A characterization of some alternating groups A_p+8 of degree p + 8 by OD

Abstract

Let $A_n$ be an alternating group of degree n. We know that $A_{10}$ is 2-fold OD-characterizable and $A_{125}$ is 6-fold OD-characterizable. In this note, we first show that $A_{189}$ and $A_{147}$ are 14-fold and 7-fold OD-characterizable, respectively, and second show that certain groups $A_{p+8}$ with that $\mathit{\pi }((p+8)!)=\mathit{\pi }(p!)$ and $p<1000$, are OD-characterizable. The first gives a negative answer to Open Problem of Kogani-Moghaddam and Moghaddamfar.

Background

For a group, it means finite, and for a simple group, it is non-abelian. If G is a group, then the set of element orders of G is denoted by $\mathit{\omega }(G)$ and the set of prime divisors of G is denoted by $\mathit{\pi }(G)$. Related to the set $\mathit{\omega }(G)$ a graph is named a prime graph of G, which is written by GK(G). The vertex set of GK(G) is written by $\mathit{\pi }(G)$, and for different primes p, q, there is an edge between the two vertices p, q if $p·q\in \mathit{\omega }(G)$, which is written by $p\sim q$. We let s(G) denote the number of connected components of the prime graph GK(G).

Moghaddamfar et al in 2005 gave the following notions which inspire some authors’ attention.

Definition 1

(Moghaddamfar et al. 2005) Let G be a finite group and $|G|=p_1^{{\mathit{\alpha }}_1}{p}_2^{{\mathit{\alpha }}_2}\cdots p_k^{{\mathit{\alpha }}_k}$, where $p_i$s are primes and ${\mathit{\alpha }}_i$s are positive integers. For $p\in \mathit{\pi }(G)$, let $deg(p):=|\{q\in \mathit{\pi }(G)|p\sim q\}|$, which we call 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.

For a given finite group M, write $h_{OD}(M)$ to denote the number of isomorphism classes of finite groups G such that (1) $|G|=|M|$ and (2) $D(G)=D(M)$.

Definition 2

(Moghaddamfar et al. 2005) A finite group M is called k-fold OD-characterizable if $h_{OD}(M)=k$. Moreover, a 1-fold OD-characterizable group is simply called an OD-characterizable group.

Up to now, some groups are proved to be k-fold OD-characterizable and we can refer to the corresponding references of Akbari and Moghaddamfar (2015).

Concerning the alternating group G with $s(G)=1$, what’s the influence of OD on the structure of group? Recently, the following results are given.

Theorem 3

The following statements hold:

1. The alternating group$A_{10}$is 2-fold OD-characterizable (see Moghaddamfar and Zokayi 2010).

2. The alternating group$A_{125}$is 6-fold OD-characterizable (see Liu and Zhang Submitted).

3. The alternating group$A_{p+3}$except$A_{10}$is OD-characterizable (see Hoseini and Moghaddamfar 2010; Kogani-Moghaddam and Moghaddamfar 2012; Liu 2015; Moghaddamfar and Rahbariyan 2011; Moghaddamfar and Zokayi 2009; Yan and Chen 2012; Yan et al. 2013; Zhang and Shi 2008; Mahmoudifar and Khosravi 2015).

4. All alternating groups$A_{p+5}$, where$p+4$is a composite and$p+6$is a prime and$5\ne p\in \mathit{\pi }(1000!)$, are OD-characterizable (see Yan et al. 2015).

In Moghaddamfar (2015), $A_{189}$is at least 14-fold OD-characterizable. In this paper, we show the results as follows.

Theorem 4

The following hold:

1. The alternating group$A_{189}$of degree 189 is 14-foldOD-characterizable.

2. The alternating group$A_{147}$of degree 147 is 7-fold OD-characterizable.

These results give negative answers to the Open Problem (Kogani-Moghaddam and Moghaddamfar 2012).

Open Problem

(Kogani-Moghaddam and Moghaddamfar 2012) All alternating groups $A_m$, with $m\ne 10$, are OD-characterizable.

We also prove that some alternating groups $A_{p+8}$ with $p<1000$ are OD-characterizable.

