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This is Problem 66 of Chapter 24- Contemporary Abstract algebra-Gallian

Let $p$ be the smallest prime that divides the order of a finite group G. If H is a Sylow-p subgroup and is cyclic, show that $N(H)=C(H)$. Here $N(H)$ is the normalizer of $H$.$C(H)$ is the centralizer of H.

I tried but not have the right idea to proceed. There is a theorem saying that $\frac{N(H)}{C(H)} \cong$ to some subgroup of $Aut(H)$. H is cyclic of order $p^{n}$. I was trying to show somehow that $|N(H)|/|C(H)|=1$. But it doesn't seem to work.

Thanks in advance for the help!!!!

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  • $\begingroup$ I have changed your title in order that later on, your question will be accessible. The former title had no interest for serching. $\endgroup$ – Jean Marie May 6 '17 at 7:48
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The group $N(H)/C(H)$ maps injectively into Aut$(H)$. But if $H$ is cyclic of order $p^k$, then Aut$(H)$ has order $p^{k-1}(p-1)$. As $p$ is the smallest prime factor of $|G|$, the image in Aut$(K)$ has $p$-power order. Unless that order is $1$, $N(H)$ will have a larger $p$-subgroup than $H$.

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$N(H)/C(H)$ is isomorphic to some.subgroup of Aut(H) where Aut(H) has order $\phi(p^k)$ but H is a subgroup of C(H) so order of $ N(H)/C(H) $ is a factor of p-1 but this can't happen u less it is 1 since p is the smallest prime divisor of o(G). Thus$ N(H)=C(H)$

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