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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.

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What is the standard example of an infinite product of amenable groups, which is not amenable? (and does it have a product in the category of amenable groups?) – Martin Brandenburg Oct 4 '10 at 14:44
@Martin: As Mark explained, the free group embeds into an infinite product of finite groups. Hence, this product is not amenable. – Andreas Thom Oct 4 '10 at 16:36
It is worth noting that there is a "residual amenabilization" of a group, and this is just $G/\cap S$. It is the largest residually amenable quotient of $G$. – Andreas Thom Oct 4 '10 at 16:38
up vote 12 down vote accepted

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.

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I don't get it; why does this disprove the existence of the amenabilization? – Martin Brandenburg Oct 4 '10 at 22:43
@Martin: Suppose that $F_2 \to A(F_2)$ is the amenabilization in the sense of Henrik. Let $G$ be any finite quotient of $F_2$. Since $G$ is amenable, there exists a unique map $A(F_2) \to G$. Taking all the finite quotients together one gets a homomorphism $A(F_2) \to \prod G$. By properties of the free group $F_2 \to \prod G$ is injective and hence $F_2 \to A(F_2)$ is injective. This is a contradiction since an amenable group does not contain free subgroups. – Andreas Thom Oct 5 '10 at 5:43
@Martin: You asked about my proof? If $G$ is a just infinite non-amenable group, then its amenabilization must be finite (since all proper homomorphic images are finite). On the other hand, since $G$ is residually finite, it has arbitrary large finite images. Hence the amenabelization is infinite, a contradiction. – Mark Sapir Oct 5 '10 at 11:56
@Martin: In the previous comment, $G$ is a lattice in higher rank semi-simple Lie group. It is non-amenable, and both just infinite and residually finite. There are many other examples (there is a book by Grigorchuk and others on that subject, see MathSci). For example, M. Ershov recently constructed an infinite finitely generated group with property (T) which is a) residually finite b)every proper finitely generated subgroup is either finite or of finite index (hence just infinite). – Mark Sapir Oct 5 '10 at 12:00

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