A new characterization of A12

Abstract

Purpose

It is well known that the conjugacy class sizes have an important influence on the structure of a group. This work is considering a different set of ‘sizes’, the number of elements of a given order.

Methods

By using the set nse(G) and the order of G, We prove that G is isomorphic to A 12 .

Results

Thompson’s conjecture is true for A 1 12 .

Conclusions

We proved that a finite group G is isomorphic to A 12 , the alternating group A 12 of degree 12 if, and only if, | G |=| A 12 | and nse(G)=nse( A 12 ).

Introduction

In this study, all groups are assumed to be finite. It is well known that the conjugacy class sizes have an important influence on the structure of a group. The relation between conjugacy class sizes and the structure of a group has been studied by many authors (for example, see [ 1 – 4 ]). In the present work, we are considering a different set of ‘sizes’, the number of elements of a given order.

Most notations are standard (see [ 5 , 6 ]). We introduce some which may be unfamiliar to the reader. Let ω ( G ) denote the set of element orders of G . Let m i ( G ):=|{ g ∈ G |the order of g is i }| ( m i for short) be the number of elements of order i , and let nse( G ):={ m i ( G )| i ∈ ω ( G )} be the set of sizes of elements with the same order. n p ( G ) denotes the number of Sylow p -subgroup of G , namely, n p ( G )=|Syl p ( G )|. Π ( G ) denotes the set of all prime divisors of | G |. A 12 is the alternating group of degree 12. We use a | b to mean that a divides b; if p is a prime, then p n || b means p n | b but p n + 1 ∤ b . N={1,2,3,4,⋯} denotes the set of positive integers. Π ( G ) denotes the set of prime divisors of | G | and | Π ( G )|, the number of the element of the set Π ( G ). nse(G) denotes the number of elements of a given order of G .

Methods

Thompson’s problem

For the set nse( G ), the most important problem is related to Thompson’s problem. In 1987, JG Thompson put forward the following problem. For each finite group G and each integer d ≥ 1, let G ( d )={ x ∈ G | x d =1}. Defining G 1 and G 2 is of the same order type if, and only if, | G 1 ( d )|=| G 2 ( d )|, d =1,2,3,⋯. Suppose G 1 and G 2 are of the same order type. If G 1 is solvable, is G 2 necessarily solvable?

Professor WJ Shi in [ 7 ] made the above problem public in 1989. Unfortunately, no one can solve it or give a counterexample until now, and it remains open. The influence of nse( G ) on the structure of finite groups was studied by some authors (see [ 8 – 11 ]). In the present work, we show that a condition related to the order type characterizes A 12 ; explicitly, we prove the following:

Main theorem

Let G be a group, and then G ≅ A 12 if, and only if , | G |=| A 12 | and nse(G)=nse ( A 12 ) .

Preliminary

Lemma 1

If G is a soluble group of order mn , where m is prime to n ([ 12 ] (p. 99)), then

  • G possesses at least one subgroup of order m ,

  • Any two subgroups of G of order m are conjugate,

  • Any subgroup of G whose order divides m belongs to some subgroup of order m ,

  • The number h m of subgroups of G of order m may be expressed as a product of factors, each of which (1) is congruent to 1 modulo some prime factor of m, (2) is a power of a prime and divides one of the chief factors of G .

Definition 1

A finite group G is called a simple K n -group if G is a simple group with | Π ( G )|= n (see [ 13 ] (p. 658) or [ 14 ]).

We will give need a description of the simple K n -groups for n ≤ 5.

Remark 1

If G is a simple K 1 -group, then G is a cyclic of prime-order.

Remark 2

If | G |= p a q b with p and q distinct primes and a, b nonnegative integers, then by Burnside’s pq -theorem (see [ 6 ] section (10.2.1)), G is soluble. In particular, there are no simple K 2 -groups.

