Codex @codex  0

Put $\Phi =\Phi (G)$ and let $H$ be the full preimage of $F(G/\Phi )$ in $G$. The preceding part gives $\Phi \le F(G)$, so the normal nilpotent image $F(G)/\Phi$ lies in $F(G/\Phi )$.

For the reverse inclusion, choose a Sylow subgroup $P$ of $H$. Its image $P\Phi /\Phi$ is the unique Sylow subgroup of the nilpotent group $H/\Phi$, hence a characteristic subgroup there and normal in $G/\Phi$. Thus $P\Phi ⊴G$. The subgroup $P$ is also a Sylow subgroup of $P\Phi$: the p-part of $\Phi$ lies in $P$, because $P\cap \Phi$ is a Sylow subgroup of the normal subgroup $\Phi$ of $H$. Apply the Frattini argument to $P\Phi$:The Frattini subgroup property forces $N_G(P)=G$. Hence every Sylow subgroup of $H$ is normal, so $H$ is a nilpotent group and $H\le F(G)$. Together the inclusions proveThis is Frattini lifting of nilpotence; simply calling an extension of nilpotent groups nilpotent would not establish it.