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BY-NC-ND 4.0 license Open Access Published by De Gruyter Open Access August 19, 2017

OD-characterization of alternating groups Ap+d

  • Yong Yang EMAIL logo , Shitian Liu and Zhanghua Zhang
From the journal Open Mathematics

Abstract

Let An be an alternating group of degree n. Some authors have proved that A10, A147 and A189 cannot be OD-characterizable. On the other hand, others have shown that A16, A23+4, and A23+5 are OD-characterizable. We will prove that the alternating groups Ap+d except A10, are OD-characterizable, where p is a prime and d is a prime or equals to 4. This result generalizes other results.

MSC 2010: 20D05; 20D06; 20D60

1 Introduction

All groups under consideration are finite and for a simple group we mean a non-abelian simple group. For a given group G, the socle of G is defined as the subgroups generated by the minimal normal subgroups of G, denoted by Soc(G). π(G) is the set of prime divisors of |G|. Sylp(G) is the set of all Sylow p-subgroups of G, where pπ(G), Sylr(G) is the set of the Sylow r-subgroups of G for rπ(G). Let Aut(G) and Out(G) denote the automorphism and outer-automorphism of G, respectively. Sn and An denote the symmetric and alternating groups of degree n, respectively. ω(G) is the set of orders of its elements of G. Associated to ω(G) a graph is named prime graph of G, which is written by GK(G). The vertex set of GK(G) is π(G), and two distinct vertices p, q are joined by an edge if pqω(G) which is denoted by pq. We use s(G) to denote the number of connected components of the prime graph GK(G). If n = pam with (p, m) = 1, then np = pa. The other symbols are standard (see [1], for instance).

Definition 1.1

([2]). Let G be a finite group and |G|=p1α1p2α2pkαk, where pis are different primes and αis are positive integers. For pπ(G), let deg(p) := |{qπ(G)|pq which we call the degree of p. We also define D(G) := (deg(p1), deg(p2), ⋯, deg(pk)), where p1 < p2 < ⋯ < pk. We call D(G) the degree pattern of G.

Given a finite group M, denote by hOD(M) the number of isomorphism classes of finite groups G such that (1) |G| = |M| and (2) D(G) = D(M).

Definition 1.2

([2]). A finite group M is called k-fold OD-characterizable if hOD(M) = k. Moreover, a 1-fold OD-characterizable group is simply called an OD-characterizable group.

For the groups which are k-fold OD-characterizable, please see Tables 1 and 2 and corresponding references of [3]. Many finite groups are k-fold OD-characterizable.

Table 1

The simple classical groups

L Lie;rankL d O |L|
Ln(q) An − 1(q) (n, q − 1) 2 df, if n ≥ 3; 1dqn(n1/2i=2n(qi1)
n − 1 df, if n = 2
Un(q) 2An − 1(q) (n, q + 1) 2 df, if n ≥ 3 1dqn(n1/2i=2n(qi(1)i)
[n/2] df, if n = 2
PSp2m(q) Cm(q) (2, q − 1) df, m ≥ 3; 1dqm2i=1m(q2i1)
m 2f, if m = 2
Ω2m+1(q) Bm (q) 2 2f 12qm2i=1m(q2i1)
q odd m
PΩ2m+(q) Dm (q) (4, qm − 1) 2df, if m ≠ 4 1dqm(m1)(qm1)Πi=1m1(q2i1)
m ≥ 3 m 6df, if m = 4
PΩ2m(q) 2Dm(q) (4, qm + 1) 2 df 1dqm(m1)(qm+1)Πi=1m1(q2i1)
m ≥ 2 m − 1

Table 2

The simple exceptional groups

L L d O |L|
G2(q) 2 1 f, if p ≠ 3 q6(q2 − 1)(q6 − 1)
2f, if p = 3
F4(q) 4 1 (2, p)f q24(q2 − 1)(q6 − 1)(q8 − 1)(q12 − 1)
E6(q) 6 (3, q − 1) 2df 1dq36i{2,5,6,8,9,12}(qi1)
E7(q) 7 (2, q − 1) df 1dq63i{2,6,8,10,12,14,18}(qi1)
E8(q) 8 1 f q120i{2,8,12,14,18,20,24,30}(qi1)
2B2(q), q = 22m+1 1 1 f q2(q2 + 1)(q − 1)}
2G2(q), q = 32m+1 1 1 f q3(q3 + 1)(q − 1)
2F4(q), q = 22m+1 2 1 f q12(q6 + 1)(q4 − 1)(q3 + 1)(q − 1)
3D4(q) 2 1 3f q12(q8 + q4 + 1)(q6 − 1)(q2 − 1)
2E6(q) 4 (3, q + 1) 2df 1dq36i{2,5,6,8,9,12}(qi(1)i)

Some authors have proved that some alternating groups with s(G) ≥ 2 are OD-characterizable.

