Research Article On Thompson s Conjecture for Alternating Groups · 2019. 7. 31. · Research...

Research ArticleOn Thompson’s Conjecture for Alternating Groups 𝐴𝑝+3

Shitian Liu and Yong Yang

School of Science, Sichuan University of Science and Engineering, Zigong, Sichuan 643000, China

Correspondence should be addressed to Shitian Liu; [email protected]

Received 14 May 2014; Revised 25 June 2014; Accepted 10 July 2014; Published 22 July 2014

Academic Editor: Yijia Tan

Copyright © 2014 S. Liu and Y. Yang. This is an open access article distributed under the Creative Commons Attribution License,which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Let 𝐺 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 twoprimes 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 connectedcomponents of the prime graph 𝐺𝐾(𝐺). Denote by𝑁(𝐺) the set of nonidentity orders of conjugacy classes of elements in 𝐺. Alaviand Daneshkhah proved that the groups, 𝐴

𝑛where 𝑛 = 𝑝, 𝑝 + 1, 𝑝 + 2 with 𝑠(𝐺) ≥ 2, are characterized by𝑁(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

.

1. Introduction

In this paper, all groups considered are finite and simplegroups are nonabelian simple groups. Let 𝐺 be a finite groupand let𝑍(𝐺) be its center. For any 1 ̸= 𝑥 ∈ 𝐺, suppose that 𝑥𝐺denotes the conjugacy classes in 𝐺 containing 𝑥 and 𝐶

𝐺(𝑥)

denotes the centralizer of 𝑥 in 𝐺. We will use𝑁(𝐺) to denotethe set {𝑛 : 𝐺 has a conjugacy class of size 𝑛}. Thompson in1987 gave the following conjecture with respect to𝑁(𝐺).

Thompson’s Conjecture (see [1], Question 12.38). If 𝐿 is a finitesimple nonabelian group, 𝐺 is a finite group with trivialcenter, and𝑁(𝐺) = 𝑁(𝐿), then 𝐺 ≅ 𝐿.

Let 𝜋(𝐺) denote the set of all prime divisors of |𝐺|. Let𝐺𝐾(𝐺) be the graphwith vertex set𝜋(𝐺) such that two primes𝑝 and 𝑞 in 𝜋(𝐺) are joined by an edge if 𝐺 has an element oforder 𝑝 ⋅ 𝑞. We set 𝑠(𝐺) to denote the number of connectedcomponents of the prime graph 𝐺𝐾(𝐺). A classificationof all finite simple groups with disconnected prime graphwas obtained in [2, 3]. Based on these results, Thompson’sconjecture was proved valid for many finite simple groupswith 𝑠(𝐺) ≥ 2 (see [4, 5]). In Ahanjideh and Xu’s works, thegroups 𝐿

𝑛(𝑞), 𝐷

𝑛(𝑞), 𝐵

𝑛, 𝐶𝑛, 2𝐷𝑁(𝑞), and 𝐸

7(𝑞) are true for

Thompson’s conjecture (see [6–10]). Alavi and Daneshkhahproved that the groups 𝐴

𝑛with 𝑛 = 𝑝, 𝑝 + 1, 𝑝 + 2 and

𝑠(𝐺) ≥ 2 are characterized by 𝑁(𝐺) (see [11]). So is therea group with connected prime graph for which Thompson’s

conjecture would be true? Recently, the groups𝐴10,𝐴16,𝐴22,

and𝐴26were proved valid for this conjecture (see [12–15]). As

a development of these topics, we will prove thatThompson’sconjecture is true for the alternating groups 𝐴

𝑝+3of degree

𝑝 + 3 with 𝑝 + 2 composite.We will introduce some notations used in the proof of the

main theorem. Let 𝐺 be a group and 𝑝 a prime. Then denoteby 𝐺𝑝the Sylow 𝑝-subgroup of 𝐺. Let Aut(𝐺) and Out(𝐺)

denote the automorphism and outer automorphism groupsof 𝐺, respectively. Let 𝜔(𝐺) denote the set of element ordersof𝐺. The other notations are standard (see [16], for instance).

2. Preliminary Results

In this section we will give some preliminary results.

Lemma 1. Let 𝑥, 𝑦 ∈ 𝐺, (|𝑥|, |𝑦|) = 1, and 𝑥𝑦 = 𝑦𝑥. Then

(1) 𝐶𝐺(𝑥𝑦) = 𝐶

𝐺(𝑥) ∩ 𝐶

𝐺(𝑦);

(2) |𝑥𝐺| divides |(𝑥𝑦)𝐺|;

(3) if |𝑥𝐺| = |(𝑥𝑦)𝐺|, then 𝐶𝐺(𝑥) ≤ 𝐶

𝐺(𝑦).

Proof. See [12, Lemma 1.2] and [7, Lemma 2.3].

Lemma 2. If 𝑃 and𝐻 are finite groups with trivial centers and𝑁(𝑃) = 𝑁(𝐻), then 𝜋(𝑃) = 𝜋(𝐻).

Hindawi Publishing Corporatione Scientific World JournalVolume 2014, Article ID 752598, 10 pageshttp://dx.doi.org/10.1155/2014/752598

2 The Scientific World Journal

Proof. See [12, Lemma 3].

Lemma 3. Suppose that 𝐺 is a finite group with trivial centerand𝑝 is a prime from𝜋(𝐺) such that𝑝2 does not divide |𝑥𝐺| forall 𝑥 in𝐺.Then a Sylow 𝑝-subgroup of𝐺 is elementary abelian.


Lemma 4. Let 𝐾 be a normal subgroup of 𝐺, and 𝐺 = 𝐺/𝐾.

(1) If 𝑥 is the image of an element 𝑥 of 𝐺 in 𝐺, then |𝑥𝐺|divides |𝑥𝐺|.

(2) If (|𝑥|, |𝐾|) = 1, then 𝐶𝐺(𝑥) = 𝐶

𝐺(𝑥)𝐾/𝐾.

(3) If 𝑦 ∈ 𝐾, then |𝑦𝐾| divides |𝑦𝐺|.


Let exp(𝑛, 𝑟) denote the nonnegative integer 𝑎 such that𝑟𝑎

| 𝑛 but 𝑟𝑎+1 ∤ 𝑛.

Lemma 5. Let 𝐴𝑝+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)/2

𝑖

] − 1; in particular,exp(|𝐴

𝑝+3|, 2) ≤ 𝑝 + 2.

