# A characterization of some finite simple groups by their character codegrees

Hung P. Tong-Viet  H.P. Tong-Viet, Department of Mathematics and Statistics,
Binghamton University, Binghamton, NY 13902-6000, USA<EMAIL_ADDRESS>
###### Abstract.

For a finite group $G$ and an irreducible complex character $\chi$ of $G$, the
codegree of $\chi$ is defined by
$\textrm{cod}(\chi)=|G:\textrm{ker}(\chi)|/\chi(1)$, where
$\textrm{ker}(\chi)$ is the kernel of $\chi$. In this paper, we show that if
$H$ is a finite simple exceptional group of Lie type or a projective special
linear group and $G$ is any finite group such that the character codegree sets
of $G$ and $H$ coincide, then $G$ and $H$ are isomorphic.

###### Key words and phrases:

codegrees; simple groups

###### 2010 Mathematics Subject Classification:

Primary 20C15; Secondary 20D05

###### Contents

  1. 1 Introduction
  2. 2 Preliminaries
  3. 3 Simple exceptional groups of Lie type
  4. 4 Projective special linear groups
  5. 5 Symplectic groups

##  1\. Introduction

Let $G$ be a finite group and let $\chi$ be an irreducible complex character
of $G$. Let $\ker(\chi)$ denote the kernel of $\chi$, that is, the set of all
elements $g\in G$ such that $\chi(g)=\chi(1)$. The positive integer $\chi(1)$
is called a character degree or a degree of $\chi$. This is the degree of the
representation of $G$ that affords the character $\chi$. In [9], Chillag and
Herzog call the number $|G|/\chi(1)$ a ‘character degree quotient’, and obtain
several interesting results concerning the structure of the finite groups $G$
by putting certain arithmetic restriction on these character degree quotient.
For example, they show that if $|G|/\chi(1)$ is a prime power for some
irreducible character $\chi$ of $G$, then $G$ is not a finite nonabelian
simple group (see [9, Theorem 1]). In a subsequent paper [10], Chillag, Mann
and Manz called the quotient $|G|/\chi(1)$ a ‘codegree of $\chi$’. Among other
results, they show that if $G$ has an irreducible character $\chi$ which
satisfies the same condition as above for a prime $p$, then $\chi$ is monomial
and $F^{*}(G)=O_{p}(G)\neq 1$, where $F^{*}(G)$ is the generalized Fitting
subgroup of $G$ and $O_{p}(G)$ is the largest normal $p$-subgroup of $G$. This
result was later reproved by Riese and Schmid in [35].

Qian, Wang and Wei [34] modified the definition of codegrees of characters by
defining it to be $|G:\ker(\chi)|/\chi(1)$. Clearly, if $\chi$ is faithful,
then the two definitions coincide. However, the new definition of codegrees
which we will use in this paper is more compatible with taking quotients in
the sense that the codegree of a quotient of a finite group $G$ is also the
codegree of $G$. Now let ${\mathrm{cd}}(G)$ and ${\mathrm{cod}}(G)$ be the set
of all degrees and codegrees of $G$, respectively. It is well known that these
two sets have strong influence on the group structure. In the late 1990s, B.
Huppert [23] conjectured that every finite nonabelian simple group is
essentially determined by the set of its character degrees, that is, if $H$ is
a finite nonabelian simple group and $G$ is a finite group such that
${\mathrm{cd}}(G)={\mathrm{cd}}(H)$, then $G\cong H\times A$ for some abelian
group $A$. This conjecture has been verified for all sporadic simple groups,
the alternating groups of degree at least $5$ and low degree finite nonabelian
simple groups of Lie type (see [6, 20, 33] and the references therein). In
[44], we are able to verify Huppert’s conjecture for the remaining simple
exceptional groups of Lie type. Motivated by Huppert’s Conjecture, Qian [27,
Problem 20.79] proposed the following:

######  Conjecture.

_(Codegrees Conjecture)._ Let $H$ be a finite nonabelian simple group and let
$G$ be a finite group such that ${\mathrm{cod}}(G)={\mathrm{cod}}(H)$. Then
$G\cong H.$

This conjecture has been verified for all $26$ sporadic simple groups, the
alternating groups, ${}^{2}{\rm B}_{2}(q^{2}),{}^{2}{\rm
G}_{2}(q^{2}),{}^{2}{\rm
F}_{4}(q^{2}),\operatorname{PSL}_{2}(q),\operatorname{PSL}_{3}(q)$,
$\textrm{PSU}_{3}(q),\operatorname{PSp}_{4}(q)$ ($q$ even) and other small
finite nonabelian simple groups in [1, 2, 11, 12, 14, 15, 30, 45, 47]. In a
recent paper, Hung and Moretó [21] show that the set of character codegrees
together with their multiplicity determines the finite nonabelian simple
groups up to isomorphism. This is analogous to a result stating that finite
nonabelian quasisimple groups are uniquely determined by the structure of
their complex group algebra or equivalently their character degrees together
with their multiplicity obtained in [5, 41, 42, 43]. The authors also show
that if $S$ and $H$ are finite nonabelian simple groups such that
${\mathrm{cod}}(S)\subseteq{\mathrm{cod}}(H)$, then $S\cong H$. This important
result allows us to complete the verification of the Codegrees conjecture for
all finite simple exceptional groups of Lie type and ${\rm PSL}_{n}(q)$ for
$n\geq 4$ and $q$ a prime power.

######  Theorem 1.

Let $H$ be a finite nonabelian simple exceptional group of Lie type or ${\rm
PSL}_{n}(q)$ with $n\geq 4$ and $q$ a prime power. Let $G$ be a finite group
such that ${\mathrm{cod}}(G)={\mathrm{cod}}(H).$ Then $G\cong H.$

We should mention that this is the first time that a conjecture in this flavor
has been confirmed for a family of finite simple groups of Lie type with
unbounded Lie rank over an arbitrary finite field.

We now describe our approach to the proof of Theorem 1. Let $H$ be one of the
finite simple groups in the statement of the theorem defined over a finite
field of size $q=p^{f}$ for some prime $p$. So $H$ is a finite nonabelian
simple group of Lie type in characteristic $p$, where $p$ is a prime. Let $G$
be a counterexample to the Codegrees conjecture with minimal order. By [21,
Theorem 8.3], $G$ has a unique minimal normal subgroup $N$ and $G/N\cong H$.
By Lemma 2.1, $N$ is an elementary abelian $r$-group of order $r^{m}$ for some
prime $r$ and $G$ is perfect. Hence, either $N=Z(G)$ or $C_{G}(N)=N$.

Since $G$ is perfect, the former case implies that $G$ is a finite quasisimple
group with $G/Z(G)\cong H$ and $|Z(G)|=r$. This case can be eliminated by
finding certain small degree faithful irreducible character $\chi$ of $G$ such
that $\chi(1)/r$ is not a degree of $H$ (see Lemma 2.2). In the latter case,
$N$ is a faithful irreducible $\mathbb{F}_{r}H$-module and $r^{2m}$ divides
$|H|$ (see Lemma 2.3). At this point, we can divide the proof into two cases
according to whether $r$ is the defining characteristic $p$ or not. For the
cross characteristic cases, we can use the well-known Landazuri-Seitz-
Zalesskii’s bounds (see [26, Table 5.3.A]) to obtain a lower bound for $m$ and
check that $r^{2m}$ is too large comparing to the order of the Sylow
$r$-subgroup of $H$ and we arrive at a contradiction. For the defining
characteristic, we first get the lower bound $m\geq Df$, where $D$ is the
dimension of the natural module or the minimal module for $H$ (see [17]). For
finite simple exceptional groups of Lie type, this bound is large enough to
guarantee that $r^{2m}\nmid|H|$ and we are done. However, for ${\rm
PSL}_{n}(q)$, this is not the case. To obtain a contradiction, we use some
results on the representation theory of finite groups of Lie type in defining
characteristic to reduce it to the case where $N$ is a natural module for
$\operatorname{SL}_{n}(q)$ and then using the fact that the second cohomology
$H^{2}(\operatorname{SL}_{n}(q),N)$ is trivial to deduce that $G\cong
N:\operatorname{SL}_{n}(q)$ is a split extension and produce a faithful
irreducible character of $G$ whose degree is not divisible by $p$, which leads
to a contradiction.

Our approach can be summarized as follows.

Step 1. For each quasisimple group $K$ with $K/Z(K)\cong H$ and $r=|Z(K)|$
being a prime, show that ${\mathrm{cod}}(K)\neq{\mathrm{cod}}(H)$. To do this,
it suffices to find a faithful irreducible character $\chi$ of $K$ such that
$\chi(1)/r$ is not a character degree of $H.$ (See Lemma 2.2.)

Step 2. Assume $r\neq p$. Use the Landazuri-Seitz-Zalesskii’s bounds and their
improvement (see [28, 36]) to get a lower bound $m\geq e(H)$, where $e(H)$ is
given in [26, Table 5.3.A], and verify that $2m\geq 2e(H)>\log_{r}(|H|_{r})$,
which violates the claim that $r^{2m}$ divides $|H|$ (see Lemma 2.3). For
those simple groups $H$ with a small $e(H)$, we can use MAGMA [8] or GAP [13]
to obtain a contradiction.

Step 3. Assume $r=p.$ Let $D$ be the dimension of the natural module for $H$
if $H$ is classical or the dimension of a minimal module if $H$ is of
exceptional type. Let $L$ be the universal cover of $H$ and let
$k=\overline{\mathbb{F}}_{p}$. We first obtain the bound $m\geq Df$ (see
[17]). For $\operatorname{PSL}_{n}(q)$, we make use of the low-degree
representations of $L$ in defining characteristic in [18, 32] together with
the Steinberg’s twisted tensor product theorem, to reduce to the case when
every $kL$-irreducible constituent of $N\otimes_{\mathbb{F}_{p}}k$ is quasi-
equivalent to the natural module for $L$.

Step 4. We can then use the action of $L$ on $N$ as well as the results on the
second cohomology group of $H$ on its natural module to obtain some
information on the sizes of the nontrivial orbits of $G$ on the set of
irreducible characters of $N$ which will lead to a contradiction.

Our approach could be used to verify the Codegree conjecture for other small
degree classical groups. However, we might need a different method to verify
the last step for large degree cases.

Organization. We collect some useful results for the proofs of the main
theorem in Section 2. We will verify the Codegrees conjecture for the finite
simple exceptional groups of Lie type in Section 3 and
$\operatorname{PSL}_{n}(q)$ in Section 4. In the last section, we verify the
first three steps for the symplectic groups $\operatorname{PSp}_{2n}(q)$.

Notation. Our notation is more or less standard. We follow [26] for the
notation of simple groups and [24, 25] for representation and character
theories of finite groups. For a finite group $G$, $M(G)$ denotes the Schur
multiplier of $G$, $F(G)$ is the Fitting subgroup of $G$, that is, the largest
nilpotent normal subgroup of $G$, $E(G)$ is the layer of $G$, that is, a
subgroup of $G$ generated by all components of $G$ and $F^{*}(G)=F(G)E(G)$ is
the generalized Fitting subgroup of $G$. A component $L$ of $G$ is a subnormal
quasisimple subgroup of $G$. A finite group $L$ is quasisimple if $L$ is
perfect and $L/Z(L)$ is a finite nonabelian simple group. We denote by
$\operatorname{Irr}(G)$ the set of all irreducible characters of $G$. For
$\chi\in\operatorname{Irr}(G)$, we denote by ${\mathrm{cod}}(\chi)$, the
codegree of $\chi$, that is, ${\mathrm{cod}}(\chi)=|G:\ker(\chi)|/\chi(1)$. If
$N$ is a normal subgroup of a finite group $G$ and
$\theta\in\operatorname{Irr}(N)$, we write $\operatorname{Irr}(G|\theta)$ for
the set of all irreducible characters of $G$ lying above $\theta.$ For a
positive integer $n$, we write ${\mathbb{Z}}_{n}$ for a cyclic group of order
$n.$ For a prime $p$ and a positive integer $n$, the $p$-part of $n$, denoted
by $n_{p}$ is the largest power of $p$ dividing $n$ and
$n_{p^{\prime}}=n/n_{p}$ is the $p^{\prime}$-part of $n.$ As usual, $Z(G)$ is
the center of $G$, $C_{G}(A)$ is the centralizer of $A$ in $G$, where $A$ is a
subgroup of $G$.

##  2\. Preliminaries

In this section, we prove several lemmas that are needed for the proofs of the
main theorem. We first deduce some properties of a minimal counterexample to
the Codegrees conjecture.

######  Lemma 2.1.

Let $H$ be a finite nonabelian simple group and let $G$ be a finite group with
${\mathrm{cod}}(G)={\mathrm{cod}}(H)$. Suppose that $G$ has a unique minimal
normal subgroup $N$ such that $G/N\cong H.$ Then $G$ is perfect and $N$ is an
elementary abelian $r$-group, for some prime $r$.

