A Characterization of the Small Suzuki Groups by the Number of the Same Element Order

A Characterization of the Small Suzuki Groups by the

Number of the Same Element Order

H. Parvizi Mosaed

1

, A. Iranmanesh

2*

, and A. Tehranian

1

1_{Department of Mathematics, Faculty of Basic Sciences, Science and Research Branch, Islamic Azad}

University, Tehran, Islamic Republic of Iran

2_{Department of Mathematics, Faculty of Mathematical Sciences, Tarbiat Modares University, Tehran,}

Islamic Republic of Iran

Received: 23 October 2014 / Revised: 3 January 2015 / Accepted: 17 May 2015

Abstract

Suppose that

G

is a finite group. Then the set of all prime divisors of

G

is denoted

by



( )

G

and the set of element orders of

G

is denoted by



_e

( )

G

. Suppose that

( )

e

k





G

. Then the number of elements of order

k

in

G

is denoted by

m

_k

and the

sizes of the set of elements with the same order is denoted by

nse G

 

; that is,

  

m k

k

:

e

( )



nse G







G

. In this paper, we prove that if

G

is a group such that

 

nse

( ( ))

ns

e

G



Sz

n

, where

n





32,128



, then

G Sz n



( )

. Here

Sz n

( )

denotes the

family of Suzuki simple groups,

n



2

2 1k

,

k





. This proves that the second and

third member of the family of Suzuki simple groups are characterizable by the set of

the number of the same element order.

Keywords:Element order; Sylow subgroup; Simple Kn-group; Suzuki group.

*_{Corresponding author: Tel: +982188009730; Fax: +982188009730; Email: [email protected]}

Introduction

Suppose that

G

is a finite simple group and

( )

G

n





, where



( )

G

denotes the number of prime numbers dividing the order of

G

. Then

G

is called a simple

K

_n-group. Suppose that

G

is a finite group. Then a Sylow

q

-subgroup of

G

is denoted by

q

P

and the number of Sylow

q

-subgroups of

G

is denoted by

n

_q and the greatest order of elements in

P

_q is denoted by

exp( )

P

_q . The Euler totient function is

that if

G

is a group such that

nse G

( )



nse Sz n

( ( ))

, where

n





32,128



, then

G Sz n



( )

.

Preliminary and Notations

In this section, we bring some lemmas that is need in the proof of main theorem.

Lemma 1.1 [5] If

G

is a simple

K

₃-group, then

G

is isomorphic to one of the following groups: 5

A

,

A

₆,

L

₂

(7)

,

L

₂

(8)

,

L

₂

(17)

,

L

₃

(3)

,

U

₃

(3)

,

4

(2)

U

.

Lemma 1.2 [8]If

G

is a simple

K

₄-group, then

G

is isomorphic to one of the following groups: (1)

A

₇,

A

₈,

A

₉,

A

₁₀.

(2)

M

₁₁,

M

₁₂,

J

₂.

(3) (a)

L r

₂

( )

, where

r

is a prime and satisfies 2

1 2 3

a b c

r

   

v

with

a



1

,

b



1

,

c



1

,

v



3

,

v

is a prime. (b)

L

₂

(2 )

m , where

m

satisfies

2

1

2

1 3

m

m b

u

t



 



 



with

m



2

,

u

,

t

are primes,

t



3

,

b



1

. (c)

L

₂

(3 )

m , where

m

satisfies

3 1 4

3 1 2

m

m c

t

u

 

 







or

3 1 4

3 1 2

m b

m

t

u

 

 







with

m



2

,

u

,

t

are odd primes,

b



1

,

c



1

. (d)

L

₂

(16)

,

L

₂

(25)

,

L

₂

(49)

,

L

₂

(81)

,

L

₃

(4)

,

3

(5)

L

,

L

₃

(7)

,

L

₃

(8)

,

L

₃

(17)

,

L

₄

(3)

,

S

₄

(4)

,

4

(5)

S

,

S

₄

(7)

,

S

₄

(9)

,

S

₆

(2)

,

O

₈

(2)

,

G

₂

(3)

, 3

(4)

U

,

U

₃

(5)

,

U

₃

(7)

,

U

₃

(8)

,

U

₃

(9)

,

U

₄

(3)

,

5

(2)

U

,

Sz

(8)

,

Sz

(32)

, 3

D

₄

(2)

, 2

F

₄

(2)



.

