conjugacy classes in the symmetric group S_n

conjugacy classes in the symmetric group

S

n

We start by providing an alternate proof that in

S

n

, every

permutation

has a cycle decomposition, and we prove that the cycle decomposition is

essentially unique

.

Theorem 1

.

Every permutation in

S

n

has a cycle decomposition that is unique up to

ordering

of the cycles and up to a

cyclic permutation

of the

elements

within each cycle.

Proof.

We construct the cycle decomposition for

σ

∈

S

n

. Let

X

=

{

1

,

…

,

n

}

, and regard

σ

as acting on

X

. Let

G

=

⟨

σ

⟩

be the

subgroup

of

S

n

generated by

σ

. Then

G

acts on

X

, so by the

orbit-stabilizer theorem

, it

partitions

X

into a unique set of orbits. In

addition

, for any orbit

G

⁢

x

, we have that

σ

i

⁢

x

↔

σ

i

⁢

G

x

is a

bijection

, where

G

x

is the

stabilizer

of

x

in

G

.

Now,

G

=

⟨

σ

⟩

is cyclic and thus

G

/

G

x

is cyclic as well; its order is the smallest power

d

of

σ

such that

σ

d

∈

G

x

. Note also that

d

=

|

G

x

|

=

[

G

:

G

x

]

. Using the explicit bijection above, we see that the unique cosets of

G

x

in

G

are

G

x

,

σ

⁢

G

x

,

…

,

σ

d

-

1

⁢

G

x

and that the elements of

G

⁢

x

are

x

,

σ

⁢

x

,

…

,

σ

d

-

1

⁢

x

Thus on any orbit of size

d

,

σ

is a

d

-cycle. This shows that a cycle decomposition exists.

Uniqueness now follows easily, since the cycle determined by

σ

on an element of order

d

is determined uniquely by construction from the element. Choosing a different element in the same orbit, say

σ

j

⁢

x

, gives instead

σ

j

⁢

x

,

σ

j

+

1

⁢

x

,

…

,

σ

d

-

1

⁢

x

,

x

,

σ

⁢

x

,

…

,

σ

j

-

1

⁢

x

which is the same cycle permuted left by

j

.
∎

We can thus write

Definition 1

.

If

σ

∈

S

n

and

σ

is written as the

product

of the

disjoint

cycles of lengths

n

1

,

…

,

n

k

with

n

i

≤

n

i

+

1

for each

i

<

k

, then

n

1

,

…

,

n

k

is the

cycle type

of

σ

.

The above

theorem

proves that the cycle type is well-defined.

Theorem 2

.

Two permutations

σ

,

τ

∈

S

n

are

conjugate

if and only if they have the same cycle type.

Proof.

Assume first that

σ

and

τ

are conjugate; say

τ

=

σ

1

⁢

σ

⁢

σ

1

-

1

. Write

σ

as a product of disjoint cycles

(

α

1

⁢

…

⁢

α

a

)

⁢

(

β

1

⁢

…

⁢

β

b

)

⁢

(

…

)

To show that

σ

and

τ

have the same cycle type, it clearly suffices to show that if

j

follows

i

in the cycle decomposition of

σ

, then

σ

1

⁢

(

j

)

follows

σ

1

⁢

(

i

)

in the cycle decomposition of

τ

. But suppose

σ

⁢

(

i

)

=

j

. Then

τ

⁢

(

σ

1

⁢

(

i

)

)

=

σ

1

⁢

σ

⁢

σ

1

-

1

⁢

(

σ

1

⁢

(

i

)

)

=

σ

1

⁢

σ

⁢

(

i

)

=

σ

1

⁢

(

j

)

and we are done.

Now suppose

σ

and

τ

have the same cycle type. Write the cycle decomposition for each permutation in such a way that the cycles are listed in nondecreasing order of their length (including cycles of length 1). We then have (for example)

σ

=

(

a

1

)

⁢

(

a

2

⁢

a

3

⁢

a

4

)

⁢

(

a

5

⁢

…

⁢

a

n

)

τ

=

(

b

1

)

⁢

(

b

2

⁢

b

3

⁢

b

4

)

⁢

(

b

5

⁢

…

⁢

b

n

)

Define

σ

1

to be the permutation that takes

a

i

to

b

i

. Clearly

σ

∈

S

n

, since each of

1

,

…

,

n

appears exactly once among the

a

i

and once among the

b

i

. But also, since the cycle types of

σ

and

τ

match, we see that

σ

1

⁢

σ

⁢

σ

1

-

1

⁢

(

b

i

)

=

σ

1

⁢

σ

⁢

(

a

i

)

=

σ

1

⁢

(

a

j

)

=

b

j

where

a

j

,

b

j

are the ”‘next”’ elements in their respective cycles. Thus

τ

=

σ

1

⁢

σ

⁢

σ

1

-

1

and we are done.
∎

Corollary 3

.

The number of

conjugacy classes

in

S

n

is the number of partitions of

n

.

Proof.

