Originally Posted by

**crushingyen** If $\displaystyle N $ is a normal subgroup of $\displaystyle G $ and if every element of $\displaystyle N $ and of $\displaystyle G/N $ has finite order, prove that every element of $\displaystyle G $ has finite order.

I started off doing this:

Let $\displaystyle g \in G $. By showing that $\displaystyle g^n = e $, $\displaystyle n \in \mathbb{Z} $

$\displaystyle (Ng)^k = Ne $ for some $\displaystyle k \in \mathbb{Z} $

$\displaystyle \Rightarrow (eg)^k \in (Ng)^k = Ne = N $

$\displaystyle \Rightarrow (eg)^k \in N $

$\displaystyle \Rightarrow (e^k)(g^k) \in N $

$\displaystyle \Rightarrow eg^k \in N $

$\displaystyle \Rightarrow g^k \in N $

$\displaystyle \Rightarrow \exists $ $\displaystyle n \in \mathbb{Z} $ such that $\displaystyle (g^k)^n = e $

I don't know how to finish off this problem after point. Any help would be greatly appreciated.