 Research
 Open Access
 Published:
ODcharacterization of the automorphism groups of simple ${K}_{3}$groups
Journal of Inequalities and Applications volume 2013, Article number: 95 (2013)
Abstract
The degree pattern of a finite group G associated to its prime graph has been introduced in (Moghaddamfar et al. in Algebra Colloq. 12(3):431442, 2005) and denoted by $D(G)$. The group G is called kfold ODcharacterizable if there exist exactly k nonisomorphic groups H satisfying conditions (1) $G=H$ and (2) $D(G)=D(H)$. Moreover, a onefold ODcharacterizable group is simply called ODcharacterizable group. In this problem, those groups with connected prime graphs are somewhat much difficult to be solved. In the present paper, we continue this investigation and show that the automorphism groups of simple ${K}_{3}$groups are characterized by their orders and degree patterns. In fact, the automorphism groups of simple ${K}_{3}$groups except ${A}_{6}$ and ${U}_{4}(2)$ are ODcharacterizable. Moreover, $Aut({A}_{6})$ is fourfold ODcharacterizable and $Aut({U}_{4}(2))$ is at least fourfold ODcharacterizable.
AMS Subject Classification:20D05.
1 Introduction
Let G be a finite group, ${\pi}_{e}(G)$ denote the set of element orders of G and $\pi (G)$ the set of all prime divisors of $G$. The prime graph of G was defined by Gruenberg and Kegel (ref. to [1]), which was denoted by $\mathrm{\Gamma}(G)$ and constructed as follows: The vertex set of this graph is $\pi (G)$ and two distinct vertices p and q are jointed by an edge if and only if $pq\in {\pi}_{e}(G)$. In this case, we say vertices p and q are adjacent and denote this fact by $p\sim q$. The number of connected components of $\mathrm{\Gamma}(G)$ is denoted by $t(G)$ and the sets of vertices of connected components of $\mathrm{\Gamma}(G)$ are denoted as ${\pi}_{i}={\pi}_{i}(G)$ ($i=1,2,\dots ,t(G)$). If $G$ is even, we always assume that $2\in {\pi}_{1}(G)$. Set $T(G)=\{{\pi}_{i}(G)i=1,2,\dots ,t(G)\}$.
Let n be a positive integer, we use $\pi (n)$ to denote the set of all prime divisors of n. If the prime graph of G is known, then $G$ can be expressed as a product of ${m}_{1},{m}_{2},\dots ,{m}_{t(G)}$, where ${m}_{i}$s are positive integers such that $\pi ({m}_{i})={\pi}_{i}$. These ${m}_{i}$s were called the order components of G by the second author, who proved a lot of finite simple groups can be uniquely determined by their order components (ref. to [2]). The set of order components of G is denoted as $\mathit{OC}(G)=\{{m}_{1},{m}_{2},\dots ,{m}_{t(G)}\}$. We also use the following notations. Given a finite group G, denote by $Soc(G)$ the socle of G which is the subgroup generated by the set of all minimal normal subgroups of G. ${Syl}_{p}(G)$ denotes the set of all Sylow psubgroups of G, where $p\in \pi (G)$. And ${P}_{r}$ denotes a Sylow rsubgroup of G for $r\in \pi (G)$. All further unexplained notations are referred to [3].
Definition 1.1 [4]
Let G be a finite group and $G={p}_{1}^{{\alpha}_{1}}{p}_{2}^{{\alpha}_{2}}\cdots {p}_{k}^{{\alpha}_{k}}$, where ${p}_{i}$s are primes and ${\alpha}_{i}$ are integers. For $p\in \pi (G)$, let $deg(p):=\{q\in \pi (G)p\sim q\}$, called the degree of p. We also define $D(G):=(deg({p}_{1}),deg({p}_{2}),\dots ,deg({p}_{k}))$, where ${p}_{1}<{p}_{2}<\cdots <{p}_{k}$. We call $D(G)$ the degree pattern of G.
Definition 1.2 [4]
A group M is called kfold ODcharacterizable if there exist exactly k nonisomorphic groups G such that $G=M$ and $D(G)=D(M)$. Moreover, a onefold ODcharacterizable group is simply called an ODcharacterizable group.
Definition 1.3 A group G is said to be almost simple related to S if and only if $S\u22b4G\le Aut(S)$ for some nonabelian simple group S.
In a series of articles such as [4–7], many finite nonabelian simple groups or almost simple groups were shown to be ODcharacterizable. For convenience, we recall some of them in the following proposition.
Proposition 1.4 A finite group G is ODcharacterizable if G is one of the following groups:

