# On the proportion of $p$-elements in a finite group, and a modular Jordan
type theorem

Gareth Tracey  School of Mathematics, University of Birmingham, Edgbaston,
Birmingham, B15 2TT
United Kingdom<EMAIL_ADDRESS>
###### Abstract.

In 1878, Jordan proved that if a finite group $G$ has a faithful
representation of dimension $n$ over $\mathbb{C}$, then $G$ has a normal
abelian subgroup with index bounded above by a function of $n$. The same
result fails if one replaces $\mathbb{C}$ by a field of positive
characteristic, due to the presence of large unipotent and/or Lie type
subgroups. For this reason, a long-standing problem in group and
representation theory has been to find the “correct analogue” of Jordan’s
theorem in characteristic $p>0$. Progress has been made in a number of
different directions, most notably by Brauer and Feit in 1966; by Collins in
2008; and by Larsen and Pink in 2011. With a 1968 theorem of Steinberg in mind
(which shows that a significant proportion of elements in a simple group of
Lie type are unipotent), we prove in this paper that if a finite group $G$ has
a faithful representation over a field of characteristic $p$, then a
significant proportion of the elements of $G$ must have $p$-power order. We
prove similar results for permutation groups, and present a general method for
counting $p$-elements in finite groups. All of our results are best possible.

###### 1991 Mathematics Subject Classification:

20D20, 20D06, 20C20

Supported by the Engineering and Physical Sciences Research Council, grant
number EP/T017619/1.

##  1\. Introduction

The study of the dimensions of the faithful representations of a finite group
$G$ (over an arbitrary field) has a long and rich history. Amongst the early
results in this direction, one of the most famous is due to Jordan [18], who
proved in 1878 that if $n$ is the dimension of a faithful
$\mathbb{C}[G]$-representation, then $G$ has an abelian normal subgroup $A$
whose index is bounded above by a function $J(n)$ of $n$. The function $J(n)$
was shown to be at most $(n+1)!$ (for large enough $n$) by Collins [5] in 2007
(this bound is best possible – realised by the symmetric group acting on the
fully deleted permutation module over $\mathbb{C}$).

If we replace $\mathbb{C}$ by a field $\mathbb{F}$ of characteristic $p>0$,
then Jordan’s theorem fails, even if $\mathbb{F}$ is algebraically closed. For
example, the quasisimple group $\operatorname{SL}_{2}(\mathbb{F}_{p^{n}})$
embeds into
$\operatorname{GL}_{2n}(\mathbb{F}_{p})\leq\operatorname{GL}_{2n}(\overline{\mathbb{F}_{p}})$,
and the index of the largest abelian normal subgroup of
$\operatorname{SL}_{2}(\mathbb{F}_{p^{n}})$ depends on the prime $p$.

For this reason, finding the “correct” modular analogue of Jordan’s theorem
has been a well-studied problem in representation theory for a number of
years. In one direction, Brauer and Feit [1] (in proving a conjecture of
Kegel) showed in 1966 that if $\mathbb{F}$ is a field of characteristic $p>0$
and $G$ is a finite subgroup of $\mathrm{GL}_{n}(\mathbb{F})$ with a Sylow
$p$-subgroup of order $p^{m}$, then $G$ has an abelian normal subgroup of
index bounded above by a function of $m$ and $n$. In another direction,
Collins [6] relaxed the requirement that the normal subgroup $A$ of $G$ be
abelian, and proved that if $\mathbb{F}$ and
$G\leq\mathrm{GL}_{n}(\mathbb{F})$ are as above, with $O_{p}(G)=1$, then $G$
has an abelian normal subgroup $A$ with $E_{p}(G)A$ having index bounded above
by a function of $n$ (the group $E_{p}(G)$ is the subgroup of $G$ generated by
the quasisimple subnormal subgroups of $G$ which are Lie type groups in
characteristic $p$). A remarkable paper by Larsen and Pink [24] later proved a
similar result, but without relying on the Classification of Finite Simple
Groups (henceforth abbreviated to CFSG). See the preprint [3] by Breuillard
for another approach to the problem.

As the discussion above alludes to, the obstacle to Jordan’s theorem in
characteristic $p>0$ is the presence of large Lie type subgroups in
characteristic $p$, and large unipotent subgroups. By a 1968 theorem of
Steinberg [36], one property that is shared by these two classes of linear
groups is that a “large” proportion of their elements have $p$-power order. Of
course, all elements in a finite unipotent linear group have $p$-power order;
Steinberg’s theorem shows that around $1/q^{f(n)}$ elements of a Lie type
group $X=X_{n}(q)$ in characteristic $p$ have $p$-power order, where $f(n)$ is
a linear function of the Lie rank $n$ of $X$ (we discuss this in more detail
below).

In this paper, we prove a new modular analogue of Jordan’s theorem by showing
that although it is not true that a significant number of elements in a finite
linear group in characteristic $p>0$ are contained in an abelian normal
subgroup, it is true that a significant number of elements have $p$-power
order. This also shows that Steinberg’s result is indicative of the general
case for finite linear groups in characteristic $p$.

To state our theorem, we fix the following notation: for a positive real
number $m$, define $h_{p}(m):=3^{(m-1)/2}$ if $p=2$; $h_{p}(m):=20^{(m-1)/4}$
if $p=3$; and $h_{p}(m):=(p-1)!^{(m-1)/(p-2)}$ if $p>3$. Our theorem can now
be stated as follows.

######  Theorem 1.

Let $G$ be a finite group, let $p$ be prime, and suppose that $n$ is the
dimension of a faithful representation of $G$ over a field $\mathbb{F}_{q}$ of
size $q=p^{f}$. Define $\alpha(q):=96$ if $q:=5$, $\alpha(q):=144$ if $q=7$,
and $\alpha(q):=600$ if $q=11$. Then the proportion of elements of $G$ of
$p$-power order is at least $1/f(n,q)$, where

$f(n,q):=\begin{cases}[3(q^{3}-1)]^{n/3}h_{p}(n/3)&\text{if $p=2$;}\\\
640^{n/4}h_{p}(n/4)&\text{if $q=3$;}\\\ [2(q^{2}-1)]^{n/2}h_{p}(n/2)&\text{if
$q\in\\{9,27\\}$;}\\\ \alpha(q)^{n/2}h_{p}(n/2)&\text{if
$q\in\\{5,7,11\\}$;}\\\ (q-1)^{n}h_{p}(n)&\text{otherwise.}\end{cases}$

In particular, $|G:O^{p^{\prime}}(G)|\leq f(n,q)$.

We remark that constructing elements of prime power order in a given finite
group $G$ is an important tool in computational group theory (see for example
[19, 20, 21]). In practice, this is done by choosing elements at random (with
an “almost” uniform distribution) and computing their orders. Theorem 1 shows
that if $G$ is constructed as a linear group and $p$ is the characteristic of
the underlying field, then one should find a $p$-element relatively quickly in
this way.

Our methods also allow us to prove the following slightly stronger result,
when the representation in question is irreducible.

######  Theorem 2.

Let $G$ be a finite group, let $p$ be prime, and suppose that $n$ is the
dimension of a faithful irreducible representation of $G$ over a field
$\mathbb{F}_{q}$ of size $q=p^{f}$. Define $\alpha(q):=96$ if $q:=5$,
$\alpha(q):=144$ if $q=7$, and $\alpha(q):=600$ if $q=11$. Then the proportion
of elements of $G$ of $p$-power order is at least $1/i(n,q)$, where

$i(n,q):=\begin{cases}[3(q^{3}-1)]^{n/3}h_{p}(n/3)&\text{if $p=2$;}\\\
640^{n/4}h_{p}(n/4)&\text{if $q=3$;}\\\ [2(q^{2}-1)]^{n/2}h_{p}(n/2)&\text{if
$q=3$ and $4\nmid n$;}\\\ [2(q^{2}-1)]^{n/2}h_{p}(n/2)&\text{if
$q\in\\{9,27\\}$;}\\\ \alpha(q)^{n/2}h_{p}(n/2)&\text{if
$q\in\\{5,7,11\\}$;}\\\ (q-1)^{n}h_{p}(n)&\text{if $q\in\\{5,7,11\\}$ and $n$
is odd;}\\\ (q-1)^{n}h_{p}(n)&\text{otherwise.}\end{cases}$

In particular, $|G:O^{p^{\prime}}(G)|\leq i(n,q)\leq f(n,q)$, where $f(n,q)$
is as in Theorem 4.

We remark that it is not immediately obvious that the function $i(n,q)$
defined in Theorem 2 is bounded above by the function $f(n,q)$ from Theorem 1.
This is however, easy to see, and will be proved in Lemma 8.1.

The bounds in Theorems 1 and 2 are best possible. To see this, define the
integer $m$ and the permutation group $S$ as follows: if $p=2$, then set
$m=3^{k}$ and let $S$ be a Sylow $3$-subgroup of $\operatorname{Sym}_{m}$. If
$p=3$, then set $m=5^{k}$, and let $S:=(C_{5}\wr C_{4})\wr\ldots\wr(C_{5}\wr
C_{4})\leq\operatorname{Sym}_{m}$ be the iterated wreath product of $k$ copies
of $C_{5}\rtimes C_{4}\leq\operatorname{Sym}_{5}$. If $p\geq 5$, then set
$m:=(p-1)^{k}$, and let
$S:=\operatorname{Sym}_{p-1}\wr\ldots\wr\operatorname{Sym}_{p-1}\leq\operatorname{Sym}_{m}$
be the iterated wreath product of $k$ copies of $\operatorname{Sym}_{p-1}$.
Also, define the integer $r$ and the irreducible linear group
$R\leq\operatorname{GL}_{r}(q)$ as follows: if $q=2$, then set $r:=3$ and
define $R$ to be the semilinear group
$R:=\Gamma\mathrm{L}_{1}(q^{r})\cong(q^{r}-1)\rtimes
r\leq\operatorname{GL}_{r}(q)$. If $q=3$ then set $r:=4$ and
$R:=2^{1+4}.(C_{5}\rtimes C_{4})\leq\operatorname{GL}_{4}(q)$. If
$q\in\\{9,27\\}$, then set $r:=2$ and $R:=\Gamma\mathrm{L}_{1}(q^{r})$. If
$q\in\\{5,7\\}$, then set $r:=4$ and $R:=(q-1)\circ
2^{1+2}.\operatorname{Sp}_{2}(2)\leq\operatorname{GL}_{4}(q)$. If $q=11$, then
set $r:=2$ and $R:=2.\operatorname{Alt}_{5}\leq\operatorname{GL}_{2}(11)$. In
all other cases, define $r:=1$ and $R:=\operatorname{GL}_{1}(q)$.

Finally, set $n:=rm$ and $G:=R\wr S\leq\operatorname{GL}_{n}(q)$. In each
case, $G$ is an irreducible $p^{\prime}$-group of order $i(n,q)\leq f(n,q)$.
It also follows from these examples that the bounds in Theorems 1 and 2 are
best possible even if one replaces “$p$-elements” by “$p$-singular elements”
in the statements of the theorems. (An element $g$ of a finite group $G$ is
said to be _$p$ -singular_ if $p$ divides the order of $g$.) 111for bounds in
Theorems 1 and 2 that are exponential in $q$, note that $(p-1)!^{1/(p-2)}\leq
p-1$.

The bounds in Theorems 1 and 2 essentially follow from two other results, one
concerning unipotent elements in primitive linear groups, and the other
concerning $p$-elements in permutation groups.

We will now state these results, beginning with the former. Recall that an
irreducible linear group $X\leq\mathrm{GL}(V)$ is said to be _imprimitive_ if
the natural $X$-module $V$ has a direct sum decomposition
$V=V_{1}\oplus\ldots\oplus V_{t}$ with the property that for each $1\leq i\leq
t$, and each $x\in X$, either $V_{i}^{x}=V_{i}$ or $V_{i}^{x}\cap V_{i}=0$. If
no such decomposition exists, then the irreducible group $X\leq\mathrm{GL}(V)$
is said to be _primitive_. Our lower bound on the proportion of unipotent
elements in a primitive linear group can now be given as follows.

######  Theorem 3.

Let $G$ be a finite group, let $p$ be prime, and suppose that $n$ is the
dimension of a faithful primitive representation of $G$ over a field
$\mathbb{F}_{q}$ of size $q=p^{f}$.

  1. (i)

If $(G,n,q)$ lies in Table 1, then the proportion of elements of $G$ of
$p$-power order is precisely $1/P(G,n,q)$, where $P(G,n,q)$ is as given in the
fourth column of the table.

  2. (ii)

If $(G,n,q)$ does not lie in Table 1, then the proportion of elements of $G$
of $p$-power order is at least $1/[n_{p^{\prime}}(q^{n}-1)]$.

Here, and throughout the paper, $n_{p^{\prime}}$ is defined to be the the
$p^{\prime}$-part of $n$.

$G$ | $n$ | $q$ | $P(G,n,q)$
---|---|---|---
$6\circ 3^{1+2}.\operatorname{Sp}_{2}(3)$ | $3$ | $7$ | $1296$
$(q-1)\circ 2^{1+2}.\operatorname{Sp}_{2}(2)$ | $2$ | $5,7$ | $96,144$
$2^{1+4}.\operatorname{Sp}_{4}(2)$ | $4$ | $7$ | $23040$
$(q-1)\circ 2^{1+4}.\operatorname{Sp}_{4}(2)$ | $4$ | $7,11,13$ | $69120,115200,138240$
$2^{1+4}.(5\rtimes 4)$ | $4$ | $3$ | $640$
$4\circ 2^{4}.(\operatorname{Sp}_{2}(2)\wr\operatorname{Sym}_{2})$ | $4$ | $5$ | $4608$
$2^{1+4}.\operatorname{Sp}_{4}(2)^{\prime}$ | $4$ | $7$ | $11520$
$6\circ 2^{1+4}.\operatorname{Sp}_{4}(2)^{\prime}$ | $4$ | $7$ | $34560$
$6\circ 2^{1+4}.\operatorname{Sym}_{5}$ | $4$ | $7$ | $11520$
$(q-1)\circ 2.\operatorname{Alt}_{5}$ | $2$ | $11,19$ | $600,1140$
$10\circ(2.\operatorname{Alt}_{5}\circ 2.\operatorname{Alt}_{5}).2$ | $4$ | $11$ | $72000$
Table 1. The triples ($G,n,q)$ from Theorem 3(i).

The lower bound in Theorem 3(ii) is best possible, as one can see by
considering a Hall $p^{\prime}$-subgroup of the semilinear group
$\Gamma\mathrm{L}_{1}(q^{n})\cong(q^{n}-1)\rtimes
n\leq\operatorname{GL}_{n}(q)$. We remark that the bound in Theorem 3 is also
best possible (up to multiplication by a “small” constant) when one considers
primitive linear groups of order divisible by $p$. For example, in the theorem
of Steinberg [36] mentioned above, it is proved that the number of elements of
$p$-power order in $\mathrm{\operatorname{GL}}_{n}(q)$ is precisely
$q^{n(n-1)}$. It follows that the proportion of elements of $p$-power order in
$\mathrm{\operatorname{GL}}_{n}(q)$ is

$\displaystyle\dfrac{q^{n(n-1)}}{|\operatorname{GL}_{n}(q)|}$
$\displaystyle=\dfrac{q^{n(n-1)/2}}{\prod_{i=1}^{n}(q^{i}-1)}=\prod_{i=1}^{n}\left(\dfrac{q^{i}}{q^{i}-1}\right)\dfrac{q^{n(n-1)/2}}{\prod_{i=1}^{n}q^{i}}=\prod_{i=1}^{n}\left(\dfrac{q^{i}}{q^{i}-1}\right)\dfrac{1}{q^{n}}.$

Since $c:=\prod_{i=1}^{n}\left(\dfrac{q^{i}}{q^{i}-1}\right)$ is at least $1$,
and $1/c=\prod_{i=1}^{n}\left(1-\dfrac{1}{q^{i}}\right)\geq 1-1/q\geq 1/2$, we
see that the proportion of elements of $\operatorname{GL}_{n}(q)$ of $p$-power
order lies between $1/q^{n}$ and $2/q^{n}$. (We refer the reader to [27, Lemma
3.1] for tighter upper and lower bounds on the constant $c$).

Before proceeding to our results on permutation groups, we state a useful
consequence of Theorems 1, 2, and 3. For a prime $p$, let
$e=\mathrm{exp}_{p^{\prime}}(G):=\mathrm{lcm}\\{|x|\text{ : }x\in G\text{,
}(|x|,p)=1\\}$. Since all modular representations of a finite group $G$ in
characteristic $p$ can be realised over the field $\mathbb{F}_{p^{e}}$, we
have:

######  Corollary 4.

Let $\mathbb{F}$ be a field of characteristic $p>0$, let $G$ be a finite
group, and suppose that $n$ is the dimension of a faithful representation
$\rho$ of $G$ over $\mathbb{F}$. Set $e:=\mathrm{exp}_{p^{\prime}}(G)$. Then:

  1. (i)

The proportion of elements of $G$ of $p$-power order is at least
$1/f(n,p^{e})$, where $f(n,p^{e})$ is as defined in Theorem 1. In particular,
$|G:O^{p^{\prime}}(G)|\leq f(n,p^{e})$.

  2. (ii)

If $\rho$ is irreducible, then the proportion of elements of $G$ of $p$-power
order is at least $1/i(n,p^{e})$, where $i(n,p^{e})$ is as defined in Theorem
1. In particular, $|G:O^{p^{\prime}}(G)|\leq i(n,p^{e})$.

  3. (iii)

If $\rho$ is primitive and $\rho(G)$ is not one of the groups in Table 1, then
the proportion of elements of $G$ of $p$-power order is at least
$1/[n_{p^{\prime}}(p^{en}-1)]$. In particular, $|G:O^{p^{\prime}}(G)|\leq
n_{p^{\prime}}(p^{en}-1)$.

Next, we state our main result for permutation groups.

######  Theorem 5.

Let $G\leq\operatorname{Sym}_{n}$ be a permutation group, and let $p$ be
prime. Then the proportion of elements of $G$ of $p$-power order is at least
$1/h_{p}(n)$, where $h_{p}(n)$ is as defined on page 2.

The bound in Theorem 5 is again best possible: one can see this by taking
$n:=m$ and $G:=S\leq\operatorname{Sym}_{n}$, where $m$ and $S$ are as defined
in the discussion after the statement of Theorem 2.

Write $\mathrm{Lie}(p)$ for the set of finite simple groups of Lie type in
characteristic $p$. As a by-product of our methods we obtain upper bounds on
the minimal centralizer orders of non-trivial $p$-elements in the
$\operatorname{Aut}(S)$-cosets of $S$. In order to state our result, we
require some standard notation and terminology: If $S\in\mathrm{Lie}(p)$, then
$S=O^{p^{\prime}}(\overline{X}_{\overline{\sigma}})$, where $\overline{X}$ is
a simple algebraic group over the algebraic closure
$\overline{\mathbb{F}_{p}}$ of $\mathbb{F}_{p}$,
$\overline{\sigma}:\overline{X}\rightarrow\overline{X}$ is a Steinberg
endomorphism of $\overline{X}$, and
$\overline{X}_{\overline{\sigma}}=\\{x\in\overline{X}\text{ :
}\overline{\sigma}(x)=x\\}$. In particular, by [11, Theorem 2.2.3], we may
write $\overline{\sigma}=\overline{\sigma_{p}}^{f}\overline{\gamma}^{i}$,
where $\overline{\sigma_{p}}$ is a Frobenius automorphism of $\overline{X}$,
and either

  1. (a)

$\overline{X}\in\\{A_{\ell},D_{\ell},E_{6}\\}$ and $\overline{\gamma}$ is a
graph automorphism of $\overline{X}$ as in [11, Theorem 1.15.2(a)];

  2. (b)

$\overline{X}\in\\{B_{2}\text{ }(p=2),G_{2}\text{ }(p=3)\text{, }F_{4}\text{
}(p=2)\\}\\}$, and $\overline{\gamma}$ is as in [11, Theorem 1.15.4(b)]; or

  3. (c)

$\overline{\gamma}=1$.

