TITLE: Show a certain group is contained in a Sylow p-group.
QUESTION [15 upvotes]: Statement: Let G be a group and p a prime that divides $|G|$. Prove that if $K\le G$ such that $|K|$ is a power of p, K is contained in at least one Sylow p-group.
I just started studying Sylow p-groups, so although I'm familiar with Sylow theorems and a couple of corollaries, I don't know how to get started with this problem. Any hint is more than welcome.
PS: I looked for something related here at Math.SE but didn't find anything. Sorry if it's a duplicate.

REPLY [4 votes]: Here's another approach. You want to show for any $p$-subgroup $K$, $K\subseteq aPa^{-1}$ where $P$ is a Sylow $p$-subgroup and $a\in G$ . Let $X=\{aP \mid a\in G\}$ be the set of left cosets of $P$ in $G$. Now let $K$ act on $X$ in the following manner $k \cdot(aP)=(ka)P$ . 
Let $|G|=p^n m$ where $p \nmid m$ , we know that $|P|=p^n$ since $P$ is a Sylow $p$-subgroup. Note that $[G:P]=|X|= \displaystyle\frac {|G|}{|P|}=m$, then $p \nmid |X|$. We also know that if a $p$-group acts on a set $X$ then $p\mid |X|-|X_f|$ where $X_f$ is the fixed set under the action (I can supply the proof if needed). But $p \nmid |X|$ then $p \nmid |X_f|$ (otherwise $p \mid |X|-|X_f|+|X_f|=|X|$ which is a contradiction) then $|X_f|\not= 0$ then there is an element $aP\in X$ such that $k(aP)=P \ \forall\ k\in K$. $kaP=aP \implies a^{-1}(ka) \in P \implies k\in aPa^{-1}$ for all $k\in K$ hence $K \subseteq aPa^{-1}$ . 
As a side note you can use this as a lemma to prove the second Sylow theorem.