# On Recognition by Order and Degree Pattern of Finite Simple Groups

B. Akbari and A. R. Moghaddamfar
Department of Mathematics, K. N. Toosi University of Technology,
P. O. Box $16315$-$1618$, Tehran, Iran
and
E-mails<EMAIL_ADDRESS>and<EMAIL_ADDRESS>
###### Abstract

Let ${\rm GK}(G)$ be the prime graph associated with a finite group $G$ and
$D(G)$ be the degree pattern of $G$. A finite group $G$ is said to be $k$-fold
OD-characterizable if there exist exactly $k$ non-isomorphic groups $H$ such
that $|H|=|G|$ and $D(H)=D(G)$. A 1-fold OD-characterizable group is simply
called OD-characterizable. The purpose of this paper is threefold. First, it
provides the reader with a few useful and efficient tools on OD-
characterizability of finite groups. Second, it lists a number of such simple
groups that have been already investigated. Third, it shows that the simple
groups $L_{6}(3)$ and $U_{4}(5)$ are OD-characterizable, too.
Keywords: prime graph, degree pattern, simple group.

footnotetext: $2010$ Mathematics Subject Classification: 20D05, 20D06.

##  1 Introduction and Previous Results

In this paper, we will consider only finite groups. The set of elements order
of a finite group $G$ is denoted by $\omega(G)$ and called the spectrum of $G$
(see [27]). This set is closed and partially ordered by the divisibility
relation; therefore, it is determined uniquely from the subset $\mu(G)$ of all
maximal elements of $\omega(G)$ with respect to divisibility. For a natural
number $n$ we denote by $\pi(n)$ the set of prime divisors of $n$ and put
$\pi(G)=\pi(|G|)$.

There are a lot of ways to associate a graph to a finite group. One of well-
known graphs associated with a finite group $G$ is called prime graph (or
Gruenberg-Kegel graph) of $G$ and denoted by ${\rm GK}(G)$, which is a simple
undirected graph and construct as follows. The vertices are all prime divisors
of $|G|$ and two distinct vertices $p$ and $q$ are adjacent if and only if
$pq\in\omega(G)$. The number of connected components of ${\rm GK}(G)$ is
denoted by $s(G)$, and the connected components are denoted by
$\pi_{i}=\pi_{i}(G)$, with $1\leq i\leq s(G)$. When the group $G$ has even
order, we will assume that $2\in\pi_{1}(G)$. Finite simple groups with
disconnected prime graph were described in [11, 24]. A complete list of these
groups (with corrected misprints) can be found in [12, Tables 1(a)-1(c)].

In [21], Suzuki studied the structure of the prime graphs associated with
finite simple groups, and found the following interesting result.

######  Proposition 1.1 (Suzuki)

([21], Theorem B) Let $G$ be a finite simple group whose prime graph ${\rm
GK}(G)$ is disconnected and let $\Delta$ be a connected component of ${\rm
GK}(G)$ whose vertex set does not contain $2$. Then $\Delta$ is a clique.

###### Remark 1.1

By a clique we shall mean a complete graph and we write $K_{n}$ to be the
clique of order $n$. Let $\Gamma=(V,E)$ be a simple graph. Note that, in
Proposition 1.1, it does not require $G$ to be simple. Moreover, it shows that
the prime graph of $G$ has the following structure:

${\rm GK}(G)={\rm GK}[\pi_{1}]\oplus K_{n_{2}}\oplus\cdots\oplus K_{n_{s}},$

where ${\rm GK}[\pi_{1}]$ denotes the induced subgraph ${\rm GK}(G)[\pi_{1}]$,
$n_{i}=|\pi_{i}|$ and $s=s(G)$. We recall that the induced subgraph ${\rm
GK}(G)[\pi_{1}]$ has the vertex set $\pi_{1}$ and it contains all edges of
${\rm GK}(G)$ which join vertices in $\pi_{1}$.