###### Proof.

We have $N\cong S^{k}$ for some finite simple group $S$ and integer $k\geq 1.$
Assume that $S$ is nonabelian. It is well-known that $S$ has a nontrivial
irreducible character $\theta$ that extends to $\operatorname{Aut}(S)$ and
thus the irreducible character $\theta^{k}\in\operatorname{Irr}(N)$ extends to
$\chi\in\operatorname{Irr}(G)$ (see [41]). Now $\chi$ is faithful by the
uniqueness of $N$. Hence $|G|/\chi(1)=|H|/\varphi(1)$ for some
$\varphi\in\operatorname{Irr}(H)$. It follows that
$\varphi(1)=\theta(1)^{k}/|S|^{k}$. However, since $S$ is a finite nonabelian
simple group, $\theta(1)<|S|$ and so $\varphi(1)=(\theta(1)/|S|)^{k}<1$, which
is a contradiction. Therefore, $S$ is abelian and hence $N$ is an elementary
abelian $r$-group for some prime $r$. The fact that $G$ is perfect follows
from [21, Theorem 2.3]. ∎

Assume the hypothesis and conclusion of Lemma 2.1. Since $N$ is a unique
minimal normal subgroup of $G$ and is elementary abelian, $N\leq
C_{G}(N)\unlhd G$. Since $G/N\cong H$ is a finite nonabelian simple group,
either $C_{G}(N)=G$ and thus $G$ is a quasisimple group with center $N$ or
$C_{G}(N)=N$. To handle the former case, we can use the following observation.

######  Lemma 2.2.

Let $H$ be a finite nonabelian simple group and let $G$ be a finite
quasisimple group with $N=Z(G)$ a cyclic group of prime order $r$ such that
$G/N\cong H$. Assume that ${\mathrm{cod}}(G)={\mathrm{cod}}(H)$. Then for any
faithful irreducible character $\chi\in\operatorname{Irr}(G)$,
$\chi(1)/r\in{\mathrm{cd}}(H)$.

###### Proof.

Let $\chi\in\operatorname{Irr}(G)$ be a faithful irreducible character of $G$.
Since ${\mathrm{cod}}(\chi)=|G|/\chi(1)\in{\mathrm{cod}}(H)$,
$|G|/\chi(1)=|H|/\varphi(1)$ for some $\varphi\in\operatorname{Irr}(H)$. Hence
$\varphi(1)=\chi(1)|H|/|G|=\chi(1)/r\in{\mathrm{cd}}(H)$. The proof is now
complete. ∎

Therefore, to show that $G$ cannot be a quasisimple group, for each prime
divisor $r$ of $|M(H)|$, where $M(H)$ is the Schur multiplier of $H$, and
every cyclic central extension $L$ of $H$ with $|Z(L)|=r$, we need to find a
faithful irreducible character $\chi$ of $L$ such that
$\chi(1)/r\not\in{\mathrm{cd}}(H)$.

The next lemma is key to our approach to verify the Codegrees conjecture.

######  Lemma 2.3.

Let $H$ be a finite nonabelian simple group and let $G$ be a finite group with
${\mathrm{cod}}(G)={\mathrm{cod}}(H)$. Suppose that $G$ has a unique minimal
normal subgroup $N$, where $N$ is an elementary abelian $r$-group of order
$r^{m}$ for some prime $r$, $G/N\cong H$ and $C_{G}(N)=N$. Then $N$ is a
nontrivial faithful irreducible $\mathbb{F}_{r}H$-module and the following
hold.

  1. $(i)$

$r^{2m}$ divides $|H|$.

  2. $(ii)$

If $\chi\in\operatorname{Irr}(G)$ is faithful, then $r^{m}\mid\chi(1)$ and
$\chi(1)/r^{m}\in{\mathrm{cd}}(H)$.

###### Proof.

Since $N$ is a minimal normal elementary abelian $r$-subgroup of $G$ and
$C_{G}(N)=N$, $H\cong G/N=G/C_{G}(N)$ is isomorphic to an irreducible subgroup
of $\operatorname{GL}(N)\cong\operatorname{GL}_{m}(\mathbb{F}_{r})$.
Therefore, $N$ is a nontrivial faithful irreducible $\mathbb{F}_{r}H$-module.

We now show that $r^{2m}\mid|H|$. We know that $G$ acts on
$\operatorname{Irr}(N)\cong N$ by conjugation and let
$\\{\mathcal{O}_{i}\\}_{i=0}^{t}$ be $G$-orbits of this action with
$\mathcal{O}_{0}=\\{1_{N}\\}.$ Since $G$ does not centralize any nontrivial
element of $N$, $G$ does not stabilize any nontrivial irreducible character of
$N$ and thus $\mathcal{O}_{0}$ is the only $G$-orbit on
$\operatorname{Irr}(N)$ that has size $1$. For any $j\geq 1$, fix
$\theta_{j}\in\mathcal{O}_{j}$ and let $T_{j}=I_{G}(\theta_{j})$. Then
$|\mathcal{O}_{j}|=|G:T_{j}|$. If $r\mid|\mathcal{O}_{j}|$ for all $j$ with
$1\leq j\leq t$, then $r\mid|N|-1=r^{m}-1$, which is impossible. Therefore,
there exists an index $j_{0}$ with $1\leq j_{0}\leq t$ such that
$r\nmid|\mathcal{O}_{j_{0}}|$. Let $\theta=\theta_{j_{0}}$ and $T=T_{j_{0}}$.

Write $\theta^{T}=\sum_{i=1}^{a}e_{i}\phi_{i}$, where
$\operatorname{Irr}(T|\theta)=\\{\phi_{1},\phi_{2},\dots,\phi_{a}\\}$ and
$e_{i}=\phi_{i}(1)$ for all $i\geq 1$. By Clifford’s correspondence theorem,
$\theta^{G}=\sum_{i=1}^{a}e_{i}\chi_{i}$, where
$\chi_{i}:=\phi_{i}^{G}\in\operatorname{Irr}(G|\theta)$. For each $1\leq k\leq
a,$ $\chi_{k}$ is faithful and thus
${\mathrm{cod}}(\chi_{k})=|G|/\chi_{k}(1)=|H|/\varphi_{k}(1)$ for some
$\varphi_{k}\in\operatorname{Irr}(H).$ It follows that
$\chi_{k}(1)=r^{m}\varphi_{k}(1)$, whence
$|G:T|\phi_{k}(1)=r^{m}\varphi_{k}(1)$. Since $r$ does not divide
$|\mathcal{O}_{j_{0}}|=|G:T|$, we deduce that $r^{m}\mid\phi_{k}(1)$ and hence
$\phi_{k}(1)=r^{m}c_{k}$ for some positive integer $c_{k}.$ We now have

$|T:N|=\theta^{T}(1)=\sum_{i=1}^{a}e_{i}\phi_{i}(1)=\sum_{i=1}^{a}\phi_{i}(1)^{2}=r^{2m}\sum_{i=1}^{a}c_{i}^{2}.$

Therefore, $r^{2m}$ divides $|H|$ as wanted.

Finally, let $\chi\in\operatorname{Irr}(G)$ be a faithful character of $G$.
Then ${\mathrm{cod}}(\chi)=|G|/\chi(1)=|H|/\varphi(1)$ for some
$\varphi\in\operatorname{Irr}(H)$. It follows that
$\chi(1)=|N|\varphi(1)=r^{m}\varphi(1)$ and (ii) follows. The proof is now
complete. ∎

Let $q$ be a prime power and let $n\geq 2$ be an integer. A prime divisor $r$
of $q^{n}-1$ is called a primitive prime divisor of $q^{n}-1$ if $r$ divides
$q^{n}-1$ but $r$ does not divide $q^{i}-1$ for any integer $i$ with $1\leq
i<n$. By Zsigmondy’s theorem [48], such a prime $r$ exists unless
$(n,q)=(6,2)$ or $n=2$ and $q$ is a Mersenne prime. For a prime $r\neq p$, we
write $d_{q}(r)$ for the order of $q$ modulo $r$. In this case, $r$ is a
primitive prime divisor of $q^{j}-1$ for $j=d_{q}(r)$ and hence $j\mid(r-1)$;
in particular, $r\geq j+1.$

We also need the following number theoretic results that will be used to
compute the orders of the Sylow subgroups of finite simple groups of Lie type.

######  Lemma 2.4.

Let $q$ be a prime power, let $d$ be a positive integer and let $r$ be a prime
divisor of $q-\epsilon$, where $\epsilon=\pm$.

  * (i)

If $r=2$, then

$(q^{d}-\epsilon)_{r}=\left\\{\begin{array}[]{ll}(q-\epsilon)_{r}&\mbox{if $d$
is odd}\\\ (q^{2}-1)_{r}(d/2)_{r}&\mbox{if $d$ is even and $\epsilon=+$}\\\
2&\mbox{if $d$ is even and $\epsilon=-$.}\end{array}\right.$

  * (ii)

If $r$ is odd, then

$\displaystyle(q^{d}-\epsilon)_{r}$
$\displaystyle=\left\\{\begin{array}[]{ll}1&\mbox{if $d$ is even and
$\epsilon=-$}\\\ (q-\epsilon)_{r}(d)_{r}&\mbox{otherwise}\end{array}\right.$
$\displaystyle(q^{d}+\epsilon)_{r}$
$\displaystyle=\left\\{\begin{array}[]{ll}(q-\epsilon)_{r}(d)_{r}&\mbox{if $d$
is even and $\epsilon=-$}\\\ 1&\mbox{otherwise.}\end{array}\right.$

###### Proof.

This is Lemma 2.1 in [7]. ∎

The orders of the Sylow $r$-subgroups of finite simple classical groups in
cross characteristic can be found in [46] and [37].

If $r\geq 2$ and $x\geq 1$ are integers, we will use the inequality $x<r^{x}$
or equivalently $\log_{r}(x)<x$ regularly. If $1\leq a_{1}\leq
a_{2}\leq\dots\leq a_{k}$ are integers with $k\geq 1$ and $q\geq 2$, then

$\prod_{i=1}^{k}(q^{a_{i}}-1)<q^{a_{1}+\dots a_{k}}.$

Moreover, if $a\geq 1$ is an integer, then $q^{a}\pm 1\geq\frac{1}{2}q^{a}$.
More generally, if $2\leq a_{1}<\dots<a_{k}$ are integers and
$\operatorname{\epsilon}_{1},\operatorname{\epsilon}_{2},\dots,\operatorname{\epsilon}_{k}\in\\{\pm
1\\}$, then

$\frac{1}{2}<\frac{(q^{a_{1}}+\operatorname{\epsilon}_{1})\dots(q^{a_{k}}+\operatorname{\epsilon}_{k})}{q^{a_{1}+\dots+a_{k}}}<2.$

For a proof, see Lemma 2.1 in [39].

##  3\. Simple exceptional groups of Lie type

Let $q=p^{f}$, where $p$ is a prime and $f\geq 1$ is an integer. We will use
the convention ${\rm E}_{6}^{+}(q)$ for ${\rm E}_{6}(q)$ and ${\rm
E}_{6}^{-}(q)$ for ${}^{2}{\rm E}_{6}(q)$. In this section, we will verify the
Codegree conjecture for the following simple exceptional groups of Lie type:

${\rm G}_{2}(q),{\rm F}_{4}(q),{\rm E}^{\epsilon}_{6}(q),{\rm E}_{7}(q),{\rm
E}_{8}(q),{}^{3}{\rm D}_{4}(q),{}^{2}{\rm B}_{2}(q),{}^{2}{\rm
F}_{4}(q),{}^{2}{\rm G}_{2}(q)$ (3.1)

We will exclude the groups ${}^{2}{\rm B}_{2}(2),{\rm G}_{2}(2),{}^{2}{\rm
G}_{2}(3)$ and ${}^{2}{\rm F}_{4}(2)$ from our list above as they are not
simple.

The order formulas for simple groups in (3.1) can be found in [26, Table
5.1.B] and the Schur multipliers of these groups can be found in [26, Theorem
5.1.4]. The simple groups with exceptional Schur multipliers are listed in
[26, Table 5.1.D]. The lower bounds for the degrees of nontrivial absolutely
irreducible projective representations of finite simple groups of Lie type in
cross characteristic can be found in [26, Table 5.3.A] and finally the minimal
modules for these simple groups can be found in [26, Table 5.4.C]. For certain
groups with nontrivial Schur multiplier, we also need some information about
the character degrees and their multiplicity available in [32]. Below is our
main theorem in this section.

######  Theorem 3.1.

Let $H$ be one of the simple groups in (3.1) and let $G$ be a finite group
such that ${\mathrm{cod}}(G)={\mathrm{cod}}(H)$. Then $G\cong H.$

We will prove Theorem 3.1 for simple groups $H$ with trivial Schur multiplier
first.