(2) exp(|𝐴𝑝+3|, 𝑟) = ∑

∞

𝑖=1[(𝑝 + 3)/𝑟

𝑖

] for each 𝑟 ∈𝜋(𝐴𝑝+3)\{2}. Furthermore, exp(|𝐴

𝑝+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, wehave that

exp (𝐴𝑝+3, 2) =

∞

∑

𝑖=1

[𝑝 + 3

2𝑖] − 1

= ([𝑝 + 3

2] + [

𝑝 + 3

22] + [

𝑝 + 3

23] + ⋅ ⋅ ⋅ ) − 1

≤ (𝑝 + 3

2+𝑝 + 4

22+𝑝 + 3

23+ ⋅ ⋅ ⋅ ) − 1

= (𝑝 + 3) (1

2+1

22+1

23+ ⋅ ⋅ ⋅ ) − 1

= 𝑝 + 2.

(1)

(2) Similarly, as (1), we have that

exp (𝐴𝑝+3, 𝑟)

≤ (𝑝 + 3) (1

𝑟+1

𝑟2+1

𝑟3+ ⋅ ⋅ ⋅ )

=𝑝 + 3

𝑟 − 1≤𝑝 + 3

2

(2)

for an odd prime 𝑟 ∈ 𝜋(𝐴𝑝+3).

If 𝑟 > [(𝑝 + 3)/2], exp(|𝐴𝑝+3|, 𝑟) = 1.

The proof is completed.

Let 𝑆𝑛be the symmetric group of degree 𝑛. Assume that

the cycle has 𝑐11-cycles, 𝑐

22-cycles, and so on, up to 𝑐

𝑘𝑘-

cycles, where 1𝑐1+ 2𝑐2+ ⋅ ⋅ ⋅ + 𝑘𝑐

𝑘= 𝑛. Then the number of

conjugacy classes in 𝑆𝑛is

𝑧 = 𝑛!(

𝑘

∏

𝑖=1

𝑖𝑐𝑖

𝑘

∏

𝑖=1

𝑐𝑖!)

−1

. (3)

Let 𝐴𝑛be the alternating group of degree 𝑛.

Lemma 6. Let 𝑥 ∈ 𝐴𝑛. Then, for the size of the conjugacy class

𝑥𝐺 of 𝑥 in 𝐴

𝑛, one has the following.

(1) If, for all even 𝑖, 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. If 𝑛 ≥ 6 is a natural number, then there are at least𝑠(𝑛) prime numbers 𝑝

𝑖such that (𝑛 + 1)/2 < 𝑝

𝑖< 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 aprime 𝑝 such that (𝑛 + 1)/2 < 𝑝 < 𝑛− 1, and, for every naturalnumber 𝑛 > 3, there exists an odd prime number 𝑝 such that𝑛 − 𝑝 < 𝑝 < 𝑛.

Proof. See Lemma 1 of [18].

Lemma 8. Let 𝑎, 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 theequation

𝑝𝑚

− 2𝑞𝑛

= ±1; 𝑝, 𝑞 prime; 𝑚, 𝑛 > 1, (4)

has exponents𝑚 = 𝑛 = 2; that is, it comes from a unit𝑝−𝑞⋅21/2

of the quadratic field 𝑄(21/2) for which the coefficients 𝑝 and 𝑞are primes.

The Scientific World Journal 3

Proof. See [20, 21].

Let 𝐿 be a nonabelian simple group and let O denote theorder of the outer automorphism group of 𝐿.

Lemma 10. Let 𝐿 be a nonabelian simple group. Then theorders and their outer automorphism of 𝐿 are as listed in Tables1, 2, and 3.

Proof. See [22].

3. Main Theorem and Its Proof

In this section, we give the main theorem and its proof.

Theorem 11. Let 𝐺 be a finite group with trivial center and𝑁(𝐺) = 𝑁(𝐴

𝑝+3) with 𝑝 + 2 composite. Then 𝐺 is isomorphic

to 𝐴𝑝+3

.

Proof. We know that 𝐴𝑝+3

are characterized by 𝑁(𝐺) if 𝑝 =7, 13, 19 (see [12–14]).Then in the following we only considerwhen 𝑝 ≥ 23.

We divide the proof into the following lemmas.

Lemma 12. Let 𝐿 = 𝐴𝑝+3

. Then the following hold.

(1) If 2 ̸= 𝑟 ≤ [(𝑝+3)/2], then we can write 𝑝+3 = 𝑘𝑟+𝑚with 0 ≤ 𝑚 < 𝑟 and conjugacy class sizes of 𝑟-elementsof 𝐿 are

(𝑝 + 3)!

(𝑝 + 3 − 𝑖𝑟)! ⋅ 𝑟𝑖 ⋅ 𝑖!(5)

for possible 𝑖 with 1 ≤ 𝑖 ≤ 𝑘 = [(𝑝 + 3)/𝑟].In particular, if 𝑟 is an odd prime divisor of |𝐺|, thenconjugacy class sizes of 𝑟-element of 𝐿 are

(𝑝 + 3)!

2 ⋅ 𝑘! ⋅ 𝑟2, (6)

where 𝑝 + 3 = 2𝑟 + 𝑘 and 0 ≤ 𝑘 < 𝑟.(2) If 𝑟 = 2, then one can write 𝑝 + 3 = 2𝑘 + 𝑚 with0 ≤ 𝑚 ≤ 1 and conjugacy class sizes of 2-elements of 𝐿are

(𝑝 + 3)!

(𝑝 + 3 − 2𝑖)! ⋅ 22𝑖 ⋅ (2𝑖)!(7)

for possible 𝑖 with 1 ≤ 𝑖 ≤ 𝑘 = [(𝑝 + 3)/2].(3) If 𝑟 > [(𝑝 + 3)/2], then one can write 𝑝 + 3 = 𝑟 + 𝑚

with 0 ≤ 𝑚 < 𝑟 and conjugacy class sizes of 𝑟-elementsof 𝐿 are

(𝑝 + 3)!

(𝑝 + 3 − 𝑟)! ⋅ 𝑟. (8)

In particular, if 𝑟 = 𝑝, then the conjugacy class size of𝑝-elements of 𝐿 is

(𝑝 + 3)!

6𝑝. (9)

(4) The following numbers from𝑁(𝐺) aremaximality withrespect to divisibility.

(a) One of the following holds:(𝑝 + 3)!/2𝑟

2 if 2 ⋅ 𝑟 = 𝑝 + 3;(𝑝 + 3)!/4𝑟

2 if 2 ⋅ 𝑟 + 2 = 𝑝 + 3;(𝑝 + 3)!/2(𝑘 − 1)𝑟

2 if 2 ⋅ 𝑟 + 𝑘 = 𝑝 + 3 and 𝑘 = 2𝑛with 𝑛 ≥ 2;

(b) (𝑝 + 3)!/6𝑝.

Proof. From (3) and Lemma 6, we get the desired results.

