Originally Posted by

**guildmage** Tell me if I got you right:

Well, $\displaystyle P$ is given to be a Sylow p-subgroup of $\displaystyle G$. I also know that each conjugate of $\displaystyle P$ is also a Sylow p-subgroup. Now $\displaystyle gP{g^{ - 1}} \subseteq N$ for all $\displaystyle g \in G$ because $\displaystyle N \triangleleft G$. Since $\displaystyle P \triangleleft N$, then $\displaystyle P$ must be the only Sylow p-subgroup of $\displaystyle N$. Since $\displaystyle P$ is a Sylow p-subgroup of $\displaystyle G$ (which is also the only Sylow p-subgroup of $\displaystyle G$), then $\displaystyle P$ must also be the only Sylow p-subgroup of $\displaystyle G$. Therefore, $\displaystyle P \triangleleft G$.

With this proof, however, I used the fact that if $\displaystyle P \triangleleft N$ then $\displaystyle P$ is unique. But the notes I have only guarantees the other way around: that if $\displaystyle P$ is the only Sylow p-subgroup of $\displaystyle N$, then $\displaystyle P \triangleleft N$. Is this statement a really biconditional? Because if it is, I would have to show it first.