The degree $\deg_{G}(p)$ of a vertex $p\in\pi(G)$ is the number of edges
incident on $p$ in ${\rm GK}(G)$. If $\pi(G)=\\{p_{1},p_{2},\ldots,p_{k}\\}$
with $p_{1}<p_{2}<\cdots<p_{k}$, then we put

${\rm
D}(G):=\big{(}\deg_{G}(p_{1}),\deg_{G}(p_{2}),\ldots,\deg_{G}(p_{k})\big{)},$

and call this $k$-tuple the degree pattern of $G$. We denote by $h_{\rm
OD}(G)$ the number of isomorphism classes of finite groups $H$ such that
$|H|=|G|$ and ${\rm D}(H)={\rm D}(G)$. Obviously, $1\leqslant h_{\rm
OD}(G)<\infty$ for any finite group $G$. In terms of function $h_{\rm
OD}(\cdot)$, the groups $G$ are classified as follows:

###### Definition 1.1

A finite group $G$ is called a $k$-fold OD-characterizable group if $h_{\rm
OD}(G)=k$. A $1$-fold OD-characterizable group is simply called an OD-
characterizable group.

The notation for sporadic and simple groups of Lie type is borrowed from [6].
Moreover, we use the notation $\mathbb{A}_{n}$ to denote an alternating of
degree $n$. There are scattered results in the literature showing that many of
simple groups have the same prime graphs, for instance, it was shown in [7],
[23], [35] that each of the following sets consist of simple groups

(a) | $\\{S_{2n}(q),O_{2n+1}(q)\\}$ |  | [23, Proposition 7.5]
---|---|---|---
(b) |  $\\{L_{2}(11),M_{11}\\}$, |  | [7, Theorem 3]
(c) |  $\\{\mathbb{A}_{5},\mathbb{A}_{6}\\}$, $\\{\mathbb{A}_{7},L_{2}(49),U_{4}(3)\\}$, $\\{\mathbb{A}_{9},J_{2},S_{6}(2),O^{+}_{8}(2)\\}$, |  | [35, Theorem]
(d) |  $\\{\mathbb{A}_{n},\mathbb{A}_{n-1}\\}$, $n$ is odd, and $n$ and $n-4$ are composite, |  |  [35, Theorem].

have the same prime graphs and so the same degree pattern. On the other hand,
the symplectic group $S_{2n}(q)$ and the orthogonal group $O_{2n+1}(q)$ have
the same orders, too, that is

$|S_{2n}(q)|=|O_{2n+1}(q)|=\frac{1}{(2,q-1)}q^{n^{2}}\prod\limits^{n}_{i=1}(q^{2i}-1).$

Therefore, in the case that $S_{2n}(q)\ncong O_{2n+1}(q)$ (that is $q$ odd and
$n>2$), we have

$h_{\rm OD}(S_{2n}(q))=h_{\rm OD}(O_{2n+1}(q))\geq 2.$

Notice that, we have $S_{2n}(2^{m})\cong O_{2n+1}(2^{m})$ and $S_{4}(q)\cong
O_{5}(q)$ (see for example [5]). There is also another pair of non-isomorphic
simple groups with the same order, that is $L_{4}(2)\cong\mathbb{A}_{8}$ and
$L_{3}(4)$. Actually, about simple groups with the same order, we have the
following proposition (see [20]).

###### Proposition 1.2

Every finite simple group can be determined up to isomorphism in the class of
finite simple groups by its order, except exactly in the following cases:

  * (a)

$L_{4}(2)\cong\mathbb{A}_{8}$ and $L_{3}(4)$ have the same order, i.e.,
$20160$,

  * (b)

$O_{2n+1}(q)$ and $S_{2n}(q)$ have the same order for $q$ odd, $n>2$.

As an immediate consequence of Proposition 1.2 we have the following
corollary.

###### Corollary 1.1

If $G$ is a finite simple group, then there exists at most one finite simple
group, non-isomorphic to $G$, with the same order and degree pattern as $G$.

It was currently shown that many finite simple groups are OD-characterizable
or $2$-fold OD-characterizable. We have listed these groups in Table 1.

Table 1. Some non-abelian simple groups $S$ with $h_{\rm OD}(S)=1$ or $2$.

