Suppose that G is a group with more than one element and G has no proper, nontrivial subgroups. Prove that |G| is prime.

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- Sep 29th 2007, 08:27 PMtttcomraderA 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, 03:36 AMRebesques
|G| must have no proper divisors, so...?

- Sep 30th 2007, 08:15 AMtttcomrader
Then the |G| can only be divided by itself or 1. So it is a prime.

This is easy, too, should have gotten it earlier.

Thanks. - Sep 30th 2007, 08:22 AMThePerfectHacker
Choose $\displaystyle x\in G$ so that $\displaystyle x\not = e$ which is possible by hypothesis. Construct the cyclic subgroup $\displaystyle \left< x\right>$. By hypothesis this must generate $\displaystyle G$ for this subgroup is nontrivial. Hence, $\displaystyle G$ is isomorphic to the cyclic group $\displaystyle \mathbb{Z}_n$. Now this group has proper non-trivial subgroups unless $\displaystyle n$ is a prime. Thus, $\displaystyle |G|=p$.