######  Proposition 3.2.

Theorem 3.1 holds for $H\cong{\rm E}_{8}(q)$, where $q=p^{f}.$

###### Proof.

Let $G$ be a counterexample to the theorem with minimal order. Then
${\mathrm{cod}}(G)={\mathrm{cod}}(H)$ but $G\not\cong H$. By [21, Theorem
8.3], we know that $G$ has a unique minimal normal subgroup $N$ and $G/N\cong
H$. By Lemma 2.1, $G$ is perfect and $N$ is an elementary abelian $r$-group of
order $r^{m}$ for some prime $r$. Then either $G$ is quasisimple with $N=Z(G)$
or $C_{G}(N)=N$. As the Schur multiplier of $\mathrm{E}_{8}(q)$ is trivial
(see [26, Theorem 5.1.4]), the former case cannot occur. Hence $C_{G}(N)=N$
and by Lemma 2.3, we have $N$ is a faithful irreducible
$\mathbb{F}_{r}H$-module and $r^{2m}\mid|H|$. It follows that
$2m\leq\log_{r}(|H|_{r})$.

Step $2$. Assume $r\neq p$. By [26, Theorem 5.3.9], $m\geq q^{27}(q^{2}-1)$.
From [26, Table 5.1.B],

$|H|=q^{120}(q^{2}-1)(q^{8}-1)(q^{12}-1)(q^{14}-1)(q^{18}-1)(q^{20}-1)(q^{24}-1)(q^{30}-1).$

Let $j=d_{q}(r)$. Then $1\leq j\leq 30$ and $j$ divides $14,18,20,24$ or $30$.

(2a) Assume $r=2$. Then $q$ is odd and $2\mid(q-1)$. So $j=1$. Now
$|H|_{2}=2^{6}(q^{2}-1)_{2}^{8}$ by Lemma 2.4 and so
$\log_{2}(|H|_{2})=6+8\log_{2}((q^{2}-1)_{2})<6+8(q^{2}-1)_{2}$. Since
$(q^{2}-1)_{2}\leq q^{2}-1$, $\log_{2}(|H|_{2})<6+8(q^{2}-1)<8q^{2}$. However,
we have that $2m\geq 2q^{27}(q^{2}-1)>8q^{2}>\log_{r}(|H|_{r})$ for any $q\geq
3,$ which is a contradiction.

(2b) Assume $r\geq 3$ is odd.

Assume $j=1$. Then $|H|_{r}=(q-1)_{r}^{8}(12\cdot 14\cdot 18\cdot 20\cdot
24\cdot 30)_{r}$ by Lemma 2.4. We see that $|H|_{r}\leq r^{5}(q-1)_{r}^{8}.$
Now $\log_{r}(|H|_{r})\leq 5+8\log_{r}((q-1)_{r})\leq 5+8(q-1)<8q.$ Clearly
$2m\geq 2q^{27}(q^{2}-1)>8q>\log_{r}(|H|_{r})$, a contradiction. Assume $2\leq
j\leq 24$. Similarly, we have $|H|_{r}\leq r^{5}(q^{j}-1)_{r}^{8}$ and hence
$\log_{r}(|H|_{r})\leq 5+8\log_{r}((q^{j}-1)_{r})<8+8(q^{j}-1)=8q^{j}\leq
8q^{24}$. Clearly, $2m\geq 2q^{27}(q^{2}-1)>8q^{24}>\log_{r}(|H|_{r})$ for any
$q\geq 2$. Finally, assume $j=30.$ Then
$\log_{r}(|H|_{r})=\log_{r}((q^{30}-1)_{r})=\log_{r}((q^{15}+1)_{r})\leq
q^{15}+1\leq q^{16}$. Now $2m\geq
2q^{27}(q^{2}-1)>q^{16}\geq\log_{r}(|H|_{r})$. Thus we have shown that if
$r\neq p$, then $r^{2m}\nmid|H|$.

Step $3$. Assume that $r=p$. By [26, Table 5.4.C], the smallest dimension of
nontrivial irreducible representations of ${\rm E}_{8}(q)$ over
$k=\overline{\mathbb{F}}_{p}$ is $248$. By [17, Lemma 4.2], $|N|=p^{m}\geq
p^{248f}$, whence $2m\geq 496f$. Now $|H|_{p}=p^{120f}$ and thus since $2m\geq
496f>120f$, $p^{2m}\nmid|H|$, which is a contradiction.

This final contradiction shows that $G\cong H$ as wanted. The proof is now
complete. ∎

######  Proposition 3.3.

Theorem 3.1 holds for $H\cong{\rm F}_{4}(q)$, where $q=p^{f}.$

###### Proof.

Let $G$ be a counterexample to the theorem with minimal order. Then
${\mathrm{cod}}(G)={\mathrm{cod}}(H)$ but $G\not\cong H$. By [21, Theorem
8.3], we know that $G$ has a unique minimal normal subgroup $N$ such that
$G/N\cong H$. By Lemma 2.1, $G$ is perfect and $N$ is an elementary abelian
group of order $r^{m}$, for some prime $r$. Then either $G$ is quasisimple
with $N=Z(G)$ or $C_{G}(N)=N$. We have that

$|H|=q^{24}(q^{2}-1)(q^{6}-1)(q^{8}-1)(q^{12}-1).$

Step $1$. If $q>2$, then the Schur multiplier of $\mathrm{F}_{4}(q)$ is
trivial (see [26, Theorem 5.1.4]). For $q=2,$ the universal cover of ${\rm
F}_{4}(2)$ is $2.{\rm F}_{4}(2)$ and so $G\cong 2.{\rm F}_{4}(2)$. By GAP
[13], $2.{\rm F}_{2}(2)$ has a faithful irreducible character $\chi$ of degree
$52$ and the smallest nontrivial character degree of $H$ is $833$. Thus
$52/2=26\not\in{\mathrm{cod}}(H)$, contradicting Lemma 2.2. Hence $C_{G}(N)=N$
and by Lemma 2.3, $r^{2m}\mid|H|$ and so $2m\leq\log_{r}(|H|_{r})$.

Step $2$. Assume that $r\neq p$. For $q=2$, we have $m\geq 52$ by [19]. Since
$|{\rm F}_{2}(2)|=2^{24}\cdot 3^{6}\cdot 5^{2}\cdot 7^{2}\cdot 13\cdot 17,$
$r^{2m}$ cannot divide $|H|$. So we assume that $q>2$. By [26, Theorem 5.3.9],
we have $m\geq q^{6}(q^{2}-1)$ if $q$ is odd and $m\geq q^{7}(q^{3}-1)(q-1)/2$
if $q>2$ is even. Note that $q^{7}(q^{3}-1)(q-1)/2>q^{6}(q^{2}-1)$. Let
$j=d_{q}(r)$. Then $j$ divides $8$ or $12$.

(2a) Assume $r=2$. Then $q$ is odd and $2\mid(q-1)$, so $j=1$. Now
$|H|_{2}=2^{3}(q^{2}-1)_{2}^{4}$ and so
$\log_{2}(|H|_{2})=3+4\log_{2}((q^{2}-1)_{2})<4+4(q^{2}-1)=4q^{2}$. Now
$2m\geq 2q^{6}(q^{2}-1)>4q^{2}\geq\log_{2}(|H|_{2})$, a contradiction.

(2b) Assume $r\geq 3$ is odd.

Case $j=1$. Then $|H|_{r}=(q-1)_{r}^{4}(6\cdot 12)_{r}\leq r^{2}(q-1)_{r}^{4}$
and so

$\log_{r}(|H|_{r})\leq 2+4\log_{r}((q-1)_{r})\leq 2+4(q-1).$

Clearly $2m\geq 2q^{6}(q^{2}-1)>2+4(q-1)>\log_{r}(|H|_{r})$, a contradiction.

Case $j=2$. Then $|H|_{r}=(q^{2}-1)_{r}^{4}(6\cdot 12)_{r}\leq
r^{2}(q^{2}-1)_{r}^{4}$ and so

$\log_{r}(|H|_{r})\leq 2+4\log_{r}((q^{2}-1)_{r})\leq 2+4(q^{2}-1).$

Clearly $2m\geq 2q^{6}(q^{2}-1)>2+4(q^{2}-1)>\log_{r}(|H|_{r})$, a
contradiction

For the remaining cases, since $j\geq 3$, $r\geq 5$, $r$ divides neither $8$
nor $12$. If $j\leq 6$, then $|H|_{r}\leq(q^{j}-1)_{r}^{3}$; therefore,
$\log_{r}(|H|_{r})\leq 3\log_{r}(q^{j}-1)\leq 3(q^{6}-1)<3q^{6}$ and hence
$2m\geq 2q^{6}(q^{2}-1)>3q^{6}\geq\log_{r}(|H|_{r})$. Finally, if $7\leq j\leq
12$, then $j=8$ or $12$ and hence

$\log_{r}(|H|_{r})\leq\log_{r}((q^{j}-1)_{r})\leq\log_{r}((q^{j/2}+1)_{r})\leq
q^{j/2}+1\leq q^{6}+1.$

Now $2m\geq q^{6}(q^{2}-1)>q^{6}+1>\log_{r}(|H|_{r})$, which is a
contradiction.

Step $3$. Assume that $r=p.$ By [26, Table 5.4.C], the smallest dimension of
nontrivial irreducible representations of ${\rm F}_{4}(q)$ over
$k=\overline{\mathbb{F}}_{p}$ is $26-\delta_{p,3}\geq 25$. By [17, Lemma 4.2],
$|N|=p^{m}\geq p^{25f}$, whence $m\geq 25f$. Now $|H|_{p}=p^{24f}$ and thus
$2m\geq 50f>24f=\log_{p}(|H|_{p})$. This final contradiction completes the
proof of the proposition. ∎

We now consider the simple groups $H\cong{\rm G}_{2}(q)$. Since ${\rm
G}_{2}(2)\cong\operatorname{U}_{3}(3).2$, we assume that $q>2.$

######  Proposition 3.4.

Theorem 3.1 holds for $H\cong{\rm G}_{2}(q)$, where $q=p^{f}>2.$

###### Proof.

Let $G$ be a counterexample to the theorem with minimal order. Then
${\mathrm{cod}}(G)={\mathrm{cod}}(H)$ but $G\not\cong H$. By [21, Theorem
8.3], we know that $G$ has a unique minimal normal subgroup $N$ such that
$G/N\cong H$. By Lemma 2.1, $G$ is perfect and $N$ is an elementary abelian
$r$-group for some prime $r$. Let $|N|=r^{m}$. Then either $G$ is quasisimple
with $N=Z(G)$ or $C_{G}(N)=N$. We have that

$|H|=q^{6}(q^{2}-1)(q^{6}-1).$

Step $1$. By [26, Theorem 5.1.4], $M({\rm G}_{2}(3))\cong{\mathbb{Z}}_{3}$,
$M({\rm G}_{2}(4))\cong{\mathbb{Z}}_{2}$ and $M({\rm G}_{2}(q))=1$ for $q\geq
5.$ By GAP [13], $3.{\rm G}_{2}(3)$ has a faithful irreducible character of
degree $27$ and $27/3=9$ is not a degree of $H\cong{\rm G}_{2}(3)$. Similarly,
$2.{\rm G}_{2}(4)$ has a faithful irreducible character of degree $12$ but
$12/2=6$ is not a character degree of ${\rm G}_{2}(4)$. Hence $C_{G}(N)=N$ and
by Lemma 2.3, we have $2m\leq\log_{r}(|H|_{r})$.

Step $2$. Assume that $r\neq p$.

Assume $q=3$. Then $m\geq 14$ by [26, Table 5.3.A]. Since $|{\rm
G}_{2}(3)|=2^{6}\cdot 3^{6}\cdot 7\cdot 13,$ $r^{2m}$ cannot divide $|H|$.
Next, assume $q=4$. Then $m\geq 12$ and since $|{\rm G}_{2}(4)|=2^{12}\cdot
3^{3}\cdot 5^{2}\cdot 7\cdot 13,$ $r^{2m}$ cannot divide $|H|$. Thus we assume
that $q\geq 5.$ By [26, Theorem 5.3.9], we have $m\geq q(q^{2}-1)$. Let
$j=d_{q}(r)$. Then $j\in\\{1,2,3,6\\}$.

(2a) Assume that $r=2$. Then $q$ is odd and $2\mid(q-1)$, so $j=1$. Now
$|H|_{2}=(q^{2}-1)_{2}^{2}$ and so
$\log_{2}(|H|_{2})=2\log_{2}((q^{2}-1)_{2})\leq 2(q^{2}-1)$. We have $2m\geq
2q(q^{2}-1)>2(q^{2}-1)\geq\log_{2}(|H|_{2})$ for any $q\geq 5$, a
contradiction.

(2b) Assume $r\geq 3$ is odd.

Case $j=1,2$. Then $|H|_{r}=(q^{j}-1)_{r}^{2}(2\cdot 6)_{r}\leq
r(q^{2}-1)_{r}^{2}$ and so

$\log_{r}(|H|_{r})\leq 1+2\log_{r}((q^{2}-1)_{r})\leq 1+2(q^{2}-1).$

Clearly $2m\geq 2q(q^{2}-1)>1+2(q^{2}-1)>\log_{r}(|H|_{r})$, a contradiction.

