# The codegree isomorphism problem
for finite simple groups

Nguyen N. Hung  Department of Mathematics, The University of Akron, Akron, OH
44325, USA<EMAIL_ADDRESS>and  Alexander Moretó  Departamento de
Matemáticas, Universidad de Valencia, 46100 Burjassot, Valencia, Spain
<EMAIL_ADDRESS>
###### Abstract.

We study the codegree isomorphism problem for finite simple groups. In
particular, we show that such a group is determined by the codegrees (counting
multiplicity) of its irreducible characters. The proof is uniform for all
simple groups and only depends on the classification by means of Artin-Tits’
simple order theorem.

###### Key words and phrases:

Character codegrees, isomorphism problem, finite simple groups, Huppert’s
conjecture

###### 2010 Mathematics Subject Classification:

Primary 20C15, 20C30, 20C33, 20D06.

Part of this work was done while the first author was visiting the Vietnam
Institute for Advanced Study in Mathematics (VIASM), and he thanks the VIASM
for its hospitality and financial support.

The research of the second author is supported by Ministerio de Ciencia e
Innovación (Grant PID2019-103854GB-I00 funded by MCIN/AEI/
10.13039/501100011033) and Generalitat Valenciana CIAICO/2021/163.

##  1\. Introduction

It has been known for a while that every finite (quasi)simple group is
determined by its complex group algebra, i.e., by the set of character degrees
with multiplicities [BNOT15]. (Recall that, by the Wedderburn theorem, the
multiset of character degrees counting multiplicities of $G$ determines and is
determined by its complex group algebra
$\mathbb{C}{G}\cong\bigoplus_{\chi\in\mathrm{Irr}(G)}\mathrm{Mat}_{\chi(1)}(\mathbb{C})$.
Here, $\mathrm{Irr}(G)$ is the set of all irreducible characters of $G$ and
$\mathrm{Mat}_{n}(\mathbb{C})$ is the $\mathbb{C}$-algebra of $n\times n$
matrices.) The version without multiplicity (and therefore stronger) is a
conjecture proposed by B. Huppert in the late nineties [Hup00]. This
conjecture has been extensively studied over the past two decades, and
progress has been made on a case-by-case basis using the classification of
finite simple groups [Hup06, NT15, BTZ17]. Further discussion on this and
other isomorphism problems of similar nature can be found in the recent survey
[Mar22].

This paper is concerned with _the codegree isomorphism problem_ for finite
simple groups. For a character $\chi$ of a finite group $G$, the _codegree_ of
$\chi$ is ${\mbox{cod}}(\chi):=|G:\ker(\chi)|/\chi(1)$. This notion was first
introduced and studied (in a slightly different form) by D. Chillag and M.
Herzog [CH89] and D. Chillag, A. Mann, and O. Manz [CMM91]. It was later
developed into the current form used nowadays by G. Qian, Y. Wang, and H. Wei
[QWW07] and has been proved to have remarkable connections with the structure
of finite groups [QWW07, Isa11, DL16, Mor21, Qia21, CN22].

Recently, there has been great interest in the codegree analogue of Huppert’s
conjecture (Problem 20.79 of the Kourovka Notebook [Kou20]), to which we will
refer to as _the codegree isomorphism conjecture_. From now on
${\mbox{cod}}(G)$ denotes the set of all the codegrees of irreducible
characters of $G$.

(CIC) Let $S$ be a finite nonabelian simple group and $G$ a finite group. Then
${\mbox{cod}}(G)={\mbox{cod}}(S)$ if and only if $G\cong S$.

The approach so far to this problem is more or less similar to Huppert’s
original method, and therefore, unfortunately, is still case-by-case [BAK21,
Aha22, GKL+22, GZY22, LY23].

Let ${\mbox{cod}}(G)=\\{c_{1}<c_{2}<...<c_{k}\\}$ and $m_{G}(c_{i})$ be the
number of irreducible characters of $G$ with codegree $c_{i}$. The multiset

$C(G):=\\{(c_{i},m_{G}(c_{i})):1\leq i\leq k\\}$

is called the group pseudo-algebra of $G$, which can be viewed as the codegree
counterpart of the aforementioned complex group algebra $\mathbb{C}G$. A
natural weaker version of (CIC) asks whether $G$ and $S$ must be isomorphic if
$C(G)=C(S)$. We confirm this in our first main result.

######  Theorem A.

Let $S$ be a finite simple group and $G$ a finite group. Then

$C(G)=C(S)\text{ if and only if }G\cong S.$

The main novelty of this paper is a more uniform approach to these codegree
problems with as little case-by-case analysis as possible. Our proof of
Theorem A, somewhat surprisingly, only relies on the classification via the
so-called _simple order theorem_ (also known as the Artin-Tits theorem
[Art55b, KLST90]), which states that two non-isomorphic finite simple groups
have the same order if and only if they are either $PSL_{4}(2)$ and
$PSL_{3}(4)$ or $\Omega_{2n+1}(q)$ and $PSp_{2n}(q)$ for some $n\geq 3$ and
odd $q$. This is perhaps the first time that a result of this type is proved
uniformly for all simple groups.

There are two key ingredients in the proof of Theorem A. We find it remarkable
that they admit strikingly elementary proofs. The first provides a
characterization of perfect groups in terms of codegrees, see Theorem 2.3.
(Recall that a group is perfect if it coincides with its derived subgroup.)
The second is an order-divisibility property involving character codegrees of
finite simple groups, see Theorem 3.1.

In general the set of codegrees ${\mbox{cod}}(G)$ of a finite group $G$ does
not determine its set of character degrees. However, this is not the case when
$G=S$ is simple. To see why, consider a prime divisor $p$ of $|S|$. Then $S$
has some nontrivial irreducible character of $p^{\prime}$-degree (by the
squares of the degrees equation). As a result, $S$ has some codegree divisible
by $|S|_{p}$ – the $p$-part of $|S|$, and this is the largest power of $p$
dividing any codegree of $S$. Therefore both the order $|S|$ and all the
character degrees of $S$ are indeed determined by ${\mbox{cod}}(S)$. This
argument also proves that if two simple groups have the same set of codegrees,
then they have the same order. Our next main result is a far stronger
statement but the proof requires the classification.

######  Theorem B.

Let $S$ and $H$ be finite simple groups such that
${\mbox{cod}}(S)\subseteq{\mbox{cod}}(H)$. Then $S\cong H$.

Theorem B in fact is the first step in proving the codegree isomorphism
conjecture (CIC). Let $G$ be any finite group and $H$ a simple group such that
${\mbox{cod}}(G)={\mbox{cod}}(H)$, and $N$ be a maximal normal subgroup of $G$
so that $S:=G/N$ is simple. In order to prove $G\cong H$, one would first need
to establish $S\cong H$, under the assumption
${\mbox{cod}}(S)\subseteq{\mbox{cod}}(G)={\mbox{cod}}(H)$. This is precisely
what we do in Theorem B (see Theorem 8.3).

######  Remark 1.

Using Theorem B, we will prove in a subsequent paper [HMT] that (CIC) holds
for all sporadic groups, alternating groups, groups of Lie type of low rank,
and, for the first time in the degree/codegree isomorphism problem, groups of
Lie type of arbitrary rank over a field of prime order. Furthermore, perhaps
unexpectedly, we reduce (CIC) to a problem on $p$-groups.

######  Remark 2.

The proof of Theorem B is fairly complicated and combines several techniques.
In particular, it essentially utilizes some deep results on the representation
theory of finite simple groups, including the classification of prime-power-
degree representations [MZ01, BBOO01], lower/upper bounds for the largest
degree of irreducible representations [VK85, LMT13], and the existence of
$p$-defect zero characters [Mic86, Wil88, GO96]. Along the way we prove an
effective and _explicit_ upper bound for the largest character degree $b(S)$
of an exceptional group $S$ of Lie type (Theorem 5.3). (See Lemma 5.4 for
previous related work on symmetric and alternating groups [VK85] and classical
groups [LMT13].)

######  Remark 3.

We need bounds for the largest character degree $b(S)$ in order to control the
behavior of $f(S):=|S|/b(S)$ – the smallest nontrivial codegree of $S$. While
the relevance of the smallest (or low-degree in general) characters of
(quasi/almost)simple groups is well-known in group representation theory (see
[MM21] for the latest results), the smallest codegree had not been studied
much before. This invariant arises naturally in the proof of Theorem B (see
Lemma 5.1) and measures the relative growth of $b(S)$ compared to $|S|$. Our
proof would be much simpler if one can show that $f$ is _divisibly increasing_
among nonabelian simple groups, by which we mean that, if $S$ and $H$ are
nonabelian simple of different orders such that $|S|$ divides $|H|$, then
$f(S)<f(H)$. We indeed confirm this phenomenon in many cases, particularly
when one of the two groups involved is alternating (see Propositions 7.1, 7.3,
and 8.1).

The layout of the paper is as follows. In Section 2, we prove some results on
prime character codegrees and provide a short proof of a theorem of Riese and
Schmid on prime-power codegrees. In Section 3 we discuss the order-
divisibility property and its consequences involving character codegrees of
finite simple groups. Using the results in the preceding sections, we prove
Theorem A in Section 4. Results on bounding the largest character degree are
presented in Section 5. Finally, the proof of Theorem B is carried out in
Sections 6, 7, and 8.

##  2\. Prime-power codegrees

In this section we prove some results on prime-power character codegrees.
These results show that, in contrast to character degrees, there are
significant restrictions on the structure of groups with faithful irreducible
characters of prime/prime-power codegree.

We mainly follow the notation from [Isa76] for character theory and [Atl1,
Car85] for finite simple groups. Throughout, for a positive integer $n$ and a
prime $p$, we write $n_{p}$ to denote the maximal $p$-power divisor of $n$ and
$n_{p^{\prime}}:=n/n_{p}$ to denote the maximal divisor not divisible by $p$
of $n$. Let $N\trianglelefteq\,G$ and $\theta\in\mathrm{Irr}(N)$. We write
$\mathrm{Irr}(G|\theta)$ for the set of irreducible constituents of
$\theta^{G}$ and $\mathrm{Irr}(G|N)$ for the set of irreducible characters of
$G$ whose kernels do not contain $N$. If $G$ is a group, $\pi(G)$ is the set
of primes that divide $|G|$. If $n$ is an integer, $\pi(n)$ is the set of
primes that divide $n$, and if $\mathcal{S}$ is a set of integers, then
$\pi(\mathcal{S})$ is the set of primes that divide some member of
$\mathcal{S}$. As usual,
${\mbox{cd}}(G):=\\{\chi(1):\chi\in\mathrm{Irr}(G)\\}$ is the set of all
irreducible character degrees of $G$. Other notation will be recalled or
defined when necessary.

We begin by collecting some known facts on character codegrees that we will
use without explicit mention.

######  Lemma 2.1.

Let $G$ be a finite group and $\chi\in{{\operatorname{Irr}}}(G)$. The
following hold:

  1. (i)

If $\chi$ is not the principal character, then ${\mbox{cod}}(\chi)>\chi(1)$.

  2. (ii)

If $N\trianglelefteq G$ and $N\leq\ker(\chi)$, then the codegree of $\chi$ as
a character of $G$ coincides with the codegree of $\chi$ viewed as a character
of $G/N$.

  3. (iii)

If $N\trianglelefteq\trianglelefteq G$ and
$\theta\in{{\operatorname{Irr}}}(N)$ lies under $\chi$, then
${\mbox{cod}}(\theta)$ divides ${\mbox{cod}}(\chi)$.

  4. (iv)

$\pi(G)=\pi({\mbox{cod}}(G))$.

  5. (v)

If $G$ is abelian, then ${\mbox{cod}}(G)=o(G)$, where $o(G)$ is the set of
orders of the elements of $G$.