Lemma 1.3 [3]Let

G

be a finite solvable group and

G mn



, where 1

1

...

rr

m p





p



,



m n

,





1

. Let



p

1

, ,

p

r









and

h

_m be the number of Hall



-subgroups of

G

. Then 1

1

...

s

m s

h



q



q



satisfies the following conditions for all

i

 



1, ,

s



:

(1) i

1  mod 





i j

q





p

, for some

p

_j.

(2) The order of some chief factor of

G

is divisible by i

i

q

 _.

Lemma 1.4 [2]Let

G

be a finite group and

m

be a positive integer dividing

| |

G

. If





( )

:

m

1

m

L G



g G g





, then

m L G

     ( )

_m .

Lemma 1.5 [10]Let

G

be a group containing more than two elements. Let

k





_e

( )

G

and

m

_k be the number of elements of order

k

in

G

. If



k

:

e

( )



s sup m k







G

is finite, then

G

is finite and

G s s



(

2



1)

.

Lemma 1.6 [7] Let

G

be a finite group and

( )

p





G

be odd. Suppose that

P

is a Sylow

p

-subgroup of

G

and

n p m



s , where

( , ) 1

p m



. If

P

is not cyclic and

s



1

, then the number of elements of order

n

is always a multiple of

p

s.

Lemma 1.7 [4]Let

G

be a solvable group and



be any set of primes. Then

(1)

G

has a Hall



-subgroup.

(2) If

H

is a Hall



-subgroup of

G

and

V

is any



-subgroup of

G

, then

V H



g for some

g G



. In particular, the Hall



-subgroups of

G

form a single conjugacy class of subgroups of

G

.

Lemma 1.8 Let

G

be a finite group which is not solvable. Then there is a normal series

1

(

N M G

(

(

such that

N

is a maximal solvable normal subgroup of

G

and

M N

is a non-abelian simple group or the direct product of isomorphic non-abelian simple groups.

Proof. Since

G

is a finite group, there is chief series

1



M

₀

(

M

₁

(

...

(

M

_n1

(

M

_n



G

. Since

G

is not solvable, there is a maximal

i

such that 1

i

subgroup of

G

and

M M

_{i i1} is a non-abelian simple group or the direct product of isomorphic non-abelian simple groups.

Lemma 1.9 Let

G

be a group such that

( )

( ( ))

nse G



nse Sz n

, where

n





32,128



. Then

G

is finite and for every

i





_e

( )

G

,

 

  

    

 

    

i

d d i

i m

i

m













and if

i 2



, then

m

_i is even.

Proof. By Lemma 1.5,

G

is a finite group. By Lemma 1.4,

    

d

d i

i



m

. We know that the number of

elements of order

i

in a cyclic group of order

i

is equal with



( )

i

. Hence

m

_i





( )

i k

, where

k

is the number of cyclic subgroups of order

i

in

G

. Thus

    

( )  

i m

_i



. We know that if

i



2

, then



( )

i

is even and since



( )  

i m

    

_i, we conclude that

m

_i is even. □

Results

In this section, we prove two theorems as the main results of our paper. The first theorem is the following theorem:

Theorem 2.1 Suppose that

G

is a group such that

( )

( (32))

nse G



nse Sz

. Then

G Sz



(32)

.

Proof. By a program written in theGAP, we have in nse(G)=nse(Sz(32))=

{1,31775,1016800,1301504,6507520, 7936000,15744000 }.