Assume next that $j=3$. Then $r\geq 5$ and
$|H|_{r}=(q^{3}-1)_{r}=(q^{2}+q+1)_{r}$, hence

$\log_{r}(|H|_{r})\leq\log_{r}((q^{2}+q+1)_{r})\leq q^{2}+q+1.$

Clearly $2m\geq 2q(q^{2}-1)>q^{2}+q+1\geq\log_{r}(|H|_{r})$. The case $j=6$ is
similar.

Step $3$. Assume that $r=p.$ By [26, Table 5.4.C], the smallest dimension of
nontrivial irreducible projective representations of ${\rm G}_{2}(q)$ over
$k=\overline{\mathbb{F}}_{p}$ is $7-\delta_{p,2}\geq 6$. By [17, Lemma 4.2],
$|N|=p^{m}\geq p^{6f}$, whence $2m\geq 12f$. Now $|H|_{p}=p^{6f}$ and thus
since $2m\geq 12f>6f$, $p^{2m}\nmid|H|$. This final contradiction completes
the proof. ∎

We now consider the simple groups $H\cong{}^{3}{\rm D}_{4}(q)$. Note that

$|{}^{3}{\rm D}_{4}(q)|=q^{12}(q^{2}-1)(q^{6}-1)(q^{8}+q^{4}+1)\text{ and }$
$q^{12}-1=(q^{4}-1)(q^{8}+q^{4}+1)\text{ and
}q^{8}+q^{4}+1=(q^{4}-q^{2}+1)(q^{2}-q+1)(q^{2}+q+1).$

######  Proposition 3.5.

Theorem 3.1 holds for $H\cong{}^{2}{\rm D}_{4}(q)$, where $q=p^{f}.$

###### Proof.

Let $G$ be a counterexample to the theorem with minimal order. Then
${\mathrm{cod}}(G)={\mathrm{cod}}(H)$ but $G\not\cong H$. By Lemma 2.1, $G$ is
perfect and $N$ is a unique minimal normal subgroup of $G$ which is an
elementary abelian $r$-group for some prime $r$. Let $|N|=r^{m}$. Then either
$G$ is quasisimple with $N=Z(G)$ or $C_{G}(N)=N$.

Step $1$. By [26, Theorem 5.1.4], $M({}^{3}{\rm D}_{4}(q))=1$. Hence
$C_{G}(N)=N$ and so by Lemma 2.3, we have $2m\leq\log_{r}(|H|_{r})$.

Step $2$. Assume $r\neq p$. By [26, Theorem 5.3.9], $m\geq q^{3}(q^{2}-1).$
Let $j=d_{q}(r)$. Then $j\in\\{1,2,3,6,12\\}$.

(2a) Assume that $r=2$. Then $q$ is odd and $j=1$. Now since $q^{8}+q^{4}+1$
is always odd, we have $|H|_{2}=(q^{2}-1)_{2}^{2}$ and so
$\log_{2}(|H|_{2})=2\log_{2}((q^{2}-1)_{2})\leq 2(q^{2}-1)$. Clearly $2m\geq
2q^{3}(q^{2}-1)>2(q^{2}-1)\geq\log_{2}(|H|_{2})$ for any $q\geq 3$, a
contradiction.

(2b) Assume $r\geq 3$ is odd.

Case $j=1,2$. Then $|H|_{r}=(q^{j}-1)_{r}^{2}(6\cdot 12)_{r}\leq
r^{2}(q^{2}-1)_{r}^{2}$ and so

$\log_{r}(|H|_{r})\leq 2+2\log_{r}((q^{2}-1)_{r})\leq 2+2(q^{2}-1).$

Clearly $2m\geq 2q^{3}(q^{2}-1)>2+2(q^{2}-1)>\log_{r}(|H|_{r})$.

Assume next that $j=3,6$. Then $r\geq 5$ and so $|H|_{r}=(q^{3}\pm
1)_{r}^{2}=(q^{2}\pm q+1)_{r}^{2}$ and so

$\log_{r}(|H|_{r})=2\log_{r}((q^{2}\pm q+1)_{r})\leq 2(q^{2}\pm q+1)\leq
2(q^{2}+q+1).$

Now $2m\geq 2q^{3}(q^{2}-1)>2(q^{2}+q+1)\geq\log_{r}(|H|_{r})$.

Assume $j=12$. Then $r\geq 13$ and
$|H|_{r}=(q^{8}+q^{4}+1)_{r}=(q^{4}-q^{2}+1)_{r}$ and so
$\log_{r}(|H|_{r})\leq q^{4}-q^{2}+1.$ Clearly $2m\geq
2q^{3}(q^{2}-1)>q^{4}-q^{2}+1>\log_{r}(|H|_{r})$.

Step $3$. Assume $r=p.$ By [26, Table 5.4.C], the smallest dimension of
nontrivial irreducible representations of ${}^{3}{\rm D}_{4}(q)$ over
$k=\overline{\mathbb{F}}_{p}$ is $8$. By [17, Lemma 4.2], $|N|=p^{m}\geq
p^{8f}$, whence $2m\geq 16f\geq$. Now $|H|_{p}=p^{12f}$ and thus since $2m\geq
16f>12f$, whence $p^{2m}\nmid|H|_{p}$. This final contradiction shows that
$G\cong H$ as wanted. The proof is now complete. ∎

For the remaining cases ${\rm E}_{6}(q),{}^{2}{\rm E}_{6}(q)$ ($q>2$) and
${\rm E}_{7}(q)$, their Schur multipliers are of order
$\gcd(3,q-1),\gcd(3,q+1)$ and $\gcd(2,q-1),$ respectively. The Schur
multiplier of ${}^{2}{\rm E}_{6}(q)$ is
${\mathbb{Z}}_{2}\times{\mathbb{Z}}_{6}$. To handle the cases when the Schur
multiplier of $H$ is nontrivial, we will need the data available on F.
Lübeck’s website [32]. For each positive integer $k$, denote by
$\Phi_{m}=\Phi_{m}(q)$ the cyclotomic polynomial of degree $m$ evaluated at
$q$.

######  Lemma 3.6.

Let $H$ be one of the simple groups ${\rm E}_{6}(q),{}^{2}{\rm E}_{6}(q)$ and
${\rm E}_{7}(q)$, where $q=p^{f}$ for some prime $p$. If $G$ is a finite
quasisimple group such that $|Z(G)|=r$ is a prime and $G/Z(G)\cong H$, then
${\mathrm{cod}}(G)\neq{\mathrm{cod}}(H).$

###### Proof.

We first consider the case when $H$ has an exceptional Schur multiplier. So,
let $H={}^{2}{\rm E}_{6}(2)$. Now
$M(H)\cong{\mathbb{Z}}_{2}\times{\mathbb{Z}}_{6}$. We only need to consider
the cases $G\cong 3.{}^{2}{\rm E}_{6}(2)$ and $G\cong 2.{}^{2}{\rm E}_{6}(2)$.
In the first case, $3.{}^{2}{\rm E}_{6}(2)$ has a faithful irreducible
character of degree $46683$ and $46683/3=15561$ is not a degree of $H$. The
smallest nontrivial degree of ${}^{2}{\rm E}_{6}(2)$ is $1938.$ Similarly,
$2.{}^{2}{\rm E}_{6}(2)$ has a faithful irreducible character of degree $2432$
and $2432/2=1216$ is not a degree of $H$ as well. Thus by Lemma 2.2,
${\mathrm{cod}}(G)\neq{\mathrm{cod}}(H)$.

Now assume that $H$ is one of the simple groups in the lemma with $q>2$ if
$H\cong{}^{2}{\rm E}_{6}(q)$. Assume also that $H$ has a nontrivial Schur
cover. Thus $|M(H)|$ is a prime which is 3 in the first two cases and $2$ in
the last case. Let $L$ be the full covering group of $H$ and let $L^{*}$ be
the dual of $L$, so $L$ is a simply connected group and $L^{*}$ is the adjoint
group. We can also assume that $H=[L^{*},L^{*}]$ and $L/Z(L)\cong H.$ Note
that $Z(L)\cong{\mathbb{Z}}_{r}$ and $|L^{*}:H|=r$.

(a) Assume that $H\cong{\rm E}_{7}(q)$. In this case, $d=\gcd(2,q-1)>1$, so
$q\geq 3$ is odd. Note that $L={\rm E}_{7}(q)_{sc}$ and $L^{*}={\rm
E}_{7}(q)_{ad}.$ By using [32], $L$ has a degree

$D=\Phi_{1}^{4}\Phi_{3}^{2}\Phi_{4}\Phi_{5}\Phi_{6}\Phi_{7}\Phi_{8}\Phi_{9}\Phi_{12}\Phi_{18}$

with multiplicity $(q-1)/2$. Now $D/2$ is not a degree of $L$, so it is not a
degree of $H$. Note that ${\mathrm{cd}}(H)\subseteq{\mathrm{cd}}(L).$ We are
done if we can show that $L$ has a faithful irreducible character of degree
$D$. Assume by contradiction that every irreducible character of $L$ of degree
$D$ is unfaithful, so they are characters of $H$. Now $L^{*}$ also have
irreducible characters of degree $D$ with multiplicity $(q-1)/2$ if
$3\nmid(q-1)$ and $(q-3)/2$ if $3\mid(q-1).$ Furthermore, $L^{*}$ has no
irreducible character of degree $2D.$ Therefore, every irreducible character
of $H$ of degree $D$ extends to $L^{*}$ but then $L^{*}$ would have at least
$2(q-1)/2=q-1$ irreducible character of degrees $D$, which is not the case.
Hence $L$ must have some faithful irreducible character of degree $D$ as
required.

The remaining two cases can be argued similarly using the degree $D$ of $L$
with the given multiplicity.

(b) Assume that $H\cong{\rm E}_{6}(q)$. Then $3\mid(q-1)$ and so $q\equiv 1,4$
(mod 6). By [32], $L$ has a degree

$D=\Phi_{3}^{2}\Phi_{6}\Phi_{9}\Phi_{12}$

with multiplicity $q-2$ and $L$ has no character of degree $D/3$. Also the
multiplicity of $D$ in $L^{*}$ is $q-4$ has $L^{*}$ has no character of degree
$3D$.

(c) Assume that $H\cong{{}^{2}}{\rm E}_{6}(q)$ with $q>2$. Then $3\mid(q+1)$
and so $q\equiv 2,5$ (mod 6). By [32], $L$ has a degree

$D=\Phi_{3}\Phi_{6}^{2}\Phi_{12}\Phi_{18}$

with multiplicity $q$ and $L$ has no character of degree $D/3$. Also the
multiplicity of $D$ in $L^{*}$ is $q-2$ has $L^{*}$ has no character of degree
$3D$. ∎

######  Proposition 3.7.

Theorem 3.1 holds for $H\cong{\rm E}_{7}(q)$, where $q=p^{f}.$

###### Proof.

Let $G$ be a counterexample to the theorem with minimal order. Then
${\mathrm{cod}}(G)={\mathrm{cod}}(H)$ but $G\not\cong H$. By Lemma 2.1, $G$ is
perfect and $N$ is a unique minimal normal subgroup of $G$ which is an
elementary abelian $r$-group for some prime $r$. Let $|N|=r^{m}$. Then either
$G$ is quasisimple with $N=Z(G)$ or $C_{G}(N)=N$. By Lemma 3.6, $G$ is not a
quasisimple group. Thus $C_{G}(N)=N$ and so by Lemma 2.3, we have
$2m\leq\log_{r}(|H|_{r})$.

Let $d=\gcd(2,q-1).$ By [26, Table 5.1.B], we have

$|H|=\frac{1}{d}q^{63}(q^{2}-1)(q^{6}-1)(q^{8}-1)(q^{10}-1)(q^{12}-1)(q^{14}-1)(q^{18}-1).$

Step $2$. Assume $r\neq p$. By [26, Theorem 5.3.9], $m\geq q^{15}(q^{2}-1).$
Let $j=d_{q}(r)$. Then

$j\in\\{1,2,3,4,5,6,7,8,9,10,12,14,18\\}.$

(2a) Assume $r=2$. Then $q$ is odd, $d=2$ and $j=1$. We have
$|H|_{2}=2^{2}(q^{2}-1)_{2}^{7}$ and so
$\log_{2}(|H|_{2})=2+7\log_{2}((q^{2}-1)_{2})\leq 7q^{2}$. We have $2m\geq
2q^{15}(q^{2}-1)>7q^{2}\geq\log_{2}(|H|_{2})$ for any $q\geq 3,$ a
contradiction.

(2b) Assume $r\geq 3$ is odd.

Case $j=1,2$. Then $|H|_{r}=(q^{j}-1)_{r}^{7}(6\cdot 10\cdot 12\cdot 14\cdot
18)_{r}\leq r^{4}(q^{2}-1)_{r}^{7}$ and so

$\log_{r}(|H|_{r})\leq 4+7\log_{r}((q^{2}-1)_{r})\leq 4+7(q^{2}-1)<7q^{2}.$

Now $2m\geq 2q^{15}(q^{2}-1)>7q^{2}>\log_{r}(|H|_{r})$, a contradiction.