Lemma 13. Let 𝐺 be a finite group with trivial center and𝑁(𝐺) = 𝑁(𝐿). Then |𝐿| | |𝐺| and 𝜋(𝐺) = 𝜋(𝐿).

Proof. Note that |𝐿| = ∏𝑛∈𝑁(𝐿)

𝑛. Since |𝑥𝐺||𝐶𝐺(𝑥)| = |𝐺|,

everymember from𝑁(𝐺) divides the order of𝐺 and |𝐿| | |𝐺|.So by Lemma 12, we have that 𝜋(𝐺) = 𝜋(𝐿).

Lemma 14. Suppose that 𝐺 is a finite group with trivial centerand𝑁(𝐺) = 𝑁(𝐿). Then the following hold.

(1) There exist different primes 𝑟1, 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 where𝑟 ∈ {𝑟

1, 𝑟2, 𝑝}. There does not exist an element of order

𝑟1⋅ 𝑟2, 𝑟1⋅ 𝑝, or 𝑟

2⋅ 𝑝.

(2) For all 𝑛 ∈ 𝑁(𝐺), if 𝑛 is divisible at most by 𝑟𝑎, then theSylow 𝑟-subgroup 𝑆 of 𝐺 is of order 𝑟𝑎.

Proof. (1) By Lemma 7, there exist different prime numbers𝑟1, 𝑟2, 𝑝 from 𝜋(𝐺) such that 𝑟

1, 𝑟2, 𝑝 > [(𝑝 + 3)/2].

From Lemmas 12 and 13, we have that the primes 𝑟1, 𝑟2, 𝑝

are prime divisors of |𝐺| and 𝑟21, 𝑟2

2, 𝑝2 donot divide |𝑥𝐺| for all

𝑥 ∈ 𝐺. Then by Lemma 3, 𝑆 is elementary abelian. Thereforeif |𝑥| = 𝑟, then |𝑥𝐺| is an 𝑟-number.

Let |𝑆| ≥ 𝑝2. Consider an element 𝑦 of 𝐺 with

𝑦𝐺=(𝑝 + 3)!

4 ⋅ 𝑟2if 2𝑟 + 2 = 𝑝 + 3,

𝑦𝐺=(𝑝 + 3)!

2 ⋅ 𝑟2if 2𝑟 = 𝑝 + 3,

(10)

or

𝑦𝐺=

(𝑝 + 3)!

2 (𝑘 − 1) 𝑟2if 2𝑟 + 𝑘 = 𝑝 + 3,

𝑘 = 2𝑛 with 𝑛 ≥ 2(11)

by Lemma 12.Assume that 𝑝 ∤ |𝑦|. Let 𝑥 be an element of 𝐶

𝐺(𝑦) having

order 𝑝. Then 𝐶𝐺(𝑥𝑦) = 𝐶

𝐺(𝑥) ∩ 𝐶

𝐺(𝑦), |𝑥𝐺| | |(𝑥𝑦)𝐺|, and


Table 1: The simple classical groups.

𝐿 Lie; rankL 𝑑 𝑂 |𝐿|

𝐿𝑛(𝑞) 𝐴

𝑛−1(𝑞) (𝑛, 𝑞 − 1) 2𝑑𝑓, if 𝑛 ≥ 3; 1

𝑑𝑞𝑛(𝑛−1)/2

𝑛

∏

𝑖=2

(𝑞𝑖

− 1)

𝑛 − 1 𝑑𝑓, if 𝑛 = 2

𝑈𝑛(𝑞)

2

𝐴𝑛−1(𝑞) (𝑛, 𝑞 + 1) 2𝑑𝑓, if 𝑛 ≥ 3 1

𝑑𝑞𝑛(𝑛−1)/2

𝑛

∏

𝑖=2

(𝑞𝑖

− (−1)𝑖

)

[𝑛/2] 𝑑𝑓, if 𝑛 = 2

𝑃𝑆𝑝2𝑚(𝑞) 𝐶

𝑚(𝑞) (2, 𝑞 − 1) 𝑑𝑓,𝑚 ≥ 3; 1

𝑑𝑞𝑚2𝑚

∏

𝑖=1

(𝑞2𝑖

− 1)

𝑚 2𝑓, if𝑚 = 2

Ω2𝑚+1

(𝑞) 𝐵𝑚(𝑞) 2 2𝑓 1

2𝑞𝑚2𝑚

∏

𝑖=1

(𝑞2𝑖

− 1)

𝑞 odd 𝑚𝑃Ω+

2𝑚(𝑞) 𝐷

𝑚(𝑞) (4, 𝑞

𝑚

− 1) 2𝑑𝑓, if𝑚 ̸= 4 1𝑑𝑞𝑚(𝑚−1)(𝑞

𝑚−1)∏𝑚−1

𝑖=1(𝑞2𝑖−1)

𝑚 ≥ 3 𝑚 6𝑑𝑓, if𝑚 = 4𝑃Ω−

2𝑚(𝑞)

2

𝐷𝑚(𝑞) (4, 𝑞

𝑚

+ 1) 2𝑑𝑓1

𝑑𝑞𝑚(𝑚−1)(𝑞

𝑚+1)∏𝑚−1

𝑖=1 (𝑞2𝑖

− 1)

𝑚 ≥ 2 𝑚 − 1

Table 2: The simple exceptional group.

𝐿 L 𝑑 𝑂 |𝐿|𝐺2(𝑞) 2 1 𝑓, if 𝑝 ̸= 3 𝑞6 (𝑞2 − 1) (𝑞6 − 1)

2𝑓, if 𝑝 = 3𝐹4(𝑞) 4 1 (2, 𝑝)𝑓 𝑞24 (𝑞2 − 1) (𝑞6 − 1) (𝑞8 − 1) (𝑞12 − 1)

𝐸6(𝑞) 6 (3, 𝑞 − 1) 2𝑑𝑓 1

𝑑𝑞36

∏

𝑖∈{2,5,6,8,9,12}

(𝑞𝑖

− 1)

𝐸7(𝑞) 7 (2, 𝑞 − 1) 𝑑𝑓 1

𝑑𝑞63

∏

𝑖∈{2,6,8,10,12,14,18}

(𝑞𝑖

− 1)

𝐸8(𝑞) 8 1 𝑓 𝑞120 ∏

𝑖∈{2,8,12,14,18,20,24,30}

(𝑞𝑖

− 1)

2

𝐵2(𝑞), 𝑞 = 2

2𝑚+1 1 1 𝑓 𝑞2 (𝑞2 + 1) (𝑞 − 1)2

𝐺2(𝑞), 𝑞 = 3

2𝑚+1 1 1 𝑓 𝑞3 (𝑞3 + 1) (𝑞 − 1)2

𝐹4(𝑞), 𝑞 = 2

2𝑚+1 2 1 𝑓 𝑞12 (𝑞6 + 1) (𝑞4 − 1) (𝑞3 + 1) (𝑞 − 1)3

𝐷4(𝑞) 2 1 3𝑓 𝑞12 (𝑞8 + 𝑞4 + 1) (𝑞6 − 1) (𝑞2 − 1)

2

𝐸6(𝑞) 4 (3, 𝑞 + 1) 2𝑑𝑓 1

𝑑𝑞36

∏

𝑖∈{2,5,6,8,9,12}

(𝑞𝑖

− (−1)𝑖

)

|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|. Onthe other hand, |𝑦𝐺| | |(𝑥𝑦)𝐺|; then we have that

(𝑝 + 3)!

