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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  • Edward Hou  On February 6, 2011 at 9:27 pm

    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

      Because G is finite.

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