For groups, does $G \times G \cong H\times H$ imply $G \cong H$? [duplicate]
By Remak's theorem, this is true for finite groups, and so any counter example would counter Remak so is probably nontrivial.
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 $G\not\cong H.$