Theorem 5

Assume thatpis a prime satisfying the following three conditions:

1. $p\ne 139$and$p\ne 181$,

2. $\mathit{\pi }((p+8)!)=\mathit{\pi }(p!)$,

3. $p\le 997$.

Then the alternating group$A_{p+8}$of degree$p+8$is OD-characterizable.

Let G be a finite group, then let $\mathrm{Soc}(G)$ denote the socle of G regarded as a subgroup which is generated by the minimal normal subgroup of G. Let ${\mathrm{Syl}}_p(G)$ be the set of all Sylow p-subgroups $G_p$ of G, where $p\in \mathit{\pi }(G)$. Let $\mathrm{Aut}(G)$ and $\mathrm{Out}(G)$ be the automorphism and outer-automorphism group of G, respectively. Let $S_n$ denote the symmetric groups of degree n. Let p be a prime divisor of a positive integer n, then the p-part of n is denoted by $n_p$, namely, $n_p\Vert n$. The other symbols are standard (see Conway et al. 1985, for instance).

Some preliminary results

In this section, some preliminary results are given to prove the main theorem.

Lemma 6

Let$S=P_1\times \cdots \times P_r$, where$P_i$’s are isomorphic non-abelian simple groups. Then$\mathrm{Aut}(S)=\mathrm{Aut}(P_1)\times \cdots \times \mathrm{Aut}(P_r).S_r$.

Proof

See Zavarnitsin (2000). $\square$

Lemma 7

Let$A_n$ (or$S_n$) be an alternating (or a symmetric group) of degreen. Then the following hold.

1. Let$p,q\in \mathit{\pi }(A_n)$be odd primes. Then$p\sim q$if and only if$p+q\le n$.

2. Let$p\in \mathit{\pi }(A_n)$be odd prime. Then$2\sim p$if and only if$p+4\le n$.

3. Let$p,q\in \mathit{\pi }(S_n)$. Then$p\sim q$if and only if$p+q\le n$.

Proof

It is easy to get from Zavarnitsin and Mazurov (1999). $\square$

Lemma 8

The number of groups of order 189 is 13.

Proof

See Western (1898). $\square$

Lemma 9

LetPbe a finite simple group and assume thatris the largest prime divisor of |P| with$50<r<1000$. Then for every prime numberssatisfying the inequality$(r-1)/2<s\le r$, the order of the factor group$\mathrm{Aut}(P)/P$is not divisible bys.

Proof

It is easy to check this results by Conway et al. (1985) and Zavarnitsine (2009). $\square$

Let $n=p_1^{{\mathit{\alpha }}_1}{p}_2^{{\mathit{\alpha }}_2}\cdots p_r^{{\mathit{\alpha }}_r}$ where $p_1,p_2,\dots ,p_r$ are different primes and ${\mathit{\alpha }}_1,{\mathit{\alpha }}_2,\dots ,{\mathit{\alpha }}_r$ are positive integers, then $exp(n,p_i)={\mathit{\alpha }}_i$ with $p_i^{{\mathit{\alpha }}_i}\mid n$ but $p_i^{{\mathit{\alpha }}_i+1}\nmid n$.

Lemma 10

Let$L:=A_{p+8}$be an alternating group of degree$p+8$with thatpis a prime and$\mathit{\pi }(p+8)!=\mathit{\pi }(p!)$. Let$|\mathit{\pi }(A_{p+8})|=d$withda positive integer. Then the following hold:

1. $deg(p)=4$and$deg(r)=d-1$for$r\in \{2,3,5,7\}$.

2. $exp(|L|,2)\le p+7$.

3. $exp(|L|,r)={\sum }_{i=1}^{\infty }[\frac{p+8}{r^i}]$for each$r\in \mathit{\pi }(L)\backslash \{2\}$. Furthermore,$exp(|L|,r)<\frac{p+8}{2}$ where $5\le r\in \mathit{\pi }(L)$. In particular, if$r>[\frac{p+8}{2}]$, then$exp(|L|,r)=1$.

Proof

By Lemma 7, it is easy to compute that for odd prime $r,p·r\in \mathit{\omega }(L)$ if and only if $p+r\le p+8$. Hence $r=3,5,7$. If $r=2$, then since $p+4\le p+8$, then $2·p\in \mathit{\omega }(L)$. This completes (1).

