# On Thompson’s Conjecture for Alternating GroupsAp+3

Authors: Shitian Liu; Yong Yang
Journal: The Scientific World Journal
 (2014)

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## Abstract

LetG be a group. Denote by π(G) the set of prime divisors of |G|. Let GK(G) be the graph with vertex set π(G) such that two primes p and q in π(G) are joined by an edge if G has an element of order p·q. We set s(G) to denote the number of connected components of the prime graph GK(G). Denote by N(G) the set of nonidentity orders of conjugacy classes of elements in G. Alavi and Daneshkhah proved that the groups, An where n=p,p+1,p+2 with s(G)≥2, are characterized by N(G). As a development of these topics, we will prove that if G is a finite group with trivial center and N(G)=N(Ap+3) with p+2 composite, then G is isomorphic to Ap+3.


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## Body

## 1. Introduction

In this paper, all groups considered are finite and simple groups are nonabelian simple groups. LetG be a finite group and let Z(G) be its center. For any 1≠x∈G, suppose that xG denotes the conjugacy classes in G containing x and CG(x) denotes the centralizer of x in G. We will use N(G) to denote the set {n:G has a conjugacy class of size n}. Thompson in 1987 gave the following conjecture with respect to N(G).Thompson’s Conjecture (see [1], Question 12.38). If L is a finite simple nonabelian group, G is a finite group with trivial center, and N(G)=N(L), then G≅L.Letπ(G) denote the set of all prime divisors of |G|. Let GK(G) be the graph with vertex set π(G) such that two primes p and q in π(G) are joined by an edge if G has an element of order p·q. We set s(G) to denote the number of connected components of the prime graph GK(G). A classification of all finite simple groups with disconnected prime graph was obtained in [2, 3]. Based on these results, Thompson’s conjecture was proved valid for many finite simple groups with s(G)≥2 (see [4, 5]). In Ahanjideh and Xu’s works, the groups Ln(q), Dn(q), Bn, Cn, 2DN(q), and E7(q) are true for Thompson’s conjecture (see [6–10]). Alavi and Daneshkhah proved that the groups An with n=p, p+1, p+2 and s(G)≥2 are characterized by N(G) (see [11]). So is there a group with connected prime graph for which Thompson’s conjecture would be true? Recently, the groups A10, A16, A22, and A26 were proved valid for this conjecture (see [12–15]). As a development of these topics, we will prove that Thompson’s conjecture is true for the alternating groups Ap+3 of degree p+3 with p+2 composite.We will introduce some notations used in the proof of the main theorem. LetG be a group and p a prime. Then denote by Gp the Sylow p-subgroup of G. Let Aut(G) and Out(G) denote the automorphism and outer automorphism groups of G, respectively. Let ω(G) denote the set of element orders of G. The other notations are standard (see [16], for instance).


## 2. Preliminary Results

In this section we will give some preliminary results.Lemma 1.
Letx,y∈G, (|x|,|y|)=1, and xy=yx. Then (1)
CG(xy)=CG(x)∩CG(y);(2)
|xG| divides |(xy)G|;(3)
if|xG|=|(xy)G|, then CG(x)≤CG(y).Proof.
See [12, Lemma 1.2] and [7, Lemma 2.3].Lemma 2.
IfP and H are finite groups with trivial centers and N(P)=N(H), then π(P)=π(H).Proof.
See [12, Lemma 3].Lemma 3.
Suppose thatG is a finite group with trivial center and p is a prime from π(G) such that p2 does not divide |xG|  for all x in G. Then a Sylow p-subgroup of G is elementary abelian.Proof.
See [12, Lemma 4].Lemma 4.
LetK be a normal subgroup of G, and G¯=G/K. (1)
Ifx¯ is the image of an element x of G in G¯, then |x¯G¯| divides |xG|.(2)
If(|x|,|K|)=1, then CG¯(x¯)=CG(x)K/K.(3)
Ify∈K, then |yK| divides |yG|.Proof.
See [12, Lemma 5].Letexp⁡(n,r) denote the nonnegative integer a such that ra∣n but ra+1∤n.Lemma 5.
LetAp+3 be the alternating group of degree p+3, where p is a prime. Then the following hold.(1)
exp⁡(|Ap+3|,2)=∑i=1∞[(p+3)/2i]-1; in particular, exp⁡(|Ap+3|,2)≤p+2.(2)
exp⁡(|Ap+3|,r)=∑i=1∞[(p+3)/ri]  for each r∈π(Ap+3)∖{2}. Furthermore, exp⁡(|Ap+3|,r)<(p-1)/2, where 3≤r∈π(Ap+3). In particular, if r>[(p+3)/2], then exp⁡(|Ap+3|,r)=1.Proof.
(1) By the definition of Gaussian integer function, we have that(1)exp⁡⁡(|Ap+3|,2)=∑i=1∞‍[p+32i]-1=([p+32]+[p+322]+[p+323]+⋯)-1≤(p+32+p+422+p+323+⋯)-1=(p+3)(12+122+123+⋯)-1=p+2.
