Decide whether the following staement is true of false. If true, prove it. If false, provide a counterexample.
Let $G$ be a finite group and $P\lhd G$ $p$-sylow. For all $H<G$, $P\cap H$ is $p$-sylow of $H.$
I think this statement is true, I thouhgt about taking $p$-sylow of $H,$ $H_1$ , extend it to $p$-sylow of G. And think about $g^{-1}H_1g$ for $g\in G$ but I don't really know how to prove it.
Please if you know how help me to prove it.
Without normality it's not true.
Take a group $G$ with two different $p$-Sylow subgroups $P_1$ and $P_2$. Then $P_1 \cap P_2 < P_1$, but the $p$ sylow subgroup of $P_1$ is $P_1$ itself.
Under the assumption that the sylow subgroup is normal, it is true. The point is that $P \cap H$ is a $p$-group, and hence contained in a sylow subgroup of $H, Q$. Since $Q$ is also a $p$-group of $G$, it must be contained in a $p$ sylow subgroup of $G$. But since the $p$-sylow subgroup of $G$ is normal, and hence unique (they are all conjugates of each other), $Q$ is contained in $P$. This implies that $Q$ is contained in $P \cap H$, so $Q = P \cap H$.