Is there a Schur-Zassenhaus-free proof that $\Phi(G)$ cannot contain a Sylow subgroup of $G$?

Question

As we know, the Frattini subgroup of a finite group G can not contain a Sylow subgroup of G, but if we want to prove this, we need the Schur-Zassenhaus theorem. What I want to know is if there is a more elementary proof of this.

Answer 1

Presumably this doesn't count as S-Z-free:

Let $P \leq \Phi(G)$ be a Sylow $p$-subgroup of $G$. Since $\Phi(G)$ is nilpotent (by the Frattini argument), $P$ is characteristic in $\Phi(G)$, so $P$ is characteristic in $G$ as well. By induction, we may assume $P$ is elementary abelian (mod out by $\Phi(P) \unlhd G$ to get a smaller example). But then, we have a non-split extension of $G/P$ by $P$, contradicting Maschke's theorem that $P$ is an injective $G/P$-module so that $H^2(G/P,P)=0$.