Existence of abelian group which has no “square-root” but whose “cube” has a “square-root”

Asked Dec 27 '18 at 01:05

Active Dec 27 '18 at 01:05

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Does there exist an abelian group $G$ such that $G \ncong H \times H$ for every abelian group $H$ but $G \times G \times G \cong K \times K$ for some abelian group $K$ ?

Also see Existence of topological space which has no "square-root" but whose "cube" has a "square-root"

asked Dec 27 '18 at 01:05

user521337

3,705

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Structure theorem for finitely generated abelian groups clearly shows that such a group cannot be finitely generated. – Crostul Dec 27 '18 at 01:15
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@RobArthan: no ... I do mean what I wrote ... – user521337 Dec 27 '18 at 01:38
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@Crostul: I'm not so sure about the torsion part ... – user521337 Dec 27 '18 at 01:39
OK. I just wanted to check that you didn't mean "some". "any" would be more idiomatic than "every" in this context, – Rob Arthan Dec 27 '18 at 23:42

Existence of abelian group which has no “square-root” but whose “cube” has a “square-root”

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