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

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Groups acting freely without inversions on trees are torsion-free

Statement

If a group acts freely and without inversions on a simplicial tree, then the group is torsion-free.

Facts & Assumptions

Given: A group $G$ acting freely and without inversions on a simplicial tree $T$ .

[L1]

A free action has no nonidentity element fixing a point. (A free group action has no nonidentity element fixing a point)

[L2]

An action without inversions sends no oriented edge to its reverse. (Edge inversions and actions without inversions)

[L3]

A finite group acting on a tree fixes a vertex after barycentric subdivision. (Finite groups acting on trees have a global fixed vertex after subdivision)

Proof

technique · direct

1.1

Let $g \in G$ have finite order. Then the cyclic subgroup $⟨ g ⟩$ is finite, so [L3] gives a fixed vertex for its action on the barycentric subdivision of $T$ .

L3given

2.1

Let $p$ be the fixed vertex from step 1.1 in the barycentric subdivision. If $p$ is an original vertex of $T$ , then [L1] forces $g = e$ . If $p$ is the midpoint of an original geometric edge, then $g$ preserves that edge setwise. Because the original action is without inversions by [L2], $g$ cannot swap its two orientations, so it fixes both endpoints of that edge. Now [L1] again gives $g = e$ . Hence no nonidentity torsion element exists.

L1L2step 1.1algebra∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

7 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

Jean-Pierre Serre, Trees (standard reference, not scraped)