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I think the question is very general and hard to answer. However I've seen a paper by Baumslag ("Wreath products and finitely presented groups", 1961) showing, as a particular case, that the lamplighter group is not finitely presented. To prove this, he gives conditions to say if a wreath product of groups is finitely presented. The question is: which ways (techniques, invariants, etc) are available to determine whether a finitely generated group is also finitely presented? For instance, is there another way to show that fact about the lamplighter group?

Thanks in advance for references and comments.

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    $\begingroup$ Maybe this is too basic, but algebraic (ex. nilpotent) and geometric (ex. hyperbolic) properties can give f.p. automatically. $\endgroup$
    – Steve D
    Commented Feb 10, 2011 at 4:48
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    $\begingroup$ Techniques to prove that a group is f.p. and that a group is not f.p. are quite different. For the second, I'd say that the general idea is to have a good understanding of some "approximating sequence" $(G_n\twoheadrightarrow)$ of truncated presentations. This applies in many examples, e.g. in Bieri-Strebel for infinitely presented metabelian groups, for wreath products, for topological-full groups of minimal self-homeos, lacunary hyperbolic groups, etc. By "good" understanding, I mean we either directly see that $G_n$ is not isomorphic to $G$, or that $G_n\to G_{n+1}$ has nontrivial kernel. $\endgroup$
    – YCor
    Commented Jun 3, 2019 at 8:31

3 Answers 3

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One general method is to consider an infinite presentation of the group, and then show that every finite subset of the set of relations defines a group with clearly different property. for example, the lamplighter group has the presentation $\langle \ldots a_{-n}, \ldots, a_1, a_2, \ldots, a_n,\ldots,t \mid a_0^2=1, [a_i,a_j]=1, ta_it^{-1}=a_{i+1}\rangle$. Every finite subpresentation defines a group that has as a quotient one of the following groups $H_n=\langle a_{-n}, \ldots, a_1, a_2, \ldots, a_n,t \mid a_0^2=1, [a_i,a_j]=1, ta_it^{-1}=a_{i+1}\rangle$ for some $n$. The group $H_n$ is an HNN extension of a finite Abelian group $\langle a_{-n},\ldots, a_n\rangle$ with the free letter $t$. Hence $H_n$ is a virtually free group, in particular, $H_n$ contains a non-Abelian free subgroup. Therefore every finite subpresentation defines a group containing a free non-Abelian subgroup, while the Lamplighter group is solvable and thus cannot contain a free non-Abelian subgroup. Similarly lacunary hyperbolic but not hyperbolic groups given by presentations satisfying small cancelation conditions or their generalizations are infinitely presented since every finite subpresentation of their presentation defines a hyperbolic group.

I need to add a very nice characterisation of finitely presented metabelian groups: Robert Bieri and Ralph Strebel, Valuations and finitely presented metabelian groups, In: Proceedings of the London Mathematical Society 3.3 (1980), pp. 439-464, https://doi.org/10.1112/plms/s3-41.3.439

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    $\begingroup$ Can this type of argument be used to show that the group has no finite presentation? (Or does it only show what it seems to show, namely that the group has no finite sub-presentation of a given infinite presentation.) $\endgroup$
    – aaron
    Commented Feb 10, 2011 at 13:26
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    $\begingroup$ If a finitely generated group $G$ is finitely presented with finite set of relators $Q$, then every infinite presentation $R$ has a finite subset that is also a presentation. Indeed, consider any proof of $Q$ using $R$. It involves only finite subset $R'$ of $R$. The relations from $R'$ define the group $G$. Indeed, let $G'$ be the group defined by $R'$. Then the identity map on the generating set induces a hom. $\phi: G'\to G$. All relations of $Q$ hold in $G'$, so the kernel of $\phi$ is trivial, and $G'=G$. $\endgroup$
    – user6976
    Commented Feb 10, 2011 at 13:36
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    $\begingroup$ ... of course, you need to use the fact that if $G$ has finite presentation with one generating set, then for any other finite generating set, it also has finite presentation (just rewrite all relations in the new generating set). $\endgroup$
    – user6976
    Commented Feb 10, 2011 at 14:27
  • $\begingroup$ IIRC, technically you can only say that if a group $G$ is f.p. and $G=<X|R>$ where $X$ is finite, then there's a finite subset R' of $R$ where $G=<X|R'>$. There's a formal proof of this in Kapovich-Drutu, for instance. I can't remember a counterexample when $X$ is infinite at the moment, but I think they exist. Anyway, in the lamplighter example, you should then start with an infinite presentation with finitely many generators if you want to show it's not f.p. (You only need $a_0,t$ to generate.) $\endgroup$
    – biringer
    Commented Oct 19, 2022 at 0:39
  • $\begingroup$ Ok, more commentary, just for posterity: Use the presentation $L=<e,t | e^2, [t^net^{-n},e]>$. To show no finite subset of relators suffices, you need to say that it's impossible to derive the relation $[t^net^{-n},e]=1$ from the relation $e^2=1$ and the relations $[t^iet^{-i},e]=1$, where $|i|\leq n$. The easiest way to do this is to map $e,t$ to the permutations $(12)$ and $(12\cdots (2n+1))^2$ in the symmetric group $S_{2n+1}$. In that group, the latter relations hold, but $[t^net^{-n},e]=1$ doesn't. $\endgroup$
    – biringer
    Commented Oct 19, 2022 at 0:49
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An often-used method is to compute $H_2$. If the group is finitely presentable then $H_2$ is of finite rank with any coefficients.

For instance, you can use this technique to show that if $q:F\to\mathbb{Z}$ is the map from the free group of rank two that sends both generators to one then the fibre product $H\subseteq F\times F$, ie $(q\times q)^{-1}$ of the diagonal, is infinitely presented.

A famous theorem of Bestvina and Brady shows that this doesn't always work: they give a similar example which is infinitely presented but has finite-rank $H_2$.

A related technique shows that this question is indeed `very hard'. Grunewald showed that the fibre product coming from a surjection $f:F\to Q$ is finitely presented if and only $Q$ is finite. It follows that you cannot in general tell if a recursively presented group is (in)finitely presented.

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  • $\begingroup$ ...with any small choice of coefficients. $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Feb 10, 2011 at 3:40
  • $\begingroup$ Mariano - right. I suppose I want the coefficient module itself to be finitely generated. $\endgroup$
    – HJRW
    Commented Feb 10, 2011 at 14:56
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A seminal work in proving finiteness properties among groups is Brown's article Finiteness properties of groups. The article itself contains applications to Thompson-Higman groups, Houghton's groups and Abels' matrix groups, but since then the strategy has been applied to many other groups.

Another useful strategy comes from Bestvina and Brady's Morse theory.

The main idea is that, if a group $G$ acts on a (contractible) cellular complex $X$, endowed with a $G$-equivariant Morse function $f : X \to \mathbb{R}$ such that $G$ acts geometrically on the pre-images of bounded intervals under $f$, then there exists a connection between the finiteness properties of $G$ and the contractibility of the ascending and descending links in $X$.

Such a strategy has been applied successfully in many situations, including some subgroups of right-angled Artin groups, Basilica Thompson group, higher dimensional Thompson groups, braided Thompson's groups. (Other applications can be found by looking at the articles citing Bestvina and Brady's paper.)

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