Given any group $G$, is there an amenable group $A(G)$ together with a morphism $G\rightarrow A(G)$, such that every other morphism $G\rightarrow A'$ to another amenable group $G'$ uniquely factorizes through $A'$?


That is the question. My approach would be to consider the set of normal subgroups with amenable quotient $S:=\{N\unlhd G|G/N $ is amenable $\}$. Then $\bigcap S$ is a normal subgroup of $G$. But I don't know, whether $G/\bigcap S$ is amenable. It embeds into the group $\prod\_{H\in S}G/H$. It is not clear, that a infinite product of amenable groups is amenable again. But maybe one can embed $G/\bigcap S$ in a smaller group.



Is there an amenabilization of groups ?
Infinite cartesian product of amenable groups is not necessarily amenable. For example, the free group is a subgroup of the infinite (cartesian) product of finite groups. In general, there are finitely generated residually finite just infinite groups (for example, lattices in semi-simple Lie groups of higher ranks, say, $SL\_3({\mathbb Z})$, by the Margulis normal subgroup theorem). These groups do not have amenabelization in your sense. 


Update: As Andreas pointed out in a comment below, any residually finite non-amenable group does not have an amenabelization. Indeed, if $G$ is this group. $A$ is the amenabelization, then every hom. $G\to S$ from $G$ to a finite group should factor through $A$. Since the homomorphisms of $G$ onto finite groups separate all elements, the natural homomorphism $G\to A$ is an embedding, but $A$ is amenable while $G$ is not, a contradiction. 


