Prove that any group of order 561 is cyclic.
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4As 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? – Dec 13 '13 at 07:17
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15 not prim but is cyclic. – ayoob Dec 13 '13 at 07:20
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@Magdiragdag : fine fine :) – Dec 13 '13 at 07:23
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What would be prime factorization of $561$? – Dec 13 '13 at 07:26
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2@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. – Dec 13 '13 at 07:27
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we shoud use this theorem – ayoob Dec 13 '13 at 07:29
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2N(H)/C(H) -----> AUT(H) normalizer - central theorem – ayoob Dec 13 '13 at 07:31
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What theorem? But let me guess, the third Sylow theorem. Still, you could at least formulate that theorem. – Magdiragdag Dec 13 '13 at 07:31
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DO you mind to edit your question to "Use N/C theorem"...? – Dec 13 '13 at 08:51
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I don't understand how this question has 6 upvotes (as of now). – Bruno Stonek Dec 26 '13 at 09:51
2 Answers
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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.
Nicky Hekster
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@Praphulla - well if a proof is given by mathematics2x2life making use of Sylow's Theorems, what exactly is the definition of advanced then? If you know Sylow Theory, then the criterium as mentioned is not far away from an actual proof. – Nicky Hekster Dec 13 '13 at 12:00
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@NickyHekster : I have not seen this statement before.. It may be advanced or not.. I can not comment anything on that.. All i could say is "thank you for giving another exercise which looks so beautiful" :) – Dec 13 '13 at 12:06
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@Praphulla - OK, it has been discussed many times here at this Math StackExchange. In fact, there is a paper of Pakianathan and Shankar (2000) which gives characterizations of the set of positive integers n such that every group of order n is (i) cyclic, (ii) abelian, or (iii) nilpotent. See also http://math.stackexchange.com/questions/67407/group-of-order-15-is-abelian/67757#67757 !! – Nicky Hekster Dec 13 '13 at 12:20
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@Praphulla - I always find it extremely intriguing that from the order only of a finite group one can read of some special properties! – Nicky Hekster Dec 13 '13 at 12:30
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1@NickyHekster This is really a neat fact. I had never heard of it either, until now. – rschwieb Dec 18 '13 at 16:05
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1@rschwieb Yes, I think so too! And the nice thing of it it is so simple and you will never forget it! – Nicky Hekster Dec 18 '13 at 16:26
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Wow! This is a very interesting theorem! Thank you for sharing it! – Tyler Clark Dec 25 '13 at 07:56
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I do not see why this fact tells us that the given group is necessarily cyclic, could you explain please? – Brain May 26 '22 at 15:32
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Well, $561=3 \cdot 11 \cdot 17$. So $\varphi(561)=2^6 \cdot 5$, whence relatively prime to $561$. And for any natural number $n$ there is always the cyclic group $C_n$ of that order…. – Nicky Hekster May 26 '22 at 17:30
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Firstly, $|G|=561=3.11.17$. By Sylow III, $n_{11}\mid 3.17$ and $n_{11}\equiv 1\pmod {11}$; so, $n_{11}=1$. Say $K$ the $11$-Sylow subgroup. Since it is normal, $K$ is the union of conjugacy classes of $G$, with as many singletons as the order of $|K\cap Z(G)|$. But, by Lagrange, $|K\cap Z(G)|=1$ or $11$, and the former option is ruled out because there isn't any union of noncentral conjugacy classes of $G$ summing to size $10$. Therefore $K$ is central, and being $G/K$ cyclic (as $3\nmid 17-1$), then $G$ is abelian and finally cyclic (take the product of any three elements of order $3$, $11$ and $17$, respectively).
citadel
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