use the fact that (Hg)^k = H(g^k).
what can you say about (Hg)^m?
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.
the number of elements in the coset Hg is always |H|. but the order of Hg in G/H is the least positive integer k with (Hg)^k = H (since H is the identity of G/H).
now if an element g of a group has the property g^k = e, what is the relationship between the order of g, and k?
Ahhh, I get it now. The order of g divides k in that case.
This same logic can be applied and the proof will revolve around the division algorithm.
Thank you for helping me figure this out. I think a lot of my problem was in not realizing the difference between the order of Hg, and the order of Hg in G / H.