Let $G$ be a group and $P\in Syl_p(G)$, $H$ is normal in $G$. I want to show that $P\cap H\in Syl_p(H)$.
So I let $P_0\in Syl_p(H)$. $P\cap H$ is a $p$ subgroup of $H$, so by Sylow 2nd Theorem, $P\cap H \leq P_0$.
And by Sylow's 2nd and 3rd theorem, I get that there exists $g\in G$ such that $P_0 \leq gPg^{-1}$.
I think I want to prove that $P_0 \leq P\cap H$ next in order to conclude that $P_0=P\cap H$ but got stuck at this part.
Best Answer
You're almost there.
So, you have chosen a $P_0\in Syl_p(H)$ such that $P\cap H\le P_0$. Then, we have $P_0\le gPg^{-1}$. Also, $P_0\le H$, so, as $gHg^{-1}=H$, it means $$g^{-1}P_0\,g\le P\cap H\,.$$ Assuming everything is finite, by calculating sizes, we are ready, as $|g^{-1}P_0\,g|=|P_0|$ and both are included in $H$, so $|P\cap H|=|P_0|$ also follows.