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>> No.11479050 [View]
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11479050

>>11479031
Is this about the product thing? Because I think I just came up with the shittiest counterexample in history.
Set [math]G=H=0[/math], the trivial group, and [math]R = \mathbb{Z}^2[/math].
For obvious reasons (unless I'm literally having vivid hallucinations), [math]R(G) \cong R[/math], and [math]G \oplus H \cong G[/math], but [math]R \oplus R \neq R[/math].
>>11479038
Yes.

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