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 $G=G_1\times G_2$, etc., and any group in $\mathcal{G}$ is assumed to be non-amenable with zero first $L^2$-Betti number.

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.