LemmaStatement: Literature-sourcedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (gpt-5.6-terra)audited 2026-08-28

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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Paradoxical groups admit no invariant mean

Statement

If a group admits a paradoxical decomposition, then it admits no left-invariant mean.

Facts & Assumptions

Given: A group $G$ with a paradoxical decomposition.

[L1]

A left-invariant mean is a positive normalized functional on $ℓ^{\infty} (G)$ invariant under left translation (Left-invariant means and amenable groups).

[L2]

In a paradoxical decomposition, disjoint pieces $A_{i}, B_{j}$ partition $G$ , and the translated families $(a_{i} A_{i})$ and $(b_{j} B_{j})$ each partition $G$ (Paradoxical decompositions of groups).

Proof

technique · direct

1.1

Assume toward contradiction that $m$ is a left-invariant mean on $G$ . By [L2] and positivity, finite additivity on indicator functions gives $1 = m (1_{G}) = \sum_{i} m (1_{A_{i}}) + \sum_{j} m (1_{B_{j}})$ .

L1L2givenassume-contra

2.1

The translated families from [L2] also partition $G$ , so left invariance gives $1 = m (1_{G}) = \sum_{i} m (1_{a_{i} A_{i}}) = \sum_{i} m (1_{A_{i}})$ and likewise $1 = \sum_{j} m (1_{B_{j}})$ . Combining these equalities with step 1.1 yields $1 = 2$ , a contradiction.

L1L2step 1.1discharge-contradiction∎

Depends on

Used by

Dependency tree · two levels

5 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

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