OD-characterization of the automorphism groups of simple $K_3$-groups

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):431-442, 2005) and denoted by $D(G)$. The group G is called k-fold OD-characterizable if there exist exactly k non-isomorphic groups H satisfying conditions (1) $|G|=|H|$ and (2) $D(G)=D(H)$. Moreover, a one-fold OD-characterizable group is simply called OD-characterizable 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 OD-characterizable. Moreover, $Aut(A_6)$ is fourfold OD-characterizable and $Aut(U_4(2))$ is at least fourfold OD-characterizable.

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 p-subgroups of G, where $p\in \pi (G)$. And $P_r$ denotes a Sylow r-subgroup 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 k-fold OD-characterizable if there exist exactly k non-isomorphic groups G such that $|G|=|M|$ and $D(G)=D(M)$. Moreover, a one-fold OD-characterizable group is simply called an OD-characterizable group.

Definition 1.3 A group G is said to be almost simple related to S if and only if $S⊴G\le Aut(S)$ for some non-abelian simple group S.

In a series of articles such as [4–7], many finite non-abelian simple groups or almost simple groups were shown to be OD-characterizable. For convenience, we recall some of them in the following proposition.

Proposition 1.4 A finite group G is OD-characterizable 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 OD-characterizable, and also some finite groups, especially the automorphism groups of some finite simple groups, have been shown not to be OD-characterizable but k-fold OD-characterizable 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 OD-characterizable.

• (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 OD-characterizable.

• (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 $(P_3⋊P_5)P_2$, where $P_r\in {Syl}_r(G)$ for each $r\in \pi (G)$. In other words, $Aut(U_4(2))$ is at least fourfold OD-characterizable.

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 2-Frobenius group;

• (b) G has a normal series $1⊴H⊴K⊴G$ such that H and $G/K$ are ${\pi }_1$-groups and $K/H$ is a non-abelian 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 2-Frobenius group if there exists a normal series $1⊴H⊴K⊴G$ 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 non-abelian 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 $\pi (Out(G))\subseteq \{2,3\}$.

Table 1 Finite non-abelian simple groups with $\mathit{\pi }\mathbf{(}\mathit{S}\mathbf{)}\mathbf{\subseteq }\mathbf{\{}\mathbf{2}\mathbf{,}\mathbf{3}\mathbf{,}\mathbf{5}\mathbf{,}\mathbf{7}\mathbf{,}\mathbf{11}\mathbf{,}\mathbf{13}\mathbf{,}\mathbf{17}\mathbf{\}}$

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) $|F|\equiv 1(mod|C|)$.

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 p-groups.

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.

Table 2 ${\mathit{K}}_{\mathbf{3}}$ -groups

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 2-Frobenius 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 $|N|-1$, it follows that $|N|=2^3\cdot 3$ and $|H|=5$. Clearly, this is impossible because $P_5$ cannot act fixed-point-freely for instance on $P_3$ as $5\nmid (3-1)$.

Now assume that G is a 2-Frobenius 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 fixed-point-freely 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 $1⊴N⊴G_1⊴G$ such that N and $G/G_1$ are ${\pi }_1$-groups and $G_1/N$ is a non-abelian 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:

$$\begin{array}{c}2\cdot A_5(\text{a non-split extension of }{Z}_2\text{ by }{A}_5);\hfill \\ 2:A_5\cong Z_2\times A_5(\text{a split extension of }{Z}_2\text{ by }{A}_5).\hfill \end{array}$$

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 OD-characterizable.

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 OD-characterizable.

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 2-group. 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 $N=N_G(〈x〉)$. By the structure of $D(G)$, C is a $\{2,5\}$-group. By N-C Theorem, $N/C\lesssim Aut(〈x〉)\cong Z_4$. 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 2-group. 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 non-abelian 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 non-abelian 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 N-C 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 $C_G(K)/K⊴G/K\cong A_6$ 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 non-split 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 $C_G(K)/K⊴G/K\cong A_6$, 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 non-split 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 non-abelian 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 N-C 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 non-split 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: $1⊴N_1⊴N_2⊴G$ such that $N_1$ is unity or a 2-group, $N_2/N_1$ is a 3-group or 5-group. While $N_2/N_1$ is a 5-group, we consider the action a 3-element $x{N}_1$ 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 3-group, it is enough to consider the action of a 5-element 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 5-element of $G/N_1$ must act fixed-point-freely 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 3-group.

Set $H=P_3{P}_5$. Since $Z(\mathrm{\Omega }(P_3))$ char $P_3⊴H$ and $|P_3|=3^4$, then $Z(\mathrm{\Omega }(P_3))⊴H$ 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 3-group of order 3⁴, as required.

Let x be an element of $P_3$ of order 3, then we have $\varphi (g〈x〉)=g〈x〉$ 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 $G=(P_3⋊P_5)P_2$, a product of $P_2$ and $P_3⋊P_5$. 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 $G=(P_3⋊P_5)\times P_2$ 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 OD-characterizable?