Alphabeta Math
PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-28
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The universal property of an HNN extension

Statement

Let

G=⟨A,t | t α(c) t−1=β(c) for c∈C⟩

be an HNN extension. Let H be a group, let f:A→H be a group homomorphism, and let h∈H satisfy

h f(α(c)) h−1=f(β(c))(c∈C).

Then there is a unique group homomorphism

f‾:G→H

whose restriction to A is f and whose value on the stable letter is f‾(t)=h.

Facts & Assumptions

Given: The HNN extension, the homomorphism f:A→H, and the element h∈H in the statement.

[L1]

An HNN extension is the group presented by adjoining a stable letter t and the relators t α(c) t−1=β(c) for every c∈C. (An HNN extension with its stable letter)

[L2]

A map on generators of a presentation extends uniquely once every defining relator evaluates to the identity. (Von Dyck's theorem: maps of generators that satisfy the relators extend uniquely from a presented group)

Proof

technique · constructive
1.1L1L2givenconstruct

Use f on the base-group generators of a presentation of A and send the stable letter t to h. The old relators from A are satisfied because f is a homomorphism, and each new HNN relator from [L1] is satisfied because the hypothesis gives h f(α(c)) h−1=f(β(c)).

2.1L2step 1.1discharge-construct∎

Therefore [L2] gives a unique homomorphism f‾:G→H extending those assignments. By construction it restricts to f on A and sends t to h, which is exactly the required universal property.

Depends on

Used by

Dependency tree · two levels

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Sources