In 1945, Alfred Tarski asked the question:

Let $G$ be a group. How to determine the groups elementarily equivalent to $G$?

We answered the question when $G$ is a divisible group in the previous note Elementary Theories of Divisible Groups. However, Tarski’s problem seemed to be dramatically difficult when $G$ is a free group; in particular, it was not known wether two non-abelian free groups of finite rank were elementarily equivalent.

Some partial results have been showed in the last half of the twentieth century, but the first complete answer was given by Z. Sela between 2001 and 2006 in a series of papers untitled Diophantine Geometry over Groups. Of course, because of its complexity, it is not possible to describe the solution here: we restrict ourself to the weaker problem

How to describe the finitely generated groups having the same universal theory as a free group of finite rank?

The solution described here is based on the paper Limit groups as limits of free groups by C. Champetier and V. Guirardel, simplifying an argument of Z. Sela. Another argument, without using the space of marked groups, can be found in Chiswell’s book, Introduction to $\Lambda$-trees.

First, notice that $\mathrm{Th}_{\forall}(\mathbb{F}_n)= \mathrm{Th}_{\forall} (\mathbb{F}_m)$ for all $n,m \geq 2$. It is a consequence of our previous note A free group contains a free group of any rank, showing that $\mathbb{F}_n \hookrightarrow \mathbb{F}_m$ and $\mathbb{F}_m \hookrightarrow \mathbb{F}_n$. So the problem becomes

How to describe the finitely generated groups $G$ satisfying $\mathrm{Th}_{\forall}(G)= \mathrm{Th}_{\forall} (\mathbb{F}_2)$ or $\mathrm{Th}_{\forall}(G)= \mathrm{Th}_{\forall} (\mathbb{Z})$?

The space of marked groups has already been defined in our previous note Cantor-Bendixson rank in group theory. The link between universal theories and the space of marked groups is made explicit by the following lemma:

Lemma 1: Let $(H_i,R_i)$ be a sequence of marked groups converging to $(G,S)$. Then

$\limsup\limits_{i \to + \infty} \mathrm{Th}_{\forall}(H_i) \subset \mathrm{Th}_{\forall}(G)$.

Conversely, if $G$ and $H$ are two finitely generated groups satisfying $\mathrm{Th}_{\forall} (H) \subset \mathrm{Th}_{\forall} (G)$ then, for every marking $S$, $(G,S)$ is the limit of $(H,R_i)$ for some markings $R_i$.

Proof. Let $(H_i,R_i)$ be a sequence of marked groups converging to $(G,S)$ and let $\sigma \notin \mathrm{Th}_{\forall} (G)$ be a universal formula. Then $\neg \sigma \in \mathrm{Th}_{\exists} (G)$ and we may write

$\neg \sigma = \exists x_1, \dots , x_m \Sigma(x_1, \dots, x_m)$,

where $\Sigma$ is a system of equations and inequations. In particular, there exist $g_1, \dots, g_m \in G$ such that $G \models \Sigma(g_1, \dots, g_m)$. For all $1 \leq i \leq m$, let $w_i$ be a word so that $g_i=w_i(S)$ in $G$.

Because $G \models \Sigma(w_1(S),\dots, w_m(S))$, we deduce that for all but finitely many $i$

$H_i \models \Sigma(w_1(R_i),\dots, w_m(R_i))$,

ie. $\neg \sigma \in \mathrm{Th}_{\exists}(H_i)$ or $\sigma \notin \mathrm{Th}_{\forall} (H_i)$. Thus, $\sigma \notin \limsup\limits_{i \to + \infty} \mathrm{Th}_{\forall} (H_i)$.

Let $G$, $H$ be two finitely generated groups satisfying $\mathrm{Th}_{\forall} (H) \subset \mathrm{Th}_{\forall} (G)$ and let $S$ be a marking of $G$. Let $\{w_i \mid i \in I\}$ be the set of reduced words of length at most $n$ over an alphabet of size $|S|=m$.

