By Remak's theorem, this is true for finite groups, and so any counter example would counter Remak so is probably nontrivial.
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@Orat I'm not familiar with it, but looking in wikipedia it's the same proof of Remak just noticing it works in greater generality, but not for all groups... – Andy May 11 '18 at 21:35
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2Do you mean the Cartesian product? For non-abelian groups it seems weird to talk about direct sums. – Arnaud D. May 11 '18 at 21:42
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1@ArnaudD. yes, i mean the product then (didn't know the notation) – Andy May 11 '18 at 22:03
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For the same question for rings, see this. – Watson May 12 '18 at 09:23
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No. According to the answers to this Math Overflow question (also this other one as pointed out in a comment by Orat), there is a countable Abelian group $A$ such that $A^2\not\cong A\cong A^3.$ Let $G=A$ and let $H=A^2.$ Then $G$ and $H$ are countable Abelian groups such that $G\times G\cong H\times H\cong H$ while $G\not\cong H.$
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1See also this question which is regarded as a duplicate of this but asking exactly the same. – Orat May 12 '18 at 03:27
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As a follow up, what if we know $G^n=H^n$ for all $n>1$? This amounts to knowing it for $n=2,n=3$. This example doesn't generalize – Andy May 12 '18 at 09:26
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Is there any concrete specification of $A$ readily available? – Hagen von Eitzen May 12 '18 at 09:59