2
$\begingroup$

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

3 Answers 3

4

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.

  • 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
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 .

  • 0
    Perhaps...is there?2012-10-18
-2

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.

  • 8
    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 21.2012-10-17