Proposition 1.3

The alternating groups Ap, Ap+1 and Ap+2, where p is a prime, are OD-characterizable [4].

But if s(G) = 1, then what is happening to the influence of OD on the structure of groups? Recently the following results have been shown.

Proposition 1.4

Let An be alternating group of degree n. Then the following statements hold.

  1. Ap+3, where p is a prime and 7 ≠ p, are OD-characterizable [5-7].

  2. A10 is 2-fold OD-characterizable [8].

  3. An with that either 5 ≤ n ≤ 100 and n ≠ 10 or n = 106, 112, is OD-characterizable [5, 7, 9-12].

  4. Ap+5 with p ≤ 1000 a prime is OD-characterizable [13].

  5. A147 and A189 are 7-fold and 14-fold OD-characterizable, respectively [14].

  6. Assume that p is a prime satisfying the following three conditions:

    (6a) p ≠ 139 and p ≠ 181,

    (6b) π((p + 8)!) = π(p!),

    (6c) p ≤ 997.

    Then Ap+8 is OD-characterizable [14].

Form Proposition 1.4(2)(3), some alternating groups cannot be OD-characterizable. So naturally, we will ask this question: which of the alternating groups is OD-characterizable? Thus in this paper, we give a new characterization of alternating group Ap+d by OD. In fact, we will prove the following result.

Main theorem 1.5

Let p be a prime and d positive. If d is equal to 2 or 4 or d is a prime, then alternating groups Ap+d except A10 are OD-characterizable.

This result confirms the conjecture [15, Conjecture 5.1] put forward by of Yan et al.

2 Some lemmas

In this section, we will give some results which will be used.

Lemma 2.1

Let An (or Sn) be an alternating (or a symmetric group) of degree n. Then the following hold.

  1. Let p, qπ(An) be odd primes. Then pq if and only if p + qn.

  2. Let pπ(An) be odd prime. Then 2 ∼ p if and only if p + 4 ≤ n.

  3. Let p, qπ(Sn). Then pq if and only if p + qn.

Proof

It is easy to get from [16].□

By [1], we have that |An| = n!/2 and |Sn| = n!.

Let exp(n, r) = a or ran, where a is a positive integer satisfying ra|n but ra+1n.

Lemma 2.2

Let Ap+k be an alternating group of degree p + k such that p > k ≥ 3 and p + i is composite, i = 3, ⋯, k, where p is a prime. Let |π(Ap + k)| = d. Then the following hold.

  1. deg(2) = deg(3) = d − 1 if k ≥ 4; deg(2) = d − 2 and deg(3) = d − 1 if k = 3.

  2. deg(p) = |π(k!)| if k ≥ 4; deg(p) = |π(k!)| − 1 if k = 3.

  3. exp(|Ap+k|,2)=i=1[p+k2i]1. In particular, exp(|Ap+k|, 2) ≤ p + k − 1.

  4. exp(|Ap+k|,r)=i=1[p+kri] for each rπ(Ap+k)\{2}. Furthermore, exp(|Ap+k|,r)<p+k2, where 3 ≤ rπ(Ap+k). In particular, if r>[p+k2],thenexp(|Ap+k|,r)=1.

Proof

  1. (1) By Lemma 2.1, we have that 2 ∼ p and 3 ∼ p. Hence deg(2) = d − 1 = deg(3).

  2. For rπ(Ap+k)\{2, p}, by Lemma 2.1, it is easy to see that rp if and only if p + rp + k. It follows that rk and rπ(k!). Hence deg(p) = |π(k!)|.

  3. By the definition of Gaussian integer function, we have that

    exp(|Ap+k|,2)=i=1[p+k2i]1=([p+k2]+[p+k22]+[p+k23]+)1(p+k2+p+k22+p+k23+)1=(p+k)(12+122+123+)1=p+k1.
  4. By the same as in proof of (3), we have that

    exp(|Ap+k|,r)(p+k)(1r+1r2+1r3+)=p+kr1p+k2

    for an odd prime rπ(Ap+k).Ifr>[p+k2],exp(|Ap+k|,r)=1.

