Sylow-Subgroups and arbitrary groups where their order contains the same prime-power.

Question

Let $|G| = p^k m$ with $p$ and $m$ being coprime. Then it is well known that there exists a subgroup $S$ of $G$ with $|S| = p^k$, the so called Sylow-$p$-subgroups. Now let $U \le G$ be some subgroup of $G$ with $|U| = p^k \cdot n$, then does there exists some Sylow-$p$-subgroup such that $S \le U$?

Answer 1

If the exponent of $p$ is the same in $|U|$ and $|G|$, a $p$-Sylow of $U$ is in particular a $p$-sylow for $G$, and so the answer is yes