In this case, we define $p^{f}$ to be the _level_ of $S$.

Our result for centralizer orders of $p$-elements in
$\operatorname{Aut}(S)$-cosets of $S$ can now be stated as follows.

######  Theorem 6.

Fix a prime $p$. Let $S$ be a group in $\mathrm{Lie}(p)$ of level $q$ and
untwisted Lie rank $\ell$, and write $S=\overline{X}_{\overline{\sigma}}$, for
a simple algebraic group $\overline{X}$ and a Steinberg endomorphism
$\overline{\sigma}:\overline{X}\rightarrow\overline{X}$, as above. Let
$\alpha$ be a $p$-element of $\operatorname{Aut}(S)$. Then there exists an
element $x$ of the coset $S\alpha$ such that one of the following holds:

  1. (i)

$|C_{\overline{X}_{\overline{\sigma}}}(x)|\leq q^{\ell}$; or

  2. (ii)

$S$ and $p$ are as in Table 2. In this case, an upper bound for
$|C_{\overline{X}_{\overline{\sigma}}}(x)|$ is given in the third column of
Table 2.

We will see in Section 4 that when $\alpha=1$, the element $x$ chosen in
Theorem 6 is a so-called _regular unipotent_ element of $S$. Moreover, the
primes involved in Theorem 6(ii) are the _bad primes_ associated to the simple
algebraic group $\overline{X}$ (see [4, page 28] for more information).

$S$ | $p$ | $|C_{\overline{X}_{\overline{\sigma}}}(x)|\leq$
---|---|---
$\operatorname{PSp}_{2\ell}(q)$ | $2$ | $2q^{\ell}$
$\operatorname{P\Omega}^{\pm}_{2\ell}(q)$ | $2$ | $2q^{\ell}$
${}^{2}B_{2}(q)$ | $2$ | $2q$
${}^{2}G_{2}(q)$ | $3$ | $3q$
${}^{3}D_{4}(q)$ | $2$ | $2q^{4}$
$G_{2}(q)$ | $2$ | $2q^{2}$
| $3$ | $3q^{2}$
${}^{2}F_{4}(q)$ | $2$ | $4q^{4}$
$F_{4}(q)$ | $2$ | $4q^{4}$
| $3$ | $3q^{4}$
$E^{\pm}_{6}(q)$ | $2,3$ | $6q^{6}$
$E_{7}(q)$ | $2$ | $4q^{7}$
| $3$ | $6q^{7}$
$E_{8}(q)$ | $2$ | $4q^{8}$
| $3$ | $3q^{8}$
| $5$ | $5q^{8}$
Table 2. Upper bounds for centralizer orders of regular unipotent elements of
certain simple groups.

We also prove the following order bound for sporadic or cross characteristic
Lie type sections of linear groups.

######  Theorem 7.

Let $p$ be a prime, and let $S$ be a nonabelian finite simple group which is
either sporadic or lies in $\mathrm{Lie}(r)$. Suppose that $S$ is a section of
$\mathrm{GL}_{n}(\mathbb{F})$ for some field $\mathbb{F}$ of characteristic
$p>0$, with $p\neq r$. Assume also that $S$ is not one of the groups
$M_{12},M_{22},M_{24},J_{2},\mathrm{Suz},Co_{1},Co_{2},\operatorname{L}_{2}(7),\operatorname{L}_{2}(8),\operatorname{L}_{2}(9),\operatorname{L}_{3}(4),\operatorname{U}_{4}(2),\operatorname{U}_{4}(3),\operatorname{PSp}_{6}(2),$
$\operatorname{PSp}_{6}(3)$, or $\operatorname{P\Omega}^{+}_{8}(2)$. Then
$|S|\leq 2^{2(\log{n})^{2}+\log_{2}{n}}$.

The bound in Theorem 7 is best possible, up to multiplication by an absolute
constant. Indeed, as we will see in Section 6, there are primitive linear
groups $G\leq\mathrm{GL}_{2^{m}}(\overline{\mathbb{F}_{p}})$, for $p$ odd,
with sections isomorphic to $\operatorname{Sp}_{2m}(2)$ (and
$\operatorname{Sp}_{2m}(2)$ has order $c2^{2m^{2}+m}$, where $1/2<c<1$ is an
absolute constant).

The layout of the paper is as follows. In Section 2, we prove two key
inductive lemmas which will enable us to obtain lower bounds on the number of
$p$-elements in any given finite group. The second of these lemmas will also
demonstrate the motivation behind Theorems 6 and 7. In Section 3, motivated by
our work in Section 2, we derive an upper bound on the minimal order of a
centralizer of a $p$-element in a finite alternating group. We then have all
the tools we need to prove Theorem 6, and we do so in Section 4. In Section 5,
we make a necessary detour to prove some results on the structure of a finite
primitive linear group. These results will allow us to prove Theorems 5 and 7,
and we do this in Section 6. Finally, we prove Theorem 3 in Section 7, and
Theorems 1 and 2 in Section 8.

Notation: Most of our notation is standard. For a prime $r$, we write $r^{m}$
and $r^{1+2m}$ for elementary abelian and extraspecial $r$-groups of order
$r^{m}$ and $r^{1+2m}$, respectively. We also occasionally write $n$ for the
cyclic group $\mathbb{Z}/n\mathbb{Z}$, when there is no danger of confusion.
For almost simple group names, we follow [22], except that we use
$\operatorname{Alt}_{n}$ and $\operatorname{Sym}_{n}$ for the alternating and
symmetric groups of degree $n$. When it is natural to do so, we will also
occasionally use Lie theoretic notation for simple groups in place of
classical notation (so $A_{\ell}(q)$ instead of $\operatorname{L}_{\ell+1}(q)$
etc.).

For a finite group $G$, we will write $Z(G)$, $\mathrm{Frat}(G)$, $F^{*}(G)$,
and $G^{\prime}$ for the centre, Frattini subgroup, generalized Fitting
subgroup, and derived subgroup of $G$, respectively. For a field $\mathbb{F}$,
we will write $\overline{\mathbb{F}}$ for the algebraic closure of
$\mathbb{F}$. Finally, for two groups $A$ and $B$, $A\circ B$ will denote the
central product of $A$ and $B$.

##  2\. Counting $p$-elements in finite groups

As mentioned above, the purpose of this section is to prove two inductive
lemmas which will allow us to count the number of $p$-elements in a given
finite group $G$. The first of these lemmas will enable us to reduce our
enumeration to the case where $G$ is characteristically simple, while the
second gives a lower bound on the number of $p$-elements of $G$ in terms of
similar numbers associated to the composition factors of $G$. The first lemma
can in fact be proved when $p$ is replaced by an arbitrary finite set of
primes, and we do prove this more general version here.

In order to state the lemmas, we require some notation. For a subset $X$ of a
finite group $G$ and a set $\pi$ of primes, write $\mathrm{Ord}(X,\pi)$ for
the set of $\pi$-elements of $X$ (that is, the set of elements $x\in X$ such
that all prime divisors of $|x|$ lie in $\pi$). If $G=S^{t}$ is the direct
power of a simple group $S$, define the function $f_{\pi}$ by $f_{\pi}(G):=1$
if $S$ is an abelian $\pi^{\prime}$-group; $f_{\pi}(G):=|G|$ if $G$ is an
abelian $\pi$-group; and

$\displaystyle f_{\pi}(G):=\min\\{|\operatorname{Ord}(G\alpha,\pi)|\text{ :
}\alpha\in\mathrm{Ord}(\operatorname{Aut}(G),\pi)\\}$

if $S$ is nonabelian. In this latter case, we are viewing $G\alpha$ as a
subset of $\operatorname{Aut}(G)$. If $\pi$ consists of a single prime $p$,
then we will write $\operatorname{Ord}(G,p)$ and $f_{p}(G)$ in place of
$\operatorname{Ord}(G,\\{p\\})$ and $f_{\\{p\\}}(G)$.

The first lemma can now be given as follows.

######  Lemma 2.1.

Let $G$ be a finite group, and let $\pi$ be a set of primes. Then
$|\operatorname{Ord}(G,\pi)|\geq\prod_{C\text{ a chief factor of
}G}f_{\pi}(C)$.

###### Proof.

Let $N$ be a minimal normal subgroup of $G$. Clearly, it will suffice to prove
that

(2.1) $\displaystyle|\operatorname{Ord}(G,\pi)|\geq
f_{\pi}(N)|\operatorname{Ord}(G/N,\pi)|.$

To this end, notice that a (right) coset $Ng$ of $N$ in $G$ contains a
$\pi$-element only if $Ng$ is a $\pi$-element of $G/N$. Thus, writing bars to
denote reduction modulo $N$, we have
$|\operatorname{Ord}(G,\pi)|=\sum_{\overline{g}\in\operatorname{Ord}(\overline{G},\pi)}|\operatorname{Ord}(Ng,\pi)|$.
Thus, to prove (2.1), it will suffice to show that if
$\overline{g}\in\operatorname{Ord}(\overline{G},\pi)$, then

(2.2) $|\operatorname{Ord}(Ng,\pi)|\geq f_{\pi}(N)$.

So fix $g\in G$ with $\overline{g}\in\operatorname{Ord}(\overline{G},\pi)$.
Since $\operatorname{Ord}(\overline{G},\pi)=\operatorname{Ord}(G,\pi)N/N$, we
may assume that $g\in\operatorname{Ord}(G,\pi)$. Set $H:=\langle
N,g\rangle=N\langle g\rangle$. Suppose first that $N$ is an elementary abelian
$r$-group, for some prime $r$. If $r\not\in\pi$, then $f_{\pi}(N)=1$; while if
$r\in\pi$, then $H$ is a $\pi$-group, so
$|\operatorname{Ord}(Ng,\pi)|=|Ng|=|N|=f_{\pi}(N)$. The assertion (2.2)
follows in each case.

Finally, assume that $N$ is non-abelian. Since $C_{H}(N)\cap N=1$ and
$\pi(|H/N|)\subseteq\pi$, the group $C_{H}(N)$ is a $\pi$-group. It follows
that $|\operatorname{Ord}(Ng,\pi)|$ is at least the number of $\pi$-elements
in the coset $\theta(Ng)$, where $\theta$ is the natural map
$\theta:H\rightarrow H/C_{H}(N)$. Since
$|\operatorname{Ord}(\theta(Ng),\pi)|\geq f_{\pi}(N)$ by definition of
$f_{\pi}$, the claim at (2.2) follows, and the proof is complete. ∎

To apply Lemma 2.1 in the case $\pi=\\{p\\}$, we will need lower bounds on the
quantity $f_{p}(N)$ when the chief factor $N$ of $G$ is nonabelian. Our next
lemma shows that such bounds can be derived from information on the
$S$-conjugacy classes in $\operatorname{Aut}(S)$-cosets of finite simple
groups $S$.

######  Lemma 2.2.

Fix a nonabelian finite simple group $S$, a positive integer $t$, and a prime
$p$. Set $N:=S^{t}$, and let $f_{p}(N)$ be as defined above. Assume that each
coset of $S$ in $\operatorname{Aut}(S)$ contains an $S$-conjugacy class of
$p$-elements of size at least $M$. Then $f_{p}(N)\geq M^{t}$.

###### Proof.

We note first that
$\operatorname{Aut}(N)\cong\operatorname{Aut}(S)\wr\operatorname{Sym}_{t}$,
and the conjugation action of $\operatorname{Aut}(N)$ on $S^{t}$ is given by
$(s_{1},\ldots,s_{t})^{(h_{1},\ldots,h_{t})}=(s_{1}^{h_{1}},\ldots,s_{t}^{h_{t}})$,
and
$(s_{1},\ldots,s_{t})^{\sigma^{-1}}=(s_{1^{\sigma}},\ldots,s_{t^{\sigma}})$,
where $h_{i},s_{i}\in S$, and $\sigma\in\operatorname{Sym}_{t}$.

Now, let $g$ be a $p$-element of $\operatorname{Aut}(N)$, and write
$g=(x_{1},\ldots,x_{t})\sigma$, where $g_{i}\in\operatorname{Aut}(S)$, and
$\sigma\in\operatorname{Sym}_{t}$. If
$P\in\operatorname{Syl}_{p}(\operatorname{Aut}(S))$ and
$Q\in\operatorname{Syl}_{p}(\operatorname{Sym}_{t})$, then $P\wr Q\cong
P^{t}\rtimes Q$ is a Sylow $p$-subgroup of $\operatorname{Aut}(N)$. Thus, we
may assume that the elements $x_{1},\ldots,x_{t},\sigma$ have $p$-power order.

Suppose that $\sigma$ has orbit lengths $t_{1},\ldots,t_{m}$, and write
$\sigma=\sigma_{1}\ldots\sigma_{m}$ as a product of disjoint cycles, where
$\sigma_{i}$ is a cycle of length $t_{i}$. Without loss of generality, assume
that $\sigma_{1}=(1,\ldots,t_{1})$, $\sigma_{2}=(t_{1}+1,\ldots,t_{1}+t_{2})$,
and so on. Set $g_{1}:=(x_{1},\ldots,x_{t_{1}})\sigma_{1}$,
$g_{2}:=(x_{t_{1}+1},\ldots,x_{t_{1}+t_{2}})\sigma_{2}$, etc. Then $g_{i}$ is
a $p$-element of $\operatorname{Aut}(S)\wr\operatorname{Sym}_{t_{i}}$ for each
$1\leq i\leq m$, and the coset $Ng$ is the cartesian product of the sets
$N_{1}g_{1}\times\ldots\times N_{m}g_{m}$, where $N_{i}=S^{t_{i}}$. It clearly
follows that $f_{p}(N)\geq f_{p}(N_{1})\ldots f_{p}(N_{m})$. Thus, we may
assume that $m=1$ and that $\sigma=(1,\ldots,t)$ is a $t$-cycle. In
particular, $t$ is a power of $p$.

Now, fix $y:=(y_{1},\ldots,y_{t})\sigma\in Ng$,
$z:=(z_{1},\ldots,z_{t})\sigma\in Ng$, and $h:=(h_{1},\ldots,h_{t})\in N$,
where $h_{i},y_{i},z_{i}\in\operatorname{Aut}(S)$. Then a routine exercise
shows that $y^{h}=z$ if and only if $h_{i}^{-1}y_{i}h_{i+1}=z_{i}$ for each
$1\leq i\leq t$, and $h_{t}^{-1}y_{t}h_{1}=z_{t}$. In particular,

(2.3) if $y^{h}=z$, then $h_{1}^{-1}(y_{1}\ldots y_{t})h_{1}=z_{1}\ldots
z_{t}$.

More generally, $h_{i}^{-1}(y_{i}y_{i+1}\ldots y_{t}y_{1}\ldots
y_{i-1})h_{1}=z_{i}z_{i+1}\ldots z_{t}z_{1}\ldots z_{i-1}$ for all $i$. We
also have $y^{t}=(y_{1}y_{2}\ldots y_{t},y_{2}y_{3}\ldots
y_{t}y_{1},\ldots,y_{t}y_{1}\ldots y_{t-1})$. In particular,

(2.4) $y$ has order $t|y_{1}\ldots y_{t}|$.

Next, let $\mathcal{C}(S)$ be a set of representatives for the $S$-conjugacy
classes of $p$-elements in the coset $Sx_{1}\ldots
x_{t}\subseteq\operatorname{Aut}(S)$. Fix $\alpha\in\mathcal{C}(S)$ with
$|\alpha^{S}|\geq M$, and write $\alpha=ax_{1}\ldots x_{t}$, with $a\in S$.
Consider the element

$\displaystyle x(\alpha):=(a,1,\ldots,1)g=(ax_{1},x_{2},\ldots,x_{t})\sigma\in
Ng.$

Since $\alpha=ax_{1}\ldots x_{t}$ is a $p$-element of $\operatorname{Aut}(S)$,
(2.4) implies that $x(\alpha)$ is a $p$-element of $Ng$ (recall that $t$ is a
power of $p$). Furthermore, (2.3) implies that if
$\alpha\neq\beta\in\mathcal{C}(S)$, then $x(\alpha)$ and $x(\beta)$ are not
$N$-conjugate. Moreover, (2.3) and the comments afterward imply that if
$(h_{1},\ldots,h_{t})\in C_{N}(x(\alpha))$, then $h_{1}\in C_{S}(ax_{1}\ldots
x_{t})$ and $h_{i}\in$ $C_{S}(x_{i}x_{i+1}\ldots x_{t}ax_{1}\ldots x_{i-1})$
for $i>1$. Since $ax_{1}\ldots x_{t}$ and $x_{i}x_{i+1}\ldots
x_{t}ax_{1}\ldots x_{i-1}$ are $\operatorname{Aut}(S)$-conjugate for $i>1$, we
deduce that $|C_{N}(x(\alpha))|\leq|C_{S}(ax_{1}\ldots x_{t})|^{t}$. Hence,
$|x(\alpha)^{N}|\geq|\alpha^{S}|^{t}\geq M^{t}$.

Finally, let $\mathcal{C}(N)$ be a set of representatives for the
$N$-conjugacy classes of $p$-elements in the coset $Ng$, so that
$|\operatorname{Ord}(Ng,p)|=\sum_{x\in\mathcal{C}(N)}|x^{N}|$. By our work in
the previous paragraph, we have

$\displaystyle|\operatorname{Ord}(Ng,p)|=\sum_{x\in\mathcal{C}(N)}|x^{N}|\geq\sum_{\alpha\in
C(S)}|x(\alpha)^{N}|\geq\sum_{\alpha\in C(S)}|\alpha^{S}|^{t}\geq M^{t}.$

This completes the proof. ∎

We can see from Lemmas 2.1 and 2.2 that lower bounds on the sizes of
$S$-conjugacy classes of $p$-elements in $\operatorname{Aut}(S)$-cosets of
nonabelian finite simple groups $S$ can be used to determine lower bounds on
the number of $p$-elements in an arbitrary finite group. This is where our
motivation for Theorems 6 and 7 come from. We will prove these theorems in
Sections 4 and 6 respectively.

We conclude this section with a useful corollary of Lemmas 2.1 and 2.2. To
state it, we require the following notation: for a finite simple group $S$ and
a prime $p$, define $M_{p}(S):=|S|$ if $S$ is a cyclic $p^{\prime}$-group;
$M_{p}(S):=1$ if $S$ is cyclic of order $p$; and if $S$ is a nonabelian finite
simple group, then define $M_{p}(S)$ to be the minimum $M$ such that whenever
$\alpha$ is a $p$-element of $\operatorname{Aut}(S)$, then the coset
$S\alpha\subseteq\operatorname{Aut}(S)$ contains a $p$-element $x$ with
$|C_{S}(x)|\leq M$. Finally, for a finite group $G$, define
$M_{p}(G)=\prod_{S\text{ a composition factor of }G}M_{p}(S)$.

We also need the following standard terminology: If $G_{1},\ldots,G_{t}$ are
finite groups, we say that $G$ is a _subdirect subgroup_ of the direct product
$X:=G_{1}\times\ldots\times G_{t}$ if $G\leq X$ and the projection map
$\pi_{i}:G\rightarrow G_{i}$ is surjective for all $1\leq i\leq t$. Our
corollary can now be stated as follows.

######  Corollary 2.3.

Fix a prime $p$, and a finite group $G$. Then:

  1. (i)

If $N\unlhd G$, then $M_{p}(G)=M_{p}(N)M_{p}(G/N)$;

  2. (ii)

$M_{p}(G)\geq|G|/\operatorname{Ord}(G,p)$; and

  3. (iii)

Suppose that $G$ is a subdirect subgroup of a direct product
$G_{1}\times\ldots\times G_{t}$ of finite groups $G_{i}$. Then
$M_{p}(G)\leq\prod_{i=1}^{t}M_{p}(G_{i})$.

