amenability
/ uh-MEE-nuh-bil-it-ee /
Can you spread a fair, finitely-additive 'fraction of the group' across an infinite group in a way that doesn't change when you shift everything by a group element? For finite groups this is trivial — just use the fraction of elements. For infinite groups it is a real question, and the groups for which the answer is yes are called amenable. Amenability is a deep dividing line between 'tame' groups, where averaging behaves reasonably, and 'wild' groups like free groups, where it breaks down catastrophically.
There are many equivalent definitions; two are most useful. (1) Invariant mean: G is amenable if there is a finitely-additive probability measure mu on all subsets of G that is left-invariant, mu(gA) = mu(A) for every g and every subset A — a way to assign a 'size in [0,1]' to every subset, consistent with translation. (2) Folner condition: G is amenable if there exist finite subsets F_n that are almost invariant, meaning for each generator s the symmetric difference |s F_n triangle F_n| / |F_n| -> 0 — sets whose boundary is small compared to their bulk, so they barely move under multiplication. These Folner sets are the geometric heart: amenability says the group has subsets with vanishingly small boundary-to-volume ratio. Finite groups, abelian groups, and more generally all groups of subexponential growth are amenable, and amenability is preserved under subgroups, quotients, extensions, and increasing unions.
Amenability is exactly the obstruction to paradoxical decompositions. A group is non-amenable if and only if it admits a paradoxical decomposition — it can be cut into finitely many pieces that, rearranged by group elements, form two copies of itself, the algebraic engine behind the Banach-Tarski paradox. The free group F_2 is non-amenable and paradoxical, which is the source of Banach-Tarski. The Tits alternative says a finitely generated linear group either contains a free subgroup F_2 (hence is non-amenable) or is virtually solvable (hence amenable), but this alternative is special to linear groups: the honest caveat is that 'non-amenable' is strictly more general than 'contains a free subgroup' — there exist non-amenable groups with no free subgroup at all (the von Neumann-Day problem, resolved by Ol'shanskii and Adian). So do not equate amenability with the absence of free subgroups in general. Also note: amenability is NOT a quasi-isometry invariant in the naive sense one might hope, though it is invariant under quasi-isometry for finitely generated groups.
The group Z is amenable, witnessed by Folner sets: take F_n = {-n, ..., n}, an interval of 2n+1 integers. Shifting by 1 sends it to {-n+1, ..., n+1}, which differs from F_n in only 2 elements, so |F_n + 1 triangle F_n| / |F_n| = 2/(2n+1) -> 0. The boundary is negligible compared to the bulk, so averaging works. The free group F_2 has no such sets — any finite subset of its tree has boundary comparable to its size — and indeed F_2 is non-amenable and admits the paradoxical decomposition behind Banach-Tarski.
Z is amenable via the Folner sets {-n,...,n}; the free group F_2 is non-amenable and paradoxical.
Amenability is not the same as 'has no free subgroup.' Every group containing F_2 is non-amenable, but the von Neumann-Day problem (solved by Ol'shanskii, Adian) shows non-amenable groups with no free subgroup exist; the equivalence holds only inside the linear world via the Tits alternative.