Let $G$ be a finite group, and $p$ a prime. Let $P \in \operatorname{Syl}_p(G)$, and let $N$ be a normal subgroup of $G$. Use the conjugacy part of Sylow's theorem to prove that $P \cap N$ is a Sylow $p$-subgroup of $N$. Deduce that $PN/N$ is a Sylow $p$-subgroup of $G/N$.
How do we know that N will contain p-subgroup of G of same order as P ? I don't understand this question could somebody explain more.
Write $|G|=p^r m$ with $p \nmid m$ for some $r \geq 0$. By assumption, $|P| = p^r$. Since $N \triangleleft G$, it follows that $P \cap N \triangleleft P$ and so by Lagrange's Theorem, $|P \cap N| = p^s$ for some $0 \leq s \leq r$. Now, we have that $$|PN| = |P| \cdot |N|/|P \cap N| \quad \implies \quad [N : P \cap N] = [PN : P].$$ Since $[G:P] = [G : PN] [PN : P]$, we must have that $[PN : P] = [N : P \cap N]$ divides $[G:P] = m$. Can you take it from here?
For the second part, use the Second Isomorphism Theorem to get $PN/N \cong P/(P \cap N)$. Show that $PN/N$ is a $p$-subgroup of $G/N$ and use the above result.
I think the point of this question is to find a proof which relies on Sylow's theorem. This is how I did:
Since $P$ is a p-subgroup, $P\cap N$ is also a p-subgroup (of $N$), so by Sylow's theorem $P\cap N\leq P'$ for some $P'\in Syl_p(N)$. But $P'$ is also a p-subgroup of $G$, so $P'\leq gPg^{-1}$ for some $g\in G$. Since $P'$ is a subgroup of the normal subgroup $N$, $P'\leq gPg^{-1}\cap N=gP\cap Ng^{-1}$.
Since we are considering finite groups, and conjugate subgroups have the same order, we must have $P\cap N=P'\in Syl_p(N)$.
(So you don't know a priori that such subgroup of $N$ exists but you start from the weaker p-subgroup condition and prove it's equal to some Sylow subgroup).
For the second part, we have $|PN/N|=|P/P\cap N|=|P|/|P\cap N|$, the latter is $p$ to some power $\alpha$. Here $\alpha$ is the power of $p$ in $|G|$ minus the power of p in $N$ (here we used the first part). This difference is exactly the power of $p$ in $|G/N|$, so it is a Sylow p-subgroup.