By Gaussian’s integer function,

$$\begin{array}{cc}\hfill exp(|L|,2)& =\sum _{i=1}^{\infty }\left[\frac{p+8}{2^i}\right]-1\hfill \\ \hfill & =\left(\left[\frac{p+8}{2}\right]+\frac{p+8}{2^2}+\left[\frac{p+8}{2^3}\right]+\cdots \right)-1\hfill \\ \hfill & \le \left(\frac{p+8}{2}+\frac{p+8}{2^2}+\frac{p+8}{2^3}+\cdots \right)-1\hfill \\ \hfill & =p+7.\hfill \end{array}$$

This proves (2). Similarly, we can get (3). $\square$

Lemma 11

Leta, mbe positive integers. If$(a,m)=1$, then the equation$a^x\equiv 1$(modm) has solutions. In particular, if the order of a modulomish(a), thenh(a) divides$\mathit{\varphi }(m)$where$\mathit{\varphi }(m)$denotes the Euler’s function ofm.

Proof

See Theorem 8.12 of Burton (2002).$\square$

Lemma 12

Letpbe a prime and$L:=A_{p+8}$be the alternating group of degree$p+8$with that$\mathit{\pi }((p+8)!)=\mathit{\pi }(p!)$. Given$P\in {\mathrm{Syl}}_p(L)$and$Q\in {\mathrm{Syl}}_q(L)$with$11\le q<p\le 1000$. Then the following results hold:

1. The order of$N_L(P)$is not divisible by$q^{s(q)}$, where$s(q)=exp(|L|,q)$.

2. If$p\in \{113,139,199,211,241,283,293,337,467,509,619,787,797,839,863,887,953,997\}$, then$|N_L(Q)|$is not divisible byp.

3. If$p\in \{181,317,409,421,523,547,577,631,661,691,709,811,829,919\}$, then there is at least a primerwith that the order of r modulopis less than$p-1$, where$11\le r<p$and$r\in \mathit{\pi }(p!)$.

Proof

By Lemma 11, the equation $q^x\equiv 1$(mod p) has solutions. Suppose the order of q modulo p is written by h(q). If $h(q)=p-1$, then q is a primitive root of modulo p. By Lemma 11, we have $h(q)\mid p-1$. By Lemma 10, we can get s(q). If $h(q)>s(q)$, then $q^{h(q)}\mid |L|$, a contradiction to the hypotheses. Then we can assume that $h(q)\le s(q)$. We can get the q and h(q) by GAP (2016) as Table 1 (Note that there is certain prime which has order $h(q)<p-1$, but $h(q)>s(q)$. Hence we do not list in this table).

Table 1.

The values of p and h(q)

p	h(q)	Condition	p	h(q)	Condition
113	$2^4.7$	None	139	2.3.23	None
181	$2^2.3^2.5$	$q\ne 19$	181	4	$q=19$
199	$2.3^2.11$	None	211	2.3.5.7	None
241	$2^4.3.5$	None	283	2.3.47	None
293	$2^2.73$	None	317	$2^2.79$	$q\ne 73$
317	4	$q=73$	337	$2^4.3.7$	None
409	$2^3.3.17$	$q\ne 31,53$	409	8	$q=31$
409	3	$q=53$	421	$2^2.3.5.7$	$q\ne 29$
421	4	$q=29$	467	2.233	None
509	$2^2.127$	none	523	$2.3^2.29$	$q\ne 11,19,61$
523	29	$q=11$	523	9	$q=19$
523	6	$q=61$	547	2.3.7.13	$q\ne 11,13,41$
547	39	$q=11$	547	21	$q=13$
547	6	$q=41$	577	$2^6.3^2$	$q\ne 23$
577	8	$q=23$	619	2.3.103	None
631	$2.3^2.5.7$	$q\ne 43$	631	3	$q=43$
661	$2^3.3.5.11$	$q\ne 11$	661	33	$q=11$
691	2.3.5.23	$q\ne 89$	691	5	$q=89$
709	$2^2.3.59$	$q\ne 227$	709	3	$q=227$
787	2.3.131	None	797	$2^2.199$	None
811	$2.3^4.5$	$q\ne 131$	811	6	$q=131$
829	$2^2.3^2.23$	$q\ne 11$	829	23	$q=11$
839	2.419	None	863	2.431	None
887	2.443	None	919	$2.3^3.17$	$q\ne 53$
919	6	$q=53$	953	$2^3.7.17$	None
997	$2^2.3.83$	None

