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ExampleConstruction: Literature-sourcedVerification: AI-generatedprecheck passverified 2026-09-24 (gpt-6-sol)
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A small-cancellation presentation gives a hyperbolic group

Example

Consider the one-relator presentation

G=⟨x1,x2,x3,x4,x5,x6,x7∣x1x2x3x4x5x6x7⟩.

This is a finite C′(1/6) presentation and hence defines a hyperbolic group.

Facts & Assumptions

Given: The displayed presentation of G.

[A1]

In the single relator x1x2x3x4x5x6x7, no nonempty subword occurs as an initial segment of two distinct cyclic conjugates or inverse cyclic conjugates, so the symmetrized presentation has no nontrivial pieces and therefore satisfies C′(1/6) vacuously.

[F1]

A presentation is the quotient of the free group by the normal closure of its relators (Group presentation by generators and relations).

[L1]

Every finite-rank free group is hyperbolic (Finite groups and free groups are hyperbolic).

Verification

technique · direct
1.1givenA1

By [A1], the displayed finite presentation satisfies the C′(1/6) condition.

1.2F1givenalgebra

The relator gives x7=(x1x2x3x4x5x6)−1 in G, so the first six generators generate G. The natural map F(x1,…,x6)→G is therefore onto. Conversely send xi to the identically named free generator for 1≤i≤6 and send x7 to (x1⋯x6)−1. The relator maps to 1, so [F1] factors this assignment through a homomorphism G→F(x1,…,x6). The two composites fix every respective generator, hence are identities. Thus G≅F6.

2.1L1step 1.1step 1.2∎

By [L1], F6 is hyperbolic, and the explicit isomorphism in step 1.2 transfers its Cayley tree to G with the corresponding generating set. Hence this finite C′(1/6) presentation defines a hyperbolic group, without importing the general linear-isoperimetric theorem.

Depends on

Used by

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Dependency tree · two levels

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Sources