1. SECTION 9 Orbits, Cycles, and the Alternating Groups Given a set A, a relation in A is defined by : For a, b  A, let a  b if and only if b =  n (a) for some n  Z. (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 Definition Let  be a permutation of a set A. The equivalence classes in A determined by the equivalence relation (1) are the orbits of .

2. 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 S 8. 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}

3. 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 S n. 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:

4. Cycle Definition A permutation  S n 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 .

5. Example In S 5, we see that Example Note: these cycles are disjoint, meaning that any integer is moved by at most one of these cycles; thus no one number appears in the notations of Two cycles.

6. Theorem Every 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:

7. Example Consider the cycles (1, 4, 5, 6) and (2, 1, 5) in S 6. 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)=

8. Even and Odd Permutations Definition A cycle of length 2 is a transposition. A computation shows that (a 1,a 2,…, a n )=(a 1,a n )(a 1,a n-1 )…(a 1,a 3 )(a 1,a 2 ), therefore any cycle is a product of transpositions. Corollary Any permutation of a finite set of at least two elements is a product of transpositions.

9. Examples Example: (1, 6)(2, 5, 3)=(1, 6)(2, 3)(2, 5) In S n for n  2, 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 S n can be expressed both as a product of an even number of transpositions and as a product of an odd number of transpositions.

10. Odd/Even permutation Definition A 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 S n is an even permutation since  =(1, 2)(2, 1). The permutation (1, 4, 5, 6)(2,1, 5) in S 6 is odd since (1, 4, 5, 6)(2,1, 5)=(1, 6)(1, 5)(1, 4)(2, 5)(2, 1)

11. The Alternating Groups Theorem If n  2, then the collection of all even permutations of {1, 2, 3,…, n} forms a subgroup of order n! / 2 of the symmetric group S n. Definition The subgroup of S n consisting of all even permutations of n letters is the alternating groups A n on n letters.