Task is:
Let $H \le G$ and $P$ be a Sylow $p$-subgroup of $H.$ Let $Q$ be a Sylow $p$-subgroup of $G$ such that $P \le Q$. Show that $P=Q \cap H$.
This seems so obvious but i don't know how to prove it.
Attempt:
Consider a prime in question, $p$. Iterating this $p$ to a power $i$ ,where $i\in\{1,2,3,…,k\}$ where $k$ is the maximal power such that $|G|=p^{k}m$, we would obtain a $p$-subgroup which is the sylow $p$ subgroup of $H$ at a certain index $a \le k$. Iterating this further to $k$, gives us the sylow $p$ subgroup of $G$.
So the sylow subgroup of $H$, group of order $p^{a}$ is a subgroup of the sylow $p$ subgroup of order $p^{k}$.
Iterating $p^{i}$ gives us all $i$ powers of $p$, which means that $Q \cap H$ will give us powers of $i \le a$.
However, that just gives us $p^{a}=|P|$. So $P=Q \cap H$
Feedback and corrections(which are certainly needed) would be appreciated. I think i have the right idea but the proof seems incomplete.
Best Answer
$P$ is certainly inside $Q\cap H$. On the other hand $Q\cap H$ is a $p$-group, being inside $Q$ (by Lagrange). Since $P$ is a maximal $p$-subgroup of $H$, you get $P=Q\cap H$.