Let $p$ be a prime. Suppose $N$ is a normal subgroup of a finite group $G$. If $n_{p,G}$ is the number of Sylow $p$-subgroups of $G$, and $n_{p,N}$ is the number of Sylow $p$-subgroups of $N$, then $n_{p,N}$ divides $n_{p,G}$.
I've been working on this problem for a while now. I have shown that if $P_G \in Syl_p(G)$, i.e., $P_G$ is a Sylow $p$-subgroup of $G$, then $P_G \cap N \in Syl_p(N)$, and that if $P_N \in Syl_p(N)$, then $P_N = P_G \cap N$ for some $P_G \in Syl_p(G)$. In conclusion, there is a surjective function $f:Syl_p(G) \to Syl_p(N)$ defined by $f(P) := P \cap N$. Therefore, $n_{p,N} \leq n_{p,G}$.
However, I cannot prove the divisibility relation.
Any help will be appreciated.