integer $k$ with $d(G)\leqslant
k\leqslant m(G),$ the group $G$ has a minimal generating set of cardinality
$k$. However, minimal generating sets for finite groups are not well
understood. In particular, while several results in the literature (e.g., [26,
42, 52]) yield good estimates for $d(G)$, very little is known about $m(G)$.
An exhaustive investigation [12, 57] was carried out for the finite symmetric
groups, proving that $m(\operatorname{S}_{n})=n-1$ for each $n$, and giving a
complete description of the minimal generating sets of $\operatorname{S}_{n}$
having cardinality $n-1$. The problem of determining $m(G)$ in general remains
open, even for finite simple groups, though partial results for certain
families of these groups are given in [58].

One natural related question is “which subsets of $G$ lie in a minimal
generating set?” For singletons, the answer is easy: an element belongs to
some minimal generating set for $G$ if and only if it is not contained in the
Frattini subgroup of $G$. Therefore, the first meaningful question is “which
pairs of distinct elements belong to a minimal generating set?” Similarly, we
can ask “which pairs of distinct elements belong to a generating set of size
$d(G)$?” A partial answer to the corresponding question about singletons is
given in [1, §6], using [8, Theorem 1].

We will call two distinct elements $x$ and $y$ of $G$ _independent_ in $G$ if
there exists a minimal generating set $X$ for $G$ with $\\{x,y\\}\subseteq X.$
Similarly, we will call $x$ and $y$ _rank-independent_ in $G$ if there exists
such an $X$ with $|X|=d(G)$. An obvious obstruction to $x$ and $y$ being
independent is that one of the two is a power of the other. We say that $G$
satisfies the _independence property_ if this is the unique obstruction, i.e.,
if two distinct elements $x$ and $y$ are independent whenever neither of $x$
and $y$ is a power of the other. Similarly, an obvious obstruction to $x$ and
$y$ being rank-independent is that they generate a cyclic subgroup, i.e., that
each of $x$ and $y$ is a power of some $z\in G$. We say that a non-cyclic
finite group $G$ satisfies the _rank-independence property_ if $\\{x,y\\}$
extends to a generating set for $G$ of size $d(G)$ whenever $\langle
x,y\rangle$ is not cyclic.

Note that we can also formulate the independence and rank-independence
properties in the context of certain graphs associated with $G$. Each graph
defined here has vertex set $G$. In the _independence graph_ of $G$, which was
introduced and investigated in [41], two distinct vertices are adjacent if and
only if they are independent, while in the _rank graph_ of $G$, two distinct
vertices are adjacent if and only if they are rank-independent. The _power
graph_ of $G$, where distinct vertices are adjacent if and only if one is a
power of the other, was introduced by Kelarev and Quinn [37] and investigated
by several authors (see for example [13, 14, 15, 16, 18]). Finally, the edges
of the _enhanced power graph_ of $G$ are the pairs $\\{x,y\\}$ of distinct
vertices such that $\langle x,y\rangle$ is cyclic. This graph was introduced
to interpolate between the power graph and the well-known commuting graph, but
has since been studied in its own right (see [17, 48, 59, 60]). The
independence property of a group is equivalent to its independence graph being
the complement of its power graph. Similarly, $G$ satisfies the rank-
independence property if and only if its rank graph is the complement of its
enhanced power graph.

In this paper, we will give a complete classification of the finite groups $G$
satisfying the independence property and those satisfying the rank-
independence property. We will see in particular that in each case $G$ is
supersoluble.

The classification of finite groups with the rank-independence property is not
particularly difficult (however, our proof relies on the classification of
finite simple groups). The description depends on whether $d(G)=2$ or
$d(G)\geqslant 3$.

######  Theorem 1.

Let $G$ be a finite group with $d(G)=2$. Then $G$ satisfies the rank-
independence property if and only if one of the following occurs:

  1. (i)

$G\cong C_{p}\times C_{p}$, with $p$ a prime;

  2. (ii)

$G\cong Q_{8}$; or

  3. (iii)

$G\cong C_{p}\rtimes C_{q^{m}}$, where $p$ and $q$ are distinct primes, $m$ is
an arbitrary positive integer, and the action of $C_{q^{m}}$ on $C_{p}$ has
kernel $C_{q^{m-1}}$.

######  Theorem 2.

Let $G$ be a finite group with $d(G)\geqslant 3.$ Then $G$ satisfies the rank-
independence property if and only if $G=P\rtimes C,$ where $P$ is an
elementary abelian $p$-subgroup of $G$ and $C$ is a cyclic group of coprime
order acting on $P$ as scalar multiplication.

In the above result, we permit $C=1$, and more generally, $G=P\times C$.

In a very recent paper, Harper [32] introduced the notion of $k$-flexible
groups, for each positive integer $k$. A finite group $G$ is _$k$ -flexible_
if, for all $g_{1},\dots,g_{k}\in G$ such that $d(\langle
g_{1},\dots,g_{k}\rangle)=k,$ there exist $g_{k+1},\dots,g_{d(G)}\in G$ such
that $\langle g_{1},\dots,g_{d(G)}\rangle=G$. In particular, the notions of
2-flexible groups and groups with the rank-independence property coincide.
Theorems 1 and 2 above correspond to a small correction of Lemma 2.7 and a
slightly more precise statement of part of Theorem 2.14 in [32], respectively,
and were proved independently.

The classification of the finite groups with the independence property is much
more difficult. To prove that a group satisfying the independence property is
supersoluble, we require several new tools that rely on classifications of the
finite simple groups and their maximal subgroups, including the following key
result. For an element $s$ of a group $G$, we write $\mathcal{M}_{G}(s)$ to
denote the set of maximal subgroups of $G$ containing $s$.

######  Theorem 3.

Let $S$ be a non-abelian finite simple group. Then there exist non-commuting
elements $s,x\in S$ such that, for each almost simple group $G$ with socle
$S$, the intersection $\bigcap_{M\in\mathcal{M}_{G}(s)}M$ contains $x$.

We shall prove Theorem 3 in Section 2 as a consequence of two stronger
theorems, which may be of interest in their own right: Theorem 4 below, which
deals with all but three choices for $S$, and Theorem 2.1, which addresses the
remaining groups.

For an almost simple group $G$ with socle $S$, we let
$\mathcal{M}^{\prime}_{G}(s)$ denote the set of maximal subgroups $M$ of $G$
with $s\in M$ and $S\not\leqslant M$. A _novelty maximal_ of $G$ is a maximal
subgroup $M$ such that $M\cap S$ is a proper non-maximal subgroup of $S$. We
shall use ATLAS notation for the names of simple groups. For example,
$\operatorname{O}^{+}_{8}(3)$ denotes the simple $8$-dimensional orthogonal
group of plus type defined over $\mathbb{F}_{3}$, while
$\mathrm{GO}^{+}_{8}(3)$ is the corresponding general orthogonal group.

######  Theorem 4.

Let $S$ be a non-abelian finite simple group, and if
$S\cong\operatorname{O}^{+}_{8}(q)$, then suppose that $q\notin\\{2,3,5\\}$.
Then $S$ contains an element $s$ such that:

  1. (i)

$N_{S}(\langle s\rangle)>C_{S}(s)$; and

  2. (ii)

for each almost simple group $G$ with socle $S$, there is at most one novelty
maximal in $\mathcal{M}^{\prime}_{G}(s)$, and no two subgroups in
$\mathcal{M}^{\prime}_{G}(s)$ are $G$-conjugate.

Even once we have proved that all finite groups satisfying the independence
property are supersoluble, it is not straightforward to classify these groups.
Indeed, the description of these groups is neither natural nor easy, as
evidenced by the following statement. Throughout, we denote the Frattini
subgroup of a group $G$ by $\operatorname{\Phi}(G)$.

######  Theorem 5.

A finite group $G$ satisfies the independence property if and only if one of
the following occurs:

  1. (i)

$G$ is a cyclic group of prime power order;

  2. (ii)

$G$ is the quaternion group $Q_{8}$ of order 8; or

  3. (iii)

$G\cong(V_{1}^{\delta_{1}}\times\dots\times{V_{r}^{\delta_{r}})}\rtimes H$,
where $H$ is abelian, $\delta_{1},\dots,\delta_{r}$ are positive integers for
some $r\geqslant 0$, $V_{1},\dots,V_{r}$ are irreducible $H$-modules on which
$H$ acts non-trivially, and the following statements hold. Here, for $h\in H$,
we write $I_{h}:=\\{i\in\\{1,\dots,r\\}\mid h\in C_{H}(V_{i})\\}$.

    1. (a)

If $\delta_{i}=1$, then $|H/C_{H}(V_{i})|$ is prime.

    2. (b)

If $|V_{i}|=|V_{j}|$, then $i=j$.

    3. (c)

$(|H|,|V_{i}|)=1$ for all $i\in\\{1,\ldots,r\\}.$

    4. (d)

For all $x,y\in H$, if $y\in\langle x\rangle\operatorname{\Phi}(H)$ and
$I_{x}\subseteq I_{y}$, then one of $x$ and $y$ is a power of the other. If in
addition $I_{x}\neq\varnothing$, then $y\in\langle x\rangle$.

In part (iii), notice that if $G$ is abelian, then $G=H$ and $r=0$, hence (d)
implies that $\operatorname{\Phi}(H)=1$.

The structure of the paper is as follows. Theorems 3 and 4 are proved in
Section 2. In Section 3, we prove that a finite group $G$ satisfying the
independence property is supersoluble. The structure of the finite
supersoluble groups satisfying the independence property is investigated in
Section 4, where Theorem 5 is proved. Finally, finite groups with the rank-
independence property are studied in Section 5, where we prove Theorems 1 and
2.

##  2\. Non-conjugate maximal subgroups of almost simple groups

In this section, we prove Theorem 3. To do so, we shall first prove several
elementary lemmas and show how Theorem 3 follows from Theorems 4 and 2.1. We
then work through the families of finite simple groups, proving Theorems 4 and
2.1: for most families, Theorem 4 follows easily from known results on
elements of $S$ that lie in few maximal subgroups. However, significantly more
work is required in the case where $S$ is orthogonal of plus type.

###  2.1. Preliminary results, and statement of Theorem 2.1

Recall the notation $\mathcal{M}_{G}(s)$ and $\mathcal{M^{\prime}}_{G}(s)$
from just before Theorems 3 and 4, respectively.

######  Theorem 2.1.

Let $S:=\operatorname{O}^{+}_{8}(q)$, with $q\in\\{2,3,5\\}$. Then $S$
contains an element $s$ such that, for any almost simple group $G$ with socle
$S$, the following statements hold.

  1. (i)

$\bigcap_{M\in\mathcal{M}_{G}(s)}M$ contains an element of $N_{S}(\langle
s\rangle)\setminus C_{S}(s)$.

  2. (ii)

If $G>S$, then there exists an $\alpha_{G}\in\operatorname{Aut}(S)$ such that
$\mathcal{M}^{\prime}_{G}(s^{\alpha_{G}})$ contains at most one novelty
maximal, and no two subgroups in $\mathcal{M}^{\prime}_{G}(s^{\alpha_{G}})$
are conjugate in $G$.

On the other hand, for each $r\in S$, the set $\mathcal{M}_{S}(r)$ contains
two $S$-conjugate subgroups, and there exists a group $R$ with
$S<R\leqslant\operatorname{Aut}(S)$ such that $\mathcal{M}^{\prime}_{R}(r)$
contains two $R$-conjugate subgroups.

This theorem suggests that the statements about the group $\Omega^{+}_{8}(5)$
from [30, p. 767] and [6, Lemma 5.15(b)] are not quite correct. We also note
that Theorem 4 implies that, when $S$ is as in that theorem, $G$-conjugate
subgroups in $\mathcal{M}_{G}(s)$ correspond to conjugate maximal subgroups of
$G/S$, and vice versa. On the other hand, the final part of Theorem 2.1 shows
that Theorem 4 does not hold when $S=\operatorname{O}^{+}_{8}(q)$ with
$q\in\\{2,3,5\\}$. In particular, there is no $s\in S$ such that we may set
$\alpha_{G}=1$ for all $G>S$.

We now state several elementary but useful results. The first of these follows
readily from the Orbit-Stabiliser Theorem, and is well known; for example, see
[30, Lemma 2.4] and the proof of [6, Lemma 5.9]. We will usually apply this
result in the special case where $H$ is a (non-normal) maximal subgroup of an
almost simple group $G$.

######  Lemma 2.2.

Let $H$ be a self-normalising subgroup of a finite group $J$, let $s\in H$,
and let $k$ be the number of $J$-conjugates of $H$ that contain $s$. Then the
following statements hold.

  1. (i)

$k\geqslant|N_{J}(\langle s\rangle):N_{H}(\langle s\rangle)|$, with equality
if all $J$-conjugates of $\langle s\rangle$ in $H$ are $H$-conjugate.

  2. (ii)

Suppose that $|C_{H}(f)|$ is constant for all $f\in s^{J}\cap H$, and let $r$
be the number of $H$-classes of elements in $s^{J}\cap H$. Then
$k=r|C_{J}(s):C_{H}(s)|$.

Our next result describes how novelty maximal subgroups of a finite almost
simple group $G$ may arise as elements of $\mathcal{M}^{\prime}_{G}(s)$, for
an element $s$ of the socle $S$ of $G$. Here, we adapt [6, Lemma 2.4(2)] and
its proof, which assume that $|G:S|$ is prime, to the general case.

######  Lemma 2.3.

Let $G$ be a finite almost simple group with socle $S$, and $s\in
S\setminus\\{1\\}$. Then each subgroup in $\mathcal{M}^{\prime}_{G}(s)$ is the
normaliser in $G$ of the intersection of one or more $G$-conjugate subgroups
in $\mathcal{M}_{S}(s)$. In particular,

$\bigcap_{M\in\mathcal{M}_{S}(s)}M\leqslant\bigcap_{M\in\mathcal{M}_{G}(s)}M.$

###### Proof.

Let $M\in\mathcal{M}^{\prime}_{G}(s)$. Then $M\cap S\leqslant L$ for some
$L\in\mathcal{M}_{S}(s)$. Since $M\cap S\trianglelefteq M$, we observe that
$s^{x^{-1}}\in M\cap S\leqslant L$ for each $x\in M$, and so $s\in L^{x}$.
Thus $X:=\bigcap_{x\in M}L^{x}$ is an intersection of $G$-conjugate subgroups
in $\mathcal{M}_{S}(s)$. It is clear that $M\leqslant N_{G}(X)<G$, and hence
$M=N_{G}(X)$. ∎

Notice that, for a given non-abelian finite simple group $S$, the final claim
in Lemma 2.3 shows that Theorem 3 holds for each finite almost simple group
$G$ with socle $S$ if and only if it holds in the special case $G=S$.

Next, observe from the Orbit-Stabiliser Theorem that if $L$ and $M$ are non-
conjugate maximal subgroups in $S$ such that $N_{G}(L)$ and $N_{G}(M)$ are
maximal in $G$, then these normalisers are not conjugate in $G$. Combining
this with Lemma 2.3 yields the following lemma.

######  Lemma 2.4.

Let $G$ be a finite almost simple group with socle $S$, and let $s$ be an
element of $S$ such that no two subgroups in $\mathcal{M}_{S}(s)$ are
$S$-conjugate. If the novelty maximals in $\mathcal{M}^{\prime}_{G}(s)$ are
pairwise non-conjugate in $G$, then all subgroups in
$\mathcal{M}^{\prime}_{G}(s)$ are pairwise non-conjugate in $G$. In
particular, if $G>S$ and $\mathcal{M}_{S}(s)$ contains at most two
$G$-conjugate subgroups, then Theorem 4(ii) holds for $G$ and $s$.

We finish this subsection by proving that Theorems 4 and 2.1 imply Theorem 3.
Note that we have included the statements about novelty maximals in the first
two of these theorems for general interest; they are not required to prove
Theorem 3.

###### Proof of Theorem 3 (assuming Theorems 4 and 2.1).

If $S\cong\operatorname{O}^{+}_{8}(q)$ with $q\in\\{2,3,5\\}$, then the result
is an immediate consequence of Theorem 2.1(i).

In the remaining cases, let $s$ be the element whose existence is guaranteed
by Theorem 4, so that there exists a suitable $x\in N_{S}(\langle
s\rangle)\setminus C_{S}(s)$. By Theorem 4(ii), no two subgroups in
$\mathcal{M}_{S}(s)$ are $S$-conjugate. Lemma 2.2(i), with $J=S$ and
$H=M\in\mathcal{M}_{S}(s)$, then shows that $N_{S}(\langle
s\rangle)=N_{M}(\langle s\rangle)$, and so

$x\in N_{S}(\langle s\rangle)\leqslant\bigcap_{M\in\mathcal{M}_{S}(s)}M.$

Finally, Lemma 2.3 shows that for each almost simple group $G$ with socle $S$,
the element $x$ lies in $\bigcap_{M\in\mathcal{M}_{G}(s)}M$. ∎

The remainder of this section is dedicated to proving Theorems 4 and 2.1. As
stated in our proofs, several of our arguments involve computations performed
via GAP [24] and Magma [3]. Our proofs also contain details about specific
elements $s$ that may be chosen.

###  2.2. Alternating, sporadic and exceptional groups

In this subsection, we prove Theorem 4 when $S$ is an alternating, sporadic or
exceptional group.

######  Proposition 2.5.

Theorem 4 holds when $S$ is an alternating group $\operatorname{A}_{n}$.

###### Proof.

Suppose first that $n\in\\{5,6,7,9\\}$, and let $s$ be any element of $S$ of
order $5$, $5$, $7$ or $15$, respectively. We deduce from the ATLAS [20] and
calculations using the GAP Character Table Library [5] that:

  1. (i)

the subgroups in $\mathcal{M}_{S}(s)$ lie in exactly one, two, two and three
$S$-conjugacy classes of subgroups, respectively; and

  2. (ii)

for each $M\in\mathcal{M}_{S}(s)$, all $S$-conjugates of $\langle s\rangle$ in
$M$ are $M$-conjugate, and $N_{M}(\langle s\rangle)=N_{S}(\langle
s\rangle)>C_{S}(s)$, so Theorem 4(i) holds for $s$.

Using Lemma 2.2(i), with $J=S$ and $H\in\mathcal{M}_{S}(s)$, we see from (ii)
that $k=1$, and so no two subgroups in $\mathcal{M}_{S}(s)$ are conjugate in
$S$. Moreover, if $|\mathcal{M}_{S}(s)|>2$, so that $n=9$, then the three
subgroups in $\mathcal{M}_{S}(s)$ are pairwise non-isomorphic, and hence
pairwise non-conjugate in $G$. Thus Theorem 4(ii) holds for $S$ by Lemma 2.4.

Next, suppose that $n\geqslant 8$ is even. Additionally, let

$s:=(1,\ldots,p)(p+1,\ldots,n)\in S,$

where $p$ is the largest prime less than $n-1$. First observe that $s^{2}$ and
$s^{-1}$ are both $\operatorname{S}_{n}$-conjugate to $s$, and so
$|N_{\operatorname{S}_{n}}(\langle s\rangle):\langle s\rangle|>2$. Thus
$N_{S}(\langle s\rangle)>\langle s\rangle=C_{S}(s)$, and Theorem 4(i) holds
for $S$. Next, Bertrand’s Postulate states that there exists a prime strictly
between $i:=n/2\geqslant 4$ and $2i-2$, so $(p,n-p)=1$. As observed in [31],
it follows that $|\mathcal{M}^{\prime}_{G}(s)|=1$. This is because the
$p$-cycle $s^{n-p}$ fixes at least $3$ points (since $n$ is even), and so [36]
(see also [56, Thm. 13.9]) implies that $\operatorname{A}_{n}$ and
$\operatorname{S}_{n}$ are the only primitive subgroups of
$\operatorname{S}_{n}$ containing $s^{n-p}$. It is also clear that an
imprimitive group preserving $j$ blocks of size $k$, with $jk=n$, cannot
contain an element of order $p>n/2\geqslant j,k$. Therefore,
$\mathcal{M}^{\prime}_{G}(s)=\\{(\operatorname{S}_{p}\times\operatorname{S}_{n-p})\cap
G\\}$, and Theorem 4(ii) holds for $S$.