###### Proof.

Part (i) is [DL16, Lem. 2.1]. Parts (ii) and (iii) are contained in [QWW07,
Lem. 2.1], and part (iv) is [QWW07, Lem. 2.4]. Now, we prove part (v). The
inclusion $o(G)\subseteq{\mbox{cod}}(G)$ follows from [DL16, Lem. 2.2].
Conversely, if $d\in{\mbox{cod}}(G)$ there exists
$\chi\in{{\operatorname{Irr}}}(G)$ such that $d=|G:\ker(\chi)|$ (note that
since $G$ is abelian, $\chi$ is linear). Since $G/\ker(\chi)$ is cyclic, we
conclude that $G$ has elements of order $d$. ∎

###  2.1. Prime codegrees: characterizing perfect groups

The goal in this subsection is to provide a characterization of perfect groups
in terms of the absence of prime codegrees.

######  Theorem 2.2.

Let $G$ be a finite group. Suppose that there exists
$\chi\in{{\operatorname{Irr}}}(G)$ faithful such that ${\mbox{cod}}(\chi)=p$
is a prime number. Then $G$ is either the cyclic group of order $p$ or a
Frobenius group with Frobenius kernel of order $p$.

###### Proof.

We argue by induction on $|G|$. Let $N\trianglelefteq G$ be minimal such that
there exists $\theta\in{{\operatorname{Irr}}}(N)$ lying under $\chi$ with
${\mbox{cod}}(\theta)=p$. Then

$\frac{|G|}{\chi(1)}=p={\mbox{cod}}(\theta)=\frac{[N:\ker(\theta)]}{\theta(1)},$

and we deduce that

$p={\mbox{cod}}(\chi)>\chi(1)=[G:N]|\ker(\theta)|\theta(1).$

In particular, $p$ does not divide any of the three factors in the right hand
side.

Suppose first that $N<G$. By the inductive hypothesis, $N/\ker(\theta)$ is
cyclic of order $p$ or a Frobenius group with Frobenius kernel
$K/\ker(\theta)$ of order $p$. In the latter case, if
$\lambda\in{{\operatorname{Irr}}}(K/\ker(\theta))$ lies under $\theta$ then
${\mbox{cod}}(\lambda)=p$. This contradicts the choice of $N$. (Note that $K$
is normal in $G$ because $K$ is characteristic in $N$.)

Hence, we may assume that $N/\ker(\theta)$ is cyclic of order $p$. By
Clifford’s theorem the faithful character $\chi_{N}$ is a sum of
$G$-conjugates of $\theta$. Let $T$ be a complete set of representatives in
$G$ for these conjugates. By [Isa76, Lem. 2.21], the intersection of
$\ker(\theta^{g})$, where $g$ runs over $T$, is trivial. We conclude that $N$
embeds into the direct product

$\prod_{g\in T}N/\ker(\theta^{g}).$

Each of the direct factors has order $p$, and so $N$ is an elementary abelian
$p$-group. Since $p$ does not divide $|\ker(\theta)|$, we conclude that $N$ is
cyclic of order $p$. As $\theta$ is linear and $\chi(1)=|G|/p$, we now have
$\chi=\theta^{G}$. It follows that $G$ is a Frobenius group with kernel $N$,
as desired.

Now, we consider the case $N=G$. Let $M$ be a maximal normal subgroup of $G$.
Since $N=G$ the codegree of any irreducible character of $N$ lying under
$\chi$ is $1$. This means that $\chi_{M}=\chi(1)\mathbf{1}_{M}$. But $\chi$ is
faithful, so we deduce that $M=1$ and $G$ is simple. If $G$ is abelian, then
it is the cyclic group of order $p$. If $G$ is not abelian, then
$|G|/p=\chi(1)<\sqrt{|G|}$ and it follows that $|G|<p^{2}$. By Sylow’s
theorems, it follows that $G$ has a normal Sylow $p$-subgroup. This
contradiction completes the proof. ∎

The following consequence of Theorem 2.2 is already mentioned in the
introduction.

######  Theorem 2.3.

A finite nontrivial group $G$ is perfect if and only if $G$ does not have any
prime character codegree.

###### Proof.

By Lemma 2.2, if $G$ has an irreducible character $\chi$ of prime codegree
then $G/\ker(\chi)$ is solvable. In particular, $G$ is not perfect.
Conversely, if $G$ is not perfect, then the abelian group $G/G^{\prime}$ has
some irreducible character of prime codegree. ∎

###  2.2. Prime power codegrees: the Riese-Schmid theorem

Chillag and Herzog proved in [CH89, Thm. 1] that a simple group does not
possess nontrivial irreducible characters of prime power codegree. The proof
relied on a case by case analysis of the simple groups, using the fact that,
most of the times, they have $p$-blocks of defect zero. This was generalized
by Riese and Schmid in [RS98, Cor. 3] to quasisimple groups, using also block
theory and the classification. We offer a short proof of this result that only
depends on an easy consequence of the classification, which is due to W.
Kimmerle, R. Lyons, R. Sandling, and D. N. Teague [KLST90, Thm. 3.6]:

######  Lemma 2.4.

For every finite simple group $S$ and prime $p$, $|S|<(|S|_{p^{\prime}})^{2}$.

The following is a restatement of [RS98, Lem. 1] in the language of codegrees.

######  Lemma 2.5.

Let $p$ be a prime. Let $G$ be a finite group and
$\chi\in{{\operatorname{Irr}}}(G)$ faithful. Then ${\mbox{cod}}(\chi)$ is a
power of $p$ if and only if $\chi$ is induced from a Sylow $p$-subgroup of
$G$.

######  Theorem 2.6.

A quasisimple group $G$ does not possess nonprincipal characters of prime
power codegree.

###### Proof.

Suppose that there exists $\mathbf{1}_{G}\neq\chi\in{{\operatorname{Irr}}}(G)$
of $p$-power codegree. Let $K:=\ker(\chi)$ and note that
$K\leq{\mathbf{Z}}(G)$. By Lemma 2.5, $\chi$ is induced from a Sylow
$p$-subgroup of $G/K$. Therefore,

$\chi(1)\geq|G/K|_{p^{\prime}}.$

By Lemma 2.4, we know that

$|G/{\mathbf{Z}}(G)|<(|G/{\mathbf{Z}}(G)|_{p^{\prime}})^{2}\leq(|G/K|_{p^{\prime}})^{2}.$

Hence, by [Isa76, Cor. 2.30],

$\chi(1)\leq|G:{\mathbf{Z}}(G)|^{1/2}<|G/K|_{p^{\prime}},$

which violates the inequality above. ∎

The next result is a restatement in terms of character codegrees of Theorem B
of [RS98]. The proof in [RS98] uses Brauer’s first and third main theorems.
Recall that if a group $G$ has trivial $p^{\prime}$-core
${\mathbf{O}}_{p^{\prime}}(G)$, then it is defined to be $p$-constrained if
the $p$-core ${\mathbf{O}}_{p}(G)$ contains its centralizer.

######  Theorem 2.7 (Riese-Schmid).

Let $G$ be a finite group and let $p$ be a prime. Suppose that
$\chi\in{{\operatorname{Irr}}}(G)$ is faithful of $p$-power codegree. Then
${\mathbf{O}}_{p^{\prime}}(G)=1$ and $G$ is $p$-constrained.

###### Proof.

By Theorem 2.6, we know that $G$ is not simple. Let $N$ be a minimal normal
subgroup of $G$ and let $\theta\in{{\operatorname{Irr}}}(N)$ lying under
$\chi$. Since $\theta$ has $p$-power codegree (by Lemma 2.1(iii)) and
${\mbox{cod}}(\theta)>1$ (note that since $\chi$ is faithful,
$\theta\neq\mathbf{1}_{N}$), we deduce that $N$ is either an elementary
abelian $p$-subgroup or a direct product of nonabelian simple groups of order
divisible by $p$. In particular, ${\mathbf{O}}_{p^{\prime}}(G)=1$.

We claim that $N$ is an elementary abelian $p$-group. Suppose that
$N=S_{1}\times\cdots\times S_{t}$, with $S_{i}\cong S$ for some nonabelian
simple group $S$ of order divisible by $p$. We wish to reach a contradiction.
Since $\theta\neq\mathbf{1}_{N}$, there exists a nonprincipal
$\psi\in{{\operatorname{Irr}}}(S_{i})$ lying under $\theta$ for some $i$. Note
that ${\mbox{cod}}(\psi)$ is a power of $p$ and this contradicts Theorem 2.6.
The claim follows.

Write $P:={\mathbf{O}}_{p}(G)$ and $C:={\mathbf{C}}_{G}(P)$. Note that $C\cap
P={\mathbf{Z}}(P)$ and ${\mathbf{O}}_{p^{\prime}}(C)=1$ (because
${\mathbf{O}}_{p^{\prime}}(G)=1$). We want to see that $C\leq P$. Assume not.
Take $K$ subnormal in $G$ such that ${\mathbf{Z}}(P)\leq K\leq C$ and
$K/{\mathbf{Z}}(P)$ is simple. Since ${\mathbf{O}}_{p^{\prime}}(C)=1$ and $G$
does not have nonabelian minimal normal subgroups, we conclude that
$K^{\prime}$ is quasisimple.

Now, take $\gamma\in{{\operatorname{Irr}}}(K^{\prime})$ lying under $\chi$.
Again, we have that $\gamma$ is not principal and ${\mbox{cod}}(\gamma)$ is a
$p$-power. This contradicts Theorem 2.6. ∎

We end this section with a variation of Theorem 2.6.

######  Theorem 2.8.

Let $G$ be a finite group. Suppose that $p$ is a prime and
$\chi\in{{\operatorname{Irr}}}(G)$ is faithful of $p$-power codegree. Then
${\mbox{cod}}(\chi)$ exceeds the $p$-part of the product of the orders of the
nonabelian composition factors in a composition series of $G$. In particular,
if $K/L$ is a non-abelian chief factor of $G$, then
${\mbox{cod}}(\chi)>|K/L|_{p}$.

###### Proof.

Let $n$ be the product of the orders of the non-abelian composition factors in
a composition series of $G$. Using again that $\chi$ is induced from a Sylow
$p$-subgroup and [KLST90], we have

${\mbox{cod}}(\chi)>\chi(1)\geq|G|_{p^{\prime}}\geq n_{p^{\prime}}>n_{p},$

as wanted. ∎

##  3\. An order-divisibility result for codegrees

The following order-divisibility result will be crucial in the proofs of our
main theorems.

######  Theorem 3.1.

Suppose that $S$ is a finite simple group and $G$ a finite group such that
${\mbox{cod}}(S)\subseteq{\mbox{cod}}(G)$. Then $|S|$ divides $|G|$.

###### Proof.

Let $d_{1},...,d_{k}$ be all the degrees of nontrivial irreducible characters
of $S$, and let $m_{i}$ ($1\leq i\leq k$) be the number of those characters of
degree $d_{i}$. By the assumption, for each $i$, there exists
$\chi_{i}\in\mathrm{Irr}(G)$ such that

$\frac{|S|}{d_{i}}=\frac{[G:\ker(\chi_{i})]}{\chi_{i}(1)}=\frac{|G|}{\chi_{i}(1)|\ker(\chi_{i})|}.$

It follows that

$\sum_{i=1}^{k}\frac{m_{i}d_{i}^{2}}{|S|^{2}}=\sum_{i=1}^{k}\frac{m_{i}\chi_{i}(1)^{2}|\ker(\chi_{i})|^{2}}{|G|^{2}},$

and thus

$\frac{\sum_{i=1}^{k}m_{i}\chi_{i}(1)^{2}|\ker(\chi_{i})|^{2}}{|G|^{2}}=\frac{\sum_{i=1}^{k}m_{i}d_{i}^{2}}{|S|^{2}}=\frac{|S|-1}{|S|^{2}}.$

Therefore $|S|^{2}$ divides $|G|^{2}(|S|-1)$, and the theorem follows. ∎

We record some consequences of Theorem 3.1 that will be needed in subsequent
sections.

