We proved in the previous note of our series that there exist 2^{\aleph_0} finitely generated groups (up to isomorphism). However, clearly there exist only \aleph_0 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 \langle X \mid R \rangle be a presentation of a group G, where X is finite and R infinite. If G is finitely presented, then there exists R' \subset R finite such that \langle X \mid R' \rangle is a presentation of G.

Roughly, it is possible to argue as follow: Let \langle X \mid W \rangle be a finite presentation. For any relation w \in W, it is possible to write a proof of w=1 using the relations R; such a proof uses only finite many relations R_w \subset R. Let R' = \bigcup\limits_{w \in W} R_w. Then \langle X \mid R' \rangle is a finite presentation of G, because any relation may be deduced from W and a fortiori from R'.

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

Proof. Let w(X) be a word over X such that w(X)=1 in G. We first prove that the equality w(X)=1 can be deduced from a finite set R_w \subset R of relations.

Let \mathcal{T}_{\text{group}} denote the theory of group, and let us define the theory

\mathcal{T} = \mathcal{T}_{\text{group}} \cup \{ r(X)=1, \ r \in R \}.

Clearly, G is a model of \mathcal{T}; let H be another model. Let g_x (resp. h_x) denote the interpretation of x \in X in G (resp. in H). Then the map g_x \mapsto h_x extends to a morphism \varphi : G \to H. We deduce that

w(h_x, \ x \in X)= \varphi (w ( g_x, \ x \in X))= \varphi(1)=1,

so H satisfies w(X)=1. Because it is true for any model, \mathcal{T} \models w(X)=1. From compactness theorem, we deduce that there exists \Delta \subset \mathcal{T} finite such that \Delta \models w(X)=1. Now, it is sufficient to set

R_w= \Delta \cap \{ r(X)=1, \ r \in R \}.

Then, let \langle Y \mid W \rangle be a finite presentation of G and \phi : \langle Y \mid W \rangle \to \langle X \mid R \rangle be an isomorphism; in particular, for any word w over Y (resp. over X), let \phi(w) (resp. \phi^{-1}(w)) denote the word over X (resp. over Y) corresponding to w via \phi (resp. via \phi^{-1}). Let

R'= \bigcup\limits_{w \in W} R_w.

To conclude, it is sufficient to notice that any relation r \in R may be deduced from the relations of R'. But this is clear, because \phi^{-1}(r)=1 may be deduced from the relations W, so r=1 may be deduced from the relations \phi(W), and so from the relations R' by construction. \square

Property: Let A,B be two finitely presented groups. The amalgamation A \underset{C}{\ast} B is finitely presented if and only if C is finitely generated.

Proof. By contradiction, suppose that C is not finitely generated and that A \underset{C}{\ast} B is finitely presented.

Let \langle X \mid R \rangle (resp. \langle Y \mid S \rangle) be a finite presentation of A (resp. B). Let \varphi_1 : C \hookrightarrow A and \varphi_2 : C \hookrightarrow B be the two monomorphisms associated to our amalgamation. Then we have the following presentation

A \underset{C}{\ast} B = \langle X,Y \mid R,S, \varphi_1(c)= \varphi_2(c) \ \forall c \in C \rangle.

Because A \underset{C}{\ast} B is finitely presented, there exist c_1, \dots, c_n \in C such that

\Pi = \langle X,Y \mid R,S, \varphi_1(c_i) = \varphi_2(c_i), 1 \leq i \leq n \rangle

is a presentation of A \underset{C}{\ast} B. Let P denote the subgroup of C generated by \{c_1, \dots, c_n \}. Then \Pi may be identified with the amalgamated product A \underset{P}{\ast} B.

Because C is not finitely generated, there exists c \in C \backslash P, and \varphi_1(c) \varphi_2(c)^{-1} \neq 1 in \Pi since \varphi_1(c) \notin \varphi_1(P) and \varphi_2(c) \notin \varphi_2(P). However, \varphi_1(c) = \varphi_2(c) in A \underset{C}{\ast} B, hence \Pi \neq A \underset{C}{\ast} B, a contradiction. \square

Property: Let A be a finitely presented group. The HNN extension \underset{C}{\ast} A is finitely presented if and only if C is finitely generated.