Suppose now that $n\geqslant 11$ is odd. Here, we let

$s:=(1,\ldots,p)(p+1,p+2)(p+3,\ldots,n)\in S,$

where $p$ is the largest prime less than $n-5$. Applying Bertrand’s Postulate
to $(n-1)/2-1$ yields $p>(n-1)/2-1$. Hence $p\geqslant(n-1)/2$, which is
greater than each proper divisor of $n$. In particular, $n-(p+3)$ is coprime
to each of $2$, $p$ and $n-(p+2)$, and so $s^{n-(p+3)}$ and $s^{-1}$ are
$\operatorname{S}_{n}$-conjugate to $s$. Therefore, $N_{S}(\langle
s\rangle)>C_{S}(s)$. We also observe, similarly to the previous case, that
$\operatorname{A}_{n}$ is the only proper transitive subgroup of
$\operatorname{S}_{n}$ containing the $p$-cycle $s^{n-(p+2)}$. Thus
$\mathcal{M}^{\prime}_{G}(s)$ consists of three intransitive subgroups, of
shape $(\operatorname{S}_{p}\times\operatorname{S}_{n-p})\cap G$,
$(\operatorname{S}_{2}\times\operatorname{S}_{n-2})\cap G$ and
$(\operatorname{S}_{p+2}\times\operatorname{S}_{n-(p+2)})\cap G$,
respectively. Since $(n-1)/2\leqslant p<n-5$, these subgroups are pairwise
non-isomorphic, and hence Theorem 4 holds for $S$. ∎

We shall treat the Tits group ${}^{2}\mathrm{F}_{4}(2)^{\prime}$ as an
exceptional group.

######  Proposition 2.6.

Theorem 4 holds when $S$ is a sporadic group.

###### Proof.

For each sporadic group $S$, [9, Table 1] describes, for at least one element
$g\in S$, the corresponding set $\mathcal{M}_{S}(g)$ (much of this information
is also given in [30, Table IV]). Suppose first that
$S\notin\\{\mathrm{M}_{12},\mathrm{Suz}\\}$, and let $s$ be an element of $S$
described in [9, Table 1], with $|s|=17$ if $S=\mathrm{He}$, and $|s|=22$ if
$S=\mathrm{Fi}_{22}$. We see in [9, Table 1] that if $\mathcal{M}_{S}(s)$
contains two isomorphic subgroups, then $|\mathcal{M}_{S}(s)|=2$ and $\langle
s\rangle$ is a Sylow subgroup of $S$. Since at least one maximal subgroup in
$\mathcal{M}_{S}(s)$ contains $N_{S}(\langle s\rangle)$, and $\langle
s\rangle$ is a Sylow subgroup, it follows from Lemma 2.2(i) that if
$\mathcal{M}_{S}(s)$ contains two isomorphic subgroups, then they are not
$S$-conjugate.

If instead $S=\mathrm{M}_{12}$, then let $s$ be any element of $S$ of order
$11$, and if $S=\mathrm{Suz}$, then let $s$ be any element of order $21$.
Lemma 2.2(i) and character table calculations in GAP show that
$\mathcal{M}_{S}(s)$ contains exactly three subgroups, which are pairwise non-
conjugate in $S$. Furthermore, precisely two of these subgroups are isomorphic
when $S=\mathrm{M}_{12}$, and they are pairwise non-isomorphic when
$S=\mathrm{Suz}$, so at most two are $G$-conjugate in each case.

For all $S$, Lemma 2.4 now yields Theorem 4(ii), and we deduce Theorem 4(i)
from the ATLAS. ∎

######  Proposition 2.7.

Theorem 4 holds when $S$ is an exceptional group.

###### Proof.

Suppose first that $S$ is isomorphic to $\mathrm{G}_{2}(3)$,
$\mathrm{G}_{2}(4)$, $\mathrm{F}_{4}(2)$ or
${}^{2}\mathrm{F}_{4}(2)^{\prime}$. Then [9, p. 566] describes an element
$s\in S$ of order $13$, $21$, $17$ or $16$, respectively, that lies in exactly
three, one, two or two maximal subgroups of $S$, respectively. In each of
these cases, we deduce from the ATLAS that $N_{S}(\langle s\rangle)>C_{S}(s)$,
and from Lemma 2.2(i) and character table calculations in GAP that no two
maximal subgroups in $\mathcal{M}_{S}(s)$ are conjugate in $S$. Additionally,
if $S=\mathrm{G}_{2}(3)$, then we see in the ATLAS that
$\mathcal{M}^{\prime}_{G}(s)$ contains no novelty maximal. Thus Lemma 2.4
implies that Theorem 4 holds for $S$.

For each remaining finite exceptional simple $S$, Propositions 6.1 and 6.2 of
[30], building on the work of Weigel [55, §4], give a semisimple $s\in S$ such
that $|\mathcal{M}^{\prime}_{G}(s)|\leqslant 2$. In fact, if
$|\mathcal{M}^{\prime}_{G}(s)|=2$, then $S$ is equal to
$\mathrm{G}_{2}(3^{e})$ or $\mathrm{F}_{4}(2^{e})$ with $e\geqslant 2$,
$|\mathcal{M}_{S}(s)|=2$, and $\mathcal{M}^{\prime}_{G}(s)$ contains at most
one novelty maximal. Moreover, the two groups in $\mathcal{M}_{S}(s)$ are not
$S$-conjugate [55, pp. 74–78]. Note that when $S=\mathrm{G}_{2}(3^{e})$,
Weigel states only that the groups in $\mathcal{M}_{S}(s)$ are members of two
$S$-conjugacy classes, with each group having shape
$\operatorname{SU}_{3}(q).2$. We can use Lemma 2.2(i) and [4, Tables 8.5, 8.6
& 8.42] to show that the element $s$ of order $q^{2}-q+1$ lies in exactly one
member of each of these conjugacy classes. Thus in all cases, Theorem 4(ii)
follows from Lemma 2.4.

It remains to show that $N_{S}(\langle s\rangle)>C_{S}(s)$ in each case. If
$S$ is not isomorphic to $\mathrm{E}_{7}(2)$ or $\mathrm{G}_{2}(q)$, then this
is clear from [27, Table 6]. If instead
$S\in\\{\mathrm{E}_{7}(2),\mathrm{G}_{2}(q)\\}$, then $\langle s\rangle$ is a
maximal torus of $S$ [19, Tables 3, 7 & 10, p. 46]. As $S$ is the set of fixed
points under a Frobenius endomorphism of a simply connected algebraic group,
it follows from [22, pp. 2011–2012] that $C_{S}(s)=\langle s\rangle$. No non-
abelian finite simple group contains a self-normalising cyclic subgroup [25],
and so $N_{S}(\langle s\rangle)>C_{S}(s)$. ∎

###  2.3. Classical groups

This subsection consists of the proof of Theorem 4 when $S$ is classical, as
well as the proof of Theorem 2.1. Let $u:=2$ if $S$ is unitary and $u:=1$
otherwise, and let the natural module for $S$ be $\mathbb{F}_{q^{u}}^{n}$. We
also let $\tilde{S}$ be a classical quasisimple subgroup of
$\operatorname{GL}_{n}(q^{u})$ such that $\tilde{S}/Z(\tilde{S})\cong S$. For
each $h\in S$ and $H\leqslant S$, we will write $\tilde{h}$ to denote a
preimage of $h$ in $S$, and $\tilde{H}$ to denote the preimage of $H$
containing $Z(\tilde{S})$.

If a classical subgroup $\tilde{H}$ of $\operatorname{GL}_{n}(q^{u})$ contains
irreducible cyclic subgroups, then a generator $\tilde{y}$ of any such
subgroup of maximal order is a _Singer cycle_ of $H$, and the _Singer
subgroup_ $\langle\tilde{y}\rangle$ is equal to
$\tilde{H}\cap\langle\tilde{g}\rangle$ for some Singer cycle $\tilde{g}$ of
$\operatorname{GL}_{n}(q^{u})$ (see [2, 34, 35]). We will also say that $y$ is
a _Singer cycle_ of $H$.

In order to prove Theorem 4 when $S$ is a classical group, we will often
define a suitable $s$ via Singer cycles of subgroups of $S$. The following
lemma details important properties of such a Singer cycle.

######  Lemma 2.8.

Suppose that $S$ is listed in Table I, and let $\tilde{s}$ be a Singer cycle
of $\tilde{S}$. Then $|\tilde{s}|$ is given in the table. Moreover,
$N_{\tilde{S}}(\langle\tilde{s}\rangle)>C_{\tilde{S}}(\tilde{s})$ and
$N_{S}(\langle s\rangle)>C_{S}(s)$.

###### Proof.

The order of $\tilde{s}$ is given in [2, Table 1], and [49, p. 615] yields
$N_{\tilde{S}}(\langle\tilde{s}\rangle)>C_{\tilde{S}}(\tilde{s})$. Hence
either $N_{S}(\langle s\rangle)>C_{S}(s)$, or $\tilde{s}$ and
$\tilde{z}\tilde{s}$ are $\tilde{S}$-conjugate for some non-identity scalar
matrix $\tilde{z}\in\tilde{S}$. However, $\tilde{z}\tilde{s}$ and $\tilde{s}$
have different $\mathbb{F}_{q^{un}}$-eigenvalues on $\mathbb{F}_{q^{u}}^{n}$,
and so $N_{S}(\langle s\rangle)>C_{S}(s)$. ∎

Table I. The order of a Singer cycle $\tilde{s}\in\tilde{S}$, for certain $S$. $S$ | $|\tilde{s}|$
---|---
$\operatorname{L}_{n}(q)$ | $(q^{n}-1)/(q-1)$
$\operatorname{U}_{n}(q)$, $n$ odd | $(q^{n}+1)/(q+1)$
$\operatorname{S}_{n}(q)$ | $q^{n/2}+1$
$\operatorname{O}^{-}_{n}(q)$ | $(q^{n/2}+1)/(2,q-1)$

Note that each finite simple classical group that contains a Singer cycle
(i.e., that contains an irreducible cyclic subgroup) appears in Table I (see
[2, p. 188]).

The arguments later in this subsection do not apply to certain small classical
groups. We therefore deal with these groups separately.

######  Proposition 2.9.

Theorem 4 holds for each classical group $S$ given in Table II.

Table II. The size of $\mathcal{M}_{S}(s)$, where $S$ is a given classical
simple group and $s$ is any element of $S$ of the given order.

$S$ | $|s|$ | $|\mathcal{M}_{S}(s)|$
---|---|---
$\operatorname{L}_{3}(4)$ | $7$ | $3$
$\operatorname{L}_{6}(2)$ | $63$ | $2$
$\operatorname{U}_{3}(3)$ | $7$ | $1$
$\operatorname{U}_{3}(5)$ | $10$ | $2$
$\operatorname{U}_{4}(2)$ | $9$ | $2$
$\operatorname{U}_{4}(3)$ | $7$ | $7$
(a)

$S$ | $|s|$ | $|\mathcal{M}_{S}(s)|$
---|---|---
$\operatorname{U}_{5}(2)$ | $11$ | $1$
$\operatorname{U}_{6}(2)$ | $11$ | $4$
$\operatorname{S}_{6}(2)$ | $15$ | $2$
$\operatorname{S}_{8}(2)$ | $17$ | $3$
$\operatorname{O}_{7}(3)$ | $14$ | $3$
$\operatorname{O}^{+}_{8}(4)$ | $65$ | $3$
(b)

###### Proof.

Let $s$ be any element of $S$ such that $|s|$ is as in Table II. We can use
Lemma 2.2(i), together with computations in Magma when
$S=\operatorname{O}^{+}_{8}(4)$, or character table calculations in GAP in the
remaining cases, to determine the set $\mathcal{M}_{S}(s)$ (in certain cases,
$\mathcal{M}_{S}(s)$, or its size, is also given in [6, §4], the proof of [30,
Prop. 6.3], or the proof of [9, Thm. 6.1]). In particular,
$|\mathcal{M}_{S}(s)|$ has the value given in Table II, and no two subgroups
in $\mathcal{M}_{S}(s)$ are $S$-conjugate.

To prove Part (ii), we first note that if $S=\operatorname{S}_{8}(2)$, then no
two subgroups in $\mathcal{M}_{S}(s)$ are isomorphic, and if
$S=\operatorname{O}_{7}(3)$, then the three subgroups in $\mathcal{M}_{S}(s)$
lie in two isomorphism classes. If instead
$S\in\\{\operatorname{L}_{3}(4),\operatorname{O}^{+}_{8}(4)\\}$, then we can
show using Magma that the intersection of any two subgroups in
$\mathcal{M}_{S}(s)$ is equal to the intersection of all three. When
$|\mathcal{M}_{S}(s)|\leqslant 3$, it now follows from Lemma 2.3 that
$\mathcal{M}^{\prime}_{G}(s)$ contains at most one novelty maximal. In
addition, we observe from the ATLAS that if $|\mathcal{M}_{S}(s)|>3$ (that is,
if $S\in\\{\operatorname{U}_{4}(3),\operatorname{U}_{6}(2)\\}$), then
$\mathcal{M}^{\prime}_{G}(s)$ contains no novelty maximals. Thus, in each
case, Lemma 2.4 implies that no two subgroups in $\mathcal{M}^{\prime}_{G}(s)$
are $G$-conjugate.

For Part (i), we deduce that $N_{S}(\langle s\rangle)>C_{S}(s)$, using Lemma
2.8 when $s$ is a Singer cycle of $S$, using Magma when
$S=\operatorname{O}^{+}_{8}(4)\cong\Omega^{+}_{8}(4)$, and using the ATLAS in
the remaining cases. ∎

In order to prove that Theorem 4 holds in the remaining cases where $S$
contains a Singer cycle, we require the following elementary observation.
Throughout the rest of this section, we shall abbreviate _primitive prime
divisor_ to _ppd_.

######  Lemma 2.10.

Let $p$ be a prime, $f$ a positive integer, and $r$ a ppd of $p^{f}-1$. Then
$r$ does not divide $f$.

###### Proof.

Suppose, for a contradiction, that $r\mid f$. Then there exists a positive
integer $k<f$ such that $p^{f}=(p^{k})^{r}\equiv p^{k}\pmod{r}$. Thus
$p^{f}-1\equiv p^{k}-1\pmod{r}$. As $r$ divides $p^{f}-1$, it therefore also
divides $p^{k}-1$. This contradicts the primitivity of $r$ as a prime divisor
of $p^{f}-1$. ∎

######  Proposition 2.11.

Theorem 4 holds when $S$ is linear, unitary of odd dimension, symplectic, or
orthogonal of minus type with $n\geqslant 8$.

###### Proof.

We may assume that $S$ has not been dealt with in Proposition 2.9, and by
Proposition 2.5 that $S$ is not isomorphic to an alternating group. We will
also assume that $n\geqslant 4$ in the symplectic case, and that $q\neq 7$ if
$n=2$; we will treat $\operatorname{L}_{2}(7)$ as $\operatorname{L}_{3}(2)$.

Let $s$ be a Singer cycle of $S$. Then $N_{S}(\langle s\rangle)>C_{S}(s)$ by
Lemma 2.8. Additionally, the main theorem of [2] shows that if
$\mathcal{M}_{S}(s)$ contains a maximal subgroup that is not an extension
field type subgroup, then $S=\operatorname{S}_{n}(q)$ with $q$ even. Moreover,
if there is more than one such maximal subgroup, then there are exactly two,
they are not conjugate in $S$, and $n=4$ [44, Thm. 1.1, p. 94].

If $S=\operatorname{L}_{2}(q)$, then our exclusions on $q$ imply that
$\mathcal{M}_{S}(s)=\\{N_{S}(\langle s\rangle)\\}$. Otherwise, we observe from
Lemma 2.8 and Zsigmondy’s Theorem (since we exclude the groups in Proposition
2.9) that $|s|$ is divisible by a ppd $r$ of $q^{un}-1$. In addition, by Lemma
2.10, $r\nmid un$. Hence [6, Lemma 2.12] implies that no two extension field
subgroups in $\mathcal{M}_{S}(s)$ are isomorphic. The result now follows from
Lemma 2.4. ∎

######  Proposition 2.12.

Theorem 4 holds when $S$ is unitary of even dimension or orthogonal of odd
dimension.

###### Proof.

We may assume that $S$ has not been dealt with in Proposition 2.9, and by
Proposition 2.11 that $n\geqslant 4$ in the unitary case and $n\geqslant 7$
otherwise.

Let $\tilde{M}$ be the stabiliser in $\tilde{S}$ of a non-degenerate
$(n-1)$-dimensional subspace of $\mathbb{F}_{q^{u}}^{n}$, of minus type if $S$
is orthogonal. Then $\tilde{M}$ contains a subgroup $H$ isomorphic to
$\operatorname{SU}_{n-1}(q)$ or $\Omega^{-}_{n-1}(q)$. Let $\tilde{s}$ be a
Singer cycle of $H$ (with respect to its action on
$\mathbb{F}_{q^{u}}^{n-1}$), and let $\tilde{K}:=Z(\tilde{S})H$. Then Lemma
2.8 (or an easy generalisation of this result in the case
$Z(\tilde{S})\not\leqslant H$) yields $N_{K}(\langle s\rangle)>C_{K}(s)$, and
hence $N_{S}(\langle s\rangle)>C_{S}(s)$. Furthermore,
$\mathcal{M}_{S}(s)=\\{M\\}$ by [44, Thm. 1.1, p. 93], and so the result
follows by Lemma 2.4. ∎

The remainder of this section concerns the case
$S=\operatorname{O}^{+}_{2m}(q)$, where without loss of generality $m\geqslant
4$, so that $\tilde{S}=\Omega^{+}_{2m}(q)$. The case $m$ odd is
straightforward.

######  Proposition 2.13.

Theorem 4 holds when $S=\operatorname{O}^{+}_{2m}(q)$, with $m\geqslant 5$
odd.

###### Proof.

Choose $\tilde{s}$ to be the element of order
$(q^{(m-1)/2}+1)(q^{(m+1)/2}+1)/(4,q-1)$ of $\tilde{S}$ described in
Proposition 5.13 of [6], so that $\tilde{s}$ is a product of two Singer cycles
of orthogonal subgroups of $\tilde{S}$, unless $q\equiv 3\pmod{4}$. Then [6,
Proposition 5.13] shows that $\tilde{s}$ lies in a unique maximal subgroup of
$\tilde{S}$, namely, the stabiliser of an $(m-1)$-dimensional subspace of
$\mathbb{F}_{q}^{2m}$ of minus type. Hence $|\mathcal{M}_{S}(s)|=1$, and Part
(ii) follows from Lemma 2.4.

It remains to show that $N_{S}(\langle s\rangle)>C_{S}(s)$. The set of
$\overline{\mathbb{F}_{q}}$-eigenvalues of $\tilde{s}$ is a subset of
$\mathbb{F}_{q^{(m-1)(m+1)/2}}$ that is closed under conjugation by the field
automorphism $\alpha\mapsto\alpha^{q}$. Thus the distinct elements
$\tilde{s}$, $\tilde{s}^{q}$ and $\tilde{s}^{q^{2}}$ all have the same
eigenvalues. Note also that $\langle\tilde{s}\rangle$ stabilises no one-
dimensional subspace of $\mathbb{F}_{q}^{2m}$; otherwise, $\mathcal{M}_{S}(s)$
would contain the stabiliser of such a subspace. Thus it follows from [54, pp.
38–39] (see also [6, Prop. 2.11] for an explicit statement) and [21, Thms
6.1.12 & 6.1.15] that $\tilde{s}$ is $\tilde{S}$-conjugate to at least one of
$\tilde{s}^{q}$ and $\tilde{s}^{q^{2}}$. As $\tilde{s}$ and $-\tilde{s}$ do
not have equal eigenvalues when $q$ is odd, we deduce that $N_{S}(\langle
s\rangle)>C_{S}(s)$. ∎

The case where $m$ is even is much more involved.

###### Proof of Theorem 2.1.

As specified below, many of the facts about almost simple groups mentioned in
this proof were deduced using Magma. In the case $q=5$, most of our
computations were performed in the matrix group $\tilde{S}=\Omega^{+}_{8}(5)$,
since this group’s maximal subgroups can be constructed using the
ClassicalMaximals function. In general, in order to show that an element $r\in
S$ lies in multiple $G$-conjugates of a given maximal subgroup
$M\in\mathcal{M}^{\prime}_{G}(r)$, it suffices by Lemma 2.2(i) to verify that
$N_{G}(\langle r\rangle)>N_{M}(\langle r\rangle)$.

