I came up with a new group.

Let $\displaystyle \mathbb{G}= \bigcup_{n=1}^{\infty}U_n$ where $\displaystyle U_n = \{z\in \mathbb{C} | z^n = 1\}$.

Then $\displaystyle \mathbb{G}$ 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)$\displaystyle \mathbb{G}$ is an abelian group.

2)$\displaystyle \mathbb{G}$ is a countable group.

3*)$\displaystyle \mathbb{G}$ has a cyclic subgroup for every finite order.

What are the answers to the following questions:

1)Is $\displaystyle \mathbb{G}$ is free abelian group?

2)Is $\displaystyle \mathbb{G}$ a finitely generated abelian group?