TITLE: Proper subgroup of normalizer of a finite group impies nilpotent.
QUESTION [5 upvotes]: So I came upon this exercise: Let $G$ be finite. If every proper subgroup $H$ of $G$ has the property $H < N_G(H)$, then $G$ is nilpotent.
I can prove the converse by induction on the nilpotency class of $G$, but I'm kind of stuck here.

REPLY [3 votes]: It suffices to show that every Sylow $p$-subgroup $P$ of $G$ is normal.  But $N_G(P) = N_G(N_G(P))$, so your hypothesis implies $N_G(P) = G$.