We proved in the previous note of our series that there exist finitely generated groups (up to isomorphism). However, clearly there exist only finitely presented groups (up to isomorphism), so we deduce that almost all finitely generated groups are not finitely presented. In this note, we give some examples of such groups.

**Lemma:** Let be a presentation of a group , where is finite and infinite. If is finitely presented, then there exists finite such that is a presentation of .

Roughly, it is possible to argue as follow: Let be a finite presentation. For any relation , it is possible to write a proof of using the relations ; such a proof uses only finite many relations . Let . Then is a finite presentation of , because any relation may be deduced from and a fortiori from .

Avoiding proof theory (and Gödel’s completeness theorem), we give below a rigorous proof based on model theory.

**Proof.** Let be a word over such that in . We first prove that the equality can be deduced from a finite set of relations.

Let denote the theory of group, and let us define the theory

.

Clearly, is a model of ; let be another model. Let (resp. ) denote the interpretation of in (resp. in ). Then the map extends to a morphism . We deduce that

,

so satisfies . Because it is true for any model, . From compactness theorem, we deduce that there exists finite such that . Now, it is sufficient to set

.

Then, let be a finite presentation of and be an isomorphism; in particular, for any word over (resp. over ), let (resp. ) denote the word over (resp. over ) corresponding to via (resp. via ). Let

.

To conclude, it is sufficient to notice that any relation may be deduced from the relations of . But this is clear, because may be deduced from the relations , so may be deduced from the relations , and so from the relations by construction.

**Property:** Let be two finitely presented groups. The amalgamation is finitely presented if and only if is finitely generated.

**Proof.** By contradiction, suppose that is not finitely generated and that is finitely presented.

Let (resp. ) be a finite presentation of (resp. ). Let and be the two monomorphisms associated to our amalgamation. Then we have the following presentation

.

Because is finitely presented, there exist such that

is a presentation of . Let denote the subgroup of generated by . Then may be identified with the amalgamated product .

Because is not finitely generated, there exists , and in since and . However, in , hence , a contradiction.

**Property:** Let be a finitely presented group. The HNN extension is finitely presented if and only if is finitely generated.

**Proof.** Suppose by contradiction that is finitely generated but that is not finitely generated. Let () denote the monomorphisms associated to our HNN extension; in particular, if is a presentation of ,

.

Because is finitely presented, there exist such that

is a presentation of . Let denote the subgroup of generated by . In particular, may be identified with the HNN extension .

Because is not finitely generated, there exists . Since (), in according to Britton’s lemma; however, in , hence , a contradiction.

For example, we saw in the note A free group contains a free group of any rank that the derived subgroup of the free group of rank two is not finitely-generated. Therefore, the following amalgamated product is not finitely-presented:

,

where and . The same thing can be done with a HNN extension.

**Property:** The lamplighter groups and are not finitely presented.

**Proof.** We give the proof only for : the other cases are similar. We have the presentation

.

In order to show that is not finitely presented, it is sufficient to show that the groups

are not isomorphic to . First, with , we get the equivalent presentation

,

where . We may recognize an HNN extension: Let

,

and two subgroups of , and the isomorphism defined by for all . Then

.

Using Britton’s lemma, we deduce that the subgroup of is isomorphic to . But is not solvable, because it contains a nonabelian free group (for example, and generate a free group of rank two; see the third proof in A free group contains a free group of any rank); so cannot be solvable, whereas so is (as a semi-direct product of two abelian groups). In particular, and cannot be isomorphic.