The majority of the proof will be divided into two cases, depending on $q$.
However, we first observe computationally that for the element $s$ specified
in each case below, $\bigcap_{M\in\mathcal{M}_{S}(s)}M$ contains an element of
$N_{S}(\langle s\rangle)\setminus C_{S}(s)$, and hence by Lemma 2.3, so does
$\bigcap_{M\in\mathcal{M}_{G}(s)}M$. Computations also show that for each
$r\in S$, there exist two $S$-conjugate subgroups in $\mathcal{M}_{S}(r)$.
Therefore it remains to prove Part (ii), and the rest of the final claim of
the theorem.

Case (a): $q=2$. Let $s$ be any element of $S$ of order $15$. If
$|G:S|\geqslant 3$, then we observe from the ATLAS that, up to $G$-conjugacy,
$G$ has a unique maximal subgroup of order divisible by $|s|$ that does not
contain $S$. In fact, we see using Magma that this maximal subgroup has no
element of order $|s|$, and so $\mathcal{M}^{\prime}_{G}(s)=\varnothing$.

If instead $|G:S|=2$, so that $G\cong\mathrm{SO}^{+}_{8}(2)$, then no novelty
maximal of $G$ has order divisible by $|s|$ . Furthermore, computations show
that there exist $\alpha,\beta\in\mathrm{Aut}(S)$ such that
$\mathcal{M}^{\prime}_{G}(s^{\alpha})=\varnothing$, while two subgroups in
$\mathcal{M}^{\prime}_{G}(s^{\beta})$ are conjugate in $G$. We also observe
via Magma that for each $h\in S$ that is not $G$-conjugate to $s^{\alpha}$,
there exist two $G$-conjugate subgroups in $\mathcal{M}^{\prime}_{G}(h)$.
Hence for each $r\in S$, there exists an $\mathrm{Aut}(S)$-conjugate $R$ of
$G$ such that $\mathcal{M}^{\prime}_{R}(r)$ contains two $R$-conjugate
subgroups.

Case (b): $q\in\\{3,5\\}$. Magma computations show that $S$ has precisely
three conjugacy classes of cyclic subgroups of order $(q^{3}+1)/2$, and for
any element $s$ of this order, the subgroups in $\mathcal{M}_{S}(s)$
isomorphic to $\Omega_{7}(q)$ are members of two $S$-conjugacy classes (with
these classes depending on the class of $\langle s\rangle$). Moreover, the
maximal subgroups of $S$ isomorphic to $\Omega_{7}(q)$ fall into six
$S$-conjugacy classes, and only one $\mathrm{Aut}(S)$-conjugacy class [39,
Table I]. We therefore deduce that there exist
$\theta_{1},\theta_{2}\in\mathrm{Aut}(S)$ such that $\langle s\rangle$,
$\langle s^{\theta_{1}}\rangle$ and $\langle s^{\theta_{2}}\rangle$ lie in
three distinct $S$-classes. Additional computations show that for each
$t\in\\{s,s^{\theta_{1}},s^{\theta_{2}}\\}$, the set $\mathcal{M}_{S}(t)$
contains precisely three subgroups that are not isomorphic to $\Omega_{7}(q)$:
the stabiliser $K_{t,1}$ in $S$ of a $6$-dimensional subspace of
$\mathbb{F}_{q}^{8}$ of minus type, and two of its images $K_{t,2}$ and
$K_{t,3}$ under triality automorphisms of $S$ (see [39, Table I]). In
particular, these three subgroups are pairwise non-conjugate in $S$.

Next, we observe using [4, Table 8.50] that if $M$ is a novelty maximal
subgroup of $G$ whose order is divisible by $|s|$, then either $M\cap
S\cong\mathrm{G}_{2}(q)$; $M\cap S$ is an extension of a $2$-group by
$\operatorname{A}_{8}$ or by $\operatorname{L}_{3}(2)$; or $M\cap S$ is
$S$-conjugate to $U_{t}:=K_{t,1}\cap K_{t,2}\cap K_{t,3}$. However, we see in
the ATLAS that $\mathrm{G}_{2}(q)$ contains no element of order $|s|$, and no
maximal subgroup in $\mathcal{M}_{S}(t)$ has order divisible by the order of
any of these extensions of $2$-groups. It therefore follows from Lemma 2.3
that any novelty maximal subgroup in $\mathcal{M}^{\prime}_{G}(t)$ is equal to
$N_{G}(U_{t})$.

We now conclude from [39, Table I] that, for each $G>S$, there exists
$\alpha_{G}\in\\{1,\theta_{1},\theta_{2}\\}$ such that, when
$t=s^{\alpha_{G}}$, any subgroup in $\mathcal{M}^{\prime}_{G}(t)$ is equal to
either $N_{G}(U_{t})$ or $N_{G}(K_{t,i})$ for some $i\in\\{1,2,3\\}$.
Moreover, since $K_{t,i}$ and $K_{t,j}$ are not $S$-conjugate when $i\neq j$,
if $N_{G}(K_{t,i})$ and $N_{G}(K_{t,j})$ are maximal in $G$ then they are not
$G$-conjugate. Therefore, no two subgroups in $\mathcal{M}^{\prime}_{G}(t)$
are $G$-conjugate.

Finally, let $r\in S$, and suppose that $G\cong\mathrm{PSO}^{+}_{8}(q)$ and
that no two subgroups in $\mathcal{M}^{\prime}_{G}(r)$ are $G$-conjugate.
Using Magma and [39, Table I], we deduce that either there exists
$\rho\in\mathrm{Aut}(S)$ such that $\mathcal{M}^{\prime}_{G^{\rho}}(r)$
contains two $G^{\rho}$-conjugate subgroups; or $q=5$, and there exists a
graph automorphism $\pi$ of $S$ (of order $2$ or $3$) such that
$\mathcal{M}^{\prime}_{\langle S,\pi\rangle}(r)$ contains two $\langle
S,\pi\rangle$-conjugate subgroups. Therefore, for each $r\in S$, there exists
a group $R$ with $S<R\leqslant\mathrm{Aut}(S)$ such that
$\mathcal{M}^{\prime}_{R}(r)$ contains two $R$-conjugate subgroups. ∎

Now, to prove Theorem 4 for $m$ even, we would like to use Proposition 5.14
and Lemma 5.15 of [6]. However, we require a slightly stronger statement than
Proposition 5.14. Additionally, the statement of Proposition 5.14 is not quite
correct when $4\mid m$, nor is the proof of Lemma 5.15(b)–(c). We therefore
prove the following result, much of whose proof is similar to [6, §5], and
which implies that [6, Lemma 5.15(b)] is in fact correct when $q\geqslant 7$.
We exclude the groups $\operatorname{O}^{+}_{8}(q)$ for $q\leqslant 5$, which
we dealt with in Theorem 2.1 and Proposition 2.9.

######  Theorem 2.14.

Suppose that $S=\operatorname{O}^{+}_{2m}(q)$, with $m\geqslant 4$ even, and
$q\geqslant 7$ if $m=4$. Then $\tilde{S}$ contains an element $\tilde{s}$ of
order $k=k(m,q)$, where

$k:=\begin{cases}(q^{2}+1)/(2,q-1),&\text{ if }m=4,\\\
(q^{(m-2)/2}+1)(q^{(m+2)/2}+1)/(4,q-1),&\text{ if }4\nmid m,\\\
(q^{(m-2)/2}+1)(q^{(m+2)/2}+1)/(q+1),&\text{ if }m=8\text{ and }q=2,\\\
(q^{(m-2)/2}+1)(q^{(m+2)/2}+1)/(q+1)^{2},&\text{ otherwise},\end{cases}$

such that the following statements hold.

  1. (i)

There are precisely three subgroups in $\mathcal{M}_{\tilde{S}}(\tilde{s})$,
which we will denote by $\tilde{K}_{1}$, $\tilde{K}_{2}$ and $\tilde{L}$.

  2. (ii)

The groups $\tilde{K}_{1}$ and $\tilde{K}_{2}$ are extension field subgroups
that are not conjugate in $\tilde{S}$. If $4<m\equiv 0\pmod{4}$, then
$\tilde{K}_{1}$ and $\tilde{K}_{2}$ are of (Aschbacher) type
$\mathrm{GU}_{m}(q)$. Otherwise, they are of type
$\mathrm{GO}^{+}_{m}(q^{2})$.

  3. (iii)

Let $V:=\mathbb{F}_{q}^{2m}$. If $m=4$, then $\tilde{L}$ is imprimitive,
stabilising the decomposition $V=U\perp U^{\perp}$, for some four-dimensional
subspace $U$ of $V$ of minus type. Otherwise, $\tilde{L}$ is the stabiliser of
an $(m-2)$-dimensional subspace of $V$ of minus type.

Note that the case $m=8$ and $q=2$ is exceptional as here $q^{m-2}-1=2^{6}-1$
has no ppd. We begin by defining the element $\tilde{s}$ that we will use in
each case.

######  Assumption 2.15.

Suppose that $S=\operatorname{O}^{+}_{2m}(q)$, with $m\geqslant 4$ even, and
$q\geqslant 7$ if $m=4$. Additionally, let $k=k(m,q)$ be as in Theorem 2.14.
If $4\nmid m$, then let $\tilde{s}\in\tilde{S}$ be as described in [6, Prop.
5.14], so that $\langle\tilde{s}\rangle$ has order $k$ and stabilises exactly
two proper non-zero subspaces of $V:=\mathbb{F}_{q}^{2m}$, namely, an
$(m-2)$-dimensional subspace of $V$ of minus type and its orthogonal
complement.

From now on, suppose that $4\mid m$. Let $\tilde{X}$ be the stabiliser in
$\tilde{S}$ of a subspace $U$ of $V$ of minus type, of dimension four if $m=4$
or dimension $m-2$ if $m>4$. Then $U^{\perp}$ is also of minus type.
Additionally, $\tilde{X}$ contains the subgroup $R_{1}\times R_{2}$, where
$R_{1}\cong\Omega(U)$ and $R_{2}\cong\Omega(U^{\perp})$.

We define $\tilde{r}_{1}$ and $\tilde{r}_{2}$ to be elements of Singer
subgroups of $R_{1}$ and $R_{2}$, respectively (corresponding to their actions
on $U$ and $U^{\perp}$), as follows; in each case, an element of the specified
order exists by Lemma 2.8. If $m=4$, then let $\tilde{r}_{1}$ and
$\tilde{r}_{2}$ both have order $k$. Then $\tilde{r}_{1}$ has eigenvalues
$\alpha$, $\alpha^{q}$, $\alpha^{q^{2}}$ and $\alpha^{q^{3}}$ on $U$, while
$\tilde{r}_{2}$ has eigenvalues $\beta$, $\beta^{q}$, $\beta^{q^{2}}$ and
$\beta^{q^{3}}$ on $U^{\perp}$, where $\alpha$ and $\beta$ are elements of
$\mathbb{F}_{q^{4}}$ of order $k$. Similarly to the proof of [6, Lemma 5.15],
we choose $\tilde{r}_{2}$ in this case so that these eight eigenvalues are all
distinct (we will place further restrictions on $\tilde{r}_{2}$ in the case
$m=4$ in the proof of Proposition 2.21). If instead $m>4$, then let
$\tilde{r}_{1}$ and $\tilde{r}_{2}$ have order $(q^{(m-2)/2}+1)/t$ and
$(q^{(m+2)/2}+1)/(q+1)$, respectively, where $t:=1$ if $m=8$ and $q=2$, and
$t:=q+1$ in all remaining cases. For all $m$ (with $4\mid m$), let
$\tilde{s}:=(\tilde{r}_{1},\tilde{r}_{2})\in\tilde{X}$. We deduce that
$|\tilde{s}|=k$.

Next, we show that when $4\mid m$, the group $\tilde{L}$ from Theorem 2.14 is
the unique subgroup in $\mathcal{M}_{\tilde{S}}(\tilde{s})$ that is not
isomorphic to the specified extension field group $\tilde{K}_{1}$. We first
state two preliminary results.

######  Lemma 2.16.

Let $\tilde{S}$, $U$ and $\tilde{s}$ be as in Assumption 2.15, with $4\mid m$.
Then $U$ and $U^{\perp}$ are the only proper non-zero subspaces of
$\mathbb{F}_{q}^{n}$ stabilised by $\langle\tilde{s}\rangle$.

###### Proof.

If $m=8$ and $q=2$, then the element $\tilde{r}_{1}$ from Assumption 2.15 is a
Singer cycle of $\Omega^{-}_{6}(2)$ by Lemma 2.8, and hence
$\langle\tilde{r}_{1}\rangle$ acts irreducibly on $U$. In the remaining cases,
$|\tilde{r}_{1}|$ is divisible by each ppd of $q^{\dim(U)}-1$, and in all
cases, $|\tilde{r}_{2}|$ is divisible by each ppd of $q^{\dim(U^{\perp})}-1$.
We therefore deduce from [33, Thm. 3.5] that
$\tilde{s}=(\tilde{r}_{1},\tilde{r}_{2})$ acts irreducibly on each of $U$ and
$U^{\perp}$.

Now, if $m=4$, then the eigenvalues of $\tilde{s}$ on $U$ are distinct from
its eigenvalues on $U^{\perp}$, and otherwise, $\dim(U)\neq\dim(U^{\perp})$.
Thus $U$ and $U^{\perp}$ are non-isomorphic irreducible
$\mathbb{F}_{q}[\langle\tilde{s}\rangle]$-modules, and the result follows. ∎

######  Lemma 2.17.

Let $m\geqslant 8$ be a multiple of $4$, and suppose that $m+3$ is a ppd of
$q^{m+2}-1$. Then the symmetric group $\operatorname{S}_{2m+2}$ contains no
element of order $k=k(m,q)$, as defined in Theorem 2.14.

###### Proof.

If $q=2$ and $m=8$, then $\operatorname{S}_{2m+2}=\operatorname{S}_{18}$
contains no element of order $k=99$. Assume therefore that $q>2$ or $m>8$, and
suppose for a contradiction that $y$ is an element of
$\operatorname{S}_{2m+2}$ of order $k$. Observe that $k$ is divisible by each
ppd of $q^{m+2}-1$, and hence by $m+3$. Similarly, $k$ is divisible by each
ppd of $q^{m-2}-1$, and such a prime is no smaller than $m-1$ (see [29, Remark
1.1]). As $(m-1)+(m+3)$ is the degree of $\operatorname{S}_{2m+2}$, we deduce
that $k$ is the product $(m-1)(m+3)$ of two primes. Now,
$(q^{(m-2)/2}+1,q^{(m+2)/2}+1)=q+1$, and it follows that
$m-1=(q^{(m-2)/2}+1)/(q+1)$ and $m+3=(q^{(m+2)/2}+1)/(q+1)$. Substituting
these equalities into $(m-1)+4=m+3$, we deduce that $4=q^{(m-2)/2}(q-1)$, a
contradiction for all $q$ and $m$. ∎

######  Proposition 2.18.

Let $\tilde{S}$, $U$ and $\tilde{s}$ be as in Assumption 2.15, with
$m\geqslant 8$ and $4\mid m$. Then the stabiliser of $U$ in $\tilde{S}$ is the
unique subgroup in $\mathcal{M}_{\tilde{S}}(\tilde{s})$ that is not an
extension field subgroup of type $\mathrm{GU}_{m}(q)$.

###### Proof.

Notice that $|\tilde{s}|$ is divisible by each ppd of $q^{m+2}-1$. The main
theorem of [29] yields all possibilities for maximal subgroups of $\tilde{S}$
whose orders are divisible by such a prime. We deduce, with the aid of [38,
Table 3.5.E] and [4, Table 2.5], that if $\mathcal{M}_{\tilde{S}}(\tilde{s})$
contains a subgroup $\tilde{M}$ that is not reducible or an extension field
subgroup, then $m+3$ is the unique ppd of $q^{m+2}-1$, and $\tilde{M}$ is an
extension of a (possibly trivial) $2$-group by an alternating or symmetric
group of degree at most $2m+2$. As $|\tilde{s}|$ is odd,
$\operatorname{S}_{2m+2}$ must then contain an element of order $|\tilde{s}|$.
However, Lemma 2.17 shows that this is not the case. Therefore, each subgroup
in $\mathcal{M}_{\tilde{S}}(\tilde{s})$ is reducible or an extension field
subgroup.

It is clear from Lemma 2.16 that the stabiliser of $U$ in $\tilde{S}$ is the
unique reducible subgroup in $\mathcal{M}_{\tilde{S}}(\tilde{s})$. Thus it
remains to show that $\mathcal{M}_{\tilde{S}}(\tilde{s})$ contains no
extension field subgroup $\tilde{Y}$ that is not of type $\mathrm{GU}_{m}(q)$.
By [29], any such $\tilde{Y}$ whose order is divisible by a ppd of $q^{m+2}-1$
is defined over $q^{2}$, and so $\tilde{Y}$ is an extension of
$\Omega^{+}_{m}(q^{2})$ by a group of order $4$ [4, Table 2.6]. The
$p^{\prime}$-part of the order of $\Omega^{+}_{m}(q^{2})$ divides
$(q^{m}-1)\prod_{i=1}^{m/2-1}(q^{4i}-1)$, and for each positive integer
$f\in\\{m+3,\ldots,2m-4\\}$, no ppd of $q^{m+2}-1$ divides $q^{f}-1$.We
conclude that $|\tilde{Y}|$ is not divisible by any ppd of $q^{m+2}-1$, and
therefore $\tilde{Y}\notin\mathcal{M}_{\tilde{S}}(\tilde{s})$. ∎

In order to prove a similar proposition when $m=4$, we consider maximal tori
of $\tilde{S}$ (or $\mathrm{SO}^{+}_{2m}(q)$ if $q$ is odd) that contain
$\tilde{s}$, via a brief discussion of related algebraic groups. We in fact
allow $m$ to be any even integer at least $4$, as the results here will also
be useful when considering extension field subgroups in
$\mathcal{M}_{\tilde{S}}(\tilde{s})$ in each case.

Let $\underline{S}$ be the algebraic group
$\mathrm{SO}_{2m}(\overline{\mathbb{F}_{q}})$ for some $m\geqslant 4$. There
exists a Frobenius endomorphism $\sigma$ of $\underline{S}$ such that the
subgroup $\hat{S}:=\underline{S}_{\sigma}$ of fixed points of $\underline{S}$
under $\sigma$ is equal to $\mathrm{SO}^{+}_{2m}(q)$ if $q$ is odd, or
$\Omega^{+}_{2m}(q)$ if $q$ is even (see [45, pp. 193–194], where
$\Omega^{+}_{2m}(q)$ is written as $\mathrm{SO}^{+}_{2m}(q)$ when $q$ is
even). By definition, each maximal torus $\hat{T}$ of $\hat{S}$ is equal to
$\underline{T}\cap\hat{S}$ for a corresponding maximal torus $\underline{T}$
of $\underline{S}$. An element $\tilde{r}$ of $\hat{T}$ is called _regular_ if
the dimension of $C_{\underline{S}}(\tilde{r})$ is no larger than the
dimension of $C_{\underline{S}}(\tilde{x})$ for any
$\tilde{x}\in\underline{S}$, or equivalently, if $\underline{T}$ is equal to
$C_{\underline{S}}(\tilde{r})^{\circ}$, the connected component of
$C_{\underline{S}}(\tilde{r})$ containing the identity [45, Defn. 14.8, Cor.
14.10]. We will continue to use this notation in the following two results and
their proofs.

######  Lemma 2.19.

Let $\tilde{r}$ be a regular element of a maximal torus $\hat{T}$ of
$\hat{S}$. If $|\tilde{r}|$ is odd, then $C_{\hat{S}}(\tilde{r})=\hat{T}$.

