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Prove that any group of order 561 is cyclic.

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    $\begingroup$ As you are a new user I would like to tell you that : It would not be enough to just write the question to get a reply... please explain what you have tried? $\endgroup$ – user87543 Dec 13 '13 at 7:17
  • $\begingroup$ 15 not prim but is cyclic. $\endgroup$ – ayoob Dec 13 '13 at 7:20
  • $\begingroup$ @Magdiragdag : fine fine :) $\endgroup$ – user87543 Dec 13 '13 at 7:23
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    $\begingroup$ @ayoob: At least you should describe your background, what did you learn and what difficulties did you encountered so that you know what hints and answers should be given. $\endgroup$ – user99914 Dec 13 '13 at 7:27
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    $\begingroup$ N(H)/C(H) -----> AUT(H) normalizer - central theorem $\endgroup$ – ayoob Dec 13 '13 at 7:31
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In general, there is only one group of order $n$ iff gcd$(n,\varphi(n))=1$. Of course such a group must be necessarily cyclic. 561 satisfies the condition.

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  • $\begingroup$ Isn't this more advanced for a beginner? $\endgroup$ – user87543 Dec 13 '13 at 8:43
  • $\begingroup$ ????????????????????????? $\endgroup$ – ayoob Dec 13 '13 at 9:44
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    $\begingroup$ @ayoob - what is your confusion? $\endgroup$ – Nicky Hekster Dec 13 '13 at 11:58
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    $\begingroup$ @NickyHekster This is really a neat fact. I had never heard of it either, until now. $\endgroup$ – rschwieb Dec 18 '13 at 16:05
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    $\begingroup$ @rschwieb Yes, I think so too! And the nice thing of it it is so simple and you will never forget it! $\endgroup$ – Nicky Hekster Dec 18 '13 at 16:26

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