The aim of this note is to prove to following result using a geometrical viewpoint:

Theorem 1: (B.H. Neumann) There exist continuously many non-isomorphic two-generator groups.

More precisely, we define the Cantor-Bendixson rank of a group thanks to the space of marked groups and then we show that $\mathbb{F}_2$ (and more generally any limit group) is a condensation group using small cancellation groups.

For precise references, we may mention:

Definition: A marked group $(G,S)$ is a finitely generated group $G$ endowed with a finite generator set $S=(s_1,\dots,s_n)$. A morphism between two marked groups $(G,S=(s_1,\dots,s_n))$ and $(H,R=(r_1,\dots,r_n))$ is a morphism $G \to H$ sending $s_i$ to $r_i$ for all $1 \leq i \leq n$.

Often, when $\langle x_1,\dots,x_n \mid R \rangle$ is a presentation of a group $G$, we identify $G$ with the marked group $(G,(x_1,\dots,x_n))$. For example, the free group $\mathbb{F}_2=\langle x,y \mid \ \rangle$ may be identified with $(\mathbb{F}_2,(x,y))$.

Let $\mathcal{G}_n$ denote the set of marked groups whose marking has cardinality $n$ up to isomorphism of marked groups. If $m>n$, we suppose that $\mathcal{G}_n \subset \mathcal{G}_m$ using the canonical injection given by $(G,(s_1,\dots,s_n)) \mapsto (G,(s_1,\dots,s_n,1,\dots,1))$.

The space of marked groups is $\mathcal{G}= \bigcup\limits_{n \geq 1} \mathcal{G}_n$.

For any marked group $(G,S=(s_1,\dots,s_n))$, we can build the Cayley graph $\Gamma(G,S)$ of $G$ with respect to $S$; here, $\Gamma(G,S)$ is oriented and edge-labelled by $\{1,\dots,n\}$ (each edge of $\Gamma(G,S)$ corresponding to the generator $s_i$ is labelled by $i$). Notice that two marked groups $(G,S)$ and $(H,R)$ are isomorphic (as marked groups) if, and only if, $\Gamma(G,S)$ and $\Gamma(H,R)$ are isomorphic (as oriented and labelled graphs).

Let us make $\mathcal{G}_n$ a metric space by defining a distance $d$ in the following way: if $(G,S),(H,R) \in \mathcal{G}_n$, we set $d((G,S),(H,R))=e^{-R}$ where $R$ is the greatest integer $m$ such that the closed balls $B(\Gamma(G,S),m)$ and $B(\Gamma(H,R),m)$ (centered at the neutral element) are isomorphic (as oriented and labelled graphs). More generally:

Property: $(\mathcal{G}_n,d)$ is a compact ultrametric space. Therefore, with the final topology associated to the inclusions $\mathcal{G}_n \hookrightarrow \mathcal{G}$, the space $\mathcal{G}$ is locally compact.

Another description of the distance is given by the following result:

Property: Let $(G,S),(H,R) \in \mathcal{G}_n$. Then $d((G,S),(H,R))=e^{-r}$ where $r$ is the greatest interger $m$ such that $w(S)=1$ in $G$ if, and only if, $w(R)=1$ in $H$ for all word $w$ of length $m$.

This property is a special case of a more general statement: Building a Cayley graph is equivalent to solving the word problem. Indeed, any word can be read on the Cayley graph and is a relation if, and only if, it is a cycle; conversely, an algorithm can be exhib to build the Cayley graph knowing when a path is a cycle. See for example Groups, Graphs and Trees, J. Meier, Cambridge University Press.

For completeness, recall some facts about Cantor-Bendixson rank in topology:

Definition: Let $X$ be a topological space. We define by transfinite induction:

• $X^{(0)}=X$,
• $X^{(\alpha+1)}$ is the set of non-isolated points of $X^{(\alpha)}$, for any ordinal $\alpha$,
• $X^{(\lambda)} = \bigcap\limits_{\beta<\lambda} X^{(\beta)}$ for any limit ordinal $\lambda$.

If $x \in X$ and if there exists an ordinal $\alpha$ such that $x \in X^{(\alpha)} \backslash X^{(\alpha+1)}$, we say that the Cantor-Bendixson rank $\mathrm{rk}(x)$ of $x$ is $\alpha$. Otherwise, we say that $x$ is a condensation point.

Property: Let $X$ be a locally compact topological space. Then $x \in X$ is a condensation point if, and only if, every neighborhood of $x$ is uncountable. In fact, its cardinality turns out to be at least $2^{\aleph_0}$.

A remarkable fact is that the Cantor-Bendixson rank of a marked group as a point of $\mathcal{G}$ does not depend on the marking: it turns out to be an algebraic property. This can be proved by transfinite induction on Cantor-Bendixson rank using the following lemma:

Lemma: Let $(G_k,S_k=(s_{1,k},\dots,s_{n,k}))$ be a sequence of marked groups converging to some $(G,S=(s_1,\dots,s_n))$. If $\tilde{S}=(\tilde{s}_1,\dots,\tilde{s}_n)$ is another marking of $G$, then there exist markings $\tilde{S}_k$ such that $(G_k,\tilde{S}_k)$ converges to $(G,\tilde{S})$.

Proof. For any $1 \leq i \leq n$, there exists a word $w_i$ such that $\tilde{s}_i=w_i(S)$. Let $\tilde{s}_{i,k}=w_i(S_k)$. We can show that $\tilde{S}_k=(\tilde{s}_{1,k},\dots,\tilde{s}_{n,k})$ works.