We prove this theorem in five steps.

Step 1.



  

G



2, 5, 31, 41



.

Since

31775

is odd, Lemma 1.9 implies that

 

2





G

and

m

₂



31775

. Assume that

 

q





G

and

q



2

, by Lemma 1.9, q    1



mq



and



q 1





 

q m     _q, which imply that



3, 5, 7,1 3, 31, 41,6507521



q . If

6507521





 

G

,

then by Lemma 1.9,

m

_6507521



6507520

. On the other hand, if

13015042 2 6507521

 





_e

 

G

, then by Lemma 1.9,





13015042      



m

_13015042 and



2 6507521 13015042



13015042     1 



m m





m

, which is a contradiction. Hence

2 6507521







_e

 

G

. Thus

P

_6507521 acts fixed point freely on the set of elements of order

2

by conjugation. Therefore

6507521 2

 

P

    

m

, which is a contradiction. So

 

6507521





G

. If

13





 

G

, then by Lemma 1.9,

m

₁₃



15744000

. On the other hand, if

 

26 2 13

  



_e

G

, then by Lemma 1.9,

 

26     m26



and

26    1





m

₂



m

₁₃



m

₂₆



, which is a contradiction. Hence

2 13

 



e

 

G

. Thus

P

13 acts fixed point freely on the set of elements of order

2

by conjugation. Therefore

P

₁₃

    

m

₂, which is a contradiction. So

13





 

G

. If

7





 

G

, then by Lemma 1.9,

m

₇



15744000

. On the other hand, if

 

14 2 7

  



e

G

, then by Lemma 1.9,

 

14     

m

14



and

14     1 





m m m

₂



₇



₁₄



, which is a contradiction. Hence

14 2 7

  



e

 

G

. Thus

P

7 acts fixed point freely on the set of elements of order

2

by conjugation. Therefore

P m

7

    

2, which is a contradiction. So

7





 

G

. Therefore we conclude that



  

G



2, 3, 5, 31, 41



.

If



2, 3, 5, 31, 41







 

G

, then by Lemma 1.9,

2

31775

m



,

m

₃



1301504

,

m

₅



1301504

,

31

15744000

m



,

m

₄₁



7936000

and

 

13 3 3 2 2

2 ,3 ,5 ,31 , 41 , 2 31, 3 41, 31 41











e

G

.

Since

2

13





e

 

G

, we conclude that

 



12



2

exp

P



2, , 2



. If

exp

 

P

₂



2

2, then by Lemma 1.4 and considering

m P



₂ , we conclude that

20 2

    2

Since

3

3





_e

 

G

, we conclude that

 

3

exp

P



3 

or

3

2. There are two cases:

Case 1. If

exp

 

P

₃



3

, then by Lemma 1.4 and considering

m P



₃ , we conclude that

P

₃



3

.

Hence

P

₃ is cyclic and

 

3 9

3

m

3

2 31 41

n







 

.

Case 2. If

exp

 

P

₃



3

2, then by Lemma 1.4 and considering

m P



₃ , we conclude that

P

₃

    3

3. If

3

3

3

P



, then

P

₃ is not cyclic. Hence by Lemma 1.6,

9

9    

m



15744000

, which is a contradiction.

Therefore

P

₃



3

2 and

 

32 9 3

3

3

2

2 5 41

m

n





  

.

Since

5

3





_e

 

G

, we conclude that

exp

 

P

₅



5

or

5

2. If

exp

 

P

₅



5

, then by Lemma 1.4 and by considering

m P



₅ , we conclude that

P

₅



5

and

 

5 8

5

m

5

2 31 41

n







 

. If

exp

 

P

₅



5

2, then

by Lemma 1.4 and considering

m P



₅ , we conclude

that

P

₅



5

2 and 5

 

52₂

2 31 41

8

5

m

n





  

.

