Here is the question I'm trying to figure out:
The following is how far I've been able to get in figuring this out. Could someone give me a point in the right direction?Let G be a group, let g in G have finite order m, and let H be normal in G. Given this, show that the order of the element Hg in G / H is finite and divides m.
I need to show that the number of elements in the coset Hg is finite and divides the order of g.
Since the order of an element in G divides the order of g, |G| = m*n
If the order of H is x, since all other elements in G / H have the same number of elements as H, and they sum to the order of G, then x * y = m * n
However, I want to show that x * y = m.