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. $\square$