Hm, it certainly cannot be both, as then it would be isomorphic to some .
ps. where do u come up with these?
I came up with a new group.
Let where .
Then under complex multiplication will form a group.
I hope this is group is not isomorphic to any of the known ones. And hopefully has some interesting properties.
We know the following:
1) is an abelian group.
2) is a countable group.
3*) has a cyclic subgroup for every finite order.
What are the answers to the following questions:
1)Is is free abelian group?
2)Is a finitely generated abelian group?
Yes. This group is neither free abelian nor finitely generated. The easiest way to see this is to consider the torsion subgroup of . It quite remarkable, I think, that because this is an infinite group and its torsion subgroup is itself! But anyways, if then that would be a contradiction because and if was finitely generated abelian group, that is then which is clearly impossible because one is finite and other oter infinite.
What do you mean?ps. where do u come up with these?