Let $G$ be a finite group and $p$ the largest prime that divides the order of $G$. Suppose that all the maximal subgroups of $G$ have a prime index. Then, the Sylow $p$-subgroup is normal in $G$. Furthermore, there is a normal subgroup of $G$ of order exactly $p$.
$\textbf{My attempt}$: I managed to show that the $p$-subgroup of Sylow $P$ is normal in G. For the second part, I thought to consider a minimal normal subgroup $N$ of $G$ contained in $P$, and show that $N$ has order $p$. But I don't know how to get out of here. It is possible to show that $N$ is elementary abelian, since $N'\trianglelefteq G$ and $N \leq P$ is solvable.