Each distinct cycle type in

S

n

represents a distinct partition of

n

, and each cycle type represents a conjugacy class. The result follows.
∎

We can give an explicit for the size of each conjugacy class in

S

n

.

Theorem 4

.

Suppose

σ

∈

S

n

, and let

m

1

,

m

2

,

…

⁢

m

r

be the

distinct

integers

(including

1

if applicable) in the cycle type of

σ

, and let there be

k

i

cycles of order

m

i

in

σ

. (Thus

∑

k

i

⁢

m

i

=

n

). Then the number of conjugates of

σ

is exactly

n

!

(

k

1

!

⁢

m

1

k

1

)

⁢

(

k

2

!

⁢

m

2

k

2

)

⁢

⋯

⁢

(

k

r

!

⁢

m

r

k

r

)

Proof.

The size of a conjugacy class is the number of cycles of the given cycle type. Choose a cycle type, and order the cycles in some order. Consider the

n

!

possible assignments of the integers from

1

to

n

into the ”‘holes”’ in the cycles. Call two such

arrangements

equivalent

if they define the same permutation. It is clear that this is an

equivalence relation

, and that the

relation

partitions the arrangements. We will determine the size of each

equivalence class

.

Consider a particular arrangement (i.e. permutation), and consider the

k

i

cycles of order

m

i

in that permutation. Each of the cycles can be written in

m

i

different ways (by starting with a different element). Also, the cycles themselves can appear in any of

k

i

!

possible orders while still representing the same permutation. Thus if, for example,

k

i

=

2

and the first cycle contains

a

1

,

…

⁢

a

l

while the second contains

b

1

,

…

,

b

l

, that is the same permutation as if the first cycle

contained

b

1

,

…

,

b

l

and the second contained

a

1

,

…

,

a

l

. So there are

k

i

!

⁢

m

i

k

i

equivalent permutations considering the cycles of order

m

i

. So the total number of permutations, considering each of the possible cycle orders, equivalent to the given permutation is

(

k

1

!

⁢

m

1

k

1

)

⁢

(

k

2

!

⁢

m

2

k

2

)

⁢

…

⁢

(

k

r

!

⁢

m

r

k

r

)

and the result follows.
∎

For example, let

n

=

5

and let’s compute the size of the conjugacy class containing the permutation

(

2

⁢

3

)

⁢

(

4

⁢

5

)

. In this case,

r

=

2

,

m

1

=

1

,

k

1

=

1

m

2

=

2

,

k

2

=

2

and the

skeleton

for the permutation is

(

⋅

)

(

⋅

⋅

)

(

⋅

⋅

)

There are

5

!

=

120

ways of placing

1

,

2

,

3

,

4

,

5

into this skeleton. However, given a particular choice, say

(

2

)

⁢

(

1

⁢

3

)

⁢

(

4

⁢

5

)

, the first

2

-cycle may be written as

(

1

⁢

3

)

or as

(

3

⁢

1

)

, while the second may be written as either

(

4

⁢

5

)

or

(

5

⁢

4

)

. In addition, either pair can appear first (

k

2

=

2

). The number of choices for ways to represent the

2

-cycles is thus

2

!

⋅

2

2

=

8

. The

formula

in the theorem indeed predicts that there are

5

!

/

(

2

!

⋅

2

2

)

=

120

/

8

=

15

permutations with this cycle type; in fact, they are (omitting the

trivial

1

-cycles):

(

1

⁢

2

)

⁢

(

3

⁢

4

)

(

1

⁢

2

)

⁢

(

3

⁢

5

)

(

1

⁢

2

)

⁢

(

4

⁢

5

)

(

1

⁢

3

)

⁢

(

4

⁢

5

)

(

2

⁢

3

)

⁢

(

4

⁢

5

)

(

1

⁢

3

)

⁢

(

2

⁢

4

)

(

1

⁢

3

)

⁢

(

2

⁢

5

)

(

1

⁢

4

)

⁢

(

2

⁢

5

)

(

1

⁢

4

)

⁢

(

3

⁢

5

)

(

2

⁢

4

)

⁢

(

3

⁢

5

)

(

1

⁢

4

)

⁢

(

2

⁢

3

)

(

1

⁢

5

)

⁢

(

2

⁢

3

)

(

1

⁢

5

)

⁢

(

2

⁢

4

)

(

1

⁢

5

)

⁢

(

3

⁢

4

)

(

2

⁢

5

)

⁢

(

3

⁢

4

)

References

1

D.S. Dummitt, R.M. Foote,

Abstract Algebra

, Wiley, 2004.

Title

conjugacy classes in the

symmetric group

S

n

Canonical name

ConjugacyClassesInTheSymmetricGroupSn

Date of creation

2013-03-22 17:16:24

Last modified on

2013-03-22 17:16:24

Owner

rm50 (10146)

Last modified by

rm50 (10146)

Numerical id

8

Author

rm50 (10146)

Entry type

Topic

Classification

msc 20B05

Classification

msc 20B30

Defines

cycle type

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