Now given, a finitely generated subgroup of . This means, by definition, . The subgroup generated by these elements is precisely where . Define, and . Thus, . Define to be the greatest common divisor of . Then, as seen in the first paragraph, . Thus, we see that must be cyclic.
Since it is an infinite cyclic group (assuming ) it is isomorphic to .Prove that this group is not isomorphic to the direct product of two copies of it.
It is trivial to show that is not isomorphic to .