every permutation has a cycle decomposition

every permutation has a cycle decomposition

In this entry, we shall show that every

permutation

of a

finite set

can be factored into a

product

of

disjoint

cycles.
To accomplish this, we shall proceed in two steps.

We begin by showing that, if

f

is a non-trivial
permutation of a set

{

x

i

∣

1

≤

i

≤

n

}

,
then there exists a cycle

(

y

1

,

…

⁢

y

m

)

where

{

y

i

∣

1

≤

i

≤

m

}

⊆

{

x

i

∣

1

≤

i

≤

n

}

and a permutation

g

of

{

x

i

∣

1

≤

i

≤

n

}

such that

f

=

(

y

1

,

…

⁢

y

m

)

∘

g

and

g

⁢

(

y

i

)

=

y

i

.

Since the permutation is not

trivial

, there exists

z

such that

f

⁢

(

z

)

≠

z

. Define a

sequence

inductively as follows:

z

1

=

z

z

k

+

1

=

f

⁢

(

z

k

)

Note that we cannot have

z

k

+

1

=

z

k

for any

k

.
This follows from a simple

induction

argument

. By
definition

f

⁢

(

z

1

)

=

f

⁢

(

z

)

≠

z

=

z

1

. Suppose
that

f

⁢

(

z

k

)

≠

z

k

but that

f

⁢

(

z

k

+

1

)

=

z

k

+

1

.
By definition,

f

⁢

(

z

k

)

=

z

k

+

1

. Since

f

is a
permutation,

f

⁢

(

z

k

)

=

z

k

+

1

and

f

⁢

(

z

k

+

1

)

=

z

k

+

1

imply that

z

k

=

z

k

+

1

, so

z

k

=

f

⁢

(

z

k

)

, which
contradicts a

hypothesis

. Hence, if

f

⁢

(

z

k

)

≠

z

k

,
then

f

⁢

(

z

k

+

1

)

≠

z

k

+

1

so, by induction,

f

⁢

(

z

k

)

≠

z

k

for all

k

.

By the

pigeonhole principle

, there must exist

p

and

q

such that

p

<

q

but

f

⁢

(

z

p

)

=

f

⁢

(

z

q

)

. Let

m

be the
least integer such that

f

⁢

(

z

p

)

=

f

⁢

(

z

p

+

m

)

but

f

⁢

(

z

p

)

≠

f

⁢

(

s

p

+

k

)

when

k

<

m

. Set

y

k

=

z

k

+

p

.
Then we have that

(

y

1

,

…

⁢

y

m

)

is a cycle.

Since

f

is a permutation and

{

y

i

∣

1

≤

i

≤

m

}

is

closed under

f

, it follows that

{

x

i

∣

1

≤

i

≤

n

}

∖

{

y

i

∣

1

≤

i

≤

m

}

is also closed under

f

. Define

g

as follows:

g

⁢

(

z

)

=

{

z

z

∈

{

y

i

∣

1

≤

i

≤

m

}

f

⁢

(

z

)

z

∈

{

x

i

∣

1

≤

i

≤

n

}

∖

{

y

i

∣

1

≤

i

≤

m

}

Then it is easily verified that

f

=

(

y

1

,

…

⁢

y

m

)

∘

g

.

We are now in a position to finish the proof that
every permutation can be decomposed into cycles.
Trivially, a permutation of a set with one

element

can be decomposed into cycles because the only
permutation of a set with one element is the

identity

permutation, which requires no cycles
to decompose. Next, suppose that any set with less
than

n

elements can be decomposed into cycles.
Let

f

be a permutation on a set with

n

elements.
Then, by what we have shown,

f

can be written as
the product of a cycle and a permutation

g

which
fixes the elements of the cycle. The

restriction

of

g

to those elements

z

such that

g

⁢

(

z

)

≠

z

is a permutation on less than

n

elements and
hence, by our supposition, can be decomposed into
cycles. Thus,

f

can also be decomposed into
cycles. By induction, we conclude that any
permutation of a finite set can be decomposed into cycles.

Title

every permutation has a cycle decomposition

Canonical name

EveryPermutationHasACycleDecomposition

Date of creation

2013-03-22 16:48:39

Last modified on

2013-03-22 16:48:39

Owner

rspuzio (6075)

Last modified by

rspuzio (6075)

Numerical id

10

Author

rspuzio (6075)

Entry type

Proof

Classification

msc 03-00

Classification

msc 05A05

Classification

msc 20F55

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