Let $P$ be a Sylow $p$-subgroup of a finite group $G$, for some prime $p$. Prove that if $H$ is a subgroup of $G$ that contains all Sylow $p$-subgroups of $G$, then $G = HN_G(P)$.
Here's what I have so far:
I know that the normalizer of $P$ in $G$ is $$N_G(P) = \left\{g \in G \,\mid\, gAg^{-1} = A\right\}$$
where $$gAg^{-1} = \left\{ gag^{-1}\,\mid\, a \in A \right\} $$
and that $HN_G(P)$ refers to the product of subgroups.
If the order of $G$ is $p^n$ for some $n \in \mathbb{N}$, then the multiplicity of $p$ is $n$. This means $G$ is a Sylow $p$-subgroup.
If we choose $P = G$, then $H$ must contain $P$, so $H = G$. $N_G(G) = G$, so $$HN_G(P) = GG = G$$
This isn't remotely a proof, only a case in which what I'm trying to prove is true. A hint would be great. Thanks in advance.