if , then it's a cyclic group with an order n, and one generator .
Does have a finite number of generators? (How do you prove \ disproof this?)
If you mean the set of elements of infinite order, then the answer is,
No, . I cannot conjour up a `neat' proof, but I can direct you here, wikipedia's article on the Circle Group, the group of complex numbers with absolue value 1. This group is isomorphic to .
Hmm.. I can understand that every element in Omega-Infinity 'looks like' a+bi, when both a,b are real (and cover all the real numbers). I think it's enough to say that it's order is bigger or equal to the order of R, which is not finitely generated, isn't it?
Yes, of course! That is a very neat proof, but also quite subtle.
What you need to prove is that is uncountably infinite, that it has cardinality greater than the natural numbers. This is sufficient because a finitely-generated group will always be countable (you should prove this - everyone who studies group theory should prove this! This bit is the subtle bit).
It really reminds me of set-theory, with Aleph-0, and Aleph (as orders of sets - naturals, integer, etc...)
I will try solving this, it looks a little 'heavy' right now, and the question this post relies on isn't 'too demanding', it only asks whether this group is finitely generated or not.
Thank you very much
It mightn't sound demanding, but I think I actually is!
Also, I meant to point out that, broadly speaking, generators and group order are un-related; there exist finitely generated groups with infinitely generated subgroups (for example, the derived subgroup of a free group is always infinitely generated).