D+F 4.5.44

Given $p$ is the smallest prime dividing $|G|$, $P\in{Syl_p(G)}$ and $P$ is cyclic, prove $$N_G(P)=C_G(P)$$ $\textit{Proof}$ : By corollary 15 p.134, $$N_G(P)/C_G(P) \lesssim Aut(P)$$ Since $P$ is cyclic, then by proposition 16 p.135, $$Aut(P) \cong \mathbb{Z}_p^* \cong \mathbb{Z}_{p-1}$$ By Lagrange,
$|N_G(P)/C_G(P)|$ divides $|G|$. Also,
$|N_G(P)/C_G(P)|$ divides $p-1$.
Since $p$ is the smallest prime dividing $|G|$, this forces $|N_G(P)/C_G(P)|=1 \Longrightarrow N_G(P)=C_G(P) \quad \blacksquare $