TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-28

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Under the ultrafilter lemma and a matching-extension principle, a group is amenable if and only if it is not paradoxical

Statement

Assume the ultrafilter lemma and the following perfect-matching extension principle for $G$ : whenever a finite set $K \subseteq G$ satisfies $∣ K F ∣ \geq 2 ∣ F ∣$ for every finite nonempty $F \subseteq G$ , the finite Hall matchings in the bipartite graph with left vertices $G \times {1, 2}$ , right vertices $G$ , and edges $(g, r) \sim k g$ for $k \in K$ extend to a bijection $Φ : G \times {1, 2} \to G$ satisfying $Φ (g, r) \in K g$ .

Under these assumptions, $G$ is amenable if and only if it is not paradoxical.

Facts & Assumptions

Given: A group $G$ , the ultrafilter lemma, and the matching-extension principle stated above.

[L1]

A paradoxical decomposition forbids a left-invariant mean (Paradoxical groups admit no invariant mean).

[A1]

Under the ultrafilter lemma, amenability is equivalent to the Folner condition (Under the ultrafilter lemma, the Folner condition is equivalent to amenability).

[A2]

The perfect-matching extension principle above upgrades the finite Hall matchings coming from a doubling set $K$ to an edge-respecting bijection $G \times {1, 2} \to G$ .

[L2]

Hall's theorem saturates a finite bipartite left part exactly when every finite subfamily has enough neighbors (Hall's marriage theorem for a finite bipartite graph).

Proof

technique · direct

1.1L1given

If $G$ is paradoxical, then [L1] says that no left-invariant mean exists, so $G$ is not amenable. This proves the forward implication of the statement.

1.2A1givenalgebra

Assume now that $G$ is not amenable. By contrapositive use of [A1], there is a finite set $K \subseteq G$ with $e \in K$ such that $∣ K F ∣ \geq 2 ∣ F ∣$ for every finite nonempty $F \subseteq G$ . Indeed, if no such $K$ existed, then for a finite test set $S$ and $ε > 0$ one could put $S_{0} = {e} \cup S \cup S^{- 1}$ , choose $m$ with $(1 + ε / 2)^{m} > 2$ , and find finite nonempty $F$ with $∣ S_{0}^{m} F ∣ < 2 ∣ F ∣$ . Among the layers $F_{j} = S_{0}^{j} F$ , some ratio $∣ F_{j + 1} ∣ / ∣ F_{j} ∣$ would then be below $1 + ε / 2$ , giving $∣ s F_{j} △ F_{j} ∣ < ε ∣ F_{j} ∣$ for every $s \in S$ and hence the Folner condition.

2.1A2L2step 1.2construct

Form the bipartite graph from [A2]. For every finite left subset $P \subseteq G \times {1, 2}$ with projection $Q \subseteq G$ , step 1.2 gives $∣ N (P) ∣ = ∣ K Q ∣ \geq 2 ∣ Q ∣ \geq ∣ P ∣$ , so [L2] produces a matching saturating $P$ . By [A2], these finite matchings extend to an edge-respecting bijection $Φ : G \times {1, 2} \to G$ . For $r \in {1, 2}$ and $k \in K$ , put $C_{r, k} = {Φ (g, r) : Φ (g, r) = k g} .$ The finitely many $C_{r, k}$ are pairwise disjoint and partition $G$ because $Φ$ is bijective. For fixed $r$ , the translated pieces $k^{- 1} C_{r, k}$ partition $G$ , because they are exactly the fibers in the domain copy $G \times {r}$ . Thus the two subfamilies $(C_{1, k})_{k \in K}$ and $(C_{2, k})_{k \in K}$ , with translators $k^{- 1}$ , form a paradoxical decomposition.

3.1step 1.1step 2.1∎

Steps 1.1 and 2.1 prove both directions, so $G$ is amenable exactly when it is not paradoxical.

Depends on

Used by

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Sources

Cornelia Drutu and Michael Kapovich, Lectures on Geometric Group Theory (standard reference, not scraped)
C. Löh, Geometric Group Theory: An Introduction (2015 course version) (standard reference, not scraped)