If $N$ is a normal subgroup of $G$, do $n_p(N)$ and $n_p(G/N)$ divide $n_p(G)$? Here $n_p(H)$ represents the number of Sylow-$p$ subgroups of $H$. Thanks in advance.
 A: Hint to get you started: If $N$ is a normal subgroup, then the Sylow $p$-subgroups of $G/N$ are of the form $PN/N$, where $P$ is a Sylow $p$-subgroup of $G$. Similarly by normality of $N$ the Sylow $p$-subgroups of $N$ are of the form $P \cap N$, where $P$ is a Sylow $p$-subgroup of $G$.

The following is way more than you need for this exercise, but I just want to mention this related fact. If $N \trianglelefteq G$, then
$$n_p(G) = n_p(N) n_p(G/N) n_p(T)$$
where $T = N_{PN}(P \cap N)/ P \cap N$. For a proof, see

Marshall Hall Jr., On the number of Sylow subgroups in a finite group, Journal of Algebra
  Volume 7, Issue 3, December 1967, Pages 363–371 DOI


Later edit: Here is a proof of the statement above. Let $P$ be a Sylow $p$-subgroup of $G$. 
We have $n_p(G) = [G:N_G(P)]$ and $n_p(G/N) = [G/N:N_G(PN)/N] = [G:N_G(PN)]$ since $PN/N$ is a Sylow $p$-subgroup of $G/N$. Thus \begin{equation}n_p(G) = n_p(G/N) \cdot [N_G(PN) : N_G(P)] = n_p(G/N) \cdot n_p(N_G(PN)).\tag{1}\end{equation}
Every Sylow $p$-subgroup of $N_G(PN)$ is contained in $PN$, so \begin{equation}n_p(N_G(PN)) = n_p(PN) = [PN : N_{PN}(P)].\tag{2}\end{equation}
Note that $PN = N_{PN}(P)N$, which gives \begin{equation}[PN:N_{PN}(P)] = [N_{PN}(P)N:N_{PN}(P)] = [N : N_{PN}(P) \cap N].\tag{3}\end{equation}
Since $N$ is normal, the intersection $P \cap N$ is a Sylow $p$-subgroup of $N$. Furthermore the normalizer of $P$ is contained in the normalizer of $P \cap N$, so $N_{PN}(P) \cap N \leq N_N(P \cap N)$ and \begin{align}[N:N_{PN}(P) \cap N] &= [N:N_{N}(P \cap N)] \cdot [N_N(P \cap N) : N_{PN}(P) \cap N]\\ &= n_p(N) \cdot [N_N(P \cap N) : N_{PN}(P) \cap N].\tag{4}\end{align}
At this point applying $(1)$, $(2)$, $(3)$, and $(4)$ shows that \begin{equation}n_p(G) = n_p(G/N) \cdot n_p(N) \cdot [N_N(P \cap N) : N_{PN}(P) \cap N].\tag{5}\end{equation}
Note that $[A \cap N : B \cap N] = [B(A \cap N) : B]$ for any subgroups $B \leq A$ of $G$. With $A = N_{PN}(P \cap N)$ and $B = N_{PN}(P)$, we have $B(A \cap N) = A$ since $A = P N_N(P \cap N)$. Thus $[A \cap N : B \cap N] = [A : B]$, that is: \begin{align}[N_N(P \cap N) : N_{PN}(P) \cap N] &= [N_{PN}(P \cap N) \cap N : N_{PN}(P) \cap N]\\ &= [N_{PN}(P \cap N) : N_{PN}(P)] \\ &= n_p(N_{PN}(P \cap N)).\end{align}
Plugging this into $(5)$ gives $n_p(G) = n_p(G/N) \cdot n_p(N) \cdot n_p(N_{PN}(P \cap N))$. 
Finally $P \cap N$ is a normal $p$-subgroup of $N_{PN}(P \cap N)$, from which it follows that $N_{PN}(P \cap N)$ and $N_{PN}(P \cap N) / P \cap N$ have the same number of Sylow $p$-subgroups. This completes the proof of the result by Hall.

Another remark: If $H < G$ is not a normal subgroup, then it is not true in general that $n_p(H) \mid n_p(G)$. For example, for $H = S_3$ in $G = A_5$ we have $n_2(H) = 3$ and $n_2(G) = 5$. 
However, it is always true that $n_p(H) \leq n_p(G)$ whenever $H \leq G$. See answers to this question.