Since

31

2

 

e

G





, by Lemma 1.4 and considering

31

m P



, we conclude that

P

₃₁



31

and

 

31 9 2

31

m

31

2 5 41

n







 

.

Since

41

2





_e

 

G

, by Lemma 1.4 and considering

41

m P



, we conclude that

P

₄₁

    41

2. Now we show that

3





 

G

.

If

3





 

G

, then by the above discussion, 9

3

2 31 41 

n

  

or

2 5 41

9

 

3 . Hence

41    

G

.

Since

3 41

 



_e

 

G

, we conclude that

P

₃ acts fixed point freely on the set of elements of order

41

by conjugation. Hence

P m

₃

    

₄₁, which is a contradiction. So

3





 

G

. Therefore

  

G

  2, 5, 31, 41







.

If



   

G



2

, then we know that nse G

 

7. Thus

exp

 

P

₂



4

. Hence

G



P

₂

    2

19. So

19 4

1m 2 , but





4 1 016800,1 301504,  6507520, 7936000, 1 5744000

m  ,

which is a contradiction.

If



  

G



2, 41



, then we know that

2

13,

 

2

41





e

G

and

P

2

    2

20,

      41

P

41 2. Hence

 



1 , 2 , ,2

12

 

41,41 2, ,41 2

12



e

G









 



.

Therefore,

| | = 2 × 41 = 32537600 + 1016800 + 1301504 + 6507520 + 7936000 + 15744000

where

0

     

k k k k k

_{1 2 3 4 5}

19

, l20 , k2. It is easy to check that this equation has no solution.

If

5





 

G

, then

n

₅

  

2 31 41

8 . We know that

n G

₅

   

. Hence

31    

G

.

Therefore in any cases we can assume that

 

31





G

.

Now we prove that



  

G



2,5,31, 41



. Since

 

31





G

, we conclude that

P

₃₁



31

and

 

31 9 2

31

m

31

2 5 41

n







 

. We know that

n

₃₁

   

G

, hence

2 5 41    

9

 

2

G

. It follows that

  

G

2,5,31, 41







.

Step 2.

G



2 5 31 41

k

  

l _{, where}

k



10

_,

2

l



.

By the above discussion

P

₃₁



31

,

P

5

    5

2. Since

62





e

 

G

, we conclude that

P

2 acts fixed point freely on the set of elements of order

31

by conjugation. Therefore

P m

₂

    

₃₁. Hence

P

₂

    2

10.

Since

1271





_e

 

G

, we conclude that

P

₄₁ acts fixed point freely on the set of elements of order

31

by conjugation. Therefore

P

₄₁

    

m

₃₁. Hence

P

₄₁



41

.

Step 3.

G

is not solvable.



-subgroup

H

, where







5,31, 41



and all the Hall



-subgroups of

G

are conjugate and the number of Hall



-subgroups of

G

is

G N H

 :

_G

 

    2

10. Since

G

is solvable, we conclude that

H

is solvable. Hence by Lemma 1.3, there are non negative integers

1

, ,

r







,



₁

, ,





_s such that

 

1 1

31

5

r

41

s

n H









,

5

i



1  mod  31  





,





41

i



1  mod  31

. Since

G



2 5 31 41

k

  

l ,

where

k



10 

,

l



2

, we conclude that

1 r

  2 









,

 



₁





_s



1

. Therefore

 

31

 

n H



1

. So 10

31

30



m G

( ) (2





30) 30720



, but we have

m G

₃₁

 



15744000

, which is a contradiction.

Step 4.

G



2

10

  

5 31 41

2 _.

