Sylow theorems, proof of

Sylow theorems, proof of

We let

G

be a group of order

p

m

⁢

k

where

p

∤

k

and prove Sylow’s

theorems

.

First, a fact which will be used several times in the proof:

Proposition 1

.

If

p

divides the size of every

conjugacy class

outside the center then

p

divides the order of the center.

Proof.

This follows from the

class equation

:

|

G

|

=

|

Z

⁢

(

G

)

|

+

∑

[

a

]

≠

Z

⁢

(

G

)

|

[

a

]

|

If

p

divides the

left hand side

, and divides all but one entry on the right hand side, it must divide every entry on the right side of the equation, so

p

|

Z

(

G

)

.
∎

Proposition 2

.

G

has a

Sylow p-subgroup

Proof.

By

induction

on

|

G

|

. If

|

G

|

=

1

then there is no

p

which divides its order, so the condition is

trivial

.

Suppose

|

G

|

=

p

m

⁢

k

,

p

∤

k

, and the holds for all groups of smaller order. Then we can consider whether

p

divides the order of the center,

Z

⁢

(

G

)

.

If it does, then by Cauchy’s theorem, there is an

element

f

of

Z

⁢

(

G

)

of order

p

, and therefore a

cyclic subgroup

generated by

f

,

⟨

f

⟩

, also of order

p

. Since this is a

subgroup

of the center, it is normal, so

G

/

⟨

f

⟩

is

well-defined

and of order

p

m

-

1

⁢

k

. By the

inductive hypothesis

, this group has a subgroup

P

/

⟨

f

⟩

of order

p

m

-

1

. Then there is a corresponding subgroup

P

of

G

which has

|

P

|

=

|

P

/

⟨

f

⟩

|

⋅

|

⟨

f

⟩

|

=

p

m

.

On the other hand, if

p

∤

|

Z

⁢

(

G

)

|

then consider the conjugacy classes not in the center. By the

proposition

above, since

Z

⁢

(

G

)

is not

divisible

by

p

, at least one conjugacy class can’t be. If

a

is a representative of this class then we have

p

∤

|

[

a

]

|

=

[

G

:

C

(

a

)

]

, and since

|

C

(

a

)

|

⋅

[

G

:

C

(

a

)

]

=

|

G

|

,

p

m

∣

|

C

(

a

)

|

. But

C

⁢

(

a

)

≠

G

, since

a

∉

Z

⁢

(

G

)

, so

C

⁢

(

a

)

has a subgroup of order

p

m

, and this is also a subgroup of

G

.
∎

Proposition 3

.

The

intersection

of a Sylow p-subgroup with the

normalizer

of a Sylow p-subgroup is the intersection of the subgroups. That is,

Q

∩

N

G

⁢

(

P

)

=

Q

∩

P

.

Proof.

If

P

and

Q

are Sylow p-subgroups, consider

R

=

Q

∩

N

G

⁢

(

P

)

. Obviously

Q

∩

P

⊆

R

. In

addition

, since

R

⊆

N

G

⁢

(

P

)

, the

second isomorphism theorem

tells us that

R

⁢

P

is a group, and

|

R

⁢

P

|

=

|

R

|

⋅

|

P

|

|

R

∩

P

|

.

P

is a subgroup of

R

⁢

P

, so

p

m

∣

|

R

P

|

. But

R

is a subgroup of

Q

and

P

is a Sylow p-subgroup, so

|

R

|

⋅

|

P

|

is a

multiple

of

p

. Then it must be that

|

R

⁢

P

|

=

p

m

, and therefore

P

=

R

⁢

P

, and so

R

⊆

P

. Obviously

R

⊆

Q

, so

R

⊆

Q

∩

P

.
∎

The following construction will be used in the

remainder

of the proof:

Given any Sylow p-subgroup

P

, consider the set of its

conjugates

C

. Then

X

∈

C

↔

X

=

x

⁢

P

⁢

x

-

1

=

{

x

⁢

p

⁢

x

-

1

|

∀

p

∈

P

}

for some

x

∈

G

. Observe that every

X

∈

C

is a Sylow p-subgroup (and we will show that
the

converse

holds as well). We let

G

act on

C

by

conjugation

:

g

⋅

X

=

g

⋅

x

⁢

P

⁢

x

-

1

=

g

⁢

x

⁢

P

⁢

x

-

1

⁢

g

-

1

=

(

g

⁢

x

)

⁢

P

⁢

(

g

⁢

x

)

-

1

This is clearly a

group action

, so we can consider the orbits of

P

under it; this remains true if we only consider elements from some subset of

G

. Of course, if all

G

is used then there is only one orbit, so we restrict the action to a Sylow p-subgroup

Q

. the orbits

O

1

,

…

,

O

s

, and let

P

1

,

…

,

P

s

be representatives of the corresponding orbits. By the

orbit-stabilizer theorem

, the size of an orbit is the index of the

stabilizer

, and under this action the stabilizer of any

P

i

is just

N

Q

⁢

(

P

i

)

=

Q

∩

N

G

⁢

(

P

i

)

=

Q

∩

P

, so

|

O

i

|

=

[

Q

:

Q

∩

P

i

]

.

There are two easy results on this construction. If

Q

=

P

i

then

|

O

i

|

=

[

P

i

:

P

i

∩

P

i

]

=

1

. If

Q

≠

P

i

then

[

Q

:

Q

∩

P

i

]

>

1

, and since the index of any subgroup of

Q

divides

Q

,

p

∣

|

O

i

|

.

Proposition 4

.

The number of conjugates of any Sylow p-subgroup of

G

is

congruent

to

1

modulo

p

In the construction above, let

Q

=

P

1

. Then

|

O

1

|

=

1

and

p

∣

|

O

i

|

for

i

≠

1

. Since the number of conjugates of

P

is the sum of the number in each orbit, the number of conjugates is of the form

1

+

k

2

⁢

p

+

k

3

⁢

p

+

⋯

+

k

s

⁢

p

, which is obviously congruent to

1

modulo

p

.

Proposition 5

.

Any two Sylow p-subgroups are conjugate

Proof.

Given a Sylow p-subgroup

P

and any other Sylow p-subgroup

Q

, consider again the construction given above. If

Q

is not conjugate to

P

then

Q

≠

P

i

for every

i

, and therefore

p

∣

|

O

i

|

for every orbit. But then the number of conjugates of

P

is divisible by

p

, contradicting the previous result. Therefore

Q

must be conjugate to

P

.
∎

Proposition 6

.

The number of subgroups of

G

of order

p

m

is congruent to

1

modulo

p

and is a factor of

k

Proof.

Since conjugates of a Sylow p-subgroup are precisely the Sylow p-subgroups, and since a Sylow p-subgroup has

1

modulo

p

conjugates, there are

1

modulo

p

Sylow p-subgroups.

Since the number of conjugates is the index of the normalizer, it must be

|

G

:

N

G

(

P

)

|

. Since

P

is a subgroup of its normalizer,

p

m

∣

N

G

(

P

)

, and therefore

|

G

:

N

G

(

P

)

|

∣

k

.
∎

Title

Sylow theorems

, proof of

Canonical

name

SylowTheoremsProofOf

Date of creation

2013-03-22 12:51:02

Last modified on

2013-03-22 12:51:02

Owner

Henry (455)

Last modified by

Henry (455)

Numerical id

12

Author

Henry (455)

Entry type

Proof

Classification

msc 20D20

Related topic

SylowPSubgroup

Related topic

SylowsThirdTheorem

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