[Math] 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

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?

Answer 1

Note that in $G/H$, an element $xH$ is the identity if and only if $x\in H$. This means that $(xH)^n = eH$ implies that $x^n \in H$, and so you can use the fact that elements in $H$ have finite order to proceed.