$\begin{array}[]{l|l|c|l}\hline\cr S&{\rm Conditions\ on}\ S&h_{\rm
OD}(S)&{\rm Refs.}\\\ \hline\cr\mathbb{A}_{n}&\ n=p,p+1,p+2\ (p\ {\rm a\
prime})&1&\cite[cite]{[\@@bibref{}{mz1}{}{}]},\cite[cite]{[\@@bibref{}{mzd}{}{}]}\\\
&\ 5\leqslant n\leqslant 100,n\neq
10&1&\cite[cite]{[\@@bibref{}{hm}{}{}]},\cite[cite]{[\@@bibref{}{kogani}{}{}]},\cite[cite]{[\@@bibref{}{banoo-
alireza}{}{}]},\\\
&&&\cite[cite]{[\@@bibref{}{mz3}{}{}]},\cite[cite]{[\@@bibref{}{zs-
new2}{}{}]}\\\ &\ n=106,\ 112&1&\cite[cite]{[\@@bibref{}{yan-chen}{}{}]}\\\ &\
n=10&2&\cite[cite]{[\@@bibref{}{mz2}{}{}]}\\\\[5.69046pt] L_{2}(q)&q\neq
2,3&1&\cite[cite]{[\@@bibref{}{mz1}{}{}]},\cite[cite]{[\@@bibref{}{mzd}{}{}]},\\\
&&&\cite[cite]{[\@@bibref{}{zshi}{}{}]}\\\\[2.84544pt] L_{3}(q)&\
|\pi(\frac{q^{2}+q+1}{d})|=1,\
d=(3,q-1)&1&\cite[cite]{[\@@bibref{}{mzd}{}{}]}\\\\[5.69046pt] U_{3}(q)&\
|\pi(\frac{q^{2}-q+1}{d})|=1,\
d=(3,q+1),q>5&1&\cite[cite]{[\@@bibref{}{mzd}{}{}]}\\\\[5.69046pt]
L_{3}(9)&&1&\cite[cite]{[\@@bibref{}{zs-new4}{}{}]}\\\\[2.84544pt]
U_{3}(5)&&1&\cite[cite]{[\@@bibref{}{zs-new5}{}{}]}\\\\[2.84544pt] L_{4}(q)&\
q\leqslant
37&1&\cite[cite]{[\@@bibref{}{bakbari}{}{}]},\cite[cite]{[\@@bibref{}{akbari-1}{}{}]},\cite[cite]{[\@@bibref{}{amr}{}{}]}\\\\[2.84544pt]
U_{4}(7)&&1&\cite[cite]{[\@@bibref{}{amr}{}{}]}\\\\[2.84544pt] L_{n}(2)&\ n=p\
{\rm or}\ p+1,\ {\rm for\ which}\ 2^{p}-1\ {\rm is\ a\
prime}&1&\cite[cite]{[\@@bibref{}{amr}{}{}]}\\\\[5.69046pt] L_{n}(2)&\
n=9,10,11&1&\cite[cite]{[\@@bibref{}{khoshravi}{}{}]},\cite[cite]{[\@@bibref{}{R-M}{}{}]}\\\\[2.84544pt]
U_{6}(2)&&1&\cite[cite]{[\@@bibref{}{LShi}{}{}]}\\\\[2.84544pt] R(q)&\
|\pi(q\pm\sqrt{3q}+1)|=1,\ q=3^{2m+1},\ m\geqslant
1&1&\cite[cite]{[\@@bibref{}{mzd}{}{}]}\\\\[5.69046pt] {\rm Sz}(q)&\
q=2^{2n+1}\geqslant
8&1&\cite[cite]{[\@@bibref{}{mz1}{}{}]},\cite[cite]{[\@@bibref{}{mzd}{}{}]}\\\\[5.69046pt]
B_{m}(q),C_{m}(q)&m=2^{f}\geqslant 4,\
|\pi\big{(}(q^{m}+1)/2\big{)}|=1,&2&\cite[cite]{[\@@bibref{}{akbarim}{}{}]}\\\\[5.69046pt]
B_{2}(q)\cong C_{2}(q)&\ |\pi\big{(}(q^{2}+1)/2\big{)}|=1,\ q\neq
3&1&\cite[cite]{[\@@bibref{}{akbarim}{}{}]}\\\\[5.69046pt] B_{m}(q)\cong
C_{m}(q)&m=2^{f}\geqslant 2,\ 2|q,\ |\pi\big{(}q^{m}+1\big{)}|=1,\
(m,q)\neq(2,2)&1&\cite[cite]{[\@@bibref{}{akbarim}{}{}]}\\\\[5.69046pt]
B_{p}(3),C_{p}(3)&|\pi\big{(}(3^{p}-1)/2\big{)}|=1,\ p\ {\rm is\ an\ odd\
prime}&2&\cite[cite]{[\@@bibref{}{akbarim}{}{}]},\cite[cite]{[\@@bibref{}{mzd}{}{}]}\\\\[5.69046pt]
B_{3}(5),C_{3}(5)&&2&\cite[cite]{[\@@bibref{}{akbarim}{}{}]}\\\\[5.69046pt]
C_{3}(4)&&1&\cite[cite]{[\@@bibref{}{moghadam}{}{}]}\\\\[2.84544pt] S&\
\mbox{A sporadic simple
group}&1&\cite[cite]{[\@@bibref{}{mzd}{}{}]}\\\\[2.84544pt] S&\ \mbox{A simple
group with}\ |\pi(S)|=4,\ \
S\neq\mathbb{A}_{10}&1&\cite[cite]{[\@@bibref{}{zs}{}{}]}\\\\[2.84544pt] S&\
\mbox{A simple group with}\ |S|\leqslant 10^{8},\ \ S\neq\mathbb{A}_{10},\
U_{4}(2)&1&\cite[cite]{[\@@bibref{}{ls}{}{}]}\\\\[2.84544pt] S&\ \mbox{A
simple $C_{2,2}$- group}&1&\cite[cite]{[\@@bibref{}{mz1}{}{}]}\\\\[5.69046pt]
\hline\cr\end{array}$

