# On the characterization of some non-abelian simple groups using codegree set

Hongning Wang<EMAIL_ADDRESS>,  Xuning Zhang  College of Science, China
Three Gorges University, Yichang, Hubei 443002, China<EMAIL_ADDRESS>,
Selina Zhang<EMAIL_ADDRESS>and  Michelle Chen
<EMAIL_ADDRESS>
###### Abstract.

Let $G$ be a finite group and $\chi\in\mbox{Irr}(G)$. The codegree of $\chi$
is defined as ${\mbox{cod}}(\chi)=\frac{|G:\ker(\chi)|}{\chi(1)}$ and
${\mbox{cod}}(G)=\\{{\mbox{cod}}(\chi)\ |\ \chi\in\mbox{Irr}(G)\\}$ is called
the set of codegrees of $G$. In this paper, we show that the set of codegrees
of $\textup{{S}}_{4}(4),\textup{{U}}_{4}(2)$, $\textup{{S}}_{4}(q)\ (q\geq
4)$, $\textup{{U}}_{4}(3)$, ${}^{2}\textup{{F}}_{4}(2)^{\prime}$,
$\textup{{J}}_{3}$, $\textup{{G}}_{2}(3)$, $\textup{{A}}_{9}$,
$\textup{{J}}_{2}$, ${\mbox{PSL}}(4,3)$, $\textup{{M}}^{c}\textup{{L}}$,
$\textup{{S}}_{4}(5)$, $\textup{{G}}_{2}(4)$, HS, O’N and $\textup{{M}}_{24}$
determines the group up to isomorphism.

###### 2010 Mathematics Subject Classification:

Primary 20C15.

##  1\. Introduction

Let $G$ be a finite group and $\mbox{Irr}(G)$ the set of all irreducible
characters of $G$. For any $\chi\in\mbox{Irr}(G)$ the codegree of $\chi$,
written ${\mbox{cod}}(\chi)$, is defined to be
${\mbox{cod}}(\chi)=\frac{|G:\ker(\chi)|}{\chi(1)}$. We denote
${\mbox{cod}}(G)=\\{{\mbox{cod}}(\chi)\ |\ \chi\in\mbox{Irr}(G)\\}$. The
concept of codegrees was originally considered in [6] where the codegree of
$\chi$ was defined as $\frac{|G|}{\chi(1)}$. The definition was later modified
in [15] so that ${\mbox{cod}}(\chi)$ is independent of $N$ when
$N\leq\ker(\chi)$. Many properties of codegrees have been studied, such as the
relationship between the codegrees and the element orders, codegrees of
$p$-groups, and groups with few codegrees.

Let ${\mbox{cd}}(G)=\\{\chi(1)\ |\ \chi\in\mbox{Irr}(G)\\}$. In the 1990s,
Bertram Huppert studied the character degrees of simple groups have a very
high degree of individuality. He conjectured a relationship between the
structure of groups and its codegrees.

Huppert’s conjecture: Let $H$ be any finite non-abelian simple group and $G$ a
finite group such that ${\mbox{cd}}(G)$ = ${\mbox{cd}}(H)$. Then $G\cong
H\times A$, where $A$ is abelian.

Recently, an alternative version of Huppert’s conjecture has been proposed by
Qian (question 20.79 in the Kourovka Notebook).

Codegree version of Huppert’s conjecture: Let $H$ be any finite non-abelian
simple group and $G$ a finite group such that
${\mbox{cod}}(G)={\mbox{cod}}(H)$. Then $G\cong H$.

This conjecture has been considered and shown to hold for ${\mbox{PSL}}(2,q)$
in [4]. In [1], the conjecture was proven for ${}^{2}B_{2}(2^{2f+1})$, where
$f\geq 1$, ${\mbox{PSL}}(3,4)$, $\mathrm{\textup{{A}}}_{7}$, and
$\textup{{J}}_{1}$. The conjecture holds in the cases where $H$ is
$\textup{{M}}_{11},\textup{{M}}_{12},\textup{{M}}_{22},\textup{{M}}_{23},{\mbox{PSL}}(3,3),{\mbox{PSL}}(3,q)$
where $4<q\not\equiv 1\ (\bmod\ 3)$, ${\mbox{PSU}}(3,q)$ where
$4<q\not\equiv-1\ (\bmod\ 3)$, $\textup{{G}}_{2}(2)^{\prime}$,
${\mbox{PSL}}(3,q)$ where $4<q\equiv 1\ (\bmod\ 3)$, ${\mbox{PSU}}(3,q)$ where
$4<q\equiv-1\ (\bmod\ 3)$, ${\mbox{PSL}}(4,2)$, or
${}^{2}\textup{{G}}_{2}(3^{2f+1}),f\geq 1$ by [8, 12, 9]. In this paper we
prove the following result which shows that the conjecture holds in the cases
where $H$ is $\textup{{S}}_{4}(4),\textup{{U}}_{4}(2)$, $\textup{{S}}_{4}(q)\
(q\geq 4)$, $\textup{{U}}_{4}(3)$, ${}^{2}\textup{{F}}_{4}(2)^{\prime}$,
$\textup{{J}}_{3}$, $\textup{{G}}_{2}(3)$, $\textup{{A}}_{9}$,
$\textup{{J}}_{2}$, ${\mbox{PSL}}(4,3)$, $\textup{{M}}^{c}\textup{{L}}$,
$\textup{{S}}_{4}(5)$, $\textup{{G}}_{2}(4)$, HS, O’N, or $\textup{{M}}_{24}$.

######  Theorem 1.1.

Assume a simple group $H$ is isomorphic to
$\textup{{S}}_{4}(4),\textup{{U}}_{4}(2),\textup{{S}}_{4}(q),\textup{{U}}_{4}(3),{{}^{2}\textup{{F}}_{2}(2)}^{\prime},\textup{{J}}_{3}$,
$\textup{{G}}_{2}(3)$, $\textup{{A}}_{9}$, $\textup{{J}}_{2}$,
${\mbox{PSL}}(4,3)$, $\textup{{M}}^{c}\textup{{L}}$, $\textup{{S}}_{4}(5)$,
$\textup{{G}}_{2}(4)$, HS, O’N, or $\textup{{M}}_{24}$. Let $G$ be a finite
group with ${\mbox{cod}}(G)={\mbox{cod}}(H)$. Then $G\cong H$.

##  2\. Lemmas

Our notation is standard and follows from Isaacs’ book [10] and Atlas [7]. The
following lemmas will be useful in our later proof.

######  Lemma 2.1.

([14, Lemma 4.2]) Let $S$ be a finite non-abelian simple group. Then there
exists $1_{S}\neq\phi\in{\mbox{Irr}}(S)$ that extends to ${\mbox{Aut}}(S)$.

######  Lemma 2.2.

([11, Theorem 4.3.34] or [5, Lemma 5]) Let $N$ be a minimal normal subgroup of
$G$ such that $N=S_{1}\times\cdots\times S_{t}$ where $S_{i}\cong S$ is a non-
abelian simple group for each $i=1,\dots,t$. If $\chi\in{\mbox{Irr}}(S)$
extends to ${\mbox{Aut}}(S)$, then
$\chi\times\cdots\times\chi\in{\mbox{Irr}}(N)$ extends to $G$.

######  Remark.

Cases for finite groups $G$ with $|{\mbox{cod}}(G)|\leq 3$ are addressed in
[2]. So the non-abelian simple group has at least four character codegrees.

######  Lemma 2.3.

Let $S$ be a non-abelian finite simple group. If $|{{\mbox{cod}}}(S)|\leq 20$,
then one of the following holds:

  1. (a)

$|{\mbox{cod}}(S)|=4$ and $S={{\mbox{PSL}}}(2,2^{f})$ for $f\geq 2$, or

  2. (b)

$|{\mbox{cod}}(S)|=5$ and $S={{\mbox{PSL}}}(2,p^{f})$, $p\neq 2,p^{f}>5$, or

  3. (c)

$|{\mbox{cod}}(S)|=6$ and $S={}^{2}B_{2}(2^{2f+1})$, $f\geq 1$ or
$G={{\mbox{PSL}}}(3,4)$, or

  4. (d)

$|{\mbox{cod}}(S)|=7$ and
$S={{\mbox{PSL}}}(3,3),\textup{{A}}_{7},\textup{{M}}_{11}$, or
$\textup{{J}}_{1}$ or

  5. (e)

$|{\mbox{cod}}(S)|=8$ and $S={{\mbox{PSL}}}(3,q)$ where $4<q\not\equiv 1\
(\bmod\ 3)$ or $S={\mbox{PSU}}(3,q)$ where $4<q\not\equiv-1\ (\bmod\ 3)$ or
$S=\textup{{G}}_{2}(2)^{\prime}$ or

  6. (f)

$|{\mbox{cod}}(S)|=9$ and $S={{\mbox{PSL}}}(3,q)$ where $4<q\equiv 1\ (\bmod\
3)$ or $S={\mbox{PSU}}(3,q)$ where $4<q\equiv-1\ (\bmod\ 3)$ or

  7. (g)

$|{\mbox{cod}}(S)|=10$ and $S=\textup{{M}}_{22}$ or

  8. (h)

$|{\mbox{cod}}(S)|=11$ and
$S={{\mbox{PSL}}}(4,2),\textup{{M}}_{12},\textup{{M}}_{23},$ or
${}^{2}\textup{{G}}_{2}(3^{2f+1}),f\geq 1$ or

  9. (i)

$|{\mbox{cod}}(S)|=12$ and $S=\textup{{S}}_{4}(4)$ or

  10. (j)

$|{\mbox{cod}}(S)|=13$ and $S=\textup{{U}}_{4}(2)$, $\textup{{S}}_{4}(q)$ for
$q=2^{f}>4$ or

  11. (k)

$|{\mbox{cod}}(S)|=14$ and
$S=\textup{{U}}_{4}(3),{{}^{2}\textup{{F}}_{4}(2)}^{\prime}$,
$\textup{{J}}_{3}$ or

  12. (l)

$|{\mbox{cod}}(S)|=15$ and $S=\textup{{G}}_{2}(3)$ or

  13. (m)

$|{\mbox{cod}}(S)|=16$ and $S=\textup{{A}}_{9}$, $\textup{{J}}_{2}$

  14. (n)

$|{\mbox{cod}}(S)|=17$ and $S={\mbox{PSL}}^{4}(3)$, or
$\textup{{M}}^{c}\textup{{L}}$ or

  15. (o)

$|{\mbox{cod}}(S)|=18$ and $S=\textup{{S}}_{4}(5)$, $\textup{{G}}_{2}(4)$ or
HS

  16. (p)

$|{\mbox{cod}}(S)|=19$ and $S={\mbox{O'N}}$ or

  17. (r)

$|{\mbox{cod}}(S)|=20$ and $S=\textup{{M}}_{24}$ or $\textup{{G}}_{2}(q)$ for
prime power $q=p^{n}\geq 7$ with $p$ prime and $q\equiv 2,3,4\pmod{6}$.

###### Proof.

For simple group $S$, each nontrivial irreducible character is faithful. Then
$|{\mbox{cod}}(S)|=|{\mbox{cd}}(S)|$ and the result is true by [3, Theorem
1.1].  $\hbox to0.0pt{$\sqcap$\hss}\sqcup$

The character degree sets for the relevant simple groups can be found via [7].
We list the relevant codegree sets in the following Lemma for easy reference.

######  Lemma 2.4.

If simple group $S$ is isomorphic to ${\mbox{PSL}}(2,k)$, ${}^{2}B_{2}(q)$,
${\mbox{PSL}}(3,4)$, $\textup{{A}}_{7}$, $\textup{{J}}_{1}$,
$\textup{{M}}_{11}$, ${\mbox{PSL}}(3,3)$, $\textup{{G}}_{2}(2)^{\prime}$,
$\textup{{M}}_{22}$, ${\mbox{PSL}}(4,2)$, $\textup{{M}}_{12}$,
$\textup{{M}}_{23}$, ${\mbox{PSL}}(3,q)$, ${\mbox{PSU}}(3,q)$,
${}^{2}\textup{{G}}_{2}(q)$, $\textup{{S}}_{4}(4)$, $\textup{{S}}_{4}(q)$,
$\textup{{U}}_{4}(2)$, $\textup{{U}}_{4}(3)$,
${{}^{2}\textup{{F}}_{2}(2)}^{\prime}$, $\textup{{J}}_{3}$,
$\textup{{G}}_{2}(3)$, $\textup{{A}}_{9}$, $\textup{{J}}_{2}$,
${\mbox{PSL}}(4,3)$, $\textup{{M}}^{c}\textup{{L}}$, $\textup{{S}}_{4}(5)$,
$\textup{{G}}_{2}(4)$, HS, O’N, or $\textup{{M}}_{24}$, then ${\mbox{cod}}(S)$
can be found in Table 1.

###### Proof.

The character degrees of simple groups appeared in the Lemma can be found in
[1, 4, 7, 17, 18] and the corresponding codegree set can be easily obtained.
$\hbox to0.0pt{$\sqcap$\hss}\sqcup$

Table 1. Codegree set for some simple groups Group $G$ | Codegree set ${\mbox{cod}}(G)$
---|---
${\mbox{PSL}}(2,k)$ ($k=2^{f}\geq 4$) | $\\{1,\ k(k-1),\ k(k+1),\ k^{2}-1\\}$
${\mbox{PSL}}(2,k)$ ($k>5$) |  $\\{1,\frac{k(k-1)}{2},\frac{k(k+1)}{2},\frac{k^{2}-1}{2},\ k(k-\epsilon(k))\\}$, $\epsilon(k)=(-1)^{(k-1)/2}$
${}^{2}B_{2}(q)$ | $\\{1,\ (q-1)(q^{2}+1),\ q^{2}(q-1),\ 2^{3f+2}(q^{2}+1),\ q^{2}(q-2r+1),$
$(q=2r^{2}=2^{2f+1})$ | $q^{2}(q+2r+1)\\}$
${\mbox{PSL}}(3,4)$ | $\\{1,\ 2^{4}{\cdot}3^{2}{\cdot}7,\ 2^{6}{\cdot}3^{2},\ 2^{6}{\cdot}5,\ 2^{6}{\cdot}7,\ 3^{2}{\cdot}5{\cdot}7\\}$
$\textup{{A}}_{7}$ | $\\{1,\ 2^{2}{\cdot}3{\cdot}5{\cdot}7,\ 2^{2}{\cdot}3^{2}{\cdot}7,\ 2^{2}{\cdot}3^{2}{\cdot}5,\ 2^{3}{\cdot}3{\cdot}7,\ 2^{3}{\cdot}3{\cdot}5,\ 2^{3}{\cdot}3^{2}\\}$
$\textup{{J}}_{1}$ |  $\\{1,\ 3{\cdot}5{\cdot}11{\cdot}19,\ 2{\cdot}3{\cdot}5{\cdot}7{\cdot}11,\ 2^{3}{\cdot}3{\cdot}5{\cdot}19,\ 7{\cdot}11{\cdot}19,$ $2^{3}{\cdot}3{\cdot}5{\cdot}11,\ 2^{3}{\cdot}3{\cdot}5{\cdot}7\\}$
$\textup{{M}}_{11}$ | $\\{1,\ 2^{3}{\cdot}3^{2}{\cdot}11,\ 2^{4}{\cdot}3^{2}{\cdot}5,\ 3^{2}{\cdot}5{\cdot}11,\ 2^{2}{\cdot}3^{2}{\cdot}5,\ 2^{4}{\cdot}11,\ 2^{4}{\cdot}3^{2}\\}$
${\mbox{PSL}}(3,3)$ | $\\{1,\ 2^{2}{\cdot}3^{2}{\cdot}13,\ 2^{4}{\cdot}3^{3},\ 3^{3}{\cdot}13,\ 2^{3}{\cdot}3^{3},\ 2^{4}{\cdot}13,\ 2^{4}{\cdot}3^{2}\\}$
$\textup{{G}}_{2}(2)^{\prime}$ | $\\{1,\ 2^{4}{\cdot}3^{2}{\cdot}7,\ 2^{5}{\cdot}3^{3},\ 2^{4}{\cdot}3^{3},\ 2^{5}{\cdot}3^{2},\ 2^{5}{\cdot}7,\ 2^{3}{\cdot}3^{3},\ 3^{3}{\cdot}7\\}$
$\textup{{M}}_{22}$ | $\\{1,\ 2^{7}{\cdot}3{\cdot}5{\cdot}11,\ 2^{7}{\cdot}7{\cdot}11,\ 2^{7}{\cdot}3^{2}{\cdot}7,\ 2^{7}{\cdot}5{\cdot}7,\ 2^{6}{\cdot}3^{2}{\cdot}5,\ 2^{6}{\cdot}3{\cdot}11,\ 2^{7}{\cdot}3{\cdot}5,\ 2^{4}{\cdot}3^{2}{\cdot}11,\ 2^{7}{\cdot}3^{2}\\}$
${\mbox{PSL}}(4,2)$ | $\\{1,\ 2^{6}{\cdot}3^{2}{\cdot}5,\ 2^{5}{\cdot}3^{2}{\cdot}5,\ 2^{4}{\cdot}3^{2}{\cdot}7,\ 2^{6}{\cdot}3{\cdot}5,\ 2^{4}{\cdot}3^{2}{\cdot}5,\ 2^{6}{\cdot}3^{2},\ 2^{6}{\cdot}7,\ 2^{3}{\cdot}3^{2}{\cdot}5,\ 3^{2}{\cdot}5{\cdot}7,\ 2^{5}{\cdot}3^{2}\\}$
$\textup{{M}}_{12}$ | $\\{1,\ 2^{6}{\cdot}3^{3}{\cdot}5,\ 2^{2}{\cdot}3^{3}{\cdot}5{\cdot}11,\ 2^{6}{\cdot}3{\cdot}11,\ 2^{5}{\cdot}5{\cdot}11,\ 2^{6}{\cdot}3^{3},\ 2^{5}{\cdot}3^{2}{\cdot}5,\ 2^{6}{\cdot}3{\cdot}5,$
| $2^{3}{\cdot}3^{2}{\cdot}11,\ 2^{2}{\cdot}3{\cdot}5{\cdot}11,\
2^{2}{\cdot}3^{3}{\cdot}5\\}$
$\textup{{M}}_{23}$ | $\\{1,\ 2^{6}{\cdot}3^{2}{\cdot}5{\cdot}7{\cdot}23,\ 2^{7}{\cdot}7{\cdot}11{\cdot}23,\ 2^{6}{\cdot}3^{2}{\cdot}7{\cdot}11,\ 2^{7}{\cdot}3{\cdot}5{\cdot}23,\ 2^{7}{\cdot}3^{2}{\cdot}5{\cdot}7,$
| $2^{6}{\cdot}3^{2}{\cdot}23,\ 3^{2}{\cdot}5{\cdot}11{\cdot}23,\
2^{6}{\cdot}7{\cdot}23,\ 2^{7}{\cdot}7{\cdot}11,\
2^{4}{\cdot}3^{2}{\cdot}5{\cdot}7\\}$
${\mbox{PSL}}(3,q)$ | $\\{1,\ (q^{2}+q+1)(q^{2}-1)(q-1),\ q^{2}(q^{2}+q+1)(q-1)^{2},$
$4<q\not\equiv 1\ (\bmod\ 3)$ | $\ q^{3}(q^{2}+q+1),\ q^{2}(q^{2}-1)(q-1),\ q^{3}(q^{2}-1),\ q^{3}(q^{2}-1)(q-1),\ q^{3}(q-1)^{2}\\}$
${\mbox{PSL}}(3,q)$ | $\\{1,\ \frac{1}{3}(q^{2}+q+1)(q+1)(q-1)^{2},\ \frac{1}{3}q^{2}(q^{2}+q+1)(q-1)^{2},\ \frac{1}{3}q^{3}(q^{2}+q+1),$
$4<q\equiv 1\ (\bmod\ 3)$ | $\frac{1}{3}q^{2}(q+1)(q-1)^{2},\ \frac{1}{3}q^{3}(q-1)(q+1),\ \frac{1}{3}q^{3}(q+1)(q-1)^{2},\ \frac{1}{3}q^{3}(q-1)^{2},\ q^{3}(q-1)^{2}\\}$
${\mbox{PSU}}(3,q)$ | $\\{1,\ (q^{2}-q+1)(q+1)^{2}(q-1),\ q^{3}(q^{2}-q+1),\ q^{2}(q^{2}-q+1)(q+1)^{2},$
$4<q\not\equiv-1\ (\bmod\ 3)$ | $q^{3}(q+1)^{2}(q-1),\ q^{3}(q+1)^{2},\ q^{2}(q+1)^{2}(q-1),\ q^{3}(q-1)(q+1)\\}$
${\mbox{PSU}}(3,q)$ | $\\{1,\ \frac{1}{3}(q^{2}-q+1)(q+1)^{2}(q-1),\ \frac{1}{3}q^{3}(q^{2}-q+1),\ \frac{1}{3}q^{2}(q^{2}-q+1)(q+1)^{2},$
$4<q\equiv-1\ (\bmod\ 3)$ | $\frac{1}{3}q^{3}(q+1)^{2}(q-1),\ \frac{1}{3}q^{3}(q+1)^{2},\ \frac{1}{3}q^{2}(q+1)^{2}(q-1),\ \frac{1}{3}q^{3}(q-1)(q+1),\ q^{3}(q+1)^{2}\\}$
${}^{2}\textup{{G}}_{2}(q)$, $q=3^{2f+1}$ | $\\{1,\ q^{3}(q^{2}-1),\ (q^{2}-1)(q^{2}-q+1),\ q^{2}(q^{2}-1),\ q^{3}(q-1),$
$(m=3^{f},f\geq 1)$ | $2{\cdot}3^{5f+3}(q+1)(q+1-3m),\ 2{\cdot}3^{5f+3}(q+1)(q+1+3m),$
| $3^{5f+3}(q^{2}-q+1),\ q^{3}(q+1),\ q^{3}(q+1-3m),\ q^{3}(q+1+3m)\\}$
$\textup{{S}}_{4}(4)$ | $\\{1,\ 2^{7}{\cdot}5^{2}{\cdot}17,\ 2^{7}{\cdot}3^{2}{\cdot}5^{2},\ 2^{7}{\cdot}3^{2}{\cdot}17,\ 2^{8}{\cdot}3{\cdot}5^{2},\ 2^{8}{\cdot}3^{2}{\cdot}5,\ 2^{8}{\cdot}5^{2},\ 2^{6}{\cdot}3{\cdot}5^{2},$
| $2^{8}{\cdot}17,\ 2^{8}{\cdot}3{\cdot}5,\ 3^{2}{\cdot}5^{2}{\cdot}17,\
2^{6}{\cdot}3^{2}\\}$
$\textup{{U}}_{4}(2)$ | $\\{1,\ 2^{6}{\cdot}3^{4},\ 2^{5}{\cdot}3^{3}{\cdot}5,\ 2^{5}{\cdot}3^{4},\ 2^{6}{\cdot}3^{3},\ 2^{4}{\cdot}3^{4},\ 2^{3}{\cdot}3^{3}{\cdot}5,$
| $2^{5}{\cdot}3^{3},\ 2^{3}{\cdot}3^{4},\ 2^{6}{\cdot}3^{2},\
2^{4}{\cdot}3^{3},\ 3^{4}{\cdot}5,\ 2^{6}{\cdot}5\\}$
$\textup{{S}}_{4}(q)$ (for $q=2^{f}>4$) | $\\{1,\ 2q^{3}(q-1)^{2}(q+1)^{2},\ 2q^{3}(q^{2}+1)(q-1)^{2},\ (q^{2}+1)(q-1)^{2}(q+1)^{2},$
| $\ 2q^{3}(q^{2}+1)(q+1)^{2},\ q^{4}(q-1)(q+1),\ q^{4}(q+1)^{2},\
q^{4}(q^{2}+1),\ q^{4}(q+1)(q-1)^{2},$
| $\ q^{4}(q-1)(q+1)^{2},\ q^{3}(q-1)^{2}(q+1),\ q^{3}(q-1)(q+1)^{2},\
q^{4}(q-1)^{2}\\}$
$\textup{{U}}_{4}(3)$ | $\\{1,\ 2^{7}{\cdot}3^{5}{\cdot}5,\ 2^{7}{\cdot}3^{6},\ 2^{6}{\cdot}3^{4}{\cdot}7,\ 2^{5}{\cdot}3^{6},\ 2^{7}{\cdot}3^{3}{\cdot}5,\ 2^{6}{\cdot}3^{5},\ 2^{4}{\cdot}3^{6},\ 2^{7}{\cdot}3^{4},\ 2^{5}{\cdot}3^{5},$
| $2^{3}{\cdot}3^{6},\ 3^{6}{\cdot}7,\ 2^{7}{\cdot}5{\cdot}7,\
3^{6}{\cdot}5\\}$
${}^{2}\textup{{F}}_{4}(2)^{\prime}$ | $\\{1,\ 2^{10}{\cdot}3^{3}{\cdot}5^{2},\ 2^{11}{\cdot}5^{2}{\cdot}13,\ 2^{10}{\cdot}3^{2}{\cdot}5^{2},\ 2^{9}{\cdot}3^{2}{\cdot}13,\ 2^{11}{\cdot}3^{3},\ 2^{11}{\cdot}5^{2},\ 2^{7}{\cdot}3^{2}{\cdot}5,\ 2^{10}{\cdot}3^{3},$
| $2^{11}{\cdot}13,\ 2^{10}{\cdot}5^{2},\ 2^{9}{\cdot}3^{3},\
2^{5}{\cdot}5^{2}{\cdot}13,\ 3^{3}{\cdot}5^{2}{\cdot}13\\}$
$\textup{{J}}_{3}$ | $\\{1,\ 2^{7}{\cdot}3^{5}{\cdot}19,\ 2^{7}{\cdot}3^{5}{\cdot}5,\ 2^{5}{\cdot}3{\cdot}5{\cdot}17{\cdot}19,\ 2^{6}{\cdot}3^{5}{\cdot}5,\ 2^{3}{\cdot}3^{4}{\cdot}5{\cdot}19,\ 2^{5}{\cdot}3^{4}{\cdot}17,$
| $2^{7}{\cdot}17{\cdot}19,\ 2^{7}{\cdot}3^{5},\ 3^{4}{\cdot}17{\cdot}19,\
2^{6}{\cdot}3^{4}{\cdot}5,\ 3^{5}{\cdot}5{\cdot}17,\
2^{6}{\cdot}3{\cdot}5{\cdot}19,\ 2^{6}{\cdot}3{\cdot}5{\cdot}17\\}$
$\textup{{G}}_{2}(3)$ | $\\{1,\ 2^{5}{\cdot}3^{6}{\cdot}13,\ 3^{6}{\cdot}7{\cdot}13,\ 2^{5}{\cdot}3^{5}{\cdot}7,\ 2^{6}{\cdot}3^{6},\ 2^{3}{\cdot}3^{6}{\cdot}7,\ 2^{3}{\cdot}3^{5}{\cdot}13,\ 2^{5}{\cdot}3^{6},\ 2^{6}{\cdot}3^{5},$
| $3^{6}{\cdot}13,\ 2^{5}{\cdot}3^{5},\ 2^{3}{\cdot}3^{6},\
2^{6}{\cdot}7{\cdot}13,\ 2^{6}{\cdot}3^{4},\ 3^{6}{\cdot}7\\}$
$\textup{{A}}_{9}$ | $\\{1,\ 2^{3}{\cdot}3^{4}{\cdot}5{\cdot}7,\ 2^{6}{\cdot}3^{3}{\cdot}5,\ 2^{6}{\cdot}3{\cdot}5{\cdot}7,\ 2^{4}{\cdot}3^{4}{\cdot}5,\ 2^{6}{\cdot}3^{4},\ 2^{5}{\cdot}3^{3}{\cdot}5,\ 2^{2}{\cdot}3^{3}{\cdot}5{\cdot}7,$
| $2^{3}{\cdot}3^{4}{\cdot}5,\ 2^{4}{\cdot}3^{3}{\cdot}5,\ 2^{6}{\cdot}3^{3},\
2^{3}{\cdot}3^{3}{\cdot}7,\ 2^{3}{\cdot}3^{3}{\cdot}5,\
2^{5}{\cdot}5{\cdot}7,\ 2^{6}{\cdot}3{\cdot}5,\ 2^{3}{\cdot}3{\cdot}5{\cdot
7}\\}$
$\textup{{J}}_{2}$ | $\\{1,\ 2^{6}{\cdot}3^{3}{\cdot}5^{2},\ 2^{7}{\cdot}3^{2}{\cdot}5^{2},\ 2^{5}{\cdot}3{\cdot}5^{2}{\cdot}7,\ 2^{7}{\cdot}3{\cdot}5^{2},\ 2^{6}{\cdot}3^{3}{\cdot}5,\ 2^{6}{\cdot}3{\cdot}5{\cdot}7,\ 2^{6}{\cdot}3{\cdot}5^{2},\ 2^{2}{\cdot}3^{3}{\cdot}5{\cdot}7,$
| $\ 2^{7}{\cdot}3^{3},\ 2^{7}{\cdot}5^{2},\ 2^{2}{\cdot}3^{3}{\cdot}5^{2},\
2^{7}{\cdot}3{\cdot}7,\ 2^{2}{\cdot}3{\cdot}5^{2}{\cdot}7,\
2^{5}{\cdot}3^{2}{\cdot}7,\ 2^{3}{\cdot}3^{2}{\cdot}5^{2}\\}$
${\mbox{PSL}}(4,3)$ | $\\{1,\ 2^{6}{\cdot}3^{6}{\cdot}5,\ 2^{7}{\cdot}3^{5}{\cdot}5,\ 2^{5}{\cdot}3^{6}{\cdot}5,\ 2^{7}{\cdot}3^{6},\ 2^{6}{\cdot}3^{4}{\cdot}13,\ 2^{6}{\cdot}3^{4}{\cdot}5,\ 2^{5}{\cdot}3^{6},\ 2^{7}{\cdot}3^{3}{\cdot}5,$
| $\ 2^{6}{\cdot}3^{5},\ 2^{2}{\cdot}3^{6}{\cdot}5,\
2^{5}{\cdot}3^{4}{\cdot}5,\ 2^{7}{\cdot}3^{4},\ 3^{6}{\cdot}13,\
2^{7}{\cdot}5{\cdot}13,\ 2^{5}{\cdot}3^{5},\ 2^{3}{\cdot}3^{6}\\}$
$\textup{{M}}^{c}L$ | $\\{1,\ 2^{6}{\cdot}3^{6}{\cdot}5^{3}{\cdot}7,\ 2^{7}{\cdot}3^{5}{\cdot}5^{3},\ 2^{5}{\cdot}3^{4}{\cdot}5^{3}{\cdot}11,\ 2^{6}{\cdot}3^{6}{\cdot}5^{2},\ 3^{6}{\cdot}5^{3}{\cdot}11,\ 2^{6}{\cdot}3^{6}{\cdot}11,$
| $2{\cdot}3^{6}{\cdot}5^{2}{\cdot}7,\ 2^{5}{\cdot}3^{4}{\cdot}7{\cdot}11,\
2^{3}{\cdot}3^{3}{\cdot}5^{3}{\cdot}7,\ 2^{7}{\cdot}5^{3}{\cdot}11,$
| $2^{4}{\cdot}3^{4}{\cdot}5^{3},\ 2^{7}{\cdot}5^{3}{\cdot}7,\
2^{6}{\cdot}3^{5}{\cdot}7,\ 2^{7}{\cdot}3^{6},\ 3^{6}{\cdot}5^{3},\
2^{7}{\cdot}3^{3}{\cdot}5^{2}\\}$
$\textup{{S}}_{4}(5)$ | $\\{1,\ 2^{6}{\cdot}3^{2}{\cdot}5^{4},\ 2^{3}{\cdot}3^{2}{\cdot}5^{3}{\cdot}13,\ 2^{6}{\cdot}3^{2}{\cdot}5^{3},\ 2^{5}{\cdot}3{\cdot}5^{4},\ 2^{5}{\cdot}5^{3}{\cdot}13,\ 2^{3}{\cdot}3^{2}{\cdot}5^{4},\ 2^{5}{\cdot}3^{2}{\cdot}5^{3},\ 2^{4}{\cdot}3{\cdot}5^{4},$
| $2^{2}{\cdot}3^{2}{\cdot}5^{4},\ 2^{3}{\cdot}3{\cdot}5^{4},\
2^{6}{\cdot}3^{3}{\cdot}5^{2},\ 2^{5}{\cdot}3{\cdot}5^{3},\
2^{3}{\cdot}3^{2}{\cdot}5^{3},\ 5^{4}{\cdot}13,\ 2^{2}{\cdot}3{\cdot}5^{4},\
2^{6}{\cdot}3^{2}{\cdot}13,\ 2^{4}{\cdot}3{\cdot}5^{3}\\}$
$\textup{{G}}_{2}(4)$ | $\\{1,\ 2^{12}{\cdot}3^{3}{\cdot}5{\cdot}7,\ 2^{11}{\cdot}3^{2}{\cdot}5^{2}{\cdot}7,\ 2^{10}{\cdot}3^{2}{\cdot}7{\cdot}13,\ 2^{11}{\cdot}3^{3}{\cdot}13,\ 2^{10}{\cdot}3^{3}{\cdot}5^{2},$
| $2^{11}{\cdot}5^{2}{\cdot}13,\ 2^{11}{\cdot}3^{3}{\cdot}7,\
2^{12}{\cdot}3{\cdot}5^{2},\ 2^{10}{\cdot}3^{3}{\cdot}7,\
2^{12}{\cdot}3^{2}{\cdot}5,\ 2^{12}{\cdot}3{\cdot}7,$
| $2^{10}{\cdot}3{\cdot}5^{2},\ 2^{12}{\cdot}3{\cdot}5,\
3^{3}{\cdot}5^{2}{\cdot}7{\cdot}13,\ 2^{6}{\cdot}3^{3}{\cdot}5{\cdot}7,\
2^{12}{\cdot}13,\ 2^{10}{\cdot}3^{2}{\cdot}5\\}$
HS | $1,\ 2^{8}{\cdot}3^{2}{\cdot}5^{3}{\cdot}7,\ 2^{9}{\cdot}3^{2}{\cdot}5^{3},\ 2^{8}{\cdot}3^{2}{\cdot}5^{3},\ 2^{9}{\cdot}3^{2}{\cdot}5{\cdot}11,\ 2^{9}{\cdot}3{\cdot}5^{3},\ 2^{9}{\cdot}5^{3},\ 2^{8}{\cdot}3^{2}{\cdot}5^{2},$
| $2^{9}{\cdot}3{\cdot}5{\cdot}7,\ 2^{2}{\cdot}3^{2}{\cdot}5^{3}{\cdot}11,\
2^{4}{\cdot}3{\cdot}5^{3}{\cdot}7,\ 2^{8}{\cdot}5^{3},\
2^{2}{\cdot}3^{2}{\cdot}5^{3}{\cdot}7,$
| $2^{8}{\cdot}3^{2}{\cdot}11,\ 2^{9}{\cdot}3^{2}{\cdot}5,\
2^{6}{\cdot}5^{2}{\cdot}11,\ 2^{8}{\cdot}3^{2}{\cdot}7,\
2^{2}{\cdot}3^{2}{\cdot}5{\cdot}7{\cdot}11\\}$
O’N | $\\{1,\ 2^{3}{\cdot}3^{2}{\cdot}5{\cdot}7^{3}{\cdot}11{\cdot}31,\ 2^{3}{\cdot}3^{4}{\cdot}5{\cdot}7^{3}{\cdot}31,\ 2^{7}{\cdot}3^{4}{\cdot}5{\cdot}7^{3},\ 2^{2}{\cdot}3^{4}{\cdot}5{\cdot}7^{3}{\cdot}31,\ 2^{9}{\cdot}3^{4}{\cdot}7^{3},\ 2^{3}{\cdot}3^{4}{\cdot}5{\cdot}7^{3}{\cdot}1,$
| $2^{7}{\cdot}3^{2}{\cdot}5{\cdot}7^{2}{\cdot}3,\
2^{9}{\cdot}3^{2}{\cdot}5{\cdot}7^{3},\
2^{9}{\cdot}3^{2}{\cdot}5{\cdot}11{\cdot}31,\ 2^{8}{\cdot}3^{4}{\cdot}7^{3},\
2^{6}{\cdot}3^{4}{\cdot}5{\cdot}11{\cdot}19,$
| $2^{9}{\cdot}3^{4}{\cdot}5{\cdot}19,\ 2^{8}{\cdot}3^{4}{\cdot}5{\cdot}31,\
2^{8}{\cdot}7^{3}{\cdot}31,\ 3^{4}{\cdot}5{\cdot}11{\cdot}19{\cdot}31,\
2^{8}{\cdot}7^{2}{\cdot}11{\cdot}19,\ 7^{3}{\cdot}11{\cdot}19{\cdot}31,\
2^{4}{\cdot}3^{4}{\cdot}7^{2}{\cdot}31\\}$
$\textup{{M}}_{24}$ | $\\{1,\ 2^{10}{\cdot}23,\ 2^{8}{\cdot}3{\cdot}5{\cdot}11,\ 2^{7}{\cdot}3{\cdot}5{\cdot}23,\ 2^{10}{\cdot}3^{2}{\cdot}5,\ 2^{4}{\cdot}3^{3}{\cdot}7{\cdot}23,\ 2^{6}{\cdot}3{\cdot}5{\cdot}7{\cdot}11,\ 2^{10}{\cdot}3{\cdot}5{\cdot}7,$
| $2^{7}{\cdot}3^{3}{\cdot}5{\cdot}7,\ 2^{10}{\cdot}3^{3}{\cdot}5,\
2^{10}{\cdot}3^{3}{\cdot}7,\ 2^{10}{\cdot}3{\cdot}7{\cdot}11,\
2^{9}{\cdot}3{\cdot}7{\cdot}23,\ 2^{9}{\cdot}3^{3}{\cdot}23,\
2^{10}{\cdot}3^{2}{\cdot}5{\cdot}11,$
| $2^{10}{\cdot}3^{3}{\cdot}5{\cdot}7,\
2^{8}{\cdot}3{\cdot}5{\cdot}11{\cdot}23,\
2^{10}{\cdot}3^{2}{\cdot}5{\cdot}23,\
2^{10}{\cdot}3{\cdot}7{\cdot}11{\cdot}23,\
2^{10}{\cdot}3^{3}{\cdot}5{\cdot}7{\cdot}11\\}$