4 ⋅ 𝑟2|(𝑥𝑦)𝐺

if 2𝑟 + 2 = 𝑝 + 3,

(𝑝 + 3)!

2 ⋅ 𝑟2|(𝑥𝑦)𝐺

if 2𝑟 = 𝑝 + 3,

(12)

or(𝑝 + 3)!

2 (𝑘 − 1) 𝑟2|(𝑥𝑦)𝐺

if 2𝑟 + 𝑘 = 𝑝 + 3,

𝑘 = 2𝑛 with 𝑛 ≥ 2.(13)

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. |𝑦𝐺| = (𝑝 + 3)!/4 ⋅ 𝑟2 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. |𝑦𝐺| = (𝑝 + 3)!/2(𝑘 − 1)𝑟2 if 2𝑟 + 𝑘 = 𝑝 + 3 and 𝑘 = 2𝑛with 𝑛 ≥ 2.


Table 3: The simple sporadic groups.

𝐿 𝑑 𝑂 |𝐿|

𝑀11

1 1 24 ⋅ 32 ⋅ 5 ⋅ 11𝑀12

2 2 26 ⋅ 33 ⋅ 5 ⋅ 11𝑀22

12 2 27 ⋅ 32 ⋅ 5 ⋅ 7 ⋅ 11𝑀23

1 1 27 ⋅ 32 ⋅ 5 ⋅ 7 ⋅ 11 ⋅ 23𝑀24

1 1 210 ⋅ 33 ⋅ 5 ⋅ 7 ⋅ 11 ⋅ 23𝐽1

1 1 23 ⋅ 3 ⋅ 5 ⋅ 7 ⋅ 11 ⋅ 19𝐽2

2 2 27 ⋅ 33 ⋅ 52 ⋅ 7𝐽3

3 2 27 ⋅ 35 ⋅ 5 ⋅ 17 ⋅ 19𝐽4

1 1 221 ⋅ 33 ⋅ 5 ⋅ 7 ⋅ 113 ⋅ 23 ⋅ 29 ⋅ 31 ⋅ 37 ⋅ 43𝐻𝑆 2 2 29 ⋅ 32 ⋅ 53 ⋅ 7 ⋅ 11𝑆𝑢𝑧 6 2 213 ⋅ 37 ⋅ 52 ⋅ 7 ⋅ 11 ⋅ 13𝑀𝑐𝐿 3 2 27 ⋅ 36 ⋅ 53 ⋅ 7 ⋅ 11𝑅𝑢 2 1 214 ⋅ 33 ⋅ 53 ⋅ 7 ⋅ 13 ⋅ 29𝐻𝑒(𝐹7) 1 2 210 ⋅ 33 ⋅ 52 ⋅ 73 ⋅ 17

𝐿𝑦 1 1 28 ⋅ 37 ⋅ 56 ⋅ 7 ⋅ 11 ⋅ 31 ⋅ 37 ⋅ 67𝑂𝑁 3 2 29 ⋅ 34 ⋅ 5 ⋅ 73 ⋅ 11 ⋅ 19 ⋅ 31𝐶𝑜1

2 1 221 ⋅ 39 ⋅ 54 ⋅ 72 ⋅ 11 ⋅ 13 ⋅ 23𝐶𝑜2

1 1 218 ⋅ 36 ⋅ 53 ⋅ 7 ⋅ 11 ⋅ 23𝐶𝑜3

1 1 210 ⋅ 37 ⋅ 53 ⋅ 7 ⋅ 11 ⋅ 23𝐹𝑖22

6 2 217 ⋅ 39 ⋅ 52 ⋅ 7 ⋅ 11 ⋅ 13𝐹𝑖23

1 1 218 ⋅ 313 ⋅ 52 ⋅ 7 ⋅ 11 ⋅ 13 ⋅ 17 ⋅ 23𝐹𝑖

243 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

In this case, 𝑟 ∤ (𝑝 + 1)(𝑝 + 2)(𝑝 + 3)/3. It follows that|𝑦𝐺

| = |(𝑥𝑦)𝐺

|. By Lemma 1, 𝐶𝐺(𝑦) ≤ 𝐶

𝐺(𝑥) and so |𝑥𝐺| |

|𝑦𝐺

|, a contradiction.Assume that 𝑝 | |𝑦|. Let |𝑦| = 𝑝 ⋅ 𝑡. Since 𝑆 is elementary

abelian, the numbers 𝑝 and 𝑡 are coprime. Let

𝑢 = 𝑦𝑝

, V = 𝑦𝑡. (14)

Then 𝑦 = 𝑢V and 𝐶𝐺(𝑢V) = 𝐶

𝐺(𝑢) ∩ 𝐶

𝐺(V). Therefore,

V𝐺|𝑦𝐺=(𝑝 + 3)!

2 ⋅ 𝑟2if 2𝑟 = 𝑝 + 3,

V𝐺|𝑦𝐺=(𝑝 + 3)!

2 ⋅ 𝑟2if 2𝑟 + 2 = 𝑝 + 3,

(15)

or

V𝐺|𝑦𝐺=

(𝑝 + 3)!

2 (𝑘 − 1) ⋅ 𝑟2if 2𝑟 + 𝑘 = 𝑝 + 3,

𝑘 = 2𝑛, with 𝑛 ≥ 2.(16)

On the other hand, the element V of 𝐺 is of order 𝑝. Sincethe Sylow 𝑝-subgroup of 𝐺 is elementary abelian, then 𝑝 ∤|V𝐺|. It follows that

V𝐺=(𝑝 + 3)!

6𝑝or(𝑝 + 1) (𝑝 + 2) (𝑝 + 3)

3(17)

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 followingthree cases.

Case 1. |𝑦𝐺| = (𝑝 + 3)!/2 ⋅ 𝑟2 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 themaximality of |𝑦𝐺| = (𝑝 + 3)!/2 ⋅ 𝑟2.

Case 2. |𝑦𝐺| = (𝑝 + 3)!/4 ⋅ 𝑟2 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 themaximality of |𝑦𝐺| = (𝑝 + 3)!/4 ⋅ 𝑟2.

