Sylow Theory: $gHg^{-1} \leq P$

Question

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 $P$?

Answer 1

Hit: consider the double coset decomposition of $G$ with respect to $H$ and $P$

Answer 2

The proof runs as follows: let $\Omega=\{gP:g \in G\}$, the set of left cosets of $P$ in $G$. $H$ acts on $\Omega$ by left multiplication. Note that since $P \in Syl_p(G)$, $|G:P|=\#\Omega$ is not divisible by $p$. Hence under the action of $H$ there must be an orbit $\mathscr{O}$ whose cardinality is not divisible by $p$. Now, $\#\mathscr{O}=|H:K|$, for some subgroup $K \leq H$ ($K$ is the stabiliser of some element of $\mathscr{O}$). In particular, $\#\mathscr{O}$ is a power of $p$. But $\#\mathscr{O}$ was also a $p'$-number hence $\mathscr{O}$ is a singleton, say $\mathscr{O}=\{gP\}$ for some $g \in G$. This means that $gP$ is fixed by the action of $H$, so if $h \in H$, then $hgP=gP$, implying that $h \in P^{g^{-1}}$, hence $H \subseteq P^{g^{-1}}$ or equivalently $g^{-1}Hg \subseteq P$.