(2) Similarly, as (1), we have that(2)exp⁡⁡(|Ap+3|,r)≤(p+3)(1r+1r2+1r3+⋯)=p+3r-1≤p+32
for an odd prime r∈π(Ap+3).
Ifr>[(p+3)/2], exp⁡(|Ap+3|,r)=1.
The proof is completed.LetSn be the symmetric group of degree n. Assume that the cycle has c1 1-cycles, c2 2-cycles, and so on, up to ck  k-cycles, where 1c1+2c2+⋯+kck=n. Then the number of conjugacy classes in Sn is
(3)z=n!(∏i=1k‍ici∏i=1k‍ci!)-1.
Let An be the alternating group of degree n.Lemma 6.
Letx∈An. Then, for the size of the conjugacy class xG of x in An, one has the following. (1)
If, for all eveni, ci=0 and, for all odd i, i∈{0,1}, then |xG|=z/2.(2)
In all other cases,|xG|=z.In particular, |xG|≥z/2.Proof.
See [17].Lemma 7.
Ifn≥6 is a natural number, then there are at least s(n) prime numbers pi such that (n+1)/2<pi<n. Here(i)
s(n)=6 for n≥48;(ii)
s(n)=5 for 42≤n≤47;(iii)
s(n)=4 for 38≤n≤41;(iv)
s(n)=3 for 18≤n≤37;(v)
s(n)=2 for 14≤n≤17;(vi)
s(n)=1 for 6≤n≤13.In particular, for every natural number n>6, there exists a prime p such that (n+1)/2<p<n-1, and, for every natural number n>3, there exists an odd prime number p such that n-p<p<n.Proof.
See Lemma 1 of [18].Lemma 8.
Leta,b, and n be positive integers such that (a,b)=1. Then there exists a prime p with the following properties:(i)
p divides an-bn,(ii)
p does not divide ak-bk for all k<n,with the following exceptions: a=2,  b=1; n=6 and a+b=2k; n=2.Proof.
See [19].Lemma 9.
With the exceptions of the relations(239)2-2(13)4=-1 and (3)5-2(11)2=1 every solution of the equation
(4)pm-2qn=±1;p,qprime;m,n>1,
has exponents m=n=2; that is, it comes from a unit p-q·21/2 of the quadratic field Q(21/2) for which the coefficients p and q are primes.Proof.
See [20, 21].LetL be a nonabelian simple group and letO denote the order of the outer automorphism group of L.Lemma 10.
LetL be a nonabelian simple group. Then the orders and their outer automorphism of L are as listed in Tables 1, 2, and 3.Table 1
The simple classical groups.
L
Lie; rankL
d
O
|
L
|
L
n
(
q
)
A
n
-
1
(
q
)
(
n
,
q
-
1
)
2
d
f, if n≥3;
1
d
q
n
(
n
-
1
)
/
2
∏
i
=
2
n
(
q
i
-
1
)
n
-
1
d
f, if n=2
U
n
(
q
)
2
A
n
-
1
(
q
)
(
n
,
q
+
1
)
2
d
f, if n≥3
1
d
q
n
(
n
-
1
)
/
2
∏
i
=
2
n
(
q
i
-
(
-
1
)
i
)
[
n
/
2
]
d
f, if n=2
P
S
p
2
m
(
q
)
C
m
(
q
)
(
2
,
q
-
1
)
d
f, m≥3;
1
d
q
m
2
∏
i
=
1
m
(
q
2
i
-
1
)
m
2
f, if m=2
Ω
2
m
+
1
(
q
)
B
m
(
q
)
2
2
f
1
2
q
m
2
∏
i
=
1
m
(
q
2
i
-
1
)
q odd
m
P
Ω
2
m
+
(
q
)
D
m
(
q
)
(
4
,
q
m
-
1
)
2
d
f, if m≠4
1
d
q
m
(
m
-
1
)
(
q
m
-
1
)
∏
i
=
1
m
-
1
‍
(
q
2
i
-
1
)
m
≥
3
m
6
d
f, if m=4
P
Ω
2
m
-
(
q
)
2
D
m
(
q
)
(
4
,
q
m
+
1
)
2
d
f
1
d
q
m
(
m
-
1
)
(
q
m
+
1
)
∏
i
=
1
m
-
1
‍
(
q
2
i
-
1
)
m
≥
2
m
-
1Table 2
The simple exceptional group.
