LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-28

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.

The edge-group presentation is equivalent to the associated-subgroup presentation

Statement

Let

$G = ⟨ A, t | t α (c) t^{- 1} = β (c) for c \in C ⟩$

be an HNN extension as in An HNN extension with its stable letter. Put

$C_{-} : = α (C), C_{+} : = β (C) \leq A,$

and define

$ϕ : = β \circ α^{- 1} : C_{-} \to C_{+} .$

Then $ϕ$ is an isomorphism, and the same group is presented by

$G = ⟨ A, t | t c t^{- 1} = ϕ (c) for every c \in C_{-} ⟩ .$

Facts & Assumptions

Given: The HNN extension in the statement.

[L1]

An HNN extension is defined by relations $t α (c) t^{- 1} = β (c)$ with $α, β$ injective group homomorphisms into the base group. (An HNN extension with its stable letter)

[L2]

A subgroup is a subset closed under the group operations and inverses. (Subgroup)

[L3]

A group isomorphism is a bijective group homomorphism. (Group isomorphisms, automorphisms and the set $Aut (G)$ )

[L4]

A group homomorphism is injective if and only if its kernel is trivial. (A group homomorphism is injective if and only if its kernel is trivial)

Proof

technique · direct

1.1L1L2L3L4given

Because $α$ and $β$ are injective by [L1], [L4] shows that both maps identify $C$ with their images $C_{-}, C_{+} \leq A$ from [L2]. Hence $α^{- 1} : C_{-} \to C$ exists, and $ϕ = β \circ α^{- 1}$ is a bijective group homomorphism $C_{-} \to C_{+}$ . So $ϕ$ is an isomorphism by [L3].

2.1L1step 1.1algebra

For each $c \in C$ , put $x = α (c) \in C_{-}$ . Then the defining relator $t α (c) t^{- 1} = β (c)$ from [L1] becomes $t x t^{- 1} = ϕ (x)$ . Conversely every $x \in C_{-}$ has the form $x = α (c)$ for a unique $c \in C$ , so every relator $t x t^{- 1} = ϕ (x)$ comes from exactly one original relator.

3.1step 2.1∎

The two presentations therefore have the same generators and the same set of defining relations after the change of notation $x = α (c)$ . Hence they present the same group.

Depends on

Used by

Dependency tree · two levels

18 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

Roger C. Lyndon and Paul E. Schupp, Combinatorial Group Theory (standard reference, not scraped)
C. Loh, Geometric Group Theory: An Introduction (2015 course version) (standard reference, not scraped)