Proving a Group is Cyclic: What is the Generator of G?

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Homework Statement



For each integer n, define f_{n} by f_{n}(x) = x + n. Let G = {f_{n} : n \in \mathbb{Z}}. Prove that G is cyclic, and indicate a generator of G.

Homework Equations



None as far as I can tell.

The Attempt at a Solution



Doesn't this require us to find one element of G such that, by applying that element over and over again e.g. f_n(f_n(...) we can produce any element of G? My main problem with this is I don't understand how one could find a way to go from positive to negative elements or vice-versa. For example if we let the generator be f(x) = x + 1 how could we generate f_{-1}(x) = x - 1? Or do I misinterpret the definition of G/requirements of a cyclic group?
 
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zooxanthellae said:
Or do I misinterpret the definition of G/requirements of a cyclic group?

Yes. Being a cyclic group means that the group can be generated by 1 element. But what does generated mean? Well, it means that if you take all compositions and all inverses, then you get the entire group.

So, in other words, a group G is cyclicly generated by x iff

G=\{...,x^{-3},x^{-2},x^{-1},e,x,x^2,x^3,...\}

So, in your example, you don't only allow compositions of the fn, but also inverses.
 
Oh, so then f_{1} = x + 1 is actually a generating group, since its inverse is f_{-1} = x - 1 and together these can account for any element of G.

Thanks much micromass, the book I'm using didn't specify (or I missed it if it did) that inverses are part of a generating group. That clears things up.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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