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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- September 29th 2007, 09:27 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.

- September 30th 2007, 04:36 AMRebesques
|G| must have no proper divisors, so...?

- September 30th 2007, 09:15 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. - September 30th 2007, 09:22 AMThePerfectHacker
Choose so that which is possible by hypothesis. Construct the cyclic subgroup . By hypothesis this must generate for this subgroup is nontrivial. Hence, is isomorphic to the cyclic group . Now this group has proper non-trivial subgroups unless is a prime. Thus, .