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Which abelian groups can be Schur Multiplier's of finitely generated groups?

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I think it is very unlikely that anyone can answer that one! Of course all finitely generated abelian groups can be, but so can some others, such as infinite elementary groups. – Derek Holt Sep 20 '12 at 8:07
It's likely that all countable abelian groups do. But I don't know if it's done anywhere (even if it's known that every countable abelian group embeds in the center of a f.g. group). A first step would be to do the case of the direct sum of countably copies of $\mathbf{Q}\times\mathbf{Q}/\mathbf{Z}$, because the latter contains all countable abelian groups. If it's the center of a (perfect?) f.g. group $G$ with trivial Schur multiplier and $A$ is any central subgroup, maybe the double of $G$ over $A$, whose center is $A$, has a trivial Schur multiplier, I don't know if it's too optimistic. – YCor Oct 7 '12 at 14:07

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