By NC Theorem, the factor group $\frac{N_L(P)}{C_L(P)}$ is isomorphic to a subgroup of $\mathrm{Aut}(P)\cong {\mathbb{Z}}_{p-1}$ where ${\mathbb{Z}}_n$ is a cyclic group of order n. It follows that the order of $\frac{N_L(P)}{C_L(P)}$ is less than or equal to $p-1$. If $11\le q<p$ and $q^{s(q)}\mid |N_L(P)|$ where $exp(|L|,q)=s(q)$, then $q\mid |C_L(P)|$. This forces $q\sim p$, a contradiction. This ends the proof of (1).

Next, assume that $p\in \{113,139,199,211,241,283,293,337,467,509,619,787,797,839,863,887,953,997\}$. If p divides the order of $N_L(Q)$, then by NC theorem and Table 1, $p\mid |C_L(Q)|$ and so $p\sim q$, a contradiction. This proves (2). (3) follows from Table 1.

This completes the proof of Lemma 12. $\square$

Proof of the main theorem

In this section, we first give the proof of Theorem 4 and second prove Theorem 5.

The proof of Theorem 4

Proof

We divides the proof into two steps.

Step 1 Let $M=A_{189}$. Assume that G is a finite group such that

$$|G|=|M|$$

and

$$D(G)=D(M).$$

By Lemma 7, the degree pattern GK(G) of G is connected, in particular, the degree pattern GK(G) is the same as the degree pattern of GK(M).

Lemma 13

LetKbe a maximal normal soluble subgroup ofG. ThenKis a$\{2,3,5,7\}$-group, in particular, Gis insoluble.

Proof

Assume the contrary. First we show that K is a ${181}^{\prime }$-group. We assume that K contains an element x of order 181. Let C be the centralizer of x in G and N be the normalizer of x in G. It is easy to see from D(G) that C is a $\{2,3,5,7,181\}$-group. By NC theorem, N/C is isomorphic to a subgroup of automorphism group $\mathrm{Aut}(⟨x⟩)\cong {\mathbb{Z}}_{2^2}\times {\mathbb{Z}}_{3^2}\times {\mathbb{Z}}_5$, where ${\mathbb{Z}}_n$ is a cyclic group of order n. Hence, C is a $\{2,3,5,7,181\}$-group. By Frattini’s arguments, $G=K{N}_G(⟨x⟩)$ and so $\{11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97,101,103,107,109,113,127,131,137,139,149,151,157,163,167,173,179,181\}\subseteq \mathit{\pi }(K)$. Since K is soluble, G has a Hall subgroup H of order $109·181$. Obviously, $109\nmid 181-1,H$ is cyclic and so $109·181\in \mathit{\omega }(G)$ contradicting $D(G)=D(M)$.

Second, show that K is a $p^{\prime }$-group, where $p\in \{11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97,101,103,107,109,113,127,131,137,139,149,151,157,163,167,173,179\}$. Let p be a prime divisor of |K| and P a Sylow p-subgroup of K. By Frattini’s arguments, $G=K{N}_G(P)$. It follows from Lemma 12, that 181 is a divisor of $|N_G(P)|$ if and only if $p=19$. If $181\nmid |\mathrm{Aut}(P)|$, then 181 divides the order of $C_G(P)$ and so there is an element of order $p·181$, a contradiction. On the other hand, $p=19$ and $181\mid |\mathrm{Aut}(P)|$, where P is the Sylow 19-subgroup of K. By Lemma 10, $exp(|L|,19)=9$ and so $|\frac{N_G(P)}{C_G(P)}|\mid {\prod }_{i=1}^9{19}^{45}·({19}^i-1)$. It is easy to get that $101\nmid \mid {\prod }_{i=1}^9{19}^{45}·({19}^i-1)$. If $101\mid |N_G(P)|$, then 101 is a prime divisor of $C_G(P)$. Set $C=C_G(P)$ and $C_{101}\in {\mathrm{Syl}}_{101}(C)$. Also $exp(|L|,101)=1$. By Frattini’s argument, $N=C{N}_N(C_{101})$ and so $p\nmid |N_N(C_{101})|$. Thus $181\mid |C|$ and so $181\sim p$, a contradiction. So $101\nmid |N_G(P)|$ and $101\in \mathit{\pi }(K)$. Let $K_{101}\in {\mathrm{Syl}}_{101}(K)$. Since $G=K{N}_G(K_{101})$, 101 divides the order of $N_G(K_{101})$, then $101\nmid |K|$, a contradiction. Therefore K is a $\{2,3,5,7\}$-group.

