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Let $G$ be a group with $|G|=455$. Show that $G$ is a cyclic group.

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3 Answers 3

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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Looks good, but surely there's a proof not reliant on Sylow? –  Gerry Myerson Oct 18 '12 at 1:55
    
Perhaps...is there? –  DonAntonio Oct 18 '12 at 2:36

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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It 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. –  Gerry Myerson Oct 18 '12 at 1:54

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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7  
Takes 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. –  Gerry Myerson Oct 17 '12 at 1:03

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