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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.

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    $\begingroup$ Well, as you justify, the answer is not obvious. The problem is non-trivial (and likely has no general answer) even it the most degenerate instance of HNN, namely that of a semidirect product $G=H\rtimes\mathbf{Z}$. This group can be finitely presented even if $H$ is not finitely presented, or even finitely generated. For instance, the Thompson group $F$ of the interval can be written in either of these ways (and also with $H$ finitely presented). Such phenomena also occur in the context of metabelian groups. $\endgroup$
    – YCor
    Commented Aug 10, 2012 at 11:27
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    $\begingroup$ @HW: probably SteveD means that ZwrZ admits a f.p. HNN extension (Baumslag's group). $\endgroup$
    – YCor
    Commented Aug 10, 2012 at 11:47
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    $\begingroup$ @HW: The sort of answer I would hope for would be "If you put these restrictions on $G$ and $K$ (and perhaps on the isomorphism between $K$ and $K^{\prime}$) then you can say something." Along the lines of the Baumslag-Tretfoff conditions for residual finiteness; sufficient, but not necessary. $\endgroup$
    – ADL
    Commented Aug 10, 2012 at 12:05
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    $\begingroup$ @HW: there are several Baumslag groups :p... view ZwrZ as the semidirect product $Z[t,t^{-1}]\rtimes \langle t\rangle$. (this is certainly not Baumslag's language, who rather likes group presentations). Consider the injective homomorphism of the free abelian group of infinite rank $Z[t,t^{-1}]$ given by multiplication by $(t-1)$. The associated ascending HNN-extension is thus the semidirect product $Z[t,t^{-1},(t-1)^{-1}]\rtimes \langle t,t-1\rangle$, where $\langle t,t-1\rangle$ is isomorphic to $Z^2$. Baumslag proved it's f.p., with a more conceptual and general proof by Bieri-Strebel. $\endgroup$
    – YCor
    Commented Aug 10, 2012 at 12:47
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    $\begingroup$ Thanks for the interesting comments, Mark. Perhaps either you or Benjamin could write an answer about L-presentations? $\endgroup$
    – HJRW
    Commented Aug 13, 2012 at 9:51

1 Answer 1

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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.

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  • $\begingroup$ You might give some link where to learn about L-presentations (or even define it, if it's short), I'm not sure it's very standard knowledge. $\endgroup$
    – YCor
    Commented Aug 15, 2012 at 23:23
  • $\begingroup$ I defined it in my answer, after the words "That is". The term was, I think, invented by Bartholdi, arxiv.org/abs/math/0007062 $\endgroup$
    – user6976
    Commented Aug 15, 2012 at 23:32
  • $\begingroup$ I doubt about your characterization, as I think you assume that the kernel of the homomorphism to $\mathbf{Z}$ is finitely generated (you don't really say which generators you take in an L-presentation but in Bartholdi's paper it's assumed to be finite). But there are ascending HNN extensions of infinitely generated groups (e.g. Thompson's group or Baumslag's group) that are finitely presented. $\endgroup$
    – YCor
    Commented Aug 16, 2012 at 10:06
  • $\begingroup$ The kernel is never finitely generated if $\phi$ is not onto because it is an infinite increasing union of isomorphic groups. I assume for simplicity that $H$ is finitely generated i.e. $X$ is finite. If $X$ is infinite, one needs to assume, in addition, that $X$ is a union of a finite number of $\phi$-orbits (or something similar). $\endgroup$
    – user6976
    Commented Aug 16, 2012 at 11:14
  • $\begingroup$ Sorry, I meant: you assume that $H$ is finitely generated (as you have guessed!). $\endgroup$
    – YCor
    Commented Aug 16, 2012 at 18:44

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