Sylow subgroups induce Sylow subgroups in normal subgroups

Let $P \in Syl_p(G)$ and assume N is normal in G. Use 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.

• The first part does not imply that $|P \cap N| = |P|$. It only implies that $[N:P \cap N]$ is not divisible by $p$. – Geoff Robinson Oct 4 '15 at 20:27

Write $|G|=p^r m$ with $p \nmid m$ for some $r \geq 1$. 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 $1 \leq s \leq r$. 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.