######  Corollary 3.2.

Suppose that $S$ and $H$ are finite simple groups such that
${\mbox{cod}}(S)={\mbox{cod}}(H)$. Then $|S|=|H|$.

###### Proof.

This directly follows from Theorem 3.1. (Or, alternatively, from the comment
in the paragraph that precedes the statement of Theorem B.) ∎

######  Lemma 3.3.

Suppose that $S$ and $H$ are finite nonabelian simple groups such that
${\mbox{cod}}(S)\subseteq{\mbox{cod}}(H)$. Let $x:=|H|/|S|$. Then
$x\in\mathbb{N}$ and $dx\in{\mbox{cd}}(H)$ for every $1\neq
d\in{\mbox{cd}}(S)$.

###### Proof.

We know that $x\in\mathbb{N}$ by Theorem 3.1. For each $1\neq
d\in{\mbox{cd}}(S)$, we have $|S|/d\in{\mbox{cod}}(H)$, and thus there exists
some $\chi\in\mathrm{Irr}(H)$ such that $|S|/d=|H|/\chi(1)$, implying that
$\chi(1)=dx$, as claimed. ∎

######  Lemma 3.4.

Let $S$ and $H$ be finite simple groups of Lie type. Suppose that the defining
characteristic of $H$ is $p$ and ${\mbox{cod}}(S)\subseteq{\mbox{cod}}(H)$.
Then $|S|_{p^{\prime}}=|H|_{p^{\prime}}$ and there exists
$\chi\in\mathrm{Irr}(S)$ such that $\chi(1)=|S|_{p}$.

###### Proof.

We first observe that $|S|$ is divisible by $p$ because otherwise every
codegree of $S$ is not divisible by $p$ but the only nontrivial codegree of
$H$ not divisible by $p$ is $|H|_{p^{\prime}}=|H|/\mathrm{St}_{H}(1)$.

By [Mic86, Wil88], $S$ has an irreducible character, say $\chi$, of $p$-defect
$0$, so that ${\mbox{cod}}(\chi)$ is coprime to $p$. Therefore we have

$|S|/\chi(1)=|H|_{p^{\prime}}.$

It follows from Theorem 3.1 that $\chi(1)|H|_{p^{\prime}}=|S|$ divides $|H|$,
implying that $|S|_{p^{\prime}}=|H|_{p^{\prime}}$ and $\chi(1)=|S|_{p}$. ∎

Remark that when $p\geq 5$, the above result holds for all $S$. This is
because, in such case, irreducible $p$-defect zero characters still exist in
alternating groups (by [GO96]) and in sporadic simple groups [Atl1].

######  Lemma 3.5.

Let $S$ be a finite simple group of Lie type and $H$ a finite nonabelian
simple group. Suppose that ${\mbox{cod}}(S)\subseteq{\mbox{cod}}(H)$ and there
are primes $p\neq r$ such that $H$ has a unique character degree divisible by
each $|H|_{p}$ and $|H|_{r}$. Then $|S|=|H|$ and
${\mbox{cd}}(S)\subseteq{\mbox{cd}}(H)$.

###### Proof.

Repeating the arguments in the proof of Lemma 3.4, we have
$|S|_{p^{\prime}}=|H|_{p^{\prime}}$ and $|S|_{r^{\prime}}=|H|_{r^{\prime}}$,
implying that $|S|=|H|$ and ${\mbox{cd}}(S)\subseteq{\mbox{cd}}(H)$. ∎

##  4\. Group pseudo-algebras of simple groups: Theorem A

In this section we prove Theorem A, using the results in the preceding
sections.

Theorem 3.1 is useful in proving results concerning codegrees of finite simple
groups. One of them is the next theorem, whose proof makes use of the simple
order theorem. Recall that the simple order theorem asserts that two non-
isomorphic finite simple groups have the same order if and only if they are
either $PSL_{4}(2)$ and $PSL_{3}(4)$ (of order $20,160$) or $\Omega_{2n+1}(q)$
and $PSp_{2n}(q)$ (odd-dimensional orthogonal and symplectic groups, of order
$(1/2)q^{n^{2}}\prod_{i=1}^{n}(q^{2i}-1)$) for some $n\geq 3$ and odd $q$ (see
[KLST90, Thm. 5.1] for instance). It was proved by E. Artin [Art55a, Art55b]
in the fifties for known families of simple groups at the time, and completed
by J. Tits for the remaining families discovered later on (see [KLST90]).
Artin’s method is to consider certain invariants associated to (orders of)
simple groups that can be computed explicitly and are able to distinguish the
groups easily. Therefore, the simple order theorem currently relies on the
classification of finite simple groups.

######  Theorem 4.1.

Suppose that $S$ and $H$ are finite simple groups such that
${\mbox{cod}}(S)={\mbox{cod}}(H)$. Then $S\cong H$.

###### Proof.

The statement is trivial when both of $S$ and $H$ are abelian. If one of the
two groups is nonabelian, then, by Corollary 2.3, so is the other. So assume
that both $S$ and $H$ are nonabelian. By Theorem 3.2, we have $|S|=|H|$ and
hence it follows that ${\mbox{cd}}(S)={\mbox{cd}}(H)$. Assume to the contrary
that $S$ is not isomorphic to $H$.

By the simple order theorem, we have

$\\{S,H\\}=\\{PSL_{4}(2),PSL_{3}(4)\\}$

or

$\\{S,H\\}=\\{\Omega_{2n+1}(q),PSp_{2n}(q)\\}$

for some odd prime power $q=p^{\ell}$ and $n\geq 3$. The former case is
eliminated using [Atl1], so we just need to show that
${\mbox{cd}}(\Omega_{2n+1}(q))\neq{\mbox{cd}}(PSp_{2n}(q))$ for indicated $n$
and $q$.

By Lusztig’s classification of ordinary characters of finite groups of Lie
type (see [DM91, Chapter 13]), irreducible characters of $G:=Sp_{2n}(q)$ are
parameterized by pairs $((s),\psi)$ where $(s)$ is the conjugacy class of a
semisimple element $s\in G^{*}:=SO_{2n+1}(q)$ and $\psi$ is a unipotent
character of the centralizer $C:=\mathbf{C}_{G^{\ast}}(s)$. Moreover, the
degree of the character $\chi_{((s),\psi)}$ associated to $((s),\psi)$ is

$\chi_{((s),\psi)}(1)=[G^{\ast}:C]_{p^{\prime}}\psi(1).$

Let $\alpha\in\\{\pm 1\\}$ such that $4\mid(q^{m}-\alpha)$ and consider a
semisimple element $s\in G^{*}$ with spectrum $\\{1,-1,...,-1\\}$ such that
$C\cong GO_{2n}^{\alpha}$ (see [Ng10, Lem. 2.2]). Such $s$ will then belong to
$\Omega_{2n+1}(q)=[G^{*},G^{*}]$, implying that the semisimple character
$\chi_{((s),\mathbf{1}_{C})}$ associated to the pair $((s),\mathbf{1}_{C})$ is
trivial on $\mathbf{Z}(Sp_{2n}(q))$, by [NT13, Lem. 4.4]. We therefore have an
irreducible character of $PSp_{2n}(q)$ of degree

$\chi_{(s)}(1)=(|SO_{2n+1}(q)|/|GO_{2n}^{\alpha}|)_{p^{\prime}}=(q^{n}+\alpha)/2.$

To see that ${\mbox{cd}}(PSp_{2n}(q))\neq{\mbox{cd}}(\Omega_{2n+1}(q))$ (for
odd $q$ and $n\geq 3$), it is enough to show that $(q^{n}+\alpha)/2$ is not
character degree of $\Omega_{2n+1}(q)$. By [TZ96, Thm. 6.1], under our
assumptions on $n$ and $q$ and the additional condition $(n,q)\neq(3,3)$, the
minimal (nontrivial) irreducible character of $Spin_{2n+1}(q)$ has degree

$d(Spin_{2n+1}(q))=\left\\{\begin{array}[]{ll}(q^{2n}-1)/(q^{2}-1)&\mathrm{if}\
q\geq 5\\\ (q^{n}-1)(q^{n}-q)/2(q+1)&\mathrm{if}\ q=3,\end{array}\right.$

(For the definition of classical groups of various isogeny types, including
the odd-dimensional spin groups, we refer the reader to [Car85, p. 40].) Note
that $Spin_{2n+1}(q)$ is a central extension of $\Omega_{2n+1}(q)$ and so
every character degree of $\Omega_{2n+1}(q)$ is one of $Spin_{2n+1}(q)$. It is
now easy to check that $d(Spin_{2n+1}(q))>(q^{n}+\alpha)/2$ for $n\geq 3$. For
the remaining case $(n,q)=(3,3)$, we note that $d(Spin_{2n+1}(q))=27$, which
is still greater than $(q^{n}+\alpha)/2=13$. ∎

Certainly, one has to do much more work to relax the hypothesis in Theorem 4.1
to ${\mbox{cod}}(S)\subseteq{\mbox{cod}}(H)$; that is, to obtain Theorem B.

######  Lemma 4.2.

Let $n\geq 3$ and $q$ be an odd prime power. Then
${\mbox{cd}}(\Omega_{2n+1}(q))\nsubseteq{\mbox{cd}}(PSp_{2n}(q))$ and
${\mbox{cd}}(PSp_{2n}(q))\nsubseteq{\mbox{cd}}(\Omega_{2n+1}(q))$.

###### Proof.

We have seen in the proof of Theorem 4.1 that $PSp_{2n}(q)$ possesses an
irreducible character of degree $(q^{n}+\alpha)/2$, where $\alpha\in\\{\pm
1\\}$ such that $4\mid(q^{n}-\alpha)$, and furthermore
$(q^{n}+\alpha)/2\notin{\mbox{cd}}(\Omega_{2n+1}(q))$. Therefore, it suffices
to show $cd(\Omega_{2n+1}(q))\nsubseteq{\mbox{cd}}(PSp_{2n}(q))$.

We claim that both

$(q^{2n}-1)/(q^{2}-1)\text{ and }q(q^{2n}-1)/(q^{2}-1)$

are elements of ${\mbox{cd}}(\Omega_{2n+1}(q))$ for $q$ odd. Let
$G:=Spin_{2n+1}(q)$, the universal cover of $\Omega_{2n+1}(q)$. The dual group
$G^{\ast}$ of $G$ (in the sense of [DM91, Def. 13.10]) is the projective
conformal symplectic group $PCSp_{2n}(q)$, which is the quotient of
$\widetilde{G}=CSp_{2n}(q)$ by its center ${\mathbf{Z}}(\widetilde{G})\simeq
C_{q-1}$. Consider a semisimple element $s\in\widetilde{G}$ with spectrum
$Spec(s)=\\{-1,-1,1,...,1\\}$ and

$\mathbf{C}_{\widetilde{G}}(s)\cong(Sp_{2}(q)\times Sp_{2n-2}(q))\cdot
C_{q-1}$

(see [Ng10, Lem. 2.4]). Let $s^{\ast}$ be the image of $s$ under the natural
homomorphism from $\widetilde{G}$ to $G^{\ast}$. It is easy to see that, by
the choice of $s$, $\mathbf{C}_{\widetilde{G}}(s)$ is the complete inverse
image of $\mathbf{C}_{G^{\ast}}(s^{\ast})$ under this homomorphism, and thus
$\mathbf{C}_{G^{\ast}}(s^{\ast})=\mathbf{C}_{\widetilde{G}}(s)/\mathbf{Z}(\widetilde{G})$
and

$[G^{\ast}:\mathbf{C}_{G^{\ast}}(s^{\ast})]_{p^{\prime}}=[\widetilde{G}:\mathbf{C}_{\widetilde{G}}(s)]_{p^{\prime}}=\frac{|Sp_{2n}(q)|_{p^{\prime}}}{|Sp_{2}(q)|_{p^{\prime}}|Sp_{2n-2}(q)|_{p^{\prime}}}=\frac{q^{2n}-1}{q^{2}-1},$

where $p$ is the defining characteristic of $G$.