Let $w$ be a word such that $w(\tilde{S})=1$ (resp. $w(\tilde{S}) \neq 1$ ) in $G$. But $(G_k,S_k \underset{k \to + \infty}{\longrightarrow} (G,S)$ and $w(\tilde{S})=w(w_1(S),\dots,w_n(S))$ so $w(\tilde{S}_k)=w(w_1(S_k),\dots,w_n(S_k))=1$ (resp. $w(\tilde{S}_k) \neq 1$) for $k$ large enough. $\square$

Consequently, we can talk about the Cantor-Bendixson rank of a group or about condensation groups without ambiguity. To prove theorem 1, the following result will show that in fact it is sufficient to prove that $\mathbb{F}_2$ is a condensation group:

Property: Let $(G,S) \in \mathcal{G}_n$ where $G$ is finitely presented. There exists a neighborhood $V$ of $(G,S)$ in $\mathcal{G}$ such that $H$ is a quotient of $G$ for any $(H,R) \in V$.

Proof. Let $G= \langle S \mid w_1,\dots,w_m \rangle$ be a finite presentation of $G$. If $r= \min\limits_{1 \leq i \leq m} \ \mathrm{lg}(w_i)$, let $V$ be the open ball of radius $e^{-r}$ centered at $(G,S)$. Then if $(H,R) \in V$ the relations $w_i=1$ hold in $H$: Consequently $H$ is a quotient of $G$. $\square$

In order to prove that $\mathbb{F}_2$ is a condensation group, recall some facts about small cancellation groups:

Definition: Let $G= \langle X \mid R \rangle$ be a presentation of group, where $X$ is finite and where each element of $R$ is freely and cyclically reduced; let $S(R)$ denote the closure under taking cyclic permutations and inverses of $R$.

A nontrivial freely reduced word $u$ over $X$ is called a piece if there exist $r_1,r_2 \in S(R)$ such that $u$ is the maximal initial segment common to $r_1$ and $r_2$.

Let $\lambda \in (0,1)$. We say that $G$ satisfies the condition $C'(\lambda)$ if for any $w \in S(R)$ containing a piece $u$, $\mathrm{lg}(u) < \lambda \cdot \mathrm{lg}(w)$. When $\lambda \leq 1/6$, we say that $G$ is a small cancellation group.

Our main result about small cancellation groups is:

Theorem: (Greendlinger) Let $\langle X \mid R \rangle$ be a presentation of group satisfying the condition $C'(\lambda)$ with $\lambda \leq 1/6$. Let $w$ be a nontrivial freely reduced word over $X$ such that $w=1$ in $G$. Then $w$ has a subword $u$ in common with a relator $r \in S(R)$ such that $\mathrm{lg}(u)>(1-3\lambda) \cdot \mathrm{lg}(r)$.

Theorem: $\mathbb{F}_2$ is a condensation group.

Proof. Suppose that there exists a small cancellation group $\langle x,y \mid w_i(x,y), \ i \geq 0 \rangle$ such that $\mathrm{lg}(w_i) \underset{i \to + \infty}{\longrightarrow} + \infty$. For any $I \subset \mathbb{N}$, let $G_I = \langle x,y \mid w_i(x,y) , \ i \in I \rangle$. Notice that $G_I$ is again a small cancellation group.

Let $I \neq J$ be two subsets of $\mathbb{N}$. Without loss of generality, take $j \in J \backslash I$; then $w_j(x,y)=1$ in $G_J$. By contradiction, suppose that $w_j(x,y)=1$ in $G_I$. According to Greendlinger theorem, there exist $k \in I$ and a subword $s$ common to $w_k(x,y)$ and $w_j(x,y)$ such that $\mathrm{lg}(s)> (1-3\lambda) \mathrm{lg}(w_k(x,y))$. But $G$ is a small cancellation group so $\mathrm{lg}(s)< \lambda \cdot \mathrm{lg}(w_k(x,y))$.

Hence $\lambda > 1- 3 \lambda$ ie. $\lambda >1/4$ whereas $\lambda \leq 1/6$ by assumption. Therefore, $w_j(x,y) \neq 1$ in $G_I$, and $G_I$ and $G_J$ are distinct marked groups. Moreover, a group has only countably many markings so the family $\{ G_I \mid I \subset \mathbb{N}\}$ contains uncountably many non-isomorphic small cancellation groups.

Notice that $d(\mathbb{F}_2, G_I)=e^{-R}$ where $R$ is the smallest length of a nontrivial relation of $G_I$. According to Greendlinger theorem, $R \geq \frac{1}{2} \min\limits_{i \in I} \ \mathrm{lg}(w_i(x,y))$. Because $\mathrm{lg}(w_i(x,y)) \underset{i \to + \infty}{\longrightarrow} + \infty$, for any neighborhood $V$ of $\mathbb{F}_2$ in $\mathcal{G}$ there exists $N \geq 1$ such that $\{ G_I \mid I \subset \{N,N+1, \dots\}\} \subset V$; in particular, $V$ is uncountable.

To conclude the proof, it is sufficient to show that $w_i(x,y)=xy^{2^{i-1}}xy^{2^{i-1}+1} \dots xy^{2^i}$ is such a family.

First, $\mathrm{lg}(w_i(x,y))=(2^{i-1}+1)(1+3 \cdot 2^{i-1})$. Then for $i,j \geq 0$, $w_i(x,y)$ and $w_j(x,y)$ have no subword in common; for $i \geq 0$, a piece between two cyclic permutations of $w_i(x,y)$ has the form $b^{2^i-2}ab^{2^i-1}$, whose length is $2^{i+1}-2$; for $j>i$, a piece between two cyclic permutations of $w_i(x,y)$ and $w_j(x,y)$ has the forme $b^{2^i-2}ab^{2^i-1}$, whose length is $2^{i+1}-2$.

Because $\displaystyle \frac{2^{i+1}-2}{(2^{i-1}+2)(1+3 \cdot 2^{i-1})} \leq \frac{