# Thread: Group of Unitary Roots

1. ## Group of Unitary Roots

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?

2. Hm, it certainly cannot be both, as then it would be isomorphic to some $\displaystyle \mathbb{Z}^k$.

ps. where do u come up with these?

3. Originally Posted by Rebesques
Hm, it certainly cannot be both, as then it would be isomorphic to some $\displaystyle \mathbb{Z}^k$.
Yes. This group is neither free abelian nor finitely generated. The easiest way to see this is to consider the torsion subgroup of $\displaystyle \mathbb{G}$. It quite remarkable, I think, that $\displaystyle T[\mathbb{G}] = \mathbb{G}$ because this is an infinite group and its torsion subgroup is itself! But anyways, if $\displaystyle \mathbb{G} \simeq \mathbb{Z}^k$ then that would be a contradiction because $\displaystyle T[\mathbb{Z}^k] = \{ 0 \}$ and if $\displaystyle \mathbb{G}$ was finitely generated abelian group, that is $\displaystyle \mathbb{G} \simeq \mathbb{Z}^k \times \mathbb{Z}_{p_1^{a_1}} \times ... \times \mathbb{Z}_{p_m^{a_m}}$ then $\displaystyle T[\mathbb{G}] \simeq \mathbb{Z}_{p_1^{a_1}} \times ... \times \mathbb{Z}_{p_m^{a_m}}$ which is clearly impossible because one is finite and other oter infinite.

ps. where do u come up with these?
What do you mean?