###### Proof.

As $\tilde{r}$ is regular,
$\underline{T}=C_{\underline{S}}(\tilde{r})^{\circ}$. In addition, it follows
from [45, pp. 70–72, Prop. 14.20] that the order of
$C_{\underline{S}}(\tilde{r})/C_{\underline{S}}(\tilde{r})^{\circ}$ divides
$(2,q-1)$, and its exponent divides $|\tilde{r}|$. Hence if $|\tilde{r}|$ is
odd, then $C_{\underline{S}}(\tilde{r})=\underline{T}$, and so
$C_{\hat{S}}(\tilde{r})=\underline{T}\cap\hat{S}=\hat{T}$. ∎

######  Lemma 2.20.

Let $\tilde{S}$ and $\tilde{s}$ be as in Assumption 2.15. Then the centraliser
$C_{\hat{S}}(\tilde{s})=C_{\mathrm{GO}^{+}_{2m}(q)}(\tilde{s})$ is a maximal
torus of $\hat{S}$ of shape $(q^{a}+1)\times(q^{b}+1)$, where $a+b=m$, with
$a:=2$ if $m=4$ and $a:=(m-2)/2$ otherwise.

###### Proof.

As $\tilde{s}$ is semisimple, it lies in a maximal torus $\hat{T}$ of
$\hat{S}$. By considering the maximal tori of $\hat{S}$ (see [11, §4–5], [53,
§9] and [51, Lemma 2]), we deduce that if $\hat{T}$ does not have shape
$(q^{a}+1)\times(q^{b}+1)$, then $m=4$ and $\hat{T}$ is a cyclic group of
order $q^{4}-1$ that stabilises a totally singular four-dimensional subspace
of $\mathbb{F}_{q}^{8}$. Lemma 2.16 shows that this second case cannot occur.

Recall that when $m=4$, the elements $\tilde{r}_{1}$ and $\tilde{r}_{2}$ from
Assumption 2.15 have distinct eigenvalues on $\mathbb{F}_{q}^{8}$.
Additionally, even though $q^{2a}-1$ has no ppd when $m=8$ and $q=2$, the
element $\tilde{r}_{1}$ in this case has order $9$, and no element of this
order lies in a proper subfield of $\mathbb{F}_{q^{2a}}=\mathbb{F}_{2^{6}}$.
We therefore deduce from [53, §9] in each case that $\tilde{s}^{2}$ is a
regular element of $\hat{T}$ (as is $\tilde{s}$). Since
$C_{\hat{S}}(\tilde{s})\leqslant C_{\hat{S}}(\tilde{s}^{2})$ and
$|\tilde{s}^{2}|$ is odd, we obtain $C_{\hat{S}}(\tilde{s})=\hat{T}$ from
Lemma 2.19.

Finally, it follows from Lemma 2.16 if $4\mid m$, or from the definition of
$\tilde{s}$ in Assumption 2.15 if $4\nmid m$, that $\langle\tilde{s}\rangle$
stabilises no one-dimensional subspace of $\mathbb{F}_{q}^{n}$. Hence
$\tilde{s}$ has no linear elementary divisor, and so [21, p. 85–86] yields
$C_{\hat{S}}(\tilde{s})=C_{\mathrm{GO}^{+}_{2m}(q)}(\tilde{s})$. ∎

We can now prove a version of Proposition 2.18 with $m=4$.

######  Proposition 2.21.

Let $\tilde{S}$ and $U$ be as in Assumption 2.15, with $m=4$ and $q\geqslant
7$. Then the element $\tilde{s}$ from Assumption 2.15 can be chosen so that
the stabiliser of the decomposition $\mathbb{F}_{q}^{2m}=U\perp U^{\perp}$ in
$\tilde{S}$ is the unique subgroup in $\mathcal{M}_{\tilde{S}}(\tilde{s})$
that is not an extension field subgroup of type $\mathrm{GO}^{+}_{m}(q^{2})$.

###### Proof.

We shall initially consider an arbitrary choice for $\tilde{s}$, and work
through the families of maximal subgroups of $\tilde{S}$, given in [4, Table
8.50] (see also [39, Table I]). Let $\tilde{L}$ be the stabiliser in
$\tilde{S}$ of the decomposition of $V:=\mathbb{F}_{q}^{8}$ as $U\perp
U^{\perp}$. Then $\tilde{L}$ contains the stabiliser in $\tilde{S}$ of $U$.
Lemma 2.16 shows that $U$ and $U^{\perp}$ are the only proper non-zero
subspaces of $V$ stabilised by $\langle\tilde{s}\rangle$. Therefore,
$\mathcal{M}_{\tilde{S}}(\tilde{s})$ contains no reducible subgroups.
Moreover, if $\langle\tilde{s}\rangle$ stabilises a decomposition
$V=W_{1}\oplus W_{2}$, where $\dim(W_{1})=\dim(W_{2})=4$, then
$\langle\tilde{s}^{2}\rangle=\langle\tilde{s}\rangle$ stabilises each of
$W_{1}$ and $W_{2}$, and hence $\\{W_{1},W_{2}\\}=\\{U,U^{\perp}\\}$. Thus
$\tilde{L}$ is the unique stabiliser of such an imprimitive decomposition that
contains $\tilde{s}$.

Next, let $\tilde{E}$ be an extension field subgroup of $\tilde{S}$ of type
$\mathrm{GU}_{4}(q)$, and recall that $\hat{S}=\mathrm{SO}^{+}_{2m}(q)$ if $q$
is odd, and $\hat{S}=\tilde{S}$ if $q$ is even. We see from [7, Construction
2.5.14, Lemma 5.3.6] that $\hat{E}:=N_{\hat{S}}(\tilde{E})$ has shape
$\mathrm{GU}_{4}(q).2$. Thus by [11, Cor. 2] (see also [51, Lemma 2]), each
(semisimple) element of $\hat{E}$ of order $k:=|\tilde{s}|$ lies in a cyclic
maximal torus of $\mathrm{GU}_{4}(q)$ of order $q^{4}-1$. However, Lemma 2.20
shows that $C_{\hat{S}}(\tilde{s})$ is not cyclic. We therefore deduce that
$\tilde{s}\notin\tilde{E}$.

For a similar argument, let $X$ be a maximal subgroup of $S$ that is
isomorphic to $\Omega_{7}(q)$ if $q$ is odd, or to $\mathrm{Sp}_{6}(q)$ if $q$
is even. Then each element of $X$ of order $|s|=k$ lies in a maximal torus of
shape $k\times(q+1)$ or $k\times(q-1)$ [11, Thms. 3–4]. Additionally,
$C_{S}(s)$ is abelian of order $k^{2}$ [11, Thms. 5 & 7]. As $k^{2}$ is
divisible by neither $q+1$ nor $q-1$ (since $q\neq 2$), we see that $s\notin
X$ and $\tilde{s}\notin\tilde{X}$.

Suppose now that $q$ is odd, and let $H$ be a maximal subgroup of $S$ of shape
$(\operatorname{L}_{2}(q)\times\operatorname{S}_{4}(q)).2$. Since
$|\operatorname{L}_{2}(q)|$ and $|s|$ are coprime, any element of $H$ of order
$k$ has a centraliser that contains the non-abelian group
$\operatorname{L}_{2}(q)$. Thus $\tilde{s}\notin\tilde{H}$.

It now follows from [39, Table I] and [4, Table 8.50] that if $\tilde{Y}$ is
any other maximal subgroup of $\tilde{S}$ whose order is divisible by $k$,
such that $\tilde{Y}$ is not an extension field subgroup of type
$\mathrm{GO}^{+}_{m}(q^{2})$, then $q$ is a square and
$Y\cong\Omega^{-}_{8}(\sqrt{q})$.

For a subspace $W$ of $V$ and an element $g\in\tilde{S}$, let $g_{W}$ denote
the set of $\mathbb{F}_{q^{8}}$-eigenvalues of $g$ on $W$. Recall from
Assumption 2.15 that
$\tilde{s}_{U}=\mathcal{A}:=\\{\alpha,\alpha^{q},\alpha^{q^{2}},\alpha^{q^{3}}\\}$
and
$\tilde{s}_{U^{\perp}}=\mathcal{B}:=\\{\beta,\beta^{q},\beta^{q^{2}},\beta^{q^{3}}\\}$,
where $\alpha$ and $\beta$ are elements of $\mathbb{F}_{q^{4}}$ of order $k$
and $\mathcal{A}\cap\mathcal{B}=\varnothing$. Let $\mathcal{Y}$ be the set of
maximal subgroups $\tilde{Y}$ of $\tilde{S}$ such that
$Y\cong\Omega^{-}_{8}(\sqrt{q})$, and for $\tilde{Y}\in\mathcal{Y}$, let
$\mathcal{J}({\tilde{Y}})$ be the set of elements $y\in\tilde{Y}$ of order
$k$, such that the proper non-zero subspaces of $V$ stabilised by $y$ are
precisely $U$ and $U^{\perp}$; the set $y_{V}$ contains $\mathcal{A}$; and
$y_{V}\setminus\mathcal{A}=\\{\gamma,\gamma^{q},\gamma^{q^{2}},\gamma^{q^{3}}\\}$
for some $\gamma\in\mathbb{F}_{q^{4}}$ of order $k$. Additionally, let

$\mathcal{E}:=\left\\{y_{V}\setminus\mathcal{A}\mid
y\in\mathcal{J}({\tilde{Y}}),\tilde{Y}\in\mathcal{Y}\right\\}.$

We will determine an upper bound for $|\mathcal{E}|$. Using this bound, we
will show that the element $\tilde{r}_{2}$ in Assumption 2.15 (equivalently,
the set $\mathcal{B}$) can be chosen so that, for all
$\tilde{Y}\in\mathcal{Y}$, the set $\mathcal{J}({\tilde{Y}})$ does not
contains $\tilde{s}$, and so $\tilde{s}\notin\tilde{Y}$.

Fix $\tilde{Y}\in\mathcal{Y}$ and $y\in\mathcal{J}({\tilde{Y}})$. Notice that
$\\{\mathcal{A},y_{V}\setminus\mathcal{A}\\}=\\{y_{U},y_{U^{\perp}}\\}$, and
that a given $c\in\langle y\rangle\cap\mathcal{J}({\tilde{Y}})$ satisfies
$c_{U}=y_{U}$ if and only if $c_{U^{\perp}}=y_{U^{\perp}}$ (additionally, at
least one of $c_{U}$ and $c_{U^{\perp}}$ is equal to $\mathcal{A}$). We also
deduce from [11, §4–5] (and the fact that
$Y\cong\mathrm{PSO}^{-}_{8}(\sqrt{q})$ if $q$ is odd) that any two cyclic
subgroups of $\tilde{Y}$ of order $k$ are conjugate in $\tilde{S}$. Thus the
fixed $\tilde{Y}\in\mathcal{Y}$ contributes two (not necessarily distinct)
sets to $\mathcal{E}$: the set $y_{U^{\perp}}$ for elements
$y\in\mathcal{J}({\tilde{Y}})$ such that $y_{U}=\mathcal{A}$, and the set
$y_{U}$ for elements $y\in\mathcal{J}({\tilde{Y}})$ such that
$y_{U^{\perp}}=\mathcal{A}$.

We observe from Rows 64–69 and Columns V and XII of [39, Table I] that the
subgroups in $\mathcal{Y}$ form exactly two
$N_{\operatorname{GL}_{8}(q)}(\tilde{S})$-conjugacy classes. Thus there exist
$\tilde{Y}_{1},\tilde{Y}_{2}\in\mathcal{Y}$ such that

$\mathcal{E}=\left\\{y_{V}\setminus\mathcal{A}\mid
y\in\mathcal{J}({\tilde{Y}_{1}})\cup\mathcal{J}({\tilde{Y}_{2}})\right\\}.$

By the previous paragraph, each of $\tilde{Y}_{1}$ and $\tilde{Y}_{2}$
contributes at most two sets to $\mathcal{E}$, and thus
$|\mathcal{E}|\leqslant 4$.

To ensure that $\tilde{s}$ does not lie in $\mathcal{J}({\tilde{Y}})$ for any
$\tilde{Y}\in\mathcal{Y}$ (and hence that $\tilde{s}\notin\tilde{Y}$), it
suffices to choose $\tilde{r}_{2}$ in Assumption 2.15 so that the
corresponding set $\mathcal{B}$ intersects trivially with each of the sets in
$\mathcal{E}$, and with $\mathcal{A}$ (as required to satisfy Assumption
2.15). Since each of these sets of eigenvalues has size four, this is possible
as long as $\varphi(k)/4\geqslant 6$, where $\varphi$ is Euler’s totient
function. It is well known (see, e.g., [47, Prop. 2]) that
$\varphi(i)\geqslant\sqrt{i/2}$ for each $i$. Hence an appropriate
$\tilde{r}_{2}$ exists if $k\geqslant 1152$. If instead $k<1152$, then the
square $q\geqslant 9$ is at most $25$, and again $\varphi(k)/4\geqslant 6$. ∎

We observe from [6, Prop. 5.14] that, when $4\nmid m$, the subgroup
$\tilde{L}$ described in Theorem 2.14 is the unique subgroup in
$\mathcal{M}_{\tilde{S}}(\tilde{s})$ that is not an extension field subgroup.
Thus to prove Theorem 2.14, it remains to show, for all even $m\geqslant 4$,
that $\mathcal{M}_{\tilde{S}}(\tilde{s})$ contains exactly two extension field
subgroups of the specified type, and that they are not conjugate in
$\tilde{S}$.

######  Proposition 2.22.

Let $\tilde{S}$ and $\tilde{s}$ be as in Assumption 2.15. Then
$\mathcal{M}_{\tilde{S}}(\tilde{s})$ contains exactly two extension field
subgroups of type $J$, where $J\cong\mathrm{GU}_{m}(q)$ if $4<m\equiv
0\pmod{4}$, and $J\cong\mathrm{GO}^{+}_{m}(q^{2})$ otherwise. Additionally,
the two subgroups are not conjugate in $\tilde{S}$.

###### Proof.

Let $a$, $b$, and $\hat{S}$ be as in Lemma 2.20, so that
$C_{\hat{S}}(\tilde{s})=C_{\mathrm{GO}^{+}_{2m}(q)}(\tilde{s})$ is a maximal
torus $\hat{T}$ of $\hat{S}$ of shape $(q^{a}+1)\times(q^{b}+1)$.
Additionally, let $\tilde{K}$ be a subgroup of $\tilde{S}$ of type $J$. Then
[7, Constructions 2.5.13–2.5.14, Lemmas 5.3.4 & Lemma 5.3.6] and [38, Tables
3.5.E & 3.5.G] imply that

$\hat{K}:=N_{\hat{S}}(\tilde{K})\cong J.\langle\phi\rangle,$

where $\phi$ is the involutory field automorphism of the subgroup
$\operatorname{SU}_{m}(q)$ or $\Omega^{+}_{m}(q^{2})$ of $J$. As all maximal
tori of $\hat{S}$ isomorphic to $\hat{T}$ are conjugate in $\hat{S}$ [53, p.
394], and as $J$ contains such a maximal torus (see [11, Cor. 2], [53, §9] and
[51, Lemma 2]), some $\hat{S}$-conjugate of $\hat{K}$ contains $\hat{T}$, and
hence contains $\tilde{s}$. Now, by [4, Tables 8.50 & Table 8.82] and [38,
Tables 3.5.E & 3.5.G], $\tilde{S}$ has exactly two conjugacy classes of
maximal subgroups that are extension field subgroups of type $J$, and these
extend to two conjugacy classes of maximal subgroups of $\hat{S}$. We have
shown that each of these two conjugacy classes has at least one subgroup
containing $\tilde{s}$.

To complete the proof, we will show that $\mathcal{M}_{\tilde{S}}(\tilde{s})$
contains exactly two extension field subgroups of type $J$. If $4\nmid m$,
then this is the case by [6, Prop. 5.14]. If instead $4\mid m$, then $|s|$ is
odd. Thus [54, p. 34, pp. 38–39] (see also [6, Prop. 2.11]) implies that any
two similar elements of $\mathrm{GO}^{+}_{2m}(q)$ are conjugate, as are any
two similar elements of $J$. Additionally, conjugating $\tilde{K}$ by an
element of $\hat{S}$ if necessary, we may assume that
$\hat{T}\leqslant\hat{K}$. Therefore, arguing as in the proof of [6, Lemma
5.9], we deduce that the elements of
$\tilde{s}^{\mathrm{GO}^{+}_{2m}(q)}\cap\hat{K}$ form exactly two
$\hat{K}$-conjugacy classes.

Now, any two extension field subgroups of $\tilde{S}$ of type $J$ extend to
conjugate subgroups of $\mathrm{GO}^{+}_{2m}(q)$ [38, Tables 3.5.E & Table
3.5.G], and so $\hat{K}$ is self-normalising in $\mathrm{GO}^{+}_{2m}(q)$.
Moreover, by considering the maximal tori of $J$ (again, see [11, Cor. 2],
[53, §9] and [51, Lemma 2]), we deduce that
$C_{\mathrm{GO}^{+}_{2m}(q)}(\tilde{f})\leqslant\hat{K}$ for each
$\tilde{f}\in\tilde{s}^{\mathrm{GO}^{+}_{2m}(q)}\cap\hat{K}$. Therefore, Lemma
2.2(ii) implies that $\tilde{s}$ lies in exactly two
$\mathrm{GO}^{+}_{2m}(q)$-conjugates of $\hat{K}$, and hence in exactly two
$\mathrm{GO}^{+}_{2m}(q)$-conjugates of $\tilde{K}$. ∎

Theorem 2.14 now follows from [6, Prop. 5.14] (with $4\nmid m$) and
Propositions 2.18, 2.21 and 2.22. We are also now able to prove the final case
of Theorem 4, and hence complete the proof of Theorem 3.

######  Proposition 2.23.

Theorem 4 holds when $S=\operatorname{O}^{+}_{2m}(q)$, with $m\geqslant 4$
even.

###### Proof.

By [23, Lemma 10], $s$ is $S$-conjugate to its inverse, and so $N_{S}(\langle
s\rangle)>C_{S}(s)$. Additionally, Theorem 2.14 shows that
$\mathcal{M}_{S}(s)$ consists of three subgroups, $K_{1}$, $K_{2}$ and $L$, no
two of which are conjugate in $S$. If $m=4$, then the intersection of any two
of these subgroups is equal to the intersection of all three [6, Lemma
5.15(e)], and hence Lemma 2.3 shows that $\mathcal{M}^{\prime}_{G}(s)$
contains at most one novelty maximal. If instead $m>4$, then $L$ is isomorphic
to neither $K_{1}$ nor $K_{2}$ (see Table 3.5.E and the corresponding results
in Chapter 4 of [38]). In each case, Lemma 2.4 yields the result. ∎

##  3\. Finite groups satisfying the independence property are supersoluble

The aim of this section is to prove that each finite group $G$ satisfying the
independence property is supersoluble. We will reduce the proof of this
statement to the case where $G$ is almost simple, and finally, in the almost
simple case, we will reach our conclusion by applying Theorem 3. Two elements
$x$ and $y$ of a group $G$ are _dependent_ if no minimal generating set for
$G$ contains $\\{x,y\\}$.

We shall assume throughout this section that $G$ is a finite group satisfying
the independence property. Our first result reduces the study of such groups
to the case where their Frattini subgroup is trivial.

######  Proposition 3.1.

The Frattini subgroup $\operatorname{\Phi}(G)\neq 1$ if and only if $G$ is
either a cyclic $p$-group or the quaternion group of order 8.

###### Proof.

Suppose that $\operatorname{\Phi}(G)\neq 1$, and let $x$ be an element of
$\operatorname{\Phi}(G)$ of prime order. Since $x$ is a non-generator of $G$,
$x$ and $y$ are dependent for each $y\in G\setminus\\{x\\}$. In particular,
since $|x|$ is prime, $\langle x\rangle\subseteq\langle y\rangle$ for every
$y\in G\setminus\\{1\\}$. Thus $\langle x\rangle$ is the unique minimal
subgroup of $G$ and therefore either $G$ is a cyclic $p$-group or $G=\langle
a,b\mid a^{2^{n-1}}=1,b^{2}=a^{2^{n-2}},a^{b}=a^{-1}\rangle$ (with $n\geqslant
3$) is a generalized quaternion group (see [50, Thm. 9.7.3]). In the second
case, $a^{2}\in\operatorname{\Phi}(G)$ and therefore $b$ and $ba^{2}$ are
dependent. Since $G$ satisfies the independence property, $\langle
b,ba^{2}\rangle$ is cyclic, and hence $[b,a^{2}]=1$. This implies that $n=3$,
as required. The converse is clear. ∎

We now present a sequence of results characterising the minimal normal
subgroups of our group $G$.