With $J= \{i \in I \mid G \models (w_i(S)=1) \}$ and $K= \{ i \in I \mid G \models (w_i(S) \neq 1) \}$ (in particular, $I= J \coprod K$), let $\Sigma(x_1,\dots, x_m)$ be the system

$\{ w_j(x_1, \dots, x_m)=1 \mid j \in J \} \cup \{ w_k(x_1,\dots, x_m) \neq 1 \mid k \in k \}$.

Then $\exists x_1, \dots, x_m \Sigma(x_1, \dots, x_m) \in \mathrm{Th}_{\exists}(G) \subset \mathrm{Th}_{\exists}(H)$ so there exist $h_1,\dots, h_m \in H$ such that $H \models \Sigma(h_1, \dots, h_m)$. Finally, with $R_n= (h_1, \dots, h_m)$, $d((H,R_n),(G,S)) \leq e^{-n}$ by construction hence $(H,R_n) \underset{n \to + \infty}{\longrightarrow} (G,S)$. $\square$

The lemma above justifies the following family of groups as a possible solution of our weak Tarski’s problem:

Definition: A finitely generated group $G$ is a limit group if there exists a marking $S$ such that $(G,S)$ is a limit of marked free groups.

We first notice that the property does not depend on the marking $S$, so that being a limit group becomes an algebraic property:

Lemma 2: Let $(H_i,R_i)$ be a sequence of marked groups converging to $(G,S)$ and let $\tilde{S}$ be a marking of $G$. Then $(G, \tilde{S})$ is the limit of $(H_i,\tilde{R}_i)$ for some marking $\tilde{R}_i$ of $H_i$.

Proof. Let $\tilde{S} = (\tilde{s}_1, \dots, \tilde{s}_m)$. Then for each $1 \leq i \leq m$, there exists a word $w_i$ such that $\tilde{s}_i=w_i(S)$. Let $\tilde{r}_{in} = w_i(R_n)$ and $\tilde{R}_n= (\tilde{r}_{1n} , \dots, \tilde{r}_{mn})$.

Notice that for any word $w$$w(\tilde{s}_1, \dots, \tilde{s}_m)=w(w_1(S), \dots, w_m(S))$, so $w(\tilde{S})=1$ in $G$ if and only if $w(\tilde{R}_n)= w(w_1(R_n), \dots, w_m(R_n))=1$ in $H_n$ for all but finitely many $n$ (because $(H_n, R_n)$ converges to $(G,S)$). Therefore, $(H_n, \tilde{R}_n)$ converges to $(G,\tilde{S})$. $\square$

In particular, lemma 1 solves (weak) Tarski’s problem for $\mathbb{Z}^m$:

Corollary: A finitely generated group $G$ has the same universal theory as $\mathbb{Z}$ if and only if it is isomorphic to $\mathbb{Z}^n$ for some $n \geq 1$.

Proof. If $G$ has the same universal theory as $\mathbb{Z}$ then $G$ is a finitely generated torsion-free abelian group, so it is isomorphic to a free abelian group of finite rank. Conversely, it is sufficient to show that $\mathbb{Z}$ and $\mathbb{Z}^m$ have the same universal theory for every $m \geq 1$.

Because $\mathbb{Z} \subset \mathbb{Z}^n$, $\mathrm{Th}_{\forall} (\mathbb{Z}^n) \subset \mathrm{Th}_{\forall}(\mathbb{Z})$. Conversely, according to lemma 1, it is sufficient to notice that $(\mathbb{Z}^m, \mathrm{can})$ (where $\mathrm{can}= \left( (1,0, \dots, 0),(0,1,\dots,0), \dots, (0,0, \dots, 1) \right)$ is the canonical marking) is the limit of $(\mathbb{Z}, S_n)$ where $S_n= \{ 1, 2^{2n}, 2^{3n}, \dots, 2^{mn} \}$. More precisely, it can be shown that $d((\mathbb{Z},S_n),(\mathbb{Z}^m, \mathrm{can})) \leq e^{1- 2^n}$. $\square$

Corollary: Let $G$ be a finitely generated group and $m \geq 1$. Then $G$ and $\mathbb{Z}^m$ are elementarily equivalent if and only if they are isomorphic.

