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Abstract Algebra - Direct Product Question

  1. Oct 31, 2006 #1
    I'm supposed to find a non-trivial group G such that G is isomorphic to G x G.

    I know G must be infinite, since if G had order n, then G x G would have order n^2. So, after some thought, I came up with the following. Z is isomorphic to Z x Z.

    My reasoning is similar to the oft-seen proof that the rationals are countable.

    Picture a grid with dots representing each element in Z x Z. Now, starting at the origin, trace a circuitous path (in any direction, but always a tight spiral) and define a map that sends 0 to (0,0), 1 to the next point, -1 to the next point, 2 to the next point, etc.

    Is it enough to describe this map in the way I have, or do I need further information (or am I wrong?)

    Thanks.
     
  2. jcsd
  3. Oct 31, 2006 #2

    0rthodontist

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    Z x Z is not isomorphic to Z. Z x Z is not cyclic. You might think about the group of integer functions on the integers.
     
    Last edited: Oct 31, 2006
  4. Oct 31, 2006 #3

    Hurkyl

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    That describes a map... (although having only a heuristic description makes it hard to prove things about it)

    But you've yet to prove that your map is a homomorphism, that it has an inverse, and that its inverse is a homomorphism.
     
  5. Oct 31, 2006 #4
    Thanks guys. Back to the drawingboard.
     
  6. Oct 31, 2006 #5

    Hurkyl

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    Hrm. I hate to give big hints like this, but...

    If a generating set of G must contain at least n elements... then (heuristically speaking) how many elements must a generating set of GxG contain?
     
  7. Oct 31, 2006 #6
    I don't like to give answers like this, but I haven't the slightest idea.
     
  8. Oct 31, 2006 #7

    Hurkyl

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    Well, how many generators does it take to generate the subgroup Gx1 of GxG?
     
  9. Oct 31, 2006 #8
    I would say n - same as for G.
     
  10. Oct 31, 2006 #9

    Hurkyl

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    And what about 1xG? So what does that suggest will be (roughly) true, if you want to generate all of GxG?
     
  11. Oct 31, 2006 #10
    You would need 2n?
     
  12. Nov 1, 2006 #11

    Hurkyl

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    Right. In particular, if G is finitely generated, then...

    (this is not a rigorous proof -- I don't know if weird things will happen that allow you to use less than 2n... but we're not looking for proofs here, we're searching for examples!)
     
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