$G$ is a finite group s.t. $(ab)^p=a^pb^p$; $P$ is its $p$-Sylow subgroup. Then there exists $N \unlhd G$ such that $N \cap P = (e)$ and $NP=G$.

The map $\phi:a\mapsto a^p$ is a homomorphism from $G$ to $G$. If $|P|=p^r$, then $\phi^s$ has kernel $P$ wherever $s\ge r$. Let $|G|=p^rm$. Choose $s\ge r$ with $p^s\equiv 1\pmod m$. Let $N$ be the image of $\phi^s$. It has order $m$, by the first isomorphism theorem. Its elements all satisfy $a^m=e$. But if $a^m=e$ then $\phi^s(a)=a$, so $N=\{a\in G:a^m=e\}$, and so is a characteristic subgroup of $G$. But obviously, $P\cap N$ is trivial, and so $PN$ must have order $p^rm=|G|$.