# Simple Consequence of Sylow Theorem

I am having a hard time understanding what should be a simple consequence of the Sylow theorems. If $P$ is the unique Sylow p-subgroup of a group $G$ then $P$ is normal in $G$. The argument goes that since $P$ is the only Sylow p-subgroup of $G$ then for all $g$ in $G$, $gPg^{-1} = P$. However, if I recall the theorem goes that all Sylow p-subgroups conjugate in G. So I thought that this would imply that there exists at least one g in G such that $gPg^{-1} = P$, but does not necessarily imply that all $g$ in $G$ achieve this task. Otherwise this would imply that we can find all Sylow p-subgroups by picking one and conjugating it with all the elements of g, each time getting another Sylow p-subgroup, which I know is correct. I just dont understand how the theorem says that. Thanks.

• The statement is much simpler than Sylow's theorems, and holds for any fixed order of subgroup (i.e. if there is some $m$ such that the group has a unique subgroup of order $m$ than that subgroup is normal). – Tobias Kildetoft Nov 25 '16 at 14:40

Show that for any $g\in G$,$|gPg^{-1}|=|P|$ and $gPg^{-1}$ is a subgroup of $P$. Hence $gPg^{-1}$ is a Sylow $p$-subgroup of $G$. Then use the hypothesis of your question.
The theorem says that if $P$ and $Q$ are both Sylow $p$-subgroups of a group $G$, you can find a $g \in G$ with $P^g=Q$. In other words, $G$ works transitively on the set of all Sylow $p$-subgroups by conjugation. Hence, if there is only one Sylow $p$-subgroup $P$, any conjugate of $P$ is this subgroup $P$ itself, whence the normality.
Re: "So I thought that this would imply that there exists at least one $g$ in $G$ such that g$Pg^{-1} = P$, but does not necessarily imply that all $g$ in $G$ achieve this task. "
It actually does imply that all $g$ achieve that.
Note that for any $g\in G$, $|gPg^{-1}|=|P|$, i.e. $gPg{-1}$ is a Sylow $p$-subgroup.
Since there is only one Sylow $p$-subgroup, $gPg^{-1}=P$, that is, $P$ is a normal subgroup.