Consider the canonical homomorphism $f:Sp_{2}(q)\times
Sp_{2n-2}(q)\hookrightarrow\mathbf{C}_{\widetilde{G}}(s)\rightarrow\mathbf{C}_{G^{\ast}}(s^{\ast})$.
Using [DM91, Prop. 13.20], we know that unipotent characters of
$Sp_{2}(q)\times Sp_{2n-2}(q)$ are of the form $\theta\circ f$ where $\theta$
runs over the unipotent characters of $\mathbf{C}_{G^{\ast}}(s^{\ast})$. In
particular, as $Sp_{2}(q)\cong SL_{2}(q)$ has unipotent characters of degrees
1 and $q$, $\mathbf{C}_{G^{\ast}}(s^{\ast})$ has two unipotent characters of
degree $1$ and $q$ as well. By the conclusion of the previous paragraph, the
Lusztig series $\mathcal{E}(G,(s^{\ast}))$ of $G$ associated to the conjugacy
class $(s^{\ast})$ of $G^{\ast}$ contains two characters of degrees
$(q^{2n}-1)/(q^{2}-1)$ and $q(q^{2n}-1)/(q^{2}-1)$.

Note that $s^{\ast}\in PSp_{2n}(q)=[G^{\ast},G^{\ast}]$ and
$|\mathbf{Z}(G)|=|G^{\ast}/(G^{\ast})^{\prime}|=2$, and therefore every
character in the Lusztig series $\mathcal{E}(G,(s^{\ast}))$ restricts
trivially to $\mathbf{Z}(G)$ (see [NT13, Lem. 4.4]), and so can be viewed as a
character of $G/\mathbf{Z}(G)=\Omega_{2n+1}(q)$. The claim is completely
proved.

Suppose first that $q>3$ and assume to the contrary that
${\mbox{cd}}(\Omega_{2n+1}(q))\subseteq{\mbox{cd}}(PSp_{2n}(q))$. Then we have
$(q^{2n}-1)/(q^{2}-1)\in{\mbox{cd}}(PSp_{2n}(q))$, so that
$(q^{2n}-1)/(q^{2}-1)\in{\mbox{cd}}(Sp_{2n}(q))$. Let
$\chi\in\mathrm{Irr}(Sp_{2n}(q))$ such that $\chi(1)=(q^{2n}-1)/(q^{2}-1)$.
Now, as $q>3$, we have $(q^{2n}-1)/(q^{2}-1)<(q^{2n}-1)/2(q+1)$, and
therefore, by the classification of irreducible characters of $Sp_{2n}(q)$ of
degrees up to $(q^{2n}-1)2(q+1)$ ([TZ96, Thm. 5.2]), we deduce that $\chi$
must be one of the Weyl characters of degree $(q^{n}\pm 1)/2$ or the minimal
unipotent one of degree $(q^{n}-1)(q^{n}-q)/2(q+1)$. A simple check reveals
that none of these degrees matches the degree of $\chi$.

Now we suppose $q=3$. (In such case, $(q^{2n}-1)/(q^{2}-1)$ is indeed a
character degree of $PSp_{2n}(q)$.) As the case of $\Omega_{7}(3)$ and
$PSp_{6}(3)$ can be checked directly, we suppose furthermore that $n\geq 4$.
We then have $q(q^{2n}-1)/(q^{2}-1)<(q^{2n}-1)(q^{n-1}-q)/2(q^{2}-1)$.
Examining the degrees up to $(q^{2n}-1)(q^{n-1}-q)/2(q^{2}-1)$ of irreducible
characters of $Sp_{2n}(q)$ available in [Ng10, Cor. 4.2], we observe that none
of them is equal to $q(q^{2n}-1)/(q^{2}-1)$. We have shown that
$q(q^{2n}-1)/(q^{2}-1)\notin{\mbox{cd}}(Sp_{2n}(q))$, and therefore, by the
above claim,
${\mbox{cd}}(\Omega_{2n+1}(q))\nsubseteq{\mbox{cd}}(PSp_{2n}(q))$, as desired.
∎

We can now prove our first main Theorem A, which in fact follows from the
following slightly stronger result. If $G$ and $H$ are groups we say that
$C(G)\subseteq C(H)$ if ${\mbox{cod}}(G)\subseteq{\mbox{cod}}(H)$ and
$m_{G}(c)\leq m_{H}(c)$ for every $c\in{\mbox{cod}}(G)$.

######  Theorem 4.3.

Let $H$ be a finite simple group and $G$ a nontrivial finite group such that
$C(G)\subseteq C(H)$. Then $G\cong H$.

###### Proof.

Suppose first that $H$ is abelian of prime order $p$. Then
${\mbox{cod}}(G)\subseteq{\mbox{cod}}(H)=\\{1,p\\}$. Therefore, by Lemma 2.4
of [DL16], $G$ is an elementary abelian $p$-group and since $k(G)\leq p$, we
conclude that $G$ is cyclic of order $p$, as wanted.

So we may assume that $H$ is nonabelian. By Corollary 2.3 and the assumption
$C(G)\subseteq C(H)$, we have that $G$ is perfect. Let $N\trianglelefteq\,G$
such that $S:=G/N$ is nonabelian simple. Now
${\mbox{cod}}(S)\subseteq{\mbox{cod}}(G)\subseteq{\mbox{cod}}(H)$ and
therefore, by Theorem 3.1, we have $|S|$ divides $|H|$.

Note that $C(S)\subseteq C(G)\subseteq C(H)$. Therefore there exists a subset
$I\subseteq\mathrm{Irr}(H)\backslash\\{\mathbf{1}_{H}\\}$ and a bijection
$f:\mathrm{Irr}(S)\backslash\\{\mathbf{1}_{S}\\}\rightarrow I$ such that

$\frac{|S|}{\chi(1)}=\frac{|H|}{f(\chi)(1)}$

for every $\chi\in\mathrm{Irr}(S)\backslash\\{\mathbf{1}_{S}\\}$. It follows
that

$\sum_{\chi\in\mathrm{Irr}(S)\backslash\\{\mathbf{1}_{S}\\}}\frac{\chi(1)^{2}}{|S|^{2}}=\sum_{\psi\in
I}\frac{\psi(1)^{2}}{|H|^{2}},$

and thus

$\frac{|S|-1}{|S|^{2}}\leq\frac{|H|-1}{|H|^{2}}.$

As the function $(x-1)/x^{2}$ is decreasing on $[2,\infty)$, we deduce that
$|S|\geq|H|$.

The conclusions of the last two paragraphs show that $|S|=|H|$. If $S\cong H$
then $C(G/N)=C(S)=C(H)\supseteq C(G)$ and so $G/N$ and $G$ have the same
number of conjugacy classes, which is possible only when $N=1$, and we are
done.

So assume by contradiction that $S\ncong H$. Using again the simple order
theorem, we have $\\{S,H\\}=\\{PSL_{4}(2),PSL_{3}(4)\\}$ or
$\\{S,H\\}=\\{\Omega_{2n+1}(q),PSp_{2n}(q)\\}$ for some $n\geq 3$ and odd $q$.
For the former pair, using the character tables of both $PSL_{4}(2)$ and
$PSL_{3}(4)$ available in [Atl1], one observes that
$7\in{\mbox{cd}}(PSL_{4}(2))\backslash{\mbox{cd}}(PSL_{3}(4))$ and
$63\in{\mbox{cd}}(PSL_{3}(4))\backslash{\mbox{cd}}(PSL_{4}(2))$, implying that
none of ${\mbox{cod}}(PSL_{3}(4))$ and ${\mbox{cod}}(PSL_{2}(4))$ contains the
other, and this violates the fact that
${\mbox{cod}}(S)\subseteq{\mbox{cod}}(H)$. The latter pair was already handled
in Lemma 4.2. ∎

##  5\. The largest character degree of finite simple groups

Let $b(G)$ denote the largest degree of an irreducible character of a finite
group $G$. Recall from the Introduction that if $S$ is a simple group, then
$f(S):=|S|/b(S)$ is the smallest nontrivial character codegree of $S$. The
following elementary fact explains the relevance of $f(S)$, and therefore
$b(S)$.

######  Lemma 5.1.

Let $S$ and $H$ be finite simple groups such that
${\mbox{cod}}(S)\subseteq{\mbox{cod}}(H)$. Then $f(S)\geq f(H)$.

###### Proof.

The hypothesis implies that $f(S)\in{\mbox{cod}}(S)\subseteq{\mbox{cod}}(H)$.
Since $f(H)$ is the smallest nontrivial member of ${\mbox{cod}}(H)$, it
follows that $f(S)\geq f(H)$. ∎

Under the hypothesis of Lemma 5.1, we showed in Theorem 3.1 that $|S|$ divides
$|H|$. We will see in later sections that, in many cases, the two conditions
$f(S)\geq f(H)$ and $|S|$ divides $|H|$ are enough to force $S\cong H$, as
stated in Theorem B.

Browsing through character tables of small-order simple groups in [Atl1], one
notices that if $|H|$ is a multiple of $|S|$ and $|H|>|S|$, then $b(H)>b(S)$.
However, it seems that the largest character degree grows slower than the
order – that is, $f(H)>f(S)$. This is not so easy to prove generally, but we
do confirm it in several cases, particularly when either $S$ or $H$ is an
alternating group (see Sections 7 and 8).

We shall need effective (both lower and upper) bounds for the largest degree
of an irreducible character of simple groups. For symmetric groups, asymptotic
and explicit bounds were obtained by A. M. Vershik and S. V. Kerov in [VK85]
which can be used to derive the corresponding bounds for alternating groups.
For a group $S$ of Lie type in characteristic $p$, an obvious (and in fact
very tight!) lower bound for $b(S)$ is the degree
${\mathsf{St}}_{S}(1)=|S|_{p}$ of the Steinberg character ${\mathsf{St}}_{S}$.
When $S$ is of classical type, explicit upper bounds have been worked out by
M. Larsen, G. Malle, and P. H. Tiep in [LMT13]. Unfortunately, upper bounds
for exceptional groups achieved in [LMT13] are only asymptotic and its proof
does not allow one to obtain an explicit bound. We obtain Theorem 5.3 below
that we believe will be useful in other applications.

######  Lemma 5.2.

Let ${\mathbf{G}}$ be a simple algebraic group over the algebraic closure of a
finite field of order $q$ in characteristic $p$,
$F:{\mathbf{G}}\rightarrow{\mathbf{G}}$ a Steinberg endomorphism, and
$G:={\mathbf{G}}^{F}$ be the corresponding finite group of Lie type. Let $r$
be the rank of ${\mathbf{G}}$. Then

$(q-1)^{r}\cdot|G|_{p}\leq|G|_{p^{\prime}}\leq q^{r}\cdot|G|_{p}.$

###### Proof.

Note that finite groups ${\mathbf{G}}^{F}$ of the same isogeny type have the
same order, so we may work with ${\mathbf{G}}$ being of simply-connected type.
The inequalities are then straightforward to verify using the order formulas
for finite groups of Lie type available in [Atl1, p. xvi]. ∎

######  Theorem 5.3.

Let $S$ be a simple exceptional group of Lie type defined over a field of
order $q$ in characteristic $p$. Then the following hold:

  1. (i)

$b(S)<256|S|_{p}$.

  2. (ii)

If $q>2$, then $b(S)<26|S|_{p}$.

###### Proof.

The Tits group can be verified easily, so we assume that
$S\neq{}^{2}F_{4}(2)^{\prime}$. We then may find a simple algebraic group
$\mathbf{G}$ of adjoint type and a Steinberg endomorphism
$F:\mathbf{G}\rightarrow\mathbf{G}$ such that $S=[G,G]$ where
$G:=\mathbf{G}^{F}$ (see [MT11, Prop. 24.21]). Clearly it suffices to show
that $b(G)<256{\mathsf{St}}_{G}(1)$.