The proof is completed.□

Lemma 2.3

Let L be a nonabelian simple group. Then the order and its outer-automorphism of L are as listed in Tables 13.

Table 3

The simple sporadic groups

L d O |L|
M11 1 1 24⋅ 32⋅ 5⋅ 11
M12 2 2 26⋅ 33⋅ 5⋅ 11
M22 12 2 27⋅ 32⋅ 5⋅ 7⋅ 11
M23 1 1 27⋅ 32⋅ 5⋅ 7⋅ 11⋅ 23
M24 1 1 210⋅ 33⋅ 5.7⋅ 11⋅ 23
J1 1 1 23⋅ 3⋅ 5⋅ 7⋅ 11⋅ 19
J2 2 2 27⋅ 33⋅ 52⋅ 7
J3 3 2 27⋅ 35⋅ 5⋅ 17⋅ 19
J4 1 1 221⋅ 33⋅ 5⋅ 7⋅ 113⋅ 23⋅ 29⋅ 31⋅ 37⋅ 43
HS 2 2 29⋅ 32⋅ 53⋅ 7⋅ 11
Suz 6 2 213⋅ 37⋅ 52⋅ 7⋅ 11⋅ 13
McL 3 2 27⋅ 36⋅ 53⋅ 7⋅ 11
Ru 2 1 214⋅ 33⋅ 53⋅ 7⋅ 13⋅ 29
He (F7) 1 2 210⋅ 33⋅ 52⋅ 73⋅ 17
Ly 1 1 28⋅ 37⋅ 56⋅ 7⋅ 11⋅ 31⋅ 37⋅ 67
ON 3 2 29⋅ 34⋅ 5⋅ 73⋅ 11⋅ 19⋅ 31
Co1 2 1 221⋅ 39⋅ 54⋅ 72⋅ 11⋅ 13⋅ 23
Co2 1 1 218⋅ 36⋅ 53⋅ 7⋅ 11⋅ 23
Co3 1 1 210⋅ 37⋅ 53⋅ 7⋅ 11⋅ 23
Fi22 6 2 217⋅ 39⋅ 52⋅ 7⋅ 11⋅ 13
Fi23 1 1 218⋅ 313⋅ 52⋅ 7⋅ 11⋅ 13⋅ 17⋅ 23
Fi24 3 2 221⋅ 316⋅ 52⋅ 73⋅ 11⋅ 13⋅ 17⋅ 23⋅ 29
HN(F5) 1 2 214⋅ 36⋅ 56⋅ 7⋅ 11⋅ 19
Th(F3) 1 1 215⋅ 310⋅ 53⋅ 72⋅ 13⋅ 19⋅ 31
BM(F2) 2 1 241⋅ 313⋅ 56⋅ 72⋅ 11⋅ 13⋅ 17⋅ 19⋅ 23⋅ 31⋅ 47
M(F1) 1 1 246⋅ 320⋅ 59⋅ 76⋅ 112⋅ 133⋅ 17⋅ 19⋅ 23⋅ 29⋅ 31⋅ 41⋅ 47⋅ 59⋅ 71

Proof

See [17].□

Lemma 2.4

If n≥ 6 is a natural number, then there are at least s(n) prime numbers pi such that n+12<pi<n.

Here

  1. s(n) = 6 for n ≥ 48;

  2. s(n) = 5 for 42 ≤ n ≤ 47;

  3. s(n) = 4 for 38 ≤ n ≤ 41;

  4. s(n) = 3 for 18 ≤ n ≤ 37;

  5. s(n) = 2 for 14 ≤ n ≤ 17;

  6. s(n) = 1 for 6 ≤ n ≤ 13.

In particular for every natural number n > 6, there exists a prime p such that n+12<p<n1, and for every natural number n > 3, there exists an odd prime number p such that np < p < n.

Proof

See Lemma 1 of [18].□

Lemma 2.5

Let G be a finite non-abelian simple group and p is the largest prime divisor of |G| with p ‖|G|. Then p||Out(G)|.

