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Studying a course on geometry and groups, I fell on the following property (which is not given as an exercise, but rather as an observation).

Let $A$ be any group, let $B \cong \langle t \rangle \cong \mathbb{Z}$ and let $\varphi_t \in Aut(A)$. Then the semi-direct product $A \rtimes_{\varphi} \mathbb{Z}$ is isomorphic to a HNN extension $A*_A$.

There are no more precision nor details. I don't really see how to prove this. Does anyone know how to?

Thanks in advance.

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    If $A$ has presentation $$ then the semidirect product has presentation $$.2011-04-15

2 Answers 2

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In this very simple case (ie where the edge group equals the vertex group and both edge maps are isomorphisms) then the HNN extension is identical to a semi-direct product with $\mathbb{Z}$. Please see the Wikipeda page and its references, probably starting with Serre's book.

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The map $\varphi:\mathbb{Z}\longrightarrow \mathrm{Aut}(A)$ above is given by $\varphi(z)=(\varphi_{t})^{z}$. You can easily verify that if $S$ is a set of generators for $A$, the map defined by $s\mapsto(s,0)$ for $s \in S$ and $t\mapsto(e,1)$ extends to an isomorphism $A*_{\varphi_t}\longrightarrow A\rtimes_{\varphi}\mathbb{Z}$.