######  Lemma 3.2.

Suppose that $\operatorname{\Phi}(G)=1$, and that $G$ has an abelian minimal
normal subgroup $N$. Then $N$ is cyclic of prime order.

###### Proof.

Let $n_{1}$ and $n_{2}$ be distinct non-trivial elements of $N$. We claim that
they are dependent. Indeed, assume for a contradiction that
$X:=\\{n_{1},n_{2},g_{1},\dots,g_{t}\\}$ is a minimal generating set for $G$
and let $H:=\langle g_{1},\dots,g_{t}\rangle.$ Since

$G=\langle n_{1},n_{2},g_{1},\dots,g_{t}\rangle\leqslant HN,$

$G=HN$ and $N$ is $H$-irreducible. But then $G=\langle
n_{1},g_{1},\dots,g_{t}\rangle$, contradicting the minimality of $X$.

Since $n_{1}$ and $n_{2}$ are dependent, one of them is a power of the other.
Since $N$ is elementary abelian, it follows that $\langle n_{1}\rangle=\langle
n_{2}\rangle.$ Therefore, $N\cong C_{p}.$ ∎

######  Lemma 3.3.

Suppose that $\operatorname{\Phi}(G)=1$, and that $G$ has a non-abelian
minimal normal subgroup $N$. If $\operatorname{O}^{+}_{8}(2)$ is not a
composition factor of $N,$ then $N$ is simple.

###### Proof.

Here, $N=S_{1}\times\dots\times S_{t}$ with $S_{i}\cong S$ a non-abelian
finite simple group. Assume for a contradiction that
$S\neq\operatorname{O}^{+}_{8}(2)$ and $t\geqslant 2.$ By [28, Cor. 7.2],
there exist $x$ and $y$ in $S$ such that $S=\langle
x^{\alpha},y^{\beta}\rangle$ for every $\alpha,\beta\in\operatorname{Aut}(S)$.
Notice that in particular $y\neq x^{\gamma}$ for all
$\gamma\in\operatorname{Aut}(S)$.

Let $n:=(x,y,1,\dots,1)\in N$ and choose $g_{1},\dots,g_{d}\in G$ such that
$G=\langle g_{1},\dots,g_{d}\rangle N.$ Set $H:=\langle
n,g_{1},\dots,g_{d}\rangle$ and denote by $\pi_{i}:N\to S_{i}$ the projection
to the $i$-th factor of $N.$ The factors $S_{1},\dots,S_{t}$ are permuted
transitively by $H$, so in particular there exists $h\in H$ such that
$S_{2}^{h}=S_{1}$. Thus $\pi_{1}(n^{h})=y^{\beta}$ for some
$\beta\in\operatorname{Aut}(S).$

It follows that $S_{1}=\langle
x,y^{\beta}\rangle\leqslant\langle\pi_{1}(n),\pi_{1}(n^{h})\rangle\leqslant\pi_{1}(H\cap
N)$, and since $H$ is transitive on $\\{S_{1},\dots,S_{t}\\}$, the image
$\pi_{i}(H\cap N)=S_{i}$ for all $i\in\\{1,\ldots,t\\}$. Therefore, there
exists a partition of $\\{1,\dots,t\\}$ into $u$ blocks of size $v:=t/u$, say
$J_{i}:=\\{j_{i1},\ldots,j_{iv}\\}$ for $1\leqslant i\leqslant u$,
corresponding elements $\alpha_{ik}\in\operatorname{Aut}(S)$, and diagonal
subgroups $\Delta_{i}:=\\{(s^{\alpha_{i1}},\dots,s^{\alpha_{iv}})\mid s\in
S\\}$ of $S_{j_{i1}}\times\dots\times S_{j_{iv}}$, such that $H\cap
N=\Delta_{1}\times\dots\times\Delta_{u}$. We may assume that $1=j_{11}\in
J_{1}.$

Since $n\in H\cap N,$ it follows that $J_{1}\subseteq\\{1,2\\}.$ If
$J_{1}=\\{1,2\\},$ then

$\Delta_{1}=\\{(s,s^{\gamma})\mid s\in S\\}\leqslant S_{1}\times S_{2},$

with $\gamma=\alpha_{11}^{-1}\alpha_{12}$, and $n\in\Delta_{1}.$ This implies
that $y=x^{\gamma}$, a contradiction. Hence $J_{1}=\\{1\\}$, so $|J_{i}|=1$
for all $i$ and $H\cap N=S_{1}\times\cdots\times S_{t}=N$. We conclude that
$G=H=\langle n,g_{1},\ldots,g_{d}\rangle$.

We have shown that $\langle n,g_{1},\ldots,g_{d}\rangle=G$ for all
$g_{1},\ldots,g_{d}\in G$ such that $\langle g_{1},\ldots,g_{d}\rangle N=G$.
Thus no minimal generating set for $G$ contains both $n$ and
$m:=(y,1,\dots,1)\in N$. Consequently, one of $n$ and $m$ is a power of the
other. This implies in particular that $\langle x,y\rangle$ is cyclic,
contradicting $\langle x,y\rangle=S.$ ∎

The restriction in the previous lemma concerning $\operatorname{O}^{+}_{8}(2)$
can be removed by a different argument. For this purpose we need the following
lemma. In the next few results, we shall denote elements $h\in
X\wr\operatorname{S}_{t}$ by $(\rho_{1}(h),\ldots,\rho_{t}(h))\sigma(h)$,
where $\rho_{i}$ is the projection from the base group to the $i$-th copy of
$X$ and $\sigma(h)\in\operatorname{S}_{t}$.

######  Lemma 3.4.

Let $X$ be a finite group, let $a$ be an element of $X$, let $t\geqslant 2$,
and let $\alpha:=(a,a^{2},1,\dots,1)$ and $\beta:=(a,1,\dots,1)$ be elements
of $X\wr\operatorname{S}_{t}$. Let $Y$ be a subgroup of
$X\wr\operatorname{S}_{t}$ and set $R:=\langle\alpha,Y\rangle$ and
$K:=\langle\alpha,\beta,Y\rangle.$ For any given $k\in K$ and
$i\in\\{1,\dots,t\\}$, there exists $r\in R$ such that
$\rho_{i}(r)=\rho_{i}(k)$ and $\sigma(r)=\sigma(k).$

###### Proof.

Let $A:=\langle\alpha\rangle$ and $B:=\langle\beta\rangle.$ Any element $k\in
K$ can be written in the form $k=z_{1}\cdots z_{\ell}$ with $z_{i}\in A\cup
B\cup Y.$ We prove the statement by induction on $\ell.$ First assume
$\ell=1.$ If $k=z_{1}\in Y\cup A,$ we set $r=k.$ If $k=z_{1}=\beta^{m}\in B$,
we set $r=\alpha^{m}$ if $i=1,$ and $r=1$ otherwise.

Now assume $\ell>1$ and set $k^{*}=z_{2}\cdots z_{\ell}.$ We have
$\sigma(k)=\sigma(z_{1})\sigma(k^{*})$ and
$\rho_{i}(k)=\rho_{i}(z_{1})\rho_{i\sigma(z_{1})}(k^{*}).$ By induction, for
each $i$ there exist $r_{1},r_{2}\in R$ such that
$\sigma(r_{1})=\sigma(z_{1}),$ $\sigma(r_{2})=\sigma(k^{*}),$
$\rho_{i}(r_{1})=\rho_{i}(z_{1})$, and
$\rho_{i\sigma(z_{1})}(r_{2})=\rho_{i\sigma(z_{1})}(k^{*}).$ The element
$r=r_{1}r_{2}$ satisfies the required properties. ∎

######  Lemma 3.5.

Suppose that $\operatorname{\Phi}(G)=1$, and that $G$ has a non-abelian
minimal normal subgroup $N$. Then $N$ is simple.

###### Proof.

We have $N=S_{1}\times\dots\times S_{t}$, with $S_{i}\cong S$ a non-abelian
finite simple group. By Lemma 3.3, we may assume that
$S=\operatorname{O}^{+}_{8}(2)$. In particular, $S$ contains an element $a$ of
order four.

Assume for a contradiction that $t\geqslant 2$, and let
$\alpha:=(a,a^{2},1,\dots,1)$ and $\beta:=(a,1,\dots,1).$ Suppose that
$G=\langle\alpha,\beta,g_{1},\dots,g_{d}\rangle$ for some
$g_{1},\ldots,g_{d}\in G$, and let $Y:=\langle g_{1},\dots,g_{d}\rangle$ and
$R:=\langle\alpha,g_{1},\dots,g_{d}\rangle$. We apply Lemma 3.4 to
$RC_{G}(N)/C_{G}(N)$ and
$G/C_{G}(N)\leqslant\operatorname{Aut}(N)=\operatorname{Aut}(S)\wr\operatorname{S}_{t}$,
with the element $(s,s,\ldots,s)C_{G}(N)\in G/C_{G}(N)$, for an arbitrary
$s\in S$, with $i=1$. We deduce that there exist
$\gamma_{2},\dots,\gamma_{t}\in\operatorname{Aut}(S)$ such that
$(s,\gamma_{2},\dots,\gamma_{t})C_{G}(N)\in RC_{G}(N)$.

Let $\pi_{1}$ be the projection from $N$ to $S_{1}$. Since $S$ is non-abelian
simple and $\alpha\in R\cap N\unlhd R$, we deduce that $\pi_{1}(R\cap N)=S$.
Since $RN=G$, the action of $R$ on $S_{1},\dots,S_{t}$ is transitive and
consequently, arguing as in the proof of Lemma 3.3, we see that $R\cap
N=\Delta_{1}\times\dots\times\Delta_{u}$, where
$\Delta_{i}:=\\{(s^{\alpha_{i1}},\dots,s^{\alpha_{iv}})\mid s\in S\\}$ is a
diagonal subgroup of $S_{j_{i_{1}}}\times\dots\times S_{j_{i_{v}}}$ and each
$\alpha_{ik}\in\operatorname{Aut}(S)$. On the other hand, $\alpha\in R\cap N$,
so the order of $a$ implies that $v=1$, hence $R\cap N=N$ and $R=RN=G.$ This
implies that $\alpha$ and $\beta$ are dependent, a contradiction since neither
of $\alpha$ and $\beta$ is a power of the other. ∎

######  Lemma 3.6.

Let $N_{1}$ and $N_{2}$ be two distinct minimal normal subgroups of a finite
group $X$, and let $a\in N_{1}$ and $b\in N_{2}.$ If $ab$ and $b$ are
independent in $X$, then there exists an isomorphism $\phi:N_{1}\to N_{2}$
such that $\phi(a)=b$, and such that $\phi$ is an $X$-isomorphism if $N_{1}$
is abelian.

###### Proof.

Assume that $ab$ and $b$ are independent, and let
$\\{ab,b,x_{1},\dots,x_{d}\\}$ be a minimal generating set for $X.$
Additionally, let $H:=\langle ab,x_{1},\dots,x_{d}\rangle$, so $H\neq X$. We
have $b=a^{-1}(ab)\in N_{2}\cap HN_{1}$ and therefore $X=\langle
H,b\rangle=HN_{1}=HN_{2}.$ Moreover $H\cap N_{1}$ is normalized by $H$ and
centralized by $N_{2},$ so $H\cap N_{1}$ is normal in $X=HN_{2}$ and therefore
$H\cap N_{1}=1.$ Similarly $H\cap N_{2}=1,$ so $H$ is a common complement of
$N_{1}$ and $N_{2}$ in $X.$ In particular, for any $n$ in $N_{1}$ there exists
a unique $n^{*}$ in $N_{2}$ such that $nn^{*}\in H$, and since
$[N_{1},N_{2}]=1$, the map $\phi:N_{1}\to N_{2}$ sending $n$ to $n^{*}$ is an
isomorphism. Now $ab\in H,$ so $\phi(a)=b.$

Suppose now that $N_{1}$ is abelian, and let $n\in N_{1}$ and $x\in X$. Then
$x=n^{\prime}h$ for some $n^{\prime}\in N_{1}$ and $h\in H$. As
$[N_{1},N_{1}]=1=[N_{1},N_{2}]$, we see that $(nn^{*})^{x}=(nn^{*})^{h}\in H$.
Thus $\phi(n^{x})=(n^{*})^{x}=\phi(n)^{x}$, and so $\phi$ is an
$X$-isomorphism. ∎

######  Lemma 3.7.

The group $G$ contains at most one non-abelian minimal normal subgroup.

###### Proof.

If $\operatorname{\Phi}(G)\neq 1$ then the result is immediate from
Proposition 3.1, so assume $\operatorname{\Phi}(G)=1.$ Assume further, for a
contradiction, that $N_{1}$ and $N_{2}$ are distinct non-abelian minimal
normal subgroups of $G$. By Lemma 3.5, $N_{1}$ and $N_{2}$ are simple.
Therefore, there exist $a\in N_{1}$ and $b\in N_{2}$ such that $|a|=|b|\neq
1.$

Since neither of $ab$ and $b$ is a power of the other, they are independent
and therefore, by Lemma 3.6, there exist a non-abelian simple group $S$ with
$N_{1}\cong N_{2}\cong S$, and an element $\phi\in\operatorname{Aut}(S)$ such
that $b=\phi(a).$ So $S$ is a non-abelian simple group such that all elements
of the same order are conjugate in its automorphism group. By [61, Thm. 3.1],
$S\in\\{\operatorname{L}_{2}(5),\operatorname{L}_{2}(7),\operatorname{L}_{2}(8),\operatorname{L}_{2}(9),\operatorname{L}_{3}(4)\\}$.

Assume that $S$ contains an element $a$ of order $p^{2}$, for some prime $p,$
and consider $(a,a^{p})$ and $(1,a^{p})$ in $N_{1}\times N_{2}.$ Neither of
these elements is a power of the other, so they are independent, contradicting
Lemma 3.6. So $S$ contains no elements of order $p^{2}$, and therefore
$S\cong\operatorname{L}_{2}(5)\cong\operatorname{A}_{5}$.

Here, we consider an element $u$ of order $5$ in $S$ and an involution $t$ in
$\mathrm{N}_{S}(\langle u\rangle)$. Then $M:=\langle t,u\rangle$ is isomorphic
to the dihedral group of order 10, and is the unique maximal subgroup of $S$
containing $u.$ Let $x:=(t,u),y:=(1,t)\in N_{1}\times N_{2}\cong S^{2}.$
Neither of $x$ and $y$ is a power of the other, so they are independent.

Let $\\{x,y,g_{1},\dots,g_{d}\\}$ be a minimal generating set for $G$, and
define $H:=\langle x,g_{1},\dots,g_{d}\rangle.$ We have $HN_{2}=G$, since
$y\in N_{2}.$ Let $X:=HN_{1}\cap N_{1}N_{2}.$ Now $(t,u)\in H$ and
$(t^{-1},1)\in N_{1}$, so $(1,u)\in HN_{1}\cap N_{1}N_{2}=X$. Furthermore,
$\langle(1,t),(1,u)\rangle N_{1}$ is the unique maximal subgroup of
$N_{1}N_{2}$ that contains $\langle(1,u),N_{1}\rangle$, so if $X\neq
N_{1}N_{2},$ then $X\leqslant\langle(1,t),(1,u)\rangle N_{1}$. But then $y$
normalizes $X$ and consequently $X\unlhd G,$ contradicting the minimality of
$N_{2}$.

Thus $X=N_{1}N_{2}\leqslant HN_{1}$ and using Dedekind’s modular law and
$HN_{2}=G$ we see that

$(H\cap N_{1}N_{2})N_{1}=N_{1}N_{2}=(H\cap N_{1}N_{2})N_{2}.$

In particular $H\cap N_{1}N_{2}$ is a subdirect product of
$N_{1}N_{2}\cong\operatorname{A}_{5}^{2}$. If $N_{1}N_{2}\leqslant H,$ then
$G=N_{1}N_{2}\langle g_{1},\dots,g_{d}\rangle\leqslant H,$ a contradiction.
Hence $H\cap N_{1}N_{2}=\\{(s,s^{\phi})\mid s\in S\\}$ for some
$\phi\in\operatorname{Aut}(S)$, but this contradicts $(t,u)\in H\cap
N_{1}N_{2}.$ ∎

In the following, we write $F(G)$ for the Fitting subgroup of $G$ (the largest
normal nilpotent subgroup of $G$), $E(G)$ for the subgroup of $G$ generated by
all quasisimple subnormal subgroups of $G$, and $F^{*}(G)$ for the generalized
Fitting subgroup of $G$ (the subgroup generated by $F(G)$ and $E(G)$).

######  Corollary 3.8.

If $G$ is insoluble, then:

  1. (i)

$E(G)$ is a non-abelian simple group; and

  2. (ii)

$G/E(G)$ is soluble.

###### Proof.

The Frattini subgroup of $G$ is trivial, otherwise, by Proposition 3.1, $G$
would be nilpotent. This implies that all quasisimple subnormal subgroups of
$G$ are simple, and so $\operatorname{soc}(G)=F^{*}(G)=E(G)\times F(G)$.
Moreover $C_{G}(\operatorname{soc}(G))=F(G),$ and we may write
$F(G)=N_{1}\times\cdots\times N_{t}$ as a product of abelian minimal normal
subgroups of $G.$

By Lemma 3.2, for $1\leqslant i\leqslant t$, $N_{i}\cong C_{p_{i}}$ for a
suitable prime $p_{i}$ and therefore

$G/C_{G}(F(G))\cong G/\bigcap_{1\leqslant i\leqslant
t}C_{G}(N_{i})\leqslant\prod_{1\leqslant i\leqslant
t}G/C_{G}(N_{i})\leqslant\prod_{1\leqslant i\leqslant
t}\operatorname{Aut}(C_{p_{i}})$

is abelian. If $E(G)=1$, then $\operatorname{soc}(G)=F(G)$, so
$C_{G}(F(G))=F(G)$ and $G$ is metabelian, a contradiction. Hence $G$ contains
a non-abelian minimal normal subgroup and so by Lemmas 3.5 and 3.7, $E(G)$ is
a non-abelian simple group $S$. Therefore,

$G/F(G)=G/C_{G}(\operatorname{soc}(G))=G/(C_{G}(S)\cap C_{G}(F(G))\leqslant
G/C_{G}(S)\times G/C_{G}(F(G)).$

Since $G/C_{G}(S)$ is almost simple with socle isomorphic to $S$, and
$G/C_{G}(F(G))$ is abelian, we conclude that $S=E(G)$ is the unique non-
abelian composition factor of $G$ and therefore $G/E(G)$ is soluble. ∎

We now show that in an insoluble group with the independence property, certain
elements are dependent, and so in particular must commute.

######  Lemma 3.9.

Suppose that $G$ is insoluble and let $S=E(G)$ be the unique non-abelian
minimal normal subgroup of $G$. Suppose in addition that there exist $x,s\in
S$ such that $xC_{G}(S)$ and $sC_{G}(S)$ are dependent in $G/C_{G}(S)$. Then
$x$ and $s$ are dependent, and in particular they commute.

###### Proof.

Let $\overline{G}:=G/C_{G}(S),$ and, for all $g\in G,$ let
$\overline{g}:=gC_{G}(S).$ Assume for a contradiction that
$\\{x,s,g_{1},\dots,g_{t}\\}$ is a minimal generating set for $G.$ We may
order the indices so that, for some $r$, the subset $\\{g_{1},\dots,g_{r}\\}$
of $\\{g_{1},\dots,g_{t}\\}$ is minimal, subject to
$\overline{G}=\langle\overline{x},\overline{s},\overline{g}_{1},\dots,\overline{g}_{r}\rangle.$
It follows that there exists $z\in\\{x,s\\}$ such that
$\overline{G}=\langle\overline{z},\overline{g}_{1},\dots,\overline{g}_{r}\rangle.$

Let $R:=\langle z,g_{1},\dots,g_{t}\rangle.$ Clearly $G=S\langle
z,g_{1},\dots,g_{t}\rangle=RS.$ If $R\cap S=1,$ then $R\cong G/S$ is soluble
by Corollary 3.8, contradicting $RC_{G}(S)/C_{G}(S)=\overline{G}$. Thus $R\cap
S$ is a non-trivial and normalized by $RC_{G}(S)=G$, and consequently $S=R\cap
S$ and $G=R,$ contradicting the fact that $\\{x,s,g_{1},\dots,g_{t}\\}$ is a
minimal generating set for $G.$ Hence $x$ and $s$ are dependent, so $\langle
x,s\rangle$ is cyclic. ∎

Recall that for a group $X$ and an element $x\in X$, we write
$\mathcal{M}_{X}(x)$ for the set of maximal subgroups of $X$ containing $s$.

