Question about $p$-Sylow subgroups of the quotient group I have been working on the following problem.

Let $G$ be a finite group, $N\trianglelefteq G$ and $p$ a prime, then
  $n_{p}(G/N)\leq n_{p}(G)$.

I have beeen trying to solve it, but it seems I couldn't find the appropiate argument. I read the Keith Conrad's Paper where he gives a proof to this proposition as Theorem 6.7, but he uses a (in my opinion) very sophisticated argument that involves prime relativity between a subgroup and it's index, I don't want to bring such a machinery for this.
The idea I had is exactly the same: To define a natural function from $Syl_{p}(G)$ to $Syl_{p}(G/N)$ and prove it's onto, I guess there is a not so difficult way of doing this.
Any help? Thanks!
 A: I guess it is what you call 'sophisticated'. Let $P$ be a Sylow $p$-subgroup and $N$ a normal subgroup of the group $G$. Write overbar $\overline{H}$ for quotient of subgroups $H$ modulo $N$ (so $\overline{H}=HN/N$). You need a couple of things:
$$\#Syl_p(G)=index[G:N_G(P)] $$
$$N_{\overline{G}}(\overline{P})=\overline{N_G(PN)}             $$
The first fact is standard Sylow theory, and the second one is a fairly standard group theory result - in general $N_{\overline{G}}(\overline{KN})=\overline{N_G(KN)}$ for any subgroup $K$ (see for example I.M. Isaacs' Finite Group Theory, page 23). So,
 $$\#Syl_p(G/N)=index[\overline{G}:N_{\overline{G}}(\overline{P})]=index[\overline{G}:N_{\overline{G}}(\overline{PN})]=index[\overline{G}:\overline{N_{G}(PN)}]=index[G/N:N_G(PN)N/N]=$$ (since $N \subseteq N_G(PN))$ = $$index[G/N:N_G(PN)/N]= index[G:N_G(PN)].$$ Since $N$ is normal we have $N_G(P) \subseteq N_G(PN)$, hence $$index[G:N_G(PN)] \mid index [G:N_G(P)]=\#Syl_p(G).$$ So note that the number of Sylow $p$-subgroups of $G/N$ even divides the number of Sylow $p$-subgroups of $G$! See also the first reference (a paper of Marshall Hall) that Jack Schmidt gave.
