LemmaStatement: Literature-sourcedProof: AI-generatedverified 2026-09-26 (gpt-6-sol)

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Cobounded proper geodesic actions produce finite generating sets

Statement

Let $G$ act geometrically on a geodesic metric space $X$ . Fix $x_{0} \in X$ , and choose $D \geq 0$ such that every point of $X$ lies within distance at most $D$ of the orbit $G \cdot x_{0}$ . Then $S : = {g \in G : d_{X} (x_{0}, g \cdot x_{0}) \leq 2 D + 1}$ is finite and generates $G$ .

Facts & Assumptions

Given: A geometric action of $G$ on a geodesic metric space $X$ , a point $x_{0} \in X$ , and a real $D \geq 0$ such that every point of $X$ lies within distance at most $D$ of the orbit $G \cdot x_{0}$ .

[L1]

A geometric action is isometric, proper, and cobounded (Geometric actions on a metric space).

[L2]

In a geodesic metric space, every two points are joined by a geodesic segment (Geodesics and geodesic metric spaces).

[L3]

The Archimedean property says that for every real $t$ there is a natural number $m$ with $t < m$ (Every complete ordered field is Archimedean).

[L4]

Every nonempty subset of $N$ has a least element (The well-ordering principle).

[L5]

A group is finitely generated when some finite subset generates it (Finitely generated groups).

Proof

technique · direct

1.1L1

The set $S$ is finite because the action is proper by [L1], the singleton ${x_{0}}$ and the ball ${y : d_{X} (x_{0}, y) \leq 2 D + 1}$ are bounded, and $S$ is exactly the transporter set from the first to the second.

1.2L2L3L4choose

Let $g \in G$ . By [L2], choose a geodesic $γ : [0, ℓ] \to X$ from $x_{0}$ to $g \cdot x_{0}$ , where $ℓ = d_{X} (x_{0}, g \cdot x_{0})$ . By [L3] and [L4], let $m$ be the least positive natural number with $ℓ \leq m$ . Thus $m \geq 1$ even when $g \cdot x_{0} = x_{0}$ . Put $p_{i} : = γ (i ℓ / m)$ for $0 \leq i \leq m$ . Then $d_{X} (p_{i - 1}, p_{i}) = ℓ / m \leq 1$ for each $i$ .

2.1givenstep 1.2choose

For each $i$ , choose $g_{i} \in G$ with $d_{X} (p_{i}, g_{i} \cdot x_{0}) \leq D$ , and arrange $g_{0} = e$ and $g_{m} = g$ . This is possible because $p_{0} = x_{0}$ and $p_{m} = g \cdot x_{0}$ .

3.1L1step 2.1algebra

Put $s_{i} : = g_{i}^{- 1} g_{i + 1}$ . Then $d_{X} (x_{0}, s_{i} \cdot x_{0}) = d_{X} (g_{i} \cdot x_{0}, g_{i + 1} \cdot x_{0}) \leq D + 1 + D = 2 D + 1$ , so every $s_{i}$ lies in $S$ . Since $g = s_{0} s_{1} \dots s_{m - 1}$ , the set $S$ generates $G$ .

4.1L5step 1.1step 3.1∎

Step 3.1 shows that $S$ is a generating set, and step 1.1 shows that it is finite. Hence [L5] makes $G$ finitely generated.

Depends on

Used by

The Svarc-Milnor lemma Theorem

Dependency tree · two levels

22 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, Sections 4.4 and 5.1 (standard reference, not scraped)
C. Drutu and M. Kapovich, Lectures on Geometric Group Theory, Chapter 5 (standard reference, not scraped)