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A characterization of some alternating groups A _{ p+8} of degree p + 8 by OD
SpringerPlus volume 5, Article number: 1128 (2016)
Abstract
Let \(A_n\) be an alternating group of degree n. We know that \(A_{10}\) is 2fold ODcharacterizable and \(A_{125}\) is 6fold ODcharacterizable. In this note, we first show that \(A_{189}\) and \(A_{147}\) are 14fold and 7fold ODcharacterizable, respectively, and second show that certain groups \(A_{p+8}\) with that \(\pi ((p+8)!)=\pi (p!)\) and \(p<1000\), are ODcharacterizable. The first gives a negative answer to Open Problem of KoganiMoghaddam and Moghaddamfar.
Background
For a group, it means finite, and for a simple group, it is nonabelian. 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 kfold ODcharacterizable if \(h_{OD}(M)=k\). Moreover, a 1fold ODcharacterizable group is simply called an ODcharacterizable group.
Up to now, some groups are proved to be kfold ODcharacterizable 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 2fold ODcharacterizable (see Moghaddamfar and Zokayi 2010).

(2)
The alternating group \(A_{125}\) is 6fold ODcharacterizable (see Liu and Zhang Submitted).

(3)
The alternating group \(A_{p+3}\) except \(A_{10}\) is ODcharacterizable (see Hoseini and Moghaddamfar 2010; KoganiMoghaddam 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 ODcharacterizable (see Yan et al. 2015).
In Moghaddamfar (2015), \(A_{189}\) is at least 14fold ODcharacterizable. In this paper, we show the results as follows.
Theorem 4
The following hold:

(1)
The alternating group \(A_{189}\) of degree 189 is 14fold ODcharacterizable.

(2)
The alternating group \(A_{147}\) of degree 147 is 7fold ODcharacterizable.
These results give negative answers to the Open Problem (KoganiMoghaddam and Moghaddamfar 2012).
Open Problem
(KoganiMoghaddam and Moghaddamfar 2012) All alternating groups \(A_m\), with \(m \ne 10\), are ODcharacterizable.
We also prove that some alternating groups \(A_{p+8}\) with \(p<1000\) are ODcharacterizable.
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 ODcharacterizable.
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 psubgroups \(G_p\) of G, where \(p\in \pi (G)\). Let \(\mathrm {Aut}(G)\) and \(\mathrm {Out}(G)\) be the automorphism and outerautomorphism 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 ppart 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 nonabelian 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 \(\deg (r)=d1\) 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,
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 \(p1\), 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 \(h(q)=p1\), then q is a primitive root of modulo p. By Lemma 11, we have \(h(q)\mid p1\). 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 \(h(q)<p1\), 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 \(\mathrm {Aut}(P)\cong \mathbb {Z}_{p1}\) 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 \(p1\). 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
and
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, \(109\nmid 1811, H\) 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 psubgroup 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 19subgroup of K. By Lemma 10, \(\exp (L,19)=9\) and so \(\frac{N_G(P)}{C_G(P)}\mid \prod \nolimits _{i=1}^{9}19^{45}\cdot (19^i1)\). It is easy to get that \(101\nmid \mid \prod \nolimits _{i=1}^{9}19^{45}\cdot (19^i1)\). 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 nonabelian 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 nonabelian 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 outerautomorphism 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 7fold ODcharacterizable.
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 ODcharacterization 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 \(A_{p+m}, A_{p+m}=(p+m)A_{p+m1}\).
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 nonprime integer. We get the results as Table 2 which contains some results of Liu and Zhang (Submitted), Moghaddamfar (2015), Mahmoufifar and Khosravi (2014).
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 ODcharacterizable.
Conclusion
In this paper, we have proved the following two results.
Result 1a: The alternating group \(A_{189}\) of degree 189 is 14fold ODcharacterizable.
Result 1b: The alternating group \(A_{147}\) of degree 147 is 7fold ODcharacterizable.
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 ODcharacterizable.
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Authors' contributions
SL and ZZ contributed this paper equally. Both authors read and approved the final manuscript.
Acknowlegements
The first author was supported by the Opening Project of Sichuan Province University Key Laborstory of Bridge Nondestruction Detecting and Engineering Computing (Grant Nos: 2013QYJ02 and 2014QYJ04); the Scientific Research Project of Sichuan University of Science and Engineering (Grant No: 2014RC02) and by the department of Sichuan Province Eduction(Grant Nos: 15ZA0235 and 16ZA0256). The authors are very grateful for the helpful suggestions of the referee.
Dedicated to Prof Gui Min Wei on the occasion of his 70th birthday.
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The authors declare that they have no competing interests.
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Shitian Liu and Zhanghua Zhang are contributed equally to this work.
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Liu, S., Zhang, Z. A characterization of some alternating groups A _{ p+8} of degree p + 8 by OD. SpringerPlus 5, 1128 (2016). https://doi.org/10.1186/s4006401627637
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Keywords
 Element order
 Alternating group
 Simple group
 Symmetric group
 Degree pattern
 Prime graph
Mathematics Subject Classification
 20D05
 20D06
 20D60