I am trying to prove a consequence of Sylow's theorems. Let $G$ be a group with $|G|=p^km$, where $p$ is a prime and $k, m$ are integers such that $p$ does not divide $m$. Let $Q$ be a $p$-subgroup of $G$, that is $|Q|=p^i$, for some $i \leq k$.
Then there esists a $p$-Sylow $P$ such that $Q \leq P$.
Here's my attempt:
Let $\Omega$ be set of $p$-Sylow of G. Let us consider the action $\Omega \times Q \to \Omega$, which sends the pair $(K,g)$ to $K^g=\{g^{-1}kg|k\in K\}$.
Since $Q$ is a $p$-group, we have that $|Fix(Q)|\equiv |\Omega|\pmod p$. Thanks to Sylow's theorems, we know that $|\Omega|\equiv 1\pmod p$, then $|Fix(Q)|\equiv 1\pmod p$.
This means that $Fix(Q) \neq \emptyset$. Let $P$ be a fixed point of $Q$.
We have that $P^x=P$, for every $x \in Q$.
Which means that $Q \leq N_G(P)$, where $N_G(P)$ is the normalizer of $P$ in $G$. It is true then that the set $PQ$ is a subgroup of $G$.
We must have that $|PQ|=\cfrac{|P||Q|}{|P \cap Q|}=\cfrac{p^kp^i}{|P \cap Q|}$.
Here, $|P|=p^k$, because $P$ is in $\Omega$.
To conclude the proof I would need to show that $|PQ|=p^k$, so that $|P \cap Q|=p^i$ and, since $P \cap Q$ is a subgroup of $Q$, we would have $Q=P \cap Q$, that is $Q \leq P$, but that is where I am struggling, what do you think?
$P$ is normal and Sylow in $N_G(P)$. If $Q \subseteq N_G(P)$, then the subgroup $PQ \subseteq N_G(P)$, and as you showed, $PQ$ is a $p$-group containing $P$. But $P$ is Sylow, so a maximal $p$-subgroup, implying $PQ=P$, whence $Q \subseteq P$.