# 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 $|G|$ and the argument you suggest we get $N_G(P)$ is a $p$-group since $([N_G(P):C_G(P),p-1)=1$. Then $P=N_G(P)$ and since $P$ is cyclic, $C_G(P)=P$. – Levent Jul 4 '16 at 2:35
• Thanks. I still I don't see the part where $N_G(P)$ is a $p$-group. Could you please expand on that? – R Maharaj Jul 4 '16 at 2:46

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 $C_G(P)=N_G(P)$.
• I am still having a problem with seeing that $([N_G(P) : C_G(P)] , p-1) =1$ – R Maharaj Jul 4 '16 at 3:17
• Well, $p$ is the smallest prime dividing $|G|$, so if you take a prime $q$ dividing $[N_G(P):C_G(P)]$, necessarily $q>p$. Therefore the index is coprime to $p$. – Levent Jul 4 '16 at 3:20
• aha! I see. On that note, how does $|G|$ relate to $|\operatorname{Aut}(G)|$? – R Maharaj Jul 4 '16 at 3:26
• All you can say is $[G:Z(G)]$ divides $|Aut(G)|$, nothing more as I know in the general case. They are not that related. – Levent Jul 4 '16 at 3:28