# A group with no subgroups

• Sep 29th 2007, 09:27 PM
A group with no subgroups
Suppose that G is a group with more than one element and G has no proper, nontrivial subgroups. Prove that |G| is prime.
• Sep 30th 2007, 04:36 AM
Rebesques
|G| must have no proper divisors, so...?
• Sep 30th 2007, 09:15 AM
Choose $x\in G$ so that $x\not = e$ which is possible by hypothesis. Construct the cyclic subgroup $\left< x\right>$. By hypothesis this must generate $G$ for this subgroup is nontrivial. Hence, $G$ is isomorphic to the cyclic group $\mathbb{Z}_n$. Now this group has proper non-trivial subgroups unless $n$ is a prime. Thus, $|G|=p$.