(1)
All sporadic simple groups and their automorphism groups except $Aut({J}_{2})$ and $Aut({M}^{c}L)$;

(2)
The alternating groups ${A}_{p}$, ${A}_{p+1}$, ${A}_{p+2}$ and the symmetric groups ${S}_{p}$ and ${S}_{p+1}$, where p is a prime;

(3)
All finite simple ${K}_{4}$groups except ${A}_{10}$;

(4)
The simple groups of Lie type ${L}_{2}(q)$, ${L}_{3}(q)$, ${U}_{3}(q)$, ${}^{2}B_{2}(q)$ and ${}^{2}G_{2}(q)$ for certain prime power q;

(5)
All finite simple ${C}_{2,2}$groups;

(6)
The alternating groups ${A}_{p+3}$, where $p+2$ is a composite number, $p+4$ is a prime and $7\ne p\in \pi (1\text{,}000!)$;

(7)
The almost simple groups of $Aut({O}_{10}^{+}(2))$, $Aut({O}_{10}^{}(2))$ and $Aut({F}_{4}(2))$.
Till now a lot of finite simple groups have been shown to be ODcharacterizable, and also some finite groups, especially the automorphism groups of some finite simple groups, have been shown not to be ODcharacterizable but kfold ODcharacterizable for some $k>1$. In this paper, we continue this topic and get the following Main Theorem.
Main Theorem Let M be a simple ${K}_{3}$group and G be a finite group such that $G=Aut(M)$ and $D(G)=D(Aut(M))$.

(1)
If M is one of the following simple ${K}_{3}$groups: ${A}_{5}$, ${A}_{6}$, ${L}_{2}(7)$, ${L}_{2}(8)$, ${U}_{3}(3)$, ${L}_{3}(3)$ and ${L}_{2}(17)$, then $G\cong Aut(M)$. In other words, $Aut(M)$ is ODcharacterizable.

(2)
If $M={A}_{6}$, then G is isomorphic to one of the following groups: $Aut({A}_{6})$, ${Z}_{2}\times {Z}_{2}\times {A}_{6}$, ${Z}_{2}\times ({Z}_{2}\cdot {A}_{6})$ and ${Z}_{4}\times {A}_{6}$. In other words, $Aut({A}_{6})$ is fourfold ODcharacterizable.

(3)
If $M={U}_{4}(2)$, then G is isomorphic to one of the following groups: ${Z}_{2}\cdot {U}_{4}(2)$, ${Z}_{2}\times {U}_{4}(2)$, $Aut({U}_{4}(2))$ and $({P}_{3}\u22ca{P}_{5}){P}_{2}$, where ${P}_{r}\in {Syl}_{r}(G)$ for each $r\in \pi (G)$. In other words, $Aut({U}_{4}(2))$ is at least fourfold ODcharacterizable.
2 Preliminaries
In this section, we give some results which will be applied to our further investigations.
Lemma 2.1 [[1], Theorem A]
Let G be a finite group with $t(G)\ge 2$, then G is one of the following groups:

(a)
G is a Frobenius or 2Frobenius group;

(b)
G has a normal series $1\u22b4H\u22b4K\u22b4G$ such that H and $G/K$ are ${\pi}_{1}$groups and $K/H$ is a nonabelian simple group, where ${\pi}_{1}$ is the prime graph component containing 2, H is a nilpotent group and $G/H\mid Aut(K/H)$. Moreover, any odd order component of G is also an odd order component of $K/H$.
Remark 2.2 A group G is a 2Frobenius group if there exists a normal series $1\u22b4H\u22b4K\u22b4G$ such that K and $G/H$ are Frobenius groups with kernels H and $K/H$, respectively.
Lemma 2.3 [3]
Let S be a finite nonabelian simple group with order having prime divisors at most 17. Then S is isomorphic to one of the simple groups listed in Table 1. In particular, if $Out(G)\ne 1$, then $\pi (Out(G))\subseteq \{2,3\}$.
Lemma 2.4 [8]
Let G be a Frobenius group with kernel F and complement C. Then the following assertions hold:

(a)
F is a nilpotent group.