Case 3. |𝑦𝐺| = (𝑝 + 3)!/2(𝑘 − 1)𝑟2 if 2𝑟 + 𝑘 = 𝑝 + 3 and 𝑘 = 2𝑛with 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 Sylow 𝑝-subgroup of 𝐺 is of order 𝑝.Similarly we can prove the other two cases.There does not exist an element of order 𝑟

1⋅ 𝑟2, 𝑟1⋅ 𝑝, or

𝑟2⋅ 𝑝.(2)Without loss of generality, we assume that 𝑛 is divisible

at most by 𝑟2.Assume that |𝑆| ≥ 𝑟3. Consider an element 𝑥 of 𝐺 such

that

𝑥𝐺=

{{{{{{{{{{

{{{{{{{{{{

{

(𝑝 + 3)!

2𝑟2if 2𝑟 = 𝑝 + 3;

(𝑝 + 3)!

4𝑟2if 2𝑟 + 2 = 𝑝 + 3;

(𝑝 + 3)!

2 (𝑘 − 1) 𝑟2if 2𝑟 + 𝑘 = 𝑝 + 3,

with 𝑘 = 2𝑛, 𝑛 ≥ 2,

(18)

by Lemma 12.Let 𝑟 ∤ |𝑥|. Then there is an element 𝑦 of 𝐺 of order 𝑟.

By Lemma 1 we have that 𝐶𝐺(𝑥𝑦) = 𝐶

𝐺(𝑥) ∩ 𝐶

𝐺(𝑦), |𝑥𝐺| |

|(𝑥𝑦)𝐺

|, and |𝑦𝐺| | |(𝑥𝑦)𝐺|.If 𝑦 is an 𝑟-central element, then 𝑆 ≤ 𝐶

𝐺(𝑦) and |𝑦𝐺| is an

𝑟-number. We consider the following three cases.

Case 1. 2𝑟 = 𝑝 + 3.In this case |𝑦𝐺| = (𝑝 + 3)!/2 ⋅ 𝑟2. Then we have that

|𝑥𝐺

| = |𝑦𝐺

| = |(𝑥𝑦)𝐺

| and so 𝑟2 || |𝐶𝐺(𝑥)| = |𝐶

𝐺(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)!/𝑟

3

⋅ 3! contradicting 2𝑟 = 𝑝 + 3). Thus 𝑟 | |𝑥𝐺| = |𝑦𝐺|since 𝑟3 | |𝐺|, a contradiction.

Case 2. 2𝑟 + 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. 2𝑟 + 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 fromthe 𝑟-number in 𝑁(𝐺)). If the former, then |𝑥𝐺| = |𝑦𝐺| =|(𝑥𝑦)𝐺

| and so 𝑟2 || |𝐶𝐺(𝑥)| = |𝐶

𝐺(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| since𝑟3

| |𝐺|, a contradiction.Then |𝑦𝐺| = (2𝑟 + 1)(2𝑟 + 2) ⋅ ⋅ ⋅ (2𝑟 +𝑘)/2(𝑘−2). But there is nomember from𝑁(𝐺) such that |𝑥𝐺| ||(𝑥𝑦)𝐺

| and |𝑦𝐺| | |(𝑥𝑦)𝐺|.If 𝑦 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 an𝑟-element 𝑤 such that 1 ̸= 𝑤 ∈ 𝐶

𝐺(𝑧), 𝐶

𝐺(𝑤𝑧) < 𝐶

𝐺(𝑤), and

𝑟 |

𝐶𝐺 (𝑤)

𝐶𝐺 (𝑤𝑧)

=

(𝑤𝑧)𝐺

𝑤𝐺

. (19)

On the other hand, we also have 𝑧 ∈ 𝐶𝐺(𝑤). Since 𝑤𝑧 = 𝑧𝑤,

then |(𝑤𝑧)𝐺| = |𝑧𝐺| and so 𝐶𝐺(𝑧) ≤ 𝐶

𝐺(𝑤). Therefore 𝑧 ∈

𝐶𝐺(𝑧) ≤ 𝐶

𝐺(𝑤). It follows that |𝐶

𝐺(𝑧)| = |𝐶

𝐺(𝑤𝑧)| since the

maximality of |𝑧𝐺|, and we get 𝑟 | |𝐶𝐺(𝑤)|/|𝐶

𝐺(𝑤𝑧)| = 1, a

contradiction.Let 𝑟 | |𝑥|; then we write |𝑥| = 𝑟𝑡. If 𝑆 is elementary

abelian, then (𝑟, 𝑡) = 1. Set

𝑢 = 𝑥𝑟

, V = 𝑥𝑡. (20)

Then 𝑥 = 𝑢V and 𝐶𝐺(𝑥) = 𝐶

𝐺(𝑢) ∩ 𝐶

𝐺(V). Hence |V𝐺| | 𝑥𝐺

and |𝑢𝐺| | |𝑥𝐺|. Since V is an element of order 𝑟, thenV𝐺

=

{{{{{{{{{{{

{{{{{{{{{{{

{

(𝑝 + 3)!

2𝑟2if 2𝑟 = 𝑝 + 3;

(𝑝 + 3)!

4𝑟2if 2𝑟 + 2 = 𝑝 + 3;

(𝑝 + 3)!

2 (𝑘 − 1) 𝑟2, or (2𝑟 + 1) (2𝑟 + 2) ⋅ ⋅ ⋅ (2𝑟 + 𝑘)

2 (𝑘 − 2)

if 2𝑟 + 𝑘 = 𝑝 + 3with 𝑘 = 2𝑛, 𝑛 ≥ 2.

(21)

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)| = |𝐶

𝐺(𝑢V)| = |𝐶

𝐺(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 in𝑁(𝐺), which contradictsthe maximality of (𝑝 + 3)!/2(𝑘 − 1)𝑟2.

Therefore 𝑆 is nonabelian. So we chose an element 𝑧 oforder 𝑝 such that 𝑟2 | |𝑧𝐺|. By hypothesis, 𝑟 | |𝐶

𝐺(𝑧)| and,

obviously, 𝑟 ∤ |𝑧|.Then by Lemma 1.2 of [4], we also have that|𝐶𝐺(𝑧)| = |𝐶

𝐺(𝑤𝑧)| since the maximality of |𝑧𝐺|, and we get

𝑟 | |𝐶𝐺(𝑤)|/|𝐶

𝐺(𝑤𝑧)| = 1, a contradiction.

The lemma is completed.

Lemma 15. Suppose that 𝐺 is a finite group with trivial centerand𝑁(𝐺) = 𝑁(𝐿). Let 𝜋 = {2, 3}. Then 𝑂

𝜋,𝜋(𝐺) = 𝑂

𝜋(𝐺). In

particular, 𝐺 is insoluble.