According to the results in Table 1, $h_{\rm OD}(\mathbb{A}_{10})=2$. In fact,
another group with the same order and degree pattern as $\mathbb{A}_{10}$ is
$\mathbb{Z}_{3}\times J_{2}$, that is

$|\mathbb{A}_{10}|=|\mathbb{Z}_{3}\times J_{2}|=2^{7}\cdot 3^{4}\cdot
5^{2}\cdot 7,\ \ \ \ D(\mathbb{A}_{10})=D(\mathbb{Z}_{3}\times
J_{2})=(2,3,2,1).$

Until recently, no example of simple group $M$ has been found with $h_{\rm
OD}(M)\geq 3$. So it would be reasonable to ask whether it might be possible
to find such a simple group. The following question has been posed in [16]:

###### Problem 1.1

Is there a simple group which is $k$-fold OD-characterizable for $k\geq 3$?

It is worthwhile to point out that according to the Corollary 1.1, if there
exists a simple group, say $S$, with $h_{\rm OD}(S)\geq 3$, then among non-
isomorphic groups with the same order and degree pattern as $S$, certainly
there will be a non-simple group.

In this paper especially we will concentrate on simple groups $L_{6}(3)$ and
$U_{4}(5)$. In fact, we will show that both groups are OD-characteizable. It
is worth noting that the prime graph of $L_{6}(3)$ is disconnected, while the
prime graph of $U_{4}(5)$ is connected.
Main Theorem. The simple groups $L_{6}(3)$ and $U_{4}(5)$ are OD-
characteizable.

##  2 OD-Characterizability of Simple Groups $L_{6}(3)$ and $U_{4}(5)$

In what follows, we will show that the projective special linear group
$L_{6}(3)$ and the projective special unitary group $U_{4}(5)$ are uniquely
determined by order and degree pattern.

###### Lemma 2.1

The following statements hold.

  * (a)

If $L$ be the finite simple group $L_{6}(3)$, then

(a.1) $|L|=2^{11}\cdot 3^{15}\cdot 5\cdot 7\cdot 11^{2}\cdot 13^{2}$;

(a.2) $\mu(L)=\\{182,121,120,104,80,78,36\\}$;

(a.3) ${\rm Out}(L)=\mathbb{Z}_{2}\times\mathbb{Z}_{2}$ and $s(L)=2$;

(a.4) $D(L)=(4,3,2,2,0,3)$.

