Sylow Theory: $gHg^{-1} \leq P$ Let $G$ be a finite group and $P$ be a Sylow $p$-subgroup of $G$. If
   $H$ is an arbitrary $p$-subgroup of $G$, then there exists a $g \in G$
  such that $gHg^{-1} \leq P$


Proof:

Using the $1$st Sylow theorem, we have that every $p$-subgroup of $G$ is contained in a Sylow $p$-subgroup. 

Also, from the $2$nd Sylow theorem, for every $P'$ Sylow $p$-subgroup there exists a $g \in G$ such that $gP'g^{-1} = P$.

Hence

$$
gHg^{-1} \subset gP'g^{-1}=P
$$

Any suggestions for showing $gHg^{-1}$ is a subgroup of $P$?
Hit: consider the double coset decomposition of $G$ with respect to $H$ and $P$
Hit: consider the double coset decomposition of $G$ with respect to $H$ and $P$