Assume that $j=3,6$. Then $r\geq 5$ and so
$|H|_{r}=(q^{j}-1)_{r}^{3}=(q^{2}\pm q+1)_{r}^{3}$ and so

$\log_{r}(|H|_{r})\leq 3\log_{r}((q^{2}\pm q+1)_{r})\leq 3(q^{2}\pm q+1)\leq
3(q^{2}+q+1)<3q^{3}.$

Now $2m\geq 2q^{15}(q^{2}-1)>3q^{3}\geq\log_{r}(|H|_{r})$.

Assume $j=4$. Then $r\geq 5$ and $|H|_{r}=(q^{4}-1)_{r}^{2}=(q^{2}+1)_{r}^{2}$
and so $\log_{r}(|H|_{r})\leq 2(q^{2}+1)<4q^{2}.$ Hence $2m\geq
2q^{15}(q^{2}-1)>4q^{2}\geq\log_{r}(|H|_{r})$.

Assume $j=5,7,9$. Then $r\geq 7$ and $|H|_{r}\leq r(q^{j}-1)_{r}$. So
$\log_{r}(|H|_{r})\leq 1+\log_{r}(q^{j}-1)\leq 1+(q^{j}-1)=q^{j}\leq q^{9}.$
We have $2m\geq 2q^{15}(q^{2}-1)>q^{9}\geq\log_{r}(|H|_{r})$.

For the remaining values of $j$, $8\leq j\leq 18$ is even and so $r\geq 7$ and
$|H|_{r}\leq r(q^{j}-1)_{r}\leq r(q^{9}+1)$ and so $\log_{r}(|H|_{r})\leq
1+(q^{9}+1)=q^{9}+2<2q^{9}.$ Hence $2m\geq
2q^{15}(q^{2}-1)>2q^{9}\geq\log_{r}(|H|_{r})$.

Step $3$. Assume that $r=p.$ By [26, Table 5.4.C], the smallest dimension of
nontrivial irreducible projective representations of ${\rm E}_{7}(q)$ over
$k=\overline{\mathbb{F}}_{p}$ is $56$. By [17, Lemma 4.2], $|N|=p^{m}\geq
p^{56f}$, whence $2m\geq 112f$. Now $|H|_{p}=p^{63f}$ and thus since $2m\geq
112f>63f$, whence $p^{2m}\nmid|H|$. This final contradiction completes the
proof. ∎

######  Proposition 3.8.

Theorem 3.1 holds for $H\cong{\rm E}^{\epsilon}_{6}(q)$, where $q=p^{f}.$

###### Proof.

Let $G$ be a counterexample to the theorem with minimal order. Then
${\mathrm{cod}}(G)={\mathrm{cod}}(H)$ but $G\not\cong H$. By Lemma 2.1, $G$ is
perfect and $N$ is a unique minimal normal subgroup of $G$ which is an
elementary abelian $r$-group for some prime $r$. Let $|N|=r^{m}$. Then either
$G$ is quasisimple with $N=Z(G)$ or $C_{G}(N)=N$. By Lemma 3.6, $G$ is not a
quasisimple group. Thus $C_{G}(N)=N$ and so by Lemma 2.3, we have
$2m\leq\log_{r}(|H|_{r})$.

Let $d=\gcd(3,q-\epsilon).$ By [26, Table 5.1.B], we have

$|H|=\frac{1}{d}q^{36}(q^{2}-1)(q^{5}-\epsilon)(q^{6}-1)(q^{8}-1)(q^{9}-\epsilon)(q^{12}-1).$

Step $2$. Assume $r\neq p$. By [26, Theorem 5.3.9], $m\geq q^{9}(q^{2}-1).$
Let $j=d_{q}(r)$. Then

$j\in\\{1,2,3,4,5,6,8,9,12,10,18\\}.$

(2a) Assume $r=2$. Then $q$ is odd, and $j=1$. We have $|H|_{2}\leq
2^{4}(q^{2}-1)_{2}^{6}$ and so
$\log_{2}(|H|_{2})=4+6\log_{2}((q^{2}-1)_{2})\leq 6q^{2}$. We have $2m\geq
2q^{9}(q^{2}-1)>6q^{2}\geq\log_{2}(|H|_{2})$ for any $q\geq 3$, a
contradiction.

(2b) Assume $r\geq 3$ is odd. Case $1\leq j\leq 6$. Then
$|H|_{r}\leq(q^{j}-1)_{r}^{6}(6\cdot 10\cdot 12\cdot 18)_{r}\leq
r^{4}(q^{j}-1)_{r}^{6}$ and so $\log_{r}(|H|_{r})\leq
4+6\log_{r}((q^{j}-1)_{r})\leq 4+6(q^{j}-1)\leq 6q^{6}.$ Now $2m\geq
2q^{9}(q^{2}-1)>6q^{6}\geq\log_{r}(|H|_{r})$, a contradiction. Assume that
$j=9$. Then $r\geq 11$ and $\epsilon=+$. So $|H|_{r}=(q^{9}-1)_{r}$ and hence
$\log_{r}(|H|_{r})\leq q^{9}-1\leq q^{9}.$ Now $2m\geq
2q^{9}(q^{2}-1)>q^{9}\geq\log_{r}(|H|_{r})$. For the remaining values of $j$,
$8\leq j\leq 18$ and $j$ is even. So $r\geq 11$ and
$|H|_{r}=(q^{j}-1)_{r}=(q^{j/2}+1)_{r}$ and so $\log_{r}(|H|_{r})\leq
q^{j/2}+1\leq q^{9}+1\leq 2q^{9}.$ Hence $2m\geq
2q^{9}(q^{2}-1)>2q^{9}\geq\log_{r}(|H|_{r})$.

Step $3$. Assume that $r=p.$ By [26, Table 5.4.C], the smallest dimension of
nontrivial irreducible projective representations of ${\rm
E}^{\epsilon}_{6}(q)$ over $k=\overline{\mathbb{F}}_{p}$ is $27$. By [17,
Lemma 4.2], $|N|=p^{m}\geq p^{36f}$, whence $2m\geq 54f$. Now
$|H|_{p}=p^{36f}$ and thus since $2m\geq 54f>36f$, whence $p^{2m}\nmid|H|$.
This final contradiction completes the proof. ∎

The remaining simple exceptional groups ${}^{2}{\rm B}_{2}(q),{}^{2}{\rm
F}_{4}(q)$ with $q=2^{2n+1},n\geq 1$ and ${}^{2}{\rm G}_{2}(q)$ with
$q=3^{2n+1},n\geq 1$ have been verified elsewhere. For completeness, we
include the proofs here as well.

######  Proposition 3.9.

Theorem 3.1 holds for ${}^{2}{\rm B}_{2}(q),{}^{2}{\rm F}_{4}(q)$ and
${}^{2}{\rm G}_{2}(q)$.

###### Proof.

Let $H$ be one of the simple groups in the proposition. Let $G$ be a
counterexample to the proposition with minimal order. Then
${\mathrm{cod}}(G)={\mathrm{cod}}(H)$ but $G\not\cong H$. By Lemma 2.1, $G$ is
perfect and $N$ is a unique minimal normal subgroup of $G$ which is an
elementary abelian $r$-group for some prime $r$. Let $|N|=r^{m}$. Then either
$G$ is quasisimple with $N=Z(G)$ or $C_{G}(N)=N$. By [26, Theorem 5.1.4], the
Schur multiplier of $H$ is trivial except for ${}^{2}{\rm B}_{2}(8)$, where
its Schur multiplier is a Klein four-group
${\mathbb{Z}}_{2}\times{\mathbb{Z}}_{2}$. By GAP [13], the quasisimple group
$2.{}^{2}{\rm B}_{2}(8)$ has a faithful irreducible character of degree $40$
but $40/2=20$ is not a degree of ${}^{2}{\rm B}_{2}(8)$. By Lemma 2.2, $G$
cannot be a $2.{}^{2}{\rm B}_{2}(8)$. Thus $C_{G}(N)=N$ and so by Lemma 2.3,
we have $2m\leq\log_{r}(|H|_{r})$.

(i) Assume $H\cong{}^{2}{\rm B}_{2}(q)$ with $q=2^{2n+1},n\geq 1$. We have

$|{}^{2}{\rm
B}_{2}(q)|=q^{2}(q-1)(q^{2}+1)=q^{2}(q-1)(q-2^{n+1}+1)(q+2^{n+1}+1).$

Note that the last three factors are pairwise coprime.

Assume that $r=2$. By [26, Table 5.4.C], the minimal module for $H$ is of
dimension $4$ and so by [17, Lemma 4.2], we have $|N|\geq q^{4}$ or $m\geq
4(2n+1)$. Hence $2m\geq 8(2n+1)>2(2n+1)=\log_{2}(|H|_{2})$.

Now assume that $r\geq 3$ is odd. Assume that $n=1$. Then $H={\rm B}_{2}(8)$
and so by [26, Table 5.3.A], $m\geq 8.$ Now $|H|=2^{6}\cdot 5\cdot 7\cdot 13$
and thus $r^{2m}$ cannot divide $|H|$ for any odd prime $r$. Thus we can
assume $n\geq 2$ and so $m\geq 2^{n}(q-1)$ by [26, Table 5.3.A]. For each odd
prime $r$ dividing $|H|$, $r$ divides exactly one of the last three factors in
the order formula for $H$ above. Hence $|H|_{r}\leq q+2^{n+1}+1.$ However, we
can check that $2m\geq 2^{n+1}(q-1)>q+2^{n+1}+1\geq\log_{r}(|H|_{r})$.

(ii) Assume $H\cong{}^{2}{\rm G}_{2}(q)$ with $q=3^{2n+1},n\geq 1$. Note that

$|{}^{2}{\rm G}_{2}(q)|=q^{3}(q-1)(q^{3}+1)=q^{3}(q-1)(q+1)(q^{2}-q+1).$

Assume that $r=3$. As in the previous case, note that the minimal module for
$H$ is of dimension $7$, so $m\geq 7(2n+1)$. Hence $2m\geq
14(2n+1)>3(2n+1)=\log_{3}(|H|_{3})$.

Assume that $r\neq 3.$ By [26, Table 5.3.A], $m\geq q(q-1)$. Assume $r=2$.
Then $|H|_{2}=(q^{2}-1)_{2}=8$ since $q^{2}\equiv 9$ (mod 16). Since $q\geq
27$, it is clear that $2m\geq 2q(q-1)>3$ and thus $2^{2m}$ does not divide
$|H|$. Now assume $r>2$. Then $r\geq 5$ and thus $r$ divides exactly one of
the factors $q-1,q+1$ and $q^{2}-q+1$. Hence $|H|_{r}\leq q^{2}-q+1$. Now
$2q(q-1)-(q^{2}-q+1)=q^{2}-q-1=q(q-1)-1>0$ since $q\geq 27$. Therefore,
$2m\geq 2q(q-1)>\log_{r}(|H|_{r})$.

(iii) Assume $H\cong{}^{2}{\rm F}_{2}(q)$ with $q=2^{2n+1},n\geq 1$. So $q\geq
8.$ We have

$|{}^{2}{\rm F}_{4}(q)|=q^{12}(q-1)(q^{3}+1)(q^{4}-1)(q^{6}+1).$

Assume that $r=2$. As in the previous case, note that the minimal module for
$H$ is of dimension $26$, so $m\geq 26(2n+1)$. Hence $2m\geq
52(2n+1)>12(2n+1)=\log_{2}(|H|_{2})$, a contradiction.

Assume that $r\geq 3.$ By [26, Table 5.3.A], $m\geq 2^{n}q^{4}(q-1)$. Let
$j=d_{q}(r)$. Then $j\in\\{1,2,4,6,12\\}.$

Case $j=1,2,4$. Then $|H|_{r}\leq(q^{j}-1)_{r}^{2}(3\cdot 6)_{r}\leq
r^{2}(q^{j}-1)_{r}^{2}$ and so

$\log_{r}(|H|_{r})\leq 2+2\log_{r}((q^{j}-1)_{r})\leq 2+2(q^{2}+1)\leq
4q^{2}.$

Now $2m\geq 2^{n+1}q^{4}(q-1)>4q^{2}\geq\log_{r}(|H|_{r})$. Assume $j=6$. Then
$r\geq 7$ and $|H|_{r}\leq(q^{3}+1)_{r}=(q^{2}-q+1)_{r}$ and so
$\log_{r}(|H|_{r})\leq q^{2}-q+1\leq 2q^{2}.$ Now $2m\geq
2^{n+1}q^{4}(q-1)>2q^{2}\geq\log_{r}(|H|_{r})$. Assume $j=12$. Then $r\geq 13$
and $|H|_{r}\leq(q^{6}+1)_{r}=(q^{4}-q^{2}+1)_{r}$ and so
$\log_{r}(|H|_{r})\leq q^{4}-q^{2}+1\leq 2q^{4}.$ Now $2m\geq
2^{n+1}q^{4}(q-1)>2q^{4}\geq\log_{r}(|H|_{r})$. The proof is now complete. ∎

###### Proof of Theorem 3.1.

The theorem follows from Propositions 3.2 \- 3.9. ∎

##  4\. Projective special linear groups

In this section, we assume that $H\cong\operatorname{PSL}_{n}(q)$ with $n\geq
4$ and $q=p^{f}$, where $p$ is a prime and $f\geq 1$. Moreover, we assume that
$H$ is not isomorphic to $\operatorname{PSL}_{4}(2)$ as
$\operatorname{PSL}_{4}(2)\cong{\mathrm{A}}_{8}$. By [26, Theorem 5.1.4], the
Schur multiplier $M(H)$ is cyclic of order $d=\gcd(n,q-1)$. Then
$L=\operatorname{SL}_{n}(q)$ is the universal cover of $H$.