  * (b)

If $U$ be the finite simple group $U_{4}(5)$, then

(b.1) $|U|=2^{7}\cdot 3^{4}\cdot 5^{6}\cdot 7\cdot 13$;

(b.2) $\mu(U)=\\{63,60,52,24\\}$;

(b.3) ${\rm Out}(U)=\mathbb{Z}_{2}\times\mathbb{Z}_{2}$ and $s(U)=1$;

(b.4) $D(U)=(3,3,2,1,1)$.

Proof. See [6]. $\square$

$5$$3$$13$$7$211Fig. 1.The prime graph of$L_{6}(3)$.$13$$2$$3$$7$5Fig. 2.The
prime graph of$U_{4}(5)$.

Let $\Gamma=(V,E)$ be a graph with vertex set $V$ and edge set $E$. A set
$I\subseteq V$ of vertices is said to be an independent set of $\Gamma$ if no
two vertices in $I$ are adjacent in $\Gamma$. The independence number of
$\Gamma$, denoted by $\alpha(\Gamma)$, is the maximum cardinality of an
independent set among all independent sets of $\Gamma$. For convenience, we
will denote $\alpha({\rm GK}(G))$ as $t(G)$ for a group $G$. Moreover, for a
vertex $r\in\pi(G)$, let $t(r,G)$ denote the maximal number of vertices in
independent sets of ${\rm GK}(G)$ containing $r$.

######  Theorem 2.1 (Theorem 1, [22])

Let $G$ be a finite group with $t(G)\geqslant 3$ and $t(2,G)\geqslant 2$, and
let $K$ be the largest normal solvable subgroup $K$ of $G$. Then the quotient
group $G/K$ is an almost simple group, i.e., there exists a finite non-abelian
simple group $S$ such that $S\leqslant G/K\leqslant{\rm Aut}(S)$.

Given a prime $p$, we denote by $\mathcal{S}_{p}$ the set of non-abelian
finite simple groups $P$ such that $\max\pi(P)=p$. Using Table 1 in [26] (see
also [14, Table 4]), we have listed the non-abelian simple groups and their
orders in $\mathcal{S}_{13}$, in Table 2.

