Prove that $n_p(N)$ divides $n_p(G)$ Let $N$ be a normal subgroup of $G$ where $G$ is finite group, then we have to prove $n_p(N)$ divides $n_p(G)$ ( here $n_p(G)$ means number of sylow $p$-subgroups of $G$)
I was able to prove that $n_p(N)$ $\leq$ $n_p(G)$ as by proving first that for any $P \in Syl_p(G)$ there is a conjugate $gPg^{-1}$ such that $gPg^{-1}\cap H \in Syl_p(H)$ for any subgroup $H$ of $G$ whether normal or not.( by considering action of $H$ on $G/P$ and then considering $stab_gP$ 
And by replacing $H$ by $N$ i.e. a normal subgroup I could show that for any $P \in Syl_p(G), P\cap N \in Syl_p(N)$ and all $p$-sylow subgroups of $N$ arise in this way.
It all follows from a paper of Conrad, but from here and this proof i can't seem to figure out how to show it will divide $n_p(G)$. Please help.
 A: @Bhaskar, I promised you a proof - did not find the time to write it down earlier. Although Mesel already gave you one, mine is of a different flavor and sticks more closely to Sylow theory. You already mentioned that you know that Sylow $p$-subgroups of a normal subgroup $N$ of a group $G$, are of the form $P \cap N$, where $P \in Syl_p(G)$. I will assume that you are familiar with the Frattini Argument and will also use the following easy lemma.
Lemma Let $G$ be a a group, $H \leq G$ and $N \trianglelefteq G$. Then for any $g \in G$ we have $(H \cap N)^g=H^g \cap N$.
Proof. Since $H \cap N \subseteq H$, $(H \cap N)^g \subseteq H^g$, and likewise $(H \cap N)^g \subseteq N^g=N$ (the latter equality follows from the fact that $N$ is normal. Hence $(H \cap N)^g \subseteq H^g \cap N$. Conversely, let $x \in H^g \cap N$, then $x \in N$ and $x=g^{-1}hg$ for some $h \in H$. Hence $h= gxg^{-1} \in N$, since $x \in N$ and $N$ is normal. So, $x=g^{-1}hg \in (H \cap N)^g$  and we are done.
Now let us get to the proof of your post.
Theorem Let $G$ be a group, $N \trianglelefteq G$, and $p$ a prime dividing the order of $G$. Then $$\#Syl_p(N) \mid\#Syl_p(G).$$
Proof. Let $P \in Syl_p(G)$, then $P \cap N \in Syl_p(N)$. Owing to the Frattini Argument we obtain $G=N_G(P\cap N)N$. Observe that $N \cap N_G(P \cap N)=N_N(P \cap N)$ and hence $$[G:N_G(P \cap N)]=[N_G(P \cap N)N:N_G(P \cap N)]=[N:N_N(P \cap N)]=\#Syl_p(N).$$
Finally let $g \in N_G(P)$, then the lemma gives $(P \cap N)^g=P^g \cap N =P \cap N.$ It follows that $N_G(P) \subseteq N_G(P \cap N)$, and thus $\#Syl_p(N)=[G:N_G(P \cap N)] \mid [G:N_G(P)]=\#Syl_p(G)$. $\square$
A: Theorem: Let $G$ act on the sets $\Omega_1,\Omega_2$ transitively, and assume that there is an onto function $\phi:\Omega_1\to \Omega_2$ such that $\phi(g.x)=g.\phi(x)$ for all $g \in G$ and $x \in \Omega_1$. Then $|\Omega_2|$ divides $|\Omega_1|$.
Proof:  Let $\alpha\in \Omega_1$ and $g\in G_{\alpha}$ (stabilizer of $\alpha$). Since $$g\phi(\alpha)=\phi(g\alpha)=\phi(\alpha),$$ we have $G_{\alpha}\leq G_{\phi(\alpha)}$. Since $|\Omega_1|=|G:G_\alpha|$ and $|\Omega_2|=|G:G_{\phi(\alpha)}|$, the result follows from this.
Now, Set $\Omega_1=Syl_p(G)$ , $\Omega_2=Syl_p(N)$ and $\phi(P)=P\cap N$. Now, actions of $G$ on the sets by conjugation are transitive and $\phi$ has desired properties so we are done.
Note: If you do not want mention the theorem, you can directly use the idea in the proof to show the result.
A: This is not so trivial to prove. Even more is true.
Theorem (M. Hall, 1967) Let $N \unlhd G$, $P \in Syl_p(G)$, then $n_p=a_pb_pc_p$, where 
$a_p = \#Syl_p(G/N)$
$b_p=\#Syl_p(N)$ and 
$c_p=\#Syl_p(N_{PN}(P\cap N)/(P \cap N))$.