Since

G

is a finite group which is not solvable, there is a normal series

 1 

(

N M G

(

(

such that

N

is a maximal solvable normal subgroup of

G

and

/

M N

is a non-abelian simple group or the direct product of isomorphic non-abelian simple groups, by Lemma 1.8. Let

M N S

/

 

₁

S

_r, where

S

₁ is a non-abelian simple group and

S

₁

  

S

_r. Since

 1 

(

N M G

(

(

and

G



2 5 31 41

k

  

l , where

k



10 

,

l



2

, we conclude that

r



1

and

/

M N

is a simple

K

₃-group or a simple

K

₄-group. If

M N

/

is a simple

K

₃-group, then by Lemma 1.1 and

G



2 5 31 41

k

  

l , where

k



10 

,

2

l



, we conclude a contradiction.

If

M N

/

is a simple

K

₄-group, then by Lemma 1.2 and

G



2 5 31 41

k

  

l _{, where}

k



10 

_,

  2

l



, we conclude that

M N Sz

/



 

32

. Hence

10 2 10 2

2

  

5 31 41



M N G

/          2

  

5 31 41

. So

G



Sz

 

32

.

Step 5.

G Sz



 

32

.

Since

1

(

N M G

(

(

,

M N Sz

/



 

32

and

 

32

G



Sz

, we can conclude

N



1 

,

 

32

G M Sz





and the proof is completed. □

The second theorem as the main result is the following theorem:

Theorem 2.2 Suppose that

G

is a group such that

 



 

128



nse G



nse Sz

. Then

G Sz



 

128

.

Proof. By a program written in theGAP, we have nse(G)=nse(Sz(128))=

{1,2080895,266354560,235126784,16

45887488,6583549952,8447918080,

16912465920}.

We prove this theorem in four steps.

Step 1.



  

G



2, 5, 29,1 13,1 27



.

Since

2080895

is odd, Lemma 1.9 implies that

 

2





G

and

m

₂



2080895

. Assume that

 

q





G

and

q



2

by Lemma 1.9,

q

    1





m

_q



and



q

 

1





 

q m

    

_q which imply that



3, 5,1 1,1 3, 29,1 13,1 27



q



. If

13





 

G

, then

by Lemma 1.9,

m

₁₃



16912465920

. On the other hand, by Lemma 1.9,

13

2





_e

 

G

. Thus





13

    1

13

P



m

. Therefore

P

₁₃



13

and

 

13

13

m

13

1409372160

n







. Since

113    

n

₁₃, we

deduce that

113





 

G

. Now by Lemma 1.9,

 

13 113







e

G

. Thus

P

13 acts fixed point freely on the set of elements of order

113

by conjugation. Therefore

P

₁₃

    

m

₁₁₃



8447918080

, which is a contradiction. So

13





 

G

. Similarly, we can prove that

11





 

G

.

If

3





 

G

, then by Lemma 1.9,





3

235126784,1 645887488, 6583549952

m



.

On the other hand, by Lemma 1.9,

3

2





_e

 

G

. Thus





3

    1

3

P



m

. Therefore

P

₃



3

and

  

3



3 m₃ 117563392, 822943744, 3291774976 n



  .



 



127

    1

127

1 16912465920

P



m

 

. Therefore

127

127

P



and

 

127

127

m

127

134225920

n







.

Since

29    

n

₁₂₇, we deduce that

29





 

G

. Now by Lemma 1.9,

3 29







_e

 

G

. Thus

P

₃ acts fixed point freely on the set of elements of order

29

by conjugation. Therefore

P m

₃

    

₂₉



1645887488

, which is a contradiction. So

3





 

G

. If



2, 5, 29,1 13,1 27







 

G

, then by Lemma 1.9,

2

2080895

m



,

5

235126784

m



,

29

1645887488

m



,

113

8447918080

m



,

127

16912465920

m



and

2 ,5 ,29 ,1 13 ,127

18 2 2 2 2

 

e

G





. Thus by

Lemma 1.4 and considering

m P



₅ , we conclude that

5

5

P



. Similarly,

P

₂₉



29

,

P

₁₁₃



113

and

127

127

P



.