###### Proof.

Part (i) follows immediately from the definition of the function $M_{p}$,
while (ii) follows from Lemmas 2.1 and 2.2. For (iii), note that if $G$ is a
subdirect subgroup of $G_{1}\times\ldots\times G_{t}$, then the intersection
of $G$ with the coordinate subgroup $\hat{G}_{1}:=\\{(g_{1},1,\ldots,1)\text{
: }g_{1}\in G_{1}\\}$ is a normal subgroup of $G_{1}$. Thus,
$M_{p}(G\cap\hat{G}_{1})\leq M_{p}(G_{1})$. Since
$M_{p}(G)=M_{p}(G\cap\hat{G}_{1})M_{p}(G/G\cap\hat{G}_{1})$, the result now
follows from an easy inductive argument. ∎

##  3\. Upper bounds on $M_{p}(S)$ when $S$ is an alternating group

To prove Theorems 1, 2, 3 and 5, we need to find lower bounds on
$\operatorname{Ord}(G,p)/|G|$ for particular linear and permutation groups
$G$. By Corollary 2.3, it will suffice to find upper bounds on $M_{p}(G)$ in
these cases. Thus, with the definition of the function $M_{p}$ in mind, we
will need to determine upper bounds on $M_{p}(S)$ for the finite nonabelian
simple groups $S$. These upper bounds will come from Theorems 6 and 7 (both of
which will be proved in Section 4) when $S$ is not an alternating group. The
purpose of this section is to determine an upper bound on $M_{p}(S)$, in terms
of $p$ and $n$, when $S$ is an alternating group of degree $n$. Our main
result reads as follows.

######  Theorem 3.1.

Fix a prime $p$, and let $S\cong\operatorname{Alt}_{n}$ be an alternating
group of degree $n\geq p$. Let $\alpha$ be a $p$-element of
$\operatorname{Aut}(S)$. Then the coset
$S\alpha\subseteq\operatorname{Aut}(S)$ contains an element $x$ with
$|C_{S}(x)|\leq p^{n/2}$. That is, $M_{p}(S)\leq p^{n/2}$.

To prove Theorem 3.1, we will need two straightforward combinatorial lemmas.

######  Lemma 3.2.

Fix a prime $p$, and a positive integer $n\geq p$. Write the base $p$
expansion of $n$ in the form $n=\sum_{i=1}^{k}a_{i}p^{b_{i}}$, where
$0<a_{i}\leq p-1$ for each $i$, and $0\leq b_{1}<\ldots<b_{k}\leq\log_{p}{n}$.
Then each of the following holds.

  1. (i)

Suppose that either $p\geq 5$; or that $p=3$ and $b_{i}>1$ for some $i$; or
that $p=2$ and $b_{i}>3$ for some $i$. Then $\sum_{i=1}^{k}(a_{i}+b_{i})\leq
n/2$;

  2. (ii)

If either $p=3$ and $b_{i}\leq 1$ for all $i$; or $p=2$ and $b_{i}\leq 3$ for
all $i$, then $\prod_{i=1}^{k}a_{i}!p^{b_{i}}\leq p^{n/2}$.

It follows in particular that $\prod_{i=1}^{k}a_{i}!p^{b_{i}}\leq p^{n/2}$.

###### Proof.

We first prove (i). So assume that either $p\geq 5$; or that $p=3$ and
$b_{i}>1$ for some $i$; or that $p=2$ and $b_{i}>3$ for some $i$. Set
$g_{p}(b_{i}):=p^{b_{i}}-2-2b_{i}$. Then $g_{p}(0)=-1$; $g_{p}(1)=p-4$;
$g_{p}(2)=p^{2}-6$; $g_{p}(3)=p^{3}-8$; and $g_{p}(4)=p^{4}-10$. Moreover,
$g_{p}(b_{i})\geq g_{p}(b_{i}-1)>0$ for $b_{i}\geq 5$. Note that $b_{i}\geq 1$
for some $i$, since $n\geq p$. It follows that $\sum_{i=1}^{k}g_{p}(b_{i})\geq
0$ if either $p\geq 5$; or $p=3$ and $b_{i}>1$ for some $i$; or $p=2$ and
$b_{i}>3$ for some $i$. Since we are assuming that these conditions hold, we
deduce that
$2\sum_{i=1}^{k}b_{i}\leq\sum_{i=1}^{k}(p^{b_{i}}-2)\leq\sum_{i=1}^{k}a_{i}(p^{b_{i}}-2)$,
and hence that
$\sum_{i=1}^{k}(a_{i}+b_{i})\leq\frac{1}{2}\sum_{i=1}^{k}a_{i}p^{b_{i}}=n/2$.
This proves (i).

For (ii), notice that if $p=3$ and $b_{i}\leq 1$ for all $i$, then $n$ has the
form $n=x+3y$, where $0\leq x\leq 2$, and $y\in\\{1,2\\}$. Hence, $n$ is at
most $8$. Similarly, if $p=2$ and $b_{i}\leq 3$ for all $i$, then $n\leq 15$.
We can then quickly check the result for each $n\geq p$ in each case. Finally,
since $\prod_{i=1}^{k}a_{i}!p^{b_{i}}\leq\prod_{i=1}^{k}p^{a_{i}+b_{i}}$, the
proof of the lemma is complete. ∎

######  Lemma 3.3.

For a positive integer $n\geq 2$, write the base $2$ expansion of $n$ in the
form $n=\sum_{i=1}^{k(n)}2^{b_{i}}$, where $0\leq
b_{1}<\ldots<b_{k}\leq\log_{2}{n}$. Let $m$ be minimal so that $k(n)$ and
$k(n-m)$ have different parities. Then $m=1$ if $n$ is odd, while $m=2$ if $n$
is even.

###### Proof.

It is clear that if $n$ is odd (so that $b_{1}=0$), then $k(n-1)=k(n)-1$,
whence $k(n)$ and $k(n-1)$ have different parities. So assume that $n$ is
even, so that $b_{1}=1$. Then $k(n-1)=k(n)$, so $k(n-1)$ and $k(n)$ have the
same parities. But the argument above implies that $k(n-1)$ and $k(n-2)$ have
different parities, since $n-1$ is odd. Thus, $k(n)$ and $k(n-2)$ have
different parities, and the proof is complete. ∎

We can now prove Theorem 3.1.

###### Proof of Theorem 3.1.

Fix a prime $p$, and a $p$-element $\alpha$ of $\operatorname{Aut}(S)$, where
$S=\operatorname{Alt}_{n}$ and $n\geq\max\\{5,p\\}$). To prove Theorem 3.1, it
will suffice to prove that there is a $p$-element $x$ in the coset $S\alpha$
with the property that $|C_{S}(x)|\leq p^{n/2}$. If $p=2$ and $n=6$, then this
can be checked using the Web Atlas [39], so we will assume throughout that we
are not in this case.

Write the base $p$ expansion of $n$ as $n=\sum_{i=1}^{k}a_{i}p^{b_{i}}$, where
$0<a_{i}\leq p-1$ for each $i$, and $0\leq b_{1}<\ldots<b_{k}\leq\log_{p}{n}$.
Let $x:=x_{p}(n)$ be any corresponding element of $\operatorname{Sym}_{n}$.
That is, $x$ is a permutation with $a_{i}$ cycles of length $p^{b_{i}}$, for
each $1\leq i\leq k$. Then
$|C_{\operatorname{Sym}_{n}}(x)|=\prod_{i=1}^{k}a_{i}!p^{b_{i}}$. It follows
from Lemma 3.2 that $|C_{S}(x)|\leq|C_{\operatorname{Sym}_{n}}(x)|\leq
p^{n/2}$.

Since $\operatorname{Aut}(S)/S$ is a $2$-group, this proves the proposition in
all cases apart from when $p=2$ and $x:=x_{2}(n)\not\in S\alpha$. So assume
that we are in this case, and write $k=k(n)$, so that
$n=\sum_{i=1}^{k(n)}2^{b_{i}}$ is the base $2$ expansion of $n$. Since we are
assuming that $n\neq 6$, the element $\alpha$ is contained in
$\operatorname{Sym}_{n}$. Notice also that $x=x_{2}(n)$ is in
$\operatorname{Alt}_{n}$ if and only if $k(n)$ is even. Choose $m\leq n$
minimal with the property that $k(n-m)$ and $k(n)$ have different parities.
Then by Lemma 3.3, either $n$ is odd and $m=1$, or $n$ is even and $m=2$. In
the former case, it is clear that $x(n-1)$ has the same
$\operatorname{Sym}_{n}$-centralizer as $x(n)$. In the latter case, we have
$|C_{\operatorname{Sym}_{n}}(x(n-2))|=2|C_{\operatorname{Sym}_{n}}(x(n))|$,
and $C_{\operatorname{Sym}_{n}}(x)$ contains a transposition. Setting
$y:=x(n-1)$ if $n$ is odd and $y:=x(n-2)$ if $n$ is even, it follows that
$y\in S\alpha$ and $|C_{S}(y)|\leq|C_{\operatorname{Sym}_{n}}(x(n))|\leq
p^{n/2}$, with the latter inequality following from Lemma 3.2 as in the
previous paragraph. This completes the proof. ∎

##  4\. The proof of Theorem 6

Having determined upper bounds for $M_{p}(S)$ when $S$ is an alternating
group, our attention now turns to the case where $S$ is a finite group of Lie
type in defining characteristic. Our upper bounds for $M_{p}(S)$ in this case
comes from Theorem 6, and the purpose of the current section is to prove this
theorem. We can do so without further discussion.

###### Proof of Theorem 6.

Let $S$ be a finite simple group in $\mathrm{Lie}(p)$, and write $q$ and
$\ell$ for the level and untwisted Lie rank of $S$, respectively. Fix a
$p$-element $\alpha$ of $\operatorname{Aut}(S)$. We need to prove that there
exists a $p$-element in the coset $S\alpha\subseteq\operatorname{Aut}(S)$ with
$S$-centralizer order less than or equal to $q^{\ell}$.

We begin by fixing some Lie theoretic notation. Write
$S=O^{p^{\prime}}(\overline{X}_{\overline{\sigma}})$, where $\overline{X}$ is
a simple algebraic group,
$\overline{\sigma}:\overline{X}\rightarrow\overline{X}$ is a Steinberg
endomorphism of $\overline{X}$, and
$\overline{X}_{\overline{\sigma}}=\\{x\in\overline{X}\text{ :
}\overline{\sigma}(x)=x\\}$. In particular, by [11, Theorem 2.2.3], we may
write $\overline{\sigma}=\overline{\sigma_{p}}^{f}\overline{\gamma}^{i}$,
where $\overline{\sigma_{p}}$ is a Frobenius automorphism of $\overline{X}$,
$q=p^{f}$, and either

  1. (a)

$\overline{X}\in\\{A_{\ell},D_{\ell},E_{6}\\}$ and $\overline{\gamma}$ is a
graph automorphism of $\overline{X}$ as in [11, Theorem 1.15.2(a)];

  2. (b)

$\overline{X}\in\\{B_{2}\text{ }(p=2),G_{2}\text{ }(p=3)\text{, }F_{4}\text{
}(p=2)\\}\\}$, and $\overline{\gamma}$ is as in [11, Theorem 1.15.4(b)]; or

  3. (c)

$\overline{X}$ is not one of the groups listed in (a) or (b) above, and
$\overline{\gamma}=1$.

Moreover, we may write $\alpha=x\sigma_{p}^{e}\gamma^{j}$, where
$x\in\operatorname{Inndiag}(X):=\overline{X}_{\overline{\sigma}}$;
$\sigma_{p}:=\overline{\sigma}_{p}\downarrow_{S}$; $0\leq e<f$; and
$\gamma:=\overline{\gamma}\downarrow_{S}$ (see [11, Section 2.5]). In fact,
since $\alpha$ is a $p$-element of $\operatorname{Aut}(S)$, and since we are
working in the coset $S\alpha$, we may assume that
$\alpha=\sigma_{p}^{e}\gamma^{j}$.

Now, as we will see later in the proof, the result will follow almost
immediately, via the theory of Shintani descent, from our analysis in the
cases (1) $\alpha=1$; and (2) $S\in\\{A_{\ell}(q),D_{\ell}(q),E_{6}(q)\\}$ and
$\alpha$ is a graph automorphism of $S$. With this in mind, assume first that
$\alpha=1$. We will show that there exists a unipotent element $x\in S$ such
that $|C_{S}(x)|\leq|C_{\overline{X}_{\overline{\sigma}}}(x)|\leq 2q^{\ell}$.
We will prove that this is the case when $x$ is chosen to be a regular
unipotent element of $S$. Such an element $x$ exists in $S$ by [4, Proposition
5.1.7(a)]. Moreover, at least one such $x$ satisfies
$|C_{S}(x)|\leq|C_{\overline{X}_{\overline{\sigma}}}(x)|\leq
q^{\ell}|\mathcal{C}_{reg,p}|$ by [4, Proposition 5.1.9], where
$|\mathcal{C}_{reg,p}|$ is the number of conjugacy classes of regular
unipotent elements in $\overline{X}_{\overline{\sigma}}$. If the prime $p$ is
_good_ for $\overline{X}$, we have $|\mathcal{C}_{reg,p}|=1$ by [4,
Proposition 5.1.7(b)] (see [4, page 28] for a list of the good and bad primes
for each $\overline{X}$). Thus, we may assume that $p$ is bad for
$\overline{X}$. We distinguish the classical and exceptional cases:

  1. (1)

$S$ is classical. The only bad prime here is $p=2$, and it only occurs in the
cases
$S\in\\{\operatorname{PSp}_{2\ell}(q),\operatorname{P\Omega}^{\epsilon}_{2\ell}(q)\text{
: }\epsilon\in\\{\pm\\}\\}$. So assume we are in one of these cases, and let
$\lambda$ be an element of $\mathbb{F}_{q}$ such that the polynomial
$x^{2}+x+\lambda$ is irreducible over $\mathbb{F}_{q}$. Write $\hat{S}$ for
the quasisimple matrix group associated to $S$, so that
$\hat{S}/Z(\hat{S})\cong S$. Then in the notation of [10, Theorem 3.1 and its
proof], and by using [26, Section 6.1], we see that the following holds:

    1. (i)

$V_{\lambda}(2\ell)$ reduces modulo $Z(\hat{S})$ to a regular unipotent
element of $S$ if $\hat{S}=\operatorname{Sp}_{2\ell}(q)$ (acting as a single
Jordan block on the natural module for $\hat{S}$);

    2. (ii)

$V(2\ell-2)+V(2)$ reduces modulo $Z(\hat{S})$ to a regular unipotent element
of $S$ if $\hat{S}=\Omega^{+}_{2\ell}(q)$; and

    3. (iii)

$V_{\lambda}(2\ell-2)+V(2)$ reduces modulo $Z(\hat{S})$ to a regular unipotent
element of $S$ if $\hat{S}=\Omega^{-}_{2\ell}(q)$.

It then follows from [10, Proposition 3.2] $|\mathcal{C}_{reg,p}|\leq 2$.
Thus, $|C_{S}(x)|\leq|C_{\overline{X}_{\overline{\sigma}}}(x)|\leq 2q^{\ell}$,
as claimed.

  2. (2)

$S$ is exceptional. The finite exceptional groups arising from bad primes for
algebraic groups are precisely those given in Table 3. Upper bounds for the
order of the centralizer of a regular unipotent element of $S$ are known in
these cases. We give these upper bounds, together with the associated
references, in Table 3. We also make two further remarks. First, suppose that
$S\not\in\\{E^{\pm}_{6}(q),E_{7}(q),E_{8}(q)\\}$. To identify the regular
unipotent elements in $S$ in each of the given references, we use [4,
Proposition 5.1.3] (which gives a regular semisimple element as a word in the
Chevalley generators for $\overline{X}$) and [4, Proposition 5.1.9] (which
tells us how many regular semisimple elements there are in
$\overline{X}_{\overline{\sigma}}$). Note also that
$S=\overline{X}_{\overline{\sigma}}$ in these cases. Our second remark
concerns the case $S\in\\{E^{\pm}_{6}(q),E_{7}(q),E_{8}(q)\\}$. In these cases
Mizuno [30, 31] does not explicitly give the order of the centralizer of a
regular semisimple element $x$: he simply gives
$|C_{\overline{X}}(x)/|C_{\overline{X}}(x)^{\circ}|$ (written as $Z(x)$ in
Mizuno’s tables). By [4, Proposition 5.1.2], the regular unipotent elements of
$\overline{X}$ form a single conjugacy class in $\overline{X}$. Then, by [11,
Theorem 2.1.5], the number $|\mathcal{C}_{reg,p}|$ of regular unipotent
classes in $\overline{X}_{\overline{\sigma}}$ is at most
$|C_{\overline{X}}(x)/|C_{\overline{X}}(x)^{\circ}|$. Thus, as mentioned
above, we have
$|C_{S}(x)|\leq|C_{\overline{X}_{\overline{\sigma}}}(x)|\leq|C_{\overline{X}}(x)/|C_{\overline{X}}(x)^{\circ}|q^{\ell}$.
For example, using [30, Theorem 6.2], we get
$|C_{\overline{X}}(x)/|C_{\overline{X}}(x)^{\circ}|\leq 6$. Thus,
$C_{S}(x)\leq 6q^{6}$. This is the upper bound given in Table 3.

$S$ | $p$ | $|C_{\overline{X}_{\overline{\sigma}}}(x)|\leq$ | Reference
---|---|---|---
${}^{2}B_{2}(q)$ | $2$ | $2q$ | [37, Propositions 1 and 18, and Theorem 6]
${}^{2}G_{2}(q)$ | $3$ | $3q$ | [38, Chapter III, paragraph 11]
${}^{3}D_{4}(q)$ | $2$ | $2q^{4}$ | [35, 0.5, page 677]
$G_{2}(q)$ | $2$ | $2q^{2}$ | [9, Table IV–1]
| $3$ | $3q^{2}$ | [8, Table VII–1]
${}^{2}F_{4}(q)$ | $2$ | $4q^{4}$ | [33, Table II]
$F_{4}(q)$ | $2$ | $4q^{4}$ | [32, Theorem 2.1]
| $3$ | $3q^{4}$ | [34, Table 6]
$E^{\pm}_{6}(q)$ | $2,3$ | $6q^{6}$ | [30, Theorem 6.2]
$E_{7}(q)$ | $2$ | $4q^{7}$ | [31, Table 9]
| $3$ | $6q^{7}$ | [31, Table 9]
$E_{8}(q)$ | $2$ | $4q^{8}$ | [31, Table 10]
| $3$ | $3q^{8}$ | [31, Table 10]
| $5$ | $5q^{8}$ | [31, Table 10]
Table 3. Upper bounds for centralizer orders of regular unipotent elements of
simple exceptional groups in bad characteristic.

This completes the proof in the case $S\alpha=S$.

Next, assume that $S\in\\{A_{\ell}(q),D_{\ell}(q),E_{6}(q)\\}$ and $\alpha$ is
a graph automorphism of $S$. We then have either $p=2$ and
$S\in\\{A_{\ell}(q),D_{\ell}(q),E_{6}(q)\\}$; or $p=3$ and $S=D_{4}(q)$. In
these cases, the $S$-conjugacy classes of unipotent elements in $S\alpha$ and
their centralizer orders are given in [25]. The details of what we need are as
follows. First, if $S=A_{\ell}(q)$ with $p=2$, then [25, Theorem 1.3] implies
that $S\alpha$ contains a $2$-element $x$ with $|C_{S}(x)|\leq 2q^{\ell/2}$ if
$\ell$ is even, and $|C_{S}(x)|\leq 4q^{(\ell+1)/2}$ if $\ell$ is odd (setting
$n=\ell+1$, and in the notation of [25, (1)], we take $x$ corresponding to the
decompositions $V\downarrow_{x}=V(2n)$ and $V\downarrow_{x}=V(2)+V(2n-2)$,
respectively). Since $\ell$ is (necessarily) at least $2$, this gives us what
we need unless $q=\ell=2$. However, in this case, we have
$S\cong\operatorname{L}_{3}(2)$ and there is an element of order $8$ in
$S\alpha$ with
$|C_{S}(\alpha)|=|C_{\overline{X}_{\overline{\sigma}}}(x)|=4=q^{\ell}$.

