Thread: 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.

|G| must have no proper divisors, so...?

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.

Originally Posted by tttcomrader

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

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, .