Hello!

My problem is as follows:

Circle group $\displaystyle T$ = \{z \in C: |z| = 1\}$, $G_n = \{z \in T: z^n = 1\}, G = \cup_{i = 1}^\infty(G_n)$$

First of all, which are cyclic? T is not as it is not countable. Gn is and G should not be since it is here described as a union of proper cyclic subgroups. But I am afraid that for some large n Gn becomes G or something weird like that. How can I assure myself this is not the case, and that Gn for every n is a proper subgroup of G?

Further I'm asked to describe the cyclic subgroups of these 3. In particular does T have a subgroup isomorphic to Z x Z? As I understand it T only has Gn as proper subgroups, but how do I prove non-existance of others? Gn is also a subgroup of G by definition, but does G have any other subgroups? This one in particular I do not know where to start with.

Further I am to show that T is generated by every neighbourhood of the identity. So basicly if we have an interval ]a,b[, where a and b are the arguments of the complex numbers in the original neighbourhood I of e (identity). So I'm thinking I^2 will be mapped onto something that has arguments in the range ]2a, 2b[. However, it feels they are kind of "thinned out" in the edges as it is only half of the numbers that map to something "new" and the rest end up in the previous interval. I know this is very handwavy etc, but I feel my knowledge of the real numbers is terribly insufficient so I can not really make a clear argument. How should I view this, and reassure myself nothing bad happens by just covering more and more of the unit circle?

Also is T countably generated? An interval of reals should not be countable right? But I have so far only shown it can be generated by an uncountable set, not non-existance of countable generating sets.

Cheers and thanks in advance,

-Sten