If $H \triangleleft G$ and $|G/H|=m$, show that $x^m \in H$ for $\forall x \in G$ 
If $H \triangleleft G$ and $|G/H|=m$, show that $x^m \in H$ for $\forall x \in G$. 

My attempt is: since $|G/H|=\frac{|G|}{|H|}=m$, we have $x^{|G|}=x^{m|H|}=e$, then I stuck here. Can anyone guide me ?

Your given that $H \triangleleft |G/H| = m.\,$ Hence, for every $\,x\,$ in a group of order of order $m$, $x^m = e$.  For the group $\,G/H\,$, this means that $\,(xH)^m = H\,$ for all $x \in G.\;\;$ Why?
But as $\,H\,$ is normal in $\,G,\;$ $(xH)^m = H \quad\implies\quad x^mH = eH = H\quad \implies x^m \in H$.
