I want to prove the HNN extension $G \ast_A$ is finitely presented $\Leftrightarrow$ $A$ is finitely generated, given that $G$ is a finitely presented group, say $G = \langle S \mid R \rangle$
The $\Leftarrow$ direction is easy I think, since if $A = \langle T \rangle$ for $T$ finite, then a finite presentation for $G \ast_A$ is given by $\langle S \cup \{ t \} \mid R \cup \{ tat^{-1} = \theta(a) : a \in T \} \rangle$ where $\theta : A \hookrightarrow G$ is the monomorphism defining the HNN extension.
The other direction is where I have trouble. If I suppose $G \ast_A = \langle T \mid R' \rangle$ is finitely generated, then I know $G \leq G \ast_A$, so I can write the $S$ generating $G$ in the $T$ generating the HNN extension and add in the $R$ relations: $G \ast_A = \langle S \cup T \mid R \cup R' \cup S_= \rangle$ where $S_=$ are the relations defining $S$ in terms of the $T$. This is now a finite presentation that looks kind of similar to the probably infinite presentation $\langle S \cup \{ t \} \mid R \cup \{ tat^{-1} = \theta(a) : a \in A \} \rangle$. I feel like I should do something involving Tietze transformations, but I'm not sure what.
Any help proving this would be appreciated.
I have seen some references to this theorem on this site e.g. Examples of non-finitely presented groups, so hopefully it is well known and easy to prove, I've just not seen how to do it.
If the group $H = \langle S \cup \{t\} \mid R \cup \{tat^{-1}=\theta(a) : a \in A\})$ was finitely presented, then it would be presented using a finite subset of the relation set in the infinite presentation. That is, we would have $H = \langle S \cup \{t\} \mid R \cup \{tat^{-1}=\theta(a) : a \in X\})$ for some finite subset $X$ of $A$.
But this presents the group $G_{*B}$, where $B = \langle X \rangle$. So if $A$ is not finitely generated, then $B$ is a proper subgroup of $A$.
But then, by Britton's Lemma on HNN extensions, if we choose $a \in A \setminus B$, then the element $tat^{-1}\theta(a)^{-1}$ of $G_{*B}$ contains no pinch, and so is not equal to the identity. But it is equal to the identity in $G_{*A}$, contradiction.