(b)
$F\equiv 1(modC)$.
Lemma 2.5 [9]
Let G be a Frobenius group of even order with H and K being its Frobenius kernel and Frobenius complement, respectively. Then $t(G)=2$ and $T(G)=\{\pi (K),\pi (H)\}$.
Lemma 2.6 [10]
Let G be a simple ${C}_{p,p}$group, where p is a prime. Then

(a)
If $p=5$, G is isomorphic to one of the following simple groups: ${A}_{5}$, ${A}_{6}$, ${A}_{7}$, ${M}_{11}$, ${M}_{22}$, ${L}_{3}(4)$, ${S}_{4}(3)$, ${S}_{4}(7)$, ${U}_{4}(3)$, $Sz(8)$, $Sz(32)$, ${L}_{2}(49)$, ${L}_{2}({5}^{m})$, ${L}_{2}(2\cdot {5}^{m}\pm 1)$, where $m\in \mathbb{N}$ and $2\cdot {5}^{m}\pm 1\in \mathbb{P}$.

(b)
If $p=7$, G is isomorphic to one of the following simple groups: ${A}_{7}$, ${A}_{8}$, ${A}_{9}$, ${M}_{22}$, ${J}_{1}$, ${J}_{2}$, HS, ${L}_{3}(4)$, ${S}_{6}(2)$, ${O}_{8}^{+}(2)$, ${G}_{2}(3)$, ${G}_{2}(13)$, ${U}_{3}(3)$, ${U}_{3}(5)$, ${U}_{3}(19)$, ${U}_{4}(3)$, ${U}_{6}(2)$, $Sz(8)$, ${L}_{2}(8)$, ${L}_{2}({7}^{m})$, ${L}_{2}(2\cdot {7}^{m}1)$, where $m\in \mathbb{N}$ and $2\cdot {7}^{m}1\in \mathbb{P}$.

(c)
If $p=13$, G is isomorphic to one of the following simple groups: ${A}_{13}$, ${A}_{14}$, ${A}_{15}$, $Suz$, $F{i}_{22}$, ${L}_{3}(3)$, ${L}_{4}(3)$, ${O}_{7}(3)$, ${S}_{4}(5)$, ${S}_{6}(3)$, ${O}_{8}^{+}(3)$, ${U}_{3}(4)$, ${U}_{3}(23)$, ${G}_{2}(4)$, ${G}_{2}(3)$, ${F}_{4}(2)$, $Sz(8)$, ${}^{2}E_{6}(2)$, ${}^{3}D_{4}(2)$, ${}^{2}F_{4}{(2)}^{\prime}$, ${L}_{2}(27)$, ${L}_{2}(25)$, ${L}_{2}({13}^{m})$, ${L}_{2}(2\cdot {13}^{m}1)$, where $m\in \mathbb{N}$ and $2\cdot {13}^{m}1\in \mathbb{P}$.

