SECTION 9 Orbits, Cycles, and the Alternating Groups

Given a set A, a relation in A is defined by :

For a, bA, let ab if and only if b = n(a) for some nZ. (1)We can check that the relation is indeed an equivalent relation.

(Reflexive, Symmetric, Transitive)

Since an equivalence relation on a set yields a natural partition of the set, then we have the following

DefinitionLet be a permutation of a set A. The equivalence classes in A

determined by the equivalence relation (1) are the orbits of .

Example

Example:• Let A={a, b, c, d}, and be the identity permutation of A. Find the

orbits of in A.Solution: The orbits of are: {a}, {b}, {c}, {d}

• Find the orbits of the permutation in S8.

Solution: The orbits containing 1 is {1, 3, 6} The orbits containing 2 is {2, 8} The orbits containing 4 is {4, 7, 5}Since these three orbits include all integers from 1 to 8, the complete list

of orbits of is {1, 3 ,6}, {2, 8}, {4, 7, 5}

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Cycles

For the remainder of this section, we suppose that A= {1, 2, 3, , n} and that we are dealing with the elements of the symmetric group Sn.

Recall has orbits : {1, 3 ,6}, {2, 8}, {4, 7, 5}, which

can be indicated graphically by using circles. Such a permutation, described graphically be a single circle, is called a cycle.

Note: we also consider the identity permutation to be a cycle.

Here is the term cycle in a mathematically precise way:

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Cycle

DefinitionA permutation Sn is a cycle if it has at most one orbit containing

more than one element. The length of a cycle is the number of elements in its largest orbit.

We also introduce a cyclic notation for a cycle. For example

Given . It has orbits {1, 3, 6}, {2}, {4}, {5}, {7}, {8}.

We can denote =(1, 3, 6).

Note: an integer not appearing in this notation for is left fixed by .

1 2 3 4 5 6 7 83 2 6 4 5 1 7 8

Example

ExampleIn S5, we see that

Example

1 2 3 4 5(1,3,5,4)

3 2 5 1 4(3,5,4,1)(5,4,1,3)(4,1,3,5)

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(1,3,6)(2,8)(4,7,5)

Note: these cycles are disjoint, meaning that any integer is moved by atmost one of these cycles; thus no one number appears in the notations ofTwo cycles.

Theorem

TheoremEvery permutation of a finite set is a product of disjoint cycles.

Note: • Permutation multiplication is in general not commutative, but the

multiplication of disjoint cycles is commutative.

Example:Given the permutation , write it as a product of disjoint

cycles.

Solution:

1 2 3 4 5 66 5 2 4 3 1

1 2 3 4 5 6(1,6)(2,5,3)

6 5 2 4 3 1

Example

Consider the cycles (1, 4, 5, 6) and (2, 1, 5) in S6.

Find (1, 4, 5, 6)(2, 1, 5) and (2, 1, 5)(1, 4, 5, 6).

Solution: (1, 4, 5, 6)(2, 1, 5)=

(2, 1, 5)(1, 4, 5, 6)=

1 2 3 4 5 66 4 3 5 2 1

1 2 3 4 5 64 1 3 2 6 5

Even and Odd Permutations

DefinitionA cycle of length 2 is a transposition.

A computation shows that (a1,a2,…, an)=(a1,an)(a1,an-1)…(a1,a3)(a1,a2), therefore any cycle is a product of transpositions.

CorollaryAny permutation of a finite set of at least two elements is a product of

transpositions.

Examples

Example:• (1, 6)(2, 5, 3)=(1, 6)(2, 3)(2, 5)• In Sn for n2, the identity permutation is the product (1, 2)(2, 1).

Note: a representation of the permutation in this way is not unique, but the number of transposition must either always be even or always be odd.

Theorem: No permutation is Sn can be expressed both as a product of an even

number of transpositions and as a product of an odd number of transpositions.

Odd/Even permutation

DefinitionA permutation of a finite set is even if it can be expressed as a product

of an even number of transpositions. A permutation of a finite set is odd if it can be expressed as a product of an odd number of transpositions.

Example:• The identity permutation in Sn is an even permutation since

=(1, 2)(2, 1).

• The permutation (1, 4, 5, 6)(2,1, 5) in S6 is odd since

(1, 4, 5, 6)(2,1, 5)=(1, 6)(1, 5)(1, 4)(2, 5)(2, 1)

The Alternating Groups

TheoremIf n2, then the collection of all even permutations of {1, 2, 3,…, n}

forms a subgroup of order n! / 2 of the symmetric group Sn.

DefinitionThe subgroup of Sn consisting of all even permutations of n letters is

the alternating groups An on n letters.