Let $G$ be a group with $|G|=455$. Show that $G$ is a cyclic group.
Let $G$ a group, with ... Show that $G$ is a cyclic group.
3 Answers
It is well-known that if $n$ is a natural number, there is only one group of order n if and only if $\gcd(n,\varphi(n))=1$. Here $\varphi$ is the Euler totient function. For $n=455$ this applies. If there is only one group of a particular order it must necessarily be cyclic.
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0It is a nice result, and deserves to be better-known. I wonder whether OP is permitted to use it, or is instead required to get hands dirty with the specific number 455. – 2012-10-18
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0Dear, Nicky! I am interested in group theory and I have never met this fact before. Where could I find the readable proof of this result? Hope the proof is not so difficult. – 2018-11-07
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0@RFZ See [here](https://groupprops.subwiki.org/wiki/Classification_of_cyclicity-forcing_numbers#.284.29_implies_.283.29), with $1.\Longleftrightarrow 5.$ and [this MO-question](https://mathoverflow.net/questions/148731/for-which-n-is-there-only-one-group-of-order-n). Try to google yourself for more references! – 2018-11-07
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0@DietrichBurde, I took a look at your link but as far as I realized it requires some knowledge about finite abelian groups, namely that any finite abelian group is the direct product of cyclic groups. And equivalence between (1) and (5) is proved through many intermediate steps – 2018-11-07
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0@RFZ This is why I said you should start searching yourself - because you know best what you will understand, and what you are willing to learn from new vocabulary - not us! – 2018-11-07
Hints for you to prove. Let $\,G\,$ be a group of order $\,455=5\cdot 7\cdot 13\,$ ,then:
1) There exists one unique Sylow $\,7-\,$subgroup $\,P_7\,$ , and one single Sylow $\,13-\,$ subgroup $\,P_{13}\,$ , which are then normal;
2) There exists a normal cyclic subgroup $\,Q\,$ of order $\,91\,$
3) If $\,P_5\,$ is any Sylow $\,5-\,$ subgroup, then we can form the semidirect product $\,Q\ltimes P_5\,$
4) As the only possible homomorphism from a group of order $\,91\,$ to a group of order $\,4\,$ is the trivial one, the above semidirect product is direct .
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0Looks good, but surely there's a proof not reliant on Sylow? – 2012-10-18
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0Perhaps...is there? – 2012-10-18
Notice $455 = 13*7*5$ and we know $13$, $7$ and $5$ are prime. Now use Lagrange theorem to show that G is cyclic. This is a hint.
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8Takes a bit more than Lagrange, doesn't it? I mean, there's no noncyclic group of order 35, but there is one of order 21. – 2012-10-17