(d)
If $p=17$, G is isomorphic to one of the following simple groups: ${A}_{17}$, ${A}_{18}$, ${A}_{19}$, ${J}_{3}$, He, ${F}_{i23}$, ${F}_{i24}^{\mathrm{\prime}}$, ${L}_{2}(q)$ ($q={2}^{4},{17}^{m},2\cdot {17}^{m}\pm 1$ which is a prime, $m\ge 1$), ${S}_{4}(4)$, ${S}_{8}(2)$, ${O}_{8}^{}(2)$, ${F}_{4}(2)$, ${}^{2}E_{6}(2)$.
Remark 2.7 Let p be a prime. A group G is called a ${C}_{p,p}$group if and only if $p\in \pi (G)$ and the centralizers of its elements of order p in G are pgroups.
3 Proof of Main Theorem
In this section, we give the proof of Main Theorem.
Remark 3.1 Let n be a positive integer and $n>1$. We say that a finite group G is a ${K}_{n}$group if and only if $\pi (G)=n$.
Proof of Main Theorem Let M be a simple ${K}_{3}$group, then M is isomorphic to one of the following simple ${K}_{3}$groups: ${A}_{5}$, ${A}_{6}$, ${L}_{2}(7)$, ${L}_{2}(8)$, ${U}_{3}(3)$, ${L}_{3}(3)$, ${L}_{2}(17)$ and ${U}_{4}(2)$. For convenience, using [3], we have tabulated $Aut(M)$, $D(Aut(M))$ and $Out(M)$ in Table 2.
Let G be a finite group satisfying (1) $G=Aut(M)$ and (2) $D(G)=D(Aut(M))$. We prove the theorem up to choice of M one by one. The proof is written in four cases.
Case 1. To prove the theorem if $M={A}_{5}$.
Evidently, $t(G)=2$. In fact, we have ${\pi}_{1}(G)=\{2,3\}$ and ${\pi}_{2}(G)=\{5\}$. We first show that G is neither Frobenius nor 2Frobenius group. Suppose $G=NH$ is a Frobenius group with kernel N and complement H, and hence $T(G)=\{\pi (N),\pi (H)\}$ by Lemma 2.5. Since $H$ divides $N1$, it follows that $N={2}^{3}\cdot 3$ and $H=5$. Clearly, this is impossible because ${P}_{5}$ cannot act fixedpointfreely for instance on ${P}_{3}$ as $5\nmid (31)$.
Now assume that G is a 2Frobenius group with kernels H and $K/H$, respectively. Since $T(G)=\{\pi (H)\cup \pi (G/K),\pi (K/H)\}$ and $2\in \pi (H)\cup \pi (G/K)$, it follows that $K/H=5$. On the other hand, $G/K\lesssim Aut(K/H)\cong {Z}_{4}$. Hence $G/K\mid 4$, which implies $\{3,5\}\subseteq \pi (K)$. In this case, we have $3\in \pi (H)$, so an element of order 5 must act fixedpointfreely on a subgroup of order 3 in H, which is clearly a contradiction by Table 2.
By Lemma 2.1, G has a normal series $1\u22b4N\u22b4{G}_{1}\u22b4G$ such that N and $G/{G}_{1}$ are ${\pi}_{1}$groups and ${G}_{1}/N$ is a nonabelian simple group, N is a nilpotent group. Note that one of the components of the prime graph of ${G}_{1}/N$ must be $\{5\}$ and ${G}_{1}/N$ is a simple ${C}_{5,5}$ group. By Lemma 2.6, ${G}_{1}/N$ can only be isomorphic to one of the following simple groups: ${A}_{5}$, ${A}_{6}$, ${A}_{7}$, ${M}_{11}$, ${M}_{22}$, ${L}_{3}(4)$, ${S}_{4}(3)$, ${S}_{4}(7)$, ${U}_{4}(3)$, $Sz(8)$, $Sz(32)$, ${L}_{2}(49)$, ${L}_{2}({5}^{m})$ and ${L}_{2}(2\cdot {5}^{m}\pm 1)$, where $m\in \mathbb{N}$ and $2\cdot {5}^{m}\pm 1\in \mathbb{P}$.
Considering the orders of the simple groups, ${G}_{1}/N$ can only be isomorphic to ${A}_{5}$, that is, ${G}_{1}/N\cong {A}_{5}$. Since $G/N\lesssim Aut({G}_{1}/N)$, we get ${A}_{5}\lesssim G/N\lesssim Aut({A}_{5})$.
If $G/N\cong Aut({A}_{5})$, and since $G=Aut({A}_{5})$, we deduce $N=1$ and $G\cong Aut({A}_{5})$.
If $G/N\cong {A}_{5}$, then $N=2$ and so $N\le Z(G)$. Therefore G is a central extension of ${Z}_{2}$ by ${A}_{5}$ and G is isomorphic to one of the following groups:
But whether G is isomorphic to $2\cdot {A}_{5}$ or $2:{A}_{5}\cong {Z}_{2}\times {A}_{5}$, it always follows that $15\in {\pi}_{e}(G)$ by [3] (see ATLAS), a contradiction.
Till now we have proved that $G\cong Aut({A}_{5})$ if $G=Aut({A}_{5})$ and $D(G)=D(Aut({A}_{5}))$, that is, $Aut({A}_{5})$ is ODcharacterizable.
Case 2. To prove the theorem holds for M, one of the following simple groups: ${L}_{2}(7)$, ${L}_{2}(8)$, ${U}_{3}(3)$, ${L}_{3}(3)$ and ${L}_{2}(17)$.
Since $G=Aut(M)$ and $D(G)=D(Aut(M))$, we have to discuss the following five cases. The method used below is the same as Case 1, so the detailed processes are omitted.