Let $(\mathbf{G}^{\ast},F^{\ast})$ be the dual pair of $(\mathbf{G},F)$, so
$\mathbf{G}^{\ast}$ will be the corresponding simple algebraic group of simply
connected type, and set $G^{\ast}:=({\mathbf{G}}^{\ast})^{F^{\ast}}$. As
mentioned before, Lusztig’s classification on complex characters of finite
groups of Lie type implies that the set of irreducible complex characters of
$G$ is partitioned into Lusztig series $\mathcal{E}(G,(s))$ associated to
various conjugacy classes $(s)$ of semisimple elements of $G^{\ast}$.
Furthermore, there is a bijection $\chi\mapsto\psi$ from $\mathcal{E}(G,(s))$
to $\mathcal{E}({\mathbf{C}}_{G^{\ast}}(s),(1))$ such that

(1)
$\chi(1)=\frac{|G|_{p^{\prime}}}{|{\mathbf{C}}_{G^{\ast}}(s)|_{p^{\prime}}}\psi(1).$

The detailed structure of centralizers of semisimple elements in a finite
exceptional groups of Lie type was determined by Carter [Car78], Deriziotis
[Der77], and Deriziotis-Liebeck [DL85]. A well-known result of Steinberg
states that the centralizer ${\mathbf{C}}_{{\mathbf{G}}^{\ast}}(s)$ of a
semisimple element $s$ is a connected reductive subgroup of maximal rank in
${\mathbf{G}}^{\ast}$. Such connected subgroup has a decomposition
${\mathbf{C}}_{{\mathbf{G}}^{\ast}}(s)=\mathbf{S}\mathbf{T}$, where
$\mathbf{S}$ is a semisimple subgroup, $\mathbf{T}$ is a central torus,
$\mathbf{S}\cap\mathbf{T}$ is finite, and
$|({\mathbf{C}}_{G^{\ast}}(s))^{F^{\ast}}|=|\mathbf{S}^{\ast}||\mathbf{T}^{\ast}|$
(see [DL85, p. 48]). When $s$ is in $G^{\ast}$, the centralizer
${\mathbf{C}}_{{\mathbf{G}}^{\ast}}(s)$ is $F^{\ast}$-stable and
${\mathbf{C}}_{G^{\ast}}(s)=({\mathbf{C}}_{{\mathbf{G}}^{\ast}}(s))^{F^{\ast}}$;
and so

$|{\mathbf{C}}_{G^{\ast}}(s)|=|S||T|$

where $S:=\mathbf{S}^{F^{\ast}}$ and $T:=\mathbf{T}^{F^{\ast}}$. Let $r$ be
the semisimple rank of ${\mathbf{G}}^{\ast}$ and $q$ (that will be a power of
$p$) the absolute value of all eigenvalues of $F$ on the character group of an
$F$-stable maximal torus of ${\mathbf{G}}$. Possible values for $|S|$ and
$|T|$ are available in [Der77, DL85]. In particular, we have

$|S|=\prod_{i}|L_{r_{i}}(q^{a_{i}})|$

and

$|T|=\prod_{j}\Phi_{j}(q),$

where $L_{r_{i}}(q^{a_{i}})$s are finite groups of Lie type (of simply-
connected type) of rank $r_{i}$ defined over a field of order $q^{a_{i}}$ and
$\Phi_{j}(q)$s are cyclotomic polynomials (and also polynomials of the forms
$q^{2}\pm\sqrt{2}q+1$, $q^{2}\pm\sqrt{3}q+1$, or
$q^{4}\pm\sqrt{2}q^{3}+q^{2}\pm\sqrt{2}q+1$ for Suzuki and Ree groups)
evaluated at $q$. As ${\mathbf{C}}_{{\mathbf{G}}^{\ast}}(s)$ has maximal rank,
we furthermore have

(2) $\sum_{i}a_{i}r_{i}+\sum_{j}\deg(\Phi_{j})=r.$

Now formula (1) implies that the typical degree of an irreducible character of
$G$ is of the form

$\chi(1)=\frac{|G|_{p^{\prime}}}{\prod_{i}|L_{r_{i}}(q^{a_{i}})|_{p^{\prime}}\prod_{j}\Phi_{j}(q)}\psi(1),$

where $\psi\in\mathcal{E}({\mathbf{C}}_{G^{\ast}}(s),(1))$, a unipotent
character of ${\mathbf{C}}_{G^{\ast}}(s)$. By [LMT13, Thm. 3.1], for any
finite group of Lie type $\mathbf{G}^{F}$, where ${\mathbf{G}}$ is a simple
algebraic group in characteristic $p$ and $F$ a Steinberg endomorphism on
${\mathbf{G}}$, the degree ${\mathsf{St}}(1)=|\mathbf{G}^{F}|_{p}$ of the
Steinberg character ${\mathsf{St}}$ of $\mathbf{G}^{F}$ is strictly larger
than the degree of any other unipotent character. Therefore, the degrees of
unipotent characters of ${\mathbf{C}}_{G^{\ast}}(s)$, which are in fact the
same as those of the semisimple group $S$, are bounded by
$\prod_{j}|L_{r_{i}}(q^{a_{i}})|_{p}$. It follows that

$b(G)\leq\frac{|G|_{p^{\prime}}}{\prod_{i}|L_{r_{i}}(q^{a_{i}})|_{p^{\prime}}\prod_{j}\Phi_{j}(q)}\prod_{j}|L_{r_{i}}(q^{a_{i}})|_{p},$

By Lemma 5.2,

$\frac{|L_{r_{i}}(q^{a_{i}})|_{p}}{|L_{r_{i}}(q^{a_{i}})|_{p^{\prime}}}\leq\frac{1}{(q^{a_{i}}-1)^{r_{i}}}\leq\frac{1}{(q-1)^{a_{i}r_{i}}}.$

Also, it is easy to see that

$\Phi_{j}(q)\geq(q-1)^{\deg\Phi_{j}}.$

We therefore deduce that

$b(G)\leq\frac{|G|_{p^{\prime}}}{(q-1)^{\sum_{i}a_{i}r_{i}+\sum_{j}\deg(\Phi_{j})}}=\frac{|G|_{p^{\prime}}}{(q-1)^{r}}.$

On the other hand, we have $|G|_{p^{\prime}}\leq|G|_{p}q^{r}$ by again Lemma
5.2, and it follows that

$\frac{b(G)}{|G|_{p}}\leq\frac{q^{r}}{(q-1)^{r}}.$

As the rank $r$ is at most $8$ for exceptional groups, the desired
inequalities follow. ∎

Bounds for $b(S)$ of alternating groups and classical groups are collected in
the following.

######  Lemma 5.4.

Let $S$ be a finite simple group, $n$ a positive integer, and $q$ a prime
power.

  1. (i)

For $S=\textup{{A}}_{n}$ with $n\geq 5$,

$\frac{1}{2}e^{-1.28255\sqrt{n}}\sqrt{n!}\leq b(S)\leq
e^{-0.11565\sqrt{n}}\sqrt{n!}.$

In particular,

$b(S)>\frac{1}{2}e^{-1.28255\sqrt{n}}(2\pi n)^{1/4}(n/e)^{n/2}$

  2. (ii)

For $S=\textup{{A}}_{n}$ with $n\geq 5$,

$b(\textup{{A}}_{n+1})\geq\frac{2(n+1)}{\sqrt{8n+1}+3}b(\textup{{A}}_{n}).$

  3. (iii)

For $S=PSL_{n}(q)$ with $n\geq 2$,

${b(S)}<13(1+\log_{q}(n+1))^{2.54}{\mathsf{St}}_{S}(1).$

  4. (iv)

For $S=PSU_{n}(q)$ with $n\geq 3$,

${b(S)}<2(2+\log_{q}(n+1)^{1.27}{\mathsf{St}}_{S}(1).$

  5. (v)

For $S\in\\{\Omega_{2n+1}(q),PSp_{2n}(q),P\Omega_{2n}^{\pm}(q)\\}$ with $n\geq
2$ and $q$ odd,

${b(S)}<38(1+\log_{q}(2n+1))^{1.27}{\mathsf{St}}_{S}(1).$

  6. (vi)

For $S\in\\{\Omega_{2n+1}(q),PSp_{2n}(q),P\Omega_{2n}^{\pm}(q)\\}$ with $n\geq
2$ and $q$ even,

${b(S)}<8(1+\log_{q}(2n+1))^{1.27}{\mathsf{St}}_{S}(1).$

###### Proof.

Part (i) follows from [VK85, Thm. 1] and Parts (iii)-(vi) follow from [LMT13,
Thm. 5.1, 5.2, and 5.3]. Let us prove Part (ii).

Let $\chi\in\mathrm{Irr}(\textup{{S}}_{n})$ such that
$\chi(1)=b(\textup{{S}}_{n})$. Let $\lambda$ be the partition of $n$
corresponding to $\chi$ and $Y_{\lambda}$ be the Young diagram associated to
$\lambda$. By the well-known branching rule, the induction
$\chi^{\textup{{S}}_{n+1}}$ of $\chi$ from $\textup{{S}}_{n}$ to
$\textup{{S}}_{n+1}$ is the sum of irreducible characters corresponding to the
partitions of $n+1$ whose associated Young diagrams are obtained from
$Y_{\lambda}$ by adding a suitable node. The number of those suitable nodes is
at most $(\sqrt{8n+1}+1)/2$ (see [HHN16, p. 1950]) and at most one of the
resulting Young diagrams is symmetric. We deduce that

$(n+1)b(\textup{{A}}_{n})\leq(n+1)\chi(1)=\chi^{\textup{{S}}_{n+1}}(1)\leq\frac{\sqrt{8n+1}+3}{2}b(\textup{{A}}_{n+1}),$

and the result follows. ∎

##  6\. Theorem B: Groups of Lie type

In this section we prove Theorem B when the groups involved are of Lie type.

In the following, for simplicity, we say that two groups have the same
defining characteristic if there is a common characteristic over which the
groups can be defined.

######  Proposition 6.1.

Let $S$ and $H$ be finite simple groups of Lie type. Suppose that
${\mbox{cod}}(S)\subseteq{\mbox{cod}}(H)$. Then $S$ and $H$ have the same
defining characteristic.

###### Proof.

Suppose that the defining characteristic of $H$ is $p$. By Lemma 3.4,
$|S|_{p^{\prime}}=|H|_{p^{\prime}}$ and there is $\chi\in\mathrm{Irr}(S)$ such
that $\chi(1)=|S|_{p}$. By Lemma 3.3, it follows that

(3) $d\cdot\frac{|H|_{p}}{\chi(1)}\in{\mbox{cd}}(H)\text{ for every
}d\in{\mbox{cd}}(S).$

Certainly if $\chi$ is the Steinberg character of $S$ then we are done. So we
assume otherwise and aim to find a contradiction or end up with a case where
$H$ can be defined in another characteristic not equal to $p$. By the
classification of prime-power-degree representations of quasi-simple groups
[MZ01, Thm. 1.1], we arrive at the following possibilities of $S$ and
$\chi(1)$.

