Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question

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 .

share|cite|improve this answer
Looks good, but surely there's a proof not reliant on Sylow? – Gerry Myerson Oct 18 '12 at 1:55 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.

share|cite|improve this answer
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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.