If



   

G



2

, then since

2

18

 

e

G





, we

conclude that



e

 

G





1, 2, 2 , ,2

2



17



. Therefore

| | = 2 = 34093383680 + 266354560

+ 235126784 + 1645887488 + 6583549952 + 8447918080 + 16912465920

where

k k k k k k

, ,  ,  , , 

_{1 2 3 4 5} and

k

₆ are non-negative integers and

0

     

k k k k k k

_{1 2 3 4 5 6}



10

. It is easy to check that this equation has no solution.

If 5

 

G , then P₅ 5 and

 

5

5 ₅ 235126784 58781696₄

m

n _   . We know that

5

   

n G

. Hence

127





 

G

. Similarly, we can prove that if

29





 

G

or

113





 

G

, then

 

127





G

. So in any cases, we can assume that

 

127





G

.

Now we prove that



  

G



2,5,29,1 13,1 27



. Since

127





 

G

, we conclude that

P

₁₂₇



127

and

 

127

127

m

127

134225920

n







. We know that

127

   

n

G

. Hence

134225920    

G

. It follows that

  

G

2,5,29,1 13,1 27







.

Step 2.

G



2 5 29 113 127

k

 





, where

13

 

k

14

.

By the above discussion

P

₅



5

,

P

₂₉



29

,

113

113

P



and

P

₁₂₇



127

.

By Lemma 1.9,

2 127







_e

 

G

. Thus

P

₂ acts fixed point freely on the set of elements of order

127

by conjugation. Therefore

P m

₂

    

₁₂₇. Hence

P

₂

    2

14 . On the other hand, since

n

₁₂₇

   

G

, we deduce that

13

2    

G

. Hence

2    

13

P

₂ .

Step 3.

G

is not solvable.

If

G

is solvable, then by Lemma 1.7,

G

has a Hall



-subgroup

H

, where







5, 29,1 13,1 27



and all the Hall



-subgroups of

G

are conjugate and the

number of Hall



-subgroups of

G

is

 

14

 :

G

    2

G N H

. Since

G

is solvable, we conclude that

H

is solvable. Hence by Lemma 1.3, there are nonnegative integers



₁

, ,





_r,



₁

, ,





_s,



₁

, ,





_t

such that

 

1 1 1

5

29

r

113

s

127

t

n H



 







  _,





29

i



1  mod  5  

_,

113

i



1  mod  5





_,





127

i



1  mod  5

.

Since

G



2 5 29 113 127

k

 





_{, where}

13

 

k

14

, we conclude that



₁





_r

  1 



,

1 s

1









,



₁

 



_t

1

. Therefore

 

5

1

n H



. So

4



m G

₅

 





2

14

 

4 65536



, but we have

m G

₅

 



235126784

, which is a contradiction.

Step 4.

G Sz



 

128

.

Since

G

is a finite group which is not solvable, there is a normal series

1

(

N M G

(

(

such that

N

is a maximal solvable normal subgroup of

G

and

/

Lemma 1.8. Let

M N S

/

 

₁

S

_r, where

S

₁ is a non-abelian simple group and

S

₁

  

S

_r. Since

1

(

N M G

(

(

and

G



2 5 29 113 127

k

 





, where

13

 

k

14

, we conclude that

r



1

and

/

M N

is a non-abelian simple group. Since

3 |  



G

, we deduce that

3 |



M N

/

. We know that the group

 

Sz q

is only non-abelian simple group such that

 

3 |



Sz q

. Hence

M N Sz

/



 

128

and since

2 5 29 113 127

k

G



 





, where

13

 

k

14

,

we deduce that

N



1

and

G M Sz





 

128

.

Acknowledgement

The authors would like to thank the referees for a very careful reading of the paper and for all their insightful comments and valuable suggestions, which improve considerably the presentation of this paper. This work was partially supported by Center of Excellence of Algebraic Hyperstructures and its Applications of Tarbiat Modares University (CEAHA).

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