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Let $A, B, C$ three groups such that $A \times C \cong B \times C$. I already know that if $A, B$ and $C$ are abelian and finite, then $A \cong B$. I think this result does not hold anymore if they are not supposed to be finite or abelian. Is it possible to give a counterexample ?

Thank you very much for the help !

Watson
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    $\mathbb Z \times (\mathbb Z \times \dots \times \mathbb Z\times \dots)\cong (\mathbb Z \times \mathbb Z)\times (\mathbb Z \times \dots \times \mathbb Z\times \dots )$. By the way, the result does hold for all finite groups, abelian or not. It is also known to hold for a much larger class of groups, but that is harder to show. – Ittay Weiss Dec 06 '14 at 20:28
  • I know that it holds for finite groups in general. I also know that the proof is not so elementary. And I definitely know that I forgot my proof. It was based on some structure theorems for finite groups that I got from "Endliche Gruppen" by Bertram Huppert. – WimC Dec 06 '14 at 20:44
  • @WimC: Vipul Naik found a completely elementary proof I find really nice. I described it there: http://math.stackexchange.com/a/427640/36434 – Seirios Dec 08 '14 at 08:21

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