(a)
If $M={L}_{2}(7)$, then $G\cong Aut({L}_{2}(7))$;

(b)
If $M={L}_{2}(8)$, then $G\cong Aut({L}_{2}(8))$;

(c)
If $M={U}_{3}(3)$, then $G\cong Aut({U}_{3}(3))$;

(d)
If $M={L}_{3}(3)$, then $G\cong Aut({L}_{3}(3))$;

(e)
If $M={L}_{2}(17)$, then $G\cong Aut({L}_{2}(17))$.
Hence all the almost simple groups $Aut({L}_{2}(7))$, $Aut({L}_{2}(8))$, $Aut({U}_{3}(3))$, $Aut({L}_{3}(3))$ and $Aut({L}_{2}(17))$ are ODcharacterizable.
Case 3. To prove the theorem holds for $M={A}_{6}$.
By Table 2, $G=Aut({A}_{6})={2}^{5}\cdot {3}^{2}\cdot 5$ and $D(G)=D(Aut({A}_{6}))=(2,1,1)$. By these facts, we immediately conclude that $\{2,3,5,6,10\}\in {\pi}_{e}(G)$ and $15\notin {\pi}_{e}(G)$. It is evident that the prime graph of G is connected since $deg(2)=2$ and $\pi (G)=3$. Moreover, it easy to see that $\mathrm{\Gamma}(G)=\mathrm{\Gamma}(Aut({A}_{6}))$. We break up the proof into a sequence of subcases.
Subcase 3.1. Let K be a maximal normal solvable subgroup of G. Then K is a 2group. In particular, G is nonsolvable.
We first prove that K is a 5^{′}group. Assume the contrary, then K possesses an element x of order 5. Set $C={C}_{G}(x)$ and $N={N}_{G}(\u3008x\u3009)$. By the structure of $D(G)$, C is a $\{2,5\}$group. By NC Theorem, $N/C\lesssim Aut(\u3008x\u3009)\cong {Z}_{4}$. Hence, N is a $\{2,5\}$group. By the Frattini argument, $G=KN$. This implies that $\{3,5\}\subseteq \pi (K)$. Since K is solvable, it possesses a Hall $\{3,5\}$subgroup L of order ${3}^{2}\cdot 5$. Clearly, L is nilpotent, and hence $15\in {\pi}_{e}(G)$, a contradiction.
Next, we show that K is a 3^{′}group. Otherwise, let ${P}_{3}\in {Syl}_{3}(K)$. Again, by the Frattini argument $G=K{N}_{G}({P}_{3})$. Hence 5 divides the order of ${N}_{G}({P}_{3})$. Then ${N}_{G}({P}_{3})$ contains a subgroup of order ${3}^{2}\cdot 5$, which leads to a contradiction as before. Therefore K is a 2group. Since $K\ne G$, it follows that G is nonsolvable. This completes the proof of Subcase 3.1.
Subcase 3.2. The quotient group $G/K$ is an almost simple group. In fact, $S\lesssim G/K\lesssim Aut(S)$, where S is a nonabelian simple group.
Let $\overline{G}:=G/K$ and $S:=Soc(\overline{G})$. Then $S={B}_{1}\times {B}_{2}\times \cdots \times {B}_{m}$, where ${B}_{i}$s are nonabelian simple groups and $S\lesssim \overline{G}\lesssim Aut(S)$. In what follows, we will prove that $m=1$.
Suppose that $m\ge 2$. It is easy to see that 5 does not divide the order of S, since otherwise $15\in {\pi}_{e}(G)$, a contradiction. On the other hand, by the order of G, we obtain that $\pi (S)\subseteq \{2,3\}$, which is impossible. Therefore $m=1$ and $S={B}_{1}$.
Subcase 3.3. $S\cong {A}_{6}$ and G is isomorphic to one of the following groups: $Aut({A}_{6})$, ${Z}_{2}\times {Z}_{2}\times {A}_{6}$, ${Z}_{2}\times ({Z}_{2}\cdot {A}_{6})$ and ${\mathbb{Z}}_{4}\times {A}_{6}$.
By Lemma 2.3 and Subcase 3.1, we may assume that $S={2}^{a}\cdot {3}^{2}\cdot 5$, where $2\le a\le 5$. Using Table 1, we see that S can only be isomorphic to the simple group ${A}_{6}$. Thus ${A}_{6}\lesssim G/K\lesssim Aut({A}_{6})$.
If $G/K\cong Aut({A}_{6})$, then by the order comparison, we obtain that $K=1$ and $G\cong Aut({A}_{6})$.