Lemma 2

If G is a simple K 3 -group, then G is isomorphic to one of the groups (see Theorem 2 of [ 15 ]): A 5 (2 2 ·3·5), A 6 (2 3 ·3 2 ·5), L 2 (7)(2 3 ·3·7), L 2 (8)(2 3 ·3 2 ·7), L 2 (17)(2 4 ·3 2 ·17), L 3 (3)(2 4 ·3 3 ·13), U 3 (3)(2 5 ·3 3 ·7), or U 4 (2)(2 6 ·3 4 ·5), where ∗(∗) means the group (the order of G ).

Lemma 3

Let G be a simple K 4 -group, and then G is isomorphic to one of the following groups (see Theorem 1 of [ 13 ] or Theorem 2 of [ 14 ]):

  • A 7 , A 8 , A 9 , or A 10 .

  • M 11 , M 12 , or J 2 .

  • One of the following:

  • L 2 ( r ), where r is a prime and r 2 −1=2 a ·3 b · v c with a ≥ 1, b ≥ 1, c ≥ 1, and v is a prime greater than 3.

  • L 2 (2 m ), where 2 m −1= u , 2 m + 1=3 t b with m ≥ 2, u , t are primes, t > 3, b ≥ 1;

  • L 2 (3 m ), where 3 m + 1=4 t , 3 m −1=2 u c or 3 m + 1=4 t b , 3 m −1=2 u , with m ≥ 2, u , t are odd primes, b ≥ 1, c ≥ 1;

  • L 2 (16), L 2 (25), L 2 (49), L 2 (81), L 3 (4), L 3 (5), L 3 (7), L 3 (8), L 3 (17), L 4 (3), S 4 (4), S 4 (5), S 4 (7), S4(9),S6(2),O8+(2),G2(3),U3(4),U3(5),U3 (7), U 3 (8), U 3 (9), U 4 (3), U 5 (2), Sz (8), Sz (32), 3 D 4 (2), or 2 F 4 (2) ″ .

Lemma 4

Each simple K 5 -group is isomorphic to one of the following simple groups (see Theorem A of [ 16 ]):

  • L 2 ( q ) where q satisfies | Π ( q 2 −1)|=4;

  • L 3 ( q ) where q satisfies | Π ( q 2 −1)( q 3 −1)|=4;

  • U 3 ( q ) where | Π ( q 2 −1)( q 3 + 1)|=4;

  • O 5 ( q ) where | Π ( q 4 −1)|=4;

  • Sz(22m+1)
    where |Π((22m+1−11)(224m+2+1))|=4 ;

  • R ( q ) where q is an odd power of 3 and | Π ( q 2 −1)|=3;

  • Following 30 simple groups: A 11 , A 12 , M 22 , J 3 , HS , He , McL , L 4 (4), L 4 (5), L 4 (7), L 5 (2), L 5 (3), L 6 (2), O 7 (3), O 9 (2), PS p 6 (3), PS p 8 (2), U 4 (4), U 4 (5), U 4 (7), U 4 (9), U 5 (3), U6(2),O8+(3),O8−(2),3D4(3),G2(4),G2(5),G2(7) , or G 2 (9).

Lemma 5

Let G be a finite group, P ∈ Sy l p ( G ), where p ∈ Π ( G ). Suppose that G has a normal series K ⊴ L ⊴ G and p ∤| K |, and then following statements hold [ 17 ]:

  • N G / K ( PK / K )= N G ( P ) K / K .

  • If P ≤ L , then | G : N G ( P )|=| L : N L ( P )|, namely, n p ( G )= n p ( L ).

  • If P ≤ L , then | L / K : N L / K ( PK / K )| t =| G : N G ( P )|=| L : N L ( P )|, namely, n p ( L / K )= t = n p ( G )= n p ( L ). In particular, | N K ( P )| t =| K |.

Lemma 6

Let G be a simple K 5 -group and 3 5 || G ||2 9 ·3 5 ·5 2 ·7·11; then G ≅ A 12 .

Proof

Assume G is isomorphic to L 2 ( q ) in Lemma 4, then 7 or 11 || L 2 ( q )|. □

Case 1

7|| L 2 ( q )|.

