Question

(1 point) Let f be a permutation on the set {1, 2, 3, 4, 5, 6, 7, 8, 9), defined as follows f= 1 2 3 4 5 6 7 8 9 1 2 5 8 3 9 4 6 7 (a) Write the permutation f7 as a product of disjoint cycles, separated by commas (e.g. (1, 2), (3,4,5),...). Do not include 1-cycles (e.g. (2) ) in your answer. (b) Determine the smallest value of k > O such that fk = (1). I

Answer 1

2 3 5 f= 7 8 2 5 3 9 4 6 f = (35) (4 8 6 9 7) Therefore cycles I is of disjoint product a 5. of and order 2 Therefore order of f - lcm (2,5) = 10 f is off) =10 10 a F-3 Io =(1) f f* = (F) I-3 -> 3 Let us find as f follows 2 7 8 6 9 8 6 3 4 5 5 8 3 2 - 3 5 4 4 f² 6 7 9 4 - 2 7 5 2 8 3 - 7 8 9 2 3 5 4 6 7 4 8 9 5 2 3 6 Then f² = f². f 9 2 3 6 2 4 5 6 7 8 3 9 4 6 hh 7 8 7 8 9 4 4 65 ma 8 :) 5 1 2 2 3 8 9 5 4 9 2 7 С. c 4 ; ) 3 5 8 7 3 3 to 2

Hence, f 3 ($) 4 3 8 2 5 . 7 8 oot 6 7 9 6 5 2 3 8 7 9. 6 5 7 4 2 3 4 9 7 8 f 5 6 2 3 7 = f' =(3 5)(4 6 7 8 9 7 9 ) Hence f (3 5), (4 6 7 8 10, hence of k so such that (6) Since order of f is the smallest smallest value f* = (1) is k = 10 k