Let $H$ be normal in $G$ and $[G:H]=m$. Prove that $x^m$ is in $H$ for all $x$ in $G$.
Since $H$ is normal its left cosets form a group. Also since $m$ is the order of that group $(xH)^m=x^mH=H$, which implies that $x^m \in H$.
I am actually not sure what my question is, it just hard to believe that if I pick any element of a group $G$ and I raise it to $m$ the resulting element will be in $H$. Anyways, is there another way one prove this? Thanks.