If q =7, then | Π ( q 2 −1)|=2, which contradicts | Π ( q 2 −1)|=4.

If q =2 m , then 7|2 2 m −1, so we have 5| m and 19 || L 2 ( q )|, which is a contradiction.

If q =3,3 2 , or 5, then | Π ( q 2 −1)| < 4, which is a contradiction.

If q =3 3 , or 5 2 , then 13|( q 2 −1)|| G |, which is a contradiction.

If q =3 4 , then 41|| L 2 ( q )|, which is a contradiction.

If q =3 5 , then 61|| L 2 ( q )|, which is a contradiction.

Case 2

11|| L 2 ( q )|.

If q =11, then | Π (11 2 −1)|=3, which contradicts | Π ( q 2 −1)|=4.

If q =2 m , then 11|2 2 m −1, so we have 5| m and 31|| L 2 ( q )||, which is a contradiction.

If q =3,3 2 ,5, or 7, then | Π ( q 2 −1)| < 4, which is a contradiction.

If q =3 3 , or 5 2 , then 13|( q 2 −1)|| G |, which is a contradiction.

If q =3 4 , then 41|| L 2 ( q )|, which is a contradiction.

If q =3 5 , then 61|| L 2 ( q )|, which is a contradiction.

Thus, G is not isomorphic to L 2 ( q ).

Similarly, G is not isomorphic to L 3 ( q ), U 3 ( q ), O 5 ( q ), Sz (2 2 m + 1 ), and R ( q ).

In view of item number 7 of Lemma 4, we get that G ≅ A 12 .

This completes the proof. □

Results and discussion

The proof of the main theorem

In this section, we will give the proof of the main theorem. We rewrite the main theorem here:

Main theorem

Let G be a group, and then G ≅ A 12 if, and only if , | G |=| A 12 | and nse(G)=nse( A 12 ) ={ 1, 63855, 570240, 2154900, 3825360, 3991680, 4809024, 6652800, 8553600, 11404800, 11975040, 13685760, 21621600, 25530120, 25945920, 26611200, 29937600, 43545600. }

Proof

If G ≅ A 12 , from [ 18 ] (pp. 91 to 92), we easily get the results. □

Then, we assume that | G |=| A 12 |, and nse( G )= nse ( A 12 ).

We prove G ≅ A 12 by first proving that G is insoluble, and then showing that it must be isomorphic to A 12 .

Step 1. G is insoluble.

If G is soluble, let H be a {3,5,7,11}-Hall subgroup of G . By item number 2 of Lemma 1, all {3,5,7,11}-Hall subgroups of G are conjugate in G ; hence, by I and Exercise 2 of [ 5 ], the number of {3,5,7,11}-Hall subgroup of G is | G : N G ( H )||2 9 .

By I and Theorem 2.9 of [ 5 ], n 11 ( H )=1,12,111,925,1442 in H . Thus, we have the following cases:

If n 11 ( H )=1, then the number m of elements of order 11 in G is 10 < m < 5120 and 10| m , but m ∉ nse( G ), which is a contradiction.

If n 11 ( H )=12, then the number m of elements of order 11 in G is 120 < m < 61440 and 10| m , but m ∉ nse( G ), which is a contradiction.

If n 11 ( H )=111, then the number m of elements of order 11 in G is 1110 < m < 568320 and 10| m , but m ∉ nse( G ), which is a contradiction.

If n 11 ( H )=925, or 1442, then the number m of elements of order 11 in G is 9250 < m < 4736000 or 14420 < m < | m . Since m ∈ nse ( G ), m =776600, 3825360, 570240, so 11 k + 1=776600,3825360, 570240, but none of these equations have solutions in N .

Thus, G is insoluble.

Step 2. G ≅ A 12 .

Through Step 1, it has been proven that G is insoluble. Since p || G |, where p ∈ {7,11}, then G has a normal series: 1 ⊴ K ⊴ L ⊴ G such that L / K is a simple K i -group where i =3, 4, or 5. Thus, we will prove this through the following three cases due to Remarks 1 and 2.