L
L
d
O
|
L
|
G
2
(
q
)
2
1
f, if p≠3
q
6
(
q
2
-
1
)
(
q
6
-
1
)
2
f, if p=3
F
4
(
q
)
4
1
(
2
,
p
)
f
q
24
(
q
2
-
1
)
(
q
6
-
1
)
(
q
8
-
1
)
(
q
12
-
1
)
E
6
(
q
)
6
(
3
,
q
-
1
)
2
d
f
1
d
q
36
∏
i
∈
{
2,5
,
6,8
,
9,12
}
(
q
i
-
1
)
E
7
(
q
)
7
(
2
,
q
-
1
)
d
f
1
d
q
63
∏
i
∈
{
2,6
,
8,10,12,14,18
}
(
q
i
-
1
)
E
8
(
q
)
8
1
f
q
120
∏
i
∈
{
2,8
,
12,14,18,20,24,30
}
(
q
i
-
1
)
2
B
2
(
q
)
,
q
=
2
2
m
+
1
1
1
f
q
2
(
q
2
+
1
)
(
q
-
1
)
2
G
2
(
q
)
,
q
=
3
2
m
+
1
1
1
f
q
3
(
q
3
+
1
)
(
q
-
1
)
2
F
4
(
q
)
,
q
=
2
2
m
+
1
2
1
f
q
12
(
q
6
+
1
)
(
q
4
-
1
)
(
q
3
+
1
)
(
q
-
1
)
3
D
4
(
q
)
2
1
3
f
q
12
(
q
8
+
q
4
+
1
)
(
q
6
-
1
)
(
q
2
-
1
)
2
E
6
(
q
)
4
(
3
,
q
+
1
)
2
d
f
1
d
q
36
∏
i
∈
{
2,5
,
6,8
,
9,12
}
(
q
i
-
(
-
1
)
i
)Table 3
The simple sporadic groups.
L
d
O
|
L
|
M
11
1
1
2
4
·
3
2
·
5
·
11
M
12
2
2
2
6
·
3
3
·
5
·
11
M
22
12
2
2
7
·
3
2
·
5
·
7
·
11
M
23
1
1
2
7
·
3
2
·
5
·
7
·
11
·
23
M
24
1
1
2
10
·
3
3
·
5
·
7
·
11
·
23
J
1
1
1
2
3
·
3
·
5
·
7
·
11
·
19
J
2
2
2
2
7
·
3
3
·
5
2
·
7
J
3
3
2
2
7
·
3
5
·
5
·
17
·
19
J
4
1
1
2
21
·
3
3
·
5
·
7
·
1
1
3
·
23
·
29
·
31
·
37
·
43
H
S
2
2
2
9
·
3
2
·
5
3
·
7
·
11
S
u
z
6
2
2
13
·
3
7
·
5
2
·
7
·
11
·
13
M
c
L
3
2
2
7
·
3
6
·
5
3
·
7
·
11
R
u
2
1
2
14
·
3
3
·
5
3
·
7
·
13
·
29
H
e
(
F
7
)
1
2
2
10
·
3
3
·
5
2
·
7
3
·
17
L
y
1
1
2
8
·
3
7
·
5
6
·
7
·
11
·
31
·
37
·
67
O
N
3
2
2
9
·
3
4
·
5
·
7
3
·
11
·
19
·
31
C
o
1
2
1
2
21
·
3
9
·
5
4
·
7
2
·
11
·
13
·
23
C
o
2
1
1
2
18
·
3
6
·
5
3
·
7
·
11
·
23
C
o
3
1
1
2
10
·
3
7
·
5
3
·
7
·
11
·
23
F
i
22
6
2
2
17
·
3
9
·
5
2
·
7
·
11
·
13
F
i
23
1
1
2
18
·
3
13
·
5
2
·
7
·
11
·
13
·
17
·
23
F
i
24
′
3
2
2
21
·
3
16
·
5
2
·
7
3
·
11
·
13
·
17
·
23
·
29
H
N
(
F
5
)
1
2
2
14
·
3
6
·
5
6
·
7
·
11
·
19
T
h
(
F
3
)
1
1
2
15
·
3
10
·
5
3
·
7
2
·
13
·
19
·
31
B
M
(
F
2
)
2
1
2
41
·
3
13
·
5
6
·
7
2
·
11
·
13
·
17
·
19
·
23
·
31
·
47
M
(
F
1
)
1
1
2
46
·
3
20
·
5
9
·
7
6
·
1
1
2
·
1
3
3
·
17
·
19
·
23
·
29
·
31
·
41
·
47
·
59
·
71Proof.
See [22].


## 3. Main Theorem and Its Proof

In this section, we give the main theorem and its proof.Theorem 11.
LetG be a finite group with trivial center and N(G)=N(Ap+3) with p+2 composite. Then G is isomorphic to Ap+3.Proof.
We know thatAp+3 are characterized by N(G) if p=7,13,19 (see [12–14]). Then in the following we only consider when p≥23.We divide the proof into the following lemmas.Lemma 12.
LetL=Ap+3. Then the following hold.(1)
If2≠r≤[(p+3)/2], then we can write p+3=kr+m with 0≤m<r and conjugacy class sizes of r-elements of L are
(5)(p+3)!(p+3-ir)!·ri·i!for possiblei with 1≤i≤k=[(p+3)/r].In particular, ifr is an odd prime divisor of |G|, then conjugacy class sizes of r-element of L are
(6)(p+3)!2·k!·r2,wherep+3=2r+k and 0≤k<r.(2)
Ifr=2, then one can write p+3=2k+m with 0≤m≤1 and conjugacy class sizes of 2-elements of L are
(7)(p+3)!(p+3-2i)!·22i·(2i)!for possiblei with 1≤i≤k=[(p+3)/2].(3)
Ifr>[(p+3)/2], then one can write p+3=r+m with 0≤m<r and conjugacy class sizes of r-elements of L are
(8)(p+3)!(p+3-r)!·r.In particular, ifr=p, then the conjugacy class size of p-elements of L is
(9)(p+3)!6p.(4)
The following numbers fromN(G) are maximality with respect to divisibility.
