Let $G$ be a finite non-abelian simple group and $p$ the largest prime divisor of $|G|$. Show that if $H < G $ then $|G : H | \geq p $.
This is from a chapter of a book about group actions, specifically a section on conjugacy classes. Just before the question it also explains a theorem showing that a finite non-abelian simple group, such as $G$, must be divisible by at least $2$ different primes.
I'm guessing the question can be answered with these facts but I'm not sure how to go about it.