Let H be a Sylow p-subgroup of G. Prove that H is the only Sylow p-subgroup of G contained in N(H).
I saw a proof online that was pretty long, but can't I just argue that if $H \subset N(H)$, then $H$ conjugates with itself. Then there is only one such p-subgroup H of G by the third Sylow Theorem?
Note that if $|G|=p^km,p\not |m$ and $H$ is $p$-sylow subgroup then $|H|=p^k$. Also $H$ is normal in $N(H)$ therefore $H$ is a normal $p$-sylow subgroup of $N(H)$ and any $p$-sylow subgroup of $N(H)$ is of the form $gHg^{-1}$ where $g\in N(H)$ by 2nd sylow theorem. But $H$ is normal in $N(H)\implies$ all conjunction of $H$ by elements of $N(H)$ is same as $H$. That is $H$ is the only normal subgroup of $N(H)$.