## With p minimal, the normalizer and centralizer of a cyclic Sylow p-subgroup are equal

Let $p$ be the smallest prime dividing $|G|$, and suppose that some $P \in \mathsf{Syl}_p(G)$ is cyclic. Prove that $N_G(P) = C_G(P)$.

Write $|G| = p^km$ where $p$ does not divide $m$. Note that $P \cong \mathbb{Z}/(p^k)$. Since $P$ is abelian, we have $P \leq C_G(P) \leq N_G(P)$. Now $N_G(P)/C_G(P) \leq \mathsf{Aut}(P) \cong (\mathbb{Z}/(p^k))^\times$. Note that $\mathsf{Aut}(P)$ is a group of order $\varphi(p^k) = p^{k-1}(p-1)$, where $\varphi$ denotes the Euler totient function. Moreover, $|N_G(P)/C_G(P)|$ divides $m$. Since every prime divisor of $m$ is greater than $p$, $N_G(P)/C_G(P) = 1$. Thus $C_G(P) = N_G(P)$.