######  Lemma 3.10.

Let $X$ be a finite group and let $x,y\in X$ be such that
$y\in\cap_{M\in\mathcal{M}_{X}(x)}M$. Then no minimal generating set for $X$
contains $\\{x,y\\}$.

###### Proof.

Suppose, for a contradiction, that there exist $g_{1},\dots,g_{d}\in X$ such
that $\\{x,y,g_{1},\dots,g_{d}\\}$ is a minimal generating set for $X.$ Then
$\langle x,g_{1},\dots,g_{d}\rangle\neq X$, so there exists
$M\in\mathcal{M}_{X}(x)$ such that $\langle
x,g_{1},\dots,g_{d}\rangle\leqslant M.$ By assumption, $y\in M$, so $X=\langle
x,y,g_{1},\dots,g_{d}\rangle\leqslant M,$ a contradiction. ∎

Finally, we reach the main result of this section. Recall that $G$ is assumed
to be a finite group that satisfies the independence property.

######  Theorem 3.11.

The group $G$ is supersoluble.

###### Proof.

First assume that $G$ is insoluble, so that $S:=E(G)$ is non-abelian simple by
Corollary 3.8. Let $H\cong G/C_{G}(S).$ By Theorem 3,
$\operatorname{soc}(H)\cong S$ contains two non-commuting elements $xC_{G}(S)$
and $sC_{G}(S)$ with the property that every maximal subgroup of $H$
containing $sC_{G}(S)$ also contains $xC_{G}(S)$. Furthermore, since $S\cap
C_{G}(S)=1$, we can choose the corresponding $x,s$ to lie in $E(G)$. By Lemma
3.10, no minimal generating set for $H$ contains $\\{xC_{G}(S),C_{G}(S)\\}$,
that is, $xC_{G}(S)$ and $yC_{G}(S)$ are dependent in $H$. But then, by Lemma
3.9, $s$ and $x$ must commute, a contradiction. So $G$ is soluble.

If $\operatorname{\Phi}(G)\neq 1$ then $G$ is nilpotent, by Proposition 3.1,
so assume otherwise. By Lemma 3.2, $F(G)=N_{1}\times\dots\times N_{u}$ is a
direct product of minimal normal subgroups of prime order, and
$G/F(G)\leqslant\prod_{1\leqslant i\leqslant u}\operatorname{Aut}(N_{i})$ is
abelian, hence $G$ is supersoluble. ∎

##  4\. Supersoluble groups with the independence property

In this section we shall determine the structure of the finite supersoluble
groups $G$ satisfying the independence property. By Proposition 3.1, we may
restrict our attention to the case $\operatorname{\Phi}(G)=1$. This implies
that the Fitting subgroup $\operatorname{Fit}(G)$ of $G$ is a direct product
of minimal normal subgroups of $G,$ and is complemented in $G$. Let $K$ be a
complement of $\operatorname{Fit}(G)$ in $G.$ Since $G$ is supersoluble,
$G^{\prime}\leqslant\operatorname{Fit}(G),$ and consequently $K$ is abelian.
Let $W$ be a complement of $Z(G)$ in $\operatorname{Fit}(G).$ Then $G=W\rtimes
H,$ with $H=\langle K,Z(G)\rangle.$ Moreover, by Lemma 3.2, each minimal
normal subgroup of $G$ is cyclic (of prime order). As a consequence, we assume
through this section that

$G=(V_{1}^{\delta_{1}}\times\dots\times{V_{r}^{\delta_{r}})}\rtimes H,$

where $H$ is abelian, $\delta_{1},\dots,\delta_{r}$ are positive integers, and
$V_{1},\dots,V_{r}$ are irreducible $H$-modules of prime order that are
pairwise non-$H$-isomorphic, such that $H$ acts non-trivially on each. (So if
$G$ is abelian then $r=0$ and $G=H$.) For each $i\in\\{1,\ldots,r\\}$, we set

$\begin{array}[]{c}p_{i}:=|V_{i}|,\quad
W_{i}:=V_{i}^{\delta_{i}},\quad\text{and}\quad
W:=V_{1}^{\delta_{1}}\times\dots\times{V_{r}^{\delta_{r}}}=W_{1}\times\dots\times
W_{r}.\end{array}$

Observe now that, for each $h$ in $H$, there exists an
$\alpha_{i}(h)\in\mathbb{F}_{p_{i}}^{\times}$ such that $w^{h}=\alpha_{i}(h)w$
for all $w\in W_{i}.$

######  Notation 4.1.

Let
$g_{1}:=h_{1}(w_{1,1},\dots,w_{1,r}),\dots,g_{t}:=h_{t}(w_{t,1},\dots,w_{t,r})$
be elements of $G$, with $h_{i}\in H$ and
$w_{i,j}:=(x_{i,j,1},\dots,x_{i,j,\delta_{j}})\in W_{j}$. For each
$j\in\\{1,\ldots,r\\}$, we define a matrix
$A^{(j)}=A^{(j)}(g_{1},\ldots,g_{t})$ whose columns are
$a^{(j)}_{1},\ldots,a^{(j)}_{t}$, where for each $i\in\\{1,\ldots,t\\}$ the
transpose of $a^{(j)}_{i}$ is

$(1-\alpha_{j}(h_{i}),x_{i,j,1},x_{i,j,2},\ldots,x_{i,j,\delta_{j}})\in\mathbb{F}_{p_{j}}^{\delta_{j}+1}.$

######  Lemma 4.2.

Let $g_{1},\ldots,g_{t}$ be as in Notation 4.1. Then $G=\langle
g_{1},\dots,g_{t}\rangle$ if and only if $\langle h_{1},\dots,h_{t}\rangle=H$
and ${\rm{rank}}(A^{(j)})=\delta_{j}+1$ for all $j\in\\{1,\ldots,r\\}$.

###### Proof.

See Propositions 2.1 and 2.2 in [43]. ∎

######  Lemma 4.3.

Suppose that $G$ satisfies the independence property. Then the following
statements hold.

  1. (i)

If $\delta_{i}=1,$ then $|H/C_{H}(V_{i})|$ is prime.

  2. (ii)

If $|V_{i}|=|V_{j}|$, then $i=j.$

  3. (iii)

$(|H|,|W|)=1.$

###### Proof.

For (i), notice first that by assumption, $|H/C_{H}(V_{i})|>1$. Assume for a
contradiction that $\delta_{i}=1$ but $|H/C_{H}(V_{i})|$ is not prime, and
choose $h\in H$ such that $|hC_{H}(V_{i})|$ is prime. Take $0\neq x\in
V_{i}=W_{i}$, and let $g_{1}:=h=h(0,0,\ldots,0)$ and
$g_{2}:=hx=h(0,\ldots,0,x,0,\ldots,0)$ (with $x$ in position $i$). Since
$h\not\in C_{H}(V_{i})$, the elements $g_{1}$ and $g_{2}$ do not commute, so
neither is a power of the other. Furthermore, $\langle g_{1},g_{2}\rangle\neq
G$. Since $G$ satisfies the independence property, $g_{1}$ and $g_{2}$ are
therefore independent, so there exist $g_{3},\dots g_{t}\in G$ such that
$\\{g_{1},g_{2},g_{3},\dots,g_{t}\\}$ is a minimal generating set for $G.$
Using Notation 4.1, since $\langle
h_{1},h_{2},h_{3},\dots,h_{t}\rangle=\langle h,h_{3},\dots,h_{t}\rangle$ and
$\langle h,C_{H}(V_{i})\rangle\neq H,$ there exists $k\geqslant 3$ such that
$h_{k}\not\in C_{H}(V_{i}).$ Since $\delta_{i}=1$, the matrix
$A^{(i)}=A^{(i)}(g_{1},\ldots,g_{t})$ has two rows. From $h_{k}\not\in
C_{H}(V_{i})$ we deduce that $1-\alpha_{i}(h_{k})\neq 0$, so there exists an
$\ell\in\\{1,2\\}$ such that the columns $a^{(i)}_{\ell}$ and $a^{(i)}_{k}$ of
$A^{(i)}$ are linearly independent. Moreover, if $j\neq i$ then
$a^{(j)}_{1}=a^{(j)}_{2}$. It follows from Lemma 4.2 that $\langle
g_{\ell},g_{3},\dots,g_{m}\rangle=G,$ a contradiction. This proves (i).

For (ii), assume for a contradiction that $|V_{i}|=|V_{j}|$ for distinct $i$
and $j$, and let $a\in V_{i}\setminus\\{0\\}$ and $b\in
V_{j}\setminus\\{0\\}$, so that $|a|=|b|$. Then $ab$ and $b$ are independent
in $G$, and hence Lemma 3.6 shows that there is a $G$-isomorphism, and hence
an $H$-isomorphism, between $V_{i}$ and $V_{j}$, a contradiction.

For (iii), assume, for a contradiction, that $p_{i}$ divides $|H|$ for some
$i\in\\{1,\dots,r\\}.$ Fix $v\in V_{i}$ and $h\in H$ with $|h|=|v|=p_{i}.$
Neither of $g_{1}:=h$ and $g_{2}:=hv$ is a power of the other, so there exist
$g_{3},\dots,g_{t}\in G$ such that $\\{g_{1},g_{2},g_{3},\dots,g_{t}\\}$ is a
minimal generating set for $G.$ The condition $|h|=|v|=p_{i}$ implies that
$h\in C_{H}(V_{i})$. With the notation of Notation 4.1, we notice that
$\langle h_{1},h_{2},h_{3},\dots,h_{t}\rangle=\langle
h,h_{3},\dots,h_{t}\rangle$, and if $j\neq i$ then $a^{(j)}_{1}=a^{(j)}_{2}$,
and $a^{(i)}_{1}$ is zero. But then $\langle g_{2},\dots,g_{t}\rangle=G,$ a
contradiction. ∎

The following example shows that three necessary conditions in Lemma 4.3 are
not sufficient to ensure that $G$ satisfies the independence property.

######  Example 4.4.

Consider the group

$G:=(V_{1}\times V_{2}^{2})\rtimes(\langle x\rangle\times\langle
y\rangle)\cong\operatorname{AGL}_{1}(3)\times(\mathbb{F}_{5}^{2}\rtimes\langle
2I_{2}\rangle)\leqslant\operatorname{AGL}_{1}(3)\times\operatorname{AGL}_{2}(5),$

so that $|V_{1}|=3,|V_{2}|=5,|x|=4,|y|=2$, $x\in C_{G}(V_{1}),$ $y\in
C_{G}(V_{2}),$ $w^{x}=2w$ for all $w\in V_{2}^{2}$ and $v^{y}=2v$ for all
$v\in V_{1}.$ Observe that $G$ satisfies Conditions (i)–(iii) of Lemma 4.3.

We claim that $x^{2}y$ and $y$ are dependent in $G$. To see this, suppose for
a contradiction that $\\{g_{1},\ldots,g_{r},x^{2}y,y\\}$ is a minimal
generating set for $G$, and let $R:=\langle g_{1},\ldots,g_{r},x^{2}y\rangle$.
It is not restrictive to assume that $g_{1},\ldots,g_{r}\in(V_{1}\times
V_{2}^{2})\langle x\rangle$. Notice that $RV_{1}\langle y\rangle=G$, and
therefore $R$ contains $xy^{k}v$ for some $k\in\mathbb{Z}$ and $v\in V_{1}$.
However, $(xy^{k}v)^{2}=x^{2}$, so $y\in R$, a contradiction. As $x^{2}y$ and
$y$ generate distinct subgroups of order $2$, the group $G$ does not satisfy
the independence property.

In the remainder of the section, we will determine some additional conditions.
For this purpose we need some definitions. Let $g_{1}$ and $g_{2}$ be as in
Notation 4.1. Then we say that $g_{1}$ and $g_{2}$ are _$j$ -independent_ for
some $j\in\\{1,\ldots,r\\}$ if either $\mathrm{rank}(A^{(j)}(g_{1},g_{2}))=2$
or, for some ordering $\lambda,\mu$ of $1$ and $2$, the column
$a^{(j)}_{\lambda}\neq 0$, $a^{(j)}_{\mu}=0$, and $\langle
h_{\lambda}\rangle\operatorname{\Phi}(H)\neq\langle
h_{1},h_{2}\rangle\operatorname{\Phi}(H).$

Similarly, $g_{1}$ and $g_{2}$ are _$\\{i,j\\}$ -independent_ for distinct
$i,j\in\\{1,\ldots,r\\}$ if $\langle
h_{1}\rangle\operatorname{\Phi}(H)=\langle h_{2}\rangle\operatorname{\Phi}(H)$
and for some ordering $\lambda,\mu$ of $1$ and $2$, the columns
$a^{(i)}_{\lambda}$ and $a^{(j)}_{\mu}$ are non-zero, and
$a^{(j)}_{\lambda}=a^{(i)}_{\mu}=0$.

######  Lemma 4.5.

Assume that $G$ satisfies the three conditions of Lemma 4.3. Two elements
$g_{1}=h_{1}(w_{1,1},\dots,w_{1,r})$ and $g_{2}=h_{2}(w_{2,1},\dots,w_{2,r})$
are independent in $G$ if and only if one of the following statements holds:

  1. (i)

$h_{1}$ and $h_{2}$ are independent in $H;$

  2. (ii)

there exists $j\in\\{1,\ldots,r\\}$ such that $g_{1}$ and $g_{2}$ are
$j$-independent; or

  3. (iii)

there exist distinct $i,j\in\\{1,\ldots,r\\}$ such that $g_{1}$ and $g_{2}$
are $\\{i,j\\}$-independent.

###### Proof.

If $G$ is abelian, i.e., if $G=H$, then (ii) and (iii) cannot hold, and (i) is
equivalent to $g_{1}$ and $g_{2}$ being independent. Thus we will assume that
$G$ is non-abelian. We first show the sufficiency of each of the three
conditions in turn, and then the necessity that at least one of them holds.
While proving the sufficiency of each condition, we may assume that $G$ is not
equal to $\langle g_{1},g_{2}\rangle$, as otherwise $g_{1}$ and $g_{2}$ are
independent.

If $h_{1}$ and $h_{2}$ are independent in $H$, then for some $a\geqslant 2$
there exist $h_{3},\dots,h_{a}$ in $H$ such that
$\\{h_{1},h_{2},h_{3},\dots,h_{a}\\}$ is a minimal generating set for $H.$
Then we take $v_{1},\dots,v_{b}\in W$ such that $\\{v_{1},\dots,v_{b}\\}$ is
minimal, subject to

$\\{g_{1},g_{2},h_{3},\dots,h_{a},v_{1},\dots,v_{b}\\}$

being a generating set for $G.$ In this way we obtain a minimal generating set
for $G$ containing $g_{1}$ and $g_{2}.$

Next suppose that $g_{1}$ and $g_{2}$ are $j$-independent for a fixed $j$.
First, let $A:=A^{(j)}(g_{1},g_{2})$, and assume that ${\rm{rank}}(A)=2$. Let
$\\{h_{3},\dots,h_{m}\\}$ be a subset of $H$ of minimal cardinality, subject
to $m\geqslant\delta_{j}+1$ and $H=\langle
h_{1},h_{2},h_{3},\dots,h_{m}\rangle.$ Since $H$ acts as scalars on $V_{j}$,
the group $H/C_{H}(V_{j})$ is cyclic, so we may choose $h_{3},\dots,h_{m}$ so
that $\langle h_{1},h_{2},h_{3},C_{H}(V_{j})\rangle=H$ and $h_{i}\in
C_{H}(V_{j})$ if $i\geqslant 4$. If $\langle
h_{1},h_{2},C_{H}(V_{j})\rangle=H$ and $m\geqslant 3,$ we can additionally
require $h_{3}\in C_{H}(V_{j}).$ Notice in particular that if $\delta_{j}=1$,
then $H/C_{H}(V_{j})$ is assumed to be of prime order, so $\langle
h_{1},h_{2},C_{H}(V_{j})\rangle=H$ and either $m=2$ or $h_{3}\in
C_{H}(V_{j})$. If $m>2$, then choose $z_{i,k}\in\mathbb{F}_{p_{j}}$ for
$i\in\\{3,\ldots,\delta_{j}+1\\}$ and $k\in\\{1,\ldots,\delta_{j}\\}$ such
that

${\rm{rank}}\begin{pmatrix}1-\alpha_{j}(h_{1})&1-\alpha_{j}(h_{2})&1-\alpha_{j}(h_{3})&0&\dots&0\\\
x_{1,j,1}&x_{2,j,1}&z_{3,1}&z_{4,1}&\dots&z_{\delta_{j+1},1}\\\
\vdots&\vdots&\vdots&\vdots&\cdots&\vdots\\\
x_{1,j,\delta_{j}}&x_{2,j,\delta_{j}}&z_{3,\delta_{j}}&z_{4,\delta_{j}}&\dots&z_{\delta_{j+1},\delta_{j}}\end{pmatrix}=\delta_{j}+1,$

which is possible since $\mathrm{rank}(A)=2$. Set
$g_{i}:=h_{i}(z_{i,1},\dots,z_{i,\delta_{j}})$ for each
$i\in\\{3,\ldots,\delta_{j}+1\\}$, and $g_{i}:=h_{i}$ for each
$i\in\\{\delta_{j}+2,\ldots,m\\}$. Additionally, let

$Z_{j}:=\underset{i\in\\{1,\ldots,r\\}\setminus\\{j\\}}{\prod}W_{i}.$

Then, by Lemma 4.2, $\\{g_{1}Z_{j},g_{2}Z_{j},g_{3}Z_{j},\dots,g_{m}Z_{j}\\}$
is a minimal generating set for $G/Z_{j}$ and therefore there exist elements
$g_{m+1},\ldots,g_{m+\ell}$ of $Z_{j}$, for some $\ell$, such that
$\\{g_{1},g_{2},g_{3},\dots,g_{m+\ell}\\}$ is a minimal generating set for
$G$.

Hence without loss of generality we may assume that $a^{(j)}_{1}\neq 0$,
$a^{(j)}_{2}=0$ and $\langle h_{1}\rangle\operatorname{\Phi}(H)\neq\langle
h_{1},h_{2}\rangle\operatorname{\Phi}(H).$ There exists a (possibly empty)
subset $\\{h_{3},\dots,h_{\ell}\\}$ of $H$ such that
$\\{h_{2},h_{3},\dots,h_{\ell}\\}$ is minimal subject to $\langle
h_{1},h_{2},h_{3},\dots,h_{\ell}\rangle=H.$ Since $H/C_{H}(V_{j})$ is cyclic,
we may choose $h_{4},\dots,h_{\ell}$ to lie in $C_{H}(V_{j})$. Let
$m:=\max\\{\ell,\delta_{j}+2\\}$ and set $h_{i}=1$ for each
$i\in\\{\ell+1,\ldots,m\\}$. If $\alpha_{j}(h_{1})=1$ then
$\alpha_{j}(h_{3})\neq 1$, since $H$ does not centralise $V_{j}$. Thus we may
choose $z_{i,k}\in\mathbb{F}_{p_{j}}$ for $i\in\\{3,\ldots,\delta_{j}+2\\}$
and $k\in\\{1,\ldots,\delta_{j}\\}$ such that

${\rm{rank}}\begin{pmatrix}1-\alpha_{j}(h_{1})&1-\alpha_{j}(h_{3})&0&\dots&0\\\
x_{1,j,1}&z_{3,1}&z_{4,1}&\dots&z_{\delta_{j+2},1}\\\
\vdots&\vdots&\vdots&\cdots&\vdots\\\
x_{1,j,\delta_{j}}&z_{3,\delta_{j}}&z_{4,\delta_{j}}&\dots&z_{\delta_{j+2},\delta_{j}}\end{pmatrix}=\delta_{j}+1.$

Set $g_{i}:=h_{i}(z_{i,1},\dots,z_{i,\delta_{j}})$ for each
$i\in\\{3,\ldots,\delta_{j}+2\\}$, and $g_{i}:=h_{i}$ for each
$i\in\\{\delta_{j}+3,\ldots,m\\}$. Then
$\\{g_{1}Z_{j},g_{2}Z_{j},g_{3}Z_{j},\dots,g_{m}Z_{j}\\}$ is a minimal
generating set for $G/Z_{j}$ and as in the previous case, there exist elements
of $Z_{j}$ that extend $\\{g_{1},g_{2},g_{3},\dots,g_{m}\\}$ to a minimal
generating set for $G$.