Obviously, $G\ne K$ and so G is insoluble. $\square$

Lemma 14

The quotient groupG / Kis an almost simple group. More precisely, there is a normal series such that$S\le G/K\le \mathrm{Aut}(S)$, whereSis isomorphic to$A_n$for$n\in \{181,182,183,184,185,186,187,188,189\}$.

Proof

Let $H=G/K$ and $S=\mathrm{Soc}(H)$. Then $S=B_1\times \cdots \times B_n$, where $B_i$’s are non-abelian simple groups and $S\le H\le \mathrm{Aut}(S)$. In what follows, we will prove that $n=1$ and $S\cong A_n$.

Suppose the contrary. Obviously, 181 does not divide the order of S, otherwise, there is an element of order $109·181$ contradicting $D(G)=D(A_{189})$. Hence, for every i, we have that $B_i\in {\mathfrak{F}}_{179}$, where ${\mathfrak{F}}_p$ is the set of non-abelian simple group S with that $p\in \mathit{\pi }(S)\subseteq \{2,3,\cdots ,p\}$ and p is a prime. But by Lemma 13, K is a $\{2,3,5,7\}$-group. Therefore $181\in \mathit{\pi }(H)\subseteq \mathit{\pi }(\mathrm{Aut}(S))$ and so 181 divides the order of $\mathrm{Out}(S)$. By Lemma 6, $\mathrm{Out}(S)=\mathrm{Out}(P_1)\times \cdots \times \mathrm{Out}(P_r)$, where the group $P_i$’s are satisfying $S\cong P_1\times \cdots \times P_r$. Therefore for some j 181 divides the order of an outer-automorphism group of a direct $P_j$ of t isomorphic simple group $B_i$. Since $B_i\in {\mathfrak{F}}_{179}$, the order of $\mathrm{Out}(B_i)$ is not divisible by 181 by Lemma 9. By Lemma 6, $|\mathrm{Aut}(P_j)|=|\mathrm{Aut}(P_j){|}^t·t!$. It means $t\ge 181$, and hence $4^{181}\mid |G|$, a contradiction. Thus $n=1$ and $S=B_1$.

By Lemma 13, we can assume that $|S|=2^a·3^b·5^c·7^d·{11}^{18}·{13}^{15}·{17}^{11}·{19}^9·{23}^8·{29}^6·{31}^6·{37}^5·{41}^4·{43}^4·{47}^4·{53}^4·{59}^3·{61}^3·{67}^2·{71}^2·{73}^2·{79}^2·{83}^2·{89}^2·97·101·107·109·113·127·131·137·139·149·151·157·163·167·173·179·181$, where $2\le a\le 182,1\le b\le 93,1\le c\le 45$ and $1\le d\le 30$. By Zavarnitsine (2009), the only possible group is isomorphic to $A_n$ with $n\in \{181,182,\dots ,189\}$.

This completes the proof. $\square$

We continue the proof of Theorem 4. By Lemma 14, S is isomorphic to $A_n$ with $n\in \{181,182,\cdots ,189\}$, and $S\le G/K\le \mathrm{Aut}(S)$.

Case 1

Let $S\cong A_{181}$.

