1

1

Cycles in Symmetric group

Chapter 14

2

Cycles in Sn

Suppose that we have permutationWe can see that α maps

1→2, 2→3, 3→1So we call that α is a cycle in S4, written as α = (123)Also we can write α = (231) or α = (312) Definition: Let a1, a2, a3, …, ak are different k numbers, if permutation α maps

a1 →a2, a2 → a3, …, ak → a1

and no other number was changed by α, then α is called a cycle, denoted as (a1a2a3…ak). Number k is called the length of the cycle.The numbers that are not in the expression (...) are either not changed or beyond our consideration.

1 2 3 42 3 1 4

α � �� �� �

=

2

3

Definition. If the length of a cycle is 2, then this cycle is called transposition. Such as α = (23) δ= (14)

Note: The identity permutation is a cycle of length 1, such as1 2 3 4

, , 1 2

,3 4

(1) (2) (3) (4)e e e e e� �� �� �

= � = = = =

Examples: 1 2 3 4 53 2 4 1 5

1 2 3 4 51 3 4 5 2

(134)

(2345)

β β

γ γ

� �� �� �

� �� �� �

= � =

= � =

We can verify that (23)2=e, (134)3=e, (2345)4=e

Theorem 1: If the length of cycle α is k , then α k =e.

This because after repeat α for k times, every number in the cycle will be mapped to itself.

4

1 2 3 4 5 63 2 6 5 4 1

β � �� �� �

=

Theorem 2: Every permutation in Sn can be decomposed as a product of several disjoint cycles

For example, if

We can pick any number to start, if we pick 1, then we see that

1 → 3 → 6 → 1, 2 → 2, and 4 → 5 → 4

So we get β = (136)(2)(45)

Omit identity (2) we have β = (136)(45)

However, we can start this procedure from number 4 to get

β = (45)(136)

Theorem 3: Disjoint cycles α, β are commutative, αβ =βα

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5

Theorem 4 : If cycles α, β are disjoint then for any integer k

(αβ)k = αkβk

So [(12)(34)]2 = (12)2(34)2 = ee = e

[(12)(345)]3= (12)3(345)3 = (12)(12)2(345)3 = (12)

Theorem 5 : If cycles α1, α2,…, αm are mutual disjoint then for any integer k

(α1α2…αm)k = α1kα2

k…αmk

Note: If cycles α, β are not disjoint then maybe αβ ≠βα

( )( ) ( )( )( ) ( )1 2 3 1 2 3 1 2 32 1 3 1 3 2 2 3 1

1 2 3 1 2 3 1 2 31 3 2 2 1 3 3 1 2

(12)(23)=

(23)(12)=

=

=

6

Theorem 6: Any cycle can be decomposed as a product of transpositions. (answer is not unique).Proof: cycle (a1a2a3…ak) maps

a1 →a2, a2 → a3, …, ak → a1

cycle (a2a3…ak) maps a1 →a1, a2 → a3, …, ak → a2

Therefore product (a1 a2)(a2a3…ak) mapsa1 →a2, a2 → a3, …, ak → a1

So (a1a2a3…ak) = (a1 a2)(a2a3…ak) If we continue this process, we can get

(a1a2a3…ak)= (a1 a2)(a2a3)…(ak−1ak)

Example: (1234) = (12)(23)(34)However, (1234) = (4123) = (41)(12)(23)

(1234) = (2341) = (23)(34)(41)

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7

Theorem 7. Every permutation is product of transpositions. (it may not be unique)

Example: 1 2 3 4 5 6

3 2 6 5 4 1

(136)(45)(13)(36)(45)

β � �= � �� �

==

Note 1: If a, b, c are three numbers then (ab)(bc)=(abc)

Note 2: If a, b, c are three numbers then

(ab)(bc)=(abc)=(bca) = (bc)(ca)

Note 3: If a, b, c are three numbers then (ab)(bc)=(ca)(ab)

8

Exercises

a) Decompose α to product of disjoint cyclesb) Decompose α to product of transpositionsc) Find inverse permutation α−1

d) Decompose α−1 to product of disjoint cyclese) Decompose α−1 to product of transpositions

1 2 3 4 5 65 3 6 1 4 2

α � �� �� �

=1. Let

2. Let α = (12)(35)(23)(41)(51)(34) Write α to be a product of disjoint cycles

3. Let β = (12)(34)(23)(41) Write β2 to be a product of disjoint cycles