Suppose next that $S=D_{2\ell}(q)$ with $p=2$. Let
$\hat{S}=\Omega^{+}_{2\ell}(q)$ be the associated quasisimple matrix group. We
may then choose an element $\hat{\alpha}$ of
$\mathrm{O}_{2\ell}^{+}(q)\setminus\Omega^{+}_{2\ell}(q)$ so that
$\hat{S}\langle\hat{\alpha}\rangle$ reduces modulo the scalar subgroup of
$\mathrm{O}_{2\ell}^{+}(q)$ to the group $S\langle\alpha\rangle$. Now, [10,
Lemma 3.3] implies that
$\hat{S}\hat{\alpha}\subseteq\mathrm{O}_{2\ell}^{+}(q)\subseteq\mathrm{Sp}_{2\ell}(q)$
contains a unipotent element $x$ ($x=V(2k)$ in the notation of [10, Lemma
3.3]) which acts as a single Jordan block on the natural module for
$\mathrm{Sp}_{2\ell}(q)$. As in case (1) of the $S\alpha=S$ case above, we can
then use [10, Proposition 3.2] to conclude that
$|C_{S}(x)|=|C_{\overline{X}_{\overline{\sigma}}}(x)|\leq|C_{\operatorname{Sp}_{2\ell}(q)}(x)|\leq
2q^{\ell}$. This gives us what we need.

Finally, if $(S,p)=(D_{4}(q),3)$ or $(S,p)=(E_{6}(q),2)$, then $S\alpha$
contains a $p$-element $x$ with
$|C_{\overline{X}_{\overline{\sigma}}}(x)|=q^{2}$ or
$|C_{\overline{X}_{\overline{\sigma}}}(x)|=2q^{4}$, respectively (see [25,
Tables 9 and 10]).

Thus, all that remains is to prove the theorem when $\alpha$ is non-trivial,
and when we are not in the case where
$S\in\\{A_{\ell}(q),D_{\ell}(q),E_{6}(q)\\}$ and $\alpha$ is a graph
automorphism of $S$. Then $\alpha$ is the restriction to
$S=O^{p^{\prime}}(\overline{X}_{\overline{\sigma}})$ of a Steinberg
endomorphism $\overline{\alpha}$ of $\overline{X}$. In these cases, the theory
of Shintani descent (see for example, [13]) gives us what we need. Indeed, by
[13, Theorem 2.1 and Lemma 2.16], the coset $S\alpha$ contains a unipotent
element $x$ with the property that
$|C_{\overline{X}_{\overline{\sigma}}}(x)|=|C_{\overline{X}_{\overline{\sigma_{1}}}}(x_{1})|$,
where

  * •

$\overline{\sigma_{1}}$ is a Steinberg endomorphism of $\overline{X}$ of the
form $\overline{\sigma_{1}}=\bar{\sigma_{p}}^{f_{1}}\bar{\gamma}^{i_{1}}$,
with $f_{1}\leq f$ and $[\overline{\sigma},\overline{\sigma_{1}}]=1$; and

  * •

$x_{1}$ is a $p$-element contained in $S_{1}\alpha_{1}$, where
$S_{1}:=O^{p^{\prime}}(\overline{X}_{\overline{\sigma_{1}}})$, and
$\alpha_{1}$ is either trivial; or
$S\in\\{A_{\ell}(q),D_{\ell}(q),E_{6}(q)\\}$ and $\alpha_{1}$ is a graph
automorphism of $S_{1}$ (see (a) in the first paragraph of the proof).

The result therefore follows from our work in the preceding paragraphs. ∎

##  5\. Results on primitive linear groups

In this section, we make a short but necessary detour to discuss primitive
linear groups. Recall that an irreducible linear group $G\leq\mathrm{GL}(V)$
is said to be _imprimitive_ if the natural $G$-module $V$ has a direct sum
decomposition $V=V_{1}\oplus\ldots\oplus V_{t}$ with $t>1$, such that for each
$1\leq i\leq t$, and each $x\in G$, either $V_{i}^{x}=V_{i}$ or $V_{i}^{x}\cap
V_{i}=0$. If no such decomposition exists, then the irreducible group
$G\leq\mathrm{GL}(V)$ is said to be _primitive_.

We also need to define two linear group properties which are weaker than
primitivity. First, recall that a linear group $N\leq\operatorname{GL}(V)$ is
said to be _homogeneous_ if $V\downarrow_{N}$ is completely reducible with
pairwise isomorphic irreducible constituents. A linear group
$G\leq\operatorname{GL}(V)$ with the property that every normal [respectively
characteristic] subgroup $N$ of $G$ is _homogeneous_ is called
_quasiprimitive_ [resp. weakly quasiprimitive]. Since the homogenous
components of a normal subgroup of an irreducible linear group
$G\leq\operatorname{GL}(V)$ form an imprimitivity decomposition for $V$, a
primitive linear group is both quasiprimitive and weakly quasiprimitive.

The following series of results give details on the structure of a weakly
quasiprimitive linear group.

######  Lemma 5.1.

Let $q$ be a power of a prime $p$, and let
$G\leq\operatorname{GL}(V)=\operatorname{GL}_{n}(q)$ be weakly quasiprimitive.
Then there exists a divisor $d$ of $n$, and a subnormal series $G_{a}\unlhd
G_{a-1}\unlhd\ldots\unlhd G_{0}=:G$, such that each of the following holds:

  1. (i)

$G_{i}/G_{i+1}$ is abelian, for each $1\leq i\leq a-1$;

  2. (ii)

$\left(\prod_{i=1}^{a-1}|G_{i}/G_{i+1}|\right)$ divides $d$.

  3. (iii)

$G_{a}$ is isomorphic to a subgroup of $\operatorname{GL}_{n/d}(q^{d})$
satisfying each of the following properties:

    1. (a)

The generalized Fitting subgroup $F^{*}(G_{a})$ of $G_{a}$ is a homogeneous
subgroup of $\operatorname{GL}_{n/d}(q^{d})$.

    2. (b)

Let $\mathcal{R}$ be the set of primes $r$ such that $O_{r}(G_{a})\not\leq
Z(G_{a})$, and let $\mathcal{S}$ be a set of representatives for the conjugacy
classes of subnormal quasisimple subgroups of $G_{a}$. Also, for
$S\in\mathcal{S}$, set $T(S):=\langle S^{g}\text{ : }g\in G_{a}\rangle$. Then
$F^{*}(G_{a})$ is a central product of $Z(G_{a})$, the $O_{r}(G_{a})$ for
$r\in\mathcal{R}$, and the $T(S)$ for $S\in\mathcal{S}$. Each $O_{r}(G)$ is
extraspecial of exponent $r(2,r)$. Furthermore, each $T(S)$ is a central
product of $t(S)\geq 1$ copies of $S$.

    3. (c)

$Z(G_{a})\leq Z(\operatorname{GL}_{n/d}(q^{d}))$.

    4. (d)

Let $V_{0}$ be the natural $\operatorname{GL}_{n/d}(q^{d})$-module. If $U$ is
an irreducible constituent of $V_{0}\downarrow_{F^{*}(G_{a})}$, then $U$ is a
tensor product of a $1$-dimensional module for $Z(G_{a})$, irreducible modules
$U(r)$ for each $O_{r}(G)$ with $r\in\mathcal{R}$, and irreducible modules
$U(T)$ for each $T=T(S)\in\mathcal{S}$. Moreover, each $U(r)$ is an absolutely
irreducible $\mathbb{F}_{q^{d}}[O_{r}(G_{a})]$-module; while each $U(T)$ is a
tensor product of $t:=t(S)$ copies of an absolutely irreducible
$\mathbb{F}_{q^{d}}[S]$-module $U(S)$.

###### Proof.

Let $G$ be a counterexample to the lemma with $n$ minimal. Note first that
since $G$ acts homogeneously on $V$, $G$ acts faithfully on each of its
irreducible constituents. Thus, we may assume that $G$ is irreducible.

Now, suppose first that $G$ has a characteristic abelian subgroup which is not
contained in $Z(\operatorname{GL}_{n}(q))$. Then by [16, Lemma 2.13], there
exists a divisor $d_{1}$ of $n$, and a characteristic subgroup $G_{1}$ of $G$,
such that $G_{1}$ is isomorphic to a weakly quasiprimitive subgroup of
$\operatorname{GL}_{n/d_{1}}(q^{d_{1}})$, and $G/G_{1}$ is abelian of order
dividing $d_{1}$. If $d_{1}>1$, then the lemma holds for $G_{1}$, by the
minimality of $G$ as a counterexample. But then the lemma clearly follows for
$G$ – a contradiction.

Thus, we may assume that all characteristic abelian subgroups of $G$ are
contained in $Z(\operatorname{GL}_{n}(q))$. Let $\mathcal{R}$ be the set of
primes $r$ such that $O_{r}(G)\not\leq Z(G)$, and let $\mathcal{S}$ be a set
of representatives for the conjugacy classes of subnormal quasisimple
subgroups of $G$. Also, for $S\in\mathcal{S}$, set $T(S):=\langle S^{g}\text{
: }g\in G\rangle$. Then by [16, Lemmas 2.14 and 2.16], $F^{*}(G)$ is a central
product of $Z(G_{a})$, the $O_{r}(G)$ for $r\in\mathcal{R}$, and the $T(S)$
for $S\in\mathcal{S}$. Each $O_{r}(G)$ is extraspecial of exponent $r(2,r)$.
Furthermore, each $T(S)$ is a central product of $t(S)\geq 1$ copies of $S$.

Now, $F^{*}(G)$ is homogeneous, since $G$ is weakly quasiprimitive. Suppose
that the irreducible constituents of $V\downarrow_{F^{*}(G)}$ are not
absolutely irreducible. Then it is proved in [15, Section2] that there exists
a divisor $d$ of $n$, and a normal subgroup $G_{1}$ of $G$ containing
$F^{*}(G)$, such that:

  * •

$G_{1}$ is isomorphic to a subgroup of $\operatorname{GL}_{n/d}(q^{d})$;

  * •

$V_{0}\downarrow_{F^{*}(G)}$ is homogeneous, with absolutely irreducible
constituents; and

  * •

$G/G_{1}$ is abelian of order dividing $d$.

(We remark that in [15, Section 2], the authors assume that $G$ is absolutely
irreducible. However, they only use this at the end of their argument to prove
that $G/G_{1}$ is non-trivial. We do not require this additional assertion
here, so we may use their arguments to conclude the points above.) Note that
since $F^{*}(G)\leq G_{1}\unlhd G$, we have $F^{*}(G)=F^{*}(G_{1})$.

Now, let $q_{1}=q^{d}$, and let $\mathbb{F}$ be a finite extension field of
$\mathbb{F}_{q_{1}}$ such that $\mathbb{F}$ is a splitting field for each of
the central factors $O_{r}(G_{1})$, $S$ of $F^{*}(G_{1})$, for
$r\in\mathcal{R}$, $S\in\mathcal{S}$. Let $U$ be an irreducible
$\mathbb{F}[F^{*}(G_{1})]$-constituent of $V_{0}\downarrow_{F^{*}(G_{1})}$.
Then by [16, Lemma 2.15], $U$ decomposes as a tensor product of a
$1$-dimensional module $U_{0}$ for $Z(G_{1})$, irreducible modules $U(r)$ for
each $O_{r}(G)$, and irreducible modules $U(T)$ for each $T(S)$. The modules
$U(r)$ are necessarily absolutely irreducible (over $\mathbb{F}$). Moreover,
by [16, Lemma 2.17], each $U(T)$ is a tensor product of $t(S)$ copies of a
faithful absolutely irreducible module $U(S)$ for $S$.

We claim that if $U^{\prime}$ is one of the modules $U^{\prime}=U(r)$ or
$U^{\prime}=U(S)$, then $U^{\prime}$ is realizable over $\mathbb{F}_{q_{1}}$.
Indeed, write $\chi$ for the Brauer character of $U$, and let $\chi_{0}$,
$\chi_{r}$, and $\chi_{S}$ be the characters of the modules $U_{0}$, $U(r)$
and $U(S)$ respectively. Then
$\chi=\chi_{0}(\prod_{r\in\mathcal{R}}\chi_{r})(\prod_{S\in\mathcal{S}}\chi_{S}^{t(S)})$.
It follows that the extension field $\mathbb{F}_{q_{1}}(\chi_{r})$ of
$\mathbb{F}_{q_{1}}$ generated by the values of $\chi_{r}$ is contained in
$\mathbb{F}_{q_{1}}(\chi)$. Similarly, $\mathbb{F}_{q}(\chi_{S})$ is contained
in $\mathbb{F}_{q_{1}}(\chi)$. Since $U$ is realizable over
$\mathbb{F}_{q_{1}}$, [7, Theorem 74.9] implies that
$\mathbb{F}_{q_{1}}(\chi)=\mathbb{F}_{q_{1}}$. It follows that the character
values of the modules $U(r)$ and $U(S)$ are all contained in
$\mathbb{F}_{q_{1}}$. Another application of [7, Theorem 74.9] then proves the
claim: the modules $U(r)$ and $U(S)$ are realizable over $\mathbb{F}_{q_{1}}$.
This completes the proof. ∎

Our next lemma gives more information on the central factors of
$F^{*}(G_{a})$, where $G_{a}$ is as in Lemma 5.1.

######  Lemma 5.2.

[16, Lemmas 2.14–2.17] Let $q$ be a power of a prime $p$, and let
$G\leq\operatorname{GL}(V)=\operatorname{GL}_{n}(q)$ be weakly quasiprimitive.
Let $d$, $G_{a}$, $\mathcal{R}$, $\mathcal{S}$, $U(r)$, and $U(T)$ be as in
Lemma 5.1. Also, set $q_{1}:=q^{d}$, and define $\mathcal{X}$ to be the set of
subgroups $X$ of $G_{a}$ which either have the form $X=O_{r}(G_{a})$ for
$r\in\mathcal{R}$, or $X:=T(S)$ with $S\in\mathcal{S}$ Then the following
assertions hold:

  1. (i)

Fix $r\in\mathcal{R}$, so that $O_{r}(G_{a})$ is extraspecial, of order
$r^{1+2m_{r}}=r^{1+2m}$ say. Then $G_{a}/C_{G_{a}}(O_{r}(G_{a}))O_{r}(G_{a})$
is isomorphic to a completely reducible subgroup $L$ of
$\operatorname{Sp}_{2m}(r)$, acting naturally on
$O_{r}(G_{a})/Z(O_{r}(G_{a}))\cong r^{2m}$. Moreover,
$G_{a}/C_{G_{a}}(O_{r}(G_{a}))$ has shape $r^{2m}.L$, and
$\dim_{\mathbb{F}_{q}}{U(r)}=r^{m}$.

  2. (ii)

Fix $S\in\mathcal{S}$, so that $T(S)$ is a central product of $t=t(S)$ copies
of $S$. Then $G_{a}/C_{G_{a}}(T(S))$ is isomorphic to a subgroup of the wreath
product $A(S)\wr\operatorname{Sym}_{t}$, where $A(S)$ is the stabilizer in
$\operatorname{Aut}(S)$ of the module $U(S)$. Moreover,
$G_{a}/C_{G_{a}}(T(S))$ contains $[S/Z(S)]^{t}$, and
$\dim_{\mathbb{F}_{q}}{U(T)}=(\dim_{\mathbb{F}_{q}}{U(S)})^{t(S)}$.

In particular, writing $s(S):=\dim_{\mathbb{F}_{q_{1}}}{U(S)}$, we see that
$\left(\prod_{r\in\mathcal{R}}r^{m_{r}}\right)\left(\prod_{S\in\mathcal{S}}s(S)^{t(S)}\right)$
divides $n$.

The following two consequences of Lemmas 5.1 and 5.2 will be useful in our
proofs of Theorems 1 and 7, respectively.

######  Corollary 5.3.

Let $q$ be a power of a prime $p$, and let
$G\leq\operatorname{GL}(V)=\operatorname{GL}_{n}(q)$ be weakly quasiprimitive.
Adopt the notation of Lemmas 5.1 and 5.2. Then
$M_{p}(G_{a}/Z(G_{a}))\leq\prod_{X\in\mathcal{X}}M_{p}(G_{a}/C_{G_{a}}(X))$.

###### Proof.

Adopting the notation in the statement of the corollary, we have
$Z(G_{a})=C_{G_{a}}(F^{*}(G_{a}))=\bigcap_{X\in\mathcal{X}}C_{G}(X)$ by
standard theory of generalized Fitting subgroups. Writing
$\mathcal{X}=\\{X_{1},\ldots,X_{m}\\}$, it follows that $G_{a}/Z(G_{a})$
embeds as a subdirect subgroup in the direct product
$G_{a}/C_{G_{a}}(X_{1})\times\ldots\times G_{a}/C_{G_{a}}(X_{m})$. The result
now follows immediately from Corollary 2.3. ∎

######  Corollary 5.4.

Let $q$ be a power of a prime $p$ and let
$G\leq\operatorname{GL}(V)=\operatorname{GL}_{n}(\mathbb{F})$ be weakly
quasiprimitive. Suppose that $N$ is a normal subgroup of $G$ contained in
$\mathrm{Frat}(G)$, with $G/N$ a nonabelian simple group. Then one of the
following holds:

  1. (i)

There exists a prime power divisor $r^{m}$ of $n$ such that $G/N$ is a section
of $\operatorname{Sp}_{2m}(r)$; or

  2. (ii)

There exists a divisor $s^{t}$ of $n$, with $s\geq 2$, such that either:

    1. (a)

$G/N$ has a projective irreducible representation of dimension $s$ over
$\overline{\mathbb{F}}$; or

    2. (b)

$G/N$ is a section of $\operatorname{Sym}_{t}$.

###### Proof.

Consider the subnormal series $G_{a}\unlhd\ldots\unlhd G_{0}=:G$ from Lemma
5.1. Since the groups $G_{i}/G_{i+1}$ are abelian and $G/N$ is nonabelian
simple, we must have $G=G_{2}N=\ldots=G_{a}N$. But then we must have
$G_{a}=G$, since $N\leq\mathrm{Frat}(G)$.

So we have proved that $G=G_{a}$. Adopt the notation of Lemmas 5.1 and 5.2,
and let $X$ be a subgroup of $G$ which either has the form $X=O_{r}(G)$ for
some $r\in\mathcal{R}$, or $X=T(S)$ for some $S\in\mathcal{S}$. Then since
$N\leq\mathrm{Frat}(G)$ and $C_{G}(X)\neq G$, we have $NC_{G}(X)\neq G$,
whence $G/NC_{G}(X)\cong G/N$. Proving the claim is now just a matter of
inspecting the nonabelian simple quotients of the groups $X$: If
$X=O_{r}(G)\cong r^{1+2m}$ for some prime $r$, then $G/C_{G}(X)\cong
r^{2m}.L$, where $L$ is a completely reducible subgroup of
$\operatorname{Sp}_{2m}(r)$, and $r^{m}$ divides $n$. It follows that $G/N$ is
a section of $\operatorname{Sp}_{2m}(r)$. If $X=T(S)$ for some quasisimple
group $S$, then $G/C_{G}(X)$ is a subgroup of $A\wr\operatorname{Sym}_{t}$
containing $[S/Z(S)]^{t}$, where $A$ is almost simple with socle $S/Z(S)$.
Since $A/S$ is soluble by Schreier’s lemma, it follows that $G/N$ is either a
section of $\operatorname{Sym}_{t}$, or isomorphic to $S$. The result now
follows from Lemma 5.2. ∎

##  6\. The proofs of Theorems 5 and 7

Having recorded the information we need on primitive linear groups in Section
5, we can now get back to the task of proving our main theorems. We remind the
reader that our main tool in proving Theorems 1 and 3 is Corollary 2.3, and
more generally, the function $M_{p}$ and its behaviour on the nonabelian
finite simple groups. We have already determined upper bounds on $M_{p}(S)$
when $S$ is an alternating group or a group of Lie type in characteristic $p$.
When $S$ is sporadic or in $\mathrm{Lie}(r)$ for $r\neq p$, our bound on
$M_{p}(S)$ will come from Theorem 7. For this reason, the second half of the
current section will be dedicated to the proof of this theorem.