Then $A_{181}\le G/K\le S_{181}$. If $G/K\cong A_{181}$, then $|K|=182·183·184·185·186·187·188·189=2^6·3^5·5·7^2·11·13·17·23·31·37·47$ and so $11,13,17,23,31,37,47\in \mathit{\pi }(K)$ contradicting to Lemma 13.

If $G/K\cong S_{181}$, we also have that 11, 13, 17 or 19 divides |K|, contradicting to Lemma 13.

Similarly we can rule out these cases “$S\cong A_n$ with $n\in \{182,183,\cdots ,187\}$”.

Case 2

Let $S\cong A_{188}$.

Then $A_{188}\le G/K\le S_{188}$. Therefore $G/K\cong A_{188}$ or $G/K\cong S_{188}$.

• (1.1)Let $G/K\cong A_{188}$. Then $|K|=7·3^3$. By Conway et al. (1985), the order of $\mathrm{Out}(A_{188})$ is 2 and the Schur multiplier of $A_{188}$ is 2. Then G is isomorphic to $K\times A_{188}$. By Lemma 8, there are 13 types of groups of order 189 satisfying that $|G|=|M|$ and $D(G)=D(M)$.
• (1.2)Let $G/K\cong S_{188}$. Since $|S_{188}{|}_2=|S_{189}{|}_2>{|A_{189}|}_2$, then we rule out this case.

Case 3

Let $S\cong A_{189}$.

Then $A_{189}\le G/K\le S_{189}$. If $G/K\cong A_{189}$, then order consideration implies that G is isomorphic to $A_{189}$. If $G/K\cong S_{189}$, then as $|S_{189}{|}_2>|A_{189}{|}_2={|G|}_2$, we rule out this case.

Step 2 Similarly as the proof of (1), the following results are given:

1. K is a maximal soluble normal $\{2,3,5,7\}$-group.

2. $S\le G/K\le \mathrm{Aut}(S)$, where S is isomorphic to one of the groups: $A_{139},A_{140},\dots ,A_{146}$ and $A_{147}$.

Case 1

Let $S\cong A_{139}$.

Then $A_{139}\le G/K\le S_{139}$. If the former, then $11\mid |K|$, a contradiction. If the latter, we also have that $11\mid |K|$ and so we rule out.

Similarly we can rule out these cases “S is isomorphic to $A_{140},A_{141},\dots ,A_{145}$”.

Case 2

Let $S\cong A_{146}$.

Then $A_{146}\le G/K\le S_{146}$. If $G/K\cong A_{146}$, then $|K|=3·7^2$. Since the order of $\mathrm{Out}(A_{147})$ is 2 and the Schur multiplier of $A_{147}$ is 2. Then G is isomorphic to $K\times A_{146}$. By GAP (2016), there are six types of groups of order 147. So there are 6 groups with the hypotheses: $|G|=|A_{147}|$ and $D(G)=D(A_{147})$. If $G/K\cong S_{147}$, then as $|S_{146}{|}_2>|A_{146}{|}_2=|A_{147}{|}_2={|G|}_2$, we rule out.

Case 3

Let $S\cong A_{147}$.

Then $A_{147}\le G/K\le S_{147}$. If the former, then $K=1$ and so $G\cong A_{147}$, the desired result. If the latter, then as $|S_{147}{|}_2>|A_{147}{|}_2={|G|}_2$, we rule out.

We also can get that $A_{147}$ is 7-fold OD-characterizable.

This completes the proof of Theorem 4. $\square$

The proof of Theorem 5

Proof

Assume that $|G|=|A_{p+8}|$ and $D(G)=D(A_{p+8})$, then by Lemma 7, the degree pattern GK(G) of G is the same as $GK(A_{p+8})$ of $A_{p+8}$. Similarly as the proof of Theorem 4, the statements are gotten:

1. Let K be a maximal soluble group. Then K is a $\{2,3,5,7\}$-group, in particular, G is insoluble.

2. There is a normal series such that $S\le G/K\le \mathrm{Aut}(S)$, where S is isomorphic to $A_{p+r}$ with that $0\le r\le 8$ and $p\in \{$113, 139, 199, 211, 241, 283, 293, 317, 337, 409, 421, 467, 509, 523, 547, 577, 619, 631, 661, 691, 709, 787, 797, 811, 829, 839, 863, 887, 919, 953, 997$\}$.