Table 2. The simple groups $S$ with $\pi(S)\subseteq\\{2,3,5,7,11,13\\}$.
$S$ $|S|$ $|{\rm Out}(S)|$ $S$ $|S|$ $|{\rm Out}(S)|$ $A_{5}$ $2^{2}\cdot
3\cdot 5$ $2$ $A_{13}$ $2^{9}\cdot 3^{5}\cdot 5^{2}\cdot 7\cdot 11\cdot 13$
$2$ $A_{6}$ $2^{3}\cdot 3^{2}\cdot 5$ $4$ $A_{14}$ $2^{10}\cdot 3^{5}\cdot
5^{2}\cdot 7^{2}\cdot 11\cdot 13$ $2$ $U_{4}(2)$ $2^{6}\cdot 3^{4}\cdot 5$ $2$
$A_{15}$ $2^{10}\cdot 3^{6}\cdot 5^{3}\cdot 7^{2}\cdot 11\cdot 13$ $2$ $A_{7}$
$2^{3}\cdot 3^{2}\cdot 5\cdot 7$ $2$ $A_{16}$ $2^{14}\cdot 3^{6}\cdot
5^{3}\cdot 7^{2}\cdot 11\cdot 13$ $2$ $A_{8}$ $2^{6}\cdot 3^{2}\cdot 5\cdot 7$
$2$ $G_{2}(2^{2})$ $2^{12}\cdot 3^{3}\cdot 5^{2}\cdot 7\cdot 13$ $2$ $A_{9}$
$2^{6}\cdot 3^{4}\cdot 5\cdot 7$ $2$ $B_{2}(2^{3})$ $2^{12}\cdot 3^{4}\cdot
5\cdot 7^{2}\cdot 13$ $6$ $A_{10}$ $2^{7}\cdot 3^{4}\cdot 5^{2}\cdot 7$ $2$
${\rm Sz}(2^{3})$ $2^{6}\cdot 5\cdot 7\cdot 13$ $3$ $B_{3}(2)$ $2^{9}\cdot
3^{4}\cdot 5\cdot 7$ $1$ $L_{2}(2^{6})$ $2^{6}\cdot 3^{2}\cdot 5\cdot 7\cdot
13$ $6$ $O_{8}^{+}(2)$ $2^{12}\cdot 3^{5}\cdot 5^{2}\cdot 7$ $6$ $L_{3}(3)$
$2^{4}\cdot 3^{3}\cdot 13$ $2$ $L_{3}(2^{2})$ $2^{6}\cdot 3^{2}\cdot 5\cdot 7$
${12}$ $L_{4}(3)$ $2^{7}\cdot 3^{6}\cdot 5\cdot 13$ $4$ $L_{2}(2^{3})$
$2^{3}\cdot 3^{2}\cdot 7$ $3$ $L_{5}(3)$ $2^{9}\cdot 3^{10}\cdot
5\cdot{11}^{2}\cdot 13$ $2$ $U_{3}(3)$ $2^{5}\cdot 3^{3}\cdot 7$ $2$
$L_{6}(3)$ $2^{11}\cdot 3^{15}\cdot 5\cdot 7\cdot{11}^{2}\cdot{13}^{2}$ $4$
$U_{4}(3)$ $2^{7}\cdot 3^{6}\cdot 5\cdot 7$ $8$ $B_{3}(3)$ $2^{9}\cdot
3^{9}\cdot 5\cdot 7\cdot 13$ $2$ $U_{3}(5)$ $2^{4}\cdot 3^{2}\cdot 5^{3}\cdot
7$ $6$ $O_{8}^{+}(3)$ $2^{12}\cdot 3^{12}\cdot 5^{2}\cdot 7\cdot 13$ $24$
$L_{2}(7)$ $2^{3}\cdot 3\cdot 7$ $2$ $G_{2}(3)$ $2^{6}\cdot 3^{6}\cdot 7\cdot
13$ $2$ $B_{2}(7)$ $2^{8}\cdot 3^{2}\cdot 5^{2}\cdot 7^{4}$ $2$ $C_{3}(3)$
$2^{9}\cdot 3^{9}\cdot 5\cdot 7\cdot 13$ $2$ $L_{2}(7^{2})$ $2^{4}\cdot 3\cdot
5^{2}\cdot 7^{2}$ $4$ $L_{3}(3^{2})$ $2^{7}\cdot 3^{6}\cdot 5\cdot 7\cdot 13$
$4$ $J_{2}$ $2^{7}\cdot 3^{3}\cdot 5^{2}\cdot 7$ $2$ $L_{2}(3^{3})$
$2^{2}\cdot 3^{3}\cdot 7\cdot 13$ $6$ $A_{11}$ $2^{7}\cdot 3^{4}\cdot
5^{2}\cdot 7\cdot 11$ $2$ $U_{4}(5)$ $2^{7}\cdot 3^{4}\cdot 5^{6}\cdot 7\cdot
13$ $4$ $A_{12}$ $2^{9}\cdot 3^{5}\cdot 5^{2}\cdot 7\cdot 11$ $2$ $B_{2}(5)$
$2^{6}\cdot 3^{2}\cdot 5^{4}\cdot 13$ $2$ $U_{5}(2)$ $2^{10}\cdot 3^{5}\cdot
5\cdot 11$ $2$ $L_{2}(5^{2})$ $2^{3}\cdot 3\cdot 5^{2}\cdot 13$ $4$ $U_{6}(2)$
$2^{15}\cdot 3^{6}\cdot 5\cdot 7\cdot 11$ $6$ $L_{2}(13)$ $2^{2}\cdot 3\cdot
7\cdot 13$ $2$ $L_{2}(11)$ $2^{2}\cdot 3\cdot 5\cdot 11$ $2$ $Suz$
$2^{13}\cdot 3^{7}\cdot 5^{2}\cdot 7\cdot 11\cdot 13$ $2$ $M_{11}$ $2^{4}\cdot
3^{2}\cdot 5\cdot 11$ $1$ $Fi_{22}$ $2^{17}\cdot 3^{9}\cdot 5^{2}\cdot 7\cdot
11\cdot 13$ $2$ $M_{12}$ $2^{6}\cdot 3^{3}\cdot 5\cdot 11$ $2$
${{}^{3}D_{4}(2)}$ $2^{12}\cdot 3^{4}\cdot 7^{2}\cdot 13$ $3$ $M_{22}$
$2^{7}\cdot 3^{2}\cdot 5\cdot 7\cdot 11$ $2$ ${{}^{2}F_{4}(2)}^{\prime}$
$2^{11}\cdot 3^{3}\cdot 5^{2}\cdot 13$ $2$ $HS$ $2^{9}\cdot 3^{2}\cdot
5^{3}\cdot 7\cdot 11$ $2$ $U_{3}(2^{2})$ $2^{6}\cdot 3\cdot 5^{2}\cdot 13$ $4$
$M^{c}L$ $2^{7}\cdot 3^{6}\cdot 5^{3}\cdot 7\cdot 11$ $2$