(i) $S=PSL_{2}(q)$, $\chi(1)\in\\{q\pm 1\\}$ or $q$ is odd and
$\chi(1)\in\\{(q\pm 1)/2\\}$. Observe that $\chi(1)$ cannot be $(q\pm 1)/2$
because otherwise, by taking $d=2\chi(1)$, we would have
$2|H|_{p}\in{\mbox{cd}}(H)$, which is impossible. So $\chi(1)=q+\alpha=p^{x}$
for some $\alpha\in\\{\pm 1\\}$ and $x\in\mathbb{N}$. Suppose first that
$q=2^{t}$ for some $t\geq 2$. Then $2^{t}+\alpha=p^{x}$. By Mihailescu’s
theorem [Mih04] (previously known as Catalan’s conjecture), either $x=1$ so
that $2^{t}+\alpha$ is a (Mersenne or Fermat) prime or $\alpha=1$ and $t=3$.
In the latter case, $p=3$ and $|H|_{3^{\prime}}=|S|_{3^{\prime}}=56$, forcing
$H$ to be ${}^{2}G_{2}(3)^{\prime}$, which turns out to be isomorphic to
$S=PSL_{2}(8)$, as desired. So it remains to consider the former case:
$q+\alpha=2^{t}+\alpha=p$ is the defining characteristic of $H$. Now
$|H|_{p^{\prime}}=|S|_{p^{\prime}}=q(q-\alpha)$. Let $p^{a}$ be the order of
the underlying field of $H$. It is clear from the order formulas of simple
groups of Lie type (see [Atl1, p. xvi]) that
$|H|_{p}<|H|_{p^{\prime}}/(p^{a}-1)\leq|H|_{p^{\prime}}/(p-1)$. We therefore
deduce that

$|H|_{p}<\frac{q(q-\alpha)}{q+\alpha-1}.$

Thus we must have $\alpha=-1$ and $|H|_{p}=p$. Now $H$ is a simple group of
Lie type in characteristic $p$ such that $|H|=p(p+1)(p+2)$. This is impossible
as for such a group $H$, one can check from the order formula that
$|H|_{p^{\prime}}<(|H|_{p})^{2}$.

Next we suppose $q\geq 5$ is odd. Then $p=2$ and $q+\alpha=|S|_{2}$. Again by
Mihailescu’s theorem, either $q$ is a prime or $\alpha=-1$ and $q=9$. The case
$q=9$ is eliminated in the same way as before. So assume that
$|H|_{2^{\prime}}=q(q-\alpha)/2$ and $q$ is a prime. Note that when $H$ is not
of type $A_{1}$, every prime divisor of $|H|_{p^{\prime}}$ is smaller than
$\sqrt{|H|_{p^{\prime}}}$. Therefore our group $H$ must be $PSL_{2}(q_{1})$
for some $2$-power $q_{1}$, implying that $q(q-\alpha)/2=q_{1}^{2}-1$. This,
however, returns no relevant solutions.

(ii) For the remaining possibilities of $S$ and $\chi$, the character $\chi$
has a decent small degree and we are able to produce an irreducible character
of $S$ whose degree is a proper multiple of $\chi(1)$. Condition (3) then
implies that a proper multiple of $|H|_{p}$ is a character degree of $H$,
which is impossible. This required character turns out to be chosen as a
unipotent character in most cases. We refer the reader to [Car85, §13.8] for
the description of unipotent characters of finite classical groups.

The next possibility of $S$ and $\chi(1)$ is $S=PSL_{n}(q)$, $q>2$, $n$ is an
odd prime, $(n,q-1)=1$, and $\chi(1)=(q^{n}-1)/(q-1)$. If $n=3$ then
$SL_{3}(q)$ has an irreducible character of degree $q^{3}-1$ (see [SF73]),
which is a proper multiple of $\chi(1)=q^{2}+q+1$. For $n\geq 5$, the
unipotent character parameterized by the partition $(2,n-2)$ with degree

$\chi^{(2,n-2)}(1)=\frac{(q^{n}-1)(q^{n-1}-q^{2})}{(q-1)(q^{2}-1)}$

fulfills our requirement.

Another possibility is $S=PSU_{n}(q)$, $n$ is an odd prime, $(n,q+1)=1$, and
$\chi(1)=(q^{n}+1)/(q+1)$. This case is handled similarly as in the previous
one. Here we note that, when $n\geq 5$ is odd, the unipotent character
parameterized by the partition $(2,n-2)$ has degree

$\chi^{(2,n-2)}(1)=\frac{(q^{n}+1)(q^{n-1}-q^{2})}{(q+1)(q^{2}-1)}.$

The next case is $S=PSp_{2n}(q)$, $n\geq 2$, $q=\ell^{t}$ with $\ell$ an odd
prime, $tn$ is a $2$-power, and $\chi(1)=(q^{n}+1)/2$. Now the unipotent
character parameterized by the symbol ${0\hskip 6.0pt1\choose n}$ has required
degree

$\chi^{{0\hskip 6.0pt1\choose n}}(1)=\frac{(q^{n}+1)(q^{n}+q)}{2(q+1)}.$

The last possibility involving a family of groups is $S=PSp_{2n}(3)$, $n\geq
3$ is a prime, and $\chi(1)=(3^{n}-1)/2$. Then $S$ has a unipotent character
with degree

$\chi^{{0\hskip 6.0pt1\hskip
6.0ptn\choose-}}(1)=\frac{(3^{n}-1)(3^{n}-3)}{8}.$

(iii) $(S,\chi(1))\in\\{(Sp_{6}(2),7)$, $(Sp_{6}(2),27)$,
$({}^{2}F_{4}(2)^{\prime},27)$, $(G_{2}(2)^{\prime},7)$,
$(G_{2}(2)^{\prime},27)$, $(G_{2}(3),64)\\}$. First assume that
$S=G_{2}(2)^{\prime}$ and $\chi(1)=27$. Then $p=3$ and
$|H|_{3^{\prime}}=|S|_{3^{\prime}}=224$. It is easy to check that there is no
such simple group $S$ of Lie type in characteristic 3 with $27\mid|S|_{3}$. In
all other cases one can find a character $\psi\in\mathrm{Irr}(S)$ such that
$\chi(1)\mid\psi(1)$ but $\chi(1)<\psi(1)$. Therefore
$\psi(1){|H|_{p}}/\chi(1)$ is a proper multiple of $|H|_{p}$ and thus cannot
be a character degree of $H$, violating condition (3). ∎

As seen in Lemma 3.4 and Proposition 6.1, we face the situation where two
simple groups $S$ and $H$ of Lie type have the same characteristics $p$ and
$|S|_{p^{\prime}}=|H|_{p^{\prime}}$. It is surprising to us that this turns
out to happen only when $|S|=|H|$ (see Proposition 6.3 below), and therefore
they are among the coincidences appeared in the simple order theorem.

We slightly modify two of the invariants in Artin’s proof of the simple order
theorem (for classical groups) to prove our results.

######  Definition 6.2.

Let $S$ be a finite group of Lie type in characteristic $p$. Let
$\omega=\omega(S)$ and $\psi=\psi(S)$ respectively denote the largest and the
second largest orders of $p$ modulo a prime divisor of $|S|_{p^{\prime}}$. We
will refer to $\omega(S)$ and $\psi(S)$ as the Artin invariants of $S$.

In fact, when $p$ is the dominant prime of $|S|$, these $\omega(S)$ and
$\psi(S)$ coincide with Artin’s invariants defined in [Art55b]. We remark that
there are only a few cases involving Mersenne and Fermat primes where $p$ is
not dominant in $|S|$, and they are listed in [KLST90, Thm. 3.3].

Assume for now that $S$ is not one of $G_{2}(2)^{\prime}$,
${}^{2}G_{2}(3)^{\prime}$, and ${}^{2}F_{4}(2)^{\prime}$. (Note that
$S_{1}=G_{2}(2)^{\prime}\cong PSU_{3}(9)$ and
$S_{2}={}^{2}G_{2}(3)^{\prime}\cong PSL_{2}(8)$ and, even though we do not
allow $S_{1}$ (or $S_{2}$) to be viewed as a Lie type group over a field of
order 2 (or 3), we do allow it to be viewed as one over the field of order 9
(or 8).) Let $q=p^{t}$ be the order of the underlying field of $S$. It is
well-known that the order $|S|$ then has the standard cyclotomic factorization
in terms of $q$ as

$|S|=\frac{1}{d}q^{k}\prod_{i}\Phi_{i}(q)^{e_{i}},$

where $d$, $k$, $e_{i}$s depend on $S$ and can be found in [Atl1, Table 6] and
[KLST90, Tables C1, C2, and C3] for instance, and $\Phi_{i}(q)$ is the value
of the $i$th cyclotomic polynomial evaluated at $q$. Replacing $q$ by $p^{n}$
and factorizing each $\Phi_{i}(x^{t})$ further into cyclotomic polynomials of
$x$, one has

$|S|=\frac{1}{d}p^{kt}\prod_{i}\Phi_{i}(p)^{f_{i}}$

for certain positive integers $f_{i}$s depending on $S$.

Using Zsigmondy’s theorem, it is not difficult to see that, except for some
‘small’ cases, the invariants $\omega(S)$ and $\psi(S)$ are precisely the
largest and second largest, respectively, index $i$ such that $\Phi_{i}(p)$
appears in the cyclotomic factorization of $|S|$ (see [KLST90, Lem. 4.6]). We
refer the reader to [KLST90, Tables A1, A2 and A3] for the list of exceptions
and the values of their Artin’s invariants, including the groups
$G_{2}(2)^{\prime}$, ${}^{2}G_{2}(3)^{\prime}$, and ${}^{2}F_{4}(2)^{\prime}$
we excluded earlier. We reproduce in Table 1 the values of $\omega(S)$ and
$\psi(S)$ for the generic case only.

Table 1. $\omega(S)$ and $\psi(S)$ for simple groups of Lie type: generic case. | $S$
---
($q=p^{t}$)
| Conditions
---
($p$ a Mersenne prime)
$\omega(S)$ | $\psi(S)$
$PSL_{n}(q),n\geq 2$ |  |  $(n,q)\neq(2,2^{6}),(3,2^{2}),(3,2^{3})$,
---
$(4,2^{2}),(6,2),(7,2),(2,p^{2}),(3,p)$
$nt$ | $(n-1)t$
$PSU_{4}(q)$ |  | $6t$ | $4t$
$PSU_{n}(q),n\geq 3$ odd | $(n,q)\neq(3,2^{3}),(5,2),(3,p)$ | $2nt$ | $2(n-2)t$
$PSU_{n}(q),n\geq 6$ even | $(n,q)\neq(6,2)$ | $2(n+1)t$ | $2(n-1)t$
$\Omega_{2n+1}(q),n\geq 2$ | $(n,q)\neq(2,2^{8}),(3,2),(4,2),(2,p)$ | $2nt$ | $2(n-1)t$
$PSp_{2n}(q),n\geq 3$ |  | $2nt$ | $2(n-1)t$
$P\Omega^{+}_{2n}(q),n\geq 4$ | $(n,q)\neq(4,2),(5,2)$ | $2(n-1)t$ | $2(n-2)t$
$P\Omega^{-}_{2n}(q),n\geq 4$ | $(n,q)\neq(4,2)$ | $2nt$ | $2(n-1)t$
${}^{2}B_{2}(2^{t}),t\geq 3$ odd | $t\equiv 3(\bmod 6)$ | $4t$ | $4t/3$
${}^{2}B_{2}(2^{t}),t\geq 3$ odd | $t\equiv\pm 1(\bmod 6)$ | $4t$ | $t$
$G_{2}(q),q\geq 3$ | $q\neq 4$ | $6t$ | $3t$
${}^{2}G_{2}(3^{t}),t\geq 3$ odd |  | $6t$ | $2t$
${}^{3}D_{4}(q)$ | $q\neq 2$ | $12t$ | $6t$
$F_{4}(q)$ |  | $12t$ | $8t$
${}^{2}F_{4}(2^{t}),t\geq 3$ odd |  | $12t$ | $6t$
$E_{6}(q)$ |  | $12t$ | $9t$
${}^{2}E_{6}(q)$ |  | $18t$ | $12t$
$E_{7}(q)$ |  | $18t$ | $14t$
$E_{8}(q)$ |  | $30t$ | $24t$

######  Proposition 6.3.

Let $p$ be a prime. Suppose that $S$ and $H$ are two non-isomorphic simple
groups of Lie type in characteristic $p$ and
$|S|_{p^{\prime}}=|H|_{p^{\prime}}$. Then
$\\{S,H\\}=\\{PSL_{4}(2),PSL_{3}(4)\\}$ or
$\\{S,H\\}=\\{\Omega_{2n+1}(q),PSp_{2n}(q)\\}$ for some $n\geq 3$ and odd $q$.
In particular, $|S|=|H|$.

