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

Let be a finite nilpotent group and a maximal subgroup. By Theorem 3, is a proper subgroup, so that (since is maximal) . Thus is normal. Now is a finite nilpotent group.

Suppose is composite, with a proper prime divisor . By this previous exercise, there is a normal subgroup . By the Lattice Isomorphism Theorem, for some normal subgroup with . Since is maximal, either or , so that is not a proper divisor of – a contradiction.

Thus is prime.

## Comments

I’m probably missing some simple fact, but how do we know the maximal subgroup has finite index?

Because is finite.