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.
 
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?
 
And what about 1xG? So what does that suggest will be (roughly) true, if you want to generate all of GxG?
 
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!)