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Let $p$ be the smallest prime dividing the order of a finite group $G$. If $P$ in $\operatorname{Syl}_p(G)$ and $P$ is cyclic, prove that $N_G(P)=C_G(P)$.

This is not homework. It is from Dummit and Foote. I'm not sure how to apply that $p$ has the smallest order.

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Perhaps this theorem can shed some insight into why the smallest $p$ is useful? groupprops.subwiki.org/wiki/… –  Ian Coley Apr 18 '13 at 17:49
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Hint: Use that $N_G(P)/C_G(P)$ is isomorphic to a subgroup of $\rm{Aut}(P)$ which has order $p-1$ which by assumption is coprime to the order of $G$. –  Tobias Kildetoft Apr 18 '13 at 17:53
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Take $\Sigma:N_G(P)\rightarrow \operatorname{Aut}(P)$ by $\Sigma(g)=\sigma_g:x\mapsto g^{-1}xg$. Then $N_G(P)/C_G(P)$ is isomorphic to $\Sigma[N_G(P)]$. Since $P$ is cyclic $\operatorname{Aut}(P)$ has order $\varphi(p^n)=p^{n-1}(p-1)$ where $p^n=|P|$. Furthermore, $P$ centralizes itself, so $\Sigma[P]=1$. All other subgroups of $N_G(P)$ must have order that does not divide $p^{n-1}(p-1)$, as by assumption all other primes are greater than $p$. Thus $\Sigma[N_G(P)]=1$. This completes the proof.

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