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Proving a Group is Cyclic

  1. Jul 14, 2011 #1
    1. The problem statement, all variables and given/known data

    For each integer n, define [tex]f_{n}[/tex] by [tex]f_{n}(x) = x + n.[/tex] Let [tex]G = {f_{n} : n \in \mathbb{Z}}.[/tex] 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 [tex]G[/tex] such that, by applying that element over and over again e.g. [tex]f_n(f_n(...)[/tex] 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 [tex]f(x) = x + 1[/tex] how could we generate [tex]f_{-1}(x) = x - 1?[/tex] Or do I misinterpret the definition of G/requirements of a cyclic group?
     
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  3. Jul 14, 2011 #2

    micromass

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    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

    [tex]G=\{...,x^{-3},x^{-2},x^{-1},e,x,x^2,x^3,...\}[/tex]

    So, in your example, you don't only allow compositions of the fn, but also inverses.
     
  4. Jul 14, 2011 #3
    Oh, so then [tex]f_{1} = x + 1[/tex] is actually a generating group, since its inverse is [tex]f_{-1} = x - 1[/tex] 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.
     
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