(a)
One of the following holds:(p+3)!/2r2 if 2·r=p+3;(p+3)!/4r2 if 2·r+2=p+3;(p+3)!/2(k-1)r2 if 2·r+k=p+3 and k=2n with n≥2;(b)
(p+3)!/6p.Proof.
From (3) and Lemma 6, we get the desired results.Lemma 13.
LetG be a finite group with trivial center and N(G)=N(L). Then |L|∣|G| and π(G)=π(L).Proof.
Note that|L|=∏n∈N(L)n. Since |xG||CG(x)|=|G|, every member from N(G) divides the order of G and |L|∣|G|. So by Lemma 12, we have that π(G)=π(L).Lemma 14.
Suppose thatG is a finite group with trivial center and N(G)=N(L). Then the following hold.(1)
There exist different primesr1,r2,p from π(L) such that r1,r2,p>[(p+3)/2]. In particular, the Sylow r-subgroup S of G is a cyclic group of order r where r∈{r1,r2,p}. There does not exist an element of order r1·r2, r1·p, or r2·p.(2)
For alln∈N(G), if n is divisible at most by ra, then the Sylow r-subgroup S of G is of order ra.Proof.
(1) By Lemma7, there exist different prime numbers r1,r2,p from π(G) such that r1,r2,p>[(p+3)/2].
From Lemmas12 and 13, we have that the primes r1,r2,p are prime divisors of |G| and r12,r22,p2 do not divide |xG| for all x∈G. Then by Lemma 3, S is elementary abelian. Therefore if |x|=r, then |xG| is an r′-number.
Let|S|≥p2. Consider an element y of G with
(10)|yG|=(p+3)!4·r2if2r+2=p+3,|yG|=(p+3)!2·r2if2r=p+3,
or
(11)|yG|=(p+3)!2(k-1)r2if2r+k=p+3,dddddddddddawik=2nwithn≥2
by Lemma 12.
Assume thatp∤|y|. Let x be an element of CG(y) having order p. Then CG(xy)=CG(x)∩CG(y), |xG|∣|(xy)G|, and |yG|∣|(xy)G| by Lemma 1. Since S is abelian, S≤CG(x). Hence, p∤|xG|. It follows that |xG| equals (p+3)!/6p or (p+1)(p+2)(p+3)/3 by Lemma 12.
If|xG| equals (p+3)!/6p, then (p+3)!/6∣|(xy)G|. On the other hand, |yG|∣|(xy)G|; then we have that
(12)(p+3)!4·r2∣|(xy)G|if2r+2=p+3,(p+3)!2·r2∣|(xy)G|if2r=p+3,
or
(13)(p+3)!2(k-1)r2∣|(xy)G|if2r+k=p+3,lllllllldldldldldldldlk=2nwithn≥2.
Obviously, there is no number from N(G) such that |xG|∣|(xy)G| and |yG|∣|(xy)G|.
Therefore|xG| equals (p+1)(p+2)(p+3)/3. In the following, we will consider the following three cases.
Case
1. |yG|=(p+3)!/2·r2 if 2r=p+3.
Obviously,r∣(p+1)(p+2)(p+3)/3. Therefore (p+3)!/r∣|(xy)G|, a contradiction, since |xG|∣|(xy)G|, |yG|∣|(xy)G|, and the maximality of |yG|=(p+3)!/2·r2.
Case
2.
|
y
G
|
=
(
p
+
3
)
!
/
4
·
r
2 if 2r+2=p+3.
Obviously,r∣(p+1)(p+2)(p+3)/3. Therefore (p+3)!/r∣|(xy)G|. Also we get a contradiction as in Case 1.
Case
3.
|
y
G
|
=
(
p
+
3
)
!
/
2
(
k
-
1
)
r
2 if 2r+k=p+3 and k=2n with n≥2.
In this case,r∤(p+1)(p+2)(p+3)/3. It follows that |yG|=|(xy)G|. By Lemma 1, CG(y)≤CG(x) and so |xG|∣|yG|, a contradiction.
Assume thatp∣|y|. Let |y|=p·t. Since S is elementary abelian, the numbers p and t are coprime. Let
(14)u=yp,v=yt.
Then y=uv and CG(uv)=CG(u)∩CG(v). Therefore,
(15)|vG|∣|yG|=(p+3)!2·r2if2r=p+3,|vG|∣|yG|=(p+3)!2·r2if2r+2=p+3,
or
(16)|vG|∣|yG|=(p+3)!2(k-1)·r2if2r+k=p+3,lklklklklkklklklkklfeeeek=2n,withn≥2.
On the other hand, the elementv of G is of order p. Since the Sylow p-subgroup of G is elementary abelian, then p∤|vG|. It follows that
(17)|vG|=(p+3)!6por(p+1)(p+2)(p+3)3
by Lemma 12.
If|vG|=(p+3)!/6p, then |vG|∣|yG|, a contradiction. Hence |vG|=(p+1)(p+2)(p+3)/3. We consider the following three cases.
