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.

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By nbloomf, on July 1, 2010 at 10:00 am, under AA:DF. Tags: finite group, maximal subgroup, nilpotent group, subgroup index. 2 Comments

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Edward Hou On February 6, 2011 at 9:27 pm
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I’m probably missing some simple fact, but how do we know the maximal subgroup has finite index?
- nbloomf On February 6, 2011 at 9:43 pm
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  Because $G$ is finite.

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