If $G/K\cong {A}_{6}$, then $K=4$ and so $K\cong {Z}_{2}\times {Z}_{2}$ or $K\cong {Z}_{4}$. Now, we divide the proof into two subcases.
Subsubcase 3.3.1. $G/K\cong {A}_{6}$ and $K\cong {Z}_{2}\times {Z}_{2}$. By NC Theorem, we know that the factor group $G/{C}_{G}(K)$ is isomorphic to a subgroup of $Aut(K)$. Thus $G/{C}_{G}(K)\mid ({2}^{2}1)({2}^{2}2)$, that is, $G/{C}_{G}(K)\mid 6$, which implies that $5\mid {C}_{G}(K)$. In particular, $K<{C}_{G}(K)$. On the other hand, we have ${C}_{G}(K)/K\u22b4G/K\cong {A}_{6}$ and hence we obtain $G={C}_{G}(K)$. So $K\le Z(G)$. Therefore G is a central extension of K by ${A}_{6}$. If G is a nonsplit extension of K by ${A}_{6}$, then $G\cong {Z}_{2}\times ({Z}_{2}\cdot {A}_{6})$ (see [3]). If G is a split extension over K, we have $G\cong {Z}_{2}\times {Z}_{2}\times {A}_{6}$.
Subsubcase 3.3.2. $G/K\cong {A}_{6}$ and $K\cong {Z}_{4}$. In this case, we have $G/{C}_{G}(K)\lesssim Aut({Z}_{4})\cong {Z}_{2}$, and so $G/{C}_{G}(K)=1\text{or}2$. If $G/{C}_{G}(K)=2$, then $K<{C}_{G}(K)$. Since ${C}_{G}(K)/K\u22b4G/K\cong {A}_{6}$, we obtain $G={C}_{G}(K)$, a contradiction. Therefore $G/{C}_{G}(K)=1$ and $K\le Z(G)$. Furthermore, G is a central extension of ${Z}_{4}$ by ${A}_{6}$. Obviously, G cannot be a nonsplit extension central extension of ${Z}_{4}$ by ${A}_{6}$ since the order of Schur multiplier of ${A}_{6}$ is 6. If G is a split extension over K, we obtain $G\cong {Z}_{4}\times {A}_{6}$. This completes the proof of Subcase 3.3 and the case.
Case 4. To prove the theorem if $M={U}_{4}(2)$.
In this case, we have $G=Aut({U}_{4}(2))={2}^{7}\cdot {3}^{4}\cdot 5$ and $D(G)=D(Aut({U}_{4}(2)))=(2,1,1)$ by Table 2. By these hypotheses, we immediately conclude that $\{2,3,5,6,10\}\in {\pi}_{e}(G)$ and $15\notin {\pi}_{e}(G)$. Clearly, the prime graph of G is connected, because the vertex 2 is adjacent to all other vertices. Moreover, it is easy to see that $\mathrm{\Gamma}(G)=\mathrm{\Gamma}(Aut({U}_{4}(2)))$. We separate the proof into a sequence of subcases.
Subcase 4.1. To prove the theorem holds while G is nonsolvable.
Let K be the maximal normal solvable subgroup of G. Then K is a $\{2,3\}$group by the same approach as that in Subcase 3.1. We assert that $G/K$ is an almost simple group. And in fact, $S\lesssim G/K\lesssim Aut(S)$.
Let $\overline{G}:=G/K$ and $S:=Soc(\overline{G})$. Then $S={B}_{1}\times {B}_{2}\times \cdots \times {B}_{m}$, where ${B}_{i}$s are nonabelian simple groups and $S\lesssim \overline{G}\lesssim Aut(S)$. It is easy to see that $m=1$ by Table 2. Therefore $S={B}_{1}$.
By Lemma 2.3, we can suppose that $S={2}^{a}\cdot {3}^{2}\cdot 5$, where $2\le a\le 7$, $1\le b\le 4$. Using Table 1, we see that S can only be isomorphic to one of the following simple groups: ${A}_{5}$, ${A}_{6}$ and ${U}_{4}(2)$.
If $S\cong {A}_{5}$, then ${A}_{5}\lesssim G/K\lesssim Aut({A}_{5})\cong {S}_{5}$. Hence $2\cdot 5\in {\pi}_{e}(G)\setminus {\pi}_{e}({S}_{5})$, a contradiction.
If $S\cong {A}_{6}$, then ${A}_{6}\lesssim G/K\lesssim Aut({A}_{6})$ and ${2}^{2}\cdot {3}^{2}$ divides the order of K.