######  Theorem 4.1.

Let $H=\operatorname{PSL}_{n}(q)$, where $n\geq 4,f\geq 1$ and
$(n,q)\neq(4,2).$ Let $G$ be a finite group such that
${\mathrm{cod}}(G)={\mathrm{cod}}(H)$. Then $G\cong H.$

###### Proof.

Let $G$ be a counterexample to the theorem with minimal order. Then
${\mathrm{cod}}(G)={\mathrm{cod}}(H)$ but $G\not\cong H$. By Lemma 2.1, $N$ is
an elementary abelian $r$-group of order $r^{m}$ and $G$ is perfect. Hence
either $G$ is a quasisimple group with center $N$ of order $r$ or
$C_{G}(N)=N$.

Step $1$. Assume that $G$ is quasisimple with center $N\cong{\mathbb{Z}}_{r}.$
Recall that the center of $\operatorname{SL}_{n}(q)$ is cyclic of order
$d=\gcd(n,q-1)$ and that $Z(\operatorname{SL}_{n}(q))$ has a unique subgroup
of order $e$ for each divisor $e\geq 1$ of $d$. Since $G$ is a central
quotient of $\operatorname{SL}_{n}(q)$, we deduce that $r\mid d$ and
$Z(\operatorname{SL}_{n}(q))$ has a unique subgroup $A$ of order $d/r$ and
that $\operatorname{SL}_{n}(q)/A\cong G$.

If $q=2$, then $d=\gcd(n,q-1)=1$. Therefore, we assume that $q>2$ and that
$d>1$. We know that $\operatorname{SL}_{n}(q)$ has $q-1$ Weil characters
$\tau_{n,q}^{i}$ (see [40]) with $0\leq i\leq q-2$ of degree $(q^{n}-1)/(q-1)$
if $1\leq i\leq q-2$ and $(q^{n}-q)/(q-1)$ if $i=0$. Using the formula for
$\tau_{n,q}^{i}$ (see [38]), we see that the kernel of $\chi=\tau_{n,q}^{r}$
is a subgroup of $Z(\operatorname{SL}_{n}(q))$ of order $d/r$, and thus $\chi$
is a faithful irreducible character of
$\operatorname{SL}_{n}(q)/\ker(\chi)\cong G$.

Since $\chi\in\operatorname{Irr}(G)$ is faithful,
${\mathrm{cod}}(\chi)=|G|/\chi(1)=|H|/\phi(1)$ for some
$\phi\in\operatorname{Irr}(H).$ It follows that $\chi(1)=r\phi(1)$. Note that
$\phi$ is nontrivial as $\chi(1)=(q^{n}-1)/(q-1)>r.$ Assume first that
$H\cong\operatorname{PSL}_{4}(3)$. Then $G\cong\operatorname{SL}_{4}(3)\cong
2.\operatorname{PSL}_{4}(3)$ and by using GAP [13], $G$ has a faithful
irreducible character of degree $40$ but $40/2=20$ is not a degree of $H$. So,
assume $(n,q)\neq(4,3).$ By [39, Theorem 1.1], we have
$\phi(1)\geq(q^{n}-q)/(q-1)$ and hence $(q^{n}-1)/(q-1)=r\phi(1)\geq
r(q^{n}-q)/(q-1)$ which implies that $(q^{n}-1)\geq r(q^{n}-q)$ or
equivalently $(r-1)(q^{n}-q)\leq q-1$, which is impossible as $q\geq 3,r\geq
2$ and $n\geq 4$.

We now have that $C_{G}(N)=N$ is an elementary abelian $r$-group and that
$G/N\cong H$ acts irreducibly on $N$. By Lemma 2.3, $N$ is an irreducible
$\mathbb{F}_{r}H$-module and $r^{2m}\mid|H|$, hence $2m\leq\log_{r}(|H|_{r})$.

Step $2$. Assume that $r\neq p$. Since $H$ has an irreducible representation
of dimension $m$ in cross characteristic $r$, by [18, Theorem 1.1], we have
$m\geq(q^{n}-q)/(q-1)-\kappa_{n}$ if $H$ is not isomorphic to
$\operatorname{PSL}_{6}(2)$ nor $\operatorname{PSL}_{6}(3)$; for the
exception, we have $m\geq 62-\kappa_{6}$ and $m\geq 363-\kappa_{6}$,
respectively, where $\kappa_{n}=1$ if $r$ divides $(q^{n}-1)/(q-1)$ and $0$
otherwise.

Assume first that $H\cong\operatorname{PSL}_{6}(2)$. We have $r>2$ and
$|H|=2^{15}\cdot 3^{4}\cdot 5\cdot 7^{2}\cdot 31$ and clearly
$2m>\log_{r}(|H|_{r})$. Similarly, for $H\cong\operatorname{PSL}_{6}(3)$,
$r\neq 3$ and $|H|=2^{11}\cdot 3^{15}\cdot 5\cdot 7\cdot 11^{2}\cdot 13^{2}$
and $2m>\log_{r}(|H|_{r})$. Thus we can assume that

$m\geq(q^{n}-q)/(q-1)-1=(q^{n}-2q+1)/(q-1)\geq q^{n-1}-1.$

(2a) Assume $r=2$. Then $q$ is odd and $2\mid(q-1)$. Assume $q\equiv 1$ mod
$4$. By [37, Table 2] or Lemma 2.4, we get
$|H|_{2}\leq(q-1)_{2}^{n-1}(n!)_{2}\leq 2^{n}(q-1)^{n}$. Hence
$\log_{2}(|H|_{2})\leq n+n\log_{2}(q-1)\leq n+n(q-1)=nq$. Since $n\geq 4,$ we
have $q^{n-2}>2n$ for any $q\geq 3.$ Hence $2m\geq 2(q^{n-1}-1)\geq
2(q-1)q^{n-2}>4n(q-1)>nq\geq\log_{2}(|H|_{2}),$ a contradiction.

Assume $q\equiv 3$ mod $4$ and $n=2m+1.$ Then

$|H|_{2}\leq|\operatorname{GL}_{n}(q)|_{2}/(q-1)_{2}\leq(m!)_{2}4^{m}(q+1)_{2}^{m}<2^{3m}(2q)^{m}=2^{4m}q^{m}.$

Hence $\log_{2}(|H|_{2})\leq 4m+m\log_{2}(q)\leq 2n+nq\leq n(q+2)$. As in the
previous case, $2m>4n(q-1)>n(q+2)\geq\log_{2}(|H|_{2})$. The remaining case is
similar.

(2b) Assume that $r>2$. Note that $r\mid\prod_{i=1}^{n}(q^{i}-1)$. Let
$d_{q}(r)$ be the order of $q$ modulo $r$ and let $j=d_{q}(r)$. Then $j$ is
the smallest positive integer such that $r\mid q^{j}-1$. We have that
$|H|_{r}=(q^{j}-1)^{k}_{r}(k!)_{r}$, where $k=\lfloor n/r\rfloor.$ Since
$r^{2m}\mid|H|$, we deduce that $r^{2m}\leq|H|_{r}=(q^{j}-1)^{k}_{r}(k!)_{r}$.
Since $m\geq(q^{n}-q)/(q-1)-\kappa_{n}$, we have

$2(q^{n-1}-1)<2[(q^{n}-q)/(q-1)-\kappa_{n}]\leq 2m.$

(i) Assume that $j=n$. Then $n\mid r-1$ and so $r\geq n+1$. Now
$|H|_{r}=(q^{n}-1)_{r}$. Clearly, $r\nmid(q-1)$ so $r\mid(q^{n}-1)/(q-1)$. It
follows that $\log_{r}(|H|_{r})\leq|H|_{r}\leq(q^{n}-1)/(q-1)$. However, we
can see that $2(q^{n}-2q+1)/(q-1)>(q^{n}-1)/(q-1)$ if and only if
$q^{n}>4q-3$, which is true since $q\geq 2$ and $n\geq 4.$ Therefore,
$2m\geq(q^{n}-2q+1)/(q-1)>(q^{n}-1)/(q-1)\geq\log_{r}(|H|_{r})$.

(ii) Assume that $n/2<j\leq n-1.$ Then $k=\lfloor n/j\rfloor=1$ and
$|H|_{r}=(q^{j}-1)_{r}.$ With the same argument as above, we get $2m\geq
2(q^{n-1}-1)>q^{n-1}-1\geq\log_{r}(|H|_{r})$, a contradiction.

(iii) Assume $1\leq j\leq n/2$. We have

$\log_{r}(|H|_{r})\leq k\log_{r}(q^{j}-1)_{r}+\log_{r}((k!)_{r})\leq
n/j(q^{j}-1)+n/j=nq^{j}/j\leq nq^{j}\leq nq^{n/2}.$

Since $2m\geq 2(q^{n-1}-1)>q^{n-1}$, it suffices to show that
$q^{n-1}>nq^{n/2}$ or equivalently $q^{n-2}>n^{2}$ for all integers $n\geq 4$
and $q\geq 2$. If $q\geq 5$, then $q^{n-2}\geq 5^{n-2}>n^{2}$ by induction on
$n\geq 4.$ So, assume $2\leq q\leq 4.$ We have $q^{n-2}>n^{2}$ except for the
following cases: $q=2$ and $5\leq n\leq 8$; $q=3$ and $n=4$ or $q=4$ and
$n=4.$ For these cases, direct calculation shows that ${2m}>\log_{r}(|H|_{r})$
for every prime divisor $r\neq p$ of $|H|$.

Step $3$. Assume that $r=p.$ Recall that $H\cong\operatorname{PSL}_{n}(q)$,
$L\cong\operatorname{SL}_{n}(q)$ and $L/Z(L)\cong H.$ Let $F=\mathbb{F}_{p}$
and $E=\mathbb{F}_{q}$ be a Galois extension of $F$ of degree $f$, where
$q=p^{f}.$ Let $k=\overline{\mathbb{F}}_{p}$ be such that $F\subseteq
E\subseteq k.$ For a field extension $F_{1}\subseteq F_{2}$, if $U$ is an
$F_{1}L$-module, then we let $U^{F_{2}}=U\otimes_{F_{1}}F_{2}$. Then
$U^{F_{2}}$ is an $F_{2}L$-module and
$\dim_{F_{2}}(U^{F_{2}})=\dim_{F_{1}}(U).$

By [26, Proposition 5.4.4], $E$ is a splitting field for $L$ and thus by [24,
Theorem VII.2.3] every irreducible $EL$-module $U$ is absolutely irreducible
and the $kL$-module $U^{k}$ is absolutely irreducible as well. Let
$V_{1},V_{2},\dots,V_{y}$ be all irreducible $EL$-modules (up to isomorphism).
Then $V_{i}^{k}$ ($1\leq i\leq y$) are all irreducible $kL$-modules (up to
isomorphism). Conversely, let $W$ be an irreducible $kL$-module. Then there
exists an irreducible $EL$-module $W_{0}$ which is isomorphic to one of the
$V_{i}^{\prime}s,(1\leq i\leq y)$, such that $W$ is isomorphic to the
$kL$-module $W_{0}^{k}$. Now $N$ is a faithful irreducible $FH$-module and
hence it is an irreducible $FL$-module, where the center of $L$ acts trivially
on $N.$ By [24, Theorem VII.1.18(c)], $N^{E}$ is a completely reducible
$EL$-module and

$N^{E}\cong W_{1}\oplus W_{2}\cdots\oplus W_{t}$

where $W_{i}^{\prime}s$ are irreducible $EL$-modules and all $W_{i}^{\prime}s$
are conjugate under the automorphisms of the Galois group
$\operatorname{Gal}(E/F)$ and every isomorphism type appears among the $W_{i}$
with multiplicity one (see [25, Theorem 9.21]) and that $t$ divides $|E:F|=f$.