In what follows, we consider the case “$p=113$”.

1. $S\cong A_{113}$.

   Then $A_{113}\le G/K\le S_{113}$. If $G/K\cong A_{113}$, then 11 divides the order of K, a contradiction. If $G/K\cong S_{113}$, then we also have that $11\mid |K|$, a contradiction. Similarly we can get a contradiction when S is isomorphic to one of $A_{114},A_{115},A_{116},A_{117},A_{118},A_{119}$, and $A_{120}$.

2. Let $S\cong A_{121}$.

   Then $A_{121}\le G/K\cong S_{121}$. If $G/K\cong A_{121}$, then $K=1$, the desired result. If $G/K\cong S_{121}$, then as $|S_{121}{|}_2{>|G|}_2={|A_{121}|}_2$, a contradiction.

Similarly we can deal with these cases “$p\in \{139,199,211,241,283,293,317,337,409,421,467,509,523,547,577,619,631,661,691,709,787,797,811,829,839,863,887,919,953,997\}$”.

This completes the proof of Theorem 5. $\square$

Non OD-characterization of some alternating groups

Assume that p is a prime and m is an integer larger than 3. If $\mathit{\pi }((p+m)!)\subseteq \mathit{\pi }(p!)$, then $GK(A_{p+m})$ is connected.

For the alternating group $A_{p+m},|A_{p+m}|=(p+m)|A_{p+m-1}|$.

We shall use the notation v(n) to denote the number of types of groups of order n where n is a positive integer. We follows the method of Moghaddamfar (2015), $h_{OD}(A_{p+m})\ge 1+v(p+m)$ where $\mathit{\pi }(A_{p+m})=\mathit{\pi }(A_p)$ and $m\ge 1$ is a non-prime integer. We get the results as Table 2 which contains some results of Liu and Zhang (Submitted), Moghaddamfar (2015), Mahmoufifar and Khosravi (2014).

Table 2.

Non OD-characterizability of alternating groups

G	p	m!	$\mathit{\pi }(p+m)$	$h_{OD}$	References
$A_{125}$	113	12!	$\{5\}$	≥6	Mahmoufifar and Khosravi (2014)
$A_{147}$	139	8!	$\{3,7\}$	≥7	Moghaddamfar (2015)
$A_{189}$	181	8!	$\{3,7\}$	≥14	Moghaddamfar (2015)
$A_{539}$	523	16!	$\{7,11\}$	≥3	Moghaddamfar (2015)
$A_{625}$	619	6!	$\{5\}$	≥16	Moghaddamfar (2015)
$A_{875}$	863	12!	$\{13,67\}$	≥6	Moghaddamfar (2015)
$A_{1029}$	1019	10!	$\{3,7\}$	≥20
$A_{1144}$	1129	15!	$\{2,11,13\}$	≥40
$A_{1274}$	1159	15!	$\{2,7,13\}$	≥11
$A_{1344}$	1319	25!	$\{2,3,7\}$	≥11,721
$A_{1352}$	1319	33!	$\{2,13\}$	≥53

Note that v(n), the number of groups of given small order n can be computed by GAP (2016). The Gap programme is as followings.

gap> SmallGroupsInformation(n);

So we have the following conjecture.

Conjecture Assume that p is a prime and $m\ge 6$ is not a prime. If $\mathit{\pi }((p+m)!)\subseteq \mathit{\pi }(p!)$ and $\mathit{\pi }(p+m)\subseteq \mathit{\pi }(m!)$, then $A_{p+m}$ is not OD-characterizable.

Conclusion

In this paper, we have proved the following two results.

Result 1a: The alternating group $A_{189}$ of degree 189 is 14-fold OD-characterizable.

Result 1b: The alternating group $A_{147}$ of degree 147 is 7-fold OD-characterizable.

Result 2: Let p be a prime with the following three conditions:

1. $p\ne 139$ and $p\ne 181$,

2. $\mathit{\pi }((p+8)!)=\mathit{\pi }(p!)$,

3. $p\le 997$.

Then the alternating group $A_{p+8}$ of degree $p+8$ is OD-characterizable.