Proof. Let $\{ w_i \mid i \in I\}$ be the set of words of the form $x_1^{\epsilon_1} \cdots x_m^{\epsilon_m}$ with $\epsilon_i \in \{ 0,1 \}$ and

$\sigma_m = \exists x_1,\dots,x_m \forall x \exists y \left( \bigvee\limits_{i \in I} x = 2y+ w_i(x_1, \dots,x_m) \right)$.

Then $\mathbb{Z}^n \models \sigma_m$ is equivalent to $| \mathbb{Z}^n / 2 \mathbb{Z}^n | \leq 2^m$, ie. $n \leq m$. Therefore, $\mathbb{Z}^n$ and $\mathbb{Z}^m$ are elementarily equivalent if and only if $n=m$. Finally, we conclude using the previous corollary. $\square$

Now, we may prove that limit groups actually solve weak Tarski’s problem:

Theorem 1: Let $G$ be a finitely generated group. Then $G$ has the same universal theory as $\mathbb{F}_2$ or $\mathbb{Z}$ if and only if $G$ is an limit group.

Lemma 3: Let $(H_i,R_i)$ be a sequence of marked groups converging to $(G,S)$, $K \subset G$ be a subgroup and $T$ be a marking of $K$. Then there exist subgroups $F_i \subset H_i$ and markings $Q_i$ of $F_i$ such that $(F_i, Q_i)$ converges to $(K,T)$.

Proof. Let $T= (t_1, \dots, t_m)$. In particular, for all $1 \leq k \leq m$, there exists a word $w_k$ such that $t_k=w_k(S)$. Let $h_{ik}=w_k(R_i)$, $F_i= \langle h_{i1}, \dots, h_{im} \rangle \leq H_i$ and $Q_i=(h_{i1}, \dots, h_{im} )$. Then for any word $w$:

$w(T)=1 \ \text{in} \ K \Leftrightarrow w(w_1(S), \dots, w_m(S))=1 \ \text{in} \ G \\ \\ \hspace{1cm} \Leftrightarrow w(w_1(R_i), \dots, w_m(R_i)) = 1 \ \text{in} \ H_i \ \text{for all but finitely many} \ i \\ \\ \hspace{1cm} \Leftrightarrow w(Q_i)=1 \ \text{in} \ F_i \ \text{for all but finitely many} \ i$

Therefore, $(F_i,Q_i)$ converges to $(K,T)$. $\square$

Lemma 4: A non-abelian $2$-generator subgroup of a limit group is isomorphic to $\mathbb{F}_2$.

Proof. Let $G$ be a limit group and let $a,b \in G \backslash \{1\}$. Using lemma 3, we deduce that $\langle a,b \rangle$ is a limit group, and using lemma 2, we deduce that there exist free groups $F_k$ and markings $(e_k,f_k)$ such that $(F_k,(e_k,f_k))$ converges to $(\langle a,b \rangle, (a,b))$.

If $\langle a,b \rangle$ were not free over $\{a,b\}$, there would exist a non-trivial relation $w(a,b)=1$ in $G$. In particular, $w(e_k,f_k)=1$ in $F_k$ for all but finitely many $k$, so $\mathrm{rank}(F_k) < 2$. Therefore, $(\langle a,b \rangle, (a,b))$ would be the limit of abelian marked groups that is $\langle a,b \rangle$ would be abelian itself: a contradiction. $\square$