Proof

We know that if a prime p divides the order of outer-automorphism of a group G, then the Sylow p-subgroups of G are noncyclic (see [19, Corollary 11.21]).□

Lemma 2.6

Let G be a simple group of Lie type in characteristic p. Assume that pk||G| but pk+1 ∤ |G| where k is a positive integer, then |G| < p3k, and if GL2(q) with q = pk, then |G|<p83k.

Proof

It is easy to get from Tables 1 and 2 of Lemma 2.3.□

3 Proof of the main theorem

In this section, we give the proof of main theorem.

For easy reading, we rewrite the main theorem.

Theorem 3.1

Let p be a prime. Assume that d is a positive integer such that either d is a prime if d is odd or d is equal to at most 4 if d is even. Then alternating groups Ap+d except A10 are OD-characterizable.

Proof

By Proposition 1.4(1), we can assume that d ≥ 4. Propositions 1.3 and 1.4(3)(5) implies that if p + d is a prime, then the alternating group is OD-characterizable, hence we always consider the case: π((p + d)!) = π(p!) and p ≥ 113. We have divided the proof of theorem into a series of lemmas. Note that p is always the largest prime divisor of |G|.

Step 1: G is insoluble.

Proof

Since n ≥ 118 and |G|=n!2, then by Lemmas 2.2 and 2.4 there exists a prime p1 such that |G|p1 = p1, and p1 ∤ (p − 1). Assume that G is soluble, then there is a subgroup H of order p1 p. Since p1 ∤ (p − 1), then H is cyclic and so G has an element of order p1 p. It follows that deg(p) > |π(d!)|, a contradiction to Lemma 2.2. Therefore G is insoluble.□

We use the notation of [20]. Let |An|=n!2=q1α1q2α2qkαkwhereα1=s=1[n2s]1andαi=s=1[nqis], 2 ≤ ik and the qi are prime numbers. If n2<p and n ≥ 113, then by Lemma 2.4, αi = 1 for i ∈ {k − 5, k − 4, k − 3, k − 2, k − 1, k}. For convenience, we set tn(1)=n2<p1np1,tn(2)=n3<p2<n2p22tn(1),, and tn(k)=i=1k(ni+1<pi<nipi)i where pj = 1 if there is no prime between nj+1andnj.

Step 2: There exists a normal series 1 < K < H < G such that sG/K ≲ Aut(S) where SH/K.

Let G = G/K and S = Soc(G). Then S = B1 × B2 × ⋯ × Bm, where the Bi are non-abelian simple groups and SG ≲ Aut(S). In the following, we will prove that m = 1.

Assume that m ≥ 2. Then there is a prime p | tn(1) such that p ∤ |S|. Otherwise, by Lemma 2.4 there is a prime p′ | tn(1) with that pp′ ∈ ω(G), in contradiction to Lemma 2.2 (since p′ ∤ (p − 1) and p′ ∉ π(K), then p′ ∈ π(CG/K(SK/K))). Thus for every i, Bi ∈ 𝔉p, where p′ < p is the second maximal prime of π(G) and 𝔉p is the set of non-abelian finite simple group S such that pπ(S){2,3,5,,p}.

We shall prove that p ∤ |K|. Otherwise, by NC theorem, NG(Kp)/CG(Kp) is isomorphic to a subgroup of Aut(Kp) where Kp is a Sylow p-subgroup of K and also a Sylow p-subgroup of G. Then there is a prime p′ that |G|p = p′ divides |K|. It follows that there exists an element of order pp′ contradicting to deg(p) = |π(d!)| by Lemma 2.2. Hence p||Out(S)|. We know that

Out(S)=Out(S1)×Out(S2)××Out(Sr),

where the groups Sj(j = 1, 2, ⋯, r) are direct products of all isomorphic Bi s such that

S=S1×S2××Sr.

Therefore for certain j, p divides the order of an outer-automorphism of a direct product Sj of t isomorphic simple groups Bi for some 1 ≤ jm. Since Bi ∈ 𝔉p, it follows from Lemma 2.5, that p ∤ |Out(Bi)|. But by [21], |Aut(S)| = |Aut(Sj)|tt!. Thus tp. Since Bj is non-abelian simple group, then 4p||Aut(Bi)|t, and hence, 22p||G| which contradicts Lemma 2.2. Hence m = 1 and S = Bi.

Let |π(n)| be the number of primes ≤ n.