At the end of this section. We introduce the result about the maximal
subgroups of $\textup{{S}}_{4}(q)$, $q>4$ and $q$ is even in Table 2. This is
obtained from [16, Table 8.14 ]. And we can easily calculate the order of
these subgroups.

######  Lemma 2.5.

Let $K$ be a maximal subgroup of $\mathrm{\textup{{S}}}_{4}(q)$ where
$q=2^{f}>4$. Then $K$ is isomorphic to one of the groups in Table 2.

Table 2. Maximal subgroups of $\mathrm{\textup{{S}}}_{4}(q)$, $q>4$, $q\ is\ even$ Subgroup | Order | Notes
---|---|---
$E_{q}^{3}:GL_{2}(q)$ | $q^{3}(q^{2}-1)(q^{2}-q)$ | point stabiliser
$[q^{4}]:{C_{q-1}}^{2}$ | $q^{4}(q-1)^{2}$ |
$Sp_{2}(q)\wr 2$ | $2q^{2}(q-1)^{2}(q+1)^{2}$ |
$Sp_{2}(q^{2}):2$ | $2q^{2}(q^{4}-1)$ |
${C_{q-1}}^{2}:D_{8}$ | $8(q-1)^{2}$ | $q\neq 4$
${C_{q+1}}^{2}:D_{8}$ | $8(q+1)^{2}$ |
$C_{q^{2}+1}:4$ | $4(q^{2}+1)$ |
$Sp_{4}(q_{0})$ | $q_{0}^{4}(q_{0}^{4}-1)(q_{0}^{2}-1)$ |  $q=q_{0}^{r}$, $r$ is a prime
$SO_{4}^{+}(q)$ | $\frac{1}{2}q^{2}(q^{2}-1)$ |
$SO_{4}^{-}(q)$ | $q^{2}(q^{2}+1)(q^{2}-1)$ |
$Sz(q)$ | $q^{2}(q^{2}+1)(q-1)$ |  $f\geq 3$ $f$ is odd

Below we list the result about the maximal subgroups of
$\mathrm{{\mbox{PSL}}}(4,3)$ in Table 3. This is obtained from [7].

######  Lemma 2.6.

Let $K$ be a maximal subgroup of $\mathrm{{\mbox{PSL}}}(4,3)$. Then $K$ is
isomorphic to one of the groups in Table 3.

Table 3. Maximal subgroups of $\mathrm{{\mbox{PSL}}}(4,3)$ Subgroup structure | Order | Index
---|---|---
$3^{3}:\mathrm{L}_{3}(3)$ | $151632$ | $40$
$3^{3}:\mathrm{L}_{3}(3)$ | $151632$ | $40$
$\mathrm{U}_{4}(2)$ | $51840$ | $117$
$\mathrm{U}_{4}(2)$ | $51840$ | $117$
$3^{4}:2(\mathrm{A}_{4}\times\mathrm{A}_{4}).2$ | $46656$ | $130$
$(4\times\mathrm{A}_{6}):2$ | $2880$ | $2106$
$\mathrm{S}_{6}$ | $720$ | $8424$
$\mathrm{S}_{4}\times\mathrm{S}_{4}$ | $576$ | $10530$

######  Lemma 2.7.

Let $G$ be a finite group. If no element of ${\mbox{cod}}(G)$ is a power of a
prime, then $G$ is a perfect group.

###### Proof.

Let $G$ be a finite group and suppose $G^{\prime}<G$. Let $N$ be a maximal
normal subgroup of $G$ containing $G^{\prime}$. Then, $G/N$ is an abelian
simple group of order $p$ for some prime $p$. Let $\chi\in\mbox{Irr}(G/N)$ be
a non-principal character. Since $G/N$ is abelian, it follows that
${\mbox{cod}}(\chi)=\frac{|G/N:\ker(\chi)|}{\chi(1)}=\frac{|G/N|}{1}=p$. Since
we can view ${\mbox{cod}}(G/N)$ as a subset of ${\mbox{cod}}(G)$, then there
exists an element of ${\mbox{cod}}(G)$ that is a power of a prime.  $\hbox
to0.0pt{$\sqcap$\hss}\sqcup$

######  Remark 2.8.

If ${\mbox{cod}}(G)={\mbox{cod}}(H)$ where $G$ is a finite group and $H$ is a
finite, non-abelian simple group, it follows that $G$ is a perfect group.
Otherwise ${\mbox{cod}}(G/G^{\prime})\subseteq{\mbox{cod}}(G)$ contains a
power of a prime, which is a contradiction by [2, Lemma 3.6].

For notation, let $N$ be a normal subgroup of $G$ and
$\mbox{Irr}(G|N):=\mbox{Irr}(G)-\mbox{Irr}(G/N)$. For any
$\lambda\in\mbox{Irr}(N)$, let $I_{G}(\lambda)$ be the inertia group of
$\lambda$ in $G$. The set of irreducible constituents of
$\lambda^{I_{G}(\lambda)}$ is denoted $\mbox{Irr}(I_{G}(\lambda)|\lambda)$.

##  3\. Main Results

######  Lemma 3.1.

Let $G$ be a finite group with
${\mbox{cod}}(G)={\mbox{cod}}(\textup{{S}}_{4}(4))$. If $N$ is a maximal
normal subgroup of $G$, then $G/N\cong\textup{{S}}_{4}(4)$.

###### Proof.

Let $N$ be a maximal normal subgroup of $G$. Since
${\mbox{cod}}(G)={\mbox{cod}}(\textup{{S}}_{4}(4))$, we see that $G$ is
perfect by our earlier remark. Then $G/N$ is a non-abelian simple group. Since
${\mbox{cod}}(G/N)\subseteq{\mbox{cod}}(G)$, we see that $|{\mbox{cod}}(G/N)|$
is either $4,5,6,7,8,9,10,11$, or $12$ by Lemma 2.3 and we analyze the cases
separately.

$|{\mbox{cod}}(G/N)|=4$. By Lemma 2.3, $G/N\cong{\mbox{PSL}}(2,k)$ where
$k=2^{f}\geq 4,$Then ${\mbox{cod}}(G/N)=\\{1,k(k-1),k(k+1),k^{2}-1\\}$. Since
$3^{2}\cdot 5^{2}\cdot 17$ is the only nontrivial odd codegree in
${\mbox{cod}}(\textup{{S}}_{4}(4))$, we have that $k^{2}-1=3^{2}\cdot
5^{2}\cdot 17$.This is a contradiction since $k\in\mathbb{Z}$.

$|{\mbox{cod}}(G/N)|=5$. By Lemma 2.3, $G/N\cong{\mbox{PSL}}(2,k)$ where $k$
is an odd prime power and
${\mbox{cod}}(G/N)=\left\\{1,\frac{k(k-1)}{2},\frac{k(k+1)}{2},\frac{k^{2}-1}{2},k(k-\epsilon(k))\right\\}$
where $\epsilon(k)=(-1)^{(k-1)/2}$. Then $k(k-\epsilon(k))/2$,
$k(k-\epsilon(k))\in{\mbox{cod}}(G)$, a contradiction for we can’t find a
codegree which is the half of another codegree in
${\mbox{cod}}(\textup{{S}}_{4}(4))$.

$|{\mbox{cod}}(G/N)|=6$. By observation, $3^{2}\cdot 5\cdot 7$ is the only
nontrivial odd codegree in ${\mbox{cod}}({\mbox{PSL}}(3,4))$. However
$3^{2}\cdot 5\cdot 7\notin{\mbox{cod}}(\textup{{S}}_{4}(4))$, so
${\mbox{cod}}({\mbox{PSL}}(3,4))\not\subset{\mbox{cod}}(\textup{{S}}_{4}(4))$.
Suppose ${\mbox{cod}}(G/N)={\mbox{cod}}({}^{2}B_{2}(q))$.
$(q-1)(q^{2}+1)=3^{2}\cdot 5^{2}\cdot 17$ since these are the only nontrivial
odd codegrees. We note that $3$ does not divide the order of the Suzuki
groups, it is a contradiction.

$|{\mbox{cod}}(G/N)|=7$. We note that the nontrivial odd codegree in codegree
sets of ${\mbox{PSL}}(3,3)$, $\textup{{A}}_{7}$, $\textup{{M}}_{11}$ and
$\textup{{J}}_{1}$ can’t be $3^{2}\cdot 5^{2}\cdot 17$.

$|{\mbox{cod}}(G/N)|=8$. If $G/N\cong{\mbox{PSU}}(3,q)$ with $4<q\not\equiv-1\
(\bmod\ 3)$. Then if $q$ is odd, we have $q^{3}(q^{2}-q+1)=3^{2}\cdot
5^{2}\cdot 17$. Since no cube of a prime divides $3^{2}\cdot 5^{2}\cdot 17$,
we have a contradiction. If $q$ is even, we have
$(q^{2}-q+1)(q+1)^{2}(q-1)=3^{2}\cdot 5^{2}\cdot 17$. Then this does not yield
an integer root for $q$, a contradiction.

If $G/N\cong{\mbox{PSL}}(3,q)$ with $4<q\not\equiv 1\ (\bmod 3)$ and if $q$ is
odd, then $q^{3}(q^{2}+q+1)=3^{2}\cdot 5^{2}\cdot 17$. We have the same
contradiction with $q$. If $q$ is even, then
$(q^{2}+q+1)(q^{2}-1)(q-1)=3^{2}\cdot 5^{2}\cdot 17$. This is a contradiction
since $q$ is an integer.

If $G/N\cong\textup{{G}}_{2}(2)^{\prime}$, we see $3^{3}\cdot
7\notin{\mbox{cod}}(\textup{{S}}_{4}(4))$, so
${\mbox{cod}}(\textup{{G}}_{2}(2)^{\prime})\not\subseteq{\mbox{cod}}(\textup{{S}}_{4}(4))$.
It is a contradiction.

$|{\mbox{cod}}(G/N)|=9$. Suppose $G/N\cong{\mbox{PSL}}(3,q)$ with $4<q\equiv
1\ (\bmod\ 3)$. If $q$ is odd, then $\frac{1}{3}q^{3}(q^{2}+q+1)=3^{2}\cdot
5^{2}\cdot 17$. This is a contradiction since $q\in\mathbb{Z}$. If $q$ is
even, then $\frac{1}{3}(q^{2}+q+1)(q+1)(q-1)^{2}=3^{2}\cdot 5^{2}\cdot 17$.
Then This also does not yield an integer root for $q$. This is a
contradiction.

$G/N\cong{\mbox{PSU}}(3,q))$ with $4<q\equiv-1\ (\bmod 3)$. If $q$ is odd,
then $\frac{1}{3}q^{3}(q^{2}-q+1)=3^{2}\cdot 5^{2}\cdot 17$. This does not
yield an integer root for $q$, a contradiction. If $q$ is even, then
$\frac{1}{3}(q^{2}-q+1)(q+1)^{2}(q-1)=3^{2}\cdot 5^{2}\cdot 17$. This also
does not yield an integer root for $q$, a contradiction.

$|{\mbox{cod}}(G/N)|=10$. Then $G/N\cong\textup{{M}}_{22}$. We note that
$2^{7}\cdot 7\cdot 11\notin{\mbox{cod}}(\textup{{S}}_{4}(4))$, we have that
${\mbox{cod}}(\textup{{M}}_{22})\not\subseteq{\mbox{cod}}(\textup{{S}}_{4}(4))$.
It is a contradiction.

$|{\mbox{cod}}(G/N)|=11$. By observation, the codegree sets of
${\mbox{PSL}}(4,2),\textup{{M}}_{12}$ and $\textup{{M}}_{23}$ contain
$3^{2}\cdot 5\cdot 7$, $2^{2}\cdot 3\cdot 5\cdot 11$ and $3^{2}\cdot 5\cdot
11\cdot 23$ respectively, none of which are elements of the codegree set of
$\textup{{S}}_{4}(4)$.

If $G/N\cong{}^{2}\textup{{G}}_{2}(q)$, $q=3^{2f+1}$. Thus
$3^{2}{\cdot}5^{2}{\cdot}17$ could equal to either $3^{5f+3}(q^{2}-q+1)$,
$q^{3}(q+1-3m)$, or $q^{3}(q+1+3m)$. By considering the power of $3$ or $5$ in
those numbers, one see that this cannot happen.

$|{\mbox{cod}}(G/N)|=12$. Then $G/N\cong\textup{{S}}_{4}(4)$. It is easily
checked that ${\mbox{cod}}(G/N)={\mbox{cod}}(G)$. Thus
$G/N\cong\textup{{S}}_{4}(4)$.  $\hbox to0.0pt{$\sqcap$\hss}\sqcup$

######  Lemma 3.2.

Let $G$ be a finite group with
${\mbox{cod}}(G)={\mbox{cod}}(\textup{{U}}_{4}(2))$. If $N$ is a maximal
normal subgroup of $G$, then $G/N\cong\textup{{U}}_{4}(2)$.

###### Proof.

Let $N$ be a maximal normal subgroup of $G$. Similarly with above lemma, we
have that $G/N$ is a non-abelian simple group and $|{\mbox{cod}}(G/N)|$ is
$4,5,6,7,8,9,10,11,12$, or $13$. Next we analyze the cases separately.

$|{\mbox{cod}}(G/N)|=4$. Then $G/N\cong{\mbox{PSL}}(2,k)$ where $k=2^{f}\geq
4$ and ${\mbox{cod}}(G/N)=\\{1,k(k-1),k(k+1),k^{2}-1\\}$. Hence
$k^{2}-1=3^{4}\cdot 5$, $k\in\mathbb{Z}$ which is a contradiction.

$|{\mbox{cod}}(G/N)|=5$. Then $G/N\cong{\mbox{PSL}}(2,k)$ where $k$ is an odd
prime power and
${\mbox{cod}}(G/N)=\left\\{1,\frac{k(k-1)}{2},\frac{k(k+1)}{2},\frac{k^{2}-1}{2},k(k-\epsilon(k))\right\\}$
where $\epsilon(k)=(-1)^{(k-1)/2}$. Then $k(k-\epsilon(k))/2$,
$k(k-\epsilon(k))\in{\mbox{cod}}(G)$, and this implies that $\frac{k(k\pm
1)}{2}=2^{3}\cdot 3^{4}$ since $2^{3}\cdot 3^{4}$ is the only element of
${\mbox{cod}}(G)$ that is half of another codegree. In either case,
$k\in\mathbb{Z}$ which is a contradiction.

$|{\mbox{cod}}(G/N)|=6$. If $G/N\cong{}^{2}B_{2}(q)$, then
$(q-1)(q^{2}+1)=3^{4}\cdot 5$ since these are the only nontrivial odd
codegrees. We note that $3$ does not divide the order of the Suzuki group, a
contradiction.