Proof. Suppose by contradiction that \underset{C}{\ast} A is finitely generated but that C is not finitely generated. Let \varphi_i : C \hookrightarrow A (i=1,2) denote the monomorphisms associated to our HNN extension; in particular, if \langle X \mid R \rangle is a presentation of A,

\underset{C}{\ast} A = \langle X,t \mid R, t \varphi_1(c) t^{-1}= \varphi_2(c) \ \forall c \in C \rangle.

Because \underset{C}{\ast} A is finitely presented, there exist c_1, \dots, c_n \in C such that

\Pi = \langle X ,t \mid R, t\varphi_1(c_i) t^{-1} = \varphi_2(c_i), 1 \leq i \leq n \rangle

is a presentation of \underset{C}{\ast} A. Let P denote the subgroup of C generated by \{ c_1, \dots, c_n \}. In particular, \Pi may be identified with the HNN extension \underset{P}{\ast} A.

Because C is not finitely generated, there exists c \in C \backslash P. Since \varphi_i(c) \notin \varphi_i(P) (i=1,2), t\varphi_1(c) t^{-1} \varphi_2(c)^{-1} \neq 1 in \Pi according to Britton’s lemma; however, t \varphi_1(c)t^{-1}= \varphi_2(c) in \underset{C}{\ast} A, hence \Pi \neq \underset{C}{\ast} A, a contradiction. \square

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 \mathbb{F}_2 of rank two is not finitely-generated. Therefore, the following amalgamated product \mathbb{F}_2 \underset{ [\mathbb{F}_2, \mathbb{F}_2] }{\ast} \mathbb{F}_2 is not finitely-presented:

\langle a,b, x,y \mid a^ib^ja^{-1} b^{-j} a^{1-i} = x^i y^j x^{-1}y^{-j} x^{1-i}, a^{-i} b^j ab^{-j}a^{i-1}= x^{-i} y^j xy^{-j} x^{i-1} \rangle,

where i > 0 and j \in \mathbb{Z} \backslash \{0\}. The same thing can be done with a HNN extension.

Property: The lamplighter groups L_n = \mathbb{Z}_n \wr \mathbb{Z} and L= \mathbb{Z} \wr \mathbb{Z} are not finitely presented.

Proof. We give the proof only for L_2 : the other cases are similar. We have the presentation

L_2= \langle a,t \mid a^2=1, [t^nat^{-n}, t^{n+1}at^{-n-1} ]=1, n \in \mathbb{Z} \rangle.

In order to show that L_2 is not finitely presented, it is sufficient to show that the groups

G_k = \langle a,t \mid a^2=1, [t^nat^{-n}, t^{n+1}at^{-n-1} ]=1, -k \leq n \leq k-1 \rangle

are not isomorphic to L_2. First, with a_n= t^nat^{-n}, we get the equivalent presentation

G_k = \langle a_{-k}, \dots, a_k,t \mid a_n^2=a_k^2=1, [a_n,a_{n+1}]=1, a_{n+1}= ta_nt^{-1} \rangle,

where -k \leq n\leq k-1. We may recognize an HNN extension: Let

A_k = \langle a_{-k}, \dots, a_k \mid a_n^2=a_k^2=1, [a_n,a_{n+1}]=1, -k \leq n \leq k-1 \rangle,

B_k = \langle a_{-k}, \dots, a_{k-1} \rangle and C_k= \langle a_{-k+1}, \dots, a_k \rangle two subgroups of A_k, and \varphi_k : B_k \to C_k the isomorphism defined by \varphi_k(a_i)=a_{i+1} for all -k \leq i \leq k-1. Then

G_k=HNN (A_k, B_k, C_k, \varphi_k).

Using Britton’s lemma, we deduce that the subgroup \langle a_0,t \rangle of G_k is isomorphic to \mathbb{Z}_2 \ast \mathbb{Z}. But \mathbb{Z}_2 \ast \mathbb{Z}  is not solvable, because it contains a nonabelian free group (for example, abab^{-1} and ab^2ab^{-2} generate a free group of rank two; see the third proof in A free group contains a free group of any rank); so G_k cannot be solvable, whereas L_2 so is (as a semi-direct product of two abelian groups). In particular, G_k and L_2 cannot be isomorphic. \square

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