Proof. Let 𝐾 = 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 aprime 𝑟 ∈ 𝜋(𝐿) \ 𝜋 with 𝑂

𝑟(𝐺) ̸= 1.

Let 𝑟 ∈ {𝑟1, 𝑟2, 𝑝} with 𝑂

𝑟(𝐺) ̸= 1. Then 𝐺 contains a Hall

{𝑟 ⋅𝑠}-subgroup of order 𝑟 ⋅𝑠with 𝑠 ∈ {𝑟1, 𝑟2, 𝑝}−{𝑟}. However,

Hall {𝑟, 𝑠}-subgroup must be cyclic contradicting Lemma 14.Let 𝑃 be a Sylow 𝑟-subgroup of 𝐺 where 𝑟 ∈ 𝜋(𝐿) \

{𝑟1, 𝑟2, 𝑝}. If 𝑂

𝑟(𝐺) ̸= 1, then 𝐴 = 𝑍(𝑂

𝑟(𝐺)) is a nontrivial


normal subgroup of 𝐺. Let 𝑥 be an element of order 𝑝 in 𝐺.So we have that |𝑥𝐺| is a divisor of

(𝑝 + 3)!

6𝑝or

(𝑝 + 1) (𝑝 + 2) (𝑝 + 3)

3. (22)

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 𝐺is of order 𝑟. Hence there is a Hall {𝑟, 𝑝}-subgroup𝐻.Since 𝐻 must be cyclic, then there is an element oforder 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+3with 𝑘 = 2𝑛 and 𝑛 ≥ 2, then the indexof 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 followsthat [𝐴, 𝑥] = 1 and 𝐴 = 𝐶

𝐴(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 aSylow 𝑝-subgroup of 𝐺. This contradicts Lemma 12.Thus 𝑂

𝑟(𝐺) = 1.

It follows that 𝑂𝑟(𝐺) = 1 for 𝑟 ∈ {5, 7, . . . , 𝑝}.

Therefore 𝑂𝜋,𝜋(𝐺) = 𝑂

𝜋(𝐺). In particular, 𝐺 is insoluble.

Lemma 16. There is a normal series 1 ≤ 𝐾 ≤ 𝐻 ≤ 𝐺 such that𝐻/𝐾 ≅ 𝐴

𝑝+3.

Proof. By Lemmas 13 and 14, |𝐺| = (𝑝 + 3)!/2.By Lemma 15, we have that𝐻/𝐾 ≤ 𝐺 ≤ Aut(𝐻/𝐾), where

𝑀 : 𝐻/𝐾 = 𝑆1×𝑆2× ⋅ ⋅ ⋅ × 𝑆

𝑘is a direct product of nonabelian

simple groups 𝑆1, 𝑆2, . . . , 𝑆

𝑘. Since 𝐺 cannot contain a Hall

{𝑟1, 𝑟2, 𝑝}-subgroup, numbers 𝑟

1, 𝑟2, and 𝑝 divide the order of

exactly one of these groups that is listed as in Lemma 10, andso we assume that they divide 𝑆

1. Since 𝑆

1⊲ 𝐺, we let 𝐺∗ and

𝑀∗ denote the factor groups 𝐺/𝑆

1and𝑀/𝑆

1, respectively. If

𝑘 > 1, then a Sylow 2-subgroup of 𝐺∗ is nontrivial and itscenter 𝑍 has a nontrivial intersection with 𝑀∗. Consider anontrivial element 𝑦 of 𝑇 = 𝑆

2× ⋅ ⋅ ⋅ × 𝑆

𝑘such 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 order𝑝, a contradiction. Thus 𝑀 = 𝑆

1, and 𝐺 is almost simple.

Therefore

𝐻

𝐾≤ 𝐺 ≤ Aut(𝐻

𝐾) . (23)

Obviously, 𝑟1, 𝑟2, 𝑝 | |𝐻/𝐾| (in fact, if 𝑟

1, 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 thefollowing, we consider 𝑆

1which is listed as in Tables 1, 2, and

3.

(i) Case 1.𝐻/𝐾 ≅ 𝐴𝑛with 𝑛 ≥ 6.

Then 𝑛 = 𝑝, 𝑝+1, . . . , 𝑝+𝑘with 𝑝+2, 𝑝+4, . . . compositeand 𝑝 + 𝑘 + 1 prime. If 𝑘 ≥ 4, then (𝑝 + 𝑘)!/2 | (𝑝 + 3)!,a contradiction. Therefore 𝐻/𝐾 is isomorphic to 𝐴

𝑝, 𝐴𝑝+1

,𝐴𝑝+2

, or 𝐴𝑝+3

.Let 𝑥 be an element of order 𝑝 in 𝐻. Then |𝑥𝐻| is 𝑝-

number since |𝐻|𝑝= 𝑝𝐻/𝐾 ≅ 𝐴

𝑝.

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 of𝑟 ⋅ 𝑝 or of order 𝑟 ⋅ 𝑝 with 3 < 𝑟 < 𝑟 < 𝑝 and 𝑟 and 𝑟 divideone of the prime divisors of the numbers 𝑝+1, 𝑝+2, or 𝑝+3,which contradicts Lemma 14. Similarly, we can rule out thesecases “𝐻/𝐾 ≅ 𝐴

𝑝+1and𝐻/𝐾 ≅ 𝐴

𝑝+2”.

Therefore𝐻/𝐾 ≅ 𝐴𝑝+3

.

(ii) Case 2.𝐻/𝐾 is not isomorphic to a sporadic simple groupaccording to Table 3.

(iii) Case 3.𝐻/𝐾 is isomorphic to a simple group of Lie type.Let 𝑞 be a prime power.

(1) 𝐻/𝐾 ≅ 𝐵𝑛(𝑞) with 𝑛 ≥ 2.

