Prove $G=N_G(P)H$ where $H \lhd G$ and $P \in Syl_p(H)$.
I definitely need some help with this one. I was thinking that maybe the problem meant to say that $P$ was a sylow subgroup of $G$, that would make this seem much more reasonable. Does anyone have any insights?
Since $H$ is normal in $G$, we have that $N_G(P)H = HN_G(P)$. We also have that $gPg^{-1} \subseteq gHg^{-1} = H$. Let’s apply Sylow’s Theorem to $H$. So we have that given any $g \in G$, there exists $x \in H $such that $ gPg^{-1} = xPx^{-1} $ which implies $x^{-1}g \in N_G(P)$ and so $g \in xN_G(P)$. Since $g$ was arbitrary we’re done.
This theorem is called Frattini’s Argument and can be used to classify finite nilpotent groups.