Let ${P}_{r}\in {Syl}_{r}(K)$ for each $r\in \pi (G)$. By the Frattini argument $G=K{N}_{G}({P}_{3})$, 5 divides the order of ${N}_{G}({P}_{3})$. Let T be a subgroup of ${N}_{G}({P}_{3})$ of order 5. By NC Theorem, the factor group ${N}_{G}({P}_{3})/{C}_{G}({P}_{3})$ is isomorphic to a subgroup of $Aut({P}_{3})$. Thus $G/{C}_{G}(K)\mid ({3}^{2}1)({3}^{2}3)$, which implies that $T\le {C}_{G}(K)$. Then $15\in {\pi}_{e}(G)$, a contradiction.
If $S\cong {U}_{4}(2)$, then ${U}_{4}(2)\lesssim G/K\lesssim Aut({U}_{4}(2))$. In this case, $G/K\cong {U}_{4}(2)$, then $K=2$ and $K\le Z(G)$. Therefore G is a central extension of K by ${U}_{4}(2)$. If G is a nonsplit extension of K by ${U}_{4}(2)$, then $G\cong {Z}_{2}\cdot {U}_{4}(2)$. If G is a split extension over K, we have $G\cong {Z}_{2}\times {U}_{4}(2)$. In the latter case $G/K\cong Aut({U}_{4}(2))$, by order comparison, we deduce that $K=1$ and $G\cong Aut({U}_{4}(2))$.
Till now we have proved that G is isomorphic to one of the following groups: ${Z}_{2}\cdot {U}_{4}(2)$, ${Z}_{2}\times {U}_{4}(2)$ and $Aut({U}_{4}(2))$ if G is nonsolvable. It is easy to see that the groups ${Z}_{2}\cdot {U}_{4}(2)$, ${Z}_{2}\times {U}_{4}(2)$ and $Aut({U}_{4}(2))$ satisfy the conditions (1) $G=Aut({U}_{4}(2))$ and (2) $D(G)=D(Aut({U}_{4}(2)))$ (see ATLAS).
Subcase 4.2. To prove the theorem holds while G is solvable.
Since G is solvable, we may take a normal series of G: $1\u22b4{N}_{1}\u22b4{N}_{2}\u22b4G$ such that ${N}_{1}$ is unity or a 2group, ${N}_{2}/{N}_{1}$ is a 3group or 5group. While ${N}_{2}/{N}_{1}$ is a 5group, we consider the action a 3element $x{N}_{1}$ on ${N}_{2}/{N}_{1}$, then we see that $G/{N}_{1}$ has an element of order 15, so does G, a contradiction. While ${N}_{2}/{N}_{1}$ is a 3group, it is enough to consider the action of a 5element of $G/{N}_{1}$ on ${N}_{2}/{N}_{1}$, a contradiction appears too if ${N}_{2}/{N}_{1}\mid {3}^{3}$. Hence, ${N}_{2}/{N}_{1}={3}^{4}$ and the 5element of $G/{N}_{1}$ must act fixedpointfreely on ${N}_{2}/{N}_{1}$. Moreover, the $\{3,5\}$Hall subgroup H of G is a Frobenius group with kernel ${P}_{3}$ and complement ${P}_{5}$, since otherwise there exists an automorphism of ${P}_{3}$ of order 5, say ϕ, such that $\varphi (x)=x$ for each $x\in {P}_{3}$. We first show that ${P}_{3}$ is an elementary abelian 3group.
Set $H={P}_{3}{P}_{5}$. Since $Z(\mathrm{\Omega}({P}_{3}))$ char ${P}_{3}\u22b4H$ and ${P}_{3}={3}^{4}$, then $Z(\mathrm{\Omega}({P}_{3}))\u22b4H$ and $Z(\mathrm{\Omega}({P}_{3}))\le {3}^{4}$. By the structure of $D(G)$, G has no elements of order 15, neither does H. Therefore, $Z(\mathrm{\Omega}({P}_{3}))$ is an elementary abelian 3group of order 3^{4}, as required.
Let x be an element of ${P}_{3}$ of order 3, then we have $\varphi (g\u3008x\u3009)=g\u3008x\u3009$ for every $g\in {P}_{3}$. Now a direct computation shows that $\varphi (g)=g\cdot {x}^{i}$, where $i=0,1,2$. Hence ${\varphi}^{3}(g)=g\cdot {x}^{3i}=g$. However, the order of ϕ is 5, a contradiction.
We have $G=({P}_{3}\u22ca{P}_{5}){P}_{2}$, a product of ${P}_{2}$ and ${P}_{3}\u22ca{P}_{5}$. It is obvious that there exists such a finite group satisfying the following conditions: (1) $G=Aut(M)$ and (2) $D(G)=D(Aut(M))$. This completes the proof of Main Theorem. □
In 1987, Shi in [11] put forward the following conjecture:
Conjecture 3.2 Let G be a group and M be a finite simple group. Then $G\cong M$ if and only if (1) $G=M$ and (2) ${\pi}_{e}(G)={\pi}_{e}(M)$.