In this situation, by Lemma 13, 𝜋(𝐺) = {2, 3, 5, 7,. . . , 𝑝} and so

1

(2, 𝑞 − 1)𝑞𝑛2𝑛

∏

𝑖=1

(𝑞2𝑖

− 1) | (𝑝 + 3)!. (24)

It follows that 𝑝 | 𝑞 or 𝑝 | ∏𝑛𝑖=1(𝑞2𝑖

− 1). If 𝑝 | 𝑞, then𝑞 is a power of 𝑝. Since |𝐺

𝑝| = 𝑝 by Lemma 14, this

is impossible as 𝑛 ≥ 2. Therefore 𝑝 | ∏𝑛𝑖=1(𝑞2𝑖

− 1). Itfollows that𝑝 | 𝑞2𝑡−1 for some 1 ≤ 𝑡 ≤ 𝑛 as𝑝 is prime.On the other hand, 𝑞𝑛

2

| 𝑟 or 𝑞𝑛2

| 𝑟𝑚. If the former,

then 𝑞 = 𝑟 and 𝑛 = 1, a contradiction. It follows that𝑞𝑛2

| 𝑟𝑚. Hence 𝑟 | 𝑞 and 𝑚 = 𝑘𝑛2 for some integer

𝑘 ≥ 1. By Lemma 5, 𝑘𝑛2 < 𝑝/2 < (𝑞2𝑛 − 1)/2, butthe equation has no solution inN. Furthermore, since𝐶𝑛(𝑞) has the same order as 𝐵

𝑛(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)∏𝑛−1

𝑖=1(𝑞2𝑖

−1) | (𝑝+3)!. Since the Sylow 𝑝-subgroupof 𝐺 is of order 𝑝, 𝑝 ∤ 𝑞 as, otherwise, 𝑞 = 𝑝 and thus𝑛 = 1, a contradiction. It follows that 𝑝 | 𝑞𝑛 − 1 or𝑝 | 𝑞2𝑡

− 1 for some integer 1 ≤ 𝑡 ≤ 𝑛 − 1.

Let 𝑝 | 𝑞𝑛 − 1. Then 𝑞𝑛(𝑛−1) | 𝑟 < 𝑝 with 𝑟 prime, andso 𝑞𝑛(𝑛−1) < 𝑝𝑛 − 1. Thus 𝑛 = 1, a contradiction.

Let 𝑝 | 𝑞2𝑡 − 1 for some integer 1 ≤ 𝑡 ≤ 𝑛 − 1. Then𝑞𝑛(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,

1

(𝑛 + 1, 𝑞 + 1)𝑞(1/2)𝑛(𝑛+1)

𝑛

∏

𝑖=1

(𝑞𝑖+1

− (−1)𝑖+1

) | (𝑝 + 3)!. (25)

Since the Sylow 𝑝-subgroup of𝐺 is of order 𝑝 and 𝑛 ≥2, we obtain that 𝑝 | 𝑞(1/2)𝑛(𝑛+1) or 𝑝 | 𝑞𝑡+1 − (−1)𝑡+1for some integer 1 ≤ 𝑡 ≤ 𝑛.If the former, then we have that 𝑞 = 𝑝 and 𝑛 = 1, acontradiction.Let 𝑝 | 𝑞𝑡+1 − (−1)𝑡+1 for some integer 1 ≤ 𝑡 ≤ 𝑛. Then𝑞(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) | 𝑟𝑚for some 𝑚 > 1 and 𝑟 | 𝑞. By Lemma 5, 𝑟 ≤ [𝑝/2] <𝑝/2 and 𝑛(𝑛 + 1)/2 ≤ 𝑚 ≤ 𝑝/2.If 𝑡 is odd, then 𝑝 | 𝑟𝑡+1−1. By Lemmas 8 and 9, 𝑞 = 2,𝑡 = 5, and 𝑟 = 2. Hence𝑝 = 3 or 7. If𝑝 = 3, we can ruleout this case since 𝑝 ≥ 5. If 𝑝 = 7, then 𝑛(𝑛 + 1) ≤ 3and so 𝑛 = 1, 2, a contradiction, since 𝑡 = 5 < 𝑛.If 𝑡 is even, then 𝑝 | 𝑟𝑡+1 + 1. So 𝑝 | 𝑟 + 1 or 𝑝 |𝑟𝑡

−𝑟𝑡−1

+⋅ ⋅ ⋅+1. If the former, then𝑝 | 𝑟+1 < [𝑝/2]+1,a contradiction. If the latter, then 𝑝 < 𝑟𝑡 +1. It followsthat 𝑝 ≤ 𝑟𝑡 − 1. Let 𝑡 = 2𝑘 with 1 < 𝑘 < [𝑝/2]. Then𝑝 ≤ 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; wealso can rule out this case as above. So 𝑝 | 𝑟𝑘 + 1. Itfollows that 𝑝 | 𝑟𝑡 − 1. Similarly, as 𝑝 | 𝑟𝑘 − 1, we canrule out this case.

(4) 𝐻/𝐾 ≅ 𝐸8(𝑞).

Therefore we have that

𝑞120

(𝑞30

− 1) (𝑞24

− 1) (𝑞20

− 1) (𝑞18

− 1)

× (𝑞14

− 1) (𝑞12

− 1) (𝑞8

− 1) (𝑞2

− 1) | (𝑝 + 3)!.

(26)

It follows that

𝑝 | 𝑞120

(𝑞30

− 1) (𝑞24

− 1) (𝑞20

− 1) (𝑞18

− 1) (𝑞14

− 1)

× (𝑞12

− 1) (𝑞8

− 1) (𝑞2

− 1) .

(27)

If 𝑝 | 𝑞120, then we can rule out this case sincethe Sylow 𝑝-subgroup of 𝐺 is of order 𝑝. Hence 𝑝 |𝑞𝑡

− 1, where 𝑡 ∈ {2, 8, 12, 14, 18, 20, 24, 30}. On theother hand, 𝑟𝑚 | 𝑞120. If 𝑚 = 1, then 𝑞 = 𝑟 and1 > 120, a contradiction. If 𝑚 > 1, then 𝑞 = 𝑟and 𝑚 ≤ 120. By Lemma 5, 120 ≤ 𝑝 and so 𝑝 ∈{5, 7, 11, 13, . . . , 101, 103, 107, 109, 113}. It is easy torule out this case by considering the orders of 𝐺.Similarly, we can exclude that 𝐻/𝐾 ≅ 𝐸

6(𝑞), 𝐸7(𝑞),

and 𝐹4(𝑞).

(5) 𝐻/𝐾 ≅ 𝐺2(𝑞).

Then we have 𝑞6(𝑞6 − 1)(𝑞2 − 1) | (𝑝 + 3)!. It followsthat 𝑞6 | 𝑝, 𝑝 | 𝑞6 − 1, or 𝑝 | 𝑞2 − 1.

If 𝑞6 | 𝑝, we rule out this case.If 𝑝 | 𝑝6 − 1, then there exists a prime 𝑟 such that𝑞6

| 𝑟𝑚 for some integer 𝑚. Therefore 𝑞 = 𝑟 and 6 ≤

𝑚 < 𝑝 by Lemma 5. It follows that 𝑝 = 5 and so wehave 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 that𝑝 | 𝑞

𝑡

− 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 casesince 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 forsome integer𝑚. It means that 36 ≤ 𝑚 ≤ 𝑝. Therefore𝑝 = 31, 29, 23, 19, 17, 13, 11, 7. We also can rule outthis case by order consideration.

(7) 𝐻/𝐾≅2𝐵2(𝑞) with 𝑞 = 22𝑚+1.