If $G/N\cong{\mbox{PSL}}(3,4)$, then $3^{2}\cdot 5\cdot
7\notin{\mbox{cod}}(\textup{{U}}_{4}(2))$ and we clearly see that
${\mbox{cod}}(G/N)\not\subset{\mbox{cod}}(G)$.

$|{\mbox{cod}}(G/N)|=7$. Then $G/N\cong{\mbox{PSL}}(3,3)$, $\textup{{A}}_{7}$,
$\textup{{M}}_{11}$ or $\textup{{J}}_{1}$. We need only to observe that none
of the codegree sets for this case are the same.

$|{\mbox{cod}}(G/N)|=8$. If $G/N\cong{\mbox{PSU}}(3,q)$ with $4<q\not\equiv-1\
(\bmod\ 3)$. Then if $q$ is odd, we have $q^{3}(q^{2}-q+1)=3^{4}\cdot 5$.
Since no cube of a prime divides $3^{4}\cdot 5$, we have a contradiction. If
$q$ is even, we have $(q^{2}-q+1)(q+1)^{2}(q-1)=3^{4}\cdot 5$. Then we can get
a contradiction which means $q$ is not an integer.

If $G/N\cong{\mbox{PSL}}(3,q)$ with $4<q\not\equiv 1\ (\bmod 3)$ and if $q$ is
odd, then $q^{3}(q^{2}+q+1)=3^{4}\cdot 5$. We have the same contradiction with
$q$. If $q$ is even, then $(q^{2}+q+1)(q^{2}-1)(q-1)=3^{4}\cdot 5$. This does
not yield an integer root for $q$, it is a contradiction.

If $G/N\cong\textup{{G}}_{2}(2)^{\prime}$ we see $3^{3}\cdot
7\notin{\mbox{cod}}(\textup{{U}}_{4}(2))$, so
${\mbox{cod}}(\textup{{G}}_{2}(2)^{\prime})\not\subseteq{\mbox{cod}}(\textup{{U}}_{4}(2))$.
It is a contradiction.

$|{\mbox{cod}}(G/N)|=9$. Suppose $G/N\cong{\mbox{PSL}}(3,q)$ with $4<q\equiv
1\ (\bmod\ 3)$. If $q$ is odd, then $\frac{1}{3}q^{3}(q^{2}+q+1)=3^{4}\cdot
5$. Then $q$ is not an integer, it is a contradiction. If $q$ is even, then
$\frac{1}{3}(q^{2}+q+1)(q+1)(q-1)^{2}=3^{4}\cdot 5$. This equation means $q$
is not an integer, a contradiction.

Suppose $G/N\cong{\mbox{PSU}}(3,q))$ with $4<q\equiv-1\ (\bmod 3)$. If $q$ is
odd, then $\frac{1}{3}q^{3}(q^{2}-q+1)=3^{4}\cdot 5$. We have the same
contradiction for $q$ is an integer. If $q$ is even, then
$\frac{1}{3}(q^{2}-q+1)(q+1)^{2}(q-1)=3^{4}\cdot 5$. $q$ is not an integer, it
is a contradiction.

$|{\mbox{cod}}(G/N)|=10$. Then $G/N\cong\textup{{M}}_{22}$. We observe that
$2^{7}\cdot 3\cdot 5\cdot 11\notin{\mbox{cod}}(\textup{{U}}_{4}(2))$. Thus we
have ${\mbox{cod}}(G/N)\not\subseteq{\mbox{cod}}(G)$.

$|{\mbox{cod}}(G/N)|=11$. If $G/N\cong{\mbox{PSL}}(4,2)$, $\textup{{M}}_{12}$
or $\textup{{M}}_{23}$. By observation, the codegree sets of
${\mbox{PSL}}(4,2),\textup{{M}}_{12}$ and $\textup{{M}}_{23}$ contain
$3^{2}\cdot 5\cdot 7$, $2^{2}\cdot 3\cdot 5\cdot 11$ and $3^{2}\cdot 5\cdot
11\cdot 23$ respectively, none of which are elements of the codegree set of
$\textup{{U}}_{4}(2)$. It is easily checked that
${\mbox{cod}}(G/N)\not\subseteq{\mbox{cod}}(G)$. It is a contradiction.

If $G/N\cong{}^{2}\textup{{G}}_{2}(q)$, ($q=3^{2f+1}$). We have that
$3^{4}\cdot 5$ equals to either $3^{5f+3}(q^{2}-q+1)$, $q^{3}(q+1-3m)$, or
$q^{3}(q+1+3m)$. This is a contradiction since $q$ is not an integer.

$|{\mbox{cod}}(G/N)|=12$. Then $G/N\cong\textup{{S}}_{4}(4)$. By observation,
$3^{2}\cdot 5^{2}\cdot 17\notin{\mbox{cod}}(\textup{{U}}_{4}(2))$. Thus
${\mbox{cod}}(G/N)\not\subseteq{\mbox{cod}}(G)$, a contradiction.

$|{\mbox{cod}}(G/N)|=13$. Then
${\mbox{cod}}(G/N)={\mbox{cod}}(\textup{{U}}_{4}(2))$ or
${\mbox{cod}}(\textup{{S}}_{4}(q))$.

${\mbox{cod}}(G/N)={\mbox{cod}}(\textup{{S}}_{4}(q))$, for $q=2^{f}>4$. We can
easily checked that $3^{4}\cdot 5=(q^{2}+1)(q+1)^{2}(q-1)^{2}$, as the are the
only nontrivial codegrees in each set. If $3^{4}\mid(q+1)^{2}(q-1)^{2}$, then
$q^{2}+1\mid 5$, this implies $q=2$, which is a contradiction for
$(2+1)^{2}(2-1)\neq 3^{4}$. If $5\mid(q+1)^{2}(q-1)^{2}$, then $q^{2}+1\mid
3^{4}$, this implies $q\not\in\mathbb{Z}$, which is a contradiction.

${\mbox{cod}}(G/N)={\mbox{cod}}(\textup{{U}}_{4}(2))$. We need only observe
that the smallest codegrees for this case are the same. Thus
$G/N\cong\textup{{U}}_{4}(2)$.  $\hbox to0.0pt{$\sqcap$\hss}\sqcup$

######  Lemma 3.3.

Let $G$ be a finite group with
${\mbox{cod}}(G)={\mbox{cod}}(\textup{{S}}_{4}(q))$, $q>4$ is even. If $N$ is
a maximal normal subgroup of $G$, then $G/N\cong\textup{{S}}_{4}(q)$.

###### Proof.

Let $N$ be a maximal normal subgroup of $G$. Similarly with above lemma, we
have that $G/N$ is a non-abelian simple group and $|{\mbox{cod}}(G/N)|$ is
$4,5,6,7,8,9,10,11,12$, or $13$. Next we analyze the cases separately.

$|{\mbox{cod}}(G/N)|=4$. By Lemma $2.3,$ Then $G/N\cong{\mbox{PSL}}(2,k)$
where $k=2^{f}\geq 4$ and ${\mbox{cod}}(G/N)=\\{1,k(k-1),k(k+1),k^{2}-1\\}$.
If $q$ is even, then $k^{2}-1=(q^{2}+1)(q-1)^{2}(q+1)^{2}$, as they are the
only nontrivial odd codegree of each set. If $k+1\mid(q-1)^{2}(q+1)^{2}$, then
$q^{2}+1\mid k-1$. Clearly $(q-1)^{2}(q+1)^{2}-(q^{2}+1)\neq 2$, this implies
that $k+1-(k-1)\neq 2$, it is a contradiction. If $k-1\mid(q-1)^{2}(q+1)^{2}$,
then $q^{2}+1\mid k+1$. Obviously $(q-1)^{2}(q+1)^{2}>q^{2}+1$, this implies
$k-1>k+1$, a contradiction.

$|{\mbox{cod}}(G/N)|=5$. Then $G/N\cong{\mbox{PSL}}(2,k)$ where $k$ is an odd
prime power and
${\mbox{cod}}(G/N)=\left\\{1,\frac{k(k-1)}{2},\frac{k(k+1)}{2},\frac{k^{2}-1}{2},k(k-\epsilon(k))\right\\}$
where $\epsilon(k)=(-1)^{(k-1)/2}$. Then $k(k-\epsilon(k))/2$,
$k(k-\epsilon(k))\in{\mbox{cod}}(G)$, a contradiction for we can’t find a
codegree which is the half of another codegree in
$|{\mbox{cod}}(\textup{{S}}_{4}(q))|$.

$|{\mbox{cod}}(G/N)|=6$. If $G/N\cong{}^{2}B_{2}(q)$, $q$ is even, then
$(s-1)(s^{2}+1)=(q^{2}+1)(q-1)^{2}(q+1)^{2}$ since these are the only
nontrivial odd codegrees. If $s-1\mid(q-1)^{2}(q+1)^{2}$, then $q^{2}+1\mid
s^{2}+1$. This implies $q$ is not an integer, a contradiction. If
$s^{2}+1\mid(q-1)^{2}(q+1)^{2}$, then $q^{2}+1\mid s-1$, we have that $q$ is
not an integer, a contradiction.

If $G/N\cong{\mbox{PSL}}(3,4)$, then ${\mbox{cod}}(G/N)=\\{1,\ 2^{4}\cdot
3^{2}\cdot 7,\ 2^{6}\cdot 3^{2},\ 2^{6}\cdot 5,\ 2^{6}\cdot 7,\ 3^{2}\cdot
5\cdot 7\\}$. If $q$ is even, it can be easily checked that $3^{2}\cdot 5\cdot
7=(q^{2}+1)(q-1)^{2}(q+1)^{2}$. It is a contradiction for $q>4$.

$|{\mbox{cod}}(G/N)|=7$. Then $G/N\cong{\mbox{PSL}}(3,3)$, $\textup{{A}}_{7}$,
$\textup{{M}}_{11}$ or $\textup{{J}}_{1}$. We need only to observe that none
of the codegree sets for this case are the same. Suppose
$G/N\cong{\mbox{PSL}}(3,3)$. Note that $3^{3}\cdot 13$ is the only nontrivial
odd codegree. If $q$ is even, $3^{3}\cdot 13=(q^{2}+1)(q-1)^{2}(q+1)^{2}$ is a
contradiction for $q>4$.

Suppose $G/N\cong\textup{{A}}_{7}$. Since every nontrivial codegree in
${\mbox{cod}}(\textup{{A}}(7))$ is even, then ${\mbox{cod}}(\textup{{A}}(7))$
is the subset of ${\mbox{cod}}(G)$ after deleting the unique nontrivial odd
codegree. If $q$ is even, then by comparing the 2-parts of the codegrees of
${\mbox{cod}}(\textup{{A}}(7))$ and ${\mbox{cod}}(G)$, we have a contradiction
since $q>4$.

Suppose $G/N\cong\textup{{M}}_{11}$. Then $G/N\cong\textup{{M}}_{11}$. If $q$
is even, then $3^{2}\cdot 5\cdot 11=(q^{2}+1)(q-1)^{2}(q+1)^{2}$, as they are
the only nontrivial odd codegree of each set. Note that $(3^{2},5\cdot 11)=1$.
If $5\cdot 11\mid(q-1)^{2}(q+1)^{2}$, then $q^{2}+1\mid 3^{2}$. It is a
contradiction since $q\in\mathbb{Z}$. If $3^{2}\mid(q-1)^{2}(q+1)^{2}$, then
$q^{2}+1\mid 5\cdot 11$. Obviously $(q-1)^{2}(q+1)^{2}>q^{2}+1$, this implies
$3^{2}>5\cdot 11$, a contradiction.

Suppose $G/N\cong\textup{{J}}_{1}$. Then $G/N\cong\textup{{J}}_{1}$. Since the
only nontrivial odd codegree in ${\mbox{cod}}\textup{{S}}_{4}(q))$ is
$(q^{2}+1)(q-1)^{2}(q+1)^{2}$. We need only note that
${\mbox{cod}}(\textup{{J}}_{1})$ has two nontrivial odd codegrees, it is a
contradiction.

$|{\mbox{cod}}(G/N)|=8$. If $G/N\cong{\mbox{PSL}}(3,s)$ with $4<s\not\equiv 1\
(\bmod 3)$. We see that $s^{3}(s+1)(s-1)^{2}$ the only nontrivial codegree
which is divided by other three codegrees in
${\mbox{cod}}({\mbox{PSL}}(3,s))$. In ${\mbox{cod}}(\textup{{S}}_{4}(q))$,
there are two nontrivial codegrees in this case, which are
$q^{4}(q+1)(q-1)^{2}$ and $q^{4}(q+1)^{2}(q-1)$. Then $s^{3}(s+1)(s-1)^{2}$
may be equal to $q^{4}(q+1)(q-1)^{2}$ or $q^{4}(q+1)^{2}(q-1)$. Furthermore
$s^{3}(s-1)^{2}=q^{4}(q+1)(q-1)$ or $q^{4}(q+1)^{2}$, as $s^{3}(s-1)$ is the
smallest codegree among three codegrees which divide same codegree in
${\mbox{cod}}({\mbox{PSL}}(3,s))$ and $q^{4}(q+1)(q-1)$ or $q^{4}(q+1)^{2}$
satisfies similar condition in ${\mbox{cod}}(G)$. This means
$q\notin\mathbb{Z}$, a contradiction.

If $G/N\cong{\mbox{PSU}}(3,s)$ with $4<s\not\equiv-1\ (\bmod\ 3)$. Similar
with the analysis above paragraph, we see that
$s^{3}(s+1)^{2}(s-1)=q^{4}(q+1)(q-1)^{2}$ or $q^{4}(q+1)^{2}(q-1)$. This means
$q\notin\mathbb{Z}$, a contradiction.

If $G/N\cong\textup{{G}}_{2}(2)^{\prime}$ we see $3^{3}\cdot
7=(q^{2}+1)(q+1)^{2}(q-1)^{2}$, as they are the only nontrivial odd codegrees
in each set. By setting this equality, we obtain no integer solution for $q$,
a contradiction.

$|{\mbox{cod}}(G/N)|=9$.

Suppose $G/N\cong{\mbox{PSL}}(3,s)$ with $4<s\equiv 1\ (\bmod\ 3)$. We have
$\frac{1}{3}s^{3}(s+1)(s-1)^{2}=q^{4}(q+1)(q-1)$ or $q^{4}(q+1)^{2}$. This
means $q\notin\mathbb{Z}$, a contradiction.

Suppose $G/N\cong{\mbox{PSU}}(3,q))$ with $4<q\equiv-1\ (\bmod 3)$. We have
$\frac{1}{3}s^{3}(s+1)^{2}(s-1)=q^{4}(q+1)(q-1)$ or $q^{4}(q+1)^{2}$. This
means $q\notin\mathbb{Z}$, a contradiction.

$|{\mbox{cod}}(G/N)|=10$. Then $G/N\cong\textup{{M}}_{22}$. We compare the
largest 2-part of each codegrees set and obtain that $q$ is not an integer, a
contradiction.

$|{\mbox{cod}}(G/N)|=11$. If $G/N\cong{\mbox{PSL}}(4,2)$, $\textup{{M}}_{12}$
or $\textup{{M}}_{23}$.

Suppose $G/N\cong{\mbox{PSL}}(4,2)$. It is easily checked that $3^{2}\cdot
5\cdot 7$ is the only nontrivial odd codegree in
${\mbox{cod}}({\mbox{PSL}}(4,2))$. Hence $3^{2}\cdot 5\cdot
7=(q^{2}+1)(q-1)^{2}(q+1)^{2}$. Note that $3^{2}\cdot 5\cdot 7=315$ is not
divisible by a square of prime, a contradiction.

If $G/N\cong{}^{2}\textup{{G}}_{2}(s)$, $s=3^{2t+1}$. We have that
$(q^{2}+1)(q-1)^{2}(q+1)^{2}$ equals to either $3^{5s+3}(s^{2}-s+1)$,
$s^{3}(s+1-3m)$, or $s^{3}(s+1+3m)$. This is a contradiction since $q$ is not
an integer.

$|{\mbox{cod}}(G/N)|=12$. Then $G/N\cong\textup{{S}}_{4}(4)$. By observation,
$s^{3}(s+1)(s-1)^{2}$ is the only nontrivial codegree which is divided by
other three codegrees in ${\mbox{cod}}(\textup{{S}}_{4}(4))$. In
${\mbox{cod}}(\textup{{S}}_{4}(q))$, there are two nontrivial codegrees in
this case, which are $q^{4}(q+1)(q-1)^{2}$ and $q^{4}(q+1)^{2}(q-1)$. Then
$s^{3}(s+1)(s-1)^{2}$ may be equal to $q^{4}(q+1)(q-1)^{2}$ or
$q^{4}(q+1)^{2}(q-1)$. Furthermore $2^{8}\cdot 5^{2}=q^{4}(q+1)(q-1)$ or
$q^{4}(q+1)^{2}$, as $2^{8}\cdot 5^{2}$ is the smallest codegree among three
codegrees which divide same codegree in ${\mbox{cod}}(\textup{{S}}_{4}(4))$
and $q^{4}(q+1)(q-1)$ or $q^{4}(q+1)^{2}$ satisfies similar condition in
${\mbox{cod}}(G)$. This means a contradiction for $q>4$.

$|{\mbox{cod}}(G/N)|=13$. We need only observe that $3^{4}\cdot
5=(q^{2}+1)(q+1)^{2}(q-1)^{2}$, as they are the only nontrivial odd codegrees
of each set. If $3^{4}\mid(q+1)^{2}(q-1)^{2}$, then $q^{2}+1\mid 5$, this
implies $q$ is not an integer, which is a contradiction. If
$5\mid(q+1)^{2}(q-1)^{2}$, then $q^{2}+1\mid 3^{4}$, this implies
$q\not\in\mathbb{Z}$, which is a contradiction.

Thus ${\mbox{cod}}(G/N)={\mbox{cod}}(\textup{{S}}_{4}(s))$ for $s=2^{t}>4$.
Comparing the smallest codegrees, we see $q=s$. Thus,
$G/N\cong\textup{{S}}_{4}(q)$.  $\hbox to0.0pt{$\sqcap$\hss}\sqcup$

######  Lemma 3.4.

Let $G$ be a finite group with
${\mbox{cod}}(G)={\mbox{cod}}(\textup{{U}}_{4}(3))$. If $N$ is a maximal
normal subgroup of $G$, then $G/N\cong\textup{{U}}_{4}(3)$.

###### Proof.

Let $N$ be a maximal normal subgroup of $G$. Then $G/N$ is a non-abelian
simple group and $|{\mbox{cod}}(G/N)|\leq 14$.

$|{\mbox{cod}}(G/N)|=4$. Then $G/N\cong{\mbox{PSL}}(2,k)$ where $k=2^{f}\geq
4$, and ${\mbox{cod}}(G/N)=\\{1,k(k-1),k(k+1),k^{2}-1\\}$. Thus
$k^{2}-1=3^{6}\cdot 7$ or $3^{6}\cdot 5$, since these are nontrivial odd
codegrees. It is a contradiction since $k\in\mathbb{Z}$.

$|{\mbox{cod}}(G/N)|=5$. Then $G/N\cong{\mbox{PSL}}(2,k)$ where $k$ is an odd
prime power and
${\mbox{cod}}(G/N)=\left\\{1,\frac{k(k-1)}{2},\frac{k(k+1)}{2},\frac{k^{2}-1}{2},k(k-\epsilon(k))\right\\}$
where $\epsilon(k)=(-1)^{(k-1)/2}$. So $k(k-\epsilon(k))/2$,
$k(k-\epsilon(k))\in{\mbox{cod}}(G)$. This implies that $\frac{k(k\pm
1)}{2}=2^{5}\cdot 3^{6}$ or $\frac{k(k\pm 1)}{2}=2^{4}\cdot 3^{6}$ since
$2^{4}\cdot 3^{6}$ is half of $2^{5}\cdot 3^{6}$, and $2^{3}\cdot 3^{6}$ is
half of $2^{4}\cdot 3^{6}$ in ${\mbox{cod}}(G)$. In either case,
$k\in\mathbb{Z}$ which is a contradiction.

$|{\mbox{cod}}(G/N)|=6$. By observation,
${\mbox{cod}}({\mbox{PSL}}(3,4))\not\subset{\mbox{cod}}(\textup{{U}}_{4}(3))$
since $3^{2}\cdot 5\cdot 7\notin{\mbox{cod}}(G)$.

Suppose ${\mbox{cod}}(G/N)={\mbox{cod}}({}^{2}B_{2}(q))$. We note that
$(q-1)(q^{2}+1)=3^{6}\cdot 7$ or $3^{6}\cdot 5$, since these are nontrivial
odd codegrees. We note that $3$ does not divide the order of the Suzuki
groups, a contradiction.

Suppose $|{\mbox{cod}}(G/N)|=7$. We note that the nontrivial odd codegree in
codegree sets of $\textup{{J}}_{1},\textup{{M}}_{11}$, and ${\mbox{PSL}}(3,3)$
are $3\cdot 5\cdot 11\cdot 19$, $3^{2}\cdot 5\cdot 11$ and $3^{3}\cdot 13$,
respectively. Also,
${\mbox{cod}}(\textup{{A}}_{7})\not\subset{\mbox{cod}}(\textup{{U}}_{4}(3))$
by an easy observation.

$|{\mbox{cod}}(G/N)|=8$.

If $G/N\cong{\mbox{PSU}}(3,q)$ with $4<q\not\equiv-1\ (\bmod\ 3)$ then if $q$
is odd we have $q^{3}(q^{2}-q+1)=3^{6}\cdot 7$ or $3^{6}\cdot 5$.This equation
means a contradiction since $q\in\mathbb{Z}$. If $q$ is even we have
$(q^{2}-q+1)(q+1)^{2}(q-1)=3^{6}\cdot 7$ or $3^{6}\cdot 5$. Then we have a
same contradiction since $q\in\mathbb{Z}$.

If $G/N\cong{\mbox{PSL}}(3,q)$ with $4<q\not\equiv 1\ (\bmod 3)$ and $q$ is
odd, then $q^{3}(q^{2}+q+1)=3^{6}\cdot 7$ or $3^{6}\cdot 5$. We have the same
contradiction with $q\in\mathbb{Z}$. If $q$ is even, then
$(q^{2}+q+1)(q^{2}-1)(q-1)=3^{6}\cdot 7$ or $3^{6}\cdot 5$. Then it is a
contradiction since Then we have a same contradiction since $q\in\mathbb{Z}$.

If $G/N\cong\textup{{G}}_{2}(2)^{\prime}$ we see
${\mbox{cod}}(\textup{{G}}_{2}(2)^{\prime})\not\subseteq{\mbox{cod}}(\textup{{U}}_{4}(3))$
since $3^{3}\cdot 7\notin{\mbox{cod}}(G)$.

$|{\mbox{cod}}(G/N)|=9$.

Suppose $G/N\cong{\mbox{PSL}}(3,q)$ with $4<q\equiv 1\ (\bmod\ 3)$. If $q$ is
odd, then $\frac{1}{3}q^{3}(q^{2}+q+1)=3^{6}\cdot 7$ or $3^{6}\cdot 5$. It is
a contradiction since $q\in\mathbb{Z}$. If $q$ is even, then
$\frac{1}{3}(q^{2}+q+1)(q+1)(q-1)^{2}=3^{6}\cdot 7$ or $3^{6}\cdot 5$. This
means $q\notin\mathbb{Z}$, a contradiction.

Suppose $G/N\cong{\mbox{PSU}}(3,q))$ with $4<q\equiv-1\ (\bmod 3)$. If $q$ is
odd, then $\frac{1}{3}q^{3}(q^{2}-q+1)=3^{6}\cdot 7$, it is a contradiction
since $q\in\mathbb{Z}$. If $q$ is even, then
$\frac{1}{3}(q^{2}-q+1)(q+1)^{2}(q-1)=3^{6}\cdot 7$ or $3^{6}\cdot 5$, which
means $q$ is not an integer, it is a contradiction.

$|{\mbox{cod}}(G/N)|=10$. We note that
${\mbox{cod}}(\textup{{M}}_{22})\not\subseteq{\mbox{cod}}(\textup{{U}}_{4}(3))$
since $2^{7}\cdot 3\cdot 5\cdot 11\notin{\mbox{cod}}(G)$.

$|{\mbox{cod}}(G/N)|=11$. We can checked that the codegree sets of
${\mbox{PSL}}(4,2),\textup{{M}}_{12}$ and $\textup{{M}}_{23}$ contain
$3^{2}\cdot 5\cdot 7$, $2^{2}\cdot 3\cdot 5\cdot 11$ and $3^{2}\cdot 5\cdot
11\cdot 23$ respectively, none of which are elements of the codegree set of
$\textup{{U}}_{4}(3)$. We note that $G/N$ cannot be isomorphic to
${\mbox{PSL}}(4,2)$, $\textup{{M}}_{12}$ or $\textup{{M}}_{23}$. It is a
contradiction.

If $G/N\cong{}^{2}\textup{{G}}_{2}(3^{2f+1})$, then by comparing the codegree
sets, we have that $3^{6}\cdot 7$ or $3^{6}\cdot 5$ can equal to
$3^{5f+3}(q^{2}-q+1)$, $q^{3}(q+1-3m)$ or $q^{3}(q+1+3m)$, since they are the
nontrivial odd codegrees in each set, it is a contradiction since $q\geq 27$.

$|{\mbox{cod}}(G/N)|=12$. Then $G/N\cong\textup{{S}}_{4}(4)$. We note that
$3^{2}\cdot 5^{2}\cdot 17\notin{\mbox{cod}}(G)$. Hence
${\mbox{cod}}(G/N)\not\subseteq{\mbox{cod}}(G)$, it is a contradiction.

$|{\mbox{cod}}(G/N)|=13$. Then
${\mbox{cod}}(G/N)={\mbox{cod}}(\textup{{U}}_{4}(2))$ or
${\mbox{cod}}(\textup{{S}}_{4}(q))$, for $q=2^{f}>4$.

Suppose ${\mbox{cod}}(G/N)={\mbox{cod}}(\textup{{U}}_{4}(2))$. We need only
observe that $3^{4}\cdot 5\notin{\mbox{cod}}(G)$ which means
${\mbox{cod}}(G/N)\not\subseteq{\mbox{cod}}(G)$. It is a contradiction.

Suppose ${\mbox{cod}}(G/N)={\mbox{cod}}(\textup{{S}}_{4}(q))$, for
$q=2^{f}>4$. Note that $(q^{2}+1)(q+1)^{2}(q-1)^{2}=3^{6}\cdot 7$ or
$3^{6}\cdot 5$, since they are nontrivial odd codegrees in each set. This has
no integer solution. A contradiction.

