Prime index subgroups are maximal

Given $H < G$ and $|G:H|=p$, $p$ prime, then $H$ is a maximal subgroup.

$\textit{Proof}$: Assume to the contrary that $H$ is not a maximal subgroup. With this, there then exists another subgroup, say, $M$ s.t. $H < M < G$. By 3.2.11, $|G:H|=|G:M| \cdot |M:H|=mn \rightarrow \leftarrow $ since $p$ is prime. $\therefore H $ is maximal $ \blacksquare$