# Sylow subgroups induce Sylow subgroups in normal subgroups

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.

• The first part does not imply that $|P \cap N| = |P|$. It only implies that $[N:P \cap N]$ is not divisible by $p$. Oct 4, 2015 at 20:27

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.