Proof of theorem 1. The case where $G$ is abelian is clear according to our previous corollaries. If $\mathrm{Th}_{\forall} (G)= \mathrm{Th}_{\forall}(\mathbb{F}_2)$, then $G$ is a limit group according to lemma 1. Conversely, suppose that $G$ is a limit group. According to lemma 1, $\mathrm{Th}_{\forall} (\mathbb{F}_2) \subset \mathrm{Th}_{\forall} (G)$. Moreover, if $a,b \in G$ do not commute, $\langle a,b \rangle \simeq \mathbb{F}_2$ according to lemma 4 hence $\mathrm{Th}_{\forall} (G) \subset \mathrm{Th}_{\forall} (\mathbb{F}_2)$. $\square$

Because of Los’ theorem, a natural way to “construct” a limit group is the following: Let $\mathbb{F}_n$ be a free group of finite rank $n \geq 2$ and let $\omega$ be a non-principal ultrafilter over $\mathbb{N}$. Then the ultrapower $\mathbb{F}_n^{\omega}$ is elementarily equivalent to $\mathbb{F}_n$; therefore, any finitely generated subgroup $L \subset \mathbb{F}_n^{\omega}$ is a limit group since $\mathrm{Th}_{\forall} ( \mathbb{F}_n) = \mathrm{Th}_{\forall} \left( \mathbb{F}_n^{\omega} \right) \subset \mathrm{Th}_{\forall} (L)$ and using lemma 1.

In fact, the converse is also true:

Theorem 2: A finitely generated group is a limit group if and only if it embeds into an ultrapower of a free group of finite rank.

Notice that it is sufficient to consider the ultrapowers $\mathbb{F}_2^{\omega}$ where $\omega$ is non-principal. Our theorem is an easy consequence of the following lemma:

Lemma 5: Let $(H_k,R_k)$ be a sequence of marked groups converging to $(G,S)$. Then for all non-principal ultrafilter $\omega$, $G$ embeds into the ultraproduct $\prod\limits_{k \geq 1} H_k / \omega$. Conversely, if $G$ is a finitely generated subgroup of an ultraproduct $\prod\limits_{k \geq 1} G_k / \omega$ (where $\omega$ is non-principal), then for every marking $S$ of $G$, there exist an increasing sequence $(i_k)$, subgroups $H_{i_k} \subset G_{i_k}$, markings $R_{i_k}$ of $H_{i_k}$ such that $(H_{i_k},R_{ik})$ converges to $(G,S)$.

Proof. Let $(H_k,R_k)$ be a sequence of marked groups converging to $(G,S)$. Notice that for any words $w_1$ and $w_2$, if $w_1(S)=w_2(S)$ then $w_1(R_k)=w_2(R_k)$ for all but finitely many $k$ hence $(w_1(R_k))=(w_2(R_k))$ in $\prod\limits_{k \geq 1} H_k / \omega$. Therefore, the morphism

$\varphi : \left\{ \begin{array}{ccc} G & \to & \prod\limits_{k \geq 1} H_k / \omega \\ w(S) & \mapsto & (w(R_k)) \end{array} \right.$ for any word $w$

is well-defined. Let $w(S) \in \mathrm{ker}(\varphi)$. Then $\{k \geq 1 \mid w(R_k)=1 \}$ belongs to $\omega$ and is infinite in particular, so there exists $k$ arbitrarily large such that $w(R_k)=1$ in $H_k$, hence $w(S)=1$. Therefore, $\varphi$ is an embedding.

Let $G$ be a finitely generated subgroup of an ultraproduct $\prod\limits_{k \geq 1} H_k / \omega$ and let $S$ be a marking of $G$. Let $S=(s_1, \dots, s_m)$ where $s_i=(r_{ik})$. Set $H_k= \langle r_{1k}, \dots, r_{mk} \rangle \leq G_k$ and $R_k= (r_{1k}, \dots, r_{mk} )$.

