Frattini Argument

Frattini Argument: Let $G$ be a finite group , let $N\trianglelefteq G$ be a normal subgroup , and let $P$ be a Sylow p-subgroup of $N$ (so $p$ divides $|N|$). Let $N_G(P)$ denote the normalizer of $P$ in $G$. Then $G = N\,N_G(P).$

Equivalently, every $g\in G$ can be written as $g=nn'$ with $n\in N$ and $n'\in N_G(P)$.

Proof sketch: For $g\in G$, the conjugate $gPg^{-1}$ is a Sylow $p$-subgroup of $gNg^{-1}=N$ (normality is used here). By Sylow's second theorem inside $N$, there exists $n\in N$ with $n(gPg^{-1})n^{-1}=P$. Then $ng\in N_G(P)$, so $g=n^{-1}(ng)\in N\,N_G(P)$.