Problem: Let $p$ be the smallest prime dividing the order of the finite group $G$. If $P \in Syl_p(G)$ and $P$ is cyclic prove that $N_G(P) = C_G(P)$.
I know the standard proof constructing an injection $\pi:N_G(P)/C_G(P)\to Aut(P)$ and calculating the order of $N_G(P)/C_G(P)$. But I remember my teacher could prove it using neither $N_G(P)/C_G(P)$ nor the injection $\pi$. Instead he proved it by constructing several (not only one, as I can vaguely remember) group actions and calculating orbits, which is quite impressive to me at that time. But I cannot remember his method now. Can anyone help reproduce it?