Now notice that $\bigcap\limits_{ \ell g(w) \leq n} \{ k \geq 1 \mid w(R_k)=1 \Leftrightarrow w(S)=1 \}$ belongs to $\omega$ and in particular is infinite. Therefore, there exists $p$ arbitrarily large such that $d((H_p,R_p),(G,S)) \leq e^{-n}$. We deduce that the sequence $(i_k)$ can be easily constructed by induction. $\square$

Finally, a purely algebraic characterization of limit groups can be stated:

Definition: A group $G$ is fully residually free if for every finite set $S \subset G \backslash \{1\}$ there exists a morphism $\varphi$ from $G$ to a free group $F$ such that $\varphi(g) \neq 1$ for all $g\in S$.

Theorem 3: A finitely generated group $G$ is a limit group if and only if it is fully residually free.

Lemma 6: Let $\omega$ be a non-principal ultrafilter and let $R$ be a finitely generated subring of $\mathbb{Z}^{\omega}$. Then $R$ is fully residually $\mathbb{Z}$, that is for every finite subset $S \subset R \backslash \{0\}$ there exists a ring homomorphism $\rho : R \to \mathbb{Z}$ such that $\rho(s) \neq 0$ for all $s \in S$.

Proof. Because $R$ is finitely generated, there exist $t_1, \dots, t_n \in \mathbb{Z}^{\omega}$ such that $R= \mathbb{Z} [t_1, \dots, t_n]$. Let $\mathbb{Z}[T_1, \dots, T_n ]$ denote the ring of $n$-variable polynomials and let

$\varphi : \mathbb{Z}[T_1, \dots, T_n] \twoheadrightarrow \mathbb{Z} [t_1, \dots, T_n] = R$

be the canonical projection. Because $\mathbb{Z} [T_1, \dots, T_n]$ is noetherian, $\mathrm{ker}(\varphi)$ is generated by finitely many elements $f_1, \dots, f_s \in \mathbb{Z}[T_1, \dots, T_n]$.

Let $r_1, \dots, r_m \in R \backslash \{0\}$ and $g_1, \dots, g_m \in \mathbb{Z}[T_1, \dots, T_n]$ such that $\varphi(g_i)=r_i$ for all $1 \leq i \leq m$. Then

$\left\{ \begin{array}{ll} f_i(t_1, \dots, t_n)=0, & 1 \leq i \leq s \\ g_i(t_1, \dots,t_n ) \neq 0, & 1 \leq i \leq m \end{array} \right.$

Therefore, for all $1 \leq i \leq s$ (resp. $1 \leq i \leq m$) and for $\omega$-almost all $k$, $f_i(t_{1k}, \dots,t_{nk})=0$ (resp. $g_i(t_{1k}, \dots, t_{nk}) \neq 0$). Thus, there exist $x_1, \dots, x_n \in \mathbb{Z}$ such that

$\left\{ \begin{array}{ll} f_i(x_1, \dots, x_n)=0, & 1 \leq i \leq s \\ g_i(x_1, \dots,x_n ) \neq 0, & 1 \leq i \leq m \end{array} \right.$

Now let $\phi : \mathbb{Z}[T_1, \dots, T_n] \to \mathbb{Z}$ be the only morphism satisfying $\phi(T_i)=x_i$ for all $i$. Because $f_i(x_1, \dots, x_n)=0$ for all $i$, $\phi$ induces a morphism $\overline{\phi} : R \to \mathbb{Z}$. Because $g_i(x_1, \dots, x_n) \neq 0$ for all $i$, $\overline{\phi}(r_i)= \phi(g_i) \neq 0$ for all $i$.

We deduce that $R$ is fully residually $\mathbb{Z}$. $\square$

Proof of theorem 3. Suppose that $G$ is fully residually free. Let $X$ be a finite generator set and let $G_n \subset G \backslash \{1\}$ be the set of non-trivial elements of length at most $n$ over $X$. For all $n$, let $\phi_n$ be a morphism from $G$ to a free group $F_n$ such that $\phi_n(g) \neq 1$ for all $g \in G_n$. Finally, let the morphism

$\varphi : \left\{ \begin{array}{ccc} G & \to & \prod\limits_{n \geq 1} F_n / \omega \\ g & \mapsto & (\phi_n(g)) \end{array} \right.$,

where $\omega$ is a non-principal ultrafilter. Let $g \in G$ such that $\ell g(g)=m \geq 1$. Then $g \in \bigcap\limits_{n \geq m} G_n$ so $\phi_n(g) \neq 1$ for all $n \geq m$. Thus, $\{n \geq 1 \mid \phi_n(g) \neq 1 \} \in \omega$ ie. $\varphi(g) \neq 1$. Therefore, $\varphi$ is injective and $G$ is a limit group according to theorem 2.