Case 1

L / K is a simple K 3 -group.

Since 7 or 11|| L / K ||2 9 ·3 5 ·5 2 ·7·11, we have known, from Lemma 2, that L / K ≅ L 2 (7), L 2 (8), or U 3 (3).

In the following, let P be the Sylow 7-subgroup of G . Then, PK / K ∈Syl 7 ( L / K ), and by Lemma 5, n 7 ( L / K ) t = n 7 ( G ) for some integer t∈N .

Subcase 1.1. L / K ≅ L 2 (7)(2 3 ·3·7).

From [ 18 ], n 7 ( L / K )= n 7 ( L 2 (7))=8. Hence, n 7 ( G )=8 t and 7∤ t , so the number of elements of order 7 in G is m =8 t ·6=48 t .

Since m ∈ nse ( G ) and 7∤ t , then m =4809024, 570240 and t =100188, 11880.

  • Let m =4809024 and t =100188. By Lemma 5 and since | G |=| G : L || L : K || K |, 100188| N K ( P )|=| K |, so 2 2 ·3 2 ·11 2 ·23|| K |. However, 23 ∉ Π ( G ), which is a contradiction.

  • Let m =570240 and t =11880. By Lemma 5 and since | G |=| G : L || L : K || K |, 11880| N K ( P )|=| K |, so 2 3 ·3 3 ·5·11|| K ||2 6 ·3 4 ·7·11, which is a contradiction.

Subcase 1.2. L / K ≅ L 2 (8)(2 3 ·3 2 ·7).

From [ 18 ], n 7 ( L / K )= n 7 ( L 2 (8))=36. Hence, n 7 ( G )=8 t and 7∤ t , so the number of elements of order 7 in G is m =36 t ·6=216 t .

Since m ∈ nse ( G ) and 7∤ t , then m =4809024,570240 and t =22264,2640.

  • Let m =4809024 and t =22264. By Lemma 5 and since | G |=| G : L || L : K || K |, 22264| N K ( P )|=| K |, so 2 3 ·11 2 ·23|| K |. However, 23 ∉ Π ( G ), which is a contradiction.

  • Let m =570240 and t =2640. By Lemma 5 and since | G |=| G : L || L : K || K |, 11880| N K ( P )|=| K |. As | K ||2 6 ·3 3 ·5 2 ·11, n 11 ( K )=1,12,45,100,144,320,540,1200,1728, so the number of elements of order 11 in G is 10, 120, 450, 1000, 1440, 3200, 12000, 17280 ∉ nse ( G ). Also, we get a contradiction.

Subcase 1.3. L / K ≅ U 3 (3)(2 5 ·3 3 ·7).

From [ 18 ], n 7 ( L / K )= n 7 ( U 3 (3))=288. Hence, n 7 ( G )=288 t and 7∤ t , so the number of elements of order 7 in G is m =288 t ·6=1728 t .

Since m ∈nse( G ) and 7∤ t , then m =4809024,570240 and t =2783,330.

  • Let m =4809024 and t =2783. By Lemma 5 and since | G |=| G : L || L : K || K |, 2783| N K ( P )|=| K |, so 11 2 ·23|| K |. However, 23 ∉ Π ( G ), which is a contradiction.

  • Let m =570240 and t =330. By Lemma 5 and since | G |=| G : L || L : K || K |, 330| N K ( P )|=| K |. As | K ||2 4 ·3 2 ·5 2 ·11, n 11 ( K )=1,12,45,100,144,1200, so the number of elements of order 11 in G is 10, 120, 450, 1000, 1440, 12000 ∉ nse( G ). Also, we get a contradiction.

Case 2

L / K is a simple K 4 -group.

Since 7 or 11 ∣| L / K |∣2 9 ·3 5 ·5 2 ·7·11, and then by Lemma 3, we have the following subcases:

Subcase 2.1

L / K ≅ A i , where i =7,8,9,10.

