## When is a finitely generated group finitely presented?

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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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This question seems like it's asking for a list. If it has no single correct answer, it should probably be Community Wiki. – HW Feb 10 2011 at 2:41

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Maybe this is too basic, but algebraic (ex. nilpotent) and geometric (ex. hyperbolic) properties can give f.p. automatically. – Steve D Feb 10 2011 at 4:48

## 2 Answers

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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...with any small choice of coefficients. – Mariano Suárez-Alvarez Feb 10 2011 at 3:40
Mariano - right. I suppose I want the coefficient module itself to be finitely generated. – HW Feb 10 2011 at 14:56

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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.

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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.) – aaron Feb 10 2011 at 13:26

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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 $G'=G$. – Mark Sapir Feb 10 2011 at 13:36
... 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). – Mark Sapir Feb 10 2011 at 14:27
@Mark: Very nice. Thanks. – aaron Feb 10 2011 at 21:27
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