Let $G$ be a finite group, $p$ a prime that divides $|G|$, and $P$ a $p$-subgroup of Sylow of $G$. Show that $P$ is the unique $p$-subgroup of Sylow that is in $N_G(P)$.
I tried this by assuming there is another $p$-subgroup of Sylow called $Q$. So by the second Sylow's theorem, $Pg=gQ$; but I don't see something to justify that $P=Q$. Any help would be appreciated.
Best Answer
If $Q$ is another $p$-Sylow subgroup of $N_G(P)$ then indeed by the Sylow theorems we have $Q=gPg^{-1}$ for some $g\in N_G(P)$. But $P$ is a normal subgroup of $N_G(P)$, and so $gPg^{-1}=P$.