Proving a Group is Cyclic

1. Jul 14, 2011

zooxanthellae

1. The problem statement, all variables and given/known data

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.

2. Relevant equations

None as far as I can tell.

3. 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?

2. Jul 14, 2011

micromass

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.

3. Jul 14, 2011

zooxanthellae

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.