# If $P \in Syl_p(G)$ and $P$ is cyclic, then $N_G(P)=C_G(P)$

Let $G$ be a group such that $|G|=p^a m$ where $p$ is the smallest prime divisor of $|G|$. If $P \in Syl_p(G)$ and $P$ is cyclic, then $N_G(P)=C_G(P)$

Proof

First, note that $C_G(P) \leq N_G(P) \leq G$. Thus, we are done if $|N_G(P)|/|C_G(P)|=1$. Since $P \leq G$, by Corollary 15, $N_G(P)/C_G(P)$ is isomorphic to a subgroup of $Aut(P)$. Since $P$ is cyclic of order $p^a$, we have $P \cong \mathbb{Z}/p^a \mathbb{Z}$. Thus, $|Aut(P)| = |Aut(\mathbb{Z}/ p^a \mathbb{Z})| = |(\mathbb{Z}/p^a \mathbb{Z})^{\times}|$ since $Aut(\mathbb{Z}/ p^a \mathbb{Z})| \cong (\mathbb{Z}/p^a \mathbb{Z})^{\times}$. Then, $|Aut(P)| = \phi(p^a) = p^{a-1}(p-1)$ where $\phi$ is Euler's totient function. Thus, $N_G(P)/C_G(P)$ divides $p^{a-1}(p-1)$.

Since $P$ is cyclic, $P$ in particular is abelian, and it follows that $P \leq C_G(P) \leq N_G(P)$. Since $P \leq N_G(P)$, there exists a positive integer $k_1, k_2$ such that $|N_G(P)|=k_1 p^a$ and $|C_G(P)|=k_2 p^a$. Since $k_1 p^a$ divides $p^a m$, $k_1$ must divide $m$. Since $|N_G(P)|/|C_G(P)|=k_1/k_2$, $|N_G(P)/C_G(P)|$ divides $m$.

Since the prime divisors of $m$ are greater than $p$, it must be that $|N_G(P)/C_G(P)|=1$. This completes the proof.

I don't really like the way argued the division of $m$. Is there a more succinct and better way to argue this?

Since $P \subseteq C_G(P) \subseteq N_G(P)$ and $P$ is a Sylow subgroup it follows that $|N_G(P):C_G(P)|$ is not divisible by $p$. So it must divide $p-1$. But $p$ is the smallest prime dividing the order $|G|$ and this means that $|N_G(P):C_G(P)|=1$.