Finally, for this direction of the proof, assume that $g_{1}$ and $g_{2}$ are
$\\{i,j\\}$-independent for distinct $i$ and $j$, so that $\langle
h_{1}\rangle\operatorname{\Phi}(H)=\langle
h_{2}\rangle\operatorname{\Phi}(H)$. Without loss of generality, $a^{(j)}_{1}$
and $a^{(i)}_{2}$ are non-zero, whilst $a^{(j)}_{2}=a^{(i)}_{1}=0.$ Let
$B:=\\{h_{3},\dots,h_{m}\\}$ be a subset of $H$ of minimal cardinality subject
to $m\geqslant\max\\{\delta_{i},\delta_{j}\\}+2$ and $H=\langle
h_{1},B\rangle=\langle h_{2},B\rangle.$ Without loss of generality, suppose
that $\delta_{i}\leqslant\delta_{j}$. Let $C:=C_{G}(V_{i})\cap C_{G}(V_{j})$.
We shall split into two cases to place various assumptions on $B$ and define
associated matrices over $\mathbb{F}_{p_{i}}$ and $\mathbb{F}_{p_{j}}$, before
concluding both cases of the proof.

#### Case (a):

$\delta_{j}\geqslant 2$, or $\delta_{i}=\delta_{j}=1$ and either $G/C$ is
cyclic or $\langle h_{1},h_{2}\rangle\not\leqslant C$. Since $H/C_{H}(V_{j})$
and $H/C_{H}(V_{i})$ are cyclic, if $\delta_{j}\geqslant 2$ then we may assume
that $h_{4},\dots,h_{m}$ lie in $C_{H}(V_{i})$, and satisfy $h_{k}\in C\cap H$
if $k>4.$ Similarly, if $\delta_{i}=\delta_{j}=1$, then $H/(C\cap H)\cong G/C$
is either cyclic or isomorphic to $C_{q}^{2}$ for some prime $q$ (this is
because $\langle h_{1},h_{2}\rangle\not\leqslant C$ is equivalent to
$h_{1},h_{2}\notin C$), so we may assume in this case that
$h_{4},\dots,h_{m}\in C\cap H$. In all three cases, we may choose elements
$z_{\ell,s,k}\in\mathbb{F}_{p_{s}}$ for $\ell\in\\{3,\ldots,m\\}$,
$s\in\\{i,j\\}$ and $k\in\\{1,\ldots,\delta_{s}\\}$ such that

${\rm{rank}}\begin{pmatrix}1-\alpha_{s}(h_{f(s)})&1-\alpha_{s}(h_{3})&\dots&1-\alpha_{s}(h_{\delta_{s}+2})\\\
x_{f(s),s,1}&z_{3,s,1}&\cdots&z_{\delta_{s}+2,s,1}\\\
\vdots&\vdots&\cdots&\vdots\\\
x_{f(s),s,\delta_{s}}&z_{3,s,\delta_{s}}&\cdots&z_{\delta_{s}+2,s,\delta_{s}}\end{pmatrix}=\delta_{s}+1$

for each $s$, where $f(j):=1$ and $f(i):=2$, and $z_{\ell,s,k}=0$
$\ell>\delta_{s}+2$.

#### Case (b):

$\delta_{i}=\delta_{j}=1$, $G/C$ is not cyclic, and $\langle
h_{1},h_{2}\rangle\leqslant C$. Here, our assumptions on $G$ imply that
$H/(C\cap H)\cong C_{q}^{2}$ for some prime $q$. Hence we may assume that
$h_{3}\in C_{H}(V_{i})\setminus C_{H}(V_{j})$, $h_{4}\in C_{H}(V_{j})\setminus
C_{H}(V_{i})$, and $h_{5},\ldots,h_{m}\in C\cap H$. Since $h_{1},h_{2}\in C$,
and $a^{(j)}_{1}$ and $a^{(i)}_{2}$ are non-zero by assumption, we observe
that

${\rm{rank}}\begin{pmatrix}1-\alpha_{j}(h_{1})&1-\alpha_{j}(h_{3})\\\
x_{1,j,1}&0\end{pmatrix}={\rm{rank}}\begin{pmatrix}1-\alpha_{i}(h_{2})&1-\alpha_{i}(h_{4})\\\
x_{2,i,1}&0\end{pmatrix}=2.$

We set $z_{k,s,1}:=0$ for all $k\in\\{3,\ldots,m\\}$ and $s\in\\{i,j\\}$.

To conclude Cases (a) and (b), we set

$g_{k}:=h_{k}(z_{k,j,1},\dots,z_{k,j,\delta_{j}})(z_{k,i,1},\dots,z_{k,i,\delta_{i}})$

for each $k\in\\{3,\ldots,m\\}$, and let $Z_{i,j}:=\prod_{k\neq i,j}W_{k}$.
Then $G/Z_{i,j}$ has minimal generating set
$\\{g_{1}Z_{i,j},g_{2}Z_{i,j},g_{3}Z_{i,j},\dots,g_{m}Z_{i,j}\\}$, and
therefore there exist elements of $Z_{i,j}$ that extend
$\\{g_{1},\dots,g_{m}\\}$ to a minimal generating set for $G$, as required.

For the converse direction, suppose that $h_{1}$, and $h_{2}$ are dependent in
$H$, and that there is no $j\in\\{1,\dots,r\\}$ such that $g_{1}$ and $g_{2}$
are $j$-independent and no 2-set $\\{i,j\\}$ such that $g_{1}$ and $g_{2}$ are
$\\{i,j\\}$-independent. Assume, for a contradiction, that
$\\{g_{1},g_{2},g_{3},\dots,g_{d}\\}$ is a minimal generating set for $G$,
with $d\geqslant 2$. We may assume that $h_{2}=h_{1}^{t}f$ with
$t\in\mathbb{Z}$ and $f\in\operatorname{\Phi}(H),$ so in particular $H=\langle
h_{1},h_{3},\dots,h_{d}\rangle.$

For each $j$, let $A^{(j)}:=A^{(j)}(g_{1},\ldots,g_{d})$ and let $B^{(j)}$ be
the matrix obtained from $A^{(j)}$ by deleting its second column. Since
$g_{1}$ and $g_{2}$ are $j$-dependent, columns $a^{(j)}_{1}$ and $a^{(j)}_{2}$
are linearly dependent. If $a^{(j)}_{2}=0$ whenever $a^{(j)}_{1}=0$, then
${\mathrm{rank}}(A^{(j)})={\mathrm{rank}}(B^{(j)})$ for all $j$, and therefore
$G=\langle g_{1},g_{3},\dots,g_{d}\rangle$ by Lemma 4.2, a contradiction. So
there exists a $k$ such that $a^{(k)}_{1}=0$ and $a^{(k)}_{2}\neq 0$. Since
$g_{1}$ and $g_{2}$ are $k$-dependent, $\langle
h_{2}\rangle\operatorname{\Phi}(H)=\langle
h_{1},h_{2}\rangle\operatorname{\Phi}(H)=\langle
h_{1},h_{1}^{t}f\rangle\operatorname{\Phi}(H)=\langle
h_{1}\rangle\operatorname{\Phi}(H).$ In particular $H=\langle
h_{2},h_{3},\dots,h_{d}\rangle$ and as before we can conclude that there
exists an $\ell$ such that $a^{(\ell)}_{2}=0$ and $a^{(\ell)}_{1}\neq 0$. But
then $g_{1}$ and $g_{2}$ are $\\{k,\ell\\}$-independent, a contradiction. ∎

In what follows, we let $I_{h}:=\\{i\in\\{1,\dots,r\\}\mid h\in
C_{H}(V_{i})\\}$ for $h\in H$, and set $F:=\operatorname{\Phi}(H)$.

######  Corollary 4.6.

Assume that $G$ satisfies the three conditions of Lemma 4.3, and let $x,y\in
H$ be such that $y\in\langle x\rangle F$ and $I_{x}\subseteq I_{y}$. Then $x$
and $y$ are dependent in $G$. If, in addition, $G$ satisfies the independence
property and $I_{x}\neq\varnothing,$ then $y\in\langle x\rangle$.

###### Proof.

If $G=H$, then the result is clear, so assume that $r\geqslant 1$. For each
$j\in\\{1,\ldots,r\\}$, let $A^{(j)}:=A^{(j)}(x,y)$. Clearly
$\mathrm{rank}(A^{(j)})\leqslant 1$ for every $j.$ Notice that $a^{(j)}_{1}=0$
if and only if $x\in C_{H}(V_{j})$, and $a^{(j)}_{2}=0$ if and only if $y\in
C_{H}(V_{j})$. In particular, since $I_{x}\subseteq I_{y}$, if $a^{(j)}_{1}=0$
then $a^{(j)}_{2}=0$. This, together with $\langle x,y\rangle F=\langle
x\rangle F,$ implies that $x$ and $y$ are $j$-dependent for all $j$, and
$\\{j_{1},j_{2}\\}$-dependent for all 2-subsets $\\{j_{1},j_{2}\\}$. The
result now follows from Lemma 4.5.

Assume now in addition that $G$ satisfies the independence property and that
$I_{x}\neq\varnothing.$ Let $k\in I_{x}\cap I_{y}$, $0\neq v\in V_{k}$, and
for $j\in\\{1,\ldots,r\\}$ let $\tilde{A}^{(j)}:=A^{(j)}(xv,y)$, with columns
$\tilde{a}^{(j)}_{1}$ and $\tilde{a}^{(j)}_{2}$. Then $\tilde{a}^{(k)}_{1}\neq
0$, $\tilde{a}^{(k)}_{2}=0$, and $\tilde{A}^{(j)}=A^{(j)}$ if $j\neq k$. Thus,
arguing as before, we conclude that $xv$ and $y$ are dependent, and
consequently one of $xv$ and $y$ is a power of the other. As $H\cap V_{k}$ is
trivial, we deduce that $y\in\langle xv\rangle$, and in fact $y\in\langle
x\rangle$.∎

Thus if $G$ satisfies the independence property, then Conditions (a)–(d) of
Theorem 5(iii) hold. The following result completes the proof of Theorem 5.

######  Proposition 4.7.

Assume that $G$ satisfies Conditions (a)–(d) of Theorem 5(iii). Then $G$
satisfies the independence property.

###### Proof.

Let $g_{1}:=h_{1}w_{1}$ and $g_{2}:=h_{2}w_{2}$ be elements, as in Notation
4.1, that are dependent in $G$. We shall prove that one of $g_{1}$ and $g_{2}$
is a power of the other.

By Lemma 4.5, the fact that $g_{1}$ and $g_{2}$ are dependent in $G$ implies
that $h_{1}$ and $h_{2}$ are dependent in $H$. We may therefore assume
throughout the proof that $h_{2}=h_{1}^{u}f$ for some $u\in\mathbb{Z}$ and
$f\in F$. If $r=0$ then $G=H$, and so $F=1$ and $g_{2}=g_{1}^{u}$, as
required. Hence we may assume that $r\geqslant 1$.

For $1\leqslant j\leqslant r,$ consider the matrix
$A^{(j)}:=A^{(j)}(g_{1},g_{2})$. We shall repeatedly use the fact that

$\mathrm{rank}(A^{(j)})<2\mbox{ for all }j\in\\{1,\ldots,r\\},$

since otherwise $g_{1}$ and $g_{2}$ are $j$-independent, and hence by Lemma
4.5 are independent, a contradiction. Note that that lemma’s proof uses
Condition (a).

By conjugating $g_{1}$ and $g_{2}$ by a common element of $W$, if necessary,
we may assume that $w_{1,j}=0$ whenever $j\notin I_{h_{1}}$. Hence if $j\notin
I_{h_{1}}$ then $w_{2,j}=0,$ and otherwise $\mathrm{rank}(A^{(j)})=2$. Thus if
$I_{h_{1}}=\varnothing$, then $g_{1}=h_{1}$ and $g_{2}=h_{2}$, and the result
follows from Condition (d). If $I_{h_{2}}=\varnothing$, then we reach the same
conclusion by a similar argument, corresponding to conjugation by a different
element of $W$. Therefore, we shall assume for the remainder of the proof that
$I_{h_{1}},I_{h_{2}}\neq\varnothing$, and that $w_{1,j}=w_{2,j}=0$ whenever
$j\notin I_{h_{1}}$. We distinguish between two possibilities.

#### Case (a):

$\langle h_{2}\rangle F\neq\langle h_{1},h_{2}\rangle F$. Since $g_{1}$ and
$g_{2}$ are $j$-dependent for all $j\in\\{1,\ldots,r\\}$, if $a^{(j)}_{1}=0$
then $a^{(j)}_{2}=0$. Fix $k\in I_{h_{1}}$, so that
$a^{(k)}_{1}=(0,w_{1,k})^{T}.$ If $w_{1,k}=0,$ then $a^{(k)}_{1}=0$ and
consequently $a^{(k)}_{2}=0$, yielding $k\in I_{h_{2}}.$ If $w_{1,k}\neq 0$
and $k\notin I_{h_{2}},$ then $\mathrm{rank}(A^{(k)})=2,$ a contradiction. So
$I_{h_{1}}\subseteq I_{h_{2}}$.

Letting $\mathcal{C}_{1}:=\\{j\in I_{h_{1}}\mid w_{1,j}\neq 0\\}$ and
$\mathcal{C}_{2}:=\\{j\in I_{h_{1}}\mid w_{2,j}\neq 0\\}$, we see that
$\mathcal{C}_{2}\subseteq\mathcal{C}_{1}$,
$w_{1}=\prod_{j\in\mathcal{C}_{1}}w_{1,j}$, and
$w_{2}=\prod_{j\in\mathcal{C}_{2}}w_{2,j}$. Since $|V_{1}|,\dots,|V_{t}|$ are
pairwise coprime by Condition (b), and since $\mathrm{rank}(A^{(j)})<2$ for
each $j$, it follows that $w_{2}=w_{1}^{\ell}$ for some $\ell\in\mathbb{Z}$.
We also deduce from Condition (d) that $h_{2}=h_{1}^{t}$ for some
$t\in\mathbb{Z}$. Since $(|W|,|H|)=1$ by Condition (c), there exists
$s\in\mathbb{Z}$ such that $s\equiv t\mod|H|$ and $s\equiv\ell\mod|W|$, and so
$g_{2}=w_{2}h_{2}=w_{1}^{\ell}h_{1}^{t}=w_{1}^{s}h_{1}^{s}=g_{1}^{s}.$

#### Case (b):

$\langle h_{1}\rangle F=\langle h_{2}\rangle F,$ so that $h_{1}\in\langle
h_{2}\rangle F$ and $h_{2}\in\langle h_{1}\rangle F$. Assume first that there
exist $j\in I_{h_{1}}\setminus I_{h_{2}}$ and $k\in I_{h_{2}}\setminus
I_{h_{1}}$. If $a^{(j)}_{1}\neq 0$ then $\mathrm{rank}(A^{(j)})=2$, a
contradiction. So $a^{(j)}_{1}=0$, and similarly, $a^{(k)}_{2}=0$. But then
$g_{1},g_{2}$ are $\\{j,k\\}$-independent, and therefore, by Lemma 4.5,
independent in $G$, a contradiction.

We may therefore assume (by swapping $g_{1}$ and $g_{2}$ throughout the proof
if necessary) that $\varnothing\neq I_{h_{1}}\subseteq I_{h_{2}}$, so that by
Condition (d), $h_{2}=h_{1}^{t}$ for some $t\in\mathbb{Z}.$ Assume that
$\Omega:=I_{h_{2}}\setminus I_{h_{1}}\neq\varnothing.$ For each
$\omega\in\Omega,$ we see that $a^{(\omega)}_{1}\neq 0$ and
$a^{(\omega)}_{2}=0$. Therefore if $k\in I_{h_{1}}$ and $w_{2,k}\neq 0,$ then
$w_{1,k}\neq 0$ (otherwise $a^{(k)}_{2}$ is non-zero, $a^{(k)}_{1}=0$, and
$g_{1}$ and $g_{2}$ are $\\{\omega,k\\}$-independent, a contradiction). We
deduce that $w_{2}=w_{1}^{\ell}$ for some $\ell\in\mathbb{Z}$, and as in Case
(a), that $g_{2}\in\langle g_{1}\rangle.$

So we may assume that $I_{h_{1}}=I_{h_{2}}\neq\varnothing$. It follows from
two applications of Condition (d) that $\langle h_{1}\rangle=\langle
h_{2}\rangle$, $g_{1}=h_{1}(\prod_{j\in I_{h_{1}}}w_{1,j})$, and
$g_{2}=h_{2}(\prod_{j\in I_{h_{1}}}w_{2,j})$. Let $\mathcal{C}_{1}:=\\{j\in
I_{h_{1}}\mid w_{1,j}\neq 0\\}$ and $\mathcal{C}_{2}:=\\{j\in I_{h_{1}}\mid
w_{2,j}\neq 0\\}$. If neither $\mathcal{C}_{1}\subseteq\mathcal{C}_{2}$ nor
$\mathcal{C}_{2}\subseteq\mathcal{C}_{1}$, then $g_{1}$ and $g_{2}$ are
$\\{i,k\\}$-independent for all $i\in\mathcal{C}_{1}\setminus\mathcal{C}_{2}$
and $k\in\mathcal{C}_{2}\setminus\mathcal{C}_{1}$, a contradiction. So without
loss of generality $\mathcal{C}_{2}\subseteq\mathcal{C}_{1}$, and we conclude
again that $g_{2}$ is a power of $g_{1}.$ ∎

##  5\. Groups satisfying the rank-independence property

The aim of this section is to classify the finite groups that satisfy the
rank-independence property, so assume throughout this section that $G$ is a
finite group. Note that the proof of the following result uses the
classification of finite simple groups.

######  Theorem 5.1 ([40]).

If $N$ is a minimal normal subgroup of $G$ with $N\neq G,$ then $d(G)\leqslant
d(G/N)+1.$

We can easily reduce to the supersoluble case. In the next proof,
$\operatorname{Fit}(G)$ denotes the Fitting subgroup of $G$.

######  Proposition 5.2.

Assume that $G$ satisfies the rank-independence property. Then $G$ is
supersoluble.

###### Proof.

If $G$ is cyclic then the result is immediate, so assume otherwise. First we
claim that $G/\operatorname{\Phi}(G)$ is not simple, so assume otherwise for a
contradiction. Since $G$ is not cyclic, $G/\operatorname{\Phi}(G)$ is a non-
abelian simple group and therefore it contains two distinct involutions
$x\operatorname{\Phi}(G)$ and $y\operatorname{\Phi}(G)$. But then $\langle
x,y\rangle$ is not cyclic, so $x$ and $y$ are rank-independent. However,
$\langle x,y\rangle\operatorname{\Phi}(G)/\operatorname{\Phi}(G)$ is dihedral,
so in particular $\langle x,y\rangle$ is a proper subgroup of $G$,
contradicting the fact that $d(G)=d(G/\operatorname{\Phi}(G))=2$. Therefore
$G/\operatorname{\Phi}(G)$ is not simple.

