D+F 4.5.35

Let $P \in{Syl_p(G)}$ and $H \leq G$. Show that $\exists g\in{G}$ s.t. $$gPg^{-1} \cap H \in{Syl_p(H)}$$ $\textit{Proof}$ : Let $Q \in Syl_p(H)$.

Then, $Q \leq H$ and $\exists g \in{G}$ and $\exists P \in{Syl_p(G)}$ s.t. $Q \leq gPg^{-1}$ by Sylow (ii). This means that $$Q \subseteq gPg^{-1} \cap H$$ We have found a $p$-group of $H$ that contains $Q$. Thus, $$Q \leq gPg^{-1} \cap H \in{Syl_p(H)} \quad \blacksquare$$