Vol 2, No 5 (2011)

International Journal of Mathematical Archive-2(5), May - 2011

, Page: 753-757

Available online through

www.ijma.info

ON SYLOW NUMBERS OF THE ALTERNATING SIMPLE GROUPS

1

Rulin Shen*

Department of Mathematics, Hubei University for Nationalities, Enshi, Hubei, P.R. China, 445000 E-mail: [email protected]

Ronghua Tan

Department of Mathematics, Hubei University for Nationalities, Enshi, Hubei, P.R. China, 445000

(Received on: 29-04-11; Accepted on: 11-05-11)

---

ABSTRACT

I

n this paper we give the formula of numbers of Sylow subgroups of alternating simple groups. This partially solves a

problem posed by Jiping Zhang.

Keywords: symmetric groups, alternating groups, Sylow numbers.

MR (2000): 20D60, 20D06.

---

1. INTRODUCTION AND LEMMAS:

The structure of normalizer of Sylow 2-subgroup of symmetric groups

S

_nis studied by P. Hall (see Lemma 4 of [1], or

[2]), he proved that Sylow 2-subgroup is self-normalized. The study in the numbers of Sylow subgroups (also called the Sylow numbers) is due to Professor Jiping Zhang [5]. He proved a conjecture of Huppert's. Then he put forward a problem in the end of his paper [5]. determine the number of Sylow

p

-subgroup of finite simple groups. In this paper

we give the formula of Sylow numbers of the alternating simple group

A

_n, which partially solves above Zhang's

problem. Let

Ω =

{1, 2,3,..., }

n

,

S

( )

Ω

or

S

_n be symmetric group on

Ω

and

P

_n be a Sylow

p

-subgroup of

symmetric group

S

_n. Then

|

| ( !)

s n( )

n p

P

=

n

=

p

, where

s n

( )

n

n

₂

n

₃

...

p

p

p

=

+

+

+

.

Suppose that r ₁ r 1

...

_{1 0}

r r

n

=

a p

+

a

−

p

−

+

+

a p

+

a

is the

p

-adic expansion of the number

n

, in which

0

≤

a

_i

≤

p

−

1

,

i

=

0,1,...,

r

and

a

_r

≠

0

. In the following paper, the number

p

is always a prime. We will prove:

Theorem: The number of Sylow

p

-subgroup of the alternating group

A n

n

(

≥

6)

is

1 2 2 ... ( )

0 1

!

! !... !

s n

(

1)

a a rar

r

n

a a

a p

p

−

+ + + .

---

Rulin Shen* and Ronghua Tan On sylow numbers of the alternating simple groups Next we will cite some lemmas.

Lemma: 1 Let

X

=

{1, 2,..., }

p

,

Y

=

{1, 2,..., , , ,...,

t i i

_{1 2}

i

_{p t}−

}

and

X

∩

Y

=

{1, 2,..., }

t

with

1

≤ ≤

t

p

−

1

.

Suppose that

a b

,

are

p

-cycles on

X

and

Y

, respectively. Then the group

a b

,

is not a

p

-group.

Proof: Suppose that

a b

,

is a

p

-group. Clearly, it is a transitive permutation on the set

X

∩

Y

, and its degree

n

is

|

X

∩

Y

| 2

=

p

−

t

. Since

1

≤ ≤

t

p

−

1

, we have

1

<

n

<

2

p

. On the other hand, the degree of

a b

,

is a

multiple of

p

since it is a

p

-group. In fact, we only need to choose a subgroup

P

of

a b

,

whose kernel is 1

and order is maximal, then

n

=

|

a b

,

:

P

|

. So that

p n

|

, which contradicts the facts that

p

<

n

<

2

p

.

Lemma: 2 The normalizer of Sylow

p

-subgroup of the symmetric group

S

_p (

p

≥

5

) is the Frobenius

group

Z

_p

:

Z

_p−₁.

Proof: Let

Z

p

=

(123... )

p

is a Sylow

p

-subgroup of

S

p. Choose an element

b

=

(12435687...)

. It is

easy to check that

(123... )

b p

p

∈

Z

, and hence

:

,

(

)

p

p S p

Z

a b

≤

N

Z

. On the other hand,

S

_p has

(

p

−

1)!

elements with order

p

exactly, so the number of subgroups of order

p

in

S

_p is

(

p

−

2)!

. Obviously, the number

|

:

(

) |

p

p S p

S

N

Z

of Sylow

p

-subgroups is also

(

p

−

2)!

$, thus

|

(

) |

(

1)

p

S p

N

Z

=

p p

−

. Therefore,

(

)

p

S p

N

Z

is

a Frobenius group with order

p p

(

−

1)

.

the wreath product of

A

and

B

, in which

B

regarded as a

2. Proof of Theorem: Denote by

permutation group. Denote by , .

It is known that the Sylow

p

-subgroup r

p

P

of r

p

S

is (see 1.6.19 in [4]). Of course, it is a transitive

group with degree ( )

r

s p

P

. If

G

acts on the set

∆

and

X

⊆ ∆

, then

N

_X stands for the global stabilizer of subset

X

in

G

. We will use the following steps to complete the proof of the Theorem.