###### Lemma 2.2

Let $G$ be a finite group, with $|G|=p_{1}^{m_{1}}p_{2}^{m_{2}}\cdots
p_{s}^{m_{s}}$, where $s,m_{1},m_{2},\ldots,m_{s}$ are positive integers and
$p_{1},p_{2},\ldots,p_{s}$ distinct primes. Let $\Delta=\\{p_{i}\in\pi(G)\ |\
m_{i}=1\\}$, and for each prime $p_{i}\in\Delta$, let
$\Delta(p_{i})=\\{p_{j}\in\Delta\ |\ j\neq i,\ p_{j}\nmid p_{i}-1\mbox{and}\
p_{i}\nmid p_{j}-1\\}$. Let $K$ be a normal slovable subgroup of $G$. Then
there hold.

  * $(1)$

If $p_{i}\in\Delta$ divides the order of $K$, then for each prime
$p_{j}\in\Delta(p_{i})$, $p_{j}\sim p_{i}$ in ${\rm GK}(G)$. In particular,
$\deg_{G}(p_{i})\geqslant|\Delta(p_{i})|$.

  * $(2)$

If $|\Delta|=s$ and for all $i=1,2,\ldots,s$, $|\Delta(p_{i})|=s-1$, then $G$
is a cyclic group of order $|G|$ and ${\rm D}(G)=(s-1,s-1,\ldots,s-1)$.

Proof. see [2]. $\square$

###### Theorem 2.2

The projective special linear group $L_{6}(3)$ is ${\rm OD}$-characterizable.

Proof. Assume that $G$ be a finite group such that

$|G|=|L_{6}(3)|=2^{11}\cdot 3^{15}\cdot 5\cdot 7\cdot{11}^{2}\cdot{13}^{2}\ \
\ \mbox{and}\ \ \ {\rm D}(G)={\rm D}(L_{6}(3))=(4,3,2,2,0,3).$

According to these conditions on $G$, the following two possibilities can
occur for the prime graph of $G$:

$5$$3$$13$$7$2$11$or$5$$13$$3$$7$2$11$Fig. 3.All possibilities for the prime
graph${\rm GK}(G)$.

Let $K$ be the largest normal solvable subgroup of $G$. We claim that $K$ is a
$\\{2,3\\}$-group. To prove this, we first observe that
$\pi(K)\cap\\{5,7\\}=\emptyset$, since otherwise by Lemma 2.2, $5\sim 7$,
which clashes with the structure of ${\rm GK}(G)$ shown in Fig. 3. Now, we
show that $\pi(K)\cap\\{11,13\\}=\emptyset$. Suppose the contrary. Assume
first that $11\in\pi(K)$ and take $P\in{\rm Syl}_{11}(K)$. Then, by Frattini
argument, we deduce that $G=KN_{G}(P)$. Therefore, the normalizer $N_{G}(P)$
contains an element of order $7$, say $x$, and so $P\langle x\rangle$ is a
subgroup of $G$ of order $11^{i}\cdot 7$, $i=1,2$, which is a cyclic group in
both cases. This shows that $7\sim 11$ in ${\rm GK}(G)$, an impossibility. By
a similar way we can show that $13\notin\pi(K)$. Finally $K$ is a
$\\{2,3\\}$-group.

