simplicity of the alternating groups

simplicity of the alternating groups

Theorem 1

.

If

n

≥

5

, then the

alternating group

on

n

symbols,

A

n

, is simple.

Throughout the proof we extensively employ the

cycle notation

, with

composition

on the left, as is usual. The

symmetric group

on

n

symbols is denoted by

S

n

.

The following observation will be useful. Let

π

be a

permutation

written as

disjoint

cycles

π

=

(

a

1

,

a

2

,

…

,

a

k

)

⁢

(

b

1

,

b

2

,

…

,

b

l

)

⁢

…

⁢

(

c

1

,

…

,

c

m

)

It is easy to check that for every permutation

σ

∈

S

n

we
have

σ

⁢

π

⁢

σ

-

1

=

(

σ

⁢

(

a

1

)

,

σ

⁢

(

a

2

)

,

…

,

σ

⁢

(

a

k

)

)

⁢

(

σ

⁢

(

b

1

)

,

σ

⁢

(

b

2

)

,

…

⁢

σ

⁢

(

b

l

)

)

⁢

…

⁢

(

σ

⁢

(

c

1

)

,

…

,

σ

⁢

(

c

m

)

)

As a

consequence

, two permutations of

S

n

are

conjugate

exactly when
they have the same

cycle type

.

Two preliminary results will also be necessary.

Lemma 2

.

The set of cycles of length

3

generates

A

n

.

Proof.

A

product

of

3

-cycles is an

even permutation

, so the

subgroup

generated by all

3

-cycles is therefore

contained

in

A

n

. For the reverse

inclusion

, by definition every even permutation is the product of

even number

of

transpositions

. Thus, it suffices to show that the product of two transpositions can be written as a product of

3

-cycles. There are two possibilities. Either the two transpositions move an

element

in common, say

(

a

,

b

)

and

(

a

,

c

)

, or the two transpositions are disjoint, say

(

a

,

b

)

and

(

c

,

d

)

. In the former case,

(

a

,

b

)

⁢

(

a

,

c

)

=

(

a

,

c

,

b

)

,

and in the latter,

(

a

,

b

)

⁢

(

c

,

d

)

=

(

a

,

b

,

d

)

⁢

(

c

,

b

,

d

)

.

This establishes the first lemma.
∎

Lemma 3

.

If a

normal subgroup

N

◁

A

n

contains a

3

-cycle, then

N

=

A

n

.

Proof.

We will show that if

(

a

,

b

,

c

)

∈

N

, then the

assumption

of normality implies that any other

(

a

′

,

b

′

,

c

′

)

∈

N

. This is easy to show, because there is some permutation in

σ

∈

S

n

that under

conjugation

takes

(

a

,

b

,

c

)

to

(

a

′

,

b

′

,

c

′

)

, that is

σ

⁢

(

a

,

b

,

c

)

⁢

σ

-

1

=

(

σ

⁢

(

a

)

,

σ

⁢

(

b

)

,

σ

⁢

(

c

)

)

=

(

a

′

,

b

′

,

c

′

)

.

In case

σ

is odd, then (because

n

≥

5

) we can choose some transposition

(

d

,

e

)

∈

A

n

disjoint from

(

a

′

,

b

′

,

c

′

)

so that

σ

⁢

(

a

,

b

,

c

)

⁢

σ

-

1

=

(

d

,

e

)

⁢

(

a

′

,

b

′

,

c

′

)

⁢

(

d

,

e

)

,

that is,

σ

′

⁢

(

a

,

b

,

c

)

⁢

σ

′

⁣

-

1

=

(

d

,

e

)

⁢

σ

⁢

(

a

,

b

,

c

)

⁢

σ

-

1

⁢

(

d

,

e

)

=

(

a

′

,

b

′

,

c

′

)

where

σ

′

is even. This means that

N

contains all

3

-cycles,
as

N

◁

A

n

. Hence, by previous lemma

N

=

A

n

as
required.
∎

Proof of theorem.

Let

N

◁

A

n

be a non-trivial normal subgroup. We will
show that

N

=

A

n

. The proof now proceeds by cases. In each
case, the normality of

N

will allow us to reduce the proof to
Lemma 2 or to one of the previous cases.

