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An HNN extension realises two chosen isomorphic subgroups as conjugate

Example

If $α, β : C \to A$ are injective, then in the HNN extension

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

the subgroups $α (C)$ and $β (C)$ become conjugate by the stable letter:

$t α (C) t^{- 1} = β (C) .$

Facts & Assumptions

Given: The HNN extension in the statement.

[L1]

The defining relations of an HNN extension identify $t α (c) t^{- 1}$ with $β (c)$ for every $c \in C$ . (An HNN extension with its stable letter)

[L2]

The stable letter is the universal element that enforces the required conjugacy relation. (The universal property of an HNN extension)

Verification

technique · direct

1.1L1given

For every $c \in C$ , [L1] gives $t α (c) t^{- 1} = β (c)$ . Hence $t α (C) t^{- 1} \subseteq β (C)$ .

2.1L1L2step 1.1algebra∎

Applying the same relation to $t^{- 1}$ shows $α (c) = t^{- 1} β (c) t$ for every $c \in C$ , so $β (C) \subseteq t α (C) t^{- 1}$ . Thus the two subgroups are conjugate exactly as [L2] predicts.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

9 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)