Case
1.
|
y
G
|
=
(
p
+
3
)
!
/
2
·
r
2 if 2r=p+3.
Obviously,r∣(p+1)(p+2)(p+3)/3. But r∤(p+3)!/2r2, a contradiction, since |xG|∣|(xy)G|, |yG|∣|(xy)G|, and the maximality of |yG|=(p+3)!/2·r2.
Case
2.
|
y
G
|
=
(
p
+
3
)
!
/
4
·
r
2 if 2r+2=p+3.
Obviously,r∣(p+1)(p+2)(p+3)/3. But r∤(p+3)!/4r2, a contradiction, since |xG|∣|(xy)G|, |yG|∣|(xy)G|, and the maximality of |yG|=(p+3)!/4·r2.
Case
3.
|
y
G
|
=
(
p
+
3
)
!
/
2
(
k
-
1
)
r
2 if 2r+k=p+3 and k=2n with n≥2.
In this case,r∤(p+1)(p+2)(p+3)/3. It follows that |yG|=|(xy)G| since the maximality of |vG|. By Lemma 1, CG(y)≤CG(x) and so |xG|∣|yG|, a contradiction.
Therefore the Sylowp-subgroup of G is of order p.
Similarly we can prove the other two cases.
There does not exist an element of orderr1·r2, r1·p, or r2·p.
(2) Without loss of generality, we assume thatn is divisible at most by r2.
Assume that|S|≥r3. Consider an element x of G such that
(18)|xG|={(p+3)!2r2if2r=p+3;(p+3)!4r2if2r+2=p+3;(p+3)!2(k-1)r2if2r+k=p+3,withk=2n,n≥2,
by Lemma 12.
Letr∤|x|. Then there is an element y of G of order r. By Lemma 1 we have that CG(xy)=CG(x)∩CG(y), |xG|∣|(xy)G|, and |yG|∣|(xy)G|.
Ify is an r-central element, then S≤CG(y) and |yG| is an r′-number. We consider the following three cases.
Case
1.
2
r
=
p
+
3.
In this case|yG|=(p+3)!/2·r2. Then we have that |xG|=|yG|=|(xy)G| and so r2∣∣|CG(x)|=|CG(y)| since y is an r-central element (in fact, if r3∣∣|CG(x)|, then there is a conjugacy class size which is a multiple of the number (p+3)!/r3·3! contradicting 2r=p+3). Thus r∣|xG|=|yG| since r3∣|G|, a contradiction.
Case
2.
2
r
+
2
=
p
+
3.
We have that|yG|=(p+3)!/4·r2. We similarly can rule out this case as “Case 1: 2r=p+3”.
Case
3.
2
r
+
k
=
p
+
3 and k=2n with n≥2.
In this case,|yG|=(p+3)!/2(k-1)r2 or (2r+1)(2r+2)⋯(2r+k)/2(k-2) (we only choose the maximality from the r′-number in N(G)). If the former, then |xG|=|yG|=|(xy)G| and so r2∣∣|CG(x)|=|CG(y)| since y is an r-central element (in fact, if r3∣∣|CG(x)|, then there is a conjugacy class size which is a multiple of the number (p+3)!/r3·3! contradicting 2r+k=p+3). Thus r∣|xG|=|yG| since r3∣|G|, a contradiction. Then |yG|=(2r+1)(2r+2)⋯(2r+k)/2(k-2). But there is no member from N(G) such that |xG|∣|(xy)G| and |yG|∣|(xy)G|.
Ify is a noncentral r-element, then we choose an element z of order p such that r2∣∣|zG|. By hypothesis, r∣|CG(z)| and, obviously, r∤|z|. Then by Lemma 1.2 of [4], there is an r-element w such that 1≠w∈CG(z), CG(wz)<CG(w), and
(19)r∣|CG(w)||CG(wz)|=|(wz)G||wG|.
On the other hand, we also have z∈CG(w). Since wz=zw, then |(wz)G|=|zG| and so CG(z)≤CG(w). Therefore z∈CG(z)≤CG(w). It follows that |CG(z)|=|CG(wz)| since the maximality of |zG|, and we get r∣|CG(w)|/|CG(wz)|=1, a contradiction.
Letr∣|x|; then we write |x|=rt. If S is elementary abelian, then (r,t)=1. Set
(20)u=xr,v=xt.
Then x=uv and CG(x)=CG(u)∩CG(v). Hence |vG|∣xG and |uG|∣|xG|. Since v is an element of order r, then
(21)|vG|={(p+3)!2r2if2r=p+3;(p+3)!4r2if2r+2=p+3;(p+3)!2(k-1)r2,or(2r+1)(2r+2)⋯(2r+k)2(k-2)if2r+k=p+3withk=2n,n≥2.
If |vG|=(p+3)!/2r2,(p+3)!/4r2,or(p+3)!/2(k-1)r2, then |(uv)G|=|vG| and so r3∣|S|∣|vG|. It follows that r∤|CG(v)|=|CG(uv)|=|CG(x)| contradicts the fact that x is an r-element. Therefore |vG|=(2r+1)(2r+2)⋯(2r+k)/2(k-2). On the other hand, |xG|=(p+3)!/2(k-1)r2. Obviously, r∣(2r+1)(2r+2)⋯(2r+k)/2(k-2) but r∤(p+3)!/2(k-1)r2. It follows that (p+3)!/2(k-1)r lies in N(G), which contradicts the maximality of (p+3)!/2(k-1)r2.
