I'm totally lost on this one.
If $G$ has a normal subgroup of index $p$, prove that $G$ has at least one element of order $p$.
Could you use Cauchy's Theorem?
Let $H$ be a normal subgroup of $G$, with index $p$ where $G$ is finite and $p$ is prime.
Then $(G:H)$ = $|G|$$/$$|H|$ = $p$, by the definition of index.
So $|G| = |H|*p$
By Cauchy's Theorem, if $G$ is a finite group, and $p$ is a prime divisor of $|G|$, then $G$ has an element of order $p$.