###### Proof.

By the assumptions, we have $\omega(S)=\omega(H)$ and $\psi(S)=\psi(H)$. First
we consider the case where both $S$ and $H$ are generic so that their
invariants $\omega$ and $\psi$ are available in Table 1. Comparing those
values, we can find all the collections of groups with equal values of
$\omega$ and $\psi$. We list these collections in Table 2 (each row in the
table is one such collection). Now one simply compare the $p^{\prime}$-parts
of orders of groups in each collection. It turns out that the only pair with
the same $p^{\prime}$-parts of orders is $\\{\Omega_{2n+1}(q),PSp_{2n}(q)\\}$
with some $n\geq 3$ and odd $q$.

Assume now that at least one of the two groups, say $S$, is non-generic. That
is, $S$ is among the exceptions listed in the second column of Table 1. The
values of the invariants $\omega$ and $\psi$ of these groups are available in
[KLST90, Tables A2 and A3]. The analysis is basically the same as in the
generic case, but more tedious. We first find all the possible groups $H$ with
$\omega(S)=\omega(H)$ and $\psi(S)=\psi(H)$, and then compare
$|S|_{p^{\prime}}$ and $|H|_{p^{\prime}}$, where $p$ is the defining
characteristic of $S$ and $H$.

Let us demonstrate the case $S=PSL_{3}(4)$ as an example. Then $\omega(S)=4$
and $\psi(S)=3$. But there are only two other simple groups of Lie type with
the same values of $\omega$ and $\psi$, namely $PSL_{4}(2)$ and
$P\Omega^{+}_{8}(2)$. However,
$|P\Omega^{+}_{8}(2)|_{2^{\prime}}\neq|PSL_{3}(4)|_{2^{\prime}}=|PSL_{4}(2)|_{2^{\prime}}$,
and so we come up with another possible pair for $\\{S,H\\}$, namely
$\\{PSL_{4}(2),PSL_{3}(4)\\}$, as stated in the theorem. ∎

Table 2. Simple groups of Lie type with the same values of $\omega$ and
$\psi$: generic case. $PSL_{n}(p^{2s})$, $\Omega_{2n+1}(p^{s})$,
$PSp_{2n}(p^{s})$, $P\Omega^{+}_{2(n+1)}(p^{s})$, $P\Omega^{-}_{2n}(p^{s})$
---
$PSL_{3}(p^{2s})$, $PSU_{4}(p^{s})$, $\Omega_{7}(p^{s})$, $PSp_{6}(p^{s})$,
$P\Omega^{+}_{8}(p^{s})$
$PSL_{2}(p^{6s})$, $\Omega_{5}(p^{3s})$, $G_{2}(p^{2s})$, ${}^{3}D_{4}(p^{s})$
$PSL_{2}(p^{3s})$, $G_{2}(p^{s})$
$PSL_{3}(p^{4s})$, $PSU_{4}(p^{2s})$, $\Omega_{7}(p^{2s})$, $PSp_{6}(p^{2s})$,
$P\Omega^{+}_{8}(p^{2s})$, $F_{4}(p^{s})$
$PSL_{2}(2^{6s})$, $\Omega_{5}(2^{3s})$, $G_{2}(2^{2s})$,
${}^{3}D_{4}(2^{s})$, ${}^{2}F_{4}(2^{s})$, $s\geq 3$ odd
$PSL_{4}(p^{3s})$, $E_{6}(p^{s})$
$PSL_{4}(p^{6s})$, $\Omega_{9}(p^{3s})$, $P\Omega_{10}^{+}(p^{3s})$,
$P\Omega_{8}^{-}(p^{3s})$, $E_{6}(p^{2s})$
$PSL_{3}(p^{6s})$, $PSU_{4}(p^{3s})$, $\Omega_{7}(p^{3s})$, $PSp_{6}(p^{3s})$,
$P\Omega^{+}_{8}(p^{3s})$, ${}^{2}E_{6}(p^{s})$
$PSL_{3}(p^{12s})$, $\Omega_{7}(p^{6s})$, $PSp_{6}(p^{6s})$,
$P\Omega^{+}_{8}(p^{6s})$, $F_{4}(p^{3s})$, ${}^{2}E_{6}(p^{2s})$
$PSL_{5}(p^{6s})$, $\Omega_{11}(p^{3s})$, $PSp_{10}(p^{3s})$,
$P\Omega_{12}^{+}(p^{3s})$, $P\Omega_{10}^{-}(p^{3s})$, $E_{8}(p^{s})$
$PSU_{n}(q)$, $PSU_{n-1}(q)$, $n\geq 7$ odd
$PSU_{3}(2^{2s})$, ${}^{2}B_{2}(2^{3s})$, $s$ odd
$PSU_{3}(3^{s})$, ${}^{2}G_{2}(3^{s})$, $s\geq 3$ odd
$PSU_{9}(p^{s})$, $PSU_{8}(p^{s})$, $E_{7}(p^{s})$
$\Omega_{2n+1}(q)$, $PSp_{2n}(q)$, $n\geq 3$, $q$ odd

The next theorem improves Theorem 4.1 when the relevant groups are of Lie
type.

######  Theorem 6.4.

Let $S$ and $H$ be finite simple groups of Lie type such that
${\mbox{cod}}(S)\subseteq{\mbox{cod}}(H)$. Then $S\cong H$.

###### Proof.

By Lemma 3.4 and Propositions 6.1 and 6.3, we have that $S$ and $H$ fall into
one of two pairs of groups concluded in Proposition 6.3. The result now
follows by Lemma 4.2. ∎

##  7\. Theorem B: The mixed case of alternating groups and groups of Lie type

In this section, we prove Theorem B in the mixed situation where the set of
codegrees of an alternating group $S$ is contained in that of a simple group
$H$ of Lie type, or vice versa.

In the following proposition we remark that the condition on $m$ is necessary,
due to the coincidences of isomorphic simple groups: $A_{5}\cong
PSL_{2}(4)\cong PSL_{2}(5)$, $A_{6}\cong PSL_{2}(9)$, and $A_{8}\cong
PSL_{4}(2)$. We also recall that $f(X):=|X|/b(X)$, where $b(X)$ is the largest
character degree of $X$.

######  Proposition 7.1.

Suppose $m=7$ or $m\geq 9$. Let $H$ be a simple group of Lie type. If
$|\textup{{A}}_{m}|$ divides $|H|$, then $f(H)>f(\textup{{A}}_{m})$. As a
consequence, ${\mbox{cod}}(\textup{{A}}_{m})\nsubseteq{\mbox{cod}}(H)$.

###### Proof.

Let $p$ be the defining characteristic and $q$, a power of $p$, the order of
the underlying field of $H$. Consider first the case $H$ being of exceptional
type. Using Lemma 5.3, we have

$f(H)>\frac{|H|}{256|H|_{p}}=\frac{1}{256}|H|_{p^{\prime}}.$

As $|\textup{{A}}_{m}|$ divides $|H|$, it follows that
$f(H)>(1/256)|\textup{{A}}_{m}|_{p^{\prime}}.$ Therefore, to prove the
theorem, it suffices to show

$b(\textup{{A}}_{m})\geq 256|\textup{{A}}_{m}|_{p}.$

Let us assume for now that that $m\geq 10$. In particular, the dominant prime
in $|\textup{{A}}_{m}|=m!/2$ is $2$. We therefore just need to show
$b(A_{m})\geq 256|\textup{{A}}_{m}|_{2}$. As $|\textup{{A}}_{m}|_{2}\leq
2^{m-2}$, for this we want to show

$b(\textup{{A}}_{m})\geq 64\cdot 2^{m}.$

Note that $b(\textup{{A}}_{19})=64,664,600>64\cdot 2^{19}$ (see [McK86] for
the degree of the largest irreducible characters and associated partitions of
symmetric groups of degree up to 75, from which one can deduce the exact value
or a good bound for the one of corresponding alternating groups). Now one just
inducts on $m$ with the help of Lemma 5.4(ii) to achieve the desired bound for
$m\geq 19$.

Suppose that $m\leq 18$ and recall that we are still dealing with exceptional
groups. When $q=2$, the proposition can be verified directly, so assume that
$q\geq 3$. In such case, $b(H)<26|H|_{p}$ by Lemma 5.3, and whence the above
estimate can be refined so that we only need to prove $b(\textup{{A}}_{m})\geq
26|\textup{{A}}_{m}|_{p}$, which turns out to be true for all $18\geq n\geq
13$. For the remaining values $m\leq 12$, the arguments go as follows. First
we are done if $f(\textup{{A}}_{m})\leq\sqrt{|H|}$, as $f(H)>\sqrt{|H|}$, so
we may assume that $|H|<f(\textup{{A}}_{m})^{2}$. For each $m\leq 12$, there
are indeed no possibilities for $H$ satisfying $|H|<f(\textup{{A}}_{m})^{2}$
and $|A_{m}|$ divides $|H|$.

Following the same idea as in the case of exceptional groups, but using Lemma
5.4 instead, we can show that in fact $f(H)>f(\textup{{A}}_{m})$ for every $H$
of classical type and $m\geq 19$. Let us present the details for only the
linear groups.

Consider $H=PSL_{n}(q)$ for some $n\geq 2$ and $q$ a prime power. By Lemma
5.4(i), we have

$f(\textup{{A}}_{m})=\frac{m!}{2b(\textup{{A}}_{m})}\leq
e^{1.28255\sqrt{m}}(m!)^{1/2}.$

Thus, if $|H|\geq e^{2.5651\sqrt{m}}m!$ then $f(H)>\sqrt{|H|}\geq
e^{1.28255\sqrt{m}}(m!)^{1/2}\geq f(\textup{{A}}_{m})$ and we would be done.
We therefore can assume that $|H|<e^{2.5651\sqrt{m}}m!$, which in particular
implies that $n<m$. Using Lemma 5.4(iii), we see that, as before, it is enough
to show that $b(\textup{{A}}_{m})\geq
13(1+\log_{q}(n+1))^{2.54}|\textup{{A}}_{m}|_{p}$. Since $m\geq n+1$ and
$|\textup{{A}}_{m}|_{p}\leq 2^{m-2}$, for this it is sufficient to show that

$b(\textup{{A}}_{m})\geq 13(1+\log_{2}m)^{2.54}2^{m-2}.$

This last inequality is indeed true for $m=20$, and therefore is true for all
$m\geq 20$, by induction and Lemma 5.4(ii). Checking directly, we see that the
inequality $b(\textup{{A}}_{m})\geq
13(1+\log_{q}(n+1))^{2.54}|\textup{{A}}_{m}|_{2}$ is still valid for $n=19$.

As for the exceptional types, we are left to consider the small cases $m\leq
18$. Again we are done if $f(\textup{{A}}_{m})\leq\sqrt{|H|}$, so we may
assume that $|H|<f(\textup{{A}}_{m})^{2}$. For each $m$, we search for
relevant $H$ satisfying $|H|<f(\textup{{A}}_{m})^{2}$ and $|A_{m}|$ divides
$|H|$ and find that, for such an $H$, the inequality
$f(H)>f(\textup{{A}}_{m})$ always holds true. ∎

We shall need the following result on $2$-defect zero and $3$-defect zero
characters of alternating groups, which easily follows from earlier work of F.
Garvan, D. Kim and D. Stanton [GKS90] on the so-called _$p$ -core partitions_.
They are partitions having no hook lengths divisible by $p$. Using Garvan-Kim-
Stanton’s result, A. Granville and K. Ono [GO96] proved the existence of
$p$-defect zero characters with $p\geq 5$ in symmetric and alternating groups.

######  Lemma 7.2.

Let $m$ be a positive integer.

  1. (i)

$\textup{{A}}_{m}$ has a $2$-defect zero irreducible character if and only if
$m=2k^{2}+k$ or $m=2k^{2}+k+2$ for some $k\in\mathbb{N}$.

