LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-09

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.

Finite normal quotients preserve ball growth

Statement

For a finite generating set $S$ of a group $G$ , finite normal $F$ , and $q : G \to G / F$ , one has $∣ B_{q S} (n) ∣ \leq ∣ B_{S} (n) ∣ \leq ∣ F ∣ ∣ B_{q S} (n) ∣$ for every integer $n \geq 0$ . Balls use generators and their inverses.

Facts & Assumptions

Given: $F ⊴ G$ is finite, $S$ is finite and generates $G$ , and $n \geq 0$ .

[F1]

Word length is the minimum number of generator or inverse letters (Word length of a group element with respect to a generating set).

[F2]

Finite generating sets give finite balls (Balls of a word metric are finite if and only if the generating set is finite).

Proof

1.1

An $S$ -word of length at most $n$ projects to a $q S$ -word of that length. Conversely, lift each letter in a $q S$ -word to a corresponding letter of $S \cup S^{- 1}$ ; their product lies in $B_{S} (n)$ and projects to its value. Only finitely many letters of this particular word need lifts. Therefore $q (B_{S} (n)) = B_{q S} (n)$ .

F1given

2.1

Each fiber of $q$ is a coset of $F$ , with exactly $∣ F ∣$ elements. Its intersection with $B_{S} (n)$ has at most $∣ F ∣$ elements and, over $B_{q S} (n)$ , at least one by step 1.1. Summing over the finite target ball yields both inequalities. At $n = 0$ both balls contain only the identity, and $1 \leq 1 \leq ∣ F ∣$ . For $F = 1$ both inequalities are equalities; the empty generating set gives the trivial group.

F2step 1.1∎

Source notes

Druţu–Kapovich, Geometric Group Theory (837-page edition), Theorem 14.26 finite-kernel reduction, printed p.511; direct fiber-count proof. The finite-quotient growth reduction in revised Theorem 14.26 is replaced by exact ball images and finite fiber cardinalities. No later quasi-isometry invariance theorem is used.

Depends on

Used by

Dependency tree · two levels

16 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

Druţu–Kapovich, Geometric Group Theory (837-page edition) (standard reference, not scraped)