Corollary 3.3 Let M be one of the following simple ${K}_{3}$groups: ${A}_{5}$, ${A}_{6}$, ${L}_{2}(7)$, ${L}_{2}(8)$, ${U}_{3}(3)$, ${L}_{3}(3)$, ${L}_{2}(17)$ and G be a finite group such that $G=Aut(M)$ and ${\pi}_{e}(G)={\pi}_{e}(Aut(M))$. Then $G\cong Aut(M)$.
Proof If ${\pi}_{e}(G)={\pi}_{e}(Aut(M))$, then G and $Aut(M)$ have the same degree pattern. Hence the result follows from Main Theorem. □
4 An example and a question
Example 4.1 According to Main Theorem, let $G=({P}_{3}\u22ca{P}_{5})\times {P}_{2}$ and $M={U}_{4}(2)$, then $G=Aut(M)$ and $D(G)=D(Aut(M))$. However, G is not isomorphic to $Aut(M)$. Hence, we put forward the following question:
Question 4.2 Is $Aut({U}_{4}(2))$ exactly fourfold ODcharacterizable?
References
 1.
Williams JS: Prime graph components of finite groups. J. Algebra 1981, 69(2):487–513. 10.1016/00218693(81)902180
 2.
Chen GY: A new characterization of sporadic simple groups. J. Algebra 1996, 3(1):49–58.
 3.
Conway JH, Curtis RT, Norton SP, Parker RA, Wilson RA: Atlas of Finite Groups. Clarendon, Oxford; 1985.
 4.
Moghaddamfar AR, Zokayi AR, Darafsheh MR: A characterization of finite simple groups by the degrees of vertices of their prime graphs. Algebra Colloq. 2005, 12(3):431–442.
 5.
Zhang LC, Shi WJ: OD characterization of simple ${K}_{4}$ groups. Algebra Colloq. 2009, 16(2):275–282.
 6.
Yan YX, Chen GY, Wang LL: OD characterization of the automorphism groups of ${O}_{10}^{\pm}(2)$ . Indian J. Pure Appl. Math. 2012, 3(43):183–195.
 7.
Yan, YX, Chen, GY: Recognizing finite groups having connected prime graphs through order and degree pattern. Chin. Ann. Math., Ser. B (to appear)
 8.
Gorenstein D: Finite Groups. Harper & Row, New York; 1980.
 9.
Chen GY: On structure of Frobenius and 2Frobenius group. J. Southwest China Norm. Univ. 1995, 20(5):485–487.
 10.
Chen ZM, Shi WJ:On ${C}_{pp}$simple groups. J. Southwest China Norm. Univ. 1993, 18(3):249–256.
 11.
Shi WJ: A new characterization of some simple groups of Lie type. Contemp. Math. 1989, 82: 171–180.
Acknowledgements
This work was supported by the Natural Science Foundation of China (Grant Nos. 11171364; 11271301); by ‘The Fundamental Research Funds for the Central Universities’ (Grant No. XDJK2012D004) and Natural Science Foundation Project of CQ CSTC (Grant No. cstc2011jjA00020); by the Fundamental Research Funds for the Central Universities (Grant No. XDJK2009C074) and GraduateInnovation Funds of Science of SWU (ky2009013).
Author information
Affiliations
Corresponding author
Additional information
Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
YY carried out the study of the alternating group of degree 5. HX carried out the study of the alternating group of degree 6. GC and LH carried out the study of the group ${U}_{4}(2)$. All authors read and approved the final manuscript.
Rights and permissions
About this article
Cite this article
Yan, Y., Xu, H., Chen, G. et al. ODcharacterization of the automorphism groups of simple ${K}_{3}$groups. J Inequal Appl 2013, 95 (2013). https://doi.org/10.1186/1029242X201395
Received:
Accepted:
Published:
Keywords
 almost simple group
 prime graph
 degree pattern
 simple ${K}_{3}$group
 degree of a vertex