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

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?

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.

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.

$$\forall\,x\in G\;\;\exists\,n_x\in\Bbb N\;\;s.t.\;\;x^{n_x}\in H\implies\,\exists m_x\in\Bbb N\;\;s.t.\;\;\left(x^{n_x}\right)^{m_x}=1$$