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Let $B$ be a group and let $A_1$ and $A_2$ subgroups of $B$ with $\phi :A_1\rightarrow A_2$ an isomorphism. Let $\left<t\right>$ be the infinite cyclic group, generated by a new element $t$. The $\mathbf{ HNN}$- extension $H$ is formed by adjoined to the free product $B* \left<t\right>$ the relations $ta t^{-1}=\phi(a)$ for all $a\in A_1$. Hence, $H$ has presentation $$H=\left<B,t\mid ta t^{-1}=\phi(a), \ a\in A_1\right>.$$ The group $B$ is called the base of $H$, $t$ is called the stable latter, and $A_1$ and $A_2$ is called the associated groups. An $\mathbf{HNN}$-extension $H$ is called an ascending $\mathbf{HNN}$-extension, if at least one of the subgroup $A_1$ and $A_2$ is equal to the base $B$. Bieri and Strebel showed that if $N$ is a normal subgroup of a finitely presented solvable group $G$ and $G/N$ is infinite cyclic, then $G$ is an ascending $\mathbf{ HNN}$-extension with a finitely generated base $B$.

My question is: if we have an ascending $\mathbf{ HNN}$-extension $H$ with a f.g. base $B$, then When $H$ is finitely presented group?

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    $\begingroup$ I doubt you can get a general answer other than just a restatement. The answer is quite subtle even in the case when $H$ is metabelian (of course, a sufficient condition is that $B$ is finitely presented, but it's not necessary). $\endgroup$
    – YCor
    Feb 11, 2016 at 14:48
  • $\begingroup$ Do you have any example for such HNN- extension? In the example that I found the group $H$ is finitely presented $\endgroup$
    – user182085
    Feb 11, 2016 at 16:34
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    $\begingroup$ You can cook up an example with $H=BS(1,2)^2$ and a well-chosen choice of $B$. Also there are examples where $H$ is Thompson's group $F$. $\endgroup$
    – YCor
    Feb 11, 2016 at 16:59

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