Conversely, suppose that $G$ is a limit group. According to theorem 2, there exists a non-principal ultrafilter $\omega$ such that $G \hookrightarrow \mathbb{F}_2^{\omega}$. The canonical projection $\mathbb{Z} \twoheadrightarrow \mathbb{Z}_p$ induces a morphism $\varphi : SL_2(\mathbb{Z}) \to SL_2(\mathbb{Z}_p)$. Then the modular group $\Gamma_p = \mathrm{ker}(\varphi)$ is a non-abelian free group.

(This fact is shown in Serre’s book, Trees. The idea is to notice that $SL_2(\mathbb{Z})$ acts on a tree in the hyperbolic plane, to deduce that $SL_2(\mathbb{Z}) \simeq \mathbb{Z}_4 \underset{\mathbb{Z}_2}{\ast} \mathbb{Z}_6$ and finally to show that a subgroup of $SL_2(\mathbb{Z})$ is free if and only if it is torsion-free.)

Thus, if $\varphi^* : SL_2(\mathbb{Z}^{\omega}) \to SL_2(\mathbb{Z}_p^{\omega})$ is induced by the canonical projection $\mathbb{Z}^{\omega} \twoheadrightarrow \mathbb{Z}_p^{\omega}$,

$G \hookrightarrow \mathbb{F}_2^{\omega} \hookrightarrow \Gamma_p^{\omega} \simeq \mathrm{ker}(\varphi^*)$.

Therefore, we may suppose without loss of generality that $G$ is a subgroup of $SL_2(\mathbb{Z}^{\omega})$. Moreover, because $G$ is finitely generated, there exists a finitely generated subring $R$ of $\mathbb{Z}^{\omega}$ such that $G \leq SL_2(R)$.

Let $g_1, \dots, g_n \in G \backslash \{1\}$ and let $a_1, \dots, a_r \in R$ denote the non-zero coefficients of the matrices $g_i - \mathrm{Id}$. According to lemma 6, there exists a ring homomorphism $\rho : R \to \mathbb{Z}$ such that $\rho(a_i) \neq 0$ for all $i$. In particular, $\rho$ induces a ring homomorphism $\overline{\rho} : SL_2(R) \to SL_2(\mathbb{Z})$.

Noticing that $\rho(G) \subset \Gamma_p$ since $G \subset \mathrm{ker}(\varphi^*)$ and that $\rho(g_i) \neq \mathrm{Id}$, we deduce that $G$ is fully residually free. $\square$

Finally, we proved:

Theorem A: Let $G$ be a finitely generated group. Then the following statements are equivalent:

1. $G$ is an abelian limit group,
2. $G$ has the same universal theory as $\mathbb{Z}$,
3. $G$ is abelian and embeds into an ultrapower $\mathbb{F}_2^{\omega}$,
4. $G$ is abelian and fully residually free,
5. $G$ is isomorphic to $\mathbb{Z}^n$ for some $n\geq 1$.

Theorem B: Let $G$ be a finitely generated group. Then the following statements are equivalent:

1. $G$ is a non-abelian limit group,
2. $G$ has the same universal theory as $\mathbb{F}_2$,
3. $G$ embeds into an ultrapower $\mathbb{F}_2^{\omega}$,
4. $G$ is non-abelian and fully residually free.