From the structure of the prime graph of $G$ as shown in Fig. 3, we conclude
that $t(2,G)\geq 2$ and $t(G)\geq 3$. Consequently, from Theorem 2.1 we
conclude that there exists a finite non-abelian simple group $S$ such that
$S\leq G/K\leq{\rm Aut}(S)$. Since $S\in{\cal S}_{13}$ and $K$ is a
$\\{2,3\\}$-group, we observe that the order of ${\rm Aut}(S)$ is divisible by
$5\cdot 7\cdot{11}^{2}\cdot{13}^{2}$. On the other hand, using Table 2 we see
that $\pi({\rm Out}(S))\subseteq\\{2,3\\}$, which forces $|S|=2^{a}\cdot
3^{b}\cdot 5\cdot 7\cdot{11}^{2}\cdot{13}^{2}$, where $2\leq a\leq 11$ and
$0\leq b\leq 15$. But then, Table 2 shows that the only possibility for $S$ is
$L_{6}(3)$ and since $|G|=|L_{6}(3)|$, we obtain $|K|=1$ and $G$ is isomorphic
to $L_{6}(3)$. $\square$

###### Theorem 2.3

The projective special unitary group $U_{4}(5)$ is ${\rm OD}$-characterizable.

Proof. Assume that $G$ is a finite group such that

$|G|=|U_{4}(5)|=2^{7}\cdot 3^{4}\cdot 5^{6}\cdot 7\cdot{13}\ \ \ \mbox{and}\ \
\ {\rm D}(G)={\rm D}(U_{4}(5))=(3,3,2,1,1).$

We have to show that $G\cong U_{4}(5)$. First of all, from the structure of
the degree pattern of $G$, it is easy to see that $7\nsim 13$ in ${\rm
GK}(G)$, since otherwise ${\rm deg}(2)\leq 2$, which is impossible. In fact,
there are only two possibilities for the prime graph of $G$ shown in Fig. 4.:

$13$$3$$2$$7$5or$13$$2$$3$$7$5Fig. 4.All possibilities for the prime
graph${\rm GK}(G)$.

Since $S\in{\cal S}_{13}$ and $\\{7,13\\}\subseteq\pi(S)$, hence we obtain
$|S|=2^{a}\cdot 3^{b}\cdot 5^{c}\cdot 7\cdot 13$, where $2\leq a\leq 7$,
$0\leq b\leq 4$ and $0\leq c\leq 6$. Comparing the orders of simple groups
listed in Table 2, we observe that the only possibility for $S$ is
$L_{2}(3^{3})$ or $U_{4}(5)$. If $S$ is isomorphic to $L_{2}(3^{3})$, then
$|K|$ is divisible by $5^{6}$, because $|{\rm Out}(S)|=4$. Let $x\in K$ be an
element of order 5 and $P\in{\rm Syl}_{5}(K)$. By Frattini argument, we deduce
that $G=KN_{G}(P)$. Hence the normalizer $N_{G}(P)$ contains an element of
order $13$, say $y$, and so $H:=P\langle y\rangle$ is a subgroup of $G$ of
order $5^{6}\cdot 13$. Now, one can easily verify that the Sylow 5-subgroup
$P$ of $H$ is a normal subgroup of $H$. As a matter of fact, $H$ is a
Frobenius group with kernel $P$ and complement $\langle y\rangle$, because $H$
does not contain an element of order $5\cdot 13$. But then we must have
$13\mid 5^{6}-1$, which is a contradiction. Therefore, $S$ is isomorphic to
$U_{4}(5)$, and since $|G|=|U_{4}(5)|$, we obtain $|K|=1$ and $G$ is
isomorphic to $U_{4}(5)$. $\square$

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