ThereforeS is nonabelian. So we chose an element z of order p such that r2∣|zG|. By hypothesis, r∣|CG(z)| and, obviously, r∤|z|. Then by Lemma 1.2 of [4], we also have that |CG(z)|=|CG(wz)| since the maximality of |zG|, and we get r∣|CG(w)|/|CG(wz)|=1, a contradiction.
The lemma is completed.Lemma 15.
Suppose thatG is a finite group with trivial center and N(G)=N(L). Let π={2,3}. Then Oπ,π′(G)=Oπ(G). In particular, G is insoluble.Proof.
LetK=Oπ(G), G¯=G/K and denote by x¯ and by H¯ the images of an element x and a subgroup H of G in G¯, respectively. Assume that the result is not true; then there is a prime r∈π(L)∖π with Or(G¯)≠1.
Letr∈{r1,r2,p} with Or(G¯)≠1. Then G¯ contains a Hall {r·s}-subgroup of order r·s with s∈{r1,r2,p}-{r}. However, Hall {r,s}-subgroup must be cyclic contradicting Lemma 14.
LetP be a Sylow r-subgroup of G¯ where r∈π(L)∖{r1,r2,p}. If Or(G¯)≠1, then A=Z(Or(G¯)) is a nontrivial normal subgroup of G¯. Let x¯ be an element of order p in G¯. So we have that |x¯G¯| is a divisor of
(22)(p+3)!6por(p+1)(p+2)(p+3)3.
By coprime action lemma, A=CA(x¯)×[A,x¯]. In the following, we consider two cases “r≤[(p+3)/2] and r≥[(p+3)/2]”.(i)
r>[(p+3)/2] and r≠r1,r2,p. In this case, by Lemma 14, we have that the Sylow r-subgroup of G is of order r. Hence there is a Hall {r,p}-subgroup H. Since H must be cyclic, then there is an element of order r·p, a contradiction by the proof of Lemma 14.(ii)
r≤[(p+3)/2] and r≠2,3. If 2r=p+3, 2r+2=p+3, or 2r+k=p+3 with k=2n and n≥2, then the index of CA(x¯) in A is at most r2. By Lemma 8, there exists a least divisor m of ϕ(p) such that p divides rm-1, and the subgroup [A,x¯]〈x¯〉 must be abelian. It follows that [A,x¯]=1 and A=CA(x¯). If y¯ is a nontrivial element of Z(P)∩A, then the order of CG¯(y¯) is a multiple of p. By Lemma 4, y lies in the center of a Sylow p-subgroup of G. This contradicts Lemma 12. Thus Or(G¯)=1.It follows that Or(G¯)=1 for r∈{5,7,…,p}.
ThereforeOπ,π′(G)=Oπ(G). In particular, G is insoluble.Lemma 16.
There is a normal series1≤K≤H≤G such that H/K≅Ap+3.Proof.
By Lemmas13 and 14, |G|=(p+3)!/2.
By Lemma15, we have that H/K≤G¯≤Aut(H/K), where M:H/K=S1×S2×⋯×Sk is a direct product of nonabelian simple groups S1,S2,…,Sk. Since G cannot contain a Hall {r1,r2,p}-subgroup, numbers r1, r2, and p divide the order of exactly one of these groups that is listed as in Lemma 10, and so we assume that they divide S1. Since S1⊲G¯, we let G* and M* denote the factor groups G¯/S1 and M/S1, respectively. If k>1, then a Sylow 2-subgroup of G* is nontrivial and its center Z has a nontrivial intersection with M*. Consider a nontrivial element y of T=S2×⋯×Sk such that its image in G¯ lies in Z. Since y centralizes S1, it lies in the center of a Sylow 2-subgroup of G¯ and centralizes an element of order p, a contradiction. Thus M=S1, and G¯ is almost simple. Therefore
(23)HK≤G¯≤Aut(HK).
Obviously, r1,r2,p∣|H/K| (in fact, if r1,r2,p∣|G/H|, then r1,r2,p∣|G/H|∣|Out(H/K)|, a contradiction, from Lemma 10; if r1,r2,p∣|K|, then there is an element of order r·p with r∈{r1,r2} contradicting Lemma 14). In the following, we always assume that r∈π(G)=π(L). In the following, we consider S1 which is listed as in Tables 1, 2, and 3.
(i) Case
1.
H
/
K
≅
A
n with n≥6.
Thenn=p,p+1,…,p+k with p+2,p+4,… composite and p+k+1 prime. If k≥4, then (p+k)!/2∣(p+3)!, a contradiction. Therefore H/K is isomorphic to Ap,  Ap+1, Ap+2, or Ap+3.
Letx be an element of order p in H. Then |xH| is p′-number since |H|p=pH/K≅Ap.