In the first half of the section, and partly because it is needed as part of
the proof of Theorem 7, we will prove Theorem 5. Our first preparatory result
in this direction is similar to Lemma 3.2, and will help us to prove Theorem 5
in the case when the permutation group in question is the full alternating or
symmetric group.

######  Lemma 6.1.

Fix a prime $p>3$ and a positive integer $n\geq 2$. Write the base $p$
expansion of $n$ as $n=\sum_{i=1}^{k}a_{i}p^{b_{i}}$, where $0<a_{i}\leq p-1$
for each $i$, and $0\leq b_{1}<\ldots<b_{k}\leq\log_{p}{n}$. Then
$\prod_{i=1}^{k}a_{i}!p^{b_{i}}\leq(p-1)!^{\frac{n-1}{p-2}}$. Furthermore, if
$a_{i}=1$ for all $i$, then
$\prod_{i=1}^{k}a_{i}!p^{b_{i}}\leq\frac{1}{2}(p-1)!^{\frac{n-1}{p-2}}$.

###### Proof.

We will begin by proving the first statement. Note that
$\prod_{i=1}^{k}a_{i}!p^{b_{i}}$ is the order of a centralizer in
$\operatorname{Sym}_{n}$, so $\prod_{i=1}^{k}a_{i}!p^{b_{i}}\leq n!$. Note
also that since $(m+1)!\leq(m+1)^{m}$, the function $m\rightarrow
m!^{\frac{1}{m-1}}$ is non-decreasing on the set of natural numbers. Hence, if
$p>n$, then we have $\prod_{i=1}^{k}a_{i}!p^{b_{i}}\leq
n!=n!^{\frac{n-1}{n-1}}\leq(p-1)!^{\frac{n-1}{p-2}}$.

Thus, we may assume that $p\leq n$. Now, since $p>3$, we have $p\leq(p-1)!$.
Since $a_{i}\leq p-1$, it follows that

(6.1)
$\displaystyle\prod_{i=1}^{k}a_{i}!p^{b_{i}}\leq(p-1)!^{\sum_{i=1}^{k}(1+b_{i})}.$

Notice also that

(6.2)
$\displaystyle\frac{n-1}{p-2}\geq\frac{n-1}{p-1}\geq\sum_{i=1}^{k}\frac{p^{b_{i}}-1}{p-1}.$

Now, $(p^{b_{i}}-1)/(p-1)=\sum_{\ell=0}^{b_{i}-1}p^{\ell}\geq(b_{i}-1)p+1$.
Since $p\geq 5$, we have $(b_{i}-1)p+1\geq b_{i}+4$ when $b_{i}\geq 2$. It
follows that if $b_{j}\geq 2$ for some $j$, then
$\sum_{i=1}^{k}(p^{b_{i}}-1)/(p-1)=\sum_{b_{i}\in\\{0,1,b_{j}\\}}(p^{b_{i}}-1)/(p-1)+\sum_{b_{i}\not\in\\{0,1,b_{j}\\}}(p^{b_{i}}-1)/(p-1)\geq
b_{j}+4+\sum_{b_{i}\not\in\\{0,1,b_{j}\\}}(1+b_{i})\geq\sum_{i=1}^{k}(1+b_{i})$.
Thus, if $b_{j}\geq 2$ for some $j$, then the result follows from (6.1) and
(6.2).

So we may assume that $n\geq p$ either has the form $n=a_{1}p$, or
$n=a_{1}+a_{2}p$. If the former case holds, then $a_{1}\leq p-1$, and so
$a_{1}!p\leq p!$. Moreover, either $a_{1}=1$ (and the bound
$a_{1}!p=p\leq(p-1)!^{\frac{p-1}{p-2}}=(p-1)!^{\frac{n-1}{p-2}}$ clearly
holds), or $a_{1}\geq 2$, and so $(p-1)!^{\frac{n-1}{p-2}}\geq(p-1)!^{2}\geq
p!$, which again gives us what we need.

Suppose next that $n=a_{1}+a_{2}p$. We need to prove that
$a_{1}!a_{2}!p\leq(p-1)!^{\frac{n-1}{p-2}}$. If $a_{2}\geq 3$, then
$(p-1)!^{\frac{n-1}{p-2}}\geq(p-1)!^{3}\geq(p-1)!p!\geq a_{1}!a_{2}!p$. If
$a_{2}=2$, then $(p-1)!^{\frac{n-1}{p-2}}\geq(p-1)!^{2}\geq 2p!\geq
a_{1}!a_{2}!p$. If $a_{2}=1$ and $a_{1}\leq p-3$, then
$a_{1}!a_{2}!p\leq(p-3)!p\leq(p-1)!\leq(p-1)!^{\frac{n-1}{p-2}}$, since $p\geq
5$. If $a_{2}=1$ and $a_{1}>p-3$, then
$(p-1)!^{\frac{n-1}{p-2}}>(p-1)!^{\frac{p-3+p-1}{p-2}}=(p-1)!^{2}\geq p!\geq
a_{1}!a_{2}!p$. The result follows in each case.

Assume finally that $a_{i}=1$ for all $i$. Since $p\geq 5$, we have
$p\leq\frac{1}{2}(p-1)!$. Then
$\prod_{i=1}^{k}a_{i}!p^{b_{i}}=p^{\sum_{i=1}^{k}b_{i}}\leq\frac{1}{2}(p-1)!^{\sum_{i=1}^{k}b_{i}}$.
Since
$(n-1)/(p-2)=(\sum_{i=1}^{k}p^{b_{i}}-1)/(p-2)\geq\sum_{i=1}^{k}(p^{b_{i}}-1)/(p-1)\geq\sum_{i=1}^{k}b_{i}$,
the result follows. ∎

We are now ready to prove Theorem 5.

###### Proof of Theorem 5.

Let $G$ be a counterexample to the theorem with $n$ minimal. Since
$M_{p}(G)\leq|G|\leq n!$, we may assume throughout that $n\geq p$. Suppose
first that $G\in\\{\operatorname{Alt}_{n},\operatorname{Sym}_{n}\\}$. If
$p=2=n$, then $M_{p}(G)=1<3^{(n-1)/2}$. If $p=2$ and $n>2$, then
$M_{p}(G)=M_{p}(\operatorname{Alt}_{n})\leq 2^{n/2}\leq 3^{(n-1)/2}$, by
Theorem 3.1. If $p=3$, then we can check that $M_{p}(G)\leq 20^{(n-1)/4}$ for
$2\leq n\leq 7$. If $n\geq 8$, then $3^{n/2}2\leq 20^{(n-1)/4}$, so
$M_{p}(G)\leq 2M_{p}(\operatorname{Alt}_{n})\leq 3^{n/2}2\leq 20^{(n-1)/4}$,
again using Theorem 3.1.

Assume now that $p>3$. If $n=6$, then we have $p=5$ and $M_{p}(G)\leq
4M_{p}(\operatorname{Alt}_{6})=20\leq(5-1)!^{(6-1)/(5-2)}$. So assume that
$n\neq 6$. Since $n\geq p\geq 5$, it follows that $M_{p}(G)\leq
2M_{p}(\operatorname{Alt}_{n})$. Now, write the base $p$ expansion of $n$ as
$n=\sum_{i=1}^{k}a_{i}p^{b_{i}}$, where $0<a_{i}\leq p-1$ for each $i$, and
$0\leq b_{1}<\ldots<b_{k}\leq\log_{p}{n}$. Let $x:=x_{p}(n)$ be any
corresponding element of $\operatorname{Sym}_{n}$. That is, $x$ is a
permutation with $a_{i}$ cycles of length $p^{b_{i}}$, for each $1\leq i\leq
k$. Set $d:=1$ if $a_{i}=1$ for all $i$, and $d:=2$ otherwise. Then
$|C_{\operatorname{Alt}_{n}}(x)|=\frac{1}{d}\prod_{i=1}^{k}a_{i}!p^{b_{i}}$.
Since $p>3$, Lemma 6.1 and the definition of the function $M_{p}$ implies that
$M_{p}(G)\leq 2M_{p}(\operatorname{Alt}_{n})\leq
2|C_{\operatorname{Alt}_{n}}(x)|\leq(p-1)!^{(n-1)/(p-2)}$.

Suppose next that $G$ is primitive, but that
$G\not\in\\{\operatorname{Alt}_{n},\operatorname{Sym}_{n}\\}$. Then by [29,
Theorem 1.1], one of the following holds:

  1. (i)

$G$ is a subgroup of $\operatorname{Sym}_{m}\wr\operatorname{Sym}_{r}$
containing $\operatorname{Alt}_{m}^{r}$ (with $m\geq 5$, $r\geq 2$), where the
action of $\operatorname{Sym}_{m}$ is on $k$-element subsets of
$\\{1,\ldots,m\\}$ (with $1\leq k\leq m/2$), and the wreath product has the
product action of degree $n=\binom{m}{k}^{r}$;

  2. (ii)

$G\in\\{M_{11},M_{12},M_{23},M_{24}\\}$ and $G$ is $4$-transitive; or

  3. (iii)

$|G|\leq n^{1+\log_{2}{n}}$.

Suppose first that we are in case (i), and let $N$ be the intersection of $G$
with the base group of the wreath product. Then $N$ is a subdirect subgroup in
a direct product of the form $N_{1}\times\ldots\times N_{r}$, where
$N_{i}\in\\{\operatorname{Alt}_{m},\operatorname{Sym}_{m}\\}$ for each $i$. It
then follows from Corollary 2.3 and our work in the previous paragraphs that
$M_{p}(G)=M_{p}(N)M_{p}(G/N)\leq(p-1)!^{(m-1)r/(p-2)}M_{p}(G/N)$. Since $G$ is
a counterexample to the theorem of minimal degree, we also have
$M_{p}(G/N)\leq(p-1)!^{(r-1)/(p-1)}$. Thus,
$M_{p}(G)\leq(p-1)!^{(mr-1)/(p-2)}\leq(p-1)!^{(n-1)/(p-2)}$, with the latter
inequality following since $mr\leq m^{r}\leq\binom{m}{k}^{r}=n$.

In case (ii), we can use direct computation to check that in each of the
listed Matheiu groups, we have $M_{p}(G)\leq(p-1)!^{(n-1)/(p-2)}$ whenever
$5\leq p\leq n$.

Assume now that we are in case (iii). As noted previously, the function
$m\rightarrow m!^{1/(m-1)}$ is increasing on $\mathbb{N}$. Thus,
$2^{n-1}\leq(p-1)!^{(n-1)/(p-2)}$ for $p\geq 2$. By [29, Corollary 1.4], we
have $M_{p}(G)\leq|G|\leq 2^{n-1}$, unless $G$ is one of the groups of degree
at most $24$ listed in [29, Corollary 1.4(i)–(iv)]. In these listed cases, we
can check that $M_{p}(G)\leq(p-1)!^{(n-1)/(p-2)}$ whenever $5\leq p\leq n$.
The result follows.

Suppose next that $G$ is imprimitive. Then $G$ may be embedded as as a
subgroup of a wreath product $R\wr\operatorname{Sym}_{s}$, where
$R\leq\operatorname{Sym}_{r}$ is primitive, $r,s\geq 2$, $rs=n$, and $G\cap
R^{s}$ is isomorphic to a subdirect subgroup of a direct product $A^{s}$ of
$s$ copies of a subnormal subgroup $A$ of $R$. Since $M_{p}(A)\leq M_{p}(R)$
by definition of the function $M_{p}$, it follows from Corollary 2.3 that
$M_{p}(G)\leq M_{p}(R)^{s}M_{p}(G/G\cap R^{s})$. Since $G$ is a counterexample
of minimal degree, we have $M_{p}(G/G\cap R^{s})\leq h_{p}(s)$ and
$M_{p}(R)\leq h_{p}(r)$. We deduce that $M_{p}(G)\leq h_{p}(r)^{s}h_{p}(s)$.
Recalling that $h_{p}(m):=3^{(m-1)/2}$ if $p=2$; $h_{p}(m):=20^{(m-1)/4}$ if
$p=3$; $h_{p}(m):=(p-1)!^{(m-1)/(p-2)}$ if $3<p\leq n$, the result follows.

Finally, assume that $G$ is intransitive. Then $G$ can be embedded as a
subgroup of $\operatorname{Sym}_{r}\times\operatorname{Sym}_{s}$, where $r\geq
2$ and $r+s=n$. Setting $N:=G\cap(\operatorname{Sym}_{r}\times 1)$, the
minimality of $G$ as a counterexample, together with Corollary 2.3, implies
that $M_{p}(S)\leq h_{p}(r)h_{p}(s)$. Since $h_{p}(r)h_{p}(s)=h_{p}(r+s)$, the
result follows. ∎

It will be useful for us to note the following corollary of Theorem 5 (and its
proof).

######  Corollary 6.2.

Let $G\leq\operatorname{Sym}_{n}$ be a permutation group of degree $n\geq 2$.
Then $M_{p}(G)\leq h_{p}(n)\leq p^{n-1}$.

###### Proof.

In the proof of Theorem 5, we in fact proved that $M_{p}(G)\leq h_{p}(n)$.
Since $(p-1)!\leq(p-1)^{p-2}$, we can see from the definition of the function
$h_{p}$ that $h_{p}(n)\leq p^{n-1}$. ∎

We now move on to our preparations for the proof of Theorem 7. These
preparations consist of the next two lemmas, both of which follow immediately
from known results on simple groups coming from the CFSG. The first concerns
minimal degrees of permutation representations, while the second concerns
minimal dimensions of projective representations.

######  Lemma 6.3.

Let $S$ be a nonabelian finite simple group in $\mathrm{Lie}(p)$, and write
$P(S)$ for the minimal degree of a non-trivial permutation representation for
$S$. Then $|S|\leq P(S)^{2\log_{2}{P(S)}}$.

###### Proof.

The values of the function $P(S)$ are well-known – see [12, Table 4], for
example. The bound in the lemma then follows immediately from inspection of
the values of $|S|$ and $P(S)$ in each case. ∎

######  Lemma 6.4.

Fix a prime $p$, and let $S$ be a nonabelian finite simple group with
$S\not\in\mathrm{Lie}(p)$. Write $n:=R_{p}(S)$ for the minimal dimension of a
non-trivial projective representation for $S$ over a field of characteristic
$p$. Then $|\operatorname{Out}(S)|\leq 6\log_{2}{n}$. Furthermore, if $S$ is
not an alternating group, and is not one of the groups
$M_{12},M_{22},M_{24},J_{2},\mathrm{Suz},Co_{1},Co_{2},\operatorname{L}_{2}(7),\operatorname{L}_{2}(8),\operatorname{L}_{2}(9),\operatorname{L}_{3}(4),\operatorname{U}_{4}(2),\operatorname{U}_{4}(3),$
$\operatorname{PSp}_{6}(2),\operatorname{PSp}_{6}(3)$, or
$\operatorname{P\Omega}^{+}_{8}(2)$, then $|S|\leq
2^{2(\log_{2}{n})^{2}+\log_{2}{n}}$.

###### Proof.

Lower bounds for the values of $R_{p}(S)$ are given in [23]. The upper bounds
in the lemma can then be deduced by inspection of these values and the orders
of the outer automorphism groups of the finite simple groups. ∎

Notice that $|\operatorname{Out}(S)|=6\log_{2}{R_{p}(S)}$ for
$S=\operatorname{L}_{3}(4)$, so the first part of Lemma 6.4 is best possible.

We are now ready to prove Theorem 7.

###### Proof of Theorem 7.

Let $X$ and $Y$ be subgroups of $\operatorname{GL}_{n}(\mathbb{F})$ with
$X/Y\cong S$. We need to prove that $|X/Y|\leq
2^{2(\log_{2}{n})^{2}+\log_{2}{n}}$. We will first reduce to the case where
$Y\leq\operatorname{Frat}(X)$ and $X$ acts irreducibly on the natural
$\operatorname{GL}_{n}(\mathbb{F})$-module $V$. Indeed, let $G$ be a subgroup
of $X$ which is minimal with the property that $GY=X$. Then $G\cap
Y\leq\operatorname{Frat}(G)$, and $G/G\cap Y\cong X/Y$. Thus, by replacing $X$
by $G$, we may assume that $Y\leq\operatorname{Frat}(X)$. Note also that since
$O_{p}(X)\leq Y$ and $X/O_{p}(X)$ embeds as a completely reducible subgroup of
$\operatorname{GL}_{n}(\mathbb{F})$, we may assume that $O_{p}(X)=1$ and that
$X$ acts completely reducibly on $V$. Next, let $W$ be an irreducible
$X$-submodule of $V$ on which $X$ acts non-trivially, and write $X^{W}$ for
the image of the induced action of $X$ on $W$. Let $\mathrm{Ker}_{X}(W)$ be
the kernel of the action of $X$ on $W$. If $Y^{W}=X^{W}$, then
$X=Y\mathrm{Ker}_{X}(W)$. But since $Y\leq\operatorname{Frat}(X)$, this
implies that $\mathrm{Ker}_{X}(W)=X$, contradicting our choice of $W$. Thus,
we must have $Y^{W}\neq X^{W}$, and hence $X^{W}/Y^{W}\cong X/Y$, since
$X^{W}/Y^{W}$ is a factor group of $X/Y$. Therefore, we may replace $X$ with
$X^{W}$ and assume that $X$ is irreducible, as claimed.

So assume that $X$ acts irreducibly on $V$, and suppose first that this action
is imprimitive. Let $V=V_{1}\oplus\ldots\oplus V_{t}$ be an imprimitivity
decomposition of $V$, and set $\Sigma:=\\{V_{1},\ldots,V_{t}\\}$. Let
$K:=\mathrm{Ker}_{X}(\Sigma)$ be the kernel of the action of $X$ on $\Sigma$.
Then since $Y\leq\operatorname{Frat}(X)$ we have (as above) $YK\neq X$. Thus,
$X/Y\cong X^{\Sigma}/Y^{\Sigma}\leq\operatorname{Sym}_{t}$. It follows from
Lemma 6.3 that $|X/Y|$ is bounded above by $P(X/Y)^{2\log_{2}{P(X/Y)}}$.
Hence, since $n\geq t\geq P(X/Y)$, we get $|X/Y|\leq
n^{2\log_{2}{n}}=2^{2(\log{n})^{2}}$. This gives us what we need.

Thus, we may assume that $X$ acts primitively on $V$. Then Corollary 5.4
implies that one of the following holds:

  1. (i)

There exists a prime power divisor $r^{m}$ of $n$ such that $X/Y$ is a section
of $\operatorname{Sp}_{2m}(r)$; or

  2. (ii)

There exists a divisor $s^{t}$ of $n$, with $s\geq 2$, such that either:

    1. (a)

$X/Y$ has a projective irreducible representation of dimension $s$ over
$\overline{\mathbb{F}}$; or

    2. (b)

$X/Y$ is a section of $\operatorname{Sym}_{t}$.

