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

Question

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

Answer 1

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.

Answer 2

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 .

Answer 3

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.