  2. (ii)

$\textup{{A}}_{m}$ has a $3$-defect zero irreducible character if and only if
there is a prime $\ell\equiv 2(\bmod 3)$ such that the the exact power of
$\ell$ dividing $3m+1$ is odd.

###### Proof.

See the discussion in [GO96, pp. 333-334]. ∎

######  Proposition 7.3.

Let $S$ be a simple group of Lie type and $8\neq m\geq 7$ an integer. Then
${\mbox{cod}}(S)\nsubseteq{\mbox{cod}}(\textup{{A}}_{m})$. In fact, if $|S|$
divides $|\textup{{A}}_{m}|$ and $m\geq 44$, then $f(S)<f(\textup{{A}}_{m})$.

###### Proof.

Assume by contradiction that
${\mbox{cod}}(S)\subseteq{\mbox{cod}}(\textup{{A}}_{m})$. By Lemma 5.1, we
then have $f(S)\geq f(\textup{{A}}_{m})$.

Suppose that the defining characteristic of $S$ is $p$. Observe that
$f(S)\leq|S|/{\mathsf{St}}_{S}(1)=|S|_{p^{\prime}}$. Furthermore,
$|S|_{p^{\prime}}<|S|_{p}^{2}$ (see [CHMN15, Proof of Thm. 12]) and
$|S|_{p}\leq|\textup{{A}}_{m}|_{p}$ by Theorem 3.1. Therefore we have
$f(S)<(|\textup{{A}}_{m}|_{p})^{2}$. Assume for a moment that $m\geq 10$ so
that $|\textup{{A}}_{m}|_{p}\leq|\textup{{A}}_{m}|_{2}\leq 2^{m-2}$. We now
have $f(S)<2^{2m-4}$. On the other hand, it is clear that
$f(\textup{{A}}_{m})>\sqrt{m!/2}$. Therefore, we would be done if $m!\geq
2^{4m-7}$. By the well-known estimate $m!>\sqrt{2\pi m}(m/e)^{m}$, this is
certainly true when $m\geq 44$. So we may now suppose that $m\leq 43$.

As mentioned above, every simple group of Lie type, and therefore $S$ in
particular, has a $2$-defect zero irreducible character, which means that $S$
has an odd codegree and so does $\textup{{A}}_{m}$ as
${\mbox{cod}}(S)\subseteq{\mbox{cod}}(\textup{{A}}_{m})$. It follows that
$m=2k^{2}+k$ or $m=2k^{2}+k+2$ for some $k\in\mathbb{N}$, by Lemma 7.2(i).
This forces $m$ to be one of $10,12,21,23,36$, or $38$. By the same reason,
$\textup{{A}}_{m}$ has a codegree not divisible by $3$ and so Lemma 7.2(ii)
further narrows down the choices for $m$: $m\in\\{10,12,21,36\\}$. In fact,
when $m=21$ or $36$, we still have
$f(\textup{{A}}_{m})>|\textup{{A}}_{m}|_{2}^{2}$, and since
$|\textup{{A}}_{m}|_{2}^{2}>f(S)$, it follows that $f(\textup{{A}}_{m})>f(S)$,
which is a contradiction.

Suppose $m=10$. The inequality
$f(\textup{{A}}_{m})<(|\textup{{A}}_{m}|_{p})^{2}$ then forces $p=2$ or $3$.
If $p=2$ then $|S|_{2^{\prime}}=|\textup{{A}}_{10}|/\chi(1)$, where
$\chi\in\mathrm{Irr}(\textup{{A}}_{10})$ is one of the two $2$-defect zero
irreducible characters of equal degree $384$, implying
$|S|_{2^{\prime}}=10!/(2\cdot 384)=4725$. It is easy to see from [Atl1] that
there is no such group of Lie type in characteristic $2$. If $p=3$ then
$|S|_{3^{\prime}}=10!/(2\cdot 567)=3200$ since $\textup{{A}}_{10}$ has a
unique $3$-defect zero character of degree $567$, which again leads to a
contradiction as there is no such group in characteristic $3$. The case $m=12$
is treated similarly and we skip the details. ∎

##  8\. Theorem B: Alternating and sporadic groups

######  Proposition 8.1.

Let $m<n$ be positive integers. Then
$f(\textup{{A}}_{m})<f(\textup{{A}}_{n})$. Consequently,
${\mbox{cod}}(\textup{{A}}_{m})\nsubseteq{\mbox{cod}}(\textup{{A}}_{m+1})$.

###### Proof.

It suffices to show that $b(\textup{{A}}_{m+1})<(m+1)b(\textup{{A}}_{m})$. Let
$\chi\in\mathrm{Irr}(\textup{{A}}_{m+1})$ such that
$\chi(1)=b(\textup{{A}}_{m+1})$. As shown in [HHN16, p. 1956], such $\chi$
must be the restriction of an irreducible character, say $\psi$, of
$\textup{{S}}_{m+1}$ whose associated partition, say $\lambda$, is not self-
conjugate. In particular, $\chi(1)=\psi(1)$. As in the proof on Lemma 5.4(ii),
let $Y_{\lambda}$ be the Young diagram associated to $\lambda$. The
restriction $\psi_{\textup{{S}}_{m}}$ of $\psi$ to $\textup{{S}}_{n}$ is the
sum of irreducible characters corresponding to the partitions of $n$ whose
associated Young diagrams are obtained from $Y_{\lambda}$ by removing a
suitable node. The number of those suitable nodes is at most
$(\sqrt{8m+9}-1)/2$, so

$b(\textup{{A}}_{m+1})=\psi(1)\leq\frac{\sqrt{8m+9}-1}{2}b(\textup{{S}}_{m}).$

Since $b(\textup{{S}}_{m})<2b(\textup{{A}}_{m})$ as already mentioned above,
it follows that

$b(\textup{{A}}_{m+1})<(\sqrt{8m+9}-1)b(\textup{{A}}_{m}),$

which implies our desired inequality
$b(\textup{{A}}_{m+1})<(m+1)b(\textup{{A}}_{m})$ for $m\geq 5$. The result is
easily checked for smaller $m$. ∎

######  Proposition 8.2.

Theorem B is true when either $S$ or $H$ is a sporadic simple group.

###### Proof.

The case where both $S$ and $H$ are sporadic simple groups can be verified by
using the available data in [Atl1].

Suppose that $S$ is a sporadic group and $H=\textup{{A}}_{m}$ for some $m\geq
5$. Let $p_{S}$ be the largest prime divisor of $|S|$. By Theorem 3.1, we have
$|S|$ divides $|\textup{{A}}_{m}|$, so $p_{S}\leq m$. By Lemma 5.1, we have
$f(S)\geq f(\textup{{A}}_{m})>\sqrt{m!/2}$. It follows that
$f(S)\geq\sqrt{p_{S}!/2}$. Again using [Atl1], it can be checked that this can
never happen.

Next we assume that $S$ is a sporadic group and $H$ is a simple group of Lie
type in characteristic $p$. Suppose first that $S$ has an irreducible
character, say $\chi$, of $p$-defect zero. Then, as argued in the proof of
Lemma 3.4, we have $|S|_{p^{\prime}}=|H|_{p^{\prime}}$ and $\chi(1)=|S|_{p}$.
In particular, $\chi(1)$ is a prime power, and therefore, [MZ01, Thm. 1.1]
yields

$(S,p,\chi(1))\in\\{(M_{11}/M_{12},11,11),(M_{11},2,16),(M_{24}/Co_{2}/Co_{3},23,23)\\}$

(We note that $M_{12}$ has another irreducible character of prime power
degree, namely $16$, but the character is not of $2$-defect zero and thus does
not fit our situation.) However, for each of these possibilities, there is no
simple group of Lie type $H$ in characteristic $p$ such that
$|H|_{p^{\prime}}=|S|_{p^{\prime}}$. Next, we suppose that $S$ has no
characters of $p$-defect zero. By [GO96, Cor. 2],

$p\in\\{2,3\\}\text{ and
}S\in\\{M_{12},M_{22},M_{24},J_{2},HS,Suz,Ru,Co_{1},Co_{3},BM\\}.$

Now we just apply Lemma 5.4 and argue similarly as in the proof of Proposition
7.1, with $S$ in place of $\textup{{A}}_{m}$, to arrive at $f(H)>f(S)$, and
thus it follows from Lemma 5.1 that
${\mbox{cod}}(S)\nsubseteq{\mbox{cod}}(H)$.

Now we consider the case where $S=\textup{{A}}_{m}$ for some $m\geq 5$ and $H$
a sporadic simple group. Using Theorem 3.1, we have $m!/2$ divides $|H|$ and
so $m$ is at most $\overline{p}_{H}-1$, where $\overline{p}_{H}$ is the
smallest prime not dividing $|H|$. This constraint is enough to ensure that
$f(A_{m})<f(H)$, and thus
${\mbox{cod}}(\textup{{A}}_{m})\nsubseteq{\mbox{cod}}(H)$ by Lemma 5.1.

Finally we consider the case where $S$ is a simple group of Lie type and $H$ a
sporadic simple group. As in the proof of Proposition 7.3 , we have
$f(H)\leq(|H|_{p})^{2}$, where $p$ is the defining characteristic of $S$. The
only possible $p$ satisfying such condition is $p=2$. Now $|S|_{2^{\prime}}$
is an odd codegree of $S$, and hence of $H$, and so
$|S|_{2^{\prime}}=|H|/\chi(1)$ for some $2$-defect zero character
$\chi\in\mathrm{Irr}(H)$. There are in fact only $16$ sporadic simple groups
having a $2$-defect zero irreducible character. For such a group and such a
character, there are no $S$ satisfying the indicated condition. This concludes
the proof. ∎

Theorem B follows from Theorem 6.4 and Propositions 7.1, 7.3, 8.1, and 8.2.

For future work on the codegree isomorphism conjecture (CIC), we record the
following immediate consequence of Theorem B.

######  Theorem 8.3.

Let $S$ be a finite nonabelian simple group. Let $G$ be a minimal
counterexample to (CIC) with respect to $S$ – that is, $G$ is minimal subject
to the conditions ${\mbox{cod}}(G)={\mbox{cod}}(S)$ and $G\ncong S$. Then $G$
has a unique minimal normal subgroup $N$ and $G/N\cong S$.

###### Proof.

Let $N$ be a maximal normal subgroup of $G$. Since
${\mbox{cod}}(G/N)\subseteq{\mbox{cod}}(G)={\mbox{cod}}(S)$ and $G/N$ is
simple, it follows from Theorem B that $G/N\cong S$. Furthermore, by the
minimality of $G$ as a counterexample, we have that $N$ is a minimal normal
subgroup of $G$. (If $G$ has a normal subgroup $M$ such that $M<N$, then
${\mbox{cod}}(G/N)\subseteq{\mbox{cod}}(G/M)\subseteq{\mbox{cod}}(G)$, forcing
${\mbox{cod}}(G/M)={\mbox{cod}}(S)$.) Also, $N$ is the unique minimal normal
subgroup of $G$ since, otherwise, $G=S\times S$, which violates the assumption
${\mbox{cod}}(G)={\mbox{cod}}(S)$. ∎

We conclude the paper with a couple of remarks. First, the group pseudo-
algebra $C(G)$ seems to better distinguish finite groups than the usual
complex group algebra $\mathbb{C}G$. For instance, while any two abelian
groups $A$ and $B$ of the same order have the same complex group algebra
$\mathbb{C}A=\mathbb{C}B$, it was shown in [Mor23] that $A\cong B$ if and only
if $C(A)=C(B)$. It has even been speculated that a finite group $G$ and an
abelian group $A$ are isomorphic if and only if $C(G)=C(A)$. This, if true,
would indicate that abelian groups have very distinctive character codegrees
(counting multiplicities). Theorem A shows that simple groups indeed have very
distinctive codegrees.

Our results are likely to remain true for quasi and/or almost simple groups.
However, at the time of this writing, we do not see yet a uniform proof for
these larger families of groups as the one presented in this paper.

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