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Baumslag proved that if $G= A \ast_{C} B$ is an amalgamated free product where $A$ and $B$ are finitely presented, $G$ is finitely presented if and only if $C$ is finitely generated.

Similarly, by using Britton's lemma, it can be shown that if $H$ is an HNN extension $\langle D ,t \mid t^{-1}bt=b, \forall b\in L \rangle$, where $L$ is a subgroup in $D$ and $D$ is finitely generated, $H$ is finitely presented if and only if $L$ is finitely generated.

I have tried to do the same by adding another relation. That is, suppose that $H= \langle D, t \mid t^{-1}h_{1}t=h_{2}, t^{-1}lt=l, \forall l\in L \rangle$, $H$ is finitely presented, and $L$ is a normal subgroup in $D$ and $h_{1},h_{2}$ are elements in $D$. Again, I am assuming that $D$ is finitely generated. Is then $L$ necessarilly finitely generated?

Attempt: Since $H$ is finitely presented and $L$ is normal in $H$, $H=\langle l_{1},\dots,l_{n},h_{1},\dots,h_{m},t\mid t^{-1}h_{1}t=h_{2},t^{-1}l_{i}t=l_{i},i\in \{1,\dots,k\}\rangle$. By using Britton's normal form, I would obtain that $\langle L, h_{1}\rangle$ is generated by $\langle l_{1},\dots,l_{k},h_{1}\rangle$, but I don't know if I could get rid of the $h_{1}$ is some way.

Does someone have any idea? Thanks in advance!

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    $\begingroup$ Note that in your second paragraph you're stating something for some very particular instance of HNN extension (HNN with respect to identity automorphism of some given subgroup). For the question, the generality is a bit hopeless, as $D$ is likely not be mapped injectively without further assumption. For instance if $h_1\in L$ and $h_1h_2^{-1}$ generates normally $D$ then $D$ is killed in $H$. $\endgroup$
    – YCor
    Commented Oct 5, 2019 at 14:50
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    $\begingroup$ @YCor but if $h_{1}\in L$, I would obtain that $L$ is generated by $\langle l_{1},\dots,l_{k},h_{1}$, so I have what I want $\endgroup$
    – J.L.
    Commented Oct 5, 2019 at 14:55
  • $\begingroup$ Ah sorry I was giving a counterexample to the converse (you wrote "I have tried to do the same", but it's not the same). $\endgroup$
    – YCor
    Commented Oct 5, 2019 at 14:58

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