$|{\mbox{cod}}(G/N)|=14$. It is easily checked that neither $3^{3}\cdot
5^{2}\cdot 13$ nor $3^{5}\cdot 5\cdot 17$ is in ${\mbox{cod}}(G)$. Thus
${\mbox{cod}}({{}^{2}\textup{{F}}_{4}(2){}}^{\prime})\not\subseteq{\mbox{cod}}(G)$
and ${\mbox{cod}}(\textup{{J}}_{3})\not\subseteq{\mbox{cod}}(G)$.

We have that $G/N\cong\textup{{U}}_{4}(3)$.  $\hbox
to0.0pt{$\sqcap$\hss}\sqcup$

######  Lemma 3.5.

Let $G$ be a finite group with
${\mbox{cod}}(G)={\mbox{cod}}(^{2}\textup{{F}}_{4}(2){}^{\prime})$. If $N$ is
a maximal normal subgroup, then
$G/N\cong{{}^{2}\textup{{F}}_{4}(2){}}^{\prime}$.

###### Proof.

Let $N$ be a maximal normal subgroup of $G$. Then $G/N$ is a non-abelian
simple group and $|{\mbox{cod}}(G/N)|$ is either $4,5,6,7,8,9,10,11,12,13$ or
$14$.

$|{\mbox{cod}}(G/N)|=4$. Then $G/N\cong{\mbox{PSL}}(2,k)$ where $k=2^{f}\geq
4$, and ${\mbox{cod}}(G/N)=\\{1,k(k-1),k(k+1),k^{2}-1\\}$. Thus
$k^{2}-1=3^{3}\cdot 5^{2}\cdot 13$ since these are the only nontrivial odd
codegrees. It is a contradiction since $k\in\mathbb{Z}$.

$|{\mbox{cod}}(G/N)|=5$. Then $G/N\cong{\mbox{PSL}}(2,k)$ where $k$ is an odd
prime power and
${\mbox{cod}}(G/N)=\left\\{1,\frac{k(k-1)}{2},\frac{k(k+1)}{2},\frac{k^{2}-1}{2},k(k-\epsilon(k))\right\\}$
where $\epsilon(k)=(-1)^{(k-1)/2}$. So $k(k-\epsilon(k))/2$,
$k(k-\epsilon(k))\in{\mbox{cod}}(G)$. This implies that $\frac{k(k\pm
1)}{2}=2^{10}\cdot 3^{3}$, $2^{9}\cdot 3^{3}$, or $\frac{k(k\pm
1)}{2}=2^{10}\cdot 5^{2}$ since they are half of another codegree
respectively. In either case, $k\not\in\mathbb{Z}$ which is a contradiction.

$|{\mbox{cod}}(G/N)|=6$. By observation,
${\mbox{cod}}({\mbox{PSL}}(3,4))\not\subset{\mbox{cod}}(^{2}\textup{{F}}_{4}(2){}^{\prime})$
since $3^{2}\cdot 5\cdot 7\notin{\mbox{cod}}(G)$.

Suppose ${\mbox{cod}}(G/N)={\mbox{cod}}({}^{2}B_{2}(q))$. We note that
$(q-1)(q^{2}+1)=3^{3}\cdot 5^{2}\cdot 13$ since these are the only nontrivial
odd codegrees. We note that $3$ does not divide the order of the Suzuki
groups, a contradiction.

$|{\mbox{cod}}(G/N)|=7$. Then $G/N\cong{\mbox{PSL}}(3,3)$, $\textup{{A}}_{7}$,
$\textup{{M}}_{11}$ or $\textup{{J}}_{1}$. We note that the nontrivial odd
codegree in codegree sets of $\textup{{J}}_{1},\textup{{M}}_{11}$,
$\textup{{A}}_{7}$ and ${\mbox{PSL}}(3,3)$ are $3\cdot 5\cdot 11\cdot 19$,
$3^{2}\cdot 5\cdot 11$, $2^{2}\cdot 3\cdot 5\cdot 7$ and $3^{3}\cdot 13$,
respectively. It is a contradiction.

$|{\mbox{cod}}(G/N)|=8$.

If $G/N\cong{\mbox{PSU}}(3,q)$ with $4<q\not\equiv-1\ (\bmod\ 3)$ then if $q$
is odd we have $q^{3}(q^{2}-q+1)=3^{3}\cdot 5^{2}\cdot 13$. Thus it does not
yield an integer root for $q$, a contradiction. If $q$ is even we have
$(q^{2}-q+1)(q+1)^{2}(q-1)=3^{3}\cdot 5^{2}\cdot 13$. This also does not yield
an integer root for $q$. This is a contradiction.

If $G/N\cong{\mbox{PSL}}(3,q)$ with $4<q\not\equiv 1\ (\bmod\ 3)$ and $q$ is
odd, then $q^{3}(q^{2}+q+1)=3^{3}\cdot 5^{2}\cdot 13$. We have the same
contradiction with $q^{3}$. If $q$ is even, then
$(q^{2}+q+1)(q^{2}-1)(q-1)=3^{3}\cdot 5^{2}\cdot 13$. It is a contradiction
since $q\in\mathbb{Z}$.

If $G/N\cong\textup{{G}}_{2}(2)^{\prime}$ we see
${\mbox{cod}}(\textup{{G}}_{2}(2)^{\prime})\not\subseteq{\mbox{cod}}({{}^{2}\textup{{F}}_{4}(2){}}^{\prime})$
since $3^{3}\cdot 7\notin{\mbox{cod}}(G)$.

$|{\mbox{cod}}(G/N)|=9$. Suppose $G/N\cong{\mbox{PSL}}(3,q)$ with $4<q\equiv
1\ (\bmod\ 3)$. If $q$ is odd, then $\frac{1}{3}q^{3}(q^{2}+q+1)=3^{3}\cdot
5^{2}\cdot 13$. It is a contradiction since $q$ is not an integer. If $q$ is
even, then $\frac{1}{3}(q^{2}+q+1)(q+1)(q-1)^{2}=3^{3}\cdot 5^{2}\cdot 13$. It
is a same contradiction since $q$ is an integer.

Suppose $G/N\cong{\mbox{PSU}}(3,q))$ with $4<q\equiv-1\ (\bmod 3)$. If $q$ is
odd, then $\frac{1}{3}q^{3}(q^{2}-q+1)=3^{3}\cdot 5^{2}\cdot 13$, it is a
contradiction since $q\in\mathbb{Z}$. If $q$ is even, then
$\frac{1}{3}(q^{2}-q+1)(q+1)^{2}(q-1)=3^{3}\cdot 5^{2}\cdot 13$. It is a
contradiction since $q$ in not an integer.

$|{\mbox{cod}}(G/N)|=10$. We note that
${\mbox{cod}}(\textup{{M}}_{22})\not\subseteq{\mbox{cod}}({{}^{2}\textup{{F}}_{4}(2){}}^{\prime})$
for $2^{7}\cdot 3\cdot 5\cdot 11\notin{\mbox{cod}}(G)$. A contradiction.

$|{\mbox{cod}}(G/N)|=11$. By observation, the codegree sets of
${\mbox{PSL}}(4,2),\textup{{M}}_{12}$ and $\textup{{M}}_{23}$ contain
$3^{2}\cdot 5\cdot 7$, $2^{2}\cdot 3\cdot 5\cdot 11$ and $3^{2}\cdot 5\cdot
11\cdot 23$ respectively, none of which are elements of the codegree set of
${{}^{2}\textup{{F}}_{4}(2){}}^{\prime}$. It is easily checked that
${\mbox{cod}}(G/N)\not\subseteq{\mbox{cod}}(G)$. We note by observation that
$G/N$ cannot be isomorphic to ${\mbox{PSL}}(4,2)$, $\textup{{M}}_{12}$ or
$\textup{{M}}_{23}$.

If $G/N\cong{}^{2}\textup{{G}}_{2}(3^{2f+1})$, then by comparing the
nontrivial odd codegree in each set, we have a contradiction since $q\geq 27$.

$|{\mbox{cod}}(G/N)|=12$. Then $G/N\cong\textup{{S}}_{4}(4)$. We note that
$3^{2}\cdot 5^{2}\cdot 17\notin{\mbox{cod}}(G)$. Hence
${\mbox{cod}}(G/N)\not\subseteq{\mbox{cod}}(G)$, it is a contradiction.

$|{\mbox{cod}}(G/N)|=13$. Then
${\mbox{cod}}(G/N)={\mbox{cod}}(\textup{{U}}_{4}(2))$ or
$\textup{{S}}_{4}(q)$, for $q=2^{f}>4$.

Suppose ${\mbox{cod}}(G/N)={\mbox{cod}}(\textup{{U}}_{4}(2))$. We need only
observe that $3^{4}\cdot 5\notin{\mbox{cod}}(G)$ which means
${\mbox{cod}}(G/N)\not\subseteq{\mbox{cod}}(G)$. It is a contradiction.

Suppose ${\mbox{cod}}(G/N)={\mbox{cod}}(\textup{{S}}_{4}(q))$, for
$q=2^{f}>4$. We have that $(q^{2}+1)(q+1)^{2}(q-1)^{2}=3^{3}\cdot 5^{2}\cdot
13$, as they are the only nontrivial odd codegrees in each set. It implies
$q\notin\mathbb{Z}$, a contradiction.

$|{\mbox{cod}}(G/N)|=14$. It is easily checked that neither $3^{6}\cdot 7$ nor
$3^{5}\cdot 5\cdot 17$ is in ${\mbox{cod}}(G)$. Thus
${\mbox{cod}}(\textup{{U}}_{4}(3))\not\subseteq{\mbox{cod}}(G)$ nor
${\mbox{cod}}(\textup{{J}}_{3})\not\subseteq{\mbox{cod}}(G)$. We have that
$G/N\cong{{}^{2}\textup{{F}}_{4}(2){}}^{\prime}$.  $\hbox
to0.0pt{$\sqcap$\hss}\sqcup$

######  Lemma 3.6.

Let $G$ be a finite group with
${\mbox{cod}}(G)={\mbox{cod}}(\textup{{J}}_{3})$. If $N$ is a maximal normal
subgroup of $G$, then $G/N\cong\textup{{J}}_{3}$.

###### Proof.

Let $N$ be a maximal normal subgroup of $G$. Then $G/N$ is a non-abelian
simple group and $|{\mbox{cod}}(G/N)|$ is either $4,5,6,7,8,9,10,11,12,13$ or
$14$.

$|{\mbox{cod}}(G/N)|=4$. Then, $G/N\cong{\mbox{PSL}}(2,k)$ where $k=2^{f}\geq
4$ and ${\mbox{cod}}(G/N)=\\{1,k(k-1),k(k+1),k^{2}-1\\}$. Hence
$k^{2}-1=3^{4}\cdot 17\cdot 19$ or $3^{5}\cdot 5\cdot 17$ since these are the
nontrivial odd codegrees. In either case $k$ is not an integer which is a
contradiction.

$|{\mbox{cod}}(G/N)|=5$. Then $G/N\cong{\mbox{PSL}}(2,k)$ where $k$ is an odd
prime power and
${\mbox{cod}}(G/N)=\left\\{1,\frac{k(k-1)}{2},\frac{k(k+1)}{2},\frac{k^{2}-1}{2},k(k-\epsilon(k))\right\\}$
where $\epsilon(k)=(-1)^{(k-1)/2}$. Then $k(k-\epsilon(k))/2$,
$k(k-\epsilon(k))\in{\mbox{cod}}(G)$, and this implies that $\frac{k(k\pm
1)}{2}=2^{6}\cdot 3^{5}\cdot 5$ since $2^{6}\cdot 3^{5}\cdot 5$ is the only
element of ${\mbox{cod}}(G)$ that is half of $2^{7}\cdot 3^{5}\cdot 5$. In
either case, $k\not\in\mathbb{Z}$ which is a contradiction.

$|{\mbox{cod}}(G/N)|=6$. By observation,
${\mbox{cod}}({\mbox{PSL}}(3,4))\not\subset{\mbox{cod}}(\textup{{J}}_{3})$ for
$3^{2}\cdot 5\cdot 7\notin{\mbox{cod}}(G)$.

Suppose ${\mbox{cod}}(G/N)={\mbox{cod}}({}^{2}B_{2}(q))$. We note that
$(q-1)(q^{2}+1)=3^{4}\cdot 17\cdot 19$ or $3^{5}\cdot 5\cdot 17$ since these
are the nontrivial odd codegrees. We note that $3$ does not divide the order
of the Suzuki groups, a contradiction.

$|{\mbox{cod}}(G/N)|=7$. We note that $\textup{{M}}_{11}$, and
${\mbox{PSL}}(3,3)$ can’t be $3^{2}\cdot 5\cdot 11\cdot 23$. That the
nontrivial odd codegree $3^{2}\cdot 5\cdot 11$, and $3^{3}\cdot 13$ in the
codegree set ${\mbox{cod}}(\textup{{M}}_{11})$ and
${\mbox{cod}}({\mbox{PSL}}(3,3))$ is not in ${\mbox{cod}}(G)$. Also,
${\mbox{cod}}(\textup{{A}}_{7})\not\subseteq{\mbox{cod}}(G)$ and
${\mbox{cod}}(\textup{{J}}_{1})\not\subseteq{\mbox{cod}}(G)$ by an easy
observation since $2^{2}\cdot 3\cdot 5\cdot 7$ nor $3\cdot 5\cdot 11\cdot
19\notin{\mbox{cod}}(G)$.

$|{\mbox{cod}}(G/N)|=8$.

If $G/N\cong{\mbox{PSU}}(3,q)$ with $4<q\not\equiv-1\ (\bmod\ 3)$ then if $q$
is odd we have $q^{3}(q^{2}-q+1)=3^{5}\cdot 5\cdot 17$ or $3^{4}\cdot 17\cdot
19$. Since no cube of a prime divides $3^{5}\cdot 5\cdot 17$ or $3^{4}\cdot
17\cdot 19$, we have a contradiction. If $q$ is even we have
$(q^{2}-q+1)(q+1)^{2}(q-1)=3^{5}\cdot 5\cdot 17$, then $(q+1)^{2}=3^{5}$. This
is a contradiction since $q\in\mathbb{Z}$. Or
$(q^{2}-q+1)(q+1)^{2}(q-1)=3^{4}\cdot 17\cdot 19$. It is a contradiction since
$q$ is not an integer.

If $G/N\cong{\mbox{PSL}}(3,q)$ with $4<q\not\equiv 1\ (\bmod 3)$ and $q$ is
odd, then $q^{3}(q^{2}+q+1)=3^{5}\cdot 5\cdot 17$ or $3^{4}\cdot 17\cdot 19$.
We have the same contradiction with $q$. If $q$ is even, then
$(q^{2}+q+1)(q^{2}-1)(q-1)=3^{5}\cdot 5\cdot 17$, which means
$q\notin\mathbb{Z}$, a contradiction. Or $(q^{2}+q+1)(q^{2}-1)(q-1)=3^{4}\cdot
17\cdot 19$. We can get the same contradiction as above.

If $G/N\cong\textup{{G}}_{2}(2)^{\prime}$ we see
${\mbox{cod}}(\textup{{G}}_{2}(2)^{\prime})\not\subseteq{\mbox{cod}}(\textup{{J}}_{3})$
since $3^{3}\cdot 7\notin{\mbox{cod}}(G)$. This is a contradiction.

$|{\mbox{cod}}(G/N)|=9$.

Suppose $G/N\cong{\mbox{PSL}}(3,q)$ with $4<q\equiv 1\ (\bmod\ 3)$. If $q$ is
odd, then $\frac{1}{3}q^{3}(q^{2}+q+1)=3^{5}\cdot 5\cdot 17$. This does not an
integer solution for $q$. It is a contradiction since. Or
$\frac{1}{3}q^{3}(q^{2}+q+1)=3^{4}\cdot 17\cdot 19$. It is a same
contradiction since $q\in\mathbb{Z}$. If $q$ is even. Then
$\frac{1}{3}(q^{2}+q+1)(q+1)(q-1)^{2}=3^{5}\cdot 5\cdot 17$. It is a
contradiction since $q\in\mathbb{Z}$. Or
$frac{1}{3}(q^{2}+q+1)(q+1)(q-1)^{2}=3^{4}\cdot 17\cdot 19$, it is a
contradiction since $q\in\mathbb{Z}$.

Suppose $G/N\cong{\mbox{PSU}}(3,q))$ with $4<q\equiv-1\ (\bmod 3)$. We have
the same contradiction with $q$ as above. If $q$ is even, then
$\frac{1}{3}(q^{2}-q+1)(q+1)^{2}(q-1)=3^{5}\cdot 5\cdot 17$. It is the same
contradiction since $q\in\mathbb{Z}$. Or
$\frac{1}{3}(q^{2}-q+1)(q+1)^{2}(q-1)=3^{4}\cdot 17\cdot 19$, it is a same
contradiction as above.

$|{\mbox{cod}}(G/N)|=10$. We note that
${\mbox{cod}}(\textup{{M}}_{22})\not\subseteq{\mbox{cod}}(\textup{{J}}_{3})$
since $2^{7}\cdot 3\cdot 5\cdot 11\notin{\mbox{cod}}(G)$.

$|{\mbox{cod}}(G/N)|=11$.

We note by observation that $G/N$ cannot be isomorphic to ${\mbox{PSL}}(4,2)$,
$\textup{{M}}_{23}$ or $\textup{{M}}_{12}$. we have
$G/N\cong{\mbox{PSL}}(4,2),$ $G/N\cong\textup{{M}}_{12},$
$G/N\cong\textup{{M}}_{23}$ or $G/N\cong{}^{2}\textup{{G}}_{2}(q)$ for
$q=3^{2f+1},f\geq 1.$ The codegree sets of
${\mbox{PSL}}(4,2),\textup{{M}}_{12}$ and $\textup{{M}}_{23}$ contain
$3^{2}\cdot 5\cdot 7$, $2^{2}\cdot 3\cdot 5\cdot 11$ and $3^{2}\cdot 5\cdot
11\cdot 23$ respectively, none of which are elements of the codegree set of
$\textup{{J}}_{3}.$

If $G/N\cong{}^{2}\textup{{G}}_{2}(3^{2f+1})$, then by comparing the
nontrivial odd codegree in each set, we can’t get an integer root for $q$
which is a contradiction.

$|{\mbox{cod}}(G/N)|=12$. Then $G/N\cong\textup{{S}}_{4}(4)$. We note that
$3^{2}\cdot 5^{2}\cdot 17\notin{\mbox{cod}}(G)$. Thus
${\mbox{cod}}(\textup{{S}}_{4}(4))\not\subseteq{\mbox{cod}}(G)$.

$|{\mbox{cod}}(G/N)|=13$. Then
${\mbox{cod}}(G/N)={\mbox{cod}}(\textup{{U}}_{4}(2))$ or
${\mbox{cod}}(\textup{{S}}_{4}(q))$, for $q=2^{f}>4$.

Suppose ${\mbox{cod}}(G/N)={\mbox{cod}}(\textup{{U}}_{4}(2))$. We have that
${\mbox{cod}}(G/N)\not\subseteq{\mbox{cod}}(G)$ by an easy observation since
$3^{4}\cdot 5\notin{\mbox{cod}}(G)$.

Suppose ${\mbox{cod}}(G/N)={\mbox{cod}}(\textup{{S}}_{4}(q))$, for
$q=2^{f}>4$. We have that $(q^{2}+1)(q+1)^{2}(q-1)^{2}$ can be equal to
$3^{5}\cdot 5\cdot 17$ or $3^{4}\cdot 17\cdot 19$. If
$(q^{2}+1)(q+1)^{2}(q-1)^{2}=3^{5}\cdot 5\cdot 17$, it is a contradiction
since $3^{5}\cdot 5\cdot 17$ is not divisible by a square of prime. If
$(q^{2}+1)(q+1)^{2}(q-1)^{2}=3^{4}\cdot 17\cdot 19$. We can’t get an integer
root for $q$. This implies a contradiction.

Suppose $|{\mbox{cod}}(G/N)|=14$. Neither $3^{6}\cdot 7$ nor $3^{3}\cdot
5^{2}\cdot 13\notin{\mbox{cod}}(G)$. We have that
${\mbox{cod}}(\textup{{U}}_{4}(3))\not\subseteq cod(G)$ nor
${\mbox{cod}}({{}^{2}\textup{{F}}_{4}(2){}}^{\prime})$.

Thus $G/N\cong\textup{{J}}_{3}$.  $\hbox to0.0pt{$\sqcap$\hss}\sqcup$

######  Lemma 3.7.

Let $G$ be a finite group with
${\mbox{cod}}(G)={\mbox{cod}}(\textup{{G}}_{2}(3))$. If $N$ is a maximal
normal subgroup of $G$, then $G/N\cong\textup{{G}}_{2}(3)$.

###### Proof.

Let $N$ be a maximal normal subgroup of $G$. Then $G/N$ is a non-abelian
simple group and $|{\mbox{cod}}(G/N)|$ is either $4,5,6,7,8,9,10,11,12,13,14$
or $15$.

$|{\mbox{cod}}(G/N)|=4$. Then, $G/N\cong{\mbox{PSL}}(2,k)$ where $k=2^{f}\geq
4$ and ${\mbox{cod}}(G/N)=\\{1,k(k-1),k(k+1),k^{2}-1\\}$. Hence
$k^{2}-1=3^{6}\cdot 7\cdot 13$, $3^{6}\cdot 7$ or $3^{6}\cdot 13$ since these
are the nontrivial odd codegrees. In either case $k$ is not an integer which
is a contradiction.

$|{\mbox{cod}}(G/N)|=5$. Then $G/N\cong{\mbox{PSL}}(2,k)$ where $k$ is an odd
prime power and
${\mbox{cod}}(G/N)=\left\\{1,\frac{k(k-1)}{2},\frac{k(k+1)}{2},\frac{k^{2}-1}{2},k(k-\epsilon(k))\right\\}$
where $\epsilon(k)=(-1)^{(k-1)/2}$. Then $k(k-\epsilon(k))/2$,
$k(k-\epsilon(k))\in{\mbox{cod}}(G)$, and this implies that $\frac{k(k\pm
1)}{2}=2^{5}\cdot 3^{6}$ or $\frac{k(k\pm 1)}{2}=2^{5}\cdot 3^{5}$ since these
are half of the other two elements. In either case, $k\not\in\mathbb{Z}$ which
is a contradiction.

$|{\mbox{cod}}(G/N)|=6$.

By observation,
${\mbox{cod}}({\mbox{PSL}}(3,4))\not\subset{\mbox{cod}}(\textup{{G}}_{2}(3))$
since $3^{2}\cdot 5\cdot 7\notin{\mbox{cod}}(G)$, a contradiction.

Suppose ${\mbox{cod}}(G/N)={\mbox{cod}}({}^{2}B_{2}(q))$. We note that
$(q-1)(q^{2}+1)=3^{6}\cdot 7\cdot 13$, $3^{6}\cdot 7$ or $3^{6}\cdot 13$ since
these are the nontrivial odd codegrees. We note that $3$ does not divide the
order of the Suzuki groups, a contradiction.

$|{\mbox{cod}}(G/N)|=7$. We note that $\textup{{M}}_{11}$, $\textup{{J}}_{1}$,
$\textup{{A}}_{7}$ and ${\mbox{PSL}}(3,3)$. None of these codegree sets are
included in ${\mbox{cod}}(G)$.

$|{\mbox{cod}}(G/N)|=8$.

If $G/N\cong{\mbox{PSU}}(3,q)$ with $4<q\not\equiv-1\ (\bmod\ 3)$, then if $q$
is odd we have $q^{3}(q^{2}-q+1)=3^{6}\cdot 7\cdot 13$, $3^{6}\cdot 7$ or
$3^{6}\cdot 13$. These equations can’t yield an integer root for $q$. We have
a contradiction. If $q$ is even, we have a contradiction for the same reason.

If $G/N\cong{\mbox{PSL}}(3,q)$ with $4<q\not\equiv 1\ (\bmod 3)$ and $q$ is
odd, then $q^{3}(q^{2}+q+1)=3^{6}\cdot 7\cdot 13$, $3^{6}\cdot 7$ or
$3^{6}\cdot 13$. We have that these equations don’t hold as above. If $q$ is
even, then $(q^{2}+q+1)(q^{2}-1)(q-1)=3^{6}\cdot 7\cdot 13$, $3^{6}\cdot 7$ or
$3^{6}\cdot 13$. We have the same contradiction.

If $G/N\cong\textup{{G}}_{2}(2)^{\prime}$ we see
${\mbox{cod}}(\textup{{G}}_{2}(2)^{\prime})\not\subseteq{\mbox{cod}}(\textup{{G}}_{2}(3))$
for $3^{3}\cdot 7\notin{\mbox{cod}}(G)$. This is a contradiction.

$|{\mbox{cod}}(G/N)|=9$.

Suppose $G/N\cong{\mbox{PSL}}(3,q)$ with $4<q\equiv 1\ (\bmod\ 3)$. If $q$ is
odd, then $\frac{1}{3}q^{3}(q^{2}+q+1)=3^{6}\cdot 7\cdot 13$, $3^{6}\cdot 7$
or $3^{6}\cdot 13$. This does not yield an integer root for $q$. It is a
contradiction. If $q$ is even. Then
$\frac{1}{3}(q^{2}+q+1)(q+1)(q-1)^{2}=3^{6}\cdot 7\cdot 13$, $3^{6}\cdot 7$ or
$3^{6}\cdot 13$. It is a same contradiction as above.

Suppose $G/N\cong{\mbox{PSU}}(3,q))$ with $4<q\equiv-1\ (\bmod 3)$. If $q$ is
odd, then $\frac{1}{3}q^{3}(q^{2}-q+1)=3^{6}\cdot 7\cdot 13$, $3^{6}\cdot 7$
or $3^{6}\cdot 13$. This does not yield an integer root for $q$, it is a
contradiction. If $q$ is even, then
$\frac{1}{3}(q^{2}-q+1)(q+1)^{2}(q-1)=3^{6}\cdot 7\cdot 13$, $3^{6}\cdot 7$ or
$3^{6}\cdot 13$. It is a same contradiction as above.

$|{\mbox{cod}}(G/N)|=10$. We note that
${\mbox{cod}}(\textup{{M}}_{22})\not\subseteq{\mbox{cod}}(\textup{{G}}_{2}(3))$
since $2^{7}\cdot 3\cdot 5\cdot 11\notin{\mbox{cod}}(G)$.

$|{\mbox{cod}}(G/N)|=11$. We have $G/N\cong{\mbox{PSL}}(4,2),$
$G/N\cong\textup{{M}}_{12},$ $G/N\cong\textup{{M}}_{23}$ or
$G/N\cong{}^{2}\textup{{G}}_{2}(q)$ for $q=3^{2f+1},f\geq 1$.