It follows that 𝑞2(𝑞2 + 1)(𝑞 − 1) | 𝑝!. Thus 𝑞2 | 𝑝,𝑝 | 𝑞2

+ 1, or 𝑝 | 𝑞 − 1.If 𝑞2 | 𝑝, then we rule out this case.If 𝑝 | 𝑞2 + 1, then there is a prime 𝑟 such that 𝑞2 |𝑟𝑚 and so 2 ≤ 𝑚 ≤ 𝑝 by Lemma 5. Hence 𝑝 = 2, acontradiction. Similarly we can rule out this case “𝑝 |𝑞 + 1”.Similarly𝐻/𝐾≇2𝐹

4(22𝑚+1

).(8) 𝐻/𝐾≅2𝐺

2(𝑞), 𝑞 = 32𝑛+1 with 𝑛 ≥ 1.

We see that 𝑞3(𝑞3 + 1)(𝑞 − 1) | 𝑝!. Since the Sylow𝑝-subgroup of 𝐺 is of order 𝑝, then 𝑝 ∤ 𝑞3. It followsthat 𝑝 | 𝑞3 + 1 or 𝑝 | 𝑞 − 1. If 𝑝 | 𝑞3 + 1, then thereexists a prime 𝑟 such that 𝑞3 | 𝑟𝑚 for some integer𝑚. If𝑚 = 1, then 1 > 3, a contradiction. Hence 3 ≤ 𝑚 ≤ 𝑝by Lemma 5, and so 𝑝 = 3, a contradiction. If 𝑝 | 𝑞−1and 𝑟 | 𝑞, then there exists a Frobenius group of 𝑟 ⋅ 𝑝with 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 that 𝑞12(𝑞8 + 𝑞4 + 1)(𝑞6 − 1)(𝑞2 − 1) | 𝑝!.In this case, since 𝐺 has a Sylow 𝑝-subgroup of order𝑝, then 𝑝 | 𝑞8 + 𝑞4 + 1, 𝑞 | 𝑞6 − 1, or 𝑝 | 𝑞2 − 1.If 𝑝 | 𝑞8 + 𝑞4 + 1, then there exists a prime 𝑟 suchthat 𝑟𝑚 | 𝑞12 and so 𝑚 ≤ 12. By Lemma 5, 𝑝 ≤ 12.It follows that 𝑝 = 5, 7, 11. Order consideration rulesout these cases “𝑝 = 5, 7, 11”. Similarly we can rule outthis case “𝑝 | 𝑞2 − 1”.

(10) 𝐻/𝐾 ≅ 𝐴𝑛(𝑞) with 𝑛 ≥ 1.

It is easy to get that

1

(𝑛 + 1, 𝑞 − 1)𝑞𝑛(𝑛+1)/2

𝑛

∏

𝑖=1

(𝑞𝑖+1

− 1) | 𝑝!. (28)


It follows that 𝑝 | 𝑞𝑛(𝑛+1)/2 or 𝑝 | ∏𝑛𝑖=1(𝑞𝑖+1

− 1). If 𝑝 |𝑞𝑛(𝑛+1)/2, then 𝑛 = 1 since the Sylow 𝑝-subgroup of 𝐺is of order 𝑝, a contradiction. Hence 𝑝 | ∏𝑛

𝑖=1(𝑞𝑖+1

−1)

and so𝑝 | 𝑞𝑡+1−1 for some integer 1 ≤ 𝑡 ≤ 𝑛. It followsthat there exists a prime 𝑟 such that 𝑞𝑛(𝑛+1)/2 | 𝑟𝑚 andso 𝑛(𝑛+1)/2 ≤ 𝑚 ≤ 𝑝/2 by Lemma 5. Since the Sylow𝑝-subgroup of 𝐺 is of order 𝑝, then 𝑛(𝑛 + 1) | 𝑝 andso 𝑛 = 1, a contradiction.

This completes the proof of the lemma.

Lemma 17. Consider the following.𝐺 ≅ 𝐴

𝑝+3.

Proof. By Lemma 16,

𝐴𝑝+3

≤ 𝐺 ≤ Aut (𝐴𝑝+3) ≅ 𝑆𝑝+3. (29)

If 𝐺 ≅ 𝑆𝑝+3

, then there exists an element 𝑥 of 𝐺 with

𝑥𝐺

=(𝑝 + 3)!

3𝑝(30)

which contradicts Lemma 12.So 𝐺 ≅ 𝐴

𝑝+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}.

If 𝐾 is a 2-group, in this case, let |𝑥| = 𝑝. Then

(𝑝 + 3)!

6𝑝|𝑥𝐺. (31)

By Lemma 12,

𝑥𝐺=𝑥𝐺=(𝑝 + 3)!

6𝑝. (32)

So 𝑥 centralizes 𝐾. It follows that there is an element of 2 ⋅ 𝑝which contradicts Lemma 12 (4).

If 𝐾 is a 3-group. Then similarly as the case “𝐾 is a 2-group”, we have that |𝑥𝐺| = |𝑥𝐺| is maximal in 𝑁(𝐺) and𝐶𝐺(𝑥) is abelian. So by Lemma 1.12 of [13],𝐾 ≤ 𝑍(𝐺) = 1.Therefore 𝐾 = 1 and 𝐺 ≅ 𝐴

𝑝+3.

This completes the proof of the lemma and also of themain theorem.

4. Some Applications and Problem

Y. Chen andG. Chen in [23] proved that the group𝐴10can be

characterized by its order and two special conjugacy classessizes. Then, obviously, we also have the following result.

Corollary 18. Let 𝐺 be a finite group with trivial center.Assume that 𝑁(𝐺) = 𝑁(𝐴

𝑝+3) and |𝐺| = |𝐴

𝑝+3|. Then 𝐺 ≅

𝐴𝑝+3

.

One knows that the alternating groups 𝐴𝑛with 𝑛 =

10, 16, 22, 26 are characterized by𝑁(𝐺). Then by [4, 5, 15, 21],one has the following.

Corollary 19. Let 𝐺 be a finite group with trivial center.Assume that 𝑁(𝐺) = 𝑁(𝐴

𝑛) with 𝑛 = 𝑝, 𝑝 + 1, 𝑝 + 2, 𝑝 + 3.

Then 𝐺 ≅ 𝐴𝑛.

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper.

Acknowledgments

This paper is supported by the Department of Education ofSichuan Province (Grant no. 12ZB291), the Opening Projectof Sichuan Province University Key Laboratory of BridgeNon-Destruction Detecting and Engineering Computing(Grant no. 2013QYJ02), and the Scientific Research Projectof Sichuan University of Science and Engineering (Grantno. 2014RC02). The authors are very grateful for the helpfulsuggestions of the referee.

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