Show a certain group is contained in a Sylow p-group. 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.
 A: 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.
A: Suppose that $|G|=p^nm$, where $p\nmid m$. Let $P$ be a Sylow $p$-subgroup of $G$, and let $\mathscr{C}$ be the set of left cosets of $P$; clearly $|\mathscr{C}|=m$. Let $G$ act on $\mathscr{C}$ in the obvious way: $g\cdot aP=(ga)P$ for each $g\in G$ and $aP\in\mathscr{C}$.
Suppose that $g\cdot aP=bP$; then $h\in\operatorname{Stab}_G(bP)$ iff $hg\cdot aP=g\cdot aP$ iff $g^{-1}hg\cdot aP=aP$ iff $g^{-1}hg\in\operatorname{Stab}_G(aP)$. In particular, $h\in\operatorname{Stab}_G(aP)$ iff $a^{-1}ha\in\operatorname{Stab}_G(P)=P$. Thus, $\operatorname{Stab}_G(aP)=aPa^{-1}$.
Now consider the restriction of this action to $K$. For each $aP\in\mathscr{C}$ we have $$\operatorname{Stab}_K(aP)=K\cap\operatorname{Stab}_G(aP)=K\cap aPa^{-1}\;.\tag{1}$$ $|K|$ is a power of $p$, as is $|aPa^{-1}|=p^n$, so
$$|\operatorname{Orb}_K(aP)|=\frac{|K|}{|\operatorname{Stab}_K(aP)|}=p^{k(aP)}$$ for some $k(aP)\in\Bbb N$. The orbits of the action of $K$ on $\mathscr{C}$ partition $\mathscr{C}$, so $$m=|\mathscr{C}|=\sum_{C\in\mathscr{C}}|\operatorname{Orb}_K(C)|\;.$$ 
Here are some questions to direct you towards finishing the argument. If you get completely stuck, I’ve finished it below the questions but left it spoiler-protected; mouse-over to see it.


*

*Why is it impossible that $k(aP)\ge 1$ for every $aP\in\mathscr{C}$?  

*If $k(aP)=0$, what does $(1)$ tell you about $K$ in relation to $aPa^{-1}$?  

*What kind of subset of $G$ is $aPa^{-1}$?



 If $k(C)$ were positive for every $C\in\mathscr{C}$, $m$ would be a multiple of $p$. It’s not, so there must be at least one $aP\in\mathscr{C}$ such that $|\operatorname{Orb}_K(aP)|=p^0=1$, and hence $K=\operatorname{Stab}_K(aP)=K\cap aPa^{-1}$ by $(1)$. But then $K\subseteq aPa^{-1}$, where $aPa^{-1}$ is a $p$-Sylow subgroup of $G$.

A: Assume $\,|G|=p^nm\,,\,(p,m)=1$:
I assume you already know the following Sylow theorems: Sylow $p$-sbgps. exist in $\,G\,$ , they all are conjugate in $\,G\,$ and their number equals $\,1\pmod p:$ let $\,K\,$ be a $p$-sbgp. and let $\,\mathcal{P}:=\{P\leq G \mid P\text{ is a Sylow }p\text{-sbgp. of }\, G\}\,$. As noted above, $\,|\mathcal{P}|\equiv 1\pmod p$.
Let now $K$ act on $\,\mathcal{P}\,$ by conjugation: $\,k\in K\,,\, P\in\mathcal{P}\Longrightarrow k\cdot P\to k^{-1}Pk=:P^k\,$ . If there is an orbit with one single element , say $\,\mathcal{O}_Q=\{Q\}\,\,,\,\,\text{for some}\,Q\in\mathcal{P}$, then $\,Q^k=Q\,\,\forall\,k\in K\Longrightarrow QK=KQ\Longrightarrow QK\,$ is a $p$-subgroup of $\,G\,$ , so if $\,K\,$ is not contained in any Sylow $p$-sbgp then $\,|QK|>|Q|=p^n\,$, which is absurd, so that all the orbit have size a power of $p$, but this means $\,|\mathcal{P}|\equiv 0\pmod p$, which of course is also absurd since, as we mentioned above, $\,|\mathcal{P}|\equiv1\pmod p\Longrightarrow \,$ it must be that $\,K\,$ is contained in some Sylow $p$-sbgp.
