Abstract Algebra - Direct Product Question

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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.
 
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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.
 
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Is it enough to describe this map in the way I have, or do I need further information (or am I wrong?)
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.
 
Thanks guys. Back to the drawingboard.
 
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?
 
I don't like to give answers like this, but I haven't the slightest idea.
 
Well, how many generators does it take to generate the subgroup Gx1 of GxG?
 
I would say n - same as for G.
 
And what about 1xG? So what does that suggest will be (roughly) true, if you want to generate all of GxG?
 
  • #10
You would need 2n?
 
  • #11
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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