If case (i) holds, then $|X/Y|\leq|\operatorname{Sp}_{2m}(r)|\leq
r^{m(2m+1)}\leq n^{2\log_{2}{n}+1}=2^{2(\log_{2}{n})^{2}+\log_{2}{n}}$; while
if (ii)(b) holds, then the same argument as in the imprimitive case above
shows that $|X/Y|\leq 2^{(\log_{2}{n})^{2}}$. So assume that case (ii)(a)
holds. Then $n\geq R_{p}(X/Y)$. The result then follows from Lemma 6.4. ∎

##  7\. The proof of Theorem 3

In this section, we prove Theorems 3. With the function $M_{p}$ continuing to
serve as our main tool, our approach to proving Theorem 3 will be to derive
upper bounds on $M_{p}(G)$, for the primitive linear group $G$ in question.
From the statement of the bound in the theorem, and with Corollary 5.3 in
mind, it will come as no surprise to the reader that the bulk of our work will
be done in the groups $G_{a}/C_{G_{a}}(X)$ for $X\in\mathcal{X}$, where
$G_{a}$ is as in Lemma 5.1 and $\mathcal{X}$ is as in Corollary 5.3.

We begin with those $X$ of the form $X=O_{r}(G_{a})\not\leq Z(G_{a})$.

######  Proposition 7.1.

Let $q$ be a power of a prime $p$, and fix a prime divisor $r$ of $q-1$.
Suppose that $H$ is a group of shape $N.L$, where $N\unlhd H$ is elementary
abelian of order $r^{2m}$, and $L\leq\operatorname{Sp}_{2m}(r)$ is completely
reducible, acting naturally on $N$. Then $M_{p}(H)\leq
r^{m}(q^{r^{m}}-1)/(q-1)$, except in the following cases:

  1. (i)

$(r,m,q)=(3,1,7)$ and $L=\operatorname{Sp}_{2}(3)$;

  2. (ii)

$(r,m,q)=(2,1,5)$ and
$L\in\\{\operatorname{Sp}_{2}(2)^{\prime},\operatorname{Sp}_{2}(2)\\}$;

  3. (iii)

$(r,m)=(2,1)$, $q\in\\{7,11\\}$, and $L=\operatorname{Sp}_{2}(2)$;

  4. (iv)

$(r,m)=(2,2)$ and either:

    1. (a)

$q=3$ and $L\in\\{C_{5}\rtimes C_{4},\operatorname{Sp}_{4}(2)\\}$;

    2. (b)

$q=5$ and $L=\operatorname{Sp}_{2}(2)\wr\operatorname{Sym}_{2}$;

    3. (c)

$q=7$ and $L\in\\{\operatorname{Alt}_{5}(\text{two
}\operatorname{Sp}_{4}(2)\text{-classes}),\operatorname{Sp}_{4}(2)^{\prime},\operatorname{Sp}_{4}(2)\\}$;
or

    4. (d)

$q\in\\{11,13\\}$ and $L=\operatorname{Sp}_{4}(2)$.

###### Proof.

Note first that since $r$ divides $q-1$ and
$|L|\leq|\operatorname{Sp}_{2m}(r)|\leq r^{2m^{2}+m}$, we have
$M_{p}(H)\leq|H|\leq r^{2m^{2}+3m}$. Since $q>r\geq 2$, we readily see that
this is at most $q^{r^{m}-1}$ if $m\geq 6$. Moreover, for $1\leq m\leq 5$, the
quantity $r^{1+2m}|\operatorname{Sp}_{2m}(r)|$ is bounded above by
$q^{r^{m}-1}$ if $q>23$ (again, using the fact that $q>r$). Since
$q^{r^{m}-1}<r^{m}(q^{r^{m}}-1)/(q-1)$, this gives us what we need in these
cases.

Suppose now that $m\leq 5$, and that $q\leq 23$. Since $r\mid q-1$, it is easy
to check that $r^{1+2m}|\operatorname{Sp}_{2m}(r)|$ is bounded above by
$q^{r^{m}-1}$ for all choices of $r,q,m$ (with the above restrictions), except
when one of the following holds:

  * •

$(r,m,q)\in\\{(3,1,4),(3,2,4),(3,1,7),(3,1,13)\\}$;

  * •

$(r,m)=(2,1)$ and $3\leq q\leq 23$ is odd;

  * •

$(r,m)=(2,2)$ and $3\leq q\leq 19$ is odd;

  * •

$(r,m)=(2,3)$ and $3\leq q\leq 13$ is odd;

  * •

$(r,m)=(2,4)$ and $3\leq q\leq 7$ is odd; or

  * •

$(r,m,q)=(2,5,3)$;

In each of the cases above, we can compute $M_{p}(L)$ for each completely
reducible subgroup $L$ of $\operatorname{Sp}_{2m}(r)$. This gives us the
precise value of $M_{p}(H)=M_{p}(N)M_{p}(L)=r^{2m}M_{p}(L)$. After doing this,
we then see that the cases (i)–(iv) in the statement of the proposition are
the only ones in which $M_{p}(H)>r^{m}(q^{r^{m}}-1)/(q-1)$. ∎

We next move on to the analogous result for those $X$ from Corollary 5.3 with
$X$ of the form $X=T(S)$, for a quasisimple group $S$. It will come as no
surprise to the reader that when computing $M_{p}(G_{a}/C_{G_{a}}(X))$, the
cases $S\in\mathrm{Lie}(p)$ and $S\not\in\mathrm{Lie}(p)$ are substantially
different. For this reason, we prove two separate propositions when dealing
with these cases. Before distinguishing our arguments in this way, we prove
the following lemma, which will be useful in both cases.

######  Lemma 7.2.

Let $q$ be a power of a prime $p$, and fix a finite simple group $S$. Suppose
that $s$ is the dimension of an absolutely irreducible
$\mathbb{F}_{q}[S]$-module $W$, and that $H$ is a subgroup of the wreath
product
$\operatorname{Stab}_{\operatorname{Aut}(S)}(W)\wr\operatorname{Sym}_{t}$
containing $S^{t}$.

  1. (i)

There exists subgroups $A_{i}$ ($1\leq i\leq t$) of
$\operatorname{Stab}_{\operatorname{Aut}(S)}(W)$ such that

$\displaystyle M_{p}(H)\leq
M_{p}(S)^{t}\left(\prod_{i=1}^{t}M_{p}(A_{i}/S)\right)M_{p}(H/N)\leq
M_{p}(S)^{t}\left(\prod_{i=1}^{t}M_{p}(A_{i}/S)\right)\min\\{t!,p^{t-1}\\}.$

  2. (ii)

Suppose that $M_{p}(A_{i})\leq q^{s-1}$ whenever $A_{i}$ is a subgroup of
$\operatorname{Stab}_{\operatorname{Aut}(S)}(W)$ containing $S$. Then
$M_{p}(H)\leq q^{s^{t}-1}$.

  3. (iii)

Suppose that $M_{p}(A_{i})\leq(q^{s}-1)/(q-1)$ whenever $A_{i}$ is a subgroup
of $\operatorname{Stab}_{\operatorname{Aut}(S)}(W)$ containing $S$. Then
$M_{p}(H)\leq(q^{s^{t}}-1)/(q-1)$.

###### Proof.

For ease of notation, set $n:=s^{t}$ and
$A:=\operatorname{Stab}_{\operatorname{Aut}(S)}(W)$. We note first that by the
hypothesis of the lemma, the intersection $N:=H\cap A^{t}$ of $H$ with the
base group of $A\wr\operatorname{Sym}_{t}$ is a subdirect subgroup in a direct
product $A_{1}\times\ldots\times A_{t}$, where each $A_{i}$ is a subgroup of
$A$ containing $S$. It then follows from Corollary 2.3 that

(7.1) $\displaystyle M_{p}(H)$
$\displaystyle\leq\left(\prod_{i=1}^{t}M_{p}(A_{i})\right)M_{p}(H/N)$
$\displaystyle\leq
M_{p}(S)^{t}\left(\prod_{i=1}^{t}M_{p}(A_{i}/S)\right)M_{p}(H/N).$

Note also that $M_{p}(H/N)\leq h_{p}(n)\leq p^{t-1}$ by Corollary 6.2. Since
$|H/N|\leq t!$, part (i) follows.

Part (ii) follows from the first upper bound in (7.1) and the fact that
$(q^{s-1})^{t}q^{t-1}=q^{st-1}\leq q^{{s^{t}}-1}$. So we just need to prove
(iii). It is a routine exercise to check that if $q$ is a prime power and
$s\geq 2$, $t\geq 1$ are positive integers, then

$\displaystyle\left(\dfrac{q^{s}-1}{q-1}\right)^{t}q^{t-1}\leq\dfrac{q^{s^{t}}-1}{q-1}$

unless either $s=t=2$ or $q=s=2$ and $t=3$. If we are not in these exceptional
cases, then the result follows from (7.1). Suppose now that $s=t=2$. Then
since $M_{p}(\operatorname{Sym}_{2})=(2,p-1)$, the first bound in (7.1) gives
$M_{p}(H)\leq(2,p-1)M_{p}(A_{1})M_{p}(A_{2})\leq(2,p-1)(q^{4}-2q^{2}+1)/(q-1)^{2}$.
It is clear that this is at most $(q^{4}-1)/(q-1)$ for all choices of $q$. The
argument in the case $q=s=2$ and $t=3$ is entirely similar. ∎

We can now derive upper bounds on $M_{p}(G_{a}/C_{G_{a}}(X))$ for those $X$
from Corollary 5.3 with $X$ of the form $X=T(S)$, for a quasisimple group
$S\not\in\mathrm{Lie}(p)$.

######  Proposition 7.3.

Let $q$ be a power of a prime $p$, and fix a finite simple group
$S\not\in\mathrm{Lie}(p)$. Suppose that $s$ is the dimension of an absolutely
irreducible $\mathbb{F}_{q}[S]$-module $W$, and that $H$ is a subgroup of the
wreath product
$\operatorname{Stab}_{\operatorname{Aut}(S)}(W)\wr\operatorname{Sym}_{t}$
containing $S^{t}$. Then
$M_{p}(H)\leq(s^{t})_{p^{\prime}}(q^{s^{t}}-1)/(q-1)$, except in the following
cases:

  1. (i)

$H=S=\operatorname{Alt}_{5}$, $(s,t)=(2,1)$, and $q\in\\{11,19\\}$; or

  2. (ii)

$H=\operatorname{Alt}_{5}\wr\operatorname{Sym}_{2}$, $(s,t)=(2,2)$, and
$q=11$.

###### Proof.

For ease of notation, set $n:=s^{t}$ and
$A:=\operatorname{Stab}_{\operatorname{Aut}(S)}(W)$. We will in fact prove
that $M_{p}(H)\leq q^{s^{t}-1}$, except in the following cases:

  1. (1)

$H=S=\operatorname{Alt}_{5}$, $(s,t)=(2,1)$, and
$q\in\\{11,19,29,31,41,59\\}$;

  2. (2)

$S=\operatorname{Alt}_{5}$, $(s,t)=(2,2)$, and $q\in\\{11,19\\}$; or

  3. (3)

$H=S=\operatorname{L}_{2}(7)$, $(s,t)=(3,1)$, and $q=11$.

Since $q^{s^{t}-1}<(s^{t})_{p^{\prime}}(q^{s^{t}}-1)/(q-1)$, this will give us
what we need in all but the three cases listed above. It is then easy to check
that $M_{p}(H)\leq(s^{t})_{p^{\prime}}(q^{s^{t}}-1)/(q-1)$ in these cases,
except when either $H=S=\operatorname{Alt}_{5}$, $(s,t)=(2,1)$, and
$q\in\\{11,19\\}$; or $H=\operatorname{Alt}_{5}\wr\operatorname{Sym}_{2}$,
$(s,t)=(2,2)$, and $q=11$.

Suppose first that $s=2$. It is well-known that the only nonabelian finite
simple group $S\not\in\mathrm{Lie}(p)$ with a projective irreducible
representation in characteristic $p$ is $\operatorname{Alt}_{5}$, and in this
case, $A=S$ (one can see this from [2, Table 8.2], for example). Moreover,
$p\not\in\\{2,5\\}$, since
$\operatorname{Alt}_{5}\cong\operatorname{L}_{2}(4)\cong\operatorname{L}_{2}(5)$.
Since $\operatorname{GL}_{2}(7)$ has no subgroup isomorphic to a cover of
$\operatorname{Alt}_{5}$, we deduce that $q\geq 9$. Assume that $t=1$, so that
$H=S$. If $p=3$, then we quickly compute that $M_{3}(H)=M_{3}(S)=3<q=q^{n-1}$.
Suppose now that $p\geq 7$. Then $M_{p}(H)=M_{p}(S)=|S|$, so we get
$M_{p}(H)\leq q=q^{n-1}$ if and only if $q\geq|S|=60$. Using [2, Table 2], we
see that if $p\geq 7$ and $q\leq 60$, then $\operatorname{Alt}_{5}$ has a
projective representation over $\mathbb{F}_{q}$ if and only if
$q\in\\{11,19,29,31,41,59\\}$. This is the case listed (1) in the first
paragraph above.

Continuing to assume that $s=2$, suppose next that $t>1$. Assume first that
either $p=3$ or $q\geq 60$. Then $M_{p}(H)\leq q^{2^{t}-1}=q^{s^{t}-1}$ by
Lemma 7.2(ii). Assume now that $7<q<60$ and $p\geq 7$. Then it follows from
Lemma 7.2(i) that $M_{p}(H)\leq 60^{t}\min\\{p^{t-1},t!\\}$. If $t\geq 3$,
then this is less than or equal to $q^{2^{t}-1}=q^{n-1}$ for all of choices of
$7<q<60$ (with $p\geq 7$). So assume that $t=2$. Then $M_{p}(H)\leq 60^{t}2$
as above, and this is less than or equal to $q^{4-1}=q^{n-1}$ unless
$q\in\\{11,13,17,19\\}$. However, we can see from [2, Table 2] that among
these values of $q$, $\operatorname{Alt}_{5}$ has a projective representation
over $\mathbb{F}_{11}$ and $\mathbb{F}_{19}$ only. This is the exception
numbered (2) in the first paragraph above.

We can analyse the case $s=3$ in an entirely similar way: From [14] we see
that
$S\in\\{\operatorname{Alt}_{6},\operatorname{Alt}_{7}\text{($q=5$)},\operatorname{L}_{2}(7)\\}$.
Furthermore, using [2, Tables 8.3–8.6], we have $A=S$ unless
$S=\operatorname{Alt}_{6}$ and $q=5$, in which case $|A/S|\leq 2$. Arguing as
in the paragraphs above then gives the bound $M_{p}(H)\leq q^{n-1}$, apart
from when $t=1$, $S=\operatorname{L}_{2}(7)$, and $q=11$.

We may assume, therefore, that $s>3$. By Lemma 7.2(ii), it suffices to
complete the proof of the proposition in the case $t=1$. Thus we may assume,
for the remainder of the proof, that $t=1$ (whence $H$ is almost simple). We
need to prove that $M_{p}(H)\leq q^{s-1}$. To this end, suppose first that if
$S$ is an alternating group, then the degree is at most $8$. We will assume
first that $s\geq 103$. Then one can quickly check that if either
$S\in\\{\operatorname{Alt}_{m}\text{ : }5\leq m\leq 27\\}$, or if $S$ is one
of the exceptions listed in Lemma 6.4, then $|S|\leq
2^{(\log_{2}{s})^{2}+\log_{2}{s}}$. Thus, $|S|\leq
2^{2(\log_{2}{s})^{2}+\log_{2}{s}}$ in all of the cases under consideration,
by Theorem 7. Furthermore, $|H/S|\leq|\operatorname{Out}(S)|\leq 6\log_{2}{s}$
by Lemma 6.4. It follows from Lemma 7.2(i) that $M_{p}(H)\leq
2^{2(\log_{2}{s})^{2}+\log_{2}{s}}(6\log_{2}{s})$. Since $s\geq 103$, one can
check that this is bounded above by $q^{s^{t}-1}=q^{n-1}$ for all values of
$q\geq 2$.

Continuing to assume that $S$ is not an alternating group of degree greater
than $27$, suppose next that $4\leq s\leq 102$. The possibilities for the
simple group $S$ are then given by [14, Tables 2 and 3]. We can then simply go
through each of these possibilities for $S$, and compute
$|\operatorname{Aut}(S)|$. We also define the set $\pi:=\pi_{S}$ and the prime
$p_{0}:=p_{0}(S)$ as follows: let $\pi$ be the set of primes for which
$S\in\mathrm{Lie}(r_{i})$. Then choose $p_{0}$ to be the smallest prime with
$p_{0}\not\in\\{r_{i}\\}_{i}$. For example, if
$S=\operatorname{L}_{2}(4)=\operatorname{L}_{2}(5)$, then $\pi=\\{2,4\\}$ and
$p_{0}=3$. We then have $q\geq p_{0}$, since we are assuming that
$S\not\in\mathrm{Lie}(p)$. Thus, if $|\operatorname{Aut}(S)|\leq p_{0}^{s-1}$,
then we have $M_{p}(H)\leq|\operatorname{Aut}(S)|\leq p_{0}^{s-1}\leq
q^{s-1}$. The groups $S$ which do not satisfy
$|\operatorname{Aut}(S/Z(S))|\leq p_{0}^{s-1}$, for some value of $s$, are as
follows:

(7.2) $\displaystyle\operatorname{L}_{2}(q_{1})\text{($q_{1}\leq 31,q_{1}\neq
16$),}\operatorname{PSp}_{4}(q_{1})\text{($q_{1}\leq
7$),}\operatorname{L}_{3}(3),\operatorname{PSp}_{6}(3),\operatorname{P\Omega}_{7}(3),\operatorname{U}_{4}(2),\operatorname{U}_{4}(3),$
$\displaystyle\operatorname{U}_{5}(2),\operatorname{PSp}_{8}(2),\operatorname{PSp}_{8}(3),G_{2}(2)^{\prime}\text{($q=5$)},G_{2}(3),M_{11},M_{12},M_{22},M_{23},M_{24},J_{1},J_{2},$
$\displaystyle
J_{3},\mathrm{Suz},HS,McL,Co_{1},Co_{2},Co_{3},Fi_{22},Ru,\operatorname{Alt}_{m}\text{($5\leq
m\leq 8$)}.$

The arguments to prove that $M_{p}(H)\leq q^{n-1}$ are almost identical for
each of the groups listed at (7.2), so we will just give the details for the
$\operatorname{L}_{2}(q_{1})$ cases. So assume that
$S\cong\operatorname{L}_{2}(q_{1})$, for some $q_{1}\leq 31$ with $q_{1}\neq
16$. Suppose first that
$S=\operatorname{L}_{2}(4)=\operatorname{L}_{2}(5)\cong\operatorname{Alt}_{5}$.
Then $\pi_{S}=\\{2,5\\}$, and we have
$M_{p}(H)\leq|\operatorname{Aut}(S)|=120\leq q^{4-1}\leq q^{s-1}$ if $q>3$. If
$q=3$, then we calculate $M_{3}(S)=3$, so $M_{3}(H)\leq 2M_{3}(S)\leq
6<3^{4-1}<3^{s-1}$. The result follows. Suppose now $q_{1}\not\in\\{4,5\\}$.
Then $n=s\geq\max\\{4,R_{p}(S)\\}$. By [23], it follows that $n=s\geq 4$ if
$q_{1}=9$, and $n=s\geq\max\\{4,(q_{1}-1)/(q_{1}-1,2)\\}$ otherwise. We then
see that the inequality $|\operatorname{Aut}(S)|\leq q^{n-1}$ holds for $q\geq
q^{\prime}$, where $q^{\prime}$ is the value associated to
$\operatorname{L}_{2}(q_{1})$ in the second column of Table 4.