The codegree sets of ${\mbox{PSL}}(4,2),\textup{{M}}_{12}$ and
$\textup{{M}}_{23}$ contain $3^{2}\cdot 5\cdot 7$, $2^{2}\cdot 3\cdot 5\cdot
11$ and $3^{2}\cdot 5\cdot 11\cdot 23$ respectively, none of which are
elements of the codegree set of $\textup{{G}}_{2}(3)$.

If $G/N\cong{}^{2}\textup{{G}}_{2}(3^{2f+1})$, then by comparing the largest
$3$-parts, this does not yield an integer root for $q$. It is a contradiction.

$|{\mbox{cod}}(G/N)|=12$. Then $G/N\cong\textup{{S}}_{4}(4)$. We note that
$3^{2}\cdot 5^{2}\cdot 17\notin{\mbox{cod}}(G)$. Thus
${\mbox{cod}}(\textup{{S}}_{4}(4))\not\subseteq{\mbox{cod}}(G)$.

$|{\mbox{cod}}(G/N)|=13$. Then
${\mbox{cod}}(G/N)={\mbox{cod}}(\textup{{U}}_{4}(2))$ or
${\mbox{cod}}(\textup{{S}}_{4}(q))$, for $q=2^{f}>4$.

${\mbox{cod}}(G/N)={\mbox{cod}}(\textup{{U}}_{4}(2))$, we have that
${\mbox{cod}}(G/N)\not\subseteq{\mbox{cod}}(G)$ since $3^{4}\cdot
5\notin{\mbox{cod}}(G)$.

${\mbox{cod}}(G/N)={\mbox{cod}}(\textup{{S}}_{4}(q))$, for $q=2^{f}>4$. Again
by setting an equality between the nontrivial odd codegrees of each set, we
obtain no integer solution in these three cases, a contradiction.

$|{\mbox{cod}}(G/N)|=14$. Then
${\mbox{cod}}(G/N)={\mbox{cod}}(\textup{{U}}_{4}(3))$,
${{}^{2}\textup{{F}}_{4}(2){}}^{\prime}$ or ${\mbox{cod}}(\textup{{J}}_{3})$.

It is easily checked that neither $3^{6}\cdot 7$, $3^{3}\cdot 5^{2}\cdot 13$
nor $3^{5}\cdot 5\cdot 17$ is in ${\mbox{cod}}(G)$. Thus
${\mbox{cod}}(\textup{{U}}_{4}(3))\not\subseteq{\mbox{cod}}(G)$,
${\mbox{cod}}({{}^{2}\textup{{F}}_{4}(2){}}^{\prime})\not\subseteq{\mbox{cod}}(G)$
nor ${\mbox{cod}}(\textup{{J}}_{3})\not\subseteq{\mbox{cod}}(G)$. This is a
contradiction.

$|{\mbox{cod}}(G/N)|=15$. We have that $G/N\cong\textup{{G}}_{2}(3)$ by an
easy observation.  $\hbox to0.0pt{$\sqcap$\hss}\sqcup$

######  Lemma 3.8.

Let $G$ be a finite group with
${\mbox{cod}}(G)={\mbox{cod}}(\textup{{A}}_{9}).$ If $N$ is the maximal normal
subgroup of $G,$ it we have $G/N\cong\textup{{A}}_{9}.$

###### Proof.

As $G/N$ is a non-abelian simple group,
${\mbox{cod}}(G/N)\subseteq{\mbox{cod}}(G)$, so we have
${\mbox{cod}}(G/N)=4,5,6,7,8,9,10,11,12,13,14,15$ or $16$.

We will show if $|{\mbox{cod}}(G/N)|\leq 16$ and if
$G/N\not\cong\textup{{A}}_{9}$, then
${\mbox{cod}}(G/N)\not\subseteq{\mbox{cod}}(G)$. Assume on the contrary that
there exists a maximal normal subgroup $N$ such that $|{\mbox{cod}}(G/N)|\leq
16$, and $G/N\not\cong\textup{{A}}_{9}$ and
${\mbox{cod}}(G/N)\subseteq{\mbox{cod}}(G)$. We analyze each case separately.

$|{\mbox{cod}}(G/N)|=4$. By Lemma 2.3, $G/N\cong{\mbox{PSL}}(2,k)$ where
$k=2^{f}\geq 4$ so $k^{2}-1\in{\mbox{cod}}(G/N)$. For all even $k$, we have
$k^{2}-1\equiv 1\pmod{2}$. However, there is no nontrivial odd codegree in
${\mbox{cod}}(\textup{{A}}_{9})$, contradicting
${\mbox{cod}}(G/N)\subseteq{\mbox{cod}}(G)$.

$|{\mbox{cod}}(G/N)|=5$. By Lemma 2.3, $G/N\cong{\mbox{PSL}}(2,k)$ where $k>5$
is the power of an odd prime number. This implies that $\displaystyle
k(k-\epsilon(k),\frac{k(k-\epsilon(k)))}{2}\in{\mbox{cod}}(G/N),$ where
$k=(-1)^{(k-1)/2}.$ We have that $\frac{k(k\pm 1)}{2}=2^{5}\cdot 3^{3}\cdot 5$
or $\frac{k(k\pm 1)}{2}=2^{3}\cdot 3^{4}\cdot 5$, since these are half of the
other two elements. Contradiction since $k\in\mathbb{Z}$.

$|{\mbox{cod}}(G/N)|=6$. By Lemma 2.3, $G/N\cong{\mbox{PSL}}(3,4)$ or
$G/N\cong{}^{2}B_{2}(q),$ with $q=2^{2f+1}$.

In the first case, notice that $3^{2}\cdot 5\cdot 7\in{\mbox{cod}}(G/N)$ is
the only nontrivial odd element, but there are no nontrivial odd elements in
${\mbox{cod}}(\textup{{A}}_{9})$, contradicting
${\mbox{cod}}(G/N)\subseteq{\mbox{cod}}(G)$.
In the second case, notice that
$(q-1)(q^{2}+1)\in{\mbox{cod}}({}^{2}B_{2}(q))$. As $q$ is even,
$(q-1)(q^{2}+1)\equiv 1\pmod{2}$. However, there are no nontrivial odd
elements in ${\mbox{cod}}(\textup{{A}}_{9})$, contradicting
${\mbox{cod}}(G/N)\subseteq{\mbox{cod}}(G)$.

$|{\mbox{cod}}(G/N)|=7$. By Lemma 2.3, we have
$G/N\cong{\mbox{PSL}}(3,3),\textup{{A}}_{7},\textup{{M}}_{11}$ or
$\textup{{J}}_{1}$. Notice that $3^{3}\cdot
13\in{\mbox{cod}}({\mbox{PSL}}(3,3))$, $2^{2}\cdot 3\cdot 5\cdot
7\in{\mbox{cod}}(\textup{{A}}_{7})$, $2^{3}\cdot 3^{2}\cdot
11\in{\mbox{cod}}(\textup{{M}}_{11})$ and $3\cdot 5\cdot 11\cdot
19\in{\mbox{cod}}(\textup{{J}}_{1})$. By inspection, we can see that none of
these elements are contained in ${\mbox{cod}}(\textup{{A}}_{9})$. Thus,
${\mbox{cod}}(G/N)\not\subseteq{\mbox{cod}}(G)$, a contradiction.

$|{\mbox{cod}}(G/N)|=8$. By Lemma 2.3, we have $G/N\cong{\mbox{PSU}}(3,q)$
with $4<q\not\equiv-1\pmod{3}$, $G/N\cong{\mbox{PSL}}(3,q)$ with
$4<q\not\equiv 1\pmod{3}$ or $G/N\cong\textup{{G}}_{2}(2)^{\prime}$. Observe
that the only nontrivial element not divisible by $3$ in
${\mbox{cod}}(\textup{{A}}_{9})$ is $2^{5}\cdot 5\cdot 7$.

In the first case, notice that
$(q^{2}-q+1)(q+1)^{2}(q-1),q^{3}(q^{2}-q+1)\in{\mbox{cod}}({\mbox{PSU}}(3,q))$.
If $q\equiv 1\mod 3$, then $q^{3}(q^{2}-q+1)\not\equiv 0\pmod{3}$, so it must
hold that $q^{3}(q^{2}-q+1)=2^{5}\cdot 5\cdot 7$. However, this does not yield
an integer root for $q$. Alternatively, if $q\equiv 0\pmod{3}$, then
$(q^{2}-q+1)(q+1)^{2}(q-1)\not\equiv 0\pmod{3}$, so it must hold that
$(q^{2}-q+1)(q+1)^{2}(q-1)=2^{5}\cdot 5\cdot 7$. This also does not yield an
integer root for $q$. Thus,
${\mbox{cod}}(G/N)\not\subseteq{\mbox{cod}}(G)={\mbox{cod}}(\textup{{A}}_{9})$,
a contradiction.

In the second case, notice that
$q^{3}(q^{2}+q+1)\in{\mbox{cod}}({\mbox{PSL}}(3,q)).$ As $q\not\equiv
1\pmod{3},q^{3}(q^{2}+q+1)\not\equiv 0\pmod{3}$, so we must have
$q^{3}(q^{2}+q+1)=2^{5}\cdot 5\cdot 7$. However, this does not an integer
solution for $q$. Thus,
${\mbox{cod}}(G/N)\not\subseteq{\mbox{cod}}(G)={\mbox{cod}}(\textup{{A}}_{9})$,
a contradiction.

In the final case, $3^{3}\cdot 7\in{\mbox{cod}}(\textup{{G}}_{2}(2)^{\prime})$
but $3^{3}\cdot 7\not\in{\mbox{cod}}(\textup{{A}}_{9})$, contradicting
${\mbox{cod}}(G/N)\subseteq{\mbox{cod}}(G)={\mbox{cod}}(\textup{{A}}_{9})$.

$|{\mbox{cod}}(G/N)|=9$. By Lemma 2.3, we have $G/N\cong{\mbox{PSL}}(3,q)$
with $4<q\equiv 1\pmod{3}$ or $G/N\cong{\mbox{PSU}}(3,q)$ with
$4<q\equiv-1\pmod{3}$.

In the first case, notice that
$\frac{1}{3}q^{3}(q^{2}+q+1)\in{\mbox{cod}}({\mbox{PSL}}(3,q))$. It can be
verified through a complete search that $q^{2}+q+1\not\equiv 0\pmod{9}$. Thus,
$\frac{1}{3}q^{3}(q^{2}+q+1)\not\equiv 0\pmod{3}$, so we must have
$\frac{1}{3}q^{3}(q^{2}+q+1)=2^{5}\cdot 5\cdot 7$. However, this yields no
integer roots for $q$, contradicting
${\mbox{cod}}(G/N)\subseteq{\mbox{cod}}(G)={\mbox{cod}}(\textup{{A}}_{9})$.

In the second case, notice that
$\frac{1}{3}q^{3}(q^{2}-q+1)\in{\mbox{cod}}({\mbox{PSU}}(3,q))$. It can be
verified through a complete search that $q^{2}-q+1\not\equiv 0\pmod{9}$, so
$\frac{1}{3}q^{3}(q^{2}-q+1)\not\equiv 0\pmod{3}$, so we must have
$\frac{1}{3}q^{3}(q^{2}-q+1)=2^{5}\cdot 5\cdot 7$. However, there are no
integer roots for $q$, contradicting
${\mbox{cod}}(G/N)\subseteq{\mbox{cod}}(G)={\mbox{cod}}(\textup{{A}}_{9})$.

$|{\mbox{cod}}(G/N)|=10$. By Lemma 2.3, we have $G/N\cong\textup{{M}}_{22}$.
However, $2^{7}\cdot 3\cdot 5\cdot 11$ is only present in
${\mbox{cod}}(\textup{{M}}_{22})$ and not in ${\mbox{cod}}(\textup{{A}}_{9})$,
contradicting ${\mbox{cod}}(G/N)\subseteq{\mbox{cod}}(G)$.

$|{\mbox{cod}}(G/N)|=11$. By Lemma 2.3, we have $G/N\cong{\mbox{PSL}}(4,2)$,
$G/N\cong\textup{{M}}_{12}$, $G/N\cong\textup{{M}}_{23}$ or
$G/N\cong{}^{2}\textup{{G}}_{2}(q)$ for $q=3^{2f+1},f\geq 1$. The codegree
sets of ${\mbox{PSL}}(4,2),\textup{{M}}_{12}$ and $\textup{{M}}_{23}$ contain
$3^{2}\cdot 5\cdot 7$, $2^{2}\cdot 3\cdot 5\cdot 11$ and $3^{2}\cdot 5\cdot
11\cdot 23$ respectively, none of which are elements of the codegree set of
$\textup{{A}}_{9}$.

Notice additionally that
$(q^{2}-1)(q^{2}-q+1)\in{\mbox{cod}}({}^{2}G_{2}(q))$. However,
$(q^{2}-1)(q^{2}-q+1)\not\equiv 0\pmod{3}$ when $q$ is a multiple of three.
This implies that we must have $(q^{2}-1)(q^{2}-q+1)=2^{5}\cdot 5\cdot 7$, yet
this does not yield an integer root for $q$. All of these cases contradicts
${\mbox{cod}}(G/N)\subseteq{\mbox{cod}}(G)={\mbox{cod}}(\textup{{A}}_{9})$.

$|{\mbox{cod}}(G/N)|=12$. By Lemma 2.3, we have $G/N\cong\textup{{S}}_{4}(4)$.
However, the element $3^{2}\cdot 5^{2}\cdot 17$ is the only nontrivial odd
codegree in the codegree set of $\textup{{S}}_{4}(4)$, it is not in the
codegree set of $\textup{{A}}_{9}$, contradicting
${\mbox{cod}}(G/N)\subseteq{\mbox{cod}}(G)$.

$|{\mbox{cod}}(G/N)|=13$. By Lemma 2.3, we have $G/N\cong\textup{{U}}_{4}(2)$
or $G/N\cong\textup{{S}}_{4}(q)$ for $q=2^{f}>4$.

In the first case, the element $3^{4}\cdot 5$ is only in the codegree set of
$\textup{{U}}_{4}(2)$ and not the codegree set of $\textup{{A}}_{9}$.

In the second case,
$2q^{3}(q-1)^{2}(q+1)^{2}\in{\mbox{cod}}(\textup{{S}}_{4}(q))$. However, if
$\nu_{2}(q)>2$, then $\nu_{2}(2q^{3}(q-1)^{2}(q+1)^{2})>7$, yet there does not
exist an element in ${\mbox{cod}}(\textup{{A}}_{9})$ that is divisible by
$2^{7}$. Both these cases contradict
${\mbox{cod}}(G/N)\subseteq{\mbox{cod}}(G)$.

$|{\mbox{cod}}(G/N)|=14$. By Lemma 2.3. If $G/N\cong\textup{{U}}_{4}(3)$. We
have that $3^{6}\cdot 5$ and $3^{6}\cdot 7$ are both the nontrivial odd
elements in ${\mbox{cod}}(\textup{{U}}_{4}(3))$, there are all the nontrivial
even elements in ${\mbox{cod}}(\textup{{A}}_{9})$, a contradiction. If
$G/N\cong{}^{2}\textup{{F}}_{4}(2)^{\prime}$ or $G/N\cong\textup{{J}}_{3}$.
the codegree sets of $\textup{{U}}_{4}(3),{}^{2}\textup{{F}}_{4}(2)^{\prime}$
and $\textup{{J}}_{3}$ contain $3^{3}\cdot 5^{2}\cdot 13$, $3^{4}\cdot 17\cdot
19$ respectively, none of which are elements of
${\mbox{cod}}(\textup{{A}}_{9})$. This contradicts
${\mbox{cod}}(G/N)\subseteq{\mbox{cod}}(G)$.

$|{\mbox{cod}}(G/N)|=15$. By Lemma 2.3, we have $G/N\cong\textup{{G}}_{2}(3)$.
However, there are three nontrivial odd elements in the codegree set of
$\textup{{G}}_{2}(3)$ and not in the codegree set of $\textup{{A}}_{9}$. This
contradicts ${\mbox{cod}}(G/N)\subseteq{\mbox{cod}}(G)$.

$|{\mbox{cod}}(G/N)|=16$. By Lemma 2.3, $G/N\cong\textup{{A}}_{9}$ or
$G/N\cong\textup{{J}}_{2}$.

Notice that $2^{6}\cdot 3^{3}\cdot 5^{2}\in{\mbox{cod}}(\textup{{J}}_{2})$ but
this element is not in ${\mbox{cod}}(\textup{{A}}_{9})$. This contradicts
${\mbox{cod}}(G/N)\subseteq{\mbox{cod}}(G)$.

Having ruled out all other cases for $G/N\equiv G^{\prime}$ with
${\mbox{cod}}(G^{\prime})\leq 16$, we conclude that
$G/N\cong\textup{{A}}_{9}$.  $\hbox to0.0pt{$\sqcap$\hss}\sqcup$

######  Lemma 3.9.

Let $G$ be a finite group with
${\mbox{cod}}(G)={\mbox{cod}}(\textup{{J}}_{2})$. If $N$ is the maximal normal
subgroup of $G$, it we have $G/N\cong\textup{{J}}_{2}$.

###### Proof.

As $G/N$ is a non-abelian simple group,
${\mbox{cod}}(G/N)\subseteq{\mbox{cod}}(G)$, so we have that
${\mbox{cod}}(G/N)=4,5,6,7,8,9,10,11,12,13,14,15$ or $16$.

Assume on the contrary that there exist a maximal normal subgroup $N$ such
that $G/N\not\cong\textup{{J}}_{2},{\mbox{cod}}(G/N)\leq 16$ and
${\mbox{cod}}(G/N)\subseteq{\mbox{cod}}(G)$. We exhaust all possibilities to
show the previous statement is false.

$|{\mbox{cod}}(G/N)|=4$. By Lemma 2.3, $G/N\cong{\mbox{PSL}}(2,k)$ where
$k=2^{f}\geq 4,$ so $k^{2}-1\in{\mbox{cod}}(G/N)$. For all even $k$, $k^{2}-1$
is odd, yet there are no nontrivial odd elements in the codegree set of
$\textup{{J}}_{2}$. This contradicts
${\mbox{cod}}(G/N)\subseteq{\mbox{cod}}(G)$.

Suppose $|{\mbox{cod}}(G/N)|=5$. By Lemma 2.3, $G/N\cong{\mbox{PSL}}(2,k)$
where $k$ is a power of an odd prime number. This implies that $\displaystyle
k(k-\epsilon(k),\frac{k(k-\epsilon(k)))}{2}\in{\mbox{cod}}(G/N)$, where
$k=(-1)^{(k-1)/2}$. Then $k(k-\epsilon(k))/2$,
$k(k-\epsilon(k))\in{\mbox{cod}}(G)$, and we have that $\frac{k(k\pm
1)}{2}=2^{6}\cdot 3\cdot 5^{2}$ since it is half of the other element. This is
a contradiction since $k\in\mathbb{Z}$.

Suppose $|{\mbox{cod}}(G/N)|=6$. By Lemma 2.3, $G/N\cong{\mbox{PSL}}(3,4)$ or
$G/N\cong{}^{2}B_{2}(q)$, with $q=2^{2n+1}$.

If $G/N\cong{\mbox{PSL}}(3,4)$ then $3^{2}\cdot 5\cdot 7\in{\mbox{cod}}(G/N)$
is odd, but $3^{2}\cdot 5\cdot
7\not\in{\mbox{cod}}(G)={\mbox{cod}}(\textup{{J}}_{2})$, contradicting
${\mbox{cod}}(G/N)\subseteq{\mbox{cod}}(G)$.

In the second case, notice that
$(q-1)(q^{2}+1)\in{\mbox{cod}}({}^{2}B_{2}(q))$. As $q$ is even,
$(q-1)(q^{2}+1)\equiv 1\pmod{2}$. However, there are no nontrivial odd
elements in ${\mbox{cod}}(\textup{{J}}_{2})$, contradicting
${\mbox{cod}}(G/N)\subseteq{\mbox{cod}}(G)$.

Suppose $|{\mbox{cod}}(G/N)|=7$. By Lemma 2.3, we have
$G/N\cong{\mbox{PSL}}(3,3),\textup{{A}}_{7},\textup{{M}}_{11}$ or
$\textup{{J}}_{1}$. We may see that $3^{3}\cdot
13\in{\mbox{cod}}({\mbox{PSL}}(3,3))$, $2^{2}\cdot 3\cdot 5\cdot
7\in{\mbox{cod}}(\textup{{A}}_{7})$, $2^{3}\cdot 3^{2}\cdot
11\in{\mbox{cod}}(\textup{{M}}_{11})$ and $3\cdot 5\cdot 11\cdot
19\in{\mbox{cod}}(\textup{{J}}_{1})$. However, none of these elements are
elements of ${\mbox{cod}}(\textup{{J}}_{2})$. Thus
${\mbox{cod}}(G/N)\not\subseteq{\mbox{cod}}(G)$, a contradiction.

Suppose $|{\mbox{cod}}(G/N)|=8$. By Lemma 2.3, we have
$G/N\cong{\mbox{PSU}}(3,q)$ with $4<q\not\equiv-1\pmod{3}$,
$G/N\cong{\mbox{PSL}}(3,q)$ with $4<q\not\equiv 1\pmod{3}$ or
$G/N\cong\textup{{G}}_{2}(2)^{\prime}$. Observe that the only nontrivial
element not divisible by $3$ in ${\mbox{cod}}(\textup{{J}}_{2})$ is
$2^{7}\cdot 5^{2}$.

If $G/N\cong{\mbox{PSU}}(3,q)$, notice that
$(q^{2}-q+1)(q+1)^{2}(q-1),q^{3}(q^{2}-q+1)\in{\mbox{cod}}({\mbox{PSU}}(3,q))$.
If $q\equiv 1\mod 3,$ then $q^{3}(q^{2}-q+1)\not\equiv 0\pmod{3}$, so it must
hold that $q^{3}(q^{2}-q+1)=2^{7}\cdot 5^{2}$. However, this does not yield an
integer root for $q$. Alternatively, if $q\equiv 0\pmod{3}$, then
$(q^{2}-q+1)(q+1)^{2}(q-1)\not\equiv 0\pmod{3}$ so it must hold that
$(q^{2}-q+1)(q+1)^{2}(q-1)=2^{7}\cdot 5^{2}$. This also does not yield an
integer root for $q$. Thus
${\mbox{cod}}(G/N)\not\subseteq{\mbox{cod}}(G)={\mbox{cod}}(\textup{{J}}_{2})$,
a contradiction.

If $G/N\cong{\mbox{PSL}}(3,q)$, we have
$q^{3}(q^{2}+q+1)\in{\mbox{cod}}({\mbox{PSL}}(3,q))$. As $q\not\equiv
1\pmod{3},q^{3}(q^{2}+q+1)\not\equiv 0\pmod{3}$ so it must hold that
$q^{3}(q^{2}+q+1)=2^{7}\cdot 5^{2}$. However, this does not lead to an integer
solution for $q$. Thus
${\mbox{cod}}(G/N)\not\subseteq{\mbox{cod}}(G)={\mbox{cod}}(\textup{{J}}_{2})$,
a contradiction.

In the final case, $3^{3}\cdot 7\in{\mbox{cod}}(\textup{{G}}_{2}(2)^{\prime})$
but $3^{3}\cdot 7\not\in{\mbox{cod}}(\textup{{J}}_{2})$, contradicting
${\mbox{cod}}(G/N)\subseteq{\mbox{cod}}(G)={\mbox{cod}}(\textup{{J}}_{2})$.

$|{\mbox{cod}}(G/N)|=9$. By Lemma 2.3, we have $G/N\cong{\mbox{PSL}}(3,q)$
with $4<q\equiv 1\pmod{3}$ or $G/N\cong{\mbox{PSU}}(3,q)$ with
$4<q\equiv-1\pmod{3}$.

Observe that by a complete search of remainders modulo $9$ that both
$q^{2}-q+1$ and $q^{2}+q+1$ cannot be divisible by $9$.

In the first scenario, we have
$\frac{1}{3}q^{3}(q^{2}+q+1)\in{\mbox{cod}}({\mbox{PSL}}(3,q))$. From the
above remark, notice that $\frac{1}{3}q^{3}(q^{2}+q+1)\not\equiv 0\pmod{3}$.
Therefore, we must have $\frac{1}{3}q^{3}(q^{2}+q+1)=2^{7}\cdot 5^{2}$, which
does not yield an integer root for $q$, a contradiction.

In the second scenario, we have
$\frac{1}{3}q^{3}(q^{2}-q+1)\in{\mbox{cod}}({\mbox{PSU}}(3,q))$. As with the
first scenario, this element is not divisible by $3$, so we must have
$\frac{1}{3}q^{3}(q^{2}-q+1)=2^{7}\cdot 5^{2}$. However, this does not yield
an integer root for $q$, a contradiction.

$|{\mbox{cod}}(G/N)|=10$. By Lemma 2.3, we have $G/N\cong\textup{{M}}_{22}$.
However, $2^{7}\cdot 3\cdot 5\cdot 11$ is only present in
${\mbox{cod}}(\textup{{M}}_{22})$ and not in ${\mbox{cod}}(\textup{{J}}_{2})$,
contradicting ${\mbox{cod}}(G/N)\subseteq{\mbox{cod}}(G)$.

$|{\mbox{cod}}(G/N)|=11$. By Lemma 2.3, we have $G/N\cong{\mbox{PSL}}(4,2)$,
$G/N\cong\textup{{M}}_{12}$, $G/N\cong\textup{{M}}_{23}$ or
$G/N\cong{}^{2}\textup{{G}}_{2}(q)$ for $q=3^{2f+1},f\geq 1$.

The codegree sets of ${\mbox{PSL}}(4,2)$, $\textup{{M}}_{12}$ and
$\textup{{M}}_{23}$ contain $3^{2}\cdot 5\cdot 7$, $2^{2}\cdot 3\cdot 5\cdot
11$ and $2^{7}\cdot 7\cdot 11$ respectively, none of which are elements of the
codegree set of $\textup{{J}}_{2}$.

If $G/N\cong{}^{2}\textup{{G}}_{2}(q)$,
$(q^{2}-1)(q^{2}-q+1)\in{\mbox{cod}}({}^{2}\textup{{G}}_{2}(q))$. However,
$(q^{2}-1)(q^{2}-q+1)\not\equiv 0\pmod{3}$ in this case, so we must have
$(q^{2}-1)(q^{2}-q+1)=2^{5}\cdot 5\cdot 7$, yet this does not yield an integer
root for $q$. This is a contradiction to
${\mbox{cod}}(G/N)\subseteq{\mbox{cod}}(G)={\mbox{cod}}(\textup{{J}}_{2})$.

