Frattini lifting of nilpotence

Let $\Phi =\Phi (G)$ in a finite group. If $H⊴G$ contains $\Phi$ and $H/\Phi$ is a nilpotent group, every Sylow subgroup $P$ of $H$ has $P\Phi ⊴G$. The Frattini argument gives $G=(P\Phi )N_G(P)=\Phi N_G(P)$, and the defining non-generator property of $\Phi$ forces $N_G(P)=G$. Hence $H$ is a nilpotent group.