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The direct product A x Z as an HNN extension

Example

If both associated subgroups equal the whole base group $A$ and the associated isomorphism is the identity, then the HNN extension is naturally isomorphic to $A \times Z$ .

Facts & Assumptions

Given: A group $A$ .

[L1]

An HNN extension is obtained by adjoining a stable letter that conjugates one chosen subgroup onto another. (An HNN extension with its stable letter)

[L2]

A homomorphism out of an HNN extension is determined by a homomorphism on the base group and the image of the stable letter, provided the conjugacy relation is respected. (The universal property of an HNN extension)

Verification

technique · direct

1.1

Take both associated subgroups to be $A$ and the associated isomorphism to be the identity. Then the defining relation in [L1] becomes $t a t^{- 1} = a$ for every $a \in A$ , so the stable letter commutes with the image of $A$ .

L1given

2.1

The map from the HNN extension to $A \times Z$ sending $A$ to $A \times {0}$ and $t$ to $(e, 1)$ satisfies the relation from step 1.1, so [L2] gives a homomorphism. The reverse map sends $(a, n)$ to $a t^{n}$ , and the commuting relation makes it a homomorphism inverse to the first one. Hence the HNN extension is $A \times Z$ .

L2step 1.1algebra∎

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

C. Loh, Geometric Group Theory: An Introduction (2015 course version) (standard reference, not scraped)