$|{\mbox{cod}}(G/N)|=12$. By Lemma 2.3, we have $G/N\cong\textup{{S}}_{4}(4)$.
However, the element $2^{7}\cdot 5^{2}\cdot 17$ is only in the codegree set of
$\textup{{S}}_{4}(4)$ and not in the codegree set of $\textup{{J}}_{2}$
,contradicting ${\mbox{cod}}(G/N)\subseteq{\mbox{cod}}(G)$.

$|{\mbox{cod}}(G/N)|=13$. By Lemma 2.3, we have $G/N\cong\textup{{U}}_{4}(2)$
or $G/N\cong\textup{{S}}_{4}(q)$ for $q=2^{f}>4$.

In the first case, the element $3^{4}\cdot
5\in{\mbox{cod}}(\textup{{U}}_{4}(2))$ but it is not in
${\mbox{cod}}(\textup{{J}}_{2})$.

In the second case,
$2q^{3}(q-1)^{2}(q+1)^{2}\in{\mbox{cod}}(\textup{{S}}_{4}(q))$. However, as
$\nu_{2}(q)>2$, then $\nu_{2}(2q^{3}(q-1)^{2}(q+1)^{2})>7$ yet there does not
exist an element in ${\mbox{cod}}(\textup{{A}}_{9})$ that is divisible by
$2^{8}$. In both of these cases,
${\mbox{cod}}(G/N)\not\subseteq{\mbox{cod}}(G)$.

$|{\mbox{cod}}(G/N)|=14$. By Lemma 2.3, we have $G/N\cong\textup{{U}}_{4}(3)$,
$G/N\cong{}^{2}\textup{{F}}_{4}(2)^{\prime}$ or $G/N\cong\textup{{J}}_{3}$.
However, the codegree sets of
$\textup{{U}}_{4}(3),{}^{2}\textup{{F}}_{4}(2)^{\prime}$ and
$\textup{{J}}_{3}$ contain $2^{7}\cdot 3^{5}\cdot 5$, $2^{10}\cdot 3^{3}\cdot
5^{2}$ and $2^{7}\cdot 3^{5}\cdot 19$ respectively, none of which are elements
of ${\mbox{cod}}(\textup{{J}}_{2})$. This contradicts
${\mbox{cod}}(G/N)\subseteq{\mbox{cod}}(G)$.

$|{\mbox{cod}}(G/N)|=15$. By Lemma 2.3, we have $G/N\cong\textup{{G}}_{2}(3)$.
However, the element $2^{5}\cdot 3^{6}\cdot 13$ is only present in the
codegree set of $\textup{{G}}_{2}(3)$ and not in the codegree set of
$\textup{{J}}_{2}$, which contradicts
${\mbox{cod}}(G/N)\subseteq{\mbox{cod}}(G)$.

$|{\mbox{cod}}(G/N)|=16$. By Lemma 2.3, we have $G/N\cong\textup{{A}}_{9}$ or
$G/N\cong\textup{{J}}_{2}$. $2^{4}\cdot 3^{3}\cdot
5\in{\mbox{cod}}(\textup{{A}}_{9})$ but is not in
${\mbox{cod}}(\textup{{J}}_{2})$, contradicting the assumption.

Having ruled out all other cases, we conclude $G/N\cong\textup{{J}}_{2}$.
$\hbox to0.0pt{$\sqcap$\hss}\sqcup$

######  Lemma 3.10.

Let $G$ be a finite group with
${\mbox{cod}}(G)={\mbox{cod}}(\textup{{M}}^{c}L)$. If $N$ is the maximal
normal subgroup of $G$, it we have $G/N\cong\textup{{M}}^{c}L$.

###### Proof.

As $G/N$ is a non-abelian simple group,
${\mbox{cod}}(G/N)\subseteq{\mbox{cod}}(G)$, so we have that
${\mbox{cod}}(G/N)=4,5,6,7,8,9,10,11,12,13,14,15$, or $16$.

Assume on the contrary that there exist a maximal normal subgroup $N$ such
that $G/N\not\cong\textup{{M}}^{c}L,{\mbox{cod}}(G/N)\leq 17$ and
${\mbox{cod}}(G/N)\subseteq{\mbox{cod}}(G)$. We exhaust all possibilities to
show the previous statement is false.

$|{\mbox{cod}}(G/N)|=4$. By Lemma 2.3, $G/N\cong{\mbox{PSL}}(2,k)$ where
$k=2^{f}\geq 4$, so $k^{2}-1\in{\mbox{cod}}(G/N)$. For all even $k$, $k^{2}-1$
is odd. Therefore, $k^{2}-1$ must equal $3^{6}\cdot 5^{3}$ or $3^{6}\cdot
5^{3}\cdot 11$ but this yields no integer roots for $k$, contradicting
${\mbox{cod}}(G/N)\subseteq{\mbox{cod}}(G)$.

$|{\mbox{cod}}(G/N)|=5$. By Lemma 2.3, $G/N\cong{\mbox{PSL}}(2,k)$ where $k$
is a power of an odd prime number. This implies that $\displaystyle
k(k-\epsilon(k),\frac{k(k-\epsilon(k)))}{2}\in{\mbox{cod}}(G/N)$, where
$k=(-1)^{(k-1)/2}.$ However, there does not exist two elements
$a,b\in{\mbox{cod}}(\textup{{M}}^{c}L)$ with $a=2b,$ contradicting
${\mbox{cod}}(G/N)\subseteq{\mbox{cod}}(G)$.

$|{\mbox{cod}}(G/N)|=6$. By Lemma 2.3, $G/N\cong{\mbox{PSL}}(3,4)$ or
$G/N\cong{}^{2}B_{2}(q)$ with $q=2^{2n+1}$.

If $G/N\cong{\mbox{PSL}}(3,4)$, then $3^{2}\cdot 5\cdot 7\in{\mbox{cod}}(G/N)$
but $3^{2}\cdot 5\cdot
7\not\in{\mbox{cod}}(G)={\mbox{cod}}(\textup{{M}}^{c}L)$, contradicting
${\mbox{cod}}(G/N)\subseteq{\mbox{cod}}(G)$.

Furthermore, notice that $(q-1)(q^{2}+1)\in{\mbox{cod}}({}^{2}B_{2}(q))$. As
$q$ is even, $(q-1)(q^{2}+1)\equiv 1\pmod{2}$. This means that
$(q-1)(q^{2}+1)=3^{6}\cdot 5^{3}$ or $3^{6}\cdot 5^{3}\cdot 11$, which does
not yield any roots. This contradicts
${\mbox{cod}}(G/N)\subseteq{\mbox{cod}}(G)$.

$|{\mbox{cod}}(G/N)|=7$. This implies that
$G/N\cong{\mbox{PSL}}(3,3),\textup{{M}}_{11},\textup{{A}}_{7}$ or
$\textup{{J}}_{1}$. By inspection, the elements $2^{2}\cdot 3^{2}\cdot 13$,
$2^{3}\cdot 3^{3}\cdot 11$, $2^{2}\cdot 3\cdot 5\cdot 7$, $3\cdot 5\cdot
11\cdot 19$ are in the codegrees of ${\mbox{PSL}}(3,3)$, $\textup{{M}}_{11}$,
$\textup{{A}}_{7}$ or $\textup{{J}}_{1}$, respectively. However, none of these
elements are in the codegree set of $\textup{{M}}^{c}L$, a contradiction.

$|{\mbox{cod}}(G/N)|=8$. By Lemma 2.3, we have $G/N\cong{\mbox{PSU}}(3,q)$
with $4<q\not\equiv-1\pmod{3}$, $G/N\cong{\mbox{PSL}}(3,q)$ with
$4<q\not\equiv 1\pmod{3}$ or $G/N\cong\textup{{G}}_{2}(2)^{\prime}$. Observe
that the only nontrivial element not divisible by $3$ in
${\mbox{cod}}(\textup{{M}}^{c}L)$ is $2^{7}\cdot 5^{3}\cdot 11$, $2^{7}\cdot
5^{3}\cdot 7$.

If $G/N\cong{\mbox{PSU}}(3,q)$, notice that
$(q^{2}-q+1)(q+1)^{2}(q-1),q^{3}(q^{2}-q+1)\in{\mbox{cod}}({\mbox{PSU}}(3,q))$.
If $q\equiv 1\mod 3$, then $q^{3}(q^{2}-q+1)\not\equiv 0\pmod{3}$, so it must
hold that $q^{3}(q^{2}-q+1)=2^{7}\cdot 5^{3}\cdot 11$ or $2^{7}\cdot
5^{3}\cdot 7$. However, this does not yield an integer root for $q$.
Alternatively, if $q\equiv 0\pmod{3}$, then
$(q^{2}-q+1)(q+1)^{2}(q-1)\not\equiv 0\pmod{3}$, so it must hold that
$(q^{2}-q+1)(q+1)^{2}(q-1)=2^{7}\cdot 5^{3}\cdot 11$ or $2^{7}\cdot 5^{3}\cdot
7$. This also does not yield an integer root for $q$. Thus
${\mbox{cod}}(G/N)\not\subseteq{\mbox{cod}}(G)$, a contradiction.

If $G/N\cong{\mbox{PSL}}(3,q)$, we have
$q^{3}(q^{2}+q+1)\in{\mbox{cod}}({\mbox{PSL}}(3,q))$. As $q\not\equiv
1\pmod{3},q^{3}(q^{2}+q+1)\not\equiv 0\pmod{3}$, so it must hold that
$q^{3}(q^{2}+q+1)=2^{7}\cdot 5^{3}\cdot 11$ or $q^{3}(q^{2}+q+1)=2^{7}\cdot
5^{3}\cdot 7$. However, this does not lead to an integer solution for $q$.
Thus ${\mbox{cod}}(G/N)\not\subseteq{\mbox{cod}}(G),$, a contradiction.

In the final case, $3^{3}\cdot 7\in{\mbox{cod}}(\textup{{G}}_{2}(2)^{\prime})$
but $3^{3}\cdot 7\not\in{\mbox{cod}}(\textup{{M}}^{c}L)$, contradicting
${\mbox{cod}}(G/N)\subseteq{\mbox{cod}}(G)$.

$|{\mbox{cod}}(G/N)|=9$. By Lemma 2.3, we have $G/N\cong{\mbox{PSL}}(3,q)$
with $4<q\equiv 1\pmod{3}$ or $G/N\cong{\mbox{PSU}}(3,q)$ with
$4<q\equiv-1\pmod{3}$. Both $q^{2}-q+1$ and $q^{2}+q+1$ are not divisible by
$9$.

In the first case, we have
$\frac{1}{3}q^{3}(q^{2}+q+1)\in{\mbox{cod}}({\mbox{PSL}}(3,q))$. Thus,
$\frac{1}{3}q^{3}(q^{2}+q+1)\not\equiv 0\pmod{3}$, so we must have
$\frac{1}{3}q^{3}(q^{2}+q+1)=2^{7}\cdot 5^{3}\cdot 11$ or $2^{7}\cdot
5^{3}\cdot 7$, which does not yield an integer root for $q$, a contradiction.

In the second scenario, we have
$\frac{1}{3}q^{3}(q^{2}-q+1)\in{\mbox{cod}}({\mbox{PSU}}(3,q))$. As with the
first scenario, this element is not divisible by $3$, so we must have
$\frac{1}{3}q^{3}(q^{2}-q+1)=2^{7}\cdot 5^{3}\cdot 11$ or $2^{7}\cdot
5^{3}\cdot 7$. However, this does not yield an integer root for $q$, a
contradiction.

$|{\mbox{cod}}(G/N)|=10$. By Lemma 2.3, we have $G/N\cong\textup{{M}}_{22}$.
However, $2^{7}\cdot 3\cdot 5\cdot 11$ is only present in
${\mbox{cod}}(\textup{{M}}_{22})$ and not in
${\mbox{cod}}(\textup{{M}}^{c}L)$, contradicting
${\mbox{cod}}(G/N)\subseteq{\mbox{cod}}(G)$.

$|{\mbox{cod}}(G/N)|=11$. By Lemma 2.3, we have $G/N\cong{\mbox{PSL}}(4,2)$,
$G/N\cong\textup{{M}}_{12}$, $G/N\cong\textup{{M}}_{23}$ or
$G/N\cong{}^{2}\textup{{G}}_{2}(q)$ for $q=3^{2f+1},f\geq 1$. The codegree
sets of ${\mbox{PSL}}(4,2),\textup{{M}}_{12}$ and $\textup{{M}}_{23}$ contain
$3^{2}\cdot 5\cdot 7$, $2^{2}\cdot 3\cdot 5\cdot 11$ and $2^{7}\cdot 7\cdot
11$ respectively, none of which are elements of the codegree set of
$\textup{{M}}^{c}L$.

In the other case, note that
$(q^{2}-1)(q^{2}-q+1)\in{\mbox{cod}}({}^{2}\textup{{G}}_{2}(q))$. However,
$(q^{2}-1)(q^{2}-q+1)\not\equiv 0\pmod{3}$ in this case, so we must have
$(q^{2}-1)(q^{2}-q+1)=2^{7}\cdot 5^{3}\cdot 7$, $2^{7}\cdot 5^{3}\cdot 11$ yet
this does not yield an integer root for $q$. This is a contradiction to
${\mbox{cod}}(G/N)\subseteq{\mbox{cod}}(G)$.

$|{\mbox{cod}}(G/N)|=12$. By Lemma 2.3, we have $G/N\cong\textup{{S}}_{4}(4)$.
However, the element $3^{2}\cdot 5^{2}\cdot 17$ is in the codegree set of
$\textup{{S}}_{4}(4)$ but not in the codegree set of $\textup{{M}}^{c}L$,
contradicting ${\mbox{cod}}(G/N)\subseteq{\mbox{cod}}(G)$.

$|{\mbox{cod}}(G/N)|=13$. By Lemma 2.3, we have $G/N\cong\textup{{U}}_{4}(2)$
or $G/N\cong\textup{{S}}_{4}(q)$ for $q=2^{f}>4$. In the first case, the
element $3^{4}\cdot 5\in{\mbox{cod}}(\textup{{U}}_{4}(2))$ but it is not in
${\mbox{cod}}(\textup{{M}}^{c}L)$.

In the second case,
$2q^{3}(q-1)^{2}(q+1)^{2}\in{\mbox{cod}}(\textup{{S}}_{4}(q))$. However, as
$\nu_{2}(q)>2$, then $\nu_{2}(2q^{3}(q-1)^{2}(q+1)^{2})>7$ yet there does not
exist an element in ${\mbox{cod}}(\textup{{M}}^{c}L)$ that is divisible by
$2^{8}$, a contradiction.

$|{\mbox{cod}}(G/N)|=14$. By Lemma 2.3, we have $G/N\cong\textup{{U}}_{4}(3)$,
$G/N\cong{}^{2}\textup{{F}}_{4}(2)^{\prime}$ or $G/N\cong\textup{{J}}_{3}$.
However, the codegree sets of $\textup{{U}}_{4}(3)$,
${}^{2}\textup{{F}}_{4}(2)^{\prime}$ and $\textup{{J}}_{3}$ contain
$2^{7}\cdot 3^{5}\cdot 5$, $2^{10}\cdot 3^{3}\cdot 5^{2}$ and $2^{7}\cdot
3^{5}\cdot 19$ respectively, but these are not in
${\mbox{cod}}(\textup{{M}}^{c}L)$. This contradicts
${\mbox{cod}}(G/N)\subseteq{\mbox{cod}}(G)$.

$|{\mbox{cod}}(G/N)|=15$. By Lemma 2.3, we have $G/N\cong\textup{{G}}_{2}(3)$.
However, the element $2^{5}\cdot 3^{6}\cdot 13$ is only present in the
codegree set of $\textup{{G}}_{2}(3)$ and not in the codegree set of
$\textup{{M}}^{c}L$, which contradicts
${\mbox{cod}}(G/N)\subseteq{\mbox{cod}}(G)$.

$|{\mbox{cod}}(G/N)|=16$. By Lemma 2.3, we have $G/N\cong\textup{{A}}_{9}$ or
$G/N\cong\textup{{J}}_{2}$. In the first scenario, $2^{4}\cdot 3^{3}\cdot
5\in{\mbox{cod}}(\textup{{A}}_{9})$ but is not in
${\mbox{cod}}(\textup{{M}}^{c}L)$, while in the second $2^{6}\cdot 3^{3}\cdot
5^{2}\in{\mbox{cod}}(\textup{{J}}_{2})$, but this element is not in
${\mbox{cod}}(\textup{{M}}^{c}L)$. This contradicts
${\mbox{cod}}(G/N)\subseteq{\mbox{cod}}(G)$.

$|{\mbox{cod}}(G/N)|=17$. Then $G/N\cong{\mbox{PSL}}(4,3)$. However, the
element $2^{6}\cdot 3^{6}\cdot 5$ is only in ${\mbox{cod}}({\mbox{PSL}}(4,3))$
and not the codegree of $\textup{{M}}^{c}L$, a contradiction.

Having ruled out all other cases, we conclude $G/N\cong\textup{{M}}^{c}L$.
$\hbox to0.0pt{$\sqcap$\hss}\sqcup$

######  Lemma 3.11.

Let $G$ be a finite group with
${\mbox{cod}}(G)={\mbox{cod}}({\mbox{PSL}}(4,3))$. If $N$ is the maximal
normal subgroup of $G$, it we have $G/N\cong{\mbox{PSL}}(4,3)$.

###### Proof.

As $G/N$ is a non-abelian simple group,
${\mbox{cod}}(G/N)\subseteq{\mbox{cod}}(G)$, so we have that
${\mbox{cod}}(G/N)=4,5,6,7,8,9,10,11,12,13,14,15,16$, or $17$.

Assume on the contrary that there exist a maximal normal subgroup $N$ such
that $G/N\not\cong{\mbox{PSL}}(4,3),{\mbox{cod}}(G/N)\leq 17$ and
${\mbox{cod}}(G/N)\subseteq{\mbox{cod}}(G)$. We exhaust all possibilities to
show the previous statement is false.

$|{\mbox{cod}}(G/N)|=4$. By Lemma 2.3, $G/N\cong{\mbox{PSL}}(2,k)$ where
$k=2^{f}\geq 4$, so $k^{2}-1\in{\mbox{cod}}(G/N)$. For all even $k$, $k^{2}-1$
is odd. Therefore, $k^{2}-1$ must equal $3^{6}\cdot 13$ but this yields no
integer roots for $k$, contradicting
${\mbox{cod}}(G/N)\subseteq{\mbox{cod}}(G)$.

$|{\mbox{cod}}(G/N)|=5$. By Lemma 2.3, $G/N\cong{\mbox{PSL}}(2,k)$ where $k$
is a power of an odd prime number. This implies that $\displaystyle
k(k-\epsilon(k),\frac{k(k-\epsilon(k)))}{2}\in{\mbox{cod}}(G/N)$, where
$k=(-1)^{(k-1)/2}$. Then $k(k-\epsilon(k))/2$,
$k(k-\epsilon(k))\in{\mbox{cod}}(G)$. By a complete search, and we have that
$\frac{k(k\pm 1)}{2}=2^{5}\cdot 3^{6}\cdot 5$, $2^{5}\cdot 3^{5}$ or
$2^{5}\cdot 3^{4}\cdot 5$ since they are half of the other three elements.
This is a contradiction since $k\in\mathbb{Z}$.

$|{\mbox{cod}}(G/N)|=6$. By Lemma 2.3, $G/N\cong{\mbox{PSL}}(3,4)$ or
$G/N\cong{}^{2}B_{2}(q)$, with $q=2^{2n+1}$.

If $G/N\cong{\mbox{PSL}}(3,4)$ then $3^{2}\cdot 5\cdot
7\in{\mbox{cod}}({\mbox{PSL}}(3,4))$ but $3^{2}\cdot 5\cdot
7\not\in{\mbox{cod}}(G)={\mbox{cod}}({\mbox{PSL}}(4,3))$, contradicting
${\mbox{cod}}(G/N)\subseteq{\mbox{cod}}(G)$.

Furthermore, notice that $(q-1)(q^{2}+1)\in{\mbox{cod}}({}^{2}B_{2}(q))$. As
$q$ is even, $(q-1)(q^{2}+1)\equiv 1\pmod{2}$. This means that
$(q-1)(q^{2}+1)=3^{6}\cdot 13$ which does not yield any roots. This
contradicts ${\mbox{cod}}(G/N)\subseteq{\mbox{cod}}(G)$.

$|{\mbox{cod}}(G/N)|=7$. By Lemma 2.3, we have $G/N\cong{\mbox{PSL}}(3,3)$,
$\textup{{M}}_{11}$, $\textup{{A}}_{7}$ or $\textup{{J}}_{1}$. By inspection,
the elements $2^{2}\cdot 3^{2}\cdot 13$, $2^{3}\cdot 3^{3}\cdot 11$,
$2^{2}\cdot 3\cdot 5\cdot 7$, $3\cdot 5\cdot 11\cdot 19$ are in the codegrees
of ${\mbox{PSL}}(3,3)$, $\textup{{M}}_{11}$, $\textup{{A}}_{7}$ or
$\textup{{J}}_{1}$ respectively. However, none of these elements are in the
codegree set of ${\mbox{PSL}}(4,3)$, a contradiction.

$|{\mbox{cod}}(G/N)|=8$. By Lemma 2.3, we have $G/N\cong{\mbox{PSU}}(3,q)$
with $4<q\not\equiv-1\pmod{3}$, $G/N\cong{\mbox{PSL}}(3,q)$ with
$4<q\not\equiv 1\pmod{3}$ or $G/N\cong\textup{{G}}_{2}(2)^{\prime}$. There are
no nontrivial elements in the codegree set of ${\mbox{PSL}}(4,3)$ that are not
divisible by $3$ except $2^{7}\cdot 5\cdot 13$.

If $G/N\cong{\mbox{PSU}}(3,q)$, notice that
$(q^{2}-q+1)(q+1)^{2}(q-1),q^{3}(q^{2}-q+1)\in{\mbox{cod}}({\mbox{PSU}}(3,q))$.
If $q\equiv 1\mod 3$, in this case, we must have $q^{3}(q^{2}-q+1)$ yet this
is the nontrivial odd codegree, this does not yield an integer root for $q$.
This is a contradiction to ${\mbox{cod}}(G/N)\subseteq{\mbox{cod}}(G)$.
Alternatively, if $q\equiv 0\pmod{3}$, then
$(q^{2}-q+1)(q+1)^{2}(q-1)\not\equiv 0\pmod{3}$. We have that
$(q^{2}-q+1)(q+1)^{2}(q-1)=2^{7}\cdot 5\cdot 13$, this does not yield an
integer root for $q$. Thus ${\mbox{cod}}(G/N)\not\subseteq{\mbox{cod}}(G)$, a
contradiction.

If $G/N\cong{\mbox{PSL}}(3,q)$, we have
$q^{3}(q^{2}+q+1)\in{\mbox{cod}}({\mbox{PSL}}(3,q))$. As $q\not\equiv
1\pmod{3},q^{3}(q^{2}+q+1)\not\equiv 0\pmod{3}$. We can also get the same
contradiction as above. Thus ${\mbox{cod}}(G/N)\not\subseteq{\mbox{cod}}(G)$.

In the final case, $3^{3}\cdot 7\in{\mbox{cod}}(\textup{{G}}_{2}(2)^{\prime})$
but $3^{3}\cdot 7\not\in{\mbox{cod}}({\mbox{PSL}}(4,3))$, contradicting
${\mbox{cod}}(G/N)\subseteq{\mbox{cod}}(G)$.

$|{\mbox{cod}}(G/N)|=9$. By Lemma 2.3, we have $G/N\cong{\mbox{PSL}}(3,q)$
with $4<q\equiv 1\pmod{3}$ or $G/N\cong{\mbox{PSU}}(3,q)$ with
$4<q\equiv-1\pmod{3}$. Both $q^{2}-q+1$ and $q^{2}+q+1$ are not divisible by
$9$.

In the first case, we have
$\frac{1}{3}q^{3}(q^{2}+q+1)\in{\mbox{cod}}({\mbox{PSL}}(3,q))$. Thus
$\frac{1}{3}q^{3}(q^{2}+q+1)\not\equiv 0\pmod{3}$, we must have
$\frac{1}{3}q^{3}(q^{2}+q+1)$ is the odd codegree, yet this does not yield an
integer root for $q$, contradiction.

In the second scenario, we have
$\frac{1}{3}q^{3}(q^{2}-q+1)\in{\mbox{cod}}({\mbox{PSU}}(3,q))$. This is not
divisible by $3$, we can get the same contradiction as above.

$|{\mbox{cod}}(G/N)|=10$. By Lemma 2.3, we have $G/N\cong\textup{{M}}_{22}.$
However, $2^{7}\cdot 3\cdot 5\cdot 11$ is only present in
${\mbox{cod}}(\textup{{M}}_{22})$ and not in
${\mbox{cod}}({\mbox{PSL}}(4,3))$, contradicting
${\mbox{cod}}(G/N)\subseteq{\mbox{cod}}(G)$.

$|{\mbox{cod}}(G/N)|=11$. By Lemma 2.3, we have $G/N\cong{\mbox{PSL}}(4,2)$,
$G/N\cong\textup{{M}}_{12}$, $G/N\cong\textup{{M}}_{23}$ or
$G/N\cong{}^{2}\textup{{G}}_{2}(q)$ for $q=3^{2f+1},f\geq 1$. The codegree
sets of ${\mbox{PSL}}(4,2)$, $\textup{{M}}_{12}$ and $\textup{{M}}_{23}$
contain $3^{2}\cdot 5\cdot 7$, $2^{2}\cdot 3\cdot 5\cdot 11$ and $2^{7}\cdot
7\cdot 11$ respectively, none of which are elements of the codegree set of
${\mbox{PSL}}(4,3)$.

If $G/N\cong{}^{2}\textup{{G}}_{2}(q)$,
$(q^{2}-1)(q^{2}-q+1)\in{\mbox{cod}}({}^{2}\textup{{G}}_{2}(q))$. However,
$(q^{2}-1)(q^{2}-q+1)\not\equiv 0\pmod{3}$ in this case, so we must have
$(q^{2}-1)(q^{2}-q+1)=2^{7}\cdot 5^{3}\cdot 7,2^{7}\cdot 5^{3}\cdot 11$ yet
this does not yield an integer root for $q$. This is a contradiction to
${\mbox{cod}}(G/N)\subseteq{\mbox{cod}}(G)$.