Note that $L$ is simply connected of Lie type over
$E=\mathbb{F}_{q}=\mathbb{F}_{p^{f}}$. By [26, Table 5.4.C], the smallest
dimension of nontrivial irreducible representations of $L$ over $k$ is $n.$
Following [17], we call the Frobenius twists of restricted modules of groups
of Lie type ‘restricted’. We now follow the proof of Lemma 4.2 in [17]. By
[17, Lemma 4.2], $|N|=p^{m}\geq q^{n}$ (we can also get this bound using the
existence of primitive prime divisor of $p^{nf}-1$ in $H$, which always exists
by Zsigmondy’s theorem except for the case $(n,q)=(6,2)$.) It follows that
$m\geq nf.$ By Lemma 2.3, we have $2m\leq\log_{p}(|H|_{p})=fn(n-1)/2$. Hence
$2nf\leq fn(n-1)/2$, which implies that $n\geq 5.$

Let $W$ be an irreducible constituent of $N^{E}$. Then $W$ is an absolutely
irreducible $\mathbb{F}_{p^{f}}L$-module. Let $p^{e}$ be the smallest power of
$p$ such that $W$ is an absolutely irreducible $\mathbb{F}_{p^{e}}L$-module,
which is realized over no proper subfield of $\mathbb{F}_{p^{e}}$. By [26,
Proposition 5.4.6(i)], there exists an irreducible $kL$-module $M$ such that

$W^{k}\cong M\otimes M^{(e)}\otimes\cdots\otimes M^{(f-e)}.$

In particular, $\dim_{k}(W)=\dim_{k}(M)^{f/e}$. Note that $\dim_{k}(M)\geq n$.
Hence

$|N|=p^{m}\geq|W|\geq p^{e\dim_{k}(M)^{f/e}}.$

Assume that $s:=f/e\geq 3$. Then $m\geq en^{s}.$ We now claim that $en^{s}\geq
n^{2}f$. This is equivalent to $n^{s-2}\geq s$, where $n\geq 5$ and $s\geq 3$.
This can be proved easily by induction on $s\geq 3.$ We now obtain that
$2fn^{2}\leq 2m\leq fn(n-1)/2$, which is equivalent to $4n\leq n-1$, which is
impossible.

Assume $s=2=f/e$. Assume $M$ is not restricted. Then $\dim_{k}(M)\geq n^{2}$
and hence $m\geq 2n^{2}f$. In this case, $2m\geq 4n^{2}f>fn(n-1)/2$. We now
assume that $M$ is restricted. We now have $W^{k}\cong M\otimes M^{(f/2)}$ and
the smallest field over which $W$ can be realized is $\mathbb{F}_{p^{f/2}}.$
Again, we get $m\geq en^{2}=n^{2}f/2$, whence $2m\geq
n^{2}f>n(n-1)f/2\geq\log_{r}(|H|_{r})$, a contradiction.

Assume $e=f.$ If $M$ is a tensor product of at least two nontrivial restricted
irreducible $kL$-modules, then $m\geq n^{2}f$ and we get a contradiction as in
the previous cases. We now assume that $M$ is a restricted irreducible
$kL$-module. Then $|W|=p^{f\dim_{k}(M)}$ and hence $m\geq f\dim_{k}(M)$.
Assume first that $\dim_{k}(M)>n$. By using [31] or [26, Table 5.4.A], we see
that $\dim_{k}(M)\geq n(n-1)/2$ and thus $2m\geq
fn(n-1)>fn(n-1)/2\geq\log_{r}(|H|_{r})$.

We are left with the case when $W^{k}\cong M$ is a restricted $kL$-module with
$\dim_{k}(W)=n$ and hence $M$ is quasi-equivalent to the natural module for
$L$. It follows that $|W|=q^{n}$ and hence $\dim_{E}(N^{E})=t\dim_{E}(W)=tn$
as $\dim_{E}(W_{i})=\dim_{E}(W)$ for all $i.$ Since
$m=\dim_{F}(N)=\dim_{E}(N^{E})$, we get $m=tn\leq fn,$ as $t$ divides
$|E:F|=f.$ However, we already know that $m\geq nf$, whence $t=f$ and $m=nf$.
Thus $|N|=p^{m}=p^{nf}=q^{n}$.

Step $4$. We have that $m=nf$ and $|N|=q^{n}$. Recall that if $X$ is a finite
group and $A$ is an $X$-module, let $H^{2}(X,A)$ be the second cohomology
group of $X$ with coefficients in $A$. It is well-known that $H^{2}(X,A)$ is
nonzero if and only if there is a nonsplit extension of $A$ by $X.$ From [3,
Table I], we know that since $n\geq 5,$ $H^{2}(\operatorname{SL}_{n}(q),V)$ is
trivial, where $V$ is the natural module for $\operatorname{SL}_{n}(q)$,
except for the case $H\cong\operatorname{SL}_{5}(2)$.

Assume that $H\cong\operatorname{SL}_{5}(2)$ and $N$ is not quasi-equivalent
to the natural module for $H$ so that $G$ is isomorphic to the nonsplit
extension $2^{5}.\operatorname{GL}_{5}(2)$ which is called the Dempwolff group
and appears as a maximal subgroup of the Thompson sporadic simple group Th. By
GAP [13], we can check that this group has a faithful irreducible character
$\chi$ of degree $248=2^{3}\cdot 31.$ Clearly $\chi(1)/2^{5}$ is not a
character degree of $H$ and thus this case cannot occur. By using GAP [13], we
can also assume that
$H\not\in\\{\operatorname{PSL}_{5}(2),\operatorname{PSL}_{6}(2)\\}$.

If $f=1$, then $N$ is quasi-equivalent to the natural module for
$\operatorname{SL}_{n}(p)$ and hence $\operatorname{SL}_{n}(p)$ has exactly
one nontrivial orbit on $N$ and since $N\cong\operatorname{Irr}(N)$,
$\operatorname{SL}_{n}(p)$ has exactly one nontrivial orbit on
$\operatorname{Irr}(N)$ of size $p^{n}-1$. Since $H$ is perfect and embeds
into $\operatorname{GL}_{n}(p)$, we see that $H$ is isomorphic to a subgroup
of $\operatorname{SL}_{n}(p)$. Since $\operatorname{SL}_{n}(p)$ is quasisimple
as $n\geq 5,$ and contains a subgroup isomorphic to
$\operatorname{PSL}_{n}(p)$, it forces $H\cong\operatorname{SL}_{n}(p)$. In
particular, $\gcd(n,p-1)=1$.

Therefore, $G=N:H_{0}$ with $H_{0}\cong H\cong\operatorname{SL}_{n}(q)$.
Furthermore, since $N\cong\operatorname{Irr}(N)$, $G$ has exactly one
nontrivial orbit $\mathcal{O}$ on $\operatorname{Irr}(N)$ of size $p^{n}-1.$
Let $\theta\in\mathcal{O}$ and $T=I_{G}(\theta)\cong N:T_{0},$ where $T_{0}$
is the stabilizer of a vector in the natural module $N$. By [26, Proposition
4.1.17], we deduce that $T_{0}\cong C:A$, where $C$ is an elementary abelian
$p$-group of order $p^{n-1}$ and $T_{0}\cong\operatorname{SL}_{n-1}(p)$. Since
$\theta$ is $T$-invariant and $T$ splits over $N$, by [22, Lemma 22.2],
$\theta$ extends to $\theta_{0}\in\operatorname{Irr}(T)$ and so since
$T_{0}/C\cong\operatorname{SL}_{n-1}(p)$, $T_{0}$ has an irreducible character
of degree $(p^{n-1}-p)/(p-1)$ (the degree of a unipotent character of
$\operatorname{SL}_{n-1}(p)$) and hence by Gallagher’s Theorem [25, Corollary
6.17], $T$ has an irreducible character $\phi$ lying over $\theta$ of degree
$(p^{n-1}-p)/(p-1)$. It follows that
$\chi=\phi^{G}\in\operatorname{Irr}(G|\theta)$ is faithful and hence by Lemma
2.3, $|N|=p^{n}$ divides $\chi(1)=|G:T|\phi(1)=(p^{n}-1)(p^{n-1}-p)/(p-1)$,
which is impossible as $n\geq 5.$ (We can also take
$\theta_{0}\in\operatorname{Irr}(I_{G}(\theta))$ that is an extension of
$\theta$ to $I_{G}(\theta)$ and thus
$\theta_{0}^{G}\in\operatorname{Irr}(G|\theta)$ with degree $p^{n}-1$ and get
a contradiction as well.)

Now assume that $f>1$. Note that $(nf,p)\neq(6,2)$ since $n\geq 5$ and $f\geq
2$. Thus a primitive prime divisor $p^{nf}-1$ exists and divides $|H|$. Since
$H\cong\operatorname{PSL}_{n}(p^{f})$ is a subgroup of
$\operatorname{SL}_{m}(p)=\operatorname{SL}_{nf}(p)$, by [4, Section 3] or
[16], $H$ must be isomorphic to a ${\mathcal{C}}_{3}$-subgroup of
$\operatorname{SL}_{m}(p)$ and thus $H\cong\operatorname{SL}_{k}(p^{e})$ with
$ke=m=nf$. It follows that $k=n$ and $e=f$ is the only possibility and thus
$H\cong\operatorname{SL}_{n}(q)$ which also forces $\gcd(n,q-1)=1$.
Furthermore, $H$ acts transitively on the nonzero vectors of $N$ as a natural
module for $\operatorname{SL}_{m}(p)$. (See [29, Appendix 1]). From the
construction of ${\mathcal{C}}_{3}$-subgroups of $\operatorname{SL}_{m}(p)$,
$N$ can be considered an $\mathbb{F}_{q}$-vector space of dimension $n$ and it
is in fact the natural module for $\operatorname{SL}_{n}(q)$. (We start with
the natural module $N$ for $\operatorname{SL}_{n}(q)$ of dimension $n$ over
$\mathbb{F}_{p^{f}}$ and then consider $N$ as an $\mathbb{F}_{p}$-module of
dimension $m=nf$.) We can then apply the same argument as in the case $f=1$ to
obtain a contradiction. The proof is now complete. ∎

The proof of Theorem 1 now follows from Theorems 3.1 and 4.1.

##  5\. Symplectic groups

Let $L=\operatorname{Sp}_{2n}(q)$, where $n\geq 2$ and $q=p^{f}$. If $n=2$, we
assume $q>2.$ Then $H=L/Z(L)\cong\operatorname{PSp}_{2n}(q)$ with
$d=|Z(L)|=\gcd(2,q-1)$. Let $D$ be the dimension of the natural module for
$L$, so $D=2n$, and it is also the smallest dimension of a nontrivial
absolutely irreducible $kL$-module, where $k=\overline{\mathbb{F}}_{p}$. Let
$\mathbb{F}=\mathbb{F}_{p}$ and $E=\mathbb{F}_{q}.$ We have

$|\operatorname{PSp}_{2n}(q)|=\frac{1}{d}q^{n^{2}}\prod_{i=1}^{n}(q^{2i}-1).$

Moreover, for $(n,q)\neq(3,2)$, we have $|M(H)|=d$ while
$M(\operatorname{PSp}_{6}(2))\cong{\mathbb{Z}}_{2}$ (see [26, Theorem 5.1.4]).
In this section, we will verify the first three steps for the finite simple
symplectic groups and as a consequence, confirming the Codegrees conjecture
for $n=2$ and $n=3.$

######  Theorem 5.1.

Let $H=\operatorname{PSp}_{2n}(q)$, where $n\geq 2$ and $(n,q)\neq(2,2).$ Let
$G$ be a counterexample to the Codegrees conjecture with minimal order. Let
$L=\operatorname{Sp}_{2n}(q)$. Then $n\geq 4$ and the following hold.

  1. (i)

$G$ is perfect and has a unique minimal normal subgroup $N$ which is an
elementary abelian $p$-group.

  2. (ii)

All the irreducible $kL$-submodules of $N\otimes_{\mathbb{F}_{p}}k$ is quasi-
equivalent to the natural module for $L$.

  3. (iii)

$|N|=q^{n}$.

###### Proof.

Let $G$ be a counterexample to the theorem with minimal order. Then
${\mathrm{cod}}(G)={\mathrm{cod}}(H)$ but $G\not\cong H$. By Lemma 2.1, $N$ is
an elementary abelian $r$-group of order $r^{m}$ and $G$ is perfect. Hence
either $G$ is a quasisimple group with center $N$ of order $r$ or
$C_{G}(N)=N$.

Step $1$. Assume that $G$ is quasisimple with center $N\cong{\mathbb{Z}}_{r}.$
Assume first that $H=\operatorname{PSp}_{6}(2)$. Then $G\cong
2.\operatorname{PSp}_{6}(2)$ and by using GAP [13], $G$ has a faithful
irreducible character of degree $8$ but $8/2=4$ is not a character degree of
$H$, so $G$ cannot be $2.\operatorname{PSp}_{6}(2)$ by Lemma 2.2.

Now assume that $(n,q)\neq(3,2)$. Then $G\cong\operatorname{Sp}_{2n}(q)$,
$r=2$ and $q$ is odd. By [40, Lemma 2.6], $\operatorname{Sp}_{2n}(q)$ always
has a faithful irreducible character of degree $(q^{n}-\alpha)/2$, where
$4\mid(q^{n}-\alpha)$ and $\alpha\in\\{\pm 1\\}$. Now $(q^{n}-\alpha)/4$ is
not a degree of $H$ since the minimal nontrivial degree of $H$ is at least
$(q^{n}-1)/2$ by [26, Table 5.3.A].