Step 3: tn(1)||S|. In particular, if n ≥ 118, then tn(9)||S|.

First we show that tn(1) divides the order |H/K| of H/K. We know that |Gp| = p, and so we can assume that p divides H/K and p ∤ |K|. Then H/K is nonabelian. If not, H/K is a group of order p. By NC theorem, G/KCG/K(H/K) a subgroup of Zp − 1 where Zn is a cyclic group of order n. As n ≥ 118, there exists another prime p′ (≠ p) dividing tn(1). Since p′ ∤ p − 1 and p′ ∉ π(K), p′ ∈ π(CG/K(H/K)) and so pp′∈ω(G) contradicting to deg(p) = |π(q)|. Thus H/K is non-abelian. If there is a prime p′ ∉ π(H/K) dividing tn(1), then p′ ∈ π(G/H). By Frattini’s argument, G = NG(P)H where P ∈ Sylp(H). Now we have NG(P)CG(P) a subgroup of Zp − 1 and so p′ ∈ π(CG(P)), whence p p′ ∈ ω(G) a contradiction since deg(p) = |π(d!)|. Therefore, tn(1) divides the order |H/K| of H/K.

Secondly, we prove tn(9)||S|. If n is ≥ 118, then there are two primes q1, q2 such that 910n<q2<q1n. Indeed, using Theorem 1 of [22],

xlogx(1+l2logx)<|π(x)|<xlogx(1+32logx).

Hence we have

|π(n)||π(910n)|>nlogn(1+l2logn)910nlog910n(1+32log910n).

By direct computation, |π(n)||π(910n)|>1 when n ≥ 18000. While 118 ≤ n < 18000, we can check it directly by [23].

Let n10<p1n2 where p1 is a prime. If p1||S|, then p1||K| or p1||G/H|. If p1||G/H|, then G has an element of order p1 p, a contradiction according to the above arguments. If p1||K|,thenp1j||K| for j ∈ {2, 3, 4, 5, 6, 7, 8, 9}. We conclude that there is a subgroup M of order ppi for i ≤ 9. Obviously, G has no element of order ppi as deg(p) = |π(d!)| and by Frattini’s argument. By Theorem 9.3.1 of [24], pi| (p − 1) for i ≤ 9. Then p1 p2p9| (p − 1). But n2q1>q2>>q9>n10>p10andhence,i=19qi>p by direct check when p ≥ 113, a contradiction. Therefore tn(9) divides the order |S| of S.

Step 4: If n ≥ 118, then tn(9) > 7.82n.

If n ≥ 118, then by Theorem 1 of [22], |π(ni)|>nilogni(1+l2logni)fori{1,2,,9} and by Corollary 2 of [22], |π(n)|<5n4logn. Thus for n ≥ 15

logtn(9)=n2<p 2nlogp1+2n3<p 9n2logp2++9n10<p 1n9logp9>(|π(n)||π(n2)|)logn2+2(|π(n2|)|π(n3)|)logn3++9(|π(n9)||π(n10)|)logn10=|π(n)|logn2+|π(n2)|(2logn3logn2)++|π(n9)|(6logn75logn6)6|π(n7)|log(n7)>nlogn(1+l2logn)log(n2)+n2logn2(1+l2logn2)log(2n32)+n9logn9(1+l2logn9)log98n10991.25n10=i=19nilogni(1+l2logni)logii1n(i+1)i1.259n10=18.9316709716.875>2.05667n

So we conclude that tn(9) > e2.05667n > (7.82)n.

Step 5: S is isomorphic to An with n = p, p + 1, p + 2, p + 3, ⋯, p + d.

According to Classification Theorem of Finite Simple Groups and Step 2, S is isomorphic to a sporadic simple group, a simple group of Lie type or an alternating group. If S is isomorphic to a sporadic simple group and p ≥ 113, then there is no group satisfying tn(9)||S| by Table 3. If S is a simple group of Lie type in characteristic p, then pn10. If not and we assume that n2<pn,thentn(1)>(|π(n)||π(n2)|)(n2)>(n2)8 (By corollary 3 of [22], |π(n)||π(n2)|>3n25logn2>8forn118). So tn(1) > n7 > p7, a contradiction to Lemma 2.6(p||S| by Step 3). If nk+1<pnk with k = 2, 3, 4, then pk||S| by Step 3 and so |S| < p3k by Lemma 2.6. Thus |S|pk<p2k<p10<(n2)10<tn(1) and hence |S| < pktn(1) < tn(9), a contradiction since tn(9) divides the order |S| of S by Step 3. If k = 5, 6, 7, 8, 9, then similarly we get |S|pk<p2k<p18<(n5)18<tn(5) and so |S| < pktn(5) < tn(9), a contradiction (when n118,|π(nk)||π(nk+1)|>1fork{5,6,7,8,9}).

