 Solution

Put $\Phi =\Phi (G)$ and choose a Sylow subgroup $P$ of $\Phi$. Since $\Phi$ is normal, the Frattini argument gives $G=\Phi N_G(P)$: for $g\in G$, the Sylow subgroups $P^g$ and $P$ of $\Phi$ are conjugate by an element of $\Phi$, leaving an element of $N_G(P)$. The non-generator property of the Frattini subgroup proved above forces $N_G(P)=G$. Thus every Sylow subgroup of $\Phi$ is normal. Distinct such Sylow subgroups commute because their group commutators lie in their trivial intersection. Hence $\Phi$ is a nilpotent normal subgroup, and