Question

abstract-algebra  

Problem 10.2. Consider the following permutations f and g in the permutation group 56: f:145, 241, 366,44 3,5 H 2,6 H4; g=(1 6 5)( 24). (1) Write f as a product of disjoint cycles. (2) Find o(g). (3) Write fg as a product of disjoint cycles. (4) Write gf as a product of disjoint cycles. (5) Write gfg as a product of disjoint cycles. Hint. All should be straightforward. Be careful though.

Answer 1

C Solution: Here for a gre permulabions in so f:1 P5, 261, 3146,463,562,665 di (16 5) (24) First note that in other notation f can be written as fa ll23 u sol 15 1 6 3 2 47. 0 cycle. now f as a product of disjoint Tf= (1 5 2) 63 64) is © O p Here q = (1 65)(29) which product of disjoint cycle. - 0(1) - Jeast common mulkple of (3,2) 0(3) = 6 1 ☺ f&= (is 2) (364) (165) (24) = 1 2 3 4 5 6 14 3 6 1 525 Tf2 = (14) (2 3 6) © 24 = (165) (24) (152) (365) =( ! 2 3 us on Ti 6 5 3 5 2 12f = (26)(354)

© First cop find Here moc o w (10 1 1 2 3 4 5 6) 16 5 3 2 15! zg'= (165)(26) Thus gf g = (af) 2 = (2 6) (3 54) (16s) (2 h) (1 2 3 4 5 6 =(2 3 5 6 14 → af 2 = (1 2 3 5) (96) Thank you.