An existence criterion for a normal maximal subgroup in a finite solvable group

Let G be a solvable group of order pm, where p is a prime not dividing m. Let P \in \mathsf{Syl}_p(G). If N_G(P) = P, prove that G has a normal subgroup of order m. Where is solvability required in the proof? [Note: This result holds for nonsolvable finite groups as well, as a special case of Burnside’s N/C-Theorem.]

By hypothesis, we have n_p(G) = m, and the Sylow p-subgroups of G intersect trivially. Thus G has (p-1)m = |G| - m elements of order p. Denote the remaining elements by H.

Let \pi denote the set of all prime divisors of |G| except p. By this previous exercise, there is a Hall \pi-subgroup M \leq G, which (by definition) has order m, and none of whose elements have order p. Thus M = H. Moreover, if N \leq G is any other subgroup of order m, then N is a Hall \pi-subgroup (again by definition) and N is conjugate to M. But N = H, so that in fact M is the unique subgroup of G of order m. Thus M is characteristic, hence normal, in G.

Solvability was needed in order to guarantee the existence and conjugateness of Hall subgroups.

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