TITLE: When is an HNN-extension finitely presented?
QUESTION [11 upvotes]: Let $G=\langle H, t; K^t=K^{\prime}\rangle$ be an HNN-extension of $H$, with $t$ inducing the isomorphism $\phi: K\rightarrow K^{\prime}$. I was wondering if the following question can be answered, and if so what is the answer,

When is $G$ finitely presented?

It seems "obvious" that the answer should be "when $H$ is finitely presentable and $K$ is finitely generated". However, Grigorchuk's group is not finitely presentable but does have a finitely-presentable HNN-extension (Lysenok’s extension).
EDIT: Now, I am not expecting the question to be answerable. What I would really like is an incomplete answer, along the lines of,

If you put these restrictions on $H$ and $K$ (and perhaps on the isomorphism $\phi$) then you can say something.

REPLY [8 votes]: Here are my  comments combined into an answer.
For ascending HNN extensions, i.e. $H=K$, $\phi\colon H\to K'$ an injective endomorphism (as in Baumslag-Remeslennikov case (see above), in the Grigorchuk case, and many others) one needs, as Ben Steinberg ponted out that $H$ has a finite L-presentation (named after Igor Lysenok, who proved that the Grigorchuk group $G_1$ has such a presentation) with respect to the endomorphism $\phi:H\to K′$. That is there are finite number of relations $r_1=1,...,r_k=1$ so that the set of relations $\{\phi^m(r_j)=1\mid m\ge 0,1\le j\le k\}$ defines $H$. Here we consider $\phi$ as a substitution $x\mapsto u_x$ where $u_x$ is any word representing $\phi(x)$ in $H$, $x$ a generator of $H$. Indeed, in this case $G$ (generated by the finite generating set $X$ of $H$ and the free letter $t$) has finite presentation consisting of relations $r_1,...,r_k$ and the HNN relations $x^t=\phi(x), x\in X$. I think that the converse statement should also be true: if the HNN extension is finitely presented then $H$ has a finite L-presentation with respect to $\phi$. 
More examples can be found in Sapir, Mark, Wise, Daniel T., Ascending HNN extensions of residually finite groups can be non-Hopfian and can have very few finite quotients. J. Pure Appl. Algebra 166 (2002), no. 1-2, 191–202 and in Olʹshanskii, Alexander Yu.; Sapir, Mark V. Non-amenable finitely presented torsion-by-cyclic groups. Publ. Math. Inst. Hautes Études Sci. No. 96 (2002), 43–169. In both cases it was crucial that the "extended" group $H$ has an L-presentation, in fact it was constructed as such. 
For arbitrary HNN extensions the situation is more difficult but not hopeless, I think that necessary and sufficient conditions can be found in many more cases.