Now let $N/\operatorname{\Phi}(G)$ be a minimal normal subgroup of
$G/\operatorname{\Phi}(G)$. By Theorem 5.1,
$d(G)=d(G/\operatorname{\Phi}(G))\leqslant d(G/N)+1.$ In particular, no two
distinct elements of $N$ are rank-independent, so every pair of such elements
generates a cyclic group. It follows that no generating set for $N$ of minimal
size contains more than one element, i.e., $N$ is cyclic. So
$\operatorname{Fit}(G)/\operatorname{\Phi}(G)$ is a direct product of cyclic
minimal normal subgroups of $G/\operatorname{\Phi}(G)$ and this implies that
$G/\operatorname{\Phi}(G)$ is supersoluble, and hence $G$ is supersoluble. ∎

######  Lemma 5.3.

Assume that $G$ satisfies the rank-independence property, and has a non-cyclic
normal $p$-subgroup $P$. Then $P$ is either elementary abelian or generalized
quaternion.

###### Proof.

If $\operatorname{\Phi}(P)=1$, then $P$ is elementary abelian, and the result
is immediate, so let $x$ be an element of $\operatorname{\Phi}(P)$ of order
$p.$ Since $\operatorname{\Phi}(P)\leqslant\operatorname{\Phi}(G),$ the
element $x$ does not lie in any minimal generating set for $G$, and so
$\langle x,y\rangle$ is cyclic for all $y\in G$. This implies in particular
that $\langle x\rangle$ is the unique minimal subgroup of $P.$ Since we are
assuming that $P$ is not cyclic, it follows from [50, Thm. 9.7.3] that $P$ is
a generalized quaternion group. ∎

######  Proposition 5.4.

Assume that $G$ is a non-cyclic nilpotent group. Then $G$ satisfies the rank-
independence property if and only if one of the following holds:

  1. (i)

$G\cong C_{p}\times C_{p}$;

  2. (ii)

$G\cong Q_{8}$;

  3. (iii)

$G\cong P\times C$, with $P$ an elementary abelian Sylow subgroup of $G$ such
that $d(P)\geqslant 3$, and $C$ cyclic (this includes the case $|C|=1$).

###### Proof.

It is easy to check that if $G$ satisfies (i), (ii) or (iii), then $G$
satisfies the rank-independence property. Conversely, suppose that $G$ has the
rank-independence property and let $d:=d(G).$ There exists a prime $p$
dividing $|G|$ such that the Sylow $p$-subgroup $P$ of $G$ satisfies $d(P)=d$.
Let $C$ be a $p$-complement in $G.$ Distinct elements of $C$ cannot belong to
a generating set for $G$ of cardinality $d$, so $C$ is cyclic. If $d\geqslant
3,$ then $P$ is an elementary abelian $p$-group by Lemma 5.3. If $d=2$, then
each proper subgroup of $G$ is cyclic, hence $G=P$ is isomorphic either to
$C_{p}\times C_{p}$ or to $Q_{8}.$ ∎

######  Proposition 5.5.

Assume that $G$ is not nilpotent, and that $d(G)\geqslant 3$. Then $G$
satisfies the rank-independence property if and only if $G=P\rtimes C$, where
$P$ is an elementary abelian Sylow $p$-subgroup of $G$ and $C$ is a cyclic
group, acting on $P$ as scalar multiplication.

###### Proof.

Assume that $G$ satisfies the rank-independence property and set
$F:=\operatorname{\Phi}(G).$ Since $G$ is supersoluble,

$G/F\cong(V_{1}^{\delta_{1}}\times\dots\times V_{r}^{\delta_{r}})\rtimes H$

where $H$ is abelian, $\delta_{1},\dots,\delta_{r}$ are positive integers, and
$V_{1},\dots,V_{r}$ are pairwise non-$H$-isomorphic, irreducible $H$-modules
on each of which $H$ acts non-trivially. Moreover, $r>0$ since $G$ is not
nilpotent, and for $i\in\\{1,\ldots,V_{i}\\}$ the group $V_{i}$ has prime
order $p_{i}$. By Lemma 4.2,

(5.1) $d:=d(G)=\max\\{d(H),\delta_{i}+1\mid 1\leqslant i\leqslant r\\}.$

Since $r\neq 0$, there exists a non-central minimal normal subgroup $N/F$ of
$G/F.$ Since $N$ is non-central, there exist $x\in N$ and $y\in G$ such that
$[x,y]\neq 1.$ Now $\langle x,y\rangle$ is not abelian, so there exist
$z_{3},\dots,z_{d}\in G$ such that $G=\langle x,y,z_{3},\dots,z_{d}\rangle.$
In particular, $G=N\langle y,z_{3},\dots,z_{d}\rangle$ and hence
$d(H)\leqslant d(G/N)\leqslant d-1$. It follows from (5.1) that
$d=\delta_{i}+1$ for some $i\in\\{1,\dots,r\\}$. We may assume that $i=1.$

Set $V:=V_{1}$, $p:=p_{1},$ and $\delta:=\delta_{1}=d-1.$ We identify each
element $w$ of $V^{{\delta}}$ with an element
$(x_{1},\dots,x_{\delta})\in\mathbb{F}_{p}^{{\delta}}$. Let $L=C/F$ be a
complement of $V^{{\delta}}$ in $G/F.$ For each $\ell$ in $L,$ there exists an
$\alpha(\ell)\in\mathbb{F}_{p}^{\times}$ such that $w^{\ell}=\alpha(\ell)w$
for all $w\in V^{{\delta}}.$ Given $g_{1},\dots,g_{d}\in G,$ we shall write
$g_{i}F=\ell_{i}(x_{i,1},\dots,x_{i,{\delta}})$, with
$x_{i,j}\in\mathbb{F}_{p}$. Consider the matrix

$A=A(g_{1},\dots,g_{d}):=\begin{pmatrix}1-\alpha(\ell_{1})&\cdots&1-\alpha(\ell_{d})\\\
x_{1,1}&\cdots&x_{d,1}\\\ \vdots&\cdots&\vdots\\\
x_{1,{\delta}}&\cdots&x_{d,{\delta}}\end{pmatrix},$

similar to Notation 4.1. It follows from Lemma 4.2 that $\langle
g_{1},\dots,g_{d}\rangle=G$ if and only if
$\langle\ell_{1},\dots,\ell_{d}\rangle=L$, and ${\rm{rank}}(A)=d.$

We now show that $C$ is cyclic. Assume for a contradiction that there exist
$g_{1},g_{2}\in C$ such that $\langle g_{1},g_{2}\rangle$ is not cyclic. Since
$G$ satisfies the rank-independence property, there exist
$g_{3},\dots,g_{d}\in G$ such that $\langle g_{1},\dots,g_{d}\rangle=G$.
However the first two columns of the matrix $A=A(g_{1},\ldots,g_{d})$ are
linearly dependent, contradicting $\mathrm{rank}(A)=d$. Since $C$ is cyclic,
$L$ is also cyclic and consequently $r=1,$ and so

$G/F\cong V^{{\delta}}\rtimes H.$

We show next that $p$ does not divide $|H|.$ Assume, for a contradiction, that
there exists $y_{1}\in G$ such that $y_{1}F$ is an element of $H$ of order
$p$, and choose $y_{2}\in G$ so that $y_{2}F$ is a non-trivial element of
$V^{{\delta}}.$ Then $K=\langle y_{1},y_{2}\rangle F\cong(C_{p}\times C_{p})F$
is a non-cyclic normal subgroup of $G$. As $G$ has the rank-independence
property, there exist $y_{3},\dots,y_{d}\in G$ such that $G=\langle
y_{1},y_{2},y_{3},\dots,y_{d}\rangle$. However, this is not possible, since
$d(G/K)=d(V^{{\delta}-1}\rtimes H/\langle y_{1}\rangle)>d-2$. Hence
$(p,|H|)=1$.

Let $P$ be a Sylow $p$-subgroup of $G$. From $(p,|H|)=1$ we deduce that $P\cap
C\leqslant F.$ Moreover, $P$ is a normal subgroup of $G$ and
$d(P)\geqslant{\delta}=d-1\geqslant 2,$ so, by Lemma 5.3, $P$ is either
elementary abelian or generalized quaternion. But in the second case, $|V|=2$,
contradicting the assumption that $H$ acts non-trivially on $V$. Thus $P$ is
elementary abelian, and consequently $P\cap C\leqslant P\cap F=1$. In
particular $P\cong V^{\delta}$ and $G\cong V^{\delta}\rtimes C$, so $p$ does
not divide $C$.

Conversely, assume $G=P\rtimes C$, with $P\cong C_{p}^{{\delta}}$,
${\delta}\geqslant 2,$ $(|C|,p)=1$ and $C$ acting on $P$ as scalar
multiplication. Since $G$ is not nilpotent, $C$ acts non-trivially on $P$.
Additionally, $d:=d(G)={\delta}+1.$ We again apply Lemma 4.2. We identify each
element of $P$ with a vector
$(y_{1},\dots,y_{\delta})\in\mathbb{F}_{p}^{\delta}$. For each $c\in C,$ there
exists an $\alpha(c)\in\mathbb{F}_{p}^{\times}$ such that $y^{c}=\alpha(c)y$
for all $y\in P.$ Let
$g_{1}:=c_{1}(y_{1,1},\dots,y_{1,{\delta}}),g_{2}:=c_{2}(y_{2,1},\dots,y_{2,{\delta}})\in
G$, and suppose $\langle x\rangle=C$. If $\langle g_{1},g_{2}\rangle$ is not
cyclic, then we may choose
$y_{3},y_{4},\ldots,y_{d}\in\mathbb{F}_{p}^{\delta}$ in such a way that
$\mathrm{rank}(A(g_{1},g_{2},xy_{3},y_{4},\ldots,y_{d}))$ is equal to $d$.
Then by Lemma 4.2, $\langle g_{1},g_{2},xy_{3},\dots,y_{d}\rangle=G$, as
required. ∎

Propositions 5.4 and 5.5 combine to prove Theorem 2. The following result
completes the proof of Theorem 1.

######  Proposition 5.6.

Assume that $G$ is not nilpotent, and that $d(G)=2$. Then $G$ satisfies the
rank-independence property if and only if $G$ is as described in Theorem
1(iii).

###### Proof.

The group $G$ has the rank-independence property if and only if all proper
subgroups of $G$ are cyclic. The conclusion follows from the description of
minimal non-abelian groups by Miller and Moreno in [46]. ∎

### Acknowledgements

We are grateful to Peter Cameron for helpful discussions, and in particular
for suggesting the problem of classifying the groups satisfying the rank-
independence property; and to anonymous referees for helpful comments. The
first author was supported by a St Leonard’s International Doctoral Fees
Scholarship and a School of Mathematics & Statistics PhD Funding Scholarship
at the University of St Andrews. The fourth author would like to thank the
Isaac Newton Institute for Mathematical Sciences, Cambridge, for support and
hospitality during the programme “Groups, Representations and Applications:
New perspectives”, where work on this paper was undertaken. This work was
supported by EPSRC grant no EP/R014604/1, and also partially supported by a
grant from the Simons Foundation.

## References

  * [1] C. Acciarri and A. Lucchini.  Graphs encoding the generating properties of a finite group.  Math. Nachr., 293(9):1644–1674, 2020.
  * [2] Á. Bereczky.  Maximal overgroups of Singer elements in classical groups.  J. Algebra, 234(1):187–206, 2000.
  * [3] W. Bosma, J. Cannon, and C. Playoust.  The Magma algebra system. I. The user language.  J. Symbolic Comput., 24(3-4):235–265, 1997.
  * [4] J. N. Bray, D. F. Holt, and C. M. Roney-Dougal.  The maximal subgroups of the low-dimensional finite classical groups, volume 407 of London Math. Soc. Lecture Note Ser. CUP, Cambridge, 2013.
  * [5] T. Breuer.  The GAP Character Table Library, Version 1.3.1, 2020\.
  * [6] T. Breuer, R. M. Guralnick, and W. M. Kantor.  Probabilistic generation of finite simple groups. II.  J. Algebra, 320(2):443–494, 2008.
  * [7] T. C. Burness and M. Giudici.  Classical groups, derangements and primes, volume 25 of Aust. Math. Soc. Lecture Ser.  CUP, Cambridge, 2016.
  * [8] T. C. Burness, R. M. Guralnick, and S. Harper.  The spread of a finite group.  Ann. of Math. (2), 193(2):619–687, 2021.
  * [9] T. C. Burness and S. Harper.  On the uniform domination number of a finite simple group.  Trans. Amer. Math. Soc., 372(1):545–583, 2019.
  * [10] S. Burris and H. P. Sankappanavar. A course in universal algebra. Grad. Texts in Math., vol. 78, Springer-Verlag, New York-Berlin, 1981.
  * [11] A. A. Buturlakin and M. A. Grechkoseeva.  The cyclic structure of maximal tori in finite classical groups.  Algebra Logika, 46(2):129–156, 2007.
  * [12] P. J. Cameron and P. Cara. Independent generating sets and geometries for symmetric groups. J. Algebra, 258(2):641–650, 2002.
  * [13] P. J. Cameron and S. Ghosh. The power graph of a finite group. Discrete Math., 311(13):1220–1222, 2011.
  * [14] P. J. Cameron. The power graph of a finite group, II. J. Group Theory, 13(6):779–783, 2010.
  * [15] P. J. Cameron, H. Guerra, and Š. Jurina. The power graph of a torsion-free group. J. Algebraic Combinatorics, 49(1):83–98, 2019.
  * [16] P. J. Cameron and S. H. Jafari, On the connectivity and independence number of power graphs of groups. Graphs Combin., 36(3):895–904, 2020.
  * [17] P. J. Cameron and B. Kuzma. Between the enhanced power graph and the commuting graph. Preprint, 2020, arXiv:2012.03789.
  * [18] P. J. Cameron, P. Manna, and R. Mehatari. On finite groups whose power graph is a cograph. J. Algebra, 591:59–74, 2022.
  * [19] R. W. Carter.  Conjugacy classes in the Weyl group.  Compositio Math., 25:1–59, 1972.
  * [20] J. H. Conway, R. T. Curtis, S. P. Norton, R. A. Parker, and R. A. Wilson.  Atlas of finite groups.  OUP, Eynsham, 1985.
  * [21] G. De Franceschi.  Centralizers and conjugacy classes in finite classical groups.  PhD thesis, University of Auckland, 2018.
  * [22] D. I. Deriziotis.  Centralizers of semisimple elements in a Chevalley group.  Comm. Algebra, 9(19):1997–2014, 1981.
  * [23] A. A. Gal’t.  Strongly real elements in finite simple orthogonal groups.  Sibirsk. Mat. Zh., 51(2):241–248, 2010.
  * [24] The GAP Group.  GAP – Groups, Algorithms, and Programming, Version 4.11.1, 2021\.
  * [25] Z. Guangxiang.  On self-normalizing cyclic subgroups.  In The Arcata Conference on Representations of Finite Groups (Arcata, Calif., 1986), volume 47 of Proc. Sympos. Pure Math., pp419–422. Amer. Math. Soc., Providence, RI, 1987.
  * [26] R. Guralnick.  On the number of generators of a finite group.  Arch. Math. (Basel), 53(6):521–523, 1989.
  * [27] R. Guralnick and G. Malle.  Products of conjugacy classes and fixed point spaces.  J. Amer. Math. Soc., 25(1):77–121, 2012.
  * [28] R. Guralnick and G. Malle. Simple groups admit Beauville structures. J. Lond. Math. Soc. (2), 85(3):694–721, 2012.
  * [29] R. Guralnick, T. Penttila, C. E. Praeger, and J. Saxl.  Linear groups with orders having certain large prime divisors.  Proc. London Math. Soc. (3), 78(1):167–214, 1999.
  * [30] R. M. Guralnick and W. M. Kantor.  Probabilistic generation of finite simple groups.  J. Algebra, 234(2):743–792, 2000.
  * [31] R. M. Guralnick and G. Tracey.  On elements of finite groups contained in a unique maximal subgroup.  In preparation.
  * [32] S. Harper.  Flexibility in generating sets of finite groups.  Arch. Math. (Basel), 118(3):231–237, 2022.
  * [33] C. Hering.  Transitive linear groups and linear groups which contain irreducible subgroups of prime order.  Geometriae Dedicata, 2:425–460, 1974.
  * [34] M. D. Hestenes.  Singer groups.  Canadian J. Math., 22:492–513, 1970.
  * [35] B. Huppert.  Singer-Zyklen in klassischen Gruppen.  Math. Z., 117:141–150, 1970.
  * [36] C. Jordan.  Sur la limite de transitivité des groupes non alternés.  Bull. Soc. Math. France, 1:40–71, 1872/73.
  * [37] A. V. Kelarev and S. J. Quinn. A combinatorial property and power graphs of groups. In Contributions to general algebra, 12 (Vienna, 1999), pages 229–235. Heyn, Klagenfurt, 2000.
  * [38] P. Kleidman and M. Liebeck.  The subgroup structure of the finite classical groups, volume 129 of London Math. Soc. Lecture Note Ser.  CUP, Cambridge, 1990.
  * [39] P. B. Kleidman.  The maximal subgroups of the finite $8$-dimensional orthogonal groups $P\Omega^{+}_{8}(q)$ and of their automorphism groups.  J. Algebra, 110(1):173–242, 1987.
  * [40] A. Lucchini. Generators and minimal normal subgroups. Arch. Math. (Basel), 64(4):273–276, 1995.
  * [41] A. Lucchini. The independence graph of a finite group. Monatsh. Math., 193(4):845–856, 2020.
  * [42] A. Lucchini. Some questions on the number of generators of a finite group. Rend. Sem. Mat. Univ. Padova, 83:201–222, 1990.
  * [43] A. Lucchini and A. Maróti. On the clique number of the generating graph of a finite group. Proc. Amer. Math. Soc., 137(10):3207–3217, 2009.
  * [44] G. Malle, J. Saxl, and T. Weigel.  Generation of classical groups.  Geom. Dedicata, 49(1):85–116, 1994.
  * [45] G. Malle and D. Testerman.  Linear algebraic groups and finite groups of Lie type, volume 133 of Cambridge Studies in Advanced Mathematics.  CUP, Cambridge, 2011.
  * [46] G. A. Miller and H. C. Moreno. Non-abelian groups in which every subgroup is abelian. Trans. Amer. Math. Soc., 4(4):398–404, 1903.
  * [47] F. Nicolas.  A simple, polynomial-time algorithm for the matrix torsion problem.  Preprint, 2009, arXiv:0806.2068.
  * [48] R. P. Panda, S. Dalal, and J. Kumar. On the enhanced power graph of a finite group. Comm. Algebra, 49(4):1697–1716, 2021.
  * [49] C. E. Praeger.  Primitive prime divisor elements in finite classical groups.  In Groups St. Andrews 1997 in Bath, II, volume 261 of London Math. Soc. Lecture Note Ser., pp605–623. CUP, Cambridge, 1999.
  * [50] W. R. Scott.  Group theory.  Dover Publications, Inc., New York, second edition, 1987.
  * [51] G. M. Seitz.  On the subgroup structure of classical groups.  Comm. Algebra, 10(8):875–885, 1982.
  * [52] G. M. Tracey.  Minimal generation of transitive permutation groups.  J. Algebra, 509:40–100, 2018.
  * [53] F. D. Veldkamp.  Regular elements in anisotropic tori.  In Contributions to algebra (collection of papers dedicated to Ellis Kolchin), pp389–424. Academic Press, New York, 1977.
  * [54] G. E. Wall.  On the conjugacy classes in the unitary, symplectic and orthogonal groups.  J. Austral. Math. Soc., 3:1–62, 1963.
  * [55] T. S. Weigel.  Generation of exceptional groups of Lie type.  Geom. Dedicata, 41(1):63–87, 1992.
  * [56] H. Wielandt.  Finite permutation groups.  Academic Press, New York-London, 1964.
  * [57] J. Whiston. Maximal independent generating sets of the symmetric group. J. Algebra, 232(1):255–268, 2000.
  * [58] J. Whiston and J. Saxl. On the maximal size of independent generating sets of $PSL_{2}(q).$ J. Algebra, 258(2):651–657, 2002.
  * [59] S. Zahirović. The power graph of a torsion-free group determines the directed power graph. Discrete Appl. Math., 305:109–118, 2021.
  * [60] S. Zahirović, I. Bošnjak, and R. Madarśz. A study of enhanced power graphs of finite groups. J. Algebra Appl., 19(4):2050062, 2020.
  * [61] J. P. Zhang. On finite groups all of whose elements of the same order are conjugate in their automorphism groups. J. Algebra, 153(1):22-–36, 1992.