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Recall that a (nontrivial) abelian group $A$ is called divisible if the multiplication by $n$ is surjection $A\to A$ for all $n\in\mathbb{Z}_{>1}$.

My question is the following.

Is there a countable discrete group $G$ in the class $\mathcal{G}$ such that $H^2(G,\mathbb{Z}G)$ is not divisible?

where the class $\mathcal{G}$ contains the following groups: groups with Kazhdan's property (T), any infinite non-amenable group $G$ of the form $G=G_1\times G_2$, etc., and any group in $\mathcal{G}$ is assumed to be non-amenable with zero first $L^2$-Betti number.

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    $\begingroup$ Could you define precisely the class $\mathcal{G}$? or your question does not make sense. $\endgroup$
    – YCor
    Commented Jul 24, 2015 at 12:18
  • $\begingroup$ Also "infinite non-amenable group $G_1\times G_2$" precisely means "infinite non-amenable group", which most likely is not what you mean. $\endgroup$
    – YCor
    Commented Jul 24, 2015 at 12:19
  • $\begingroup$ @YCor, I do not have a precise description of the class of $\mathcal{G}$, but you can think this is the class of the group which satisfies $\mathbb{T}$-valued co cycle super-rigidity for its Bernoulli shift action. So, if possible, I expect an example from (T) groups. $\endgroup$
    – Jiang
    Commented Jul 24, 2015 at 12:59
  • $\begingroup$ There is a topological Interpretation of that cohomology: when G is the fundamental group of a compact space M, then $$H^*(G,ZG)=H_c^*(\widetilde{M})$$ $\endgroup$
    – ThiKu
    Commented Jul 24, 2015 at 13:18
  • $\begingroup$ where the right hand side means cohomology with Compact Support of the universal covering. Not sure whether that helps, though. $\endgroup$
    – ThiKu
    Commented Jul 24, 2015 at 13:19

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