Exercise 4.5.44 (Sufficient condition to $N_G(P) = C_G(P)$)

Proof. Let $p$ be the smallest prime dividing the order of the finite group $G$, and consider some subgroup $P\in \mathit{𝑆𝑦}{l}_p(G)$, where $P$ is cyclic. Then $|P|=p^{\alpha }$, where

Since $C_G(P)\subseteq N_G(P)$, we want to prove $|N_G(P):C_G(P)|=1$. Put $d=|N_G(P):C_G(P)|$.

By the normalizer-centralizer theorem (Corollary 15)

Since $P$ is cyclic of order $p^{\alpha }$, then

Therefore

Moreover $d$ is a divisor of $|N_G(P)|$, thus $d$ is a divisor of $|G|=p^{\alpha }m$:

If $q$ is a prime divisor of $d$, then by (1), $q=p$ or $q\mid p-1$. In the latter case, $q<p$, but $q$ is a divisor of $|G|$ by (2), and $p$ is the smallest prime dividing $G$: this is contradiction, which proves $q=p$. So $p$ is the only prime divisor of $d$ (if any), therefore there is some integer $k$ such that

Note that $P\subseteq C_G(P)$ since $P$ is abelian, so

hence

Therefore $p^{\alpha +k}\mid p^{\alpha }m=|G|$, where $g.c.d(m,p)=1$, thus $p^{\alpha +k}\mid p^{\alpha }$. This implies $k=0$, thus $|N_G(P):C_G(P)|=p^k=1$, so

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