normal subgroups of the symmetric groups

normal subgroups of the symmetric groups

Theorem 1

.

For

n

≥

5

,

A

n

is the only proper nontrivial

normal subgroup

of

S

n

.

Proof.

This is essentially a

corollary

of the simplicity of the

alternating groups

A

n

for

n

≥

5

. Let

N

⊴

S

n

be normal. Clearly

N

∩

A

n

⊴

A

n

. But

A

n

is simple, so

N

∩

A

n

=

A

n

or

N

∩

A

n

=

{

e

}

. In the first case, either

N

=

A

n

, or else

N

also contains an odd (

http://planetmath.org/SignatureOfAPermutation

)

permutation

, in which case

N

=

S

n

. In the second case, either

N

=

{

e

}

or else

N

consists solely of one or more

odd permutations

in

addition

to

{

e

}

. But if

N

contains two distinct odd permutations,

σ

and

τ

, then either

σ

2

≠

e

or

σ

⁢

τ

≠

e

, and both

σ

2

and

σ

⁢

τ

are even (

http://planetmath.org/SignatureOfAPermutation

), contradicting the

assumption

that

N

contains only odd nontrivial permutations. Thus

N

must be of order

2

, consisting of a single odd permutation of order 2 together with the

identity

.

It is easy to see, however, that such a

subgroup

cannot be normal. An odd permutation of order

2

,

σ

, has as its cycle decomposition one or more (an

odd number

, in fact, though this does not matter here) of

disjoint

transpositions

. Suppose wlog that

(

1

⁢

2

)

is one of these transpositions. Then

τ

=

(

1

⁢

3

)

⁢

σ

⁢

(

1

⁢

3

)

=

(

1

⁢

3

)

⁢

(

1

⁢

2

)

⁢

(

…

)

⁢

(

1

⁢

3

)

takes

2

to

3

and thus is neither

σ

nor

e

. So this group is not normal.
∎

If

n

=

1

,

S

1

is the

trivial group

, so it has no nontrivial [normal] subgroups.

If

n

=

2

,

S

2

=

C

2

, the unique group on

2

elements

, so it has no nontrivial [normal] subgroups.

If

n

=

3

,

S

3

has one nontrivial proper normal subgroup, namely the group generated by

(

1

⁢

2

⁢

3

)

.

S

4

is the most interesting case for

n

≤

5

. The

arguments

in the

theorem

above do not apply since

A

4

is not simple. Recall that a normal subgroup must be a union of

conjugacy classes

of elements, and that

conjugate elements

in

S

n

have the same

cycle type

. If we examine the sizes of the various conjugacy classes of

S

4

, we get

Cycle Type

Size

4

6

3,1

8

2,2

3

2,1,1

6

1,1,1,1

1

A subgroup of

S

4

must be of order

1

,

2

,

3

,

4

,

6

,

8

, or

12

(the factors of

|

S

4

|

=

24

). Since each subgroup must contain

{

e

}

, it is easy to see that the only possible nontrivial normal subgroups have orders

4

and

12

. The order

4

subgroup is

H

=

{

e

,

(

1

⁢

2

)

⁢

(

3

⁢

4

)

,

(

1

⁢

3

)

⁢

(

2

⁢

4

)

,

(

1

⁢

4

)

⁢

(

2

⁢

3

)

}

, while the order

12

subgroup is

A

4

.

A

4

is obviously normal, being of index

2

, and one can easily check that

H

≅

V

4

is also normal in

S

4

. So these are the only two nontrivial proper normal subgroups of

S

4

.

Title

normal subgroups of the

symmetric groups

Canonical

name

NormalSubgroupsOfTheSymmetricGroups

Date of creation

2013-03-22 17:31:38

Last modified on

2013-03-22 17:31:38

Owner

rm50 (10146)

Last modified by

rm50 (10146)

Numerical id

8

Author

rm50 (10146)

Entry type

Theorem

Classification

msc 20B35

Classification

msc 20E07

Classification

msc 20B30

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