Step: 1 If

n

=

p

r, then

Let

_X

₌

{1, 2,...,

_p

r

}

_,

{ (

1) 1, (

1) 2,..., (

1)

}

i

X

=

p i

−

+

p i

−

+

p i

−

+

p

, where

1

_i

_p

r−1

≤ ≤

. Set

D

=

{

X X

1

,

2

,...,

X

_pr−1

}

. Then

Clearly,

S X

( )

_D must include a Sylow

p

-subgroup

P

_pr of

S X

( )

.

1

Rulin Shen* and Ronghua Tan On sylow numbers of the alternating simple groups

the set of the fundamental subgroups in

S X

( )

. Set

( )

(

r

) {

|

( ),

r

(

)}.

x x x

S X p p p

Fun

P

K

x

S X

K

P

Syl

K

∆ =

=

∈

∩

∈

By the definition of

∆

, we have

{

|1

r 1

}

i

K

≤ ≤

i

p

−

⊆ ∆

. Suppose that there exists an element

g

∈

S X

( )

such

that

K

g

∈ ∆

, and

K

g

≠

K

_i for each

i

. Now we let g

:

( (

₁

))

K

=

S g X

. of course

g X

(

₁

)

≠

K

_i for each

i

. By the

definition of

∆

, we know that r

( ( ))

g

p p

K

∩

P

∈

Syl S X

. Then there exists a Sylow

p

-subgroup

P

of

K

g such

that

P

≤

P

_pr. Without loss of generality, we can assume that

a

=

(123... )

p

∈

K

∩

P

_n, then

a P

,

≤

P

_pr. Suppose

that

X

i

∩

g X

(

₁

)

≠ ∅

for some

i

, and set

|

X

i

∩

g X

(

1

) |

=

t

, where

1

≤ ≤

t

p

−

1

. Then we can assume

that

g X

(

₁

) {1, 2,..., , , ,...,

=

t i i

_{1 2}

i

_{p t}−

}

. By the Lemma 1, for any

p

-cycle

b

on

g X

(

1

)

, we have

a b

,

is not a

p

-group. Then such

P

does not exist, contradicts. So

X

i

∩

g X

(

₁

)

= ∅

for any

i

, that

is

(

₁

)

(

₁

)

(

pr₁1

)

i i

g X

∩

X

=

g X

∩ ∪

=−

X

= ∅

,which contradicts the fact that

( )

g

K

≤

S X

.

Thus

{

|1

r1

}

i

K

≤ ≤

i

p

−

= ∆

. Moreover, by the definition of

∆

, we know that r

(

)

x x

p p

K

∩

P

∈

Syl

K

for

any

K

x

∈ ∆

. If _{S X( )}

(

r

)

p

g

∈

N

P

, i.e., r r

g

p p

P

=

P

, then r

(

)

xg xg

p p

K

∩

P

∈

Syl

K

, and hence

K

xg

∈ ∆

.

Thus

g

∈

N

_{S X( )}

( )

∆

, then

Without loss of generality, we suppose that r

p

P

is a Sylow

p

-subgroup of of course, Now we

denote by

1

1

:

r

,

:

r

p

p _p

A

=

S

−

B

=

S

− , _A r

,

_B r

p p

P

=

P

∩

A P

=

P

∩

B

. Since

G

=

A B

:

, we have

N

_B

(

P

_B

)

≅

/

(

r

/ )

(

r

) /

(

r

) /

(

r

)

(

r

) /

(

r

)

G A _p G _p G _p G _p G _p A _p

N

P A A

=

N

P

A A

≅

N

P

A

∩

N

P

=

N

P

N

P

, so that

(

r

)

(

r

) :

(

)

G p A p B B

N

P

=

N

P

N

P

. In the sequel, we determine the structure of

N

_A

(

P

_pr

)

. Obviously,

Choose an element

x

=

( ,

x x

1 2

,...,

x

_pr−1

;1)

∈

N

_A

(

P

_pr

)

, then we have r

x p

g

∈

P

for any element

g

=

( ,

g g

1 2

,...,

g

_pr−1

;

g

0

)

∈

P

_pr, where That is

Since

g

₀ can be chosen randomly, we can get

(

)

p

i S p

Rulin Shen* and Ronghua Tan On sylow numbers of the alternating simple groups

is a transitive group with degree

p

s p( r−1) , then for any hence _i g₀

i

x

=

x

. Furthermore,

pairs

,

{1, 2,...,

r 1

}

i j

∈

p

− , there exists an element such that

_i

g0

₌

_j

_{, and}

hence

x

1

=

x

2

=

...

=

x

_pr−1. Conversely, if

x

=

( ,

x x

1 2

,...,

x

_pr−1

;1)

and ₁

(

)

p

S p

x

∈

N

Z

, then

x

∈

N

_A

(

P

_pr

)

. Thus

1 1

(

r

)

r

:

p

A p p p

N

P

=

Z

−

Z

− . So we have

N

_A

(

P

_pr

)

=

N

_A

(

P

_pr

) :

N

_B

(

P

_B

)

≅

1 1

(

pr

:

) :

(

)

p p B B

Z

−

Z

−

N

P

. Therefore,

1

1 1 1

(

r

)

(

r

:

) :

(

r

)

r r

p p

p

S p p p S p

N

P

Z

−

Z

N

P

−

−

−

≅

. By the induction for

r

, we can obtain

Step: 2 If

n

=

a p

_r r

+

k

, then where

0

_≤

_k

_<

_p

r_.

