Let G=Sym(n) .Let C be a cyclic group. Suppose that then how to show that the generators for in G all have the same cycle type and so are conjugate in G.

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- March 20th 2009, 12:43 AMdimukCyclic groups
Let G=Sym(n) .Let C be a cyclic group. Suppose that then how to show that the generators for in G all have the same cycle type and so are conjugate in G.

- March 20th 2009, 08:36 AMThePerfectHacker
I am assuming that the cyclic group is finite.

If generates then generates where is relatively prime with the order of because a homomorphic image of a cyclic group is cyclic. Let for disjoint cycles . The order of is the least common multiple of the lengths of all these cycles. Therefore, divides for all . Now if is an integer so that then certainly . Therefore, if is an -cycle then still stays as an -cycle. Thus, we have where is same cycle length as which means and have the same cycle type and consequently they are conjugate.