Since|Ap|∣(p+3)!, then 3∣|K|. We have |xH|=(p-1)!/2. On the other hand, |xG|=(p+3)!/6p. It follows that |xK|∣(p+1)(p+2)(p+3)/3 and so there is an element of r·p or of order r′·p with 3<r′<r<p and r and r′ divide one of the prime divisors of the numbers p+1, p+2, or p+3, which contradicts Lemma 14. Similarly, we can rule out these cases “H/K≅Ap+1 and H/K≅Ap+2”.
ThereforeH/K≅Ap+3.
(ii) Case
2.
H
/
K is not isomorphic to a sporadic simple group according to Table 3.
(iii) Case
3.
H
/
K is isomorphic to a simple group of Lie type.
Letq be a prime power.(1)
H/K≅Bn(q) with n≥2.In this situation, by Lemma13, π(G)={2,3,5,7,…,p} and so
(24)1(2,q-1)qn2∏i=1n‍(q2i-1)∣(p+3)!.It follows thatp∣q or p∣∏i=1n‍(q2i-1). If p∣q, then q is a power of p. Since |Gp|=p by Lemma 14, this is impossible as n≥2. Therefore p∣∏i=1n‍(q2i-1). It follows that p∣q2t-1 for some 1≤t≤n as p is prime. On the other hand, qn2∣r or qn2∣rm. If the former, then q=r and n=1, a contradiction. It follows that qn2∣rm. Hence r∣q and m=kn2 for some integer k≥1. By Lemma 5, kn2<p/2<(q2n-1)/2, but the equation has no solution in N. Furthermore, since Cn(q) has the same order as Bn(q), we also can rule out.(2)
H/K≅Dn(q) with n≥4.Therefore we have that(1/(4,qn-1))qn(n-1)(qn-1)∏i=1n-1‍(q2i-1)∣(p+3)!. Since the Sylow p-subgroup of G is of order p, p∤q as, otherwise, q=p and thus n=1, a contradiction. It follows that p∣qn-1 or p∣q2t-1 for some integer 1≤t≤n-1.Letp∣qn-1. Then qn(n-1)∣r<p with r prime, and so qn(n-1)<pn-1. Thus n=1, a contradiction.Letp∣q2t-1 for some integer 1≤t≤n-1. Then qn(n-1)∣r<p with r prime, and so qn(n-1)-1≤q2t-1<q2n-1. It follows that n=1,2,3, a contradiction.(3)
H/K≅2An(q) with n≥2.In this situation,(25)1(n+1,q+1)q(1/2)n(n+1)∏i=1n‍(qi+1-(-1)i+1)∣(p+3)!.Since the Sylowp-subgroup of G is of order p and n≥2, we obtain that p∣q(1/2)n(n+1) or p∣qt+1-(-1)t+1 for some integer 1≤t≤n.If the former, then we have thatq=p and n=1, a contradiction.Letp∣qt+1-(-1)t+1 for some integer 1≤t≤n. Then q(1/2)n(n+1)∣rm for some m. If m=1, then q=r and n=1, a contradiction. It follows that q(1/2)n(n+1)∣rm for some m>1 and r∣q. By Lemma 5, r≤[p/2]<p/2 and n(n+1)/2≤m≤p/2.Ift is odd, then p∣rt+1-1. By Lemmas 8 and 9, q=2, t=5, and r=2. Hence p=3 or 7. If p=3, we can rule out this case since p≥5. If p=7, then n(n+1)≤3 and so n=1,2, a contradiction, since t=5<n.Ift is even, then p∣rt+1+1. So p∣r+1 or p∣rt-rt-1+⋯+1. If the former, then p∣r+1<[p/2]+1, a contradiction. If the latter, then p<rt+1. It follows that p≤rt-1. Let t=2k with 1<k<[p/2]. Then p≤rk+1 or p≤rk-1 and so p∣rk+1 or p∣rk-1. If p∣rk-1, then, by Lemma 8, r=2 and t=5; we also can rule out this case as above. So p∣rk+1. It follows that p∣rt-1. Similarly, as p∣rk-1, we can rule out this case.(4)
H/K≅E8(q).Therefore we have that(26)q120(q30-1)(q24-1)(q20-1)(q18-1)×(q14-1)(q12-1)(q8-1)(q2-1)∣(p+3)!.It follows that(27)p∣q120(q30-1)(q24-1)(q20-1)(q18-1)(q14-1)×(q12-1)(q8-1)(q2-1).Ifp∣q120, then we can rule out this case since the Sylow p-subgroup of G is of order p. Hence p∣qt-1, where t∈{2,8,12,14,18,20,24,30}. On the other hand, rm∣q120. If m=1, then q=r and 1>120, a contradiction. If m>1, then q=r and m≤120. By Lemma 5, 120≤p and so p∈{5,7,11,13,…,101,103,107,109,113}. It is easy to rule out this case by considering the orders of G. Similarly, we can exclude that H/K≅E6(q),E7(q), and F4(q).(5)
H/K≅G2(q).Then we haveq6(q6-1)(q2-1)∣(p+3)!. It follows that q6∣p, p∣q6-1, or p∣q2-1.Ifq6∣p, we rule out this case.Ifp∣p6-1, then there exists a prime r such that q6∣rm for some integer m. Therefore q=r and 6≤m<p by Lemma 5. It follows that p=5 and so we have a contradiction by considering the order of G. Similarly, we also can rule out this case “p∣q2-1”.(6)
