I translate a question that I asked in mathoverflow here:
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 different 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!
I don't think this is true. Let $D = \langle a,b \rangle$ be free of rank $2$, and take $h_1=h_2=a$, and $L = \langle b^G \rangle$, the normal closure of $b$ in $D$.
Then $L$ is not finitely generated, but $H = \langle a,b,t \mid a^t=a, b^t=b \rangle$, which is just a direct product $\langle t \rangle \times D$.