# Thread: Normal Subgroup and Quotient Group

1. ## Normal Subgroup and Quotient Group

Here is the question I'm trying to figure out:

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.
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?

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.

2. use the fact that (Hg)^k = H(g^k).

what can you say about (Hg)^m?

3. Originally Posted by Deveno
use the fact that (Hg)^k = H(g^k).

what can you say about (Hg)^m?
(Hg)^m = Hg^m = H * e = H

Maybe I am thinking about the order of Hg in the wrong way.
The order of Hg is the number of elements in the set Hg, but is it also the least positive integer x such that (Hg)^x = H?

4. 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?

5. 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.