## Maximal subgroups of a finite nilpotent group have prime index

Prove that a maximal subgroup of a finite nilpotent group has prime index.

Let $G$ be a finite nilpotent group and $M \leq G$ a maximal subgroup. By Theorem 3, $M < N_G(M)$ is a proper subgroup, so that (since $M$ is maximal) $N_G(M) = G$. Thus $M \leq G$ is normal. Now $G/M$ is a finite nilpotent group.

Suppose $[G:M]$ is composite, with a proper prime divisor $q$. By this previous exercise, there is a normal subgroup $\overline{H} \leq G/M$. By the Lattice Isomorphism Theorem, $\overline{H} = H/M$ for some normal subgroup $H \leq G$ with $M \leq H$. Since $M$ is maximal, either $H = M$ or $H = G$, so that $q$ is not a proper divisor of $[G:M]$– a contradiction.

Thus $[G:M]$ is prime.

Because $G$ is finite.