TheoremStatement: Literature-sourcedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (gpt-5.6-terra)audited 2026-08-29

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The fundamental group acts without inversions on its Bass-Serre tree

Statement

Let $Γ = π_{1} (G, T)$ and let $\tilde{X}$ be its Bass-Serre tree. Then:

$Γ$ acts on $\tilde{X}$ by left multiplication on cosets.
The action is without inversions.
The quotient graph is the original underlying graph $X$ .
The stabilizer of a vertex coset $γ G_{v}$ is $γ G_{v} γ^{- 1}$ , and the stabilizer of an edge coset $γ G_{e}$ is $γ G_{e} γ^{- 1}$ .

Facts & Assumptions

Given: A graph of groups $G$ , a maximal subtree $T$ , and $Γ = π_{1} (G, T)$ .

[L1]

The Bass-Serre graph has vertices and edges given by left cosets, with the stated origin and terminus maps. (The Bass-Serre tree of a graph of groups)

[L2]

That coset graph is a tree. (The Bass-Serre coset graph is a tree)

[L3]

Points in the same orbit have conjugate stabilizers. (If $y = g \cdot x$ , then $G_{y} = g G_{x} g^{- 1}$ )

Proof

technique · direct

1.1

Left multiplication $γ^{'} \cdot (γ G_{v}) = (γ^{'} γ) G_{v}$ and $γ^{'} \cdot (γ G_{e}) = (γ^{'} γ) G_{e}$ is well defined on the cosets of [L1], and the formulas for origin and terminus in [L1] are preserved by that multiplication. So $Γ$ acts by graph automorphisms on $\tilde{X}$ .

L1given

2.1

The orbit of the base vertex coset $G_{v}$ is the set of all cosets $γ G_{v}$ , so the quotient vertices are identified with the original vertices $v$ ; the same holds for edges. Hence the quotient graph is exactly $X$ . An inversion would send some oriented edge coset to its reverse, which would force the two opposite orientations of one edge of $X$ into the same orbit; that does not happen in the quotient description.

L1step 1.1

2.2

The stabilizer of the base vertex coset $G_{v}$ is $G_{v}$ itself, and similarly for $G_{e}$ . Therefore [L3] gives $Stab (γ G_{v}) = γ G_{v} γ^{- 1}$ and $Stab (γ G_{e}) = γ G_{e} γ^{- 1}$ for arbitrary cosets in the same orbits.

L3step 1.1

3.1

Combining steps 1.1, 2.1, and 2.2 with [L2] proves all four claims.

L2step 1.1step 2.1step 2.2∎

Depends on

Used by

The fundamental group of a graph with trivial groups is free Corollary
A graph of finite groups giving a virtually free group Example
The Bass-Serre tree of a free product Example
FALSE: every tree action is free False statement
FALSE: the quotient graph determines the acting group without stabilizer data False statement
FALSE: vertex stabilizers are literally the chosen vertex groups without conjugacy ambiguity False statement
Different maximal trees give isomorphic graph-of-groups fundamental groups Theorem
Kurosh subgroup theorem Theorem

Dependency tree · two levels

10 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)
C. Loh, Geometric Group Theory: An Introduction (2015 course version) (standard reference, not scraped)