## 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.