H/K≅2E6(q).It is easy to see that(1/(3,q+1))q36(q12-1)(q9+1)(q8-1)(q6-1)(q5+1)(q2-1)∣(p+3)!. It follows that p∣qt-1 with t=12,8,6,2, p∣qk+1 with k=9,5, or p∣q36. If p∣q36, then we rule out this case since the Sylow p-subgroup of G is of order p. So p∣qt-1 with t=12,8,6,2, p∣qk+1 with k=9,5, and so there exists a prime r such that q36∣rm for some integer m. It means that 36≤m≤p. Therefore p=31,29,23,19,17,13,11,7. We also can rule out this case by order consideration.(7)
H/K≅2B2(q) with q=22m+1.It follows thatq2(q2+1)(q-1)∣p!. Thus q2∣p, p∣q2+1, or p∣q-1.Ifq2∣p, then we rule out this case.Ifp∣q2+1, then there is a prime r such that q2∣rm and so 2≤m≤p by Lemma 5. Hence p=2, a contradiction. Similarly we can rule out this case “p∣q+1”.SimilarlyH/K≇2F4(22m+1).(8)
H/K≅2G2(q), q=32n+1 with n≥1.We see thatq3(q3+1)(q-1)∣p!. Since the Sylow p-subgroup of G is of order p, then p∤q3. It follows that p∣q3+1 or p∣q-1. If p∣q3+1, then there exists a prime r such that q3∣rm for some integer m. If m=1, then 1>3, a contradiction. Hence 3≤m≤p by Lemma 5, and so p=3, a contradiction. If p∣q-1 and r∣q, then there exists a Frobenius group of r·p with a kernel of order r and a complement of order p, respectively, and so there is an element of order r·p, which contradicts Lemma 14.(9)
H/K≅3D4(q).We have thatq12(q8+q4+1)(q6-1)(q2-1)∣p!. In this case, since G has a Sylow p-subgroup of order p, then p∣q8+q4+1, q∣q6-1, or p∣q2-1. If p∣q8+q4+1, then there exists a prime r such that rm∣q12 and so m≤12. By Lemma 5, p≤12. It follows that p=5,7,11. Order consideration rules out these cases “p=5,7,11”. Similarly we can rule out this case “p∣q2-1”.(10)
H/K≅An(q) with n≥1.It is easy to get that(28)1(n+1,q-1)qn(n+1)/2∏i=1n‍(qi+1-1)∣p!.It follows thatp∣qn(n+1)/2 or p∣∏i=1n‍(qi+1-1). If p∣qn(n+1)/2, then n=1 since the Sylow p-subgroup of G is of order p, a contradiction. Hence p∣∏i=1n‍(qi+1-1) and so p∣qt+1-1 for some integer 1≤t≤n. It follows that there exists a prime r such that qn(n+1)/2∣rm and so n(n+1)/2≤m≤p/2 by Lemma 5. Since the Sylow p-subgroup of G is of order p, then n(n+1)∣p and so n=1, a contradiction.This completes the proof of the lemma.Lemma 17.
Consider the following.
G
≅
A
p
+
3.Proof.
By Lemma16,
(29)Ap+3≤G¯≤Aut(Ap+3)≅Sp+3.
If G¯≅Sp+3, then there exists an element x¯ of G¯ with
(30)x¯G¯=(p+3)!3p
which contradicts Lemma 12.
SoG¯≅Ap+3. Then we define the normal series 1≤K≤G into the chief ones. We prove that K=1. By Lemma 15, π(K)⊆{2,3}.
IfK is a 2-group, in this case, let |x¯|=p. Then
(31)(p+3)!6p∣|x¯G¯|.
By Lemma 12,
(32)|xG|=|x¯G¯|=(p+3)!6p.
So x centralizes K. It follows that there is an element of 2·p which contradicts Lemma 12 (4).
IfK is a 3-group. Then similarly as the case “K is a 2-group”, we have that |xG|=|x¯G¯| is maximal in N(G) and CG(x) is abelian. So by Lemma 1.12 of [13], K≤Z(G)=1.
ThereforeK=1 and G≅Ap+3.
This completes the proof of the lemma and also of the main theorem.


## 4. Some Applications and Problem

Y. Chen and G. Chen in [23] proved that the group A10 can be characterized by its order and two special conjugacy classes sizes. Then, obviously, we also have the following result.Corollary 18.
LetG be a finite group with trivial center. Assume that N(G)=N(Ap+3) and |G|=|Ap+3|. Then G≅Ap+3.One knows that the alternating groupsAn with n=10,16,22,26 are characterized by N(G). Then by [4, 5, 15, 21], one has the following.Corollary 19.
LetG be a finite group with trivial center. Assume that N(G)=N(An) with n=p,p+1,p+2,p+3. Then G≅An.



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*Source: 752598-2014-07-22.xml*