1. Permutation Groups Part 1

2. Definition A permutation of a set A is a function from A to A that is both one to one and onto.

3. Array notation Let A = {1, 2, 3, 4} Here are two permutations of A:

4. Composition in Array Notation 1

5. 1 4

6. 1 4 2

7. 1 4 2 3

8. 1 4 2 3

9. Definition A permutation group of a set A is a set of permutations of A that forms a group under function composition.

10. Example The set of all permutations on {1,2,3} is called the symmetric group on three letters, denoted S 3 There are 6 permutations possible:

11. S3S3 The permutations of {1,2,3}:

12. Is S 3 a group? Composition of functions is always associative. Identity is . Since permutations are one to one and onto, there exist inverses (which are also permutations. Therefore, S 3 is group.

13. Computations in S 3

14. Simplified computations in S 3                    Double the exponent of  when switching with .  You can simplify any expression in S 3 !

15. Symmetric groups, S n Let A = {1, 2, … n}. The symmetric group on n letters, denoted S n, is the group of all permutations of A under composition. S n is a group for the same reasons that S 3 is group. |S n | = n!

16. Symmetries of a square, D 4 1 23 4 D 4 ≤ S 4

17. Why do we care? Every group turns out to be a permutation group on some set! (To be proved later).

18. Cycle Notation

19. Disjoint cycles Two permutations are disjoint if the sets of elements moved by the permutations are disjoint. Every permutation can be represented as a product of disjoint cycles.

20. Algorithm for disjoint cycles Let permutation π be given. Let a be the identity permutation, represented by an empty list of cycles. while there exists n with π(n) ≠ a(n): start a new cycle with n let b = n while

21. Compostion in cycle notation  = (1 2 3)(1 2)(3 4) = (1 3 4)(2) = (1 3 4)  = (1 2)(3 4)(1 2 3) = (1)(2 4 3) = (2 4 3)

22. Compostion in cycle notation  = (1 2 3)(1 2)(3 4) = (1  = (1 2)(3 4)(1 2 3) 1  1  2  3