From theorems A and B, we may deduce the following properties of limit groups:

Property 1: The following statements hold:

1. A limit group is torsion-free,
2. A limit group is commutative-transitive (ie. if $y \neq 1$ commute with $x$ and $z$, then $x$ and $y$ commute),
3. A limit group is residually finite,
4. A finitely-generated subgroup of a limit group is a limit group,
5. A non-trivial $2$-generator subgroup of a limit group is isomorphic to either $\mathbb{Z}$, $\mathbb{Z}^2$ or $\mathbb{F}_2$,
6. Let $G_1, \dots, G_n$ be finitely generated groups. Then $G_1 \ast \cdots \ast G_n$ is a limit group if and only if $G_1, \dots, G_n$ are limit groups.

We also have the following property as a consequence of lemma 7:

Property 2: A limit group (and more generally, any residually free group) is hopfian.

Lemma 7: Let $G_1 \overset{\varphi_1}{\twoheadrightarrow} G_2 \overset{\varphi_2}{\twoheadrightarrow} G_3 \overset{\varphi_2}{\twoheadrightarrow} \dots$ be a sequence of epimorphisms between finitely generated residually free groups. Then all but finitely many epimorphisms are isomorphisms.

Indeed, if $\varphi : G \to G$ is an epimorphism but not an isomorphism (where $G$ is finitely generated and residually free), then the sequence $G \overset{\varphi}{\twoheadrightarrow} G \overset{\varphi}{\twoheadrightarrow} G \overset{\varphi}{\twoheadrightarrow} \dots$ contradicts the lemma.

Proof: Let $S_1=(s_1^{(1)}, \dots, s_n^{(1)})$ be a generator set of $G_1$; then $S_k:= \varphi_k(S_1)= (s_1^{(k)}, \dots, s_n^{(k)})$ is a generator set of $G_k$ by surjectivity of $\varphi_k$. Let $V_k$ be the set of $(M_1, \dots, M_n) \in SL_2(\mathbb{C})^n$ such that $r(M_1, \dots, M_n)= \mathrm{Id}$ for every relation $r$ of $G_k$.

Notice that $V_k$ is an affine algebraic variety in $\mathbb{C}^{4n}$ and $V_1 \supset V_2 \supset \cdots$, so the sequence $(V_k)$ is eventually constant. To conclude, it is sufficient to show that if $\varphi_k : G_k \to G_{k+1}$ is not an isomorphism, then $V_{k+1} \subsetneq V_k$.

Let $r$ be a word such that $r(S_{k+1})=1$ in $G_{k+1}$ but $r(S_k) \neq 1$ in $G_k$. $G_k$ being residually free, there exists a morphism $\rho : G_k \to \mathbb{F}_2$ such that $\rho(r) \neq 1$. Because $\mathbb{F}_2$ is a subgroup of $SL_2(\mathbb{C})$, we may suppose that $\rho : G_k \to SL_2(\mathbb{C})$.

Let $N_i = \rho(s_i^{(k)}) \in SL_2(\mathbb{C})$. Then $r(N_1, \dots, N_n)= \rho(r) \neq 1$ so $(N_1, \dots, N_n) \notin V_{k+1}$. If $w$ is a word over $S_k$ such that $w$ is a relation of $G_k$, then $w(N_1, \dots, N_n) = \rho(w) = \rho(1)=1$ so $(N_1, \dots, N_n) \in V_k$. $\square$

A difficult result about limit groups is:

Property: A limit group is finitely presented.

See for example V. Guirardel’s article, Limit groups and groups acting freely on $\mathbb{R}^n$-trees, for a proof based on actions on $\Lambda$-trees.

To conclude this note, let us mention a result giving examples of non-free limit groups:

Theorem 4: Let $F$ be a free group and $C \leq F$ be a cyclic subgroup closed under taking roots. Then $F \underset{C}{\ast} F$ is a limit group.

A proof can be found in Henry Wilton’s document, An introduction to limit groups (proprosition 2.9). In particular, the fundamental group of a surface $\Sigma$ is a limit group when $\Sigma$ is not a non-orientable surface whose Euler characteristic -1, 0 or 1.