If L / K ≅ A 7 , then from [ 18 ], n 7 ( L / K )= n 7 ( A 7 )=120. Hence, n 7 ( G )=120 t and 7∤ t . The number of elements of order 7 in G is m =120 t ·6=720 t .

Since m ∈nse( G ) and 7∤ t , then m =570240,8553600,11404800,13685760 and t =792, 11880, 15840, 19008.

Let m =570240 and t =792. By Lemma 5 and since | G |=| G : L || L : K || K |, 792| N K ( P )|=| K |. As | K ||2 6 ·3 3 ·5·11, n 11 ( K )=1,12,45,100,144,320,540,1728, so the number of elements of order 11 in G is 10, 120, 450, 1000, 1440, 3200, 5400, 17280 ∉ nse( G ). Also, we get a contradiction.

For m =8553600, 11404800,13685760 and t =11880,15840, 19008, the proofs are similar to m =570240 and t =792. We get contradictions.

For A i , where i =8,9, or 10, we also get a contradiction.

Subcase 2.2

L / K ≅ M 11 , M 12 , J 2 .

The proof is similar to Subcase 2.1.

Subcase 2.3

L / K is isomorphic to one of the group as listed in item number 3 of Lemma 3.

The proof is similar to Subcase 2.1.

Case 3

L / K is a simple K 5 -group.

In this case, Π ( L / K )={2,3,5,7,11}. From Lemma 6, we have G ≅ A 12 .

This completes the proof. □

Conclusions

We proved that a finite group G is isomorphic to A 12 , the alternating group A 12 of degree 12 if, and only if, | G |=| A 12 | and nse(G)=nse( A 12 ).

Acknowledgements

The object is partially supported by the Department of Education of Sichuan Province (grant nos. 12ZB085 and 12ZB291). The authors are very grateful for the helpful suggestions of the referee.

Competing interests

The authors declare that they have no competing interests.

Authors’ contributions

SL and RZ carried out the studies and participated in drafting the manuscript. Both authors read and approved the final manuscript.

References

  1. Chillag and Herzog (1990) On the length of the conjugacy classes of finite groups 131(1) (pp. 110-125) 10.1016/0021-8693(90)90168-N
  2. Conway and Wang (1999) Remarks on the length of conjugacy classes of finite groups 27(9) (pp. 4347-4353) 10.1080/00927879908826701
  3. Dolfi et al. (2009) The groups with exactly one class of size a multiple of p (pp. 219-234) 10.1515/JGT.2008.074
  4. Guo et al. (2012) A new characterization of alternating group A13 62(1) (pp. 15-28)
  5. Gorenstein (1980) Chelsea Publishing Company
  6. Kurzweil and Stellmacher (2004) Springer-Verlag 10.1007/b97433
  7. Shi (1989) A new characterization of the sporadic simple groups in group theory (pp. 531-540) Walter de Gruyter
  8. Khatami et al. (2011) A new characterization for some linear groups (pp. 39-50) 10.1007/s00605-009-0168-1
  9. Shao and Jiang (2008) A new characterization of A11 24(2) (pp. 11-14 (in Chinese))
  10. Shao et al. (2009) A characterization of simple K3-groups 38(3) (pp. 327-330 (in Chinese))
  11. Shen et al. (2010) A new characterization of A5 (pp. 337-341) 10.1007/s00605-008-0083-x
  12. Hall (1928) A note on soluble groups (pp. 98-105) 10.1112/jlms/s1-3.2.98
  13. Bugeaud et al. (2001) On simple K4-groups (pp. 658-668) 10.1006/jabr.2000.8742
  14. Shi (1991) On simple K4-groups 36(17) (pp. 1281-1283 (in Chinese))
  15. Herzog (1968) Finite simple groups divisible by only three primes (pp. 383-388) 10.1016/0021-8693(68)90088-4
  16. Jafarzadeh et al. (2007) On simple Kn-groups for n= 5, 6 (pp. 668-680) Cambridge University Press
  17. Shao and Jiang (2010) A new charcterization of Mathieu groups (pp. 13-23)
  18. Conway et al. (1985) Clarendon Press