- #1

quasar_4

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## Homework Statement

If N is a normal subgroup of a group G and every element of N and G/N has finite order, prove that every element of G has finite order.

## Homework Equations

Um, a finite group element is given such that g^k = e (identity) for some integer k.

We also know that the quotient of a group by a group is itself a group, with cosets as elements. If G/N is the quotient space, N is the identity element.

## The Attempt at a Solution

The elements of N, call them n, have some finite order, so for some integer k we can say that n^k=e (the identity). The elements of the quotient group, G/N are the distinct cosets of N in G, so if they have some finite order m, then (Ng)^m=e=Ne (because N is the identity in the quotient group).

I was hoping to do something like saying

(ng)^m = ne ==> (n^m)*(g^m)= ne ==> n^(m-1)*g^m=e ==> somehow g^something =e.

First of all, I am not sure if this is general enough to prove this for ALL cosets of G/N. Second, can I use little n instead of big N (we usually denote the cosets Ng. I think I can do the little one, though, because we're taking an element of N and operating on the left by some group element g). I don't know what else to do!