If $G$ and $H$ are both groups and $G \times G \cong H \times H$, is it required that $G \cong H$?
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1I disagree with the duplicate vote. This is a much more general question. This one is answered in a comment to the answer on the one mentioned as duplicate, but not in any detail. – Tobias Kildetoft Aug 18 '13 at 19:19
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My mistake, I've retracted the vote. For reference, this is a simpler, but related question. @anon thank you for courteously pointing out my error – Ben Grossmann Aug 18 '13 at 19:21
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3To get the information directly here: Arturo remarks on the answer to that question that there is an abelian torsion free (not finitely generated) group $G$, such that $G\oplus G\oplus G\simeq G$ but $G\not\simeq G\oplus G$. So for $H = G\oplus G$, this gives a counterexample. – Tobias Kildetoft Aug 18 '13 at 19:24
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@TobiasKildetoft Can more be said how to find such $G$? – Hagen von Eitzen Aug 18 '13 at 19:48
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@HagenvonEitzen Unfortunately, I am just quoting what Arturo wrote the other place. I have no idea how one constructs such a group myself. – Tobias Kildetoft Aug 18 '13 at 19:50
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Another proof by Corner is freely available here. – Brian M. Scott Aug 18 '13 at 20:21