Centralizer of a cyclic sylow $p$-subgroup Let $G$ be a group where $|G|=p^nm$ and $p$ is the smallest prime dividing $|G|$. Suppose that $P \in \operatorname{Syl}_p(G)$ is cylic. Then $C_G(P)=N_G(P)$
Now I know $|N_G(P)/C_G(P)|$ must divide $|\operatorname{Aut}_G(P)| = p^{n-1}(p-1)$ but I can't seem to show why $|N_G(P)/C_G(P)|$ divides $m$?

Since $p$ is the smallest prime dividing $|G|$, we get $([N_G(P):C_G(P)],p-1)=1$, so $[N_G(P):C_G(P)]$ divides $p^{n-1}$. Clearly, $P\subseteq C_G(P)$ since $P$ is abelian, then $[N_G(P):C_G(P)]$ cannot be divisible by $p$. These two facts together implies $C_G(P)=N_G(P)$.