Let

Ω =

{1, 2,3,..., }

n

,

{

i1

,

i 1

1,...,

i 1 r

1}

i

X

=

p

−

p

−

+

p

−

+

p

−

in which

1

≤ ≤

i

a

_r and ₀

\

ar₁

i i

X

= Ω ∪

=

X

. Set

D

=

{

X X X

0, 1

,

2

,...,

X

_pr−1

}

. Then

Clearly,

S

( )

Ω

_D includes a Sylow

p

-subgroup

P

_n of

S

( )

Ω

. Denote by

K

_i

:

=

S X

(

_i

),

1

≤ ≤

i

a

_r, and

K

:

=

K

₁.

Set _{( )}

( ) {

x

|

( ),

x

(

x

)}.

S n n p

Fun

Ω

P

K

x

S

K

P

Syl K

∆ =

=

∈

Ω

∩

∈

In this case,

∆

is a maximal set of pair-wise commuting fundamental subgroups of

K

in

S

_pr. It is clear

that

|1

r 1

}

i

K

≤ <

i

p

−

⊆ ∆

. If there exists an element

g

∈

S

( )

Ω

such that

K

g

∈ ∆

and

K

g

≠

K

_i for each

i

.

Denote by

_K

g

:

₌

_{S g X}

( (

₁

))

. Obviously,

g X

(

₁

)

≠

K

_i for each

i

. Since r

( ( ))

g

p p

K

∩

P

∈

Syl S

Ω

, then there is a

Sylow

p

-subgroup

P

of

K

g such that

P

≤

P

_n. Now we can assume that

a

=

(123... )

p

∈

K

∩

P

_n ,

then

a P

,

≤

P

_n. Suppose that

X

_i

∩

g X

(

₁

)

≠ ∅

for some

i

, and set

|

X

_i

∩

g X

(

₁

) |

=

t

, where

1

≤ ≤

t

p

−

1

.

Without loss of generality, we can assume that

g X

(

₁

) {1, 2,..., ,

=

t

i i

₁

, ,...,

₂

i

_{p t}−

}

. Similarly, for any

p

-cycle

b

on

g X

(

₁

)

, we have

a b

,

is not a

p

-group by Lemma 1. Then such

P

does not exist, contradicts. So

1

(

)

i

X

∩

g X

= ∅

for any

i

, then

g X

(

₁

)

⊆

X

₀, which contradicts the fact that

| (

g X

₁

) | |

>

X

₀

|

. Therefore,

1

{

|1

r

}

i

K

≤ ≤

i

p

−

= ∆

. Then

Rulin Shen* and Ronghua Tan On sylow numbers of the alternating simple groups

It is easy to see that r r

a

n k p

P

=

P

×

P

, so

Step: 3 Now we continue to decompose

k

into the sum

a

_r−₁

p

r−1

+

k

₁, in which

1 1

0

r

k

p

−

≤

<

$. Next we repeat to

decompose

k

₁, and so on. At last we can get the

p

-adic expansion r ₁ r 1

...

_{1 0}

r r

a p

+

a

−

p

−

+

+

a p

+

a

by

r

steps.

Using the induction for

r

by the conclusions of step 1 and 2, we can obtain the result of the normalizer for general

n

:

In the case of

p

=

2

, it is not hard to see

|

( ) | |

|

n

S n n

N

P

=

P

since

a

_i

=

0

or 1,

then

S

_n has a self-normalized Sylow 2-subgroup. Since

|

S

_n

:

A

_n

| 2

=

, we have the Sylow

p

-number of

S

_n is

same as the one of

A

_n for

p

>

2

. If

p

=

2

, we use the result of Kondrat'ev’s [3]:

A

_n has a self-normalized

Sylow 2-subgroup for

n

≥

6

.

So the number of Sylow

p

-subgroups of the alternating group

A n

_n

(

≥

6)

is

1 2 2 ... ( )

0 1

!

! !... !

s n

(

1)

a a rar

r

n

a a

a p

p

−

+ + + .

REFERENCE:

[1] Carter R., Fong P., The Sylow 2-subgroups of the finite classical groups, J. Algebra, 1(1964) 139-151.

[2] Dmitruk, Yu. V., Sushchanskii V. I., Structure of Sylow 2-subgroups of the alternating groups and normalizers of Sylow subgroups in the symmetric and alternating groups, Ukrainian Mathematical Journal, 3(1981)235-241.

[3] Kondrat’ev A. S., Normalizers of the Sylow 2-Subgroups in Finite Simple Groups, Mathematical Notes 78(34), 338-346.

[4] Robinson D. J. S., A Course in the Theory of Groups, Springer-Verlag, 1982.

[5] Zhang J., Sylow numbers of finite groups, J. Algebra., 176(1995), 111-12.

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