So someone gave me the hint to prove that if $K$ is a normal subgroup of $G$ and $P \in$ Syl$_p(G)$ then $N_G(PK) = N_G(P)K$. I was easily able to prove the hint using the Frattini argument, but I don't see how it helps me solve the problem. I was thinking that the trick might be to come up with an appropriate chose for $K$, and the only thing I can think of that makes sense for $K$ would be to take the product of the intersection of all Syl$_q$ groups for each prime $q|n$ since I know this will be a normal subgroup. But I just don't know what to do with this.
I also know that we can imbed $G$ into $S_n$ by letting $G$ act on all the cossets of $N_G(P)$, but not sure how to use this.
Please give a hint as apposed to a solution.
Theorem Let $N \unlhd G$, then $n_p(G/N) \mid n_p(G)$. And equality holds if and only if $N \subseteq N_G(P)$.
Proof $n_p(G/N)=|G/N:N_{G/N}(PN/N)|=|G/N:N_G(P)N/N|=|G:N_G(P)N|$, and the theorem follows. $\square$
Corollary Let $G$ be a finite group, then there is a subgroup of $S_{n_p(G)}$ having the same number of Sylow subgroups as $G$.
$\textbf{Proof}$ If $P \in Syl_p(G)$. $G$ acts by left multiplication on the left cosets of $N_G(P)$ and then (see for example 1.1 Theorem in M.I. Isaacs, Finite Group Theory), $G/core_G(N_G(P))$ embeds homomorphically into $S_{n_p(G))}$. But $n_p(G/core_G(N_G(P))=n_p(G)$ according to the previous Theorem. $\square$