Let pl||S|. If p ≥ 3, then by Lemmas 2.6 and Step 4,

7.82n<tn(9)|S|pl<p2l(pi=1[npi])2=(pnp1)2<(3231)n=3n

(if x ≥ 3 and f(x)=x1x1, then the differentiate f′(x) of f(x) is < 0), a contradiction since 7. 82 ≰ 3. If p = 2 and SL2(2l), then 7.82n<tn(9)<|S|2l<25l3(2i=1[npi])53=(253)n and so 7. 82 < 25/3 < 22 = 4, a contradiction. If SL2(2l), then S contains maximal cyclic subgroups of orders pl + 1 and pl − 1 and so S contains at most two prime factors which are larger than n/2, a contradiction to the fact that |π(n)||π(n2)|>8 for n ≥ 118.

Since p is the largest prime divisor of |G| and by Step 3, p divides |S|, then S is isomorphic to An for n = p, p + 1, ⋯, p + d.

Step 6: G is isomorphic to Ap+d.

The proof of Step 5 implies that S is isomorphic to An with n = p, p + 1, ⋯, p + d. By Steps 2 and 5, we have that AnG/KSn.

If H/KAp, then ApG/KSp. If G/KAp, then |K| = (p + 1)(p + 2) ⋯ (p + d). It is easy to see that there is a prime divisor p′ of |K| such that p′ > d and p′ ∤ p − 1. Note that p is the largest prime of π(G) and d ≥ 4. Then there is cyclic subgroup of order pp′ and so p′ is adjacent to p. It follows that pp′ ∈ ω(G) and so deg(p) > |π(d!)|, a contradiction. If G/KSp, then |K| = (p + 1)(p + 2) ⋯ (p + d)/2. In this case, we also get that there exists a prime p′ of |K| with the properties: p′ > d and p′ ∤ p − 1. We similarly get that pp′ ∈ ω(G) and so we have a contradiction as deg(p) = |π(d!)|.

Similarly to the above arguments, we also can rule out these cases: H/KAn with n ∈ {p + 1, p + 2, ⋯, p + d − 1}.

If H/KAp+d, then Ap+dG/KSp+d. If G/KAp+d, then K = 1, the desired result. If G/KSp+d, then |G|2 < |Sp+d|2, a contradiction.

Steps 1-6 complete the proof of the Maintheorem 1.5.□

By Propositions 1.3 and 1.4, some alternating groups are OD-characterizable and by our main theorem 1.5 we have the following.

Theorem 3.2

Let p be a prime. Then the alternating group Ap+d except A10 where d is a prime or equals to 4, is OD-characterizable.

The conjecture in [25] was proved true by some joint works of many mathematicians and the last important part of the proof was given by V.D.Mazurov, Chen, SHI, etc. (see [26] and related references). Therefore we can get the following theorem which is also a conjecture.

Theorem 3.3

Let G be a group and H a finite simple group. Then GH if and only if (a) ω(G) = ω(H) and (b) |G| = |H|.

Corollary 3.4

Let G be a group and p ≥ 5 a prime. Assume that d is either a prime or 4. Then GAp+d if and only if ω(G) = ω(Ap+d) and |G| = |Ap+d|.

Acknowledgement

The corresponding author was supported by the Department of Education of Sichuan Province (Grant No: 15ZA0235) and by the Opening Project of Sichuan Province University Key Laborstory of Bridge Nondestruction Detecting and Engineering Computing (Grant Nos: 2016QYJ01 and 2017QZJ01). The authors are very grateful for the helpful suggestions of the referee.

  1. Conflict of interest

    The conflict of interest disclosure: The authors declare that there is no conflict of interest regarding the publication of this paper.

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Received: 2016-11-2
Accepted: 2017-6-20
Published Online: 2017-8-19

© 2017 Yang et al.

This work is licensed under the Creative Commons Attribution-NonCommercial-NoDerivatives 4.0 License.

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