$|{\mbox{cod}}(G/N)|=12$. By Lemma 2.3, we have $G/N\cong\textup{{S}}_{4}(4)$.
However, $17$ is a factor of $|{\mbox{cod}}(\textup{{S}}_{4}(4))|$ and not
$|{\mbox{cod}}({\mbox{PSL}}(4,3))|$, contradicting
${\mbox{cod}}(G/N)\subseteq{\mbox{cod}}(G)$.

$|{\mbox{cod}}(G/N)|=13$. By Lemma 2.3, we have $G/N\cong\textup{{U}}_{4}(2)$
or $G/N\cong\textup{{S}}_{4}(q)$ for $q=2^{f}>4$.

In the first case, the element $3^{4}\cdot
5\in{\mbox{cod}}(\textup{{U}}_{4}(2))$ but it is not in
${\mbox{cod}}({\mbox{PSL}}(4,3))$.

In the second case,
$2q^{3}(q-1)^{2}(q+1)^{2}\in{\mbox{cod}}(\textup{{S}}_{4}(q))$. However, as
$\nu_{2}(q)>2$, then $\nu_{2}(2q^{3}(q-1)^{2}(q+1)^{2})>7$ yet there does not
exist an element in ${\mbox{cod}}(\textup{{M}}^{c}L)$ that is divisible by
$2^{8}$, a contradiction.

$|{\mbox{cod}}(G/N)|=14$. By Lemma 2.3, we have $G/N\cong\textup{{U}}_{4}(3)$,
$G/N\cong{}^{2}\textup{{F}}_{4}(2)^{\prime}$ or $G/N\cong\textup{{J}}_{3}$.
However, the codegree sets of $\textup{{U}}_{4}(3)$,
${}^{2}\textup{{F}}_{4}(2)^{\prime}$ and $\textup{{J}}_{3}$ have $2^{4}\cdot
3^{6}$, $2^{10}\cdot 3^{3}\cdot 5^{2}$ and $2^{7}\cdot 3^{5}\cdot 19$
respectively, but these are not in ${\mbox{cod}}({\mbox{PSL}}(4,3))$. This
contradicts ${\mbox{cod}}(G/N)\subseteq{\mbox{cod}}(G)$.

$|{\mbox{cod}}(G/N)|=15$. By Lemma 2.3, we have $G/N\cong\textup{{G}}_{2}(3)$.
However, the element $2^{5}\cdot 3^{6}\cdot 13$ is only present in the
codegree set of $\textup{{G}}_{2}(3)$ and not in the codegree set of
${\mbox{PSL}}(4,3)$, which contradicts
${\mbox{cod}}(G/N)\subseteq{\mbox{cod}}(G)$.

$|{\mbox{cod}}(G/N)|=16$. By Lemma 2.3, we have $G/N\cong\textup{{A}}_{9}$ or
$G/N\cong\textup{{J}}_{2}$. In the $\textup{{A}}_{9}$ case, $2^{4}\cdot
3^{3}\cdot 5\in{\mbox{cod}}(\textup{{A}}_{9})$ but is not in
${\mbox{cod}}({\mbox{PSL}}(4,3))$, while in the $\textup{{J}}_{2}$ case
$2^{6}\cdot 3^{3}\cdot 5^{2}\in{\mbox{cod}}(\textup{{J}}_{2})$ but this
element is not in ${\mbox{cod}}({\mbox{PSL}}(4,3))$. This contradicts
${\mbox{cod}}(G/N)\subseteq{\mbox{cod}}(G)$.

$|{\mbox{cod}}(G/N)|=17$. Then $G/N\cong\textup{{M}}^{c}L$. However, the
element $2^{7}\cdot 5^{3}\cdot 11$ is only in
${\mbox{cod}}(\textup{{M}}^{c}L)$ and not the codegree of ${\mbox{PSL}}(4,3)$,
a contradiction.

Having ruled out all other cases, we conclude $G/N\cong{\mbox{PSL}}(4,3)$.
$\hbox to0.0pt{$\sqcap$\hss}\sqcup$

######  Lemma 3.12.

Let $G$ be a finite group with
${\mbox{cod}}(G)={\mbox{cod}}(\textup{{S}}_{4}(5))$. If $N$ is the maximal
normal subgroup of $G$, it we have $G/N\cong\textup{{S}}_{4}(5)$.

###### Proof.

As $G/N$ is a non-abelian simple group,
${\mbox{cod}}(G/N)\subseteq{\mbox{cod}}(G)$, so we have that
${\mbox{cod}}(G/N)=4,5,6,7,8,9,10,11,12,13,14,15,16,17$, or $18$.

Assume on the contrary that there exist a maximal normal subgroup $N$ such
that $G/N\not\cong\textup{{S}}_{4}(5),{\mbox{cod}}(G/N)\leq 18$ and
${\mbox{cod}}(G/N)\subseteq{\mbox{cod}}(G)$. We exhaust all possibilities.

Suppose $|{\mbox{cod}}(G/N)|=4$. By Lemma 2.3, $G/N\cong{\mbox{PSL}}(2,k)$
where $k=2^{f}\geq 4$, so $k^{2}-1\in{\mbox{cod}}(G/N)$. For all even $k$,
$k^{2}-1$ is odd. $k^{2}-1$ must equal $5^{4}\cdot 13$ but this yields no
integer roots for $k$, contradicting
${\mbox{cod}}(G/N)\subseteq{\mbox{cod}}(G)$.

Suppose $|{\mbox{cod}}(G/N)|=5$. By Lemma 2.3, $G/N\cong{\mbox{PSL}}(2,k)$
where $k$ is a power of an odd prime number. This implies that $\displaystyle
k(k-\epsilon(k),\frac{k(k-\epsilon(k)))}{2}\in{\mbox{cod}}(G/N)$, where
$k=(-1)^{(k-1)/2}$. By a complete search, and we have that $\frac{k(k\pm
1)}{2}=2^{4}\cdot 3\cdot 5^{4}$, $2^{4}\cdot 3\cdot 5^{3}$ or $2^{2}\cdot
3^{2}\cdot 5^{4}$ since they are half of the other three elements. This is a
contradiction since $k\in\mathbb{Z}$.

$|{\mbox{cod}}(G/N)|=6$. By Lemma 2.3, $G/N\cong{\mbox{PSL}}(3,4)$ or
$G/N\cong{}^{2}B_{2}(q)$, with $q=2^{2n+1}$.

If $G/N\cong{\mbox{PSL}}(3,4)$ then $3^{2}\cdot 5\cdot
7\in{\mbox{cod}}({\mbox{PSL}}(3,4))$, but $3^{2}\cdot 5\cdot
7\not\in{\mbox{cod}}(\textup{{S}}_{4}(5))$, contradicting
${\mbox{cod}}(G/N)\subseteq{\mbox{cod}}(G)$. Furthermore, notice that
$(q-1)(q^{2}+1)\in{\mbox{cod}}({}^{2}B_{2}(q))$. As $q$ is even,
$(q-1)(q^{2}+1)\equiv 1\pmod{2}$. This means that $(q-1)(q^{2}+1)=5^{4}\cdot
13$ which does not yield any roots, contradicting
${\mbox{cod}}(G/N)\subseteq{\mbox{cod}}(G)$.

$|{\mbox{cod}}(G/N)|=7$. By Lemma 2.3, we have $G/N\cong{\mbox{PSL}}(3,3)$,
$\textup{{M}}_{11}$, $\textup{{A}}_{7}$ or $\textup{{J}}_{1}$. The elements
$2^{2}\cdot 3^{2}\cdot 13$, $2^{3}\cdot 3^{3}\cdot 11$, $2^{2}\cdot 3\cdot
5\cdot 7$, $3\cdot 5\cdot 11\cdot 19$ are in the codegrees of
${\mbox{PSL}}(3,3)$, $\textup{{M}}_{11}$, $\textup{{A}}_{7}$ or
$\textup{{J}}_{1}$ respectively. However, none of these elements are in the
codegree set of $\textup{{S}}_{4}(5)$, a contradiction.

$|{\mbox{cod}}(G/N)|=8$. By Lemma 2.3, we have $G/N\cong{\mbox{PSU}}(3,q)$
with $4<q\not\equiv-1\pmod{3}$, $G/N\cong{\mbox{PSL}}(3,q)$ with
$4<q\not\equiv 1\pmod{3}$ or $G/N\cong\textup{{G}}_{2}(2)^{\prime}$.

The only nontrivial element of $\textup{{S}}_{4}(5)$ not divisible by $3$ is
$5^{4}\cdot 13$.

If $G/N\cong{\mbox{PSU}}(3,q)$, notice that
$(q^{2}-q+1)(q+1)^{2}(q-1),q^{3}(q^{2}-q+1)\in{\mbox{cod}}({\mbox{PSU}}(3,q))$.
If $q\equiv 1\mod 3$, then $q^{3}(q^{2}-q+1)\not\equiv 0\pmod{3}$.
Alternatively, if $q\equiv 0\pmod{3}$, then
$(q^{2}-q+1)(q+1)^{2}(q-1)\not\equiv 0\pmod{3}$.
However,$(q^{2}-q+1)(q+1)^{2}(q-1)=5^{4}\cdot 13$ yield no integer roots for
$q$. Thus, ${\mbox{cod}}(G/N)\not\subseteq{\mbox{cod}}(G)$, a contradiction.

If $G/N\cong{\mbox{PSL}}(3,q)$, we have
$q^{3}(q^{2}+q+1)\in{\mbox{cod}}({\mbox{PSL}}(3,q))$. As $q\not\equiv
1\pmod{3},q^{3}(q^{2}+q+1)\not\equiv 0\pmod{3}$. However,
$q^{3}(q^{2}+q+1)=5^{4}\cdot 13$ has no integer roots. Thus,
${\mbox{cod}}(G/N)\not\subseteq{\mbox{cod}}(G)$, a contradiction.

In the final case, $3^{3}\cdot 7\in{\mbox{cod}}(\textup{{G}}_{2}(2)^{\prime})$
but $3^{3}\cdot 7\not\in{\mbox{cod}}(\textup{{S}}_{4}(5))$, contradicting
${\mbox{cod}}(G/N)\subseteq{\mbox{cod}}(G)$.

$|{\mbox{cod}}(G/N)|=9$. By Lemma 2.3, we have $G/N\cong{\mbox{PSL}}(3,q)$
with $4<q\equiv 1\pmod{3}$ or $G/N\cong{\mbox{PSU}}(3,q)$ with
$4<q\equiv-1\pmod{3}$.

Both $q^{2}-q+1$ and $q^{2}+q+1$ are not divisible by $9$.

In the first case, we have
$\frac{1}{3}q^{3}(q^{2}+q+1)\in{\mbox{cod}}({\mbox{PSL}}(3,q))$. Thus,
$\frac{1}{3}q^{3}(q^{2}+q+1)\not\equiv 0\pmod{3}$. Furthermore, we have
$\frac{1}{3}q^{3}(q^{2}-q+1)\in{\mbox{cod}}({\mbox{PSU}}(3,q))$. This is not
divisible by $3$.

The equation $\frac{1}{3}q^{3}(q^{2}\pm q+1)=5^{4}\cdot 13$ has no integer
roots, contradicting ${\mbox{cod}}(G/N)\subseteq{\mbox{cod}}(G)$.

$|{\mbox{cod}}(G/N)|=10$. By Lemma 2.3, we have $G/N\cong\textup{{M}}_{22}$.
However, $2^{7}\cdot 3\cdot 5\cdot 11$ is only present in
${\mbox{cod}}(\textup{{M}}_{22})$ and not in
${\mbox{cod}}(\textup{{S}}_{4}(5))$, contradicting
${\mbox{cod}}(G/N)\subseteq{\mbox{cod}}(G)$.

$|{\mbox{cod}}(G/N)|=11$. By Lemma 2.3, we have $G/N\cong{\mbox{PSL}}(4,2)$,
$G/N\cong\textup{{M}}_{12}$, $G/N\cong\textup{{M}}_{23}$ or
$G/N\cong{}^{2}\textup{{G}}_{2}(q)$ for $q=3^{2f+1},f\geq 1$. The codegree
sets of ${\mbox{PSL}}(4,2)$, $\textup{{M}}_{12}$ and $\textup{{M}}_{23}$
contain $3^{2}\cdot 5\cdot 7$, $2^{2}\cdot 3\cdot 5\cdot 11$ and $2^{7}\cdot
7\cdot 11$, respectively, none of which are elements of the codegree set of
$\textup{{S}}_{4}(5)$.

If $G/N\cong{}^{2}\textup{{G}}_{2}(q)$,
$(q^{2}-1)(q^{2}-q+1)\in{\mbox{cod}}({}^{2}\textup{{G}}_{2}(q))$. However,
$(q^{2}-1)(q^{2}-q+1)\not\equiv 0\pmod{3}$. In this case, so we must have
$(q^{2}-1)(q^{2}-q+1)=5^{4}\cdot 13$ yet this does not yield an integer root
for $q$. This is a contradiction to
${\mbox{cod}}(G/N)\subseteq{\mbox{cod}}(G)$.

$|{\mbox{cod}}(G/N)|=12$. By Lemma 2.3, we have $G/N\cong\textup{{S}}_{4}(4)$.
However, $17$ is a factor of $|{\mbox{cod}}(\textup{{S}}_{4}(4))|$ and not
$|{\mbox{cod}}(\textup{{S}}_{4}(5))|$, contradicting
${\mbox{cod}}(G/N)\subseteq{\mbox{cod}}(G)$.

$|{\mbox{cod}}(G/N)|=13$. By Lemma 2.3, we have $G/N\cong\textup{{U}}_{4}(2)$
or $G/N\cong\textup{{S}}_{4}(q)$ for $q=2^{f}>4$.

In the first case, the element $3^{4}\cdot
5\in{\mbox{cod}}(\textup{{S}}_{4}(q))$ but it is not in
${\mbox{cod}}(\textup{{S}}_{4}(5))$.

In the second case,
$2q^{3}(q-1)^{2}(q+1)^{2}\in{\mbox{cod}}(\textup{{U}}_{4}(2))$. However, as
$\nu_{2}(q)>2$, then $\nu_{2}(2q^{3}(q-1)^{2}(q+1)^{2})>7$ yet there does not
exist an element in ${\mbox{cod}}(\textup{{M}}^{c}L)$ that is divisible by
$2^{8}$, a contradiction.

$|{\mbox{cod}}(G/N)|=14$. By Lemma 2.3, we have $G/N\cong\textup{{U}}_{4}(3)$,
$G/N\cong{}^{2}\textup{{F}}_{4}(2)^{\prime}$ or $G/N\cong\textup{{J}}_{3}$.
However, the codegree sets of $\textup{{U}}_{4}(3)$,
${}^{2}\textup{{F}}_{4}(2)^{\prime}$ and $\textup{{J}}_{3}$ have $2^{4}\cdot
3^{6}$, $2^{10}\cdot 3^{3}\cdot 5^{2}$ and $2^{7}\cdot 3^{5}\cdot 19$,
respectively, but these are not in ${\mbox{cod}}(\textup{{S}}_{4}(5))$. This
contradicts ${\mbox{cod}}(G/N)\subseteq{\mbox{cod}}(G)$.

$|{\mbox{cod}}(G/N)|=15$. By Lemma 2.3, we have $G/N\cong\textup{{G}}_{2}(3)$.
However, the element $2^{5}\cdot 3^{6}\cdot 13$ is only present in the
codegree set of $\textup{{G}}_{2}(3)$ and not in the codegree set of
$\textup{{S}}_{4}(5)$, which contradicts
${\mbox{cod}}(G/N)\subseteq{\mbox{cod}}(G)$.

$|{\mbox{cod}}(G/N)|=16$. By Lemma 2.3, we have $G/N\cong\textup{{A}}_{9}$ or
$G/N\cong\textup{{J}}_{2}$.

In the $\textup{{A}}_{9}$ case, $7$ is a factor of
$|{\mbox{cod}}(\textup{{A}}_{9})|$ and not
$|{\mbox{cod}}(\textup{{S}}_{4}(5))|$, contradicting
${\mbox{cod}}(G/N)\subseteq{\mbox{cod}}(G)$.

while in the $\textup{{J}}_{2}$ case, this situation is the same as the
contradiction above. $7$ is not the factor of
$|{\mbox{cod}}(\textup{{S}}_{4}(5))|$.

$|{\mbox{cod}}(G/N)|=17$. Then $G/N\cong\textup{{M}}^{c}L$ or
${\mbox{PSL}}(4,3)$ However, there are three nontrivial odd codegrees in
${\mbox{cod}}\textup{{M}}^{c}L$, there is only a nontrivial odd element in the
set of $\textup{{S}}_{4}(5)$, a contradiction. Additionally, the element
$3^{6}\cdot 13$ is only in ${\mbox{cod}}({\mbox{PSL}}(4,3))$ and not
${\mbox{cod}}(\textup{{S}}_{4}(5))$, a contradiction.

$|{\mbox{cod}}(G/N)|=18$. We have $G/N\cong\textup{{G}}_{2}(4)$ or HS. If
$G/N\cong\textup{{G}}_{2}(4)$, then $2^{12}\cdot 3^{3}\cdot 5\cdot
7\in{\mbox{cod}}(G/N)$. If $G/N\in{\mbox{HS}}$, then $11$ is not the factor of
$|{\mbox{cod}}(\textup{{S}}_{4}(5))|$, a contradiction.

Having ruled out all other cases, we conclude $G/N\cong\textup{{S}}_{4}(5)$.
$\hbox to0.0pt{$\sqcap$\hss}\sqcup$

######  Lemma 3.13.

Let $G$ be a finite group with
${\mbox{cod}}(G)={\mbox{cod}}(\textup{{G}}_{2}(4))$. If $N$ is the maximal
normal subgroup of $G$, it we have $G/N\cong\textup{{G}}_{2}(4)$.

###### Proof.

As $G/N$ is a non-abelian simple group,
${\mbox{cod}}(G/N)\subseteq{\mbox{cod}}(G)$, so we have that
${\mbox{cod}}(G/N)=4,5,6,7,8,9,10,11,12,13,14,15,16,17,$ or $18$.

Assume on the contrary that there exist a maximal normal subgroup $N$ such
that $G/N\not\cong\textup{{G}}_{2}(4),{\mbox{cod}}(G/N)\leq 18$ and
${\mbox{cod}}(G/N)\subseteq{\mbox{cod}}(G)$. We exhaust all possibilities.

$|{\mbox{cod}}(G/N)|=4$. By Lemma 2.3, $G/N\cong{\mbox{PSL}}(2,k)$ where
$k=2^{f}\geq 4$, so $k^{2}-1\in{\mbox{cod}}(G/N)$. For all even $k$, $k^{2}-1$
is odd. $k^{2}-1$ must equal $3^{3}\cdot 5^{2}\cdot 7\cdot 13$ but this yields
no integer roots for $k$, contradicting
${\mbox{cod}}(G/N)\subseteq{\mbox{cod}}(G)$.

$|{\mbox{cod}}(G/N)|=5$. By Lemma 2.3, $G/N\cong{\mbox{PSL}}(2,k)$ where $k$
is a power of an odd prime number. This implies that $\displaystyle
k(k-\epsilon(k)\in{\mbox{cod}}(G/N)$ ,where $k=(-1)^{(k-1)/2}$. we have that
$\frac{k(k\pm 1)}{2}=2^{10}\cdot 3^{3}\cdot 7$, since it is half of
$2^{11}\cdot 3^{3}\cdot 7$. This is a contradiction since $k\in\mathbb{Z}$.

$|{\mbox{cod}}(G/N)|=6$. By Lemma 2.3, $G/N\cong{\mbox{PSL}}(3,4)$ or
$G/N\cong{}^{2}B_{2}(q)$, with $q=2^{2n+1}$.

If $G/N\cong{\mbox{PSL}}(3,4)$ then $3^{2}\cdot 5\cdot
7\in{\mbox{cod}}({\mbox{PSL}}(3,4))$ but $3^{2}\cdot 5\cdot
7\not\in{\mbox{cod}}(\textup{{G}}_{2}(4))$, contradicting
${\mbox{cod}}(G/N)\subseteq{\mbox{cod}}(G)$.

If $G/N\cong{}^{2}B_{2}(q)$, notice that
$(q-1)(q^{2}+1)\in{\mbox{cod}}({}^{2}B_{2}(q))$. As $q$ is even,
$(q-1)(q^{2}+1)\equiv 1\pmod{2}$. This means that $(q-1)(q^{2}+1)=3^{3}\cdot
5^{2}\cdot 7\cdot 13$ which does not yield any roots, contradicting
${\mbox{cod}}(G/N)\subseteq{\mbox{cod}}(G)$.

$|{\mbox{cod}}(G/N)|=7$. By Lemma 2.3, we have $G/N\cong{\mbox{PSL}}(3,3)$,
$\textup{{M}}_{11}$, $\textup{{A}}_{7}$ or $\textup{{J}}_{1}$. The elements
$2^{2}\cdot 3^{2}\cdot 13$, $2^{3}\cdot 3^{3}\cdot 11$, $2^{2}\cdot 3\cdot
5\cdot 7$, $3\cdot 5\cdot 11\cdot 19$ are in the codegrees of
${\mbox{PSL}}(3,3),\textup{{M}}_{11},\textup{{A}}_{7}$ or $\textup{{J}}_{1}$,
respectively. However, none of these elements are in
${\mbox{cod}}(\textup{{G}}_{2}(4))$, a contradiction.

$|{\mbox{cod}}(G/N)|=8$. By Lemma 2.3, we have $G/N\cong{\mbox{PSU}}(3,q)$
with $4<q\not\equiv-1\pmod{3}$, $G/N\cong{\mbox{PSL}}(3,q)$ with
$4<q\not\equiv 1\pmod{3}$ or $G/N\cong\textup{{G}}_{2}(2)^{\prime}$.

The only nontrivial elements of $\textup{{G}}_{2}(4)$ not divisible by $3$ are
$2^{11}\cdot 5^{2}\cdot 13$ and $2^{12}\cdot 13$.

If $G/N\cong{\mbox{PSU}}(3,q)$, then
$(q^{2}-q+1)(q+1)^{2}(q-1),q^{3}(q^{2}-q+1)\in{\mbox{cod}}({\mbox{PSU}}(3,q))$.
If $q\equiv 1\mod 3,$ then $q^{3}(q^{2}-q+1)\not\equiv 0\pmod{3}$.
Alternatively, if $q\equiv 0\pmod{3}$, then
$(q^{2}-q+1)(q+1)^{2}(q-1)\not\equiv 0\pmod{3}$. However, setting either to be
equivalent to $2^{11}\cdot 5^{2}\cdot 13$ or $2^{12}\cdot 13$ yield no integer
roots for $q$. Thus, ${\mbox{cod}}(G/N)\not\subseteq{\mbox{cod}}(G)$, a
contradiction.

If $G/N\cong{\mbox{PSL}}(3,q)$, we have
$q^{3}(q^{2}+q+1)\in{\mbox{cod}}({\mbox{PSL}}(3,q))$. As $q\not\equiv
1\pmod{3},q^{3}(q^{2}+q+1)\not\equiv 0\pmod{3}$. However,
$q^{3}(q^{2}+q+1)=2^{11}\cdot 5^{2}\cdot 13$ or $2^{12}\cdot 13$ has no
integer roots. Thus, ${\mbox{cod}}(G/N)\not\subseteq{\mbox{cod}}(G)$, a
contradiction.

In the final case, $3^{3}\cdot 7\in{\mbox{cod}}(\textup{{G}}_{2}(2)^{\prime})$
but $3^{3}\cdot 7\not\in{\mbox{cod}}(\textup{{G}}_{2}(4))$, contradicting
${\mbox{cod}}(G/N)\subseteq{\mbox{cod}}(G)$.

$|{\mbox{cod}}(G/N)|=9$. By Lemma 2.3, we have $G/N\cong{\mbox{PSL}}(3,q)$
with $4<q\equiv 1\pmod{3}$ or $G/N\cong{\mbox{PSU}}(3,q)$ with
$4<q\equiv-1\pmod{3}$.

Both $q^{2}-q+1$ and $q^{2}+q+1$ are not divisible by $9$.

In the first case, we have
$\frac{1}{3}q^{3}(q^{2}+q+1)\in{\mbox{cod}}({\mbox{PSL}}(3,q))$. Thus,
$\frac{1}{3}q^{3}(q^{2}+q+1)\not\equiv 0\pmod{3}$. Furthermore, we have
$\frac{1}{3}q^{3}(q^{2}-q+1)\in{\mbox{cod}}({\mbox{PSU}}(3,q))$. This is not
divisible by $3$. The equations $\frac{1}{3}q^{3}(q^{2}\pm q+1)=2^{11}\cdot
5^{2}\cdot 13$ and $\frac{1}{3}q^{3}(q^{2}\pm q+1)=2^{12}\cdot 13$ have no
integer roots, contradicting ${\mbox{cod}}(G/N)\subseteq{\mbox{cod}}(G)$.

$|{\mbox{cod}}(G/N)|=10$. By Lemma 2.3, we have $G/N\cong\textup{{M}}_{22}$.
However, $2^{7}\cdot 3\cdot 5\cdot 11$ is only present in
${\mbox{cod}}(\textup{{M}}_{22})$ and not in
${\mbox{cod}}(\textup{{G}}_{2}(4))$, contradicting
${\mbox{cod}}(G/N)\subseteq{\mbox{cod}}(G)$.

$|{\mbox{cod}}(G/N)|=11$. By Lemma 2.3, we have $G/N\cong{\mbox{PSL}}(4,2)$,
$G/N\cong\textup{{M}}_{12}$, $G/N\cong\textup{{M}}_{23}$ or
$G/N\cong{}^{2}\textup{{G}}_{2}(q)$ for $q=3^{2f+1},f\geq 1$.

The codegree sets of ${\mbox{PSL}}(4,2)$, $\textup{{M}}_{12}$ and
$\textup{{M}}_{23}$ contain $3^{2}\cdot 5\cdot 7$, $2^{2}\cdot 3\cdot 5\cdot
11$ and $2^{7}\cdot 7\cdot 11$, respectively, none of which are elements of
the codegree set of $\textup{{G}}_{2}(4)$.

Furthermore, $(q^{2}-1)(q^{2}-q+1)\in{\mbox{cod}}({}^{2}\textup{{G}}_{2}(q))$.
However, $(q^{2}-1)(q^{2}-q+1)\not\equiv 0\pmod{3}$ in this case, so we must
have $(q^{2}-1)(q^{2}-q+1)=2^{11}\cdot 5^{2}\cdot 13$ or $2^{12}\cdot 13$, yet
this does not yield an integer root for $q$, contradicting
${\mbox{cod}}(G/N)\subseteq{\mbox{cod}}(G)$.

