Sylow subgroups of a finite group which are contained in some other subgroup are also Sylow there

Let $G$ be a finite group and $p$ a prime. Prove that if $P \in \mathsf{Syl}_p(G)$ and $P \leq H \leq G$, then $P \in \mathsf{Syl}_p(H)$. Give an example to show that, in general, a Sylow $p$-subgroup of if a subgroup of $G$ need not be a Sylow $p$-subgroup of $G$.

If $P \leq H \leq G$ is a Sylow $p$-subgroup of $G$, then $p$ does not divide $[G : P]$. Now $[G : P] = [G : H] [H : P]$, so that $p$ does not divide $[H : P]$; hence $P$ is a Sylow $p$-subgroup of $H$.

A trivial counterexample to the converse statement is the subgroup $\langle r \rangle \leq D_8$; $\langle r \rangle$ is a Sylow 2-subgroup of itself, but clearly not a Sylow 2-subgroup of $D_8$ since $|D_8| = 2^3$.

I have another example for you, I’m not 100% this works, though: consider $G = Z_{3} \times Z_{3} \times Z_{2}$, and $H = Z_{3} \times Z_{2} \leq G$. Then $Z_{3}$ can be identified inside $H$ as the Sylow 3-subgroup of $H$, however the order of $Z_3$ disqualifies it from being the 3-subgroup of $G$.