$S$ | $q^{\prime}$
---|---
$\operatorname{L}_{2}(7)$ | $7$
$\operatorname{L}_{2}(8)$ | $4$
$\operatorname{L}_{2}(9)$ | $13$
$\operatorname{L}_{2}(11)$ | $7$
$S$ | $q^{\prime}$
---|---
$\operatorname{L}_{2}(13)$ | $5$
$\operatorname{L}_{2}(17)$ | $4$
$\operatorname{L}_{2}(19)$ | $4$
$\operatorname{L}_{2}(23)$ | $3$
$S$ | $q^{\prime}$
---|---
$\operatorname{L}_{2}(25)$ | $3$
$\operatorname{L}_{2}(27)$ | $3$
$\operatorname{L}_{2}(29)$ | $3$
$\operatorname{L}_{2}(31)$ | $3$

Table 4. $|\operatorname{Aut}(S)|\leq q^{n-1}$ for $q\geq q^{\prime}$ as
above.

Thus, we just need to prove that $M_{p}(X)\leq q^{n-1}$ whenever
$q<q^{\prime}$ and $p\not\in\pi_{S}$. We do this by computing the precise
value of $M_{p}(S)$ in each case, and then applying the bound $M_{p}(H)\leq
M_{p}(S)M_{p}(\operatorname{Out}(S))$. For example, when
$S=\operatorname{L}_{2}(11)$, we need to check that $M_{p}(H)\leq q^{s-1}$
when $q\in\\{2,3,4,5\\}$. We have $M_{2}(S)=12$; $M_{3}(S)=6$; and
$M_{5}(S)=5$. Also, $s\geq R_{p}(S)=(11-1)/2=5$. The computations above then
give $M_{p}(S)\leq q^{5-1}\leq q^{n-1}$ in each case. This argument works in
all cases, except when either $S=\operatorname{L}_{2}(7)$ and $q=5$; or
$S=\operatorname{L}_{2}(9)\cong\operatorname{Alt}_{6}$ and $q\in\\{7,11\\}$.
In each of these cases, we can use the Atlas of Brauer Characters [17] to
check that since $s\geq 4$, we must in fact have $s\geq 5$. We then see that
$M_{p}(H)\leq|\operatorname{Aut}(S)|\leq q^{5-1}$ in each case.

Finally, assume that $S$ is an alternating group of degree $m\geq 9$. Then
$R_{p}(H)=m-2$ if $p\mid m$, and $R_{p}(H)=m-1$ otherwise (see [22,
Proposition 5.3.7], for example). If $p\geq m$, then it follows that
$M_{p}(H)\leq|\operatorname{Sym}_{m}|\leq 6p^{m-3}\leq p^{m-2}\leq q^{n-1}$.
If $p<m$, then Theorem 3.1 yields $M_{p}(H)\leq 2q^{m/2}$. Since $m\geq 9$
implies that $m/2\leq m-4$, we deduce that $M_{p}(H)\leq 2q^{m-4}\leq
q^{m-3}\leq q^{n-1}$. This gives us what we need, and completes the proof. ∎

Continuing with our analysis of the function $M_{p}$ and its values on those
$G_{a}/C_{G_{a}}(X)$ with $G_{a}$, $X$ as in Corollary 5.3, we next deal with
the case $X=T(S)$ with $S\in\mathrm{Lie}(p)$.

######  Proposition 7.4.

Let $q=p^{f}$ be a power of a prime $p$, and fix a finite simple group
$S\in\mathrm{Lie}(p)$. Write $\hat{S}$ for the Schur cover of $S$. Suppose
that $s$ is the dimension of an absolutely irreducible
$\mathbb{F}_{q}[\hat{S}]$-module $W$, and that $H$ is a subgroup of the wreath
product
$\operatorname{Stab}_{\operatorname{Aut}(S)}(W)\wr\operatorname{Sym}_{t}$
containing $S^{t}$. Suppose also that $W$ is not realizable over any proper
subfield of $\mathbb{F}_{q}$. Then $M_{p}(H)\leq(q^{s^{t}}-1)/(q-1)$, except
in the following cases:

  1. (i)

$S=\operatorname{L}_{n}(q)$, $t=1$, and $W$ is quasiequivalent to the natural
module for $\hat{S}$. In this case, $M_{p}(H)\leq q^{s-1}s_{p^{\prime}}$.

  2. (ii)

$S=\operatorname{L}_{2}(5)$, $(s,t,q)=(2,2,5)$, and
$H=\mathrm{PGL}_{2}(5)\wr\operatorname{Sym}_{2}$. In this case,
$M_{p}(H)=200<4(5^{4}-1)/(5-1)$.

###### Proof.

As in the proof of Proposition 7.3, we will set $n:=s^{t}$ and
$A:=\operatorname{Stab}_{\operatorname{Aut}(S)}(W)$. We will also adopt
similar notation to that set out in the comments preceding the statement of
Theorem 6. Indeed, write $S=O^{p^{\prime}}(\overline{X}_{\overline{\sigma}})$
where $\overline{X}$ is a simple algebraic group and
$\overline{\sigma}:\overline{X}\rightarrow\overline{X}$ is a Steinberg
endomorphism of $\overline{S}$. As before, we will also write
$\overline{\sigma}=\overline{\sigma_{p}}^{e}\overline{\gamma}^{i}$, where
$\overline{\sigma_{p}}$ is a Frobenius automorphism of $\overline{X}$, and
either

  1. (a)

$\overline{X}\in\\{A_{\ell},D_{\ell},E_{6}\\}$ and $\overline{\gamma}$ is a
graph automorphism of $\overline{X}$ as in [11, Theorem 1.15.2(a)];

  2. (b)

$\overline{X}\in\\{B_{2}\text{ }(p=2),G_{2}\text{ }(p=3)\text{, }F_{4}\text{
}(p=2)\\}\\}$, and $\overline{\gamma}$ is as in [11, Theorem 1.15.4(b)]; or

  3. (c)

$\overline{X}$ is not one of the groups in (a) or (b) above, and
$\overline{\gamma}=1$.

Writing $q_{1}:=p^{e}$ for the level of $S$; $\ell$ for the untwisted Lie rank
of $S$; and $d$ for the order of $\gamma^{i}$, we will also shorten our
notation and write $S={}^{d}S_{\ell}(q_{1})$. Recall that $q=p^{f}$, and that
$\operatorname{Inndiag}(S)$ is defined as
$\operatorname{Inndiag}(S):=\overline{X}_{\overline{\sigma}}$.

We first discuss the structure of the group $A$. It is well-known that
$\operatorname{Inndiag}(S)$ $\leq A$ (see for example [2, Proposition
5.1.9(i)]). We now determine the group $A/\operatorname{Inndiag}(S)$. The
information we require is in [22, Proposition 5.4.6] (and the comments
afterwards): Recalling that $d\in\\{1,2,3\\}$, [22, Proposition 5.4.6] implies
that there exists an irreducible $\mathbb{F}_{q}[\hat{S}]$-module $M$ such
that one of the following holds:

  1. (1)

$f\mid e$ and $s=\dim{W}=(\dim{M})^{e/f}$.

  2. (2)

$d\in\\{2,3\\}$, $f\nmid e$, $f\mid de$, and $s=\dim{W}=(\dim{M})^{de/f}$.

Moreover (see [2, Proposition 5.1.9(i)] and [22, Propositions 5.4.2, 5.4.3 and
5.4.6]), if we are in case (1) then $|A/\operatorname{Inndiag}(S)|=g(e/f)$,
where $g:=2$ if $W$ is self-dual, and $g:=1$ otherwise. If we are in case (2),
then $|A/\operatorname{Inndiag}(S)|=g(e/f)$ where $g\leq d$.

We are now ready to prove the proposition. Assume first that
$S=A_{\ell}(q_{1})$ and $t=1$. Then $\dim{M}\geq R_{p}(S)=\ell+1$ by [22,
Proposition 5.4.13]. Also, $M_{p}(S)\leq q_{1}^{\ell}$, by Theorem 6. It
follows from the paragraph above and Corollary 2.3 that $M_{p}(H)\leq
q_{1}^{\ell}(q_{1}-1,\ell+1)2(e/f)\leq q_{1}^{\ell}(\ell+1)2(e/f)$. Now,
assume that $W$ is not quasiequivalent to the natural $(\ell+1)$-dimensional
module for $\hat{S}$ over $\mathbb{F}_{q_{1}}$. Then either $e/f>1$; or
$e/f=1$ and $\dim{M}>\ell+1$. Suppose first that $e/f>1$. Then $M_{p}(H)\leq
q_{1}^{\ell}(\ell+1)2(e/f)$ as above, and this is less than
$(q^{(\ell+1)^{e/f}}-1)/(q-1)\leq(q^{s}-1)/(q-1)$ as long as
$(q,e/f,\ell)\neq(2,2,1)$. If $(q,e/f,\ell)=(2,2,1)$, then
$S=\operatorname{L}_{2}(4)\cong\operatorname{Alt}_{5}$. Since $p=2$ and
$M_{2}(\operatorname{Alt}_{5})=4$, we get $M_{p}(H)=M_{p}(S)=4$ in this case,
and this gives us what we need. Suppose next that $e/f=1$ and
$\dim{M}>\ell+1$. Note that $q_{1}=q$ in this case, so $q\geq 4$ if $\ell=1$.
Moreover, [22, Proposition 5.4.11] implies that one of the following holds:

  1. (1)

$\ell>1$ and $\dim{M}\geq\ell(\ell+1)/2$; or

  2. (2)

$\ell=1$ and $\dim{M}\geq(\ell+1)(\ell+2)/2=3$.

Assume first that case (1) holds. Then since
$q_{1}^{\ell}(\ell+1)2=q^{\ell}(\ell+1)2$ is less than
$(q^{\frac{1}{2}\ell(\ell+1)}-1)/(q-1)\leq(q^{s}-1)/(q-1)$ when either $q\geq
5$ or $\ell\geq 3$, the bound $M_{p}(H)\leq q_{1}^{\ell}(\ell+1)2$ from above
gives us what we need in these cases. Suppose now that $\ell=2$ and
$q\in\\{2,3,4\\}$, then the bound $M_{p}(H)\leq q^{\ell}(q-1,\ell+1)2=2$ gives
the result in each case. Assume next that case (2) holds. Then $q\geq 4$,
since $S$ is nonabelian simple. Also, $M_{p}(H)\leq 2M_{p}(S)\leq 2q$, by
Theorem 6. Since $2q\leq(q^{3}-1)/(q-1)\leq(q^{s}-1)/(q-1)$ when $q\geq 4$,
the result again follows.

Thus, in the case $S=A_{\ell}(q_{1})$ and $W$ is not quasiequivalent to the
natural $\mathbb{F}_{q_{1}}[\hat{S}]$-module, we have proved that
$M_{p}(H)\leq(q^{s}-1)/(q-1)$ if $t=1$. By Lemma 7.2(iii), it follows that
$M_{p}(H)\leq(q^{s^{t}}-1)/(q-1)$ for all $t\geq 1$ in this case

Suppose next that $S=A_{\ell}(q_{1})$, but that $W$ _is_ quasiequivalent to
the natural $\mathbb{F}_{q_{1}}[\hat{S}]$-module. Then we have $q_{1}=q$ and
$A=\operatorname{Inndiag}(S)$. Thus, $M_{p}(A_{i})\leq
M_{p}(S)M_{p}(A_{i}/S)\leq q^{\ell}(q-1,\ell+1)\leq q^{\ell}(\ell+1)$ whenever
$S\leq A_{i}\leq A$, using Theorem 6. It follows in particular that if $t=1$,
then $M_{p}(H)\leq q^{\ell}(\ell+1)_{p^{\prime}}=q^{s-1}s_{p^{\prime}}$. This
is case (i) in the statement of the proposition.

We may assume, therefore, that $t>1$. Suppose first that $(\ell,t)\neq(1,2)$.
By Lemma 7.2(i), we have $M_{p}(H)\leq q^{\ell
t}(\ell+1)^{t}q^{t-1}=q^{t(\ell+1)-1}(\ell+1)^{t}$. Since this is less than
$(q^{(\ell+1)^{t}}-1)/(q-1)=(q^{s^{t}}-1)/(q-1)$ for all choices of $q$,
$\ell$, and $t\geq 2$ with $(\ell,t)\neq(1,2)$, the result follows in this
case. Assume now that $(\ell,t)=(1,2)$. Then by Lemma 7.2(i), we have
$M_{p}(H)\leq q^{2\ell}(\ell+1)^{2}2=8q^{2}$. This is less than
$(q^{(\ell+1)^{2}}-1)/(q-1)=(q^{4}-1)/(q-1)$ if $q\geq 7$. So assume that
$q\leq 5$. Then $q\in\\{4,5\\}$, since $S$ is nonabelian simple. It follows
that $S\cong\operatorname{Alt}_{5}$, and $M_{2}(S)=4$,
$M_{5}(\operatorname{Alt}_{5})=5$. Thus, if $q=4$ then we have
$M_{p}(H)=M_{p}(S)\leq 4^{2}=16<(4^{4}-1)/(4-1)$. If $q=5$, then $H$ has shape
$\operatorname{Alt}_{5}^{2}.K$, where $K$ is a $2$-group of order at most $8$.
It follows that $M_{p}(H)=25\times|K|$. We then get
$M_{p}(H)\leq(5^{4}-1)/(5-1)$ if and only if $|K|<8$. Thus,
$M_{p}(H)>(5^{4}-1)/(5-1)$ if and only if
$H=\operatorname{GL}_{2}(5)\wr\operatorname{Sym}_{2}\leq\operatorname{GL}_{4}(5)$,
and this gives us what we need.

This completes the proof of the proposition in the case $S=A_{\ell}(q)$. We
may assume therefore, for the remainder of the proof, that $S\neq
A_{\ell}(q)$. By Lemma 7.2(iii), we may also assume that $t=1$ (whence $H$ is
almost simple). Suppose first that $S\neq{}^{2}A_{\ell}(q_{1})$, and let
$m:=\dim{M}$, so that $s\geq m^{e/f}$. If
$S=\operatorname{PSp}_{4}(2)^{\prime}\cong\operatorname{Alt}_{6}$, then
$M_{2}(S)=4$, $R_{p}(S)=3$ (see [22, Proposition 5.4.13]), and
$\operatorname{Aut}(S)/S$ is a $2$-group. Thus,
$M_{2}(H)=M_{2}(S)\leq(q^{3}-1)/(q-1)\leq(q^{s}-1)/(q-1)$ as needed. So we may
assume that $S\neq\operatorname{PSp}_{4}(2)^{\prime}$. Then inspecting the
upper bounds for $M_{p}(S)$ given by Theorem 6, and the values of $R_{p}(S)$
given by [22, Proposition 5.4.13], we quickly see that $M_{p}(S)\leq
2q_{1}^{m/2}$. Since $|\operatorname{Inndiag}(S)/S|\leq 4$ in these cases, it
follows (see the second paragraph above) that $M_{p}(H)\leq
8q_{1}^{m/2}(e/f)3=24q_{1}^{m/2}(e/f)=24q^{em/2f}(e/f)$. This is less than
$(q^{m^{e/f}}-1)/(q-1)\leq(q^{s}-1)/(q-1)$ if either $e/f\geq 4$ or $m\geq
10$. So assume that $e/f\in\\{1,2,3\\}$ and $m=\dim{M}\leq 9$. We can then do
more specific analysis according to the isomorphism class of $S$. For example,
if $S=C_{\ell}(q)$ (with $\ell\geq 2$), then $m\leq 9$ implies that $\ell\leq
4$, by [22, Proposition 5.4.13]. Also, $|A/S|$ divides $(2,p-1)(e/f)(2,p)$ by
the second paragraph above, and $M_{p}(S)\leq(2,p)q_{1}^{m/2}=(2,p)q^{em/2f}$
by Theorem 6. Thus, in either of the cases $p=2$ or $p\neq 2$, we get
$M_{p}(H)\leq 2q^{em/2f}(e/f)$, by definition of the function $M_{p}$. Since
$m\geq 2\ell\geq 4$ and $2q^{em/2f}(e/f)\leq(q^{m^{e/f}}-1)/(q-1)$ for $m\geq
4$, the result follows. The arguments for dealing with the remaining groups
$S$, with $S\neq{}^{2}A_{\ell}(q_{1})$, are entirely similar.

Assume finally that $S={}^{2}A_{\ell}(q_{1})$ with $\ell\geq 2$. Then
$M_{p}(S)\leq q_{1}^{\ell}=q^{e\ell/f}$ by Theorem 6. By arguing as in the
second paragraph above, we see that $M_{p}(H)\leq
q^{e\ell/f}(q_{1}+1,\ell+1)2(e/f)\leq q^{e\ell/f}(\ell+1)2(e/f)$. If $e/f>1$,
then this is less than $(q^{(\ell+1)^{e/f}}-1)/(q-1)\leq(q^{s}-1)/(q-1)$, so
assume that $e=f$. Then by [22, Proposition 5.4.6], the module $M$ is self-
dual. Also, $|A/\operatorname{Inndiag}(S)|=2$ (see the second paragraph
above). On the other hand, since $\ell\geq 2$, the natural
$\mathbb{F}_{q}[\hat{S}]$-module is not self-dual (see [22, Proposition 5.4.2
and 5.4.3]). Thus, by [22, Proposition 5.4.11] we have
$s=\dim{W}=\dim{M}\geq\frac{1}{2}\ell(\ell+1)$. Arguing as above, and since
$|A/\operatorname{Inndiag}(S)|=2$, we deduce that $M_{p}(H)\leq
q^{\ell}(q+1,\ell+1)2\leq 2q^{\ell}(\ell+1)$ if $p>2$, and $M_{p}(H)\leq
q^{\ell}(q+1,\ell+1)\leq q^{\ell}(\ell+1)$ if $p=2$. These upper bounds give
$M_{p}(H)\leq(q^{\frac{1}{2}\ell(\ell+1)}-1)/(q-1)\leq(q^{s}-1)/(q-1)$ if
$\ell>2$. So assume that $\ell=2$. Then $s\geq 6$ by [28, Theorem 4.4], and
the result follows as above. ∎

Before proceeding to the proof of Theorem 3, it will be useful for us to state
the following consequence of Propositions 7.1, 7.3, and 7.4.

######  Corollary 7.5.

Let $q$ be a power of a prime $p$, and suppose that $H$ is a finite group
satisfying one of the following:

  1. (a)

$H$ has shape $N.L$, where $N\unlhd H$ is elementary abelian of order
$r^{2m}$, and $L\leq\operatorname{Sp}_{2m}(r)$ is completely reducible, acting
naturally on $N$; or

  2. (b)

there exists a finite simple group $S$ with an $s$-dimensional absolutely
irreducible projective $\mathbb{F}_{q}[S]$-module $W$, such that $H$ is a
subgroup of the wreath product
$\operatorname{Stab}_{\operatorname{Aut}(S)}(W)\wr\operatorname{Sym}_{t}$
containing $S^{t}$.

In case (a), set $d:=r^{m}$, and in case (b) set $d:=s^{t}$. Then
$M_{p}(H)\leq d_{p^{\prime}}(q^{d}-1)/(q-1)$, unless $H$ lies in Table 5. In
these latter cases, the value of $M_{p}(H)$ is given in the fourth column of
Table 5.

$H$ | $d$ | $q$ | $M_{p}(H)$
---|---|---|---
$3^{2}.\operatorname{Sp}_{2}(3)$ | $3$ | $7$ | $216$
$2^{2}.\operatorname{Sp}_{2}(2)$ | $2$ | $5,7$ | $24$
$2^{4}.\operatorname{Sp}_{4}(2)$ | $4$ | $3$ | $288$
$2^{4}.\operatorname{Sp}_{4}(2)$ | $4$ | $7,11,13$ | $11520$
$2^{4}.(5\rtimes 4)$ | $4$ | $3$ | $320$
$2^{4}.(\operatorname{Sp}_{2}(2)\wr\operatorname{Sym}_{2})$ | $4$ | $5$ | $1152$
$2^{4}.\operatorname{Sp}_{4}(2)^{\prime}$ | $4$ | $7$ | $5760$
$2^{4}.\operatorname{Sym}_{5}$ | $4$ | $7$ | $1920$
$\operatorname{Alt}_{5}$ | $2$ | $11,19$ | $60$
$\operatorname{Alt}_{5}\wr\operatorname{Sym}_{2}$ | $4$ | $11$ | $7200$
Table 5. Exceptions from Corollary 7.5

We are almost ready to prove Theorem 3. To facilitate a tidier proof of the
theorem, we first prove the following easy lemma.

