= 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
$$
\boxed{\Phi(G)\le F(G).}
$$
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