Case 1.

Suppose that there exists a

π

∈

N

that, when written as disjoint
cycles, has a cycle of length at least

4

, say

π

=

(

a

1

,

a

2

,

a

3

,

a

4

,

…

)

⁢

…

Upon conjugation by

(

a

1

,

a

2

,

a

3

)

∈

A

n

, we obtain

π

′

=

(

a

1

,

a

2

,

a

3

)

⁢

π

⁢

(

a

3

,

a

2

,

a

1

)

=

(

a

2

,

a

3

,

a

1

,

a

4

,

…

)

⁢

…

Hence,

π

′

∈

N

, and hence

π

′

⁢

π

-

1

=

(

a

1

,

a

2

,

a

4

)

∈

N

also. Notice that the rest of the cycles cancel. By Lemma

3

,

N

=

A

n

.

Case 2.

Suppose that there exists a

π

∈

N

whose disjoint cycle
decomposition has at least two cycles of length 3, say

π

=

(

a

,

b

,

c

)

⁢

(

d

,

e

,

f

)

⁢

…

Conjugation by

(

c

,

d

,

e

)

∈

A

n

implies that

N

also contains

π

′

=

(

c

,

d

,

e

)

⁢

π

⁢

(

e

,

d

,

c

)

=

(

a

,

b

,

d

)

⁢

(

e

,

c

,

f

)

⁢

…

Hence,

N

also contains

π

′

⁢

π

=

(

a

,

d

,

c

,

b

,

f

)

⁢

…

. This reduces
the proof to Case 1.

Case 3.

Suppose that there exists a

π

∈

N

whose disjoint cycle
decomposition consists of exactly one

3

-cycle and an even (possibly
zero) number of transpositions. Hence,

π

⁢

π

is a

3

-cycle.
Lemma

3

can then be applied to

complete

the proof.

Case 4.

Suppose there exists a

π

∈

N

of the form

π

=

(

a

,

b

)

⁢

(

c

,

d

)

.
Conjugating by

(

a

,

e

,

b

)

with

e

distinct from

a

,

b

,

c

,

d

(at least
one such

e

exists, as

n

≥

5

) yields

π

′

=

(

a

,

e

,

b

)

⁢

π

⁢

(

b

,

e

,

a

)

=

(

a

,

e

)

⁢

(

c

,

d

)

∈

N

.

Hence

π

′

⁢

π

=

(

a

,

b

,

e

)

∈

N

. Again, Lemma

3

applies.

Case 5.

Suppose that

N

contains a permutation of the form

π

=

(

a

1

,

b

1

)

⁢

(

a

2

,

b

2

)

⁢

(

a

3

,

b

3

)

⁢

(

a

4

,

b

4

)

⁢

…

This time we conjugate by

(

a

2

,

b

1

)

⁢

(

a

3

,

b

2

)

.

π

′

=

(

a

2

,

b

1

)

⁢

(

a

3

,

b

2

)

⁢

π

⁢

(

a

3

,

b

2

)

⁢

(

a

2

,

b

1

)

=

(

a

1

,

a

2

)

⁢

(

a

3

,

b

1

)

⁢

(

b

2

,

b

3

)

⁢

(

a

4

,

b

4

)

⁢

…

.

Observe that

π

′

⁢

π

=

(

a

1

,

a

3

,

b

2

)

⁢

(

a

2

,

b

3

,

b

1

)

⁢

…

,

which reduces the proof to Case 2.

Since there exists at least one non-identity

π

∈

N

, and since this

π

is covered by one of the above cases, we conclude that

N

=

A

n

, as was to be shown.

Title

simplicity of the alternating groups

Canonical name

SimplicityOfTheAlternatingGroups

Date of creation

2013-03-22 13:07:57

Last modified on

2013-03-22 13:07:57

Owner

rmilson (146)

Last modified by

rmilson (146)

Numerical id

16

Author

rmilson (146)

Entry type

Result

Classification

msc 20D06

Classification

msc 20E32

Related topic

ExamplesOfFiniteSimpleGroups

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