conjugacy in A_n

conjugacy in

A

n

Recall that

conjugacy classes

in the

symmetric group

S

n

are determined solely by

cycle type

. In the

alternating group

A

n

, however, this is not always true. A single conjugacy class in

S

n

that is

contained

in

A

n

may split into two distinct classes when considered as a subset of

A

n

. For example, in

S

3

,

(

1

⁢

2

⁢

3

)

and

(

1

⁢

3

⁢

2

)

are conjugate, since

(

2

⁢

3

)

⁢

(

1

⁢

2

⁢

3

)

⁢

(

2

⁢

3

)

=

(

1

⁢

3

⁢

2

)

but these two are not conjugate in

A

3

(note that

(

2

⁢

3

)

∉

A

3

).

Note in particular that the fact that

conjugacy

in

S

n

is determined by cycle type means that if

σ

∈

A

n

then all of its conjugates in

S

n

also lie in

A

n

.

The following

theorem

fully characterizes the behavior of conjugacy classes in

A

n

:

Theorem 1

.

A conjugacy class in

S

n

splits into two distinct conjugacy classes under the action of

A

n

if and only if its cycle type consists of

distinct

odd integers

. Otherwise, it remains a single conjugacy class in

A

n

.

Thus, for example, in

S

7

, the elements of the conjugacy class of

(

1

⁢

2

⁢

3

⁢

4

⁢

5

)

are all conjugate in

A

7

, while the elements of the conjugacy class of

(

1

⁢

2

⁢

3

)

⁢

(

4

⁢

5

⁢

6

)

split into two distinct conjugacy classes in

A

7

since there are two cycles of length

3

. Similarly, any conjugacy class containing an even-length cycle, such as

(

1

⁢

2

⁢

3

⁢

4

)

⁢

(

5

⁢

6

)

, splits in

A

7

.

We will prove the above theorem by proving the following statements:

•

A conjugacy class in

S

n

consisting solely of

even permutations

(i.e. that is contained in

A

n

) either is a single conjugacy class or is the

disjoint union

of two equal-sized conjugacy classes when considered under the action of

A

n

.

•

If

σ

∈

A

n

, then the elements of the conjugacy class of

σ

in

S

n

(which is just all elements of the same cycle type as

σ

) are conjugate in

A

n

if and only if

σ

commutes with some

odd permutation

.

•

σ

∈

S

n

does not commute with an odd permutation if and only if the cycle type of

σ

consists of

distinct

odd integers.

Throughout, we will denote by

𝒞

S

⁢

(

σ

)

the conjugacy class of

σ

under the action of

S

n

.

To prove the first statement, note that conjugacy is a transitive action. By the theorem that orbits of a

normal subgroup

are equal in size when the full group acts transitively, we see that if

σ

∈

A

n

, then

𝒞

S

⁢

(

σ

)

splits into

|

S

n

:

A

n

⁢

C

S

n

⁢

(

σ

)

|

classes under the action of

A

n

(recall that

C

G

⁢

(

x

)

, the

centralizer

of

x

, is simply the

stabilizer

of

x

under the conjugation action of

G

on itself). But since

|

S

n

:

A

n

|

is either

1

or

2

, we see that the conjugacy class of

σ

either remains a single class in

A

n

or splits into two classes.

Note also that the elements of

𝒞

S

⁢

(

σ

)

are all conjugate in

A

n

if and only if

A

n

⁢

C

S

n

⁢

(

σ

)

=

S

n

, which happens if and only if

C

S

n

⁢

(

σ

)

⊈

A

n

, which in turn is the case if and only if some odd permutation is in the centralizer of

σ

, which means precisely that

σ

commutes with some odd permutation. This proves the second statement.

To prove the third statement, suppose first that

σ

does not commute with an odd permutation. Clearly

σ

commutes with any cycle in its own cycle decomposition, so if

σ

contains a cycle of even length, that is an odd permutation with which

σ

commutes. So

σ

must consist solely of [

disjoint

] cycles of odd length. If two of these cycles have the same length, say

(

a

1

⁢

a

2

⁢

…

⁢

a

2

⁢

k

+

1

)

and

(

b

1

⁢

b

2

⁢

…

⁢

b

2

⁢

k

+

1

)

, then

(

(

a

1

⁢

b

1

)

⁢

…

⁢

(

a

2

⁢

k

+

1

⁢

b

2

⁢

k

+

1

)

)

⁢

(

a

1

⁢

a

2

⁢

…

⁢

a

2

⁢

k

+

1

)

⁢

(

b

1

⁢

b

2

⁢

…

⁢

b

2

⁢

k

+

1

)

⁢

(

(

a

1

⁢

b

1

)

⁢

…

⁢

(

a

2

⁢

k

+

1

⁢

b

2

⁢

k

+

1

)

)

-

1

=

(

a

1

⁢

a

2

⁢

…

⁢

a

2

⁢

k

+

1

)

⁢

(

b

1

⁢

b

2

⁢

…

⁢

b

2

⁢

k

+

1

)

so the

product

of

(

a

1

⁢

a

2

⁢

…

⁢

a

2

⁢

k

+

1

)

and

(

b

1

⁢

b

2

⁢

…

⁢

b

2

⁢

k

+

1

)

, and thus

σ

, commutes with the product of

2

⁢

k

+

1

transpositions

, which is an odd permutation. Thus all the cycles in the cycle decomposition of

σ

must have different [odd] lengths.

To prove the

converse

, we show that if the cycles in the cycle decomposition all have distinct lengths, then

σ

commutes precisely with the group generated by its cycles. It follows then that if all the distinct lengths are odd, then

σ

commutes only with these

permutations

, which are all even. Choose

σ

with distinct cycle lengths in its cycle decomposition, and suppose that

σ

commutes with some

element

τ

∈

S

n

.

Conjugation

preserves cycle length, so since

τ

commutes with

σ

and

σ

has all its cycles of distinct lengths, each cycle in

τ

must commute with each cycle in

σ

individually.

Now, choose a nontrivial cycle

τ

1

of

τ

, and choose

j

∈

τ

such that

σ

moves

j

(we can do this, since

σ

can have at most one cycle of length

1

and the cycle length of

τ

is greater than

1

). Let

σ

1

be the cycle of

σ

containing

j

. Then

τ

1

commutes with

σ

1

since

τ

commutes with

σ

, so

τ

1

is in the centralizer of

σ

1

, and it is not disjoint from

σ

1

. But the centralizer of a

k

-cycle

ρ

consists of products of powers of

ρ

and cycles disjoint from

ρ

. Thus

τ

1

is a power of

σ

1

. So each cycle in

τ

is a power of a cycle in

σ

, and we are done.

Title

conjugacy in

A

n

Canonical name

ConjugacyInAn

Date of creation

2013-03-22 17:18:04

Last modified on

2013-03-22 17:18:04

Owner

rm50 (10146)

Last modified by

rm50 (10146)

Numerical id

6

Author

rm50 (10146)

Entry type

Theorem

Classification

msc 20M30

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