TITLE: If every element of G/H has finite order and every element of H has finite order, then every element of G has finite order
QUESTION [11 upvotes]: Let $G$ a group with normal subgroup $H$. If every element of $G/H$ has
  finite order and  every element of $H$ has finite order, then every
  element of $G$ has finite order

Proof:
Let G be a group with normal subgroup H. Suppose that every element of G/H has finite order and that every element of H has finite order.
We want to show $G$ has finite order.
Let $x \in G$ then by coset and quotient group definition, $Hx \in G/H$ has $Hx$ has finite order $n$ or in other words $(Hx)^n=e$. Also for some $h \in H$, it also has a finite order where $h^m=e$
I'm stuck on how to link it together. Any input?

REPLY [9 votes]: Here is a way of putting things together:
Let $g \in G$. Consider the coset $gH$. Since every element in $ G/H$ has ﬁnite order, there exists $n \in \Bbb N$ such that $g^nH = (gH)^n = eH = H$, i.e. $g^n\in H$ . Now we use the fact that every element in $H$ has ﬁnite order, so there exists $m \in \Bbb N$ such that $(g^n)^m = e$, i.e. $g^{mn}= e$. Thus $g$ has ﬁnite order.