$|{\mbox{cod}}(G/N)|=12$. By Lemma 2.3, we have $G/N\cong\textup{{S}}_{4}(4)$.
However, $17$ is a factor of $|{\mbox{cod}}(\textup{{S}}_{4}(4))|$ and not
$|{\mbox{cod}}(\textup{{G}}_{2}(4)|$, contradicting
${\mbox{cod}}(G/N)\subseteq{\mbox{cod}}(G)$.

$|{\mbox{cod}}(G/N)|=13$. By Lemma 2.3, we have $G/N\cong\textup{{U}}_{4}(2)$
or $G/N\cong\textup{{S}}_{4}(q)$ for $q=2^{f}>4$.

In the first case, the element $3^{4}\cdot
5\in{\mbox{cod}}(\textup{{U}}_{4}(2))$ but it is not in
${\mbox{cod}}(\textup{{G}}_{2}(4))$.

In the second case, $q^{4}(q-1)^{2}\in{\mbox{cod}}(\textup{{S}}_{4}(q))$.
However, by a complete search there is no integer $q$ such that this element
becomes equal to an element in ${\mbox{cod}}(\textup{{G}}_{2}(4))$, a
contradiction.

$|{\mbox{cod}}(G/N)|=14$. By Lemma 2.3, we have $G/N\cong\textup{{U}}_{4}(3)$,
$G/N\cong{}^{2}\textup{{F}}_{4}(2)^{\prime}$ or $G/N\cong\textup{{J}}_{3}$.

However, the codegree sets of $\textup{{U}}_{4}(3)$,
${}^{2}\textup{{F}}_{4}(2)^{\prime}$ and $\textup{{J}}_{3}$ have $2^{4}\cdot
3^{6}$, $3^{3}\cdot 5^{2}\cdot 13$ and $2^{7}\cdot 3^{5}\cdot 19$,
respectively, but these are not in ${\mbox{cod}}(\textup{{G}}_{2}(4))$. This
contradicts ${\mbox{cod}}(G/N)\subseteq{\mbox{cod}}(G)$.

$|{\mbox{cod}}(G/N)|=15$. By Lemma 2.3, we have $G/N\cong\textup{{G}}_{2}(3)$.
However, the element $2^{5}\cdot 3^{6}\cdot 13$ is only present in the
codegree set of $\textup{{G}}_{2}(3)$ and not in the codegree set of
$\textup{{G}}_{2}(4)$, which contradicts
${\mbox{cod}}(G/N)\subseteq{\mbox{cod}}(G)$.

$|{\mbox{cod}}(G/N)|=16$. By Lemma 2.3, we have $G/N\cong\textup{{A}}_{9}$ or
$G/N\cong\textup{{J}}_{2}$. In the $\textup{{A}}_{9}$ case, $2^{4}\cdot
3^{3}\cdot 5\in{\mbox{cod}}(\textup{{A}}_{9})$ but is not in
${\mbox{cod}}(\textup{{G}}_{2}(4))$, while in the $\textup{{J}}_{2}$ case
$2^{6}\cdot 3^{3}\cdot 5^{2}\in{\mbox{cod}}(\textup{{J}}_{2})$ but this
element is not in ${\mbox{cod}}(\textup{{G}}_{2}(4))$. This contradicts
${\mbox{cod}}(G/N)\subseteq{\mbox{cod}}(G)$.

$|{\mbox{cod}}(G/N)|=17$. Then $G/N\cong\textup{{M}}^{c}L$ or
${\mbox{PSL}}(4,3)$.

However, there are three nontrivial odd elements in
${\mbox{cod}}(\textup{{M}}^{c}L)$ and only a nontrivial odd codegree of
${\mbox{cod}}(\textup{{G}}_{2}(4))$, a contradiction.

If $G/N={\mbox{PSL}}(4,3)$, the element $3^{6}\cdot
13\in{\mbox{cod}}({\mbox{PSL}}(4,3))$ but it is not in
${\mbox{cod}}(\textup{{G}}_{2}(4))$, a contradiction.

$|{\mbox{cod}}(G/N)|=18$. We have $G/N\cong\textup{{G}}_{2}(4)$,
$\textup{{S}}_{4}(5)$ or HS. If $G/N\cong\textup{{S}}_{4}(5)$, then
$5^{4}\cdot 13\in{\mbox{cod}}(G/N)$. If $G/N\in{\mbox{HS}}$, then $2^{2}\cdot
3^{2}\cdot 5\cdot 7\cdot 11\in{\mbox{cod}}(G/N)$. Neither is in
${\mbox{cod}}(\textup{{G}}_{2}(4))$, a contradiction. Having ruled out all
other cases, we conclude $G/N\cong\textup{{G}}_{2}(4)$.  $\hbox
to0.0pt{$\sqcap$\hss}\sqcup$

######  Lemma 3.14.

Let $G$ be a finite group with ${\mbox{cod}}(G)={\mbox{cod}}({\mbox{HS}})$. If
$N$ is the maximal normal subgroup of $G$, we have $G/N\cong{\mbox{HS}}$.

###### Proof.

As $G/N$ is a non-abelian simple group,
${\mbox{cod}}(G/N)\subseteq{\mbox{cod}}(G)$, so we have that
${\mbox{cod}}(G/N)=4,5,6,7,8,9,10,11,12,13,14,15,16,17,18$, or $19$.

$|{\mbox{cod}}(G/N)|=4$. By Lemma 2.3, $G/N\cong{\mbox{PSL}}(2,k)$ where
$k=2^{f}\geq 4$, so $k^{2}-1\in{\mbox{cod}}(G/N)$. For all even $k$, $k^{2}-1$
is odd. $k^{2}-1$ is the only nontrivial odd codegree in ${\mbox{cod}}(G/N)$,
but None of the elements in the ${\mbox{cod}}({\mbox{HS}})$ is odd.
Contradicting ${\mbox{cod}}(G/N)\subseteq{\mbox{cod}}(G)$.

$|{\mbox{cod}}(G/N)|=5$. By Lemma 2.3, $G/N\cong{\mbox{PSL}}(2,k)$ where $k$
is a power of an odd prime number. This implies that $\displaystyle
k(k-\epsilon(k)\in{\mbox{cod}}(G/N)$, where $k=(-1)^{(k-1)/2}$. By a complete
search, $k(k+1)$ cannot be equal to $2^{8}\cdot 3^{3}\cdot 5^{3}$ or
$2^{8}\cdot 5^{3}$ in ${\mbox{cod}}({\mbox{HS}})$ since it is half of another
nontrivial codegrees, a contradiction since $k\in\mathbb{Z}$.

$|{\mbox{cod}}(G/N)|=6$. By Lemma 2.3, $G/N\cong{\mbox{PSL}}(3,4)$ or
$G/N\cong{}^{2}B_{2}(q)$, with $q=2^{2n+1}$.

If $G/N\cong{\mbox{PSL}}(3,4)$, then $3^{2}\cdot 5\cdot
7\in{\mbox{cod}}({\mbox{PSL}}(3,4))$ but $3^{2}\cdot 5\cdot
7\not\in{\mbox{cod}}({\mbox{HS}})$, contradicting
${\mbox{cod}}(G/N)\subseteq{\mbox{cod}}(G)$.

If $G/N\cong{}^{2}B_{2}(q)$, notice that
$(q-1)(q^{2}+1)\in{\mbox{cod}}({}^{2}B_{2}(q))$. As $q$ is even,
$(q-1)(q^{2}+1)\equiv 1\pmod{2}$. This means that
$(q-1)(q^{2}+1)\notin{\mbox{cod}}(HS)$ since none of these are nontrivial
codegrees. Contradicting ${\mbox{cod}}(G/N)\subseteq{\mbox{cod}}(G)$.

$|{\mbox{cod}}(G/N)|=7$. By Lemma 2.3, we have $G/N\cong{\mbox{PSL}}(3,3)$,
$\textup{{M}}_{11}$, $\textup{{A}}_{7}$ or $\textup{{J}}_{1}$. The elements
$2^{2}\cdot 3^{2}\cdot 13$, $2^{3}\cdot 3^{3}\cdot 11$, $2^{2}\cdot 3\cdot
5\cdot 7$, $3\cdot 5\cdot 11\cdot 19$ are in the codegrees of
${\mbox{PSL}}(3,3)$, $\textup{{M}}_{11}$, $\textup{{A}}_{7}$ or
$\textup{{J}}_{1}$, respectively. However, none of these elements are in
${\mbox{cod}}(\textup{{G}}_{2}(4))$, a contradiction.

$|{\mbox{cod}}(G/N)|=8$. By Lemma 2.3, we have $G/N\cong{\mbox{PSU}}(3,q)$
with $4<q\not\equiv-1\pmod{3}$, $G/N\cong{\mbox{PSL}}(3,q)$ with
$4<q\not\equiv 1\pmod{3}$ or $G/N\cong\textup{{G}}_{2}(2)^{\prime}$. There are
three nontrivial elements of $\textup{{G}}_{2}(4)$ not divisible by $3$ are
$2^{9}\cdot 5^{3}$, $2^{6}\cdot 5^{2}\cdot 11$ and $2^{8}\cdot 5^{3}$.

If $G/N\cong{\mbox{PSU}}(3,q)$, then
$(q^{2}-q+1)(q+1)^{2}(q-1),q^{3}(q^{2}-q+1)\in{\mbox{cod}}({\mbox{PSU}}(3,q))$.
If $q\equiv 1\mod 3,$ then $q^{3}(q^{2}-q+1)\not\equiv 0\pmod{3}$.
Alternatively, if $q\equiv 0\pmod{3}$, then
$(q^{2}-q+1)(q+1)^{2}(q-1)\not\equiv 0\pmod{3}$. However, setting either to be
equivalent to $2^{9}\cdot 5^{3}$, $2^{6}\cdot 5^{2}\cdot 11$ or $2^{8}\cdot
5^{3}$ yield no integer roots for $q$. Thus,
${\mbox{cod}}(G/N)\not\subseteq{\mbox{cod}}(G)$, a contradiction.

If $G/N\cong{\mbox{PSL}}(3,q)$, we have
$q^{3}(q^{2}+q+1)\in{\mbox{cod}}({\mbox{PSL}}(3,q))$. As $q\not\equiv
1\pmod{3},q^{3}(q^{2}+q+1)\not\equiv 0\pmod{3}$. However,
$q^{3}(q^{2}+q+1)=2^{9}\cdot 5^{3}$, $2^{6}\cdot 5^{2}\cdot 11$ or $2^{8}\cdot
5^{3}$ has no integer roots. Thus,
${\mbox{cod}}(G/N)\not\subseteq{\mbox{cod}}(G)$, a contradiction.

In the final case, $3^{3}\cdot 7\in{\mbox{cod}}(\textup{{G}}_{2}(2)^{\prime})$
but $3^{3}\cdot 7\not\in{\mbox{cod}}({\mbox{HS}})$, contradicting
${\mbox{cod}}(G/N)\subseteq{\mbox{cod}}(G)$.

$|{\mbox{cod}}(G/N)|=9$. By Lemma 2.3, we have $G/N\cong{\mbox{PSL}}(3,q)$
with $4<q\equiv 1\pmod{3}$ or $G/N\cong{\mbox{PSU}}(3,q)$ with
$4<q\equiv-1\pmod{3}$. Both $q^{2}-q+1$ and $q^{2}+q+1$ are not divisible by
$9$.

In the first case, we have
$\frac{1}{3}q^{3}(q^{2}+q+1)\in{\mbox{cod}}({\mbox{PSL}}(3,q))$. Thus,
$\frac{1}{3}q^{3}(q^{2}+q+1)\not\equiv 0\pmod{3}$. Furthermore, we have
$\frac{1}{3}q^{3}(q^{2}-q+1)\in{\mbox{cod}}({\mbox{PSU}}(3,q))$. This is not
divisible by $3$.

The equations $\frac{1}{3}q^{3}(q^{2}\pm q+1)=2^{9}\cdot 5^{3}$, $2^{6}\cdot
5^{2}\cdot 11$ or $2^{8}\cdot 5^{3}$, and $\frac{1}{3}q^{3}(q^{2}\pm
q+1)=2^{9}\cdot 5^{3}$, $2^{6}\cdot 5^{2}\cdot 11$ or $2^{8}\cdot 5^{3}$ have
no integer roots, contradicting ${\mbox{cod}}(G/N)\subseteq{\mbox{cod}}(G)$.

$|{\mbox{cod}}(G/N)|=10$. By Lemma 2.3, we have $G/N\cong\textup{{M}}_{22}$.
However, $2^{7}\cdot 3\cdot 5\cdot 11$ is only present in
${\mbox{cod}}(\textup{{M}}_{22})$ and not in ${\mbox{cod}}({\mbox{HS}}))$,
contradicting ${\mbox{cod}}(G/N)\subseteq{\mbox{cod}}(G)$.

$|{\mbox{cod}}(G/N)|=11$. Lemma 2.3, we have $G/N\cong{\mbox{PSL}}(4,2)$,
$G/N\cong\textup{{M}}_{12}$, $G/N\cong\textup{{M}}_{23}$ or
$G/N\cong{}^{2}\textup{{G}}_{2}(q)$ for $q=3^{2f+1},f\geq 1$.

${\mbox{cod}}({\mbox{PSL}}(4,2))$, ${\mbox{cod}}(\textup{{M}}_{12})$ and
${\mbox{cod}}(\textup{{M}}_{23})$ contain $3^{2}\cdot 5\cdot 7$, $2^{2}\cdot
3\cdot 5\cdot 11$ and $2^{7}\cdot 7\cdot 11,$ respectively, none of which are
elements of the codegree set of HS.

Furthermore, $(q^{2}-1)(q^{2}-q+1)\in{\mbox{cod}}({}^{2}\textup{{G}}_{2}(q))$.
However, $(q^{2}-1)(q^{2}-q+1)\not\equiv 0\pmod{3}$ in this case, so we must
have $(q^{2}-1)(q^{2}-q+1)=2^{8}\cdot 5^{3}$, $2^{6}\cdot 5^{2}\cdot 11$ or
$2^{9}\cdot 5^{3}$, yet this does not yield an integer root for $q$,
contradicting ${\mbox{cod}}(G/N)\subseteq{\mbox{cod}}(G)$.

$|{\mbox{cod}}(G/N)|=12$. By Lemma 2.3, we have $G/N\cong\textup{{S}}_{4}(4)$.
However, $17$ is a factor of $|{\mbox{cod}}(\textup{{S}}_{4}(4))|$ and not
$|{\mbox{cod}}({\mbox{HS}})|$, contradicting
${\mbox{cod}}(G/N)\subseteq{\mbox{cod}}(G)$.

$|{\mbox{cod}}(G/N)|=13$. By Lemma 2.3, we have $G/N\cong\textup{{U}}_{4}(2)$
or $G/N\cong\textup{{S}}_{4}(q)$ for $q=2^{f}>4$.

In the first case, the element $3^{4}\cdot
5\in{\mbox{cod}}(\textup{{U}}_{4}(2))$ but it is not in
${\mbox{cod}}({\mbox{HS}}).$

In the second case, $q^{4}(q-1)^{2}\in{\mbox{cod}}(\textup{{S}}_{4}(q))$.
However, by a complete search there is no integer $q$ such that
$q^{4}(q-1)^{2}$ equals an element in ${\mbox{cod}}({\mbox{HS}})$, a
contradiction.

$|{\mbox{cod}}(G/N)|=14$. By Lemma 2.3, we have $G/N\cong\textup{{U}}_{4}(3)$,
$G/N\cong{}^{2}\textup{{F}}_{4}(2)^{\prime}$ or $G/N\cong\textup{{J}}_{3}$.
However, the codegree sets of $\textup{{U}}_{4}(3)$,
${}^{2}\textup{{F}}_{4}(2)^{\prime}$ and $\textup{{J}}_{3}$ have $2^{4}\cdot
3^{6}$, $3^{3}\cdot 5^{2}\cdot 13$ and $2^{7}\cdot 3^{5}\cdot 19$,
respectively, but these are not in ${\mbox{cod}}({\mbox{HS}})$. This
contradicts ${\mbox{cod}}(G/N)\subseteq{\mbox{cod}}(G)$.

$|{\mbox{cod}}(G/N)|=15$. By Lemma 2.3, we have $G/N\cong\textup{{G}}_{2}(3)$.
However, the element $2^{5}\cdot 3^{6}\cdot 13$ is only present in the
codegree set of $\textup{{G}}_{2}(3)$ and not in the codegree set of HS, which
contradicts ${\mbox{cod}}(G/N)\subseteq{\mbox{cod}}(G)$.

$|{\mbox{cod}}(G/N)|=16$. By Lemma 2.3, we have $G/N\cong\textup{{A}}_{9}$ or
$G/N\cong\textup{{J}}_{2}$. In the $\textup{{A}}_{9}$ case, $2^{4}\cdot
3^{3}\cdot 5\in{\mbox{cod}}(\textup{{A}}_{9})$ but is not in
${\mbox{cod}}({\mbox{HS}})$, while in the $\textup{{J}}_{2}$ case $2^{6}\cdot
3^{3}\cdot 5^{2}\in{\mbox{cod}}(\textup{{J}}_{2})$ but $2^{6}\cdot 3^{3}\cdot
5^{2}\not\in{\mbox{cod}}({\mbox{HS}})$. This contradicts
${\mbox{cod}}(G/N)\subseteq{\mbox{cod}}(G)$.

$|{\mbox{cod}}(G/N)|=17$. Then $G/N\cong\textup{{M}}^{c}L$ or
${\mbox{PSL}}(4,3)$.

Notice that the element $2^{7}\cdot 5^{3}\cdot 11$ is only in
${\mbox{cod}}(\textup{{M}}^{c}L)$ and not the codegree of
${\mbox{cod}}({\mbox{HS}})$, a contradiction.

If $G/N={\mbox{PSL}}(4,3)$, the element $2^{3}\cdot
3^{6}\in{\mbox{cod}}({\mbox{PSL}}(4,3))$ but it is not in
${\mbox{cod}}({\mbox{HS}})$.

$|{\mbox{cod}}(G/N)|=18$. We have $G/N\cong\textup{{G}}_{2}(4)$ or
$\textup{{S}}_{4}(5)$. If $G/N\cong\textup{{S}}_{4}(5)$, then $5^{4}\cdot
13\in{\mbox{cod}}(G/N)$. If $G/N\in\textup{{G}}_{2}(4)$, then $2^{12}\cdot
3^{3}\cdot 5\cdot 7\in{\mbox{cod}}(G/N)$. Neither is in
${\mbox{cod}}({\mbox{HS}})$, a contradiction.

Having ruled out all other cases, we conclude $G/N\cong{\mbox{HS}}$.  $\hbox
to0.0pt{$\sqcap$\hss}\sqcup$

######  Lemma 3.15.

Let $G$ be a finite group with ${\mbox{cod}}(G)={\mbox{cod}}({\mbox{O'N}})$.
If $N$ is the maximal normal subgroup of $G$, it we have
$G/N\cong{\mbox{O'N}}$.

###### Proof.

As $G/N$ is a non-abelian simple group,
${\mbox{cod}}(G/N)\subseteq{\mbox{cod}}(G)$, so we have that
${\mbox{cod}}(G/N)=4,5,6,7,8,9,10,11,12,13,14,15,16,17$ or $18$.

$|{\mbox{cod}}(G/N)|=4$. By Lemma 2.3, $G/N\cong{\mbox{PSL}}(2,k)$ where
$k=2^{f}\geq 4$. The element $k^{2}-1\in{\mbox{cod}}(G/N)$. For all even $k$,
$k^{2}-1$ is odd. $k^{2}-1$ must equal $7^{3}\cdot 11\cdot 19\cdot 31$ or
$3^{4}\cdot 5\cdot 11\cdot 19\cdot 31$ but this yields no integer roots for
$k$, contradicting ${\mbox{cod}}(G/N)\subseteq{\mbox{cod}}(G)$.

$|{\mbox{cod}}(G/N)|=5$. By Lemma 2.3, $G/N\cong{\mbox{PSL}}(2,k)$ where $k$
is a power of an odd prime number. We have $\displaystyle
k(k-\epsilon(k)\in{\mbox{cod}}(G/N)$, where $k=(-1)^{(k-1)/2}$. By a complete
search, $k(k+1)$ cannot be equal to any of the nontrivial elements in
${\mbox{cod}}({\mbox{O'N}})$ for any integer $k$, a contradiction.

$|{\mbox{cod}}(G/N)|=6$. By Lemma 2.3, $G/N\cong{\mbox{PSL}}(3,4)$ or
$G/N\cong{}^{2}B_{2}(q)$, with $q=2^{2n+1}$.

If $G/N\cong{\mbox{PSL}}(3,4)$ then $3^{2}\cdot 5\cdot
7\in{\mbox{cod}}({\mbox{PSL}}(3,4))$ but $3^{2}\cdot 5\cdot
7\not\in{\mbox{cod}}({\mbox{O'N}})$, contradicting
${\mbox{cod}}(G/N)\subseteq{\mbox{cod}}(G)$.

If $G/N\cong{}^{2}B_{2}(q)$, notice that
$(q-1)(q^{2}+1)\in{\mbox{cod}}({}^{2}B_{2}(q))$. As $q$ is even,
$(q-1)(q^{2}+1)\equiv 1\pmod{2}$. This means that $(q-1)(q^{2}+1)=7^{3}\cdot
11\cdot 19\cdot 31$ or $3^{4}\cdot 5\cdot 11\cdot 19\cdot 31$ which does not
yield any roots, contradicting ${\mbox{cod}}(G/N)\subseteq{\mbox{cod}}(G)$.

$|{\mbox{cod}}(G/N)|=7$. By Lemma 2.3, we have $G/N\cong{\mbox{PSL}}(3,3)$,
$\textup{{M}}_{11}$, $\textup{{A}}_{7}$ or $\textup{{J}}_{1}$.

The elements $2^{2}\cdot 3^{2}\cdot 13$, $2^{3}\cdot 3^{3}\cdot 11$,
$2^{2}\cdot 3\cdot 5\cdot 7$, $3\cdot 5\cdot 11\cdot 19$ are in the codegrees
of ${\mbox{PSL}}(3,3)$, $\textup{{M}}_{11}$, $\textup{{A}}_{7}$ or
$\textup{{J}}_{1}$, respectively. However, none of these are elements of
${\mbox{cod}}({\mbox{O'N}})$, a contradiction.

$|{\mbox{cod}}(G/N)|=8$. By Lemma 2.3, we have $G/N\cong{\mbox{PSU}}(3,q)$
with $4<q\not\equiv-1\pmod{3}$, $G/N\cong{\mbox{PSL}}(3,q)$ with
$4<q\not\equiv 1\pmod{3}$ or $G/N\cong\textup{{G}}_{2}(2)^{\prime}$.

If $G/N\cong{\mbox{PSU}}(3,q)$, then
$q^{3}(q^{2}-q+1)\in{\mbox{cod}}({\mbox{PSU}}(3,q))$. By a complete search,
$q^{3}(q^{2}-q+1)$ does not equal an element of ${\mbox{cod}}({\mbox{O'N}})$
for any integer $q$.

If $G/N\cong{\mbox{PSL}}(3,q)$, we have
$q^{3}(q^{2}+q+1)\in{\mbox{cod}}({\mbox{PSL}}(3,q))$. By a complete search,
$q^{3}(q^{2}+q+1)$ does not equal an element of ${\mbox{cod}}({\mbox{O'N}})$
for any integer $q$. Thus, ${\mbox{cod}}(G/N)\not\subseteq{\mbox{cod}}(G)$, a
contradiction.

In the final case, $3^{3}\cdot 7\in{\mbox{cod}}(\textup{{G}}_{2}(2)^{\prime})$
but $3^{3}\cdot 7\not\in{\mbox{cod}}({\mbox{O'N}})$, contradicting
${\mbox{cod}}(G/N)\subseteq{\mbox{cod}}(G)$.

$|{\mbox{cod}}(G/N)|=9$. By Lemma 2.3, we have $G/N\cong{\mbox{PSL}}(3,q)$
with $4<q\equiv 1\pmod{3}$ or $G/N\cong{\mbox{PSU}}(3,q)$ with
$4<q\equiv-1\pmod{3}$.

In the first case, we have
$\frac{1}{3}q^{3}(q^{2}+q+1)\in{\mbox{cod}}({\mbox{PSL}}(3,q))$. Furthermore,
we have $\frac{1}{3}q^{3}(q^{2}-q+1)\in{\mbox{cod}}({\mbox{PSU}}(3,q))$.
Regardless, by a complete search, these expressions do not equal an element of
${\mbox{cod}}({\mbox{O'N}})$ for any integer $q$, contradicting
${\mbox{cod}}(G/N)\subseteq{\mbox{cod}}(G)$.

$|{\mbox{cod}}(G/N)|=10$. By Lemma 2.3, we have $G/N\cong\textup{{M}}_{22}$.
However, $2^{7}\cdot 3\cdot 5\cdot 11$ is only present in
${\mbox{cod}}(\textup{{M}}_{22})$ and not in ${\mbox{cod}}({\mbox{O'N}})$,
contradicting ${\mbox{cod}}(G/N)\subseteq{\mbox{cod}}(G)$.

$|{\mbox{cod}}(G/N)|=11$. Lemma 2.3, we have $G/N\cong{\mbox{PSL}}(4,2)$,
$G/N\cong\textup{{M}}_{12}$, $G/N\cong\textup{{M}}_{23}$ or
$G/N\cong{}^{2}\textup{{G}}_{2}(q)$ for $q=3^{2f+1},f\geq 1$.

${\mbox{cod}}({\mbox{PSL}}(4,2))$, ${\mbox{cod}}(\textup{{M}}_{12})$ and
${\mbox{cod}}(\textup{{M}}_{23})$ contain $3^{2}\cdot 5\cdot 7$, $2^{2}\cdot
3\cdot 5\cdot 11$ and $2^{7}\cdot 7\cdot 11$, respectively, none of which are
elements of the codegree set of O’N.

Furthermore, $(q^{2}-1)(q^{2}-q+1)\in{\mbox{cod}}({}^{2}\textup{{G}}_{2}(q))$.
By a complete search, this element does not equal an element of
${\mbox{cod}}({\mbox{O'N}})$ for any integer root for $q$, contradicting
${\mbox{cod}}(G/N)\subseteq{\mbox{cod}}(G)$.

$|{\mbox{cod}}(G/N)|=12$. By Lemma 2.3, we have $G/N\cong\textup{{S}}_{4}(4)$.
However, $2^{7}\cdot 5^{2}\cdot 17$ is in of
${\mbox{cod}}(\textup{{S}}_{4}(4))$ and not in ${\mbox{cod}}({\mbox{O'N}})$,
contradicting ${\mbox{cod}}(G/N)\subseteq{\mbox{cod}}(G)$.