We now have that $C_{G}(N)=N$ is an elementary abelian $r$-group and that
$G/N\cong H$ acts irreducibly on $N$. By Lemma 2.3, $N$ is an irreducible
$\mathbb{F}_{r}H$-module and $r^{2m}$ divides $|H|$. Hence
$2m\leq\log_{r}(|H|_{r})$.

Step $2$. Assume $r\neq p$.

Assume that $H\cong\operatorname{PSp}_{6}(2)$. Then $m\geq 7$ by [26, Table
5.3.A]. We have $|\operatorname{PSp}_{6}(2)|=2^{9}\cdot 3^{4}\cdot 5\cdot 7$.
We can see that $r^{2m}$ cannot divide $|H|$ for any odd prime $r$. Therefore,
we might assume that $H\not\cong\operatorname{PSp}_{6}(2).$ By [26, Table
5.3A]), we have $m\geq(q^{n}-1)/2$ if $q$ is odd and $m\geq
q^{n-1}(q^{n-1}-1)(q-1)/2$ if $n$ is even. In both cases, we have $2m\geq
q^{n}-1.$

(2a) Assume $r=2$. So $q$ is odd and
$|H|_{2}=(q^{2}-1)_{2}^{n}(n!)_{2}/2<(q^{2}-1)_{2}^{n}2^{n-1}$ and so
$\log_{2}(|H|_{2})<n-1+n(q^{2}-1)=nq^{2}-1$. If $n\geq 3$, then $2m\geq
q^{n}-1\geq nq^{2}-1>\log_{2}(|H|_{2})$ for any $q\geq 3$ as $q^{n-2}\geq
3^{n-2}\geq n$. Assume $n=2.$ Then $|H|_{2}\leq(q^{2}-1)_{2}^{2}$ and so
$\log_{2}(|H|_{2})<2\log_{2}(q^{2}-1)$. Thus we need to show that $2m\geq
q^{2}-1>2\log_{2}(q^{2}-1)$ which is always true since $q^{2}-1\geq 8.$

(2b) Assume $r\geq 3$ is odd. Let $j:=d_{q}(r)$ be the order of $q$ modulo
$r$. Then $j$ is the smallest positive integer such that $r\mid q^{j}-1$ and
$r\geq j+1$.

(i) Assume $j$ is odd. By [37, Table 1],
$|H|_{r}=|\operatorname{GL}_{n}(q)|_{r}\leq(q^{j}-1)_{r}^{k}r^{k/(r-1)}$ and
hence $\log_{r}(|H|_{r})\leq k/(r-1)+k\log_{r}(q^{j}-1)$.

Assume $j=n$ is odd. Then $r\geq n+1$ and so $|H|_{r}=(q^{n}-1)_{r}$. Since
$r\nmid(q-1)$, we have $\log_{r}(|H|_{r})\leq(q^{n}-1)/(q-1)$. If $q>2$, then
$2m\geq q^{n}-1>(q^{n}-1)/(q-1)\geq\log_{r}(|H|_{r})$. If $q=2$, then $2m\geq
2^{n-1}(2^{n-1}-1)>2^{n}-1\geq\log_{r}(|H|_{r})$.

Assume $n-1\geq j>n/2$. Then $k=\lfloor n/j\rfloor=1$ and thus
$|H|_{r}\leq(q^{j}-1)_{r}$. Hence $\log_{r}(|H|_{r})\leq q^{j}-1\leq
q^{n-1}-1$. Now $2m\geq q^{n}-1>q^{n-1}-1\geq\log_{r}(|H|_{r})$.

Assume $1\leq j\leq n/2$. Then
$|H|_{r}\leq(q^{j}-1)^{k}_{r}(k!)_{r}\leq(q^{j}-1)_{r}^{k}r^{k/(r-1)}$ and
thus $\log_{r}(|H|_{r})<k(q^{j}-1)+k=kq^{j}\leq nq^{n/2}.$

Assume that $q\geq 3$. Then $2m\geq q^{n}-1>q^{n}/2$. Observe that
$q^{n}/2\geq nq^{n/2}$ is equivalent to $q^{n}\geq 4n^{2}$ which is always
true for any $q\geq 3$ and $n\geq 4.$ Therefore, $2m>q^{n}/2\geq
nq^{n/2}\geq\log_{r}(|H|_{r})$. Assume $n=2$. Then $j=1$ and
$\log_{r}(|H|_{r})=2\log_{r}((q-1)_{r})\leq 2(q-1)$ and so $2m\geq
q^{2}-1>2(q-1)\geq\log_{r}(|H|_{r})$. Similarly, if $n=3$, then $j=1$ and
$\log_{r}(|H|_{r})\leq 1+3\log_{r}((q-1)_{r})\leq 1+3(q-1)<3q$ and so $2m\geq
q^{3}-1>3q\geq\log_{r}(|H|_{r})$.

Assume $q=2$. Then $n\geq 4$ and $2m\geq
2^{n-1}(2^{n-1}-1)/2=2^{n-2}(2^{n-1}-1)$. Now $2^{n-2}(2^{n-1}-1)>n2^{n/2}$ if
and only $2^{(n-4)/2}(2^{n-1}-1)>n$. Since $n\geq 4,$ we can check that
$2^{n-1}>n$. Therefore, $2m\geq
2^{n-2}(2^{n-1}-1)>n2^{n/2}\geq\log_{r}(|H|_{r})$, a contradiction.

(ii) Assume $j$ is even. So $(q^{j}-1)_{r}=(q^{j_{0}}+1)_{r}$. By [37, Table
1],
$|H|_{r}=|\operatorname{GL}_{2n}(q)|_{r}\leq(q^{j}-1)_{r}^{k}(k!)_{r}\leq(q^{j}-1)^{k}_{r}r^{k}$,
where $k=\lfloor 2n/j\rfloor$ and hence $\log_{r}(|H|_{r})\leq
k+k\log_{r}(q^{j}-1)_{r}$.

(ii) (1) Assume $j=2n.$ Then $r\mid(q^{n}+1)$ and $r\geq 2n+1$ so
$|H|_{r}=(q^{2n}-1)_{r}=(q^{n}+1)_{r}$. Assume $q$ is odd. Then
$(q^{n}+1)_{r}=((q^{n}+1)/2)_{r}$ since $r$ is odd. Thus $2m\geq
q^{n}-1>(q^{n}+1)/2\geq\log_{r}(|H|_{r})$. Assume $q$ is even. If $q=2$, then
$2m\geq
2^{n-1}(2^{n-1}-1)/2=2^{n-2}(2^{n-1}-1)>(2^{n}+1)\geq\log_{r}(|H|_{r})$ since
$n\geq 4.$ Assume $q\geq 4$ is even. Then $2m\geq
q^{n-1}(q^{n-1}-1)(q-1)/2>q^{n-1}(q^{n-1}-1)>q^{n}+1\geq\log_{r}(|H|_{r})$.

(ii)(2) Assume $2n>j>n$. Then $k=1$. We have
$|H|_{r}=(q^{j}-1)_{r}=(q^{j_{0}}+1)_{r}$ and so
$\log_{r}(|H|_{r})=\log_{r}(q^{j_{0}}+1)\leq q^{n-1}+1$. Now $2m\geq
q^{n}-1>q^{n-1}-1\geq\log_{r}(|H|_{r}).$

(ii)(3) Assume $n\geq j\geq 2.$ Then $k\leq n$ and
$|H|_{r}\leq(q^{j}-1)_{r}^{k}(k!)_{r}\leq(q^{j_{0}}+1)^{k}_{r}r^{k}$ and hence
$\log_{r}(|H|_{r})\leq k+k(q^{j_{0}}+1)\leq k(q^{n/2}+2)\leq n(q^{n/2}+2)\leq
2nq^{n/2}$ as $n\geq 2$ and $q\geq 2$.

Assume first that $q$ is even. So $2m\geq q^{n-1}(q^{n-1}-1)>q^{2n-2}/2$. Now
observe that $q^{n}\geq 4n$ for any $n\geq 4$ and $q\geq 2$. Hence
$q^{2n-2}/2=q^{n}q^{n-2}/2\geq q^{n}/2\geq 2nq^{n-2}\geq 2nq^{n/2}$ and thus
$2m>\log_{r}(|H|_{r})$. Assume that $n=2$ or $3$. Then $q\geq 4$ and clearly
$q^{n}\geq 4^{n}\geq 4n$ and again $2m>\log_{r}(|H|_{r})$ as above.

Assume $q$ is odd. Assume first that $n=2$. Then $j=2$ and $2m\geq q^{2}-1$.
Now $\log_{r}(|H|_{r})=2\log_{r}((q^{2}-1)_{r})\leq 2(q+1)$. However, we see
that $2m\geq q^{2}-1>2(q+1)\geq\log_{r}(|H|_{r})$. Assume next that $n=3$.
Again, $j=2$, $2m\geq q^{3}-1$ and
$\log_{r}(|H|_{r})=1+3\log_{r}((q^{2}-1)_{r})\leq 1+3(q+1)<q^{3}\leq 2m.$
Assume that $n=4$. Then $j=2$ or $4$ and $2m\geq q^{4}-1$. If $j=2$, then
$\log_{r}(|H|_{r})\leq 1+4(q+1)<q^{4}-1\leq 2m.$ If $j=4$, then
$\log_{r}(|H|_{r})\leq 2(q^{2}+1)<q^{4}-1\leq 2m.$ Assume $n\geq 5.$ By
induction on $n\geq 5$, $q^{n}>4n^{2}$ for $q\geq 3$ and thus $q^{n/2}>2n$.
Finally $q^{n}-1>2nq^{n/2}$ is equivalent to $q^{n/2}(q^{n/2}-2n)>1$ which is
true by the previous claim. Thus $2m\geq
q^{n}-1>2nq^{n/2}\geq\log_{r}(|H|_{r})$.

Step $3$. Assume $r=p$. In this case, we can use the same argument as in Step
3 of the proof of Theorem 4.1. Let $L\cong\operatorname{Sp}_{2n}(q)$ with
center $Z(L)$ of order $\gcd(2,q-1)$ and $L/Z(L)\cong H.$ Again, let
$E=\mathbb{F}_{q}$ and $\mathbb{F}=\mathbb{F}_{p}$. Let $W$ be an irreducible
constituent of $N^{E}$ and let $p^{e}$ be the smallest power of $p$ such that
$W$ is an absolutely irreducible $\mathbb{F}_{p^{e}}L$-module, which is
realized over no proper subfield of $\mathbb{F}_{p^{e}}$. Let
$N^{k}=N\otimes_{F}k$. By [17, Lemma 4.2], $m\geq 2nf$, where $D=2n$ is the
smallest dimension of irreducible representation of
$L=\operatorname{Sp}_{2n}(q)$ over $k=\overline{\mathbb{F}}_{p}$. As
$2m\geq\log_{p}(|H|_{p})$, we get $2m\leq n^{2}f$ and thus $n^{2}f\geq 4nf$ so
$n\geq 4.$

Therefore $D=2n\geq 8$ and we can apply [17, Lemma 4.2] directly and hence
cases (i), (ii) or $(v)$ holds. If (v) holds. then $m\geq 2fD^{2}\geq
8n^{2}f$. If $(ii)$ holds, then $f$ is even and $m\geq(f/2)(2n)^{2}=2n^{2}f$.
Finally, if (i) holds and $N^{k}$ is a tensor product of at least two
restricted irreducible $kL$-modules, then $m\geq 4n^{2}f$. For these cases, we
see that $m\geq 2fn^{2}$ and so $2m\geq 4n^{2}f>n^{2}f$ and thus $p^{2m}$
cannot divide $|H|$. Therefore, we are left with the case when $N^{k}$ is a
restricted $kL$-module and $f=e$. Assume $n\geq 12$. By [31], the dimension of
a restricted $kL$-module is either $2n$ or at least $2n^{2}-n-2$. If the
latter case holds, then $2m\geq 2(2n^{2}-n-2)f\geq 2n^{2}f>n^{2}f$ and thus
$p^{2m}\nmid|H|$. Therefore, $m=2nf$ and $N^{k}$ is quasi-equivalent to the
natural module for $L$. Hence $m=2nf$ and $|N|=q^{2n}$.

We now consider the case $4\leq n\leq 11.$ For $4\leq n\leq 6$ and $q$ even,
$L$ has a spin module of dimension $2^{n}$ (see [26, Table 5.4.A]). So, if
$n\geq 7$ or $q$ is odd, then either $m=2nf$ or $m\geq(2n^{2}-n-2)f$ and as
above, the former case holds.

Assume $q$ is even and $4\leq n\leq 6$. If $m\geq 2^{n}f$, then $2m\geq
2^{n+1}f$ and we can check that $2^{n+1}>n^{2}$ for these values of $n$ and
hence $2m>n^{2}f$. Thus $m=2nf$ as wanted. ∎

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