Question

4. Let T = (1,3,4,6)(2,3,5) in Ss. Find the index of <T> in So. S61 Solution: If we let H = (r), then we are looking for (S. : H) However, we cannot simply claim that H = 12 because the cycle decomposition for T is not disjoint. Therefore, we need to figure out what the actual disjoint cycle decomposition is. To do that, we need to figure out what is as a permutation, not the given cycles. To do this, just multiply the permutations out 3 4 5 6 1 2 3 4 5 6 3 5 4 2 6 4 5 6 2 1 TE 2 = (1,3,4,6)(2,3,5) = (338 56 1 3 1) Following the process we did in problem 3 to break down into its cycle decomposition, we see that T=(1,3,5, 2, 1,6) and is itself a cycle of length 6. Therefore, |M| -6 and (S6: H) 6! 120. 6

Can you explain to me how this works? Specifically, how does the permutation multiplication work. How does (1,3,4,6)(2,3,5) become the 2 permutations multiplied together. I guess I am lost on all of it.

Answer 1

If you doubt in any step please comment down i will try to explain that step further and if you were able to understand the explanation please give feedback
5 2 3 4 5 6 3 ) Multiplication of two permutations element next to ff is Colemanten en proteins Here erz (1,3,4,6)(2,3,5) 'element} HSBty, in Ist permutation I mapped to 3, 328 mapped to 4,4 R8 mapped to 6, 6 is mapped to 1 and 2,5 are tef self mapped and in 2nd permutation, a Ro mapped to 3,398 mapped to and 5 po mapped to 1 and 1,4,6 are self-mapped. 9=(1,3,4,6)(2,3,5) - 23.4 5 5 2 3 4 4 6 5 01 1 5 4 1 6 this is obtained by placing the image or mapping the element just below it. Now, check, Pn and permutation 1 maps to 1 and then the 1st permutation, I maps to 3, so in total 1 maps to 3 another, in and permutation 2 maps to 3 and then in 1st. permutation, 3 maps to 4, 80 total 2 maps to 4 and similarly, you can check for 3,4,5 and 6 So, na (1,3,4,6)(2,3,5) = 1 2 3 4 5 6 3 4 5 6 2 1 (1 35 246) of ( )

by r) and for the index of eryth S6. we need to find 141 where H=<i> (subgroup generated sihice & Po not multiplication of disthet Cycles thus to calculate Order of r , we multiplied the two permutations and get >= (1,3,5, 2,4,6) clearly, it is a cycle of 6. So H1 = 407=6. Now, Index of <ry in So Isol THI (as Bn) =n:) = 6! 6 = 645X4X3 X2x) f = 26 = 120