######  Lemma 7.6.

Let $q_{1}$ be a power of a prime $p$, and let
$(M_{1},d_{1}),\ldots,(M_{k},d_{k})$ be pairs of natural numbers, with
$d_{\geq}2$ for all $i$. Number the rows in Table 5 from $1$ to $10$. Suppose
that for each $i$, the pair $(M_{i},d_{i})$ satisfies one of the following:

  1. (i)

$M_{i}:=(d_{i})_{p^{\prime}}(q_{1}^{d_{i}}-1)/(q_{1}-1)$; or

  2. (ii)

there exists $f(i)$ with $1\leq f(i)\leq 10$ such that $d_{i},q_{1}$ and
$M_{i}$ lie in the third, fourth and fifth column of Row $f(i)$ in Table 5,
respectively.

Assume also that each of the following conditions hold:

  * •

if $(M_{i},d_{i})$ and $(M_{j},d_{j})$ satisfy (ii) above, with $i\neq j$ and
$f(i)=f(j)$, then $f(i)\in\\{9,10\\}$.

  * •

if $(M_{i},d_{i})$ and $(M_{j},d_{j})$ satisfy (ii) above with $i\neq j$ and
$2\leq f(i)\leq 8$, then $f(j)\in\\{1,9,10\\}$.

  * •

if $k=1$, then $(M_{1},d_{1})$ satisfies (i).

Then $(q_{1}-1)\prod_{i=1}^{k}M_{i}\leq(n_{1})_{p^{\prime}}(q_{1}^{n_{1}}-1)$,
where $n_{1}:=\prod_{i=1}^{k}d_{i}$.

###### Proof.

We prove the claim by induction on $k$, with the case $k=1$ being trivial. So
assume that $k>1$, and that the lemma holds for all smaller $k$.

Suppose first that $k=2$, and that both $(M_{1},d_{1})$ and $(M_{2},d_{2})$
satisfy (ii) in the statement of the lemma. By assumption, the triples
$(d_{i},q_{1},M_{i})$ cannot both come from rows 2–8 in Table 5. With this in
mind, we can now prove that
$(q_{1}-1)M_{1}M_{2}\leq(n_{1})_{p^{\prime}}(q_{1}^{n_{1}}-1)$ via a quick
calculation. For example, if $q_{1}=7$, then
$\\{f(1),f(2)\\}=\\{1,2\\},\\{1,4\\},\\{1,7\\}$ or $\\{1,8\\}$. It then
follows, respectively, that $n_{1}=6$, $12$, $12$, or $12$; and that
$(7-1)M_{1}M_{2}$ is at most $(7-1)\times 216\times 24$, $(7-1)\times
216\times 11520$, $(7-1)\times 216\times 5760$, or $(7-1)\times 216\times
1920$. This yields $(7-1)M_{1}M_{2}<6(7^{6}-1)$ in the first case; and
$(7-1)M_{1}M_{2}<12(7^{12}-1)$ in the next three cases. This gives us what we
need in each case. Thus, we may assume that if $k=2$, then at least one of the
pairs $(M_{i},d_{i})$, say $(M_{1},d_{1})$, satisfies (i) in the statement of
the lemma. The inductive hypothesis then implies that
$(q_{1}-1)\prod_{i=1}^{k-1}M_{i}\leq(d)_{p^{\prime}}(q_{1}^{d}-1)$, where
$d:=\prod_{i=1}^{k-1}d_{i}$. So it will suffice to prove that
$(d)_{p^{\prime}}(q_{1}^{d}-1)M_{k}\leq(dd_{k})_{p^{\prime}}(q_{1}^{dd_{k}}-1)$.
If $M_{k}\leq(d_{k})_{p^{\prime}}(q_{1}^{d_{k}}-1)$, then this is clear, since
each $d_{i}$ is at least $2$. So assume that there exists $f(k)$, with $1\leq
f(k)\leq 10$, such that $d_{k},q_{1}$ and $M_{k}$ lie in the third, fourth and
fifth column of Row $f(k)$ in Table 5, respectively. It is then just a matter
of going through each of the possibilities for $(d_{k},q_{1},M_{k})$ from
Table 5, and deriving the required upper bound. For example, if $f(k)=4$, so
that $q_{1}\in\\{7,11,13\\}$, $d_{k}=4$, and $M_{k}=11520$, then
$(d)_{p^{\prime}}(q_{1}^{d}-1)M_{k}=11520(d)_{p^{\prime}}(q_{1}^{d}-1)$. Note
that $11520(q_{1}^{d}-1)$ is less than $4(q_{1}^{4d}-1)$ for all choices of
$d\geq 2$, and $q_{1}\in\\{7,11,13\\}$. This gives us what we need. The
remaining possibilities for $(d_{k},q_{1},M_{k})$ from Table 5 can be dealt
with in an entirely similar way. ∎

We are now ready to prove Theorem 3.

###### Proof of Theorem 3.

We will in fact prove that if $G\leq\operatorname{GL}_{n}(q)$ is weakly
quasiprimitive; $G$ is not the group
$2^{1+4}.\operatorname{Sp}_{4}(2)\leq\operatorname{GL}_{4}(3)$ (with
$(n,q)=(4,3)$); and $G$ is not one of the exceptions listed in part (i) of the
statement of the theorem, then

(7.3) $\displaystyle M_{p}(G_{a})\leq(n/d)_{p^{\prime}}(q^{n}-1),$

where $d$ and $G_{a}\leq\operatorname{GL}_{n/d}(q^{d})$ are as in Lemma 5.1.
Since $M_{p}(G)\leq d_{p^{\prime}}M_{p}(G_{a})$ by Lemma 5.1 and the
definition of the function $M_{p}$, the theorem will follow from (7.3) when
$(G,n,q)$ is not $(2^{1+4}.\operatorname{Sp}_{4}(2),4,3)$. In this latter
case, we can compute directly that the proportion of $3$-elements in $G$ is at
least $1/[4(3^{4}-1)]$.

So assume that $G$ is not one of the exceptions listed in the statement of the
theorem, and that $(G,n,q)$ is not $(2^{1+4}.\operatorname{Sp}_{4}(2),4,3)$.
Let $d$ and $G_{a}$ be as above. Also, let $\mathcal{R}$, $\mathcal{S}$,
$q_{1}=q^{d}$, $m_{r}$, $s(S)$, $t(S)$, and $\mathcal{X}$ be as defined in
Lemmas 5.1 and 5.2. For $X\in\mathcal{X}$, define $d_{X}:=r^{m_{r}}$ if
$X=O_{r}(G_{a})$ for some $r\in\mathcal{R}$; and define $d_{X}:=s(S)^{t(S)}$
if $X=T(S)$ for some $S\in\mathcal{S}$. By Corollary 7.5, either
$M_{p}(G_{a}/C_{G_{a}}(X))\leq(d_{X})_{p^{\prime}}(q_{1}^{d_{X}}-1)/(q_{1}-1)$,
or $G_{a}/C_{G_{a}}(X)$ lies in Table 5. In the former case, define
$M_{X}:=(d_{X})_{p^{\prime}}(q_{1}^{d_{X}}-1)/(q_{1}-1)$. In the latter case,
define $M_{X}$ to be the upper bound for $M_{p}(G_{a}/C_{G_{a}}(X))$ given in
the fourth column of Table 5.

Suppose first that $|\mathcal{X}|=1$ and that
$(G_{a}/C_{G_{a}}(X),d_{X},q_{1})$ lies in Table 5, where
$\mathcal{X}=\\{X\\}$. Since $q_{1}$ is prime in each of these cases, we must
have $d=1$ and $G=G_{a}$. If $X=O_{r}(G)$ for some $r\in\mathcal{R}$, then it
follows that $G$ has shape $Z\circ r^{1+2m}.(H/O_{r}(H))$, where $Z$ is a
subgroup of $Z(\operatorname{GL}_{n}(q))$, and $H:=G/C_{G}(X)$. We can then
read off the possibilities for $G$ from Table 5, and these give us the
exceptions listed in the first nine rows of Table 1. We do the same thing in
the case $X=T(S)$, with $S\in\mathcal{S}$: we see that $G$ is one of the
groups in the final two rows of Table 1. Note here that despite the presence
of the group $2^{4}.\operatorname{Sp}_{4}(2)$ in Table 5, the associated
primitive group
$G:=2^{1+4}.\operatorname{Sp}_{4}(2)\leq\operatorname{GL}_{4}(3)$ does have
the property that the proportion of $3$-elements in $G$ is at least
$1/[4(3^{4}-1)]$, as mentioned above.

So we may assume that

  * •

if $|\mathcal{X}|=1$, then
$M_{X}=(d_{X})_{p^{\prime}}(q_{1}^{d_{X}}-1)/(q_{1}-1)$, where
$\mathcal{X}=\\{X\\}$.

Now, notice that Rows 2–8 of Table 5 are all possibilities for
$G_{a}/C_{G_{a}}(X)$ when $X=O_{2}(G_{a})$, while Row 1 is a possibility for
$G_{a}/C_{G_{a}}(X)$ when $X=O_{3}(G_{a})$. It follows that:

  * •

if $X_{i},X_{j}\in\mathcal{X}$ with $X_{i}\neq X_{j}$, then
$G_{a}/C_{G_{a}}(X_{i})$ and $G_{a}/C_{G_{a}}(X_{j})$ cannot both lie in Rows
2–8 in Table 5.

  * •

if $X_{i},X_{j}\in\mathcal{X}$ with $X_{i}\neq X_{j}$, and
$G_{a}/C_{G_{a}}(X_{i})\cong G_{a}/C_{G_{a}}(X_{j})$ lie in the same row in
Table 5, then that row must be either Row 9, or Row 10.

It now follows from Lemma 7.6 that
$(q_{1}-1)\prod_{X\in\mathcal{X}}M_{X}\leq(n_{1})_{p^{\prime}}(q_{1}^{n_{1}}-1)$,
where $n_{1}:=\prod_{X\in\mathcal{X}}d_{X}$. Since $n_{1}$ divides $n/d$ by
Lemma 5.2, and

(7.4) $\displaystyle
M_{p}(G_{a})\leq(q_{1}-1)\left(\prod_{X\in\mathcal{X}}M_{p}(G_{a}/C_{G_{a}}(X))\right)\leq(q_{1}-1)\prod_{X\in\mathcal{X}}M_{X}$

by Corollary 5.3, the upper bound at (7.3), and hence the theorem, follows. ∎

We conclude the section with the following corollary of the proof of Theorem
3, which will be useful in our proof of Theorem 1.

######  Corollary 7.7.

Let $q$ be a power of a prime $p$, and let $G\leq\operatorname{GL}_{n}(q)$ be
primitive. If $G$ is not one of those groups in Table 1, and
$(G,n,q)\neq(2^{1+2}.\operatorname{Sp}_{4}(2),4,3)$, then $M_{p}(G)\leq
n_{p^{\prime}}(q^{n}-1)$. If $G$ is one of the groups in Table 1, then
$M_{p}(G)\leq f(G,n,q)$, where $f(G,n,q)$ is as in the fourth column of the
table. Finally, if $(G,n,q)=(2^{1+2}.\operatorname{Sp}_{4}(2),4,3)$, then
$M_{p}(G)=576$.

###### Proof.

The proof of Theorem 3 actually proves the statement in the corollary (by
Corollary 2.3, proving that $M_{p}(G)\leq M$ is stronger than proving that the
proportion of elements of $p$-power order in $G$ is at least $1/M$). As can be
seen therein, the primitive group
$G=2^{1+4}.\operatorname{Sp}_{4}(2)\leq\operatorname{GL}_{4}(3)$ has
$M_{p}(G)=576$, and so does not satisfy $M_{p}(G)\leq 4(3^{4}-1)$. (Since the
proportion of $3$-elements in $G$ can be computed to be at least
$1/[4(3^{4}-1)]$, this group is not an exception to Theorem 3). ∎

##  8\. The proof of Theorems 1 and 2

In this final section, we prove Theorems 1 and 2. Before doing so, we require
the following easy lemma. Recall that the function
$h_{p}:\mathbb{R}\rightarrow\mathbb{R}$ is defined as $h_{p}(m):=3^{(m-1)/2}$
if $p=2$; $h_{p}(m):=20^{(m-1)/4}$ if $p=3$; and
$h_{p}(m):=(p-1)!^{(m-1)/(p-2)}$ if $p>3$.

######  Lemma 8.1.

Let $m$ and $t$ be positive integers, and let $q$ be a power of a prime $p$.
Then each of the following assertions hold.

  1. (i)

if $p=2$, then
$[m_{2^{\prime}}(q^{m}-1)]^{t}h_{p}(t)\leq[3(q^{3}-1)]^{mt/3}h_{p}(mt/3)$.

  2. (ii)

if $q=3$, then $[m_{2^{\prime}}(q^{m}-1)]^{t}h_{p}(t)\leq
640^{mt/4}h_{p}(mt/4)$.

  3. (iii)

if $q=3$ and $4\nmid n$, then
$[m(q^{m}-1)]^{t}h_{p}(t)\leq[2(q^{2}-1)]^{mt/2}h_{p}(mt/2)$.

  4. (iv)

if $q\in\\{9,27\\}$, then
$[m(q^{m}-1)]^{t}h_{p}(t)\leq[2(q^{2}-1)]^{mt/2}h_{p}(mt/2)$.

  5. (v)

if $q\in\\{5,7,11\\}$, then
$[m(q^{m}-1)]^{t}h_{p}(t)\leq\alpha(q)^{mt/2}h_{p}(mt/2)$, where
$\alpha(5):=96$, $\alpha(7):=144$, and $\alpha(11):=600$.

  6. (vi)

if $q\in\\{5,7,11\\}$ and $n$ is odd, then
$[m_{p^{\prime}}(q^{m}-1)]^{t}h_{p}(t)\leq(q-1)^{mt}h_{p}(mt)$.

  7. (vii)

if $p>2$ and $q\not\in\\{3,5,7,9,11,27\\}$, then
$[m_{p^{\prime}}(q^{m}-1)]^{t}h_{p}(t)\leq(q-1)^{mt}h_{p}(mt)$. Moreover, if
$q\in\\{13,19\\}$ and $m(q)$ divides $n$, where $m(13):=4$ and $m(19):=2$,
then $\alpha(q)^{n/m(q)}h_{p}(n/m(q))\leq(q-1)^{n}h_{p}(n)$, where
$\alpha(13):=138240$ and $\alpha(19):=1140$.

###### Proof.

The arguments to prove (i)–(vii) are entirely similar, so we will just prove
(vi). So assume that $p>2$ and that $q\not\in\\{3,5,7,9,11,27\\}$. Clearly, we
may also assume that $m\geq 2$. Define $k(p):=20^{1/4}$ if $p=3$, and
$k(p):=(p-1)!^{1/(p-2)}$ if $p>3$. To prove the first part of (vi), it
suffices to show that

(8.1)
$\displaystyle\left(\dfrac{q^{m}-1}{(q-1)^{m}}\right)^{t}\leq\dfrac{k(p)^{t(m-1)}}{m^{t}}.$

To this end, suppose first that $p>3$. Then since $q\geq 11$, we have
$(q^{m}-1)/(q-1)^{m}\leq q^{m}/(q-1)^{m}\leq(11/10)^{m}$. Note also that
$(11/10)^{m}\leq 24^{(m-1)/3}/m$ for $m\geq 3$. Thus, since the function
$n\rightarrow n!^{1/(n-1)}$ is non-decreasing for $n\in\mathbb{N}$, we have

$\displaystyle\dfrac{q^{m}-1}{(q-1)^{m}}\leq\left(\dfrac{5}{4}\right)^{m}\leq\dfrac{24^{(m-1)/3}}{m}\leq\dfrac{(p-1)!^{(m-1)/(p-2)}}{m}$

for $m\geq 2$. The bound at (8.1) follows. The case $p=3$ and $q\geq 81$ is
entirely similar.

Finally assume that $q\in\\{13,19\\}$, and let $m(q)$ and $\alpha(q)$ be as
defined in (vi). Then one can check that $\alpha(q)^{1/m(q)}/(q-1)$ is less
than $(p-1)!^{(1-1/m(q))/(p-2)}$ in each case, and this gives us what we need.
∎

We are now ready to prove Theorems 1 and 2.

###### Proof of Theorems 1 and 2.

Let $G$ be a subgroup [respectively irreducible subgroup] of
$\operatorname{GL}_{n}(q)=\operatorname{GL}(V)$. By Corollary 2.3, it will
suffice to prove that

(8.2) $\displaystyle M_{p}(G)\leq f(n,q)\text{ [resp. $i(n,q)$],}$

where $f(n,q)$ [resp. $i(n,q)$] is as defined in the statement of Theorem 1
[resp. Theorem 2].

To this end, let $0=V_{1}<\ldots<V_{d}=V$ be a $G$-composition series for $V$.
Then it is well-known that $G/O_{p}(G)$ is isomorphic to a completely
reducible subgroup of $\operatorname{GL}(W)\cong\operatorname{GL}_{n}(q)$,
where $W$ is the $\mathbb{F}_{q}[G]$-module $V_{1}\oplus
V_{2}/V_{1}\oplus\ldots\oplus V_{d}/V_{d-1}$. Since
$M_{p}(G)=M_{p}(G/O_{p}(G))$, we may therefore assume, for the remainder of
the proof, that $G$ is a completely reducible subgroup of
$\operatorname{GL}(V)$.

Suppose first that $G$ is irreducible. Then $G$ may be embedded as a subgroup
of a wreath product $R\wr\operatorname{Sym}_{t}$, where
$R\leq\mathrm{GL}_{m}(q)$ is primitive, $mt=n$, and $G\cap R^{t}$ is a
subdirect product in a direct product $A^{t}$ of $t$ copies of a normal
subgroup $A$ of $R$ (we allow the case $t=1$, in which case $G\cong R$ is
primitive). We now define a function $g$, as follows: if $R$ is not one of the
groups from Theorem 3(i), and $(R,m,q)$ is not
$(2^{1+4}.\operatorname{Sp}_{4}(2),4,3)$, then define
$g(R):=m_{p^{\prime}}(q^{m}-1)$. If $R$ is one of the groups from Theorem
3(i), then define $g(R):=P(G,m,q)$, where $P(G,m,q)$ is in the fourth column
of Table 1. Finally, if $(R,m,q)=(2^{1+4}.\operatorname{Sp}_{4}(2),4,3)$, then
define $g(R):=M_{3}(R)=576$. Then by Corollary 7.7, $M_{p}(R)\leq g(R)$ in
each case. It then follows from Corollary 2.3 that $M_{p}(G)\leq
g(R)^{t}M_{p}(G/G\cap R^{t})\leq g(R)^{t}h_{p}(t)$, where the last inequality
follows from Corollary 6.2. One can now deduce that $M_{p}(G)\leq i(n,q)\leq
f(n,q)$ by inspecting the values of $g(R)$, and applying Lemma 8.1 in each
case.

We can now complete the proof of (8.2) by induction on $n$. The irreducible
case above can serve as the base case for induction, so assume that $G$ is
reducible. Then $G$ embeds as a subgroup of
$\operatorname{GL}_{r}(q)\times\operatorname{GL}_{s}(q)$, where $r,s\geq 1$,
and $r+s=n$. Let $N:=G\cap(\operatorname{GL}_{r}(q)\times 1)$. Then the
inductive hypothesis implies that $M_{p}(N)\leq f(r,q)$ and $M_{p}(G/N)\leq
f(s,q)$. Since $M_{p}(G)\leq M_{p}(N)M_{p}(G/N)$ by Corollary 2.3, and
$f(r,q)f(s,q)=f(r+s,q)$, the claim (8.2) follows. ∎

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