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 $\pi ((p+8)!)=\pi (p!)$ 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 $\omega (G)$ and the set of prime divisors of G is denoted by $\pi (G)$. Related to the set $\omega (G)$ a graph is named a prime graph of G, which is written by GK(G). The vertex set of GK(G) is written by $\pi (G)$, and for different primes p, q, there is an edge between the two vertices p, q if $p\cdot q\in \omega (G)$, 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 $|G|=p_{1}^{\alpha _{1}}p_{2}^{\alpha _{2}}\cdots p_{k}^{\alpha _{k}}$, where $p_{i}$s are primes and $\alpha _{i}$s are positive integers. For $p\in \pi (G)$, let $\deg (p):=|\{q\in \pi (G)|p\sim q\}|$, which we call the degree of p. We also define $D(G):=(\deg (p_{1}),\deg (p_{2}),\ldots ,\deg (p_{k}))$, 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 $5\ne p\in \pi (1000!)$, 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-fold OD-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 that p is a prime satisfying the following three conditions:

• (1) $p\ne 139$ and $p\ne 181$,

• (2) $\pi ((p+8)!)=\pi (p!)$,

• (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 $p\in \pi (G)$. 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 degree n. Then the following hold.

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

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

• (3) Let $p,q\in \pi (S_{n})$. 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

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

Proof

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

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

Lemma 10

Let $L:=A_{p+8}$ be an alternating group of degree $p+8$ with that p is a prime and $\pi (p+8)!=\pi (p!)$. Let $|\pi (A_{p+8})|=d$ with d a 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) $\exp (|L|,r)=\sum \nolimits _{i=1}^{\infty }[\frac{p+8}{r^i}]$ for each $r\in \pi (L)\backslash \{2\}$. Furthermore, $\exp (|L|,r)<\frac{p+8}{2}$ where $5\le r\in \pi (L)$. 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 $r, p\cdot r\in \omega (L)$ 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 $2\cdot p\in \omega (L)$. This completes (1).

By Gaussian’s integer function,

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

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

Lemma 11

Let a, m be positive integers. If $(a,m)=1$, then the equation $a^x\equiv 1$(mod m) has solutions. In particular, if the order of a modulo m is h(a), then h(a) divides $\phi (m)$ where $\phi (m)$ denotes the Euler’s function of m.

Proof

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

Lemma 12

Let p be a prime and $L:=A_{p+8}$ be the alternating group of degree $p+8$ with that $\pi ((p+8)!)=\pi (p!)$. 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 by p.

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

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).

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. $\square$

Table 1 The values of p and h(q)

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

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

Proof

Assume the contrary. First we show that K is a $181'$-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 $\mathrm {Aut}(\langle x\rangle)\cong \mathbb {Z}_{2^2}\times \mathbb {Z}_{3^2}\times \mathbb {Z}_5$, 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, $G=KN_G(\langle x\rangle)$ and so $\{11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181\}\subseteq \pi (K)$. Since K is soluble, G has a Hall subgroup H of order $109\cdot 181$. Obviously, $109\nmid 181-1, H$ is cyclic and so $109\cdot 181\in \omega (G)$ contradicting $D(G)=D(M)$.

Second, show that K is a $p'$-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, $G=KN_G(P)$. 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 $p\cdot 181$, 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 $|\frac{N_G(P)}{C_G(P)}|\mid \prod \nolimits _{i=1}^{9}19^{45}\cdot (19^i-1)$. It is easy to get that $101\nmid \mid \prod \nolimits _{i=1}^{9}19^{45}\cdot (19^i-1)$. 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, $N=CN_N(C_{101})$ 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 $101\in \pi (K)$. Let $K_{101}\in \mathrm {Syl}_{101}(K)$. Since $G=KN_G(K_{101})$, 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 group G / K is an almost simple group. More precisely, there is a normal series such that $S\le G/K\le \mathrm {Aut}(S)$, where S is 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 $109\cdot 181$ 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 $p\in \pi (S)\subseteq \{2,3,\cdots ,p\}$ and p is a prime. But by Lemma 13, K is a $\{2,3,5,7\}$-group. Therefore $181\in \pi (H)\subseteq \pi (\mathrm {Aut}(S))$ 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, $|\mathrm {Aut}(P_j)|=|\mathrm {Aut}(P_j)|^t\cdot t!$. 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 $|S|=2^a\cdot 3^b\cdot 5^c\cdot 7^{d}\cdot 11^{18}\cdot 13^{15}\cdot 17^{11}\cdot 19^9\cdot 23^8\cdot 29^6\cdot 31^6\cdot 37^5\cdot 41^4\cdot 43^4\cdot 47^4\cdot 53^4\cdot 59^3\cdot 61^3\cdot 67^2\cdot 71^2\cdot 73^2\cdot 79^2\cdot 83^2\cdot 89^2\cdot 97\cdot 101\cdot 107\cdot 109\cdot 113\cdot 127\cdot 131\cdot 137\cdot 139\cdot 149\cdot 151\cdot 157\cdot 163\cdot 167\cdot 173\cdot 179\cdot 181$, 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 $n\in \{181, 182, \ldots , 189\}$.

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 $|K|=182\cdot 183\cdot 184\cdot 185\cdot 186\cdot 187\cdot 188\cdot 189=2^6\cdot 3^5\cdot 5\cdot 7^2\cdot 11\cdot 13\cdot 17\cdot 23\cdot 31\cdot 37\cdot 47$ and so $11,13,17,23,31,37,47\in \pi (K)$ 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 $|K|=7\cdot 3^3$. 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: $A_{139}, A_{140}, \ldots , A_{146}$ 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 $A_{140}, A_{141}, \ldots , A_{145}$”.

Case 2

Let $S\cong A_{146}$.

Then $A_{146}\le G/K\le S_{146}$. If $G/K\cong A_{146}$, then $|K|=3\cdot 7^2$. 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 $\pi ((p+m)!)\subseteq \pi (p!)$, 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 $\pi (A_{p+m})=\pi (A_p)$ 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

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 $\pi ((p+m)!)\subseteq \pi (p!)$ and $\pi (p+m)\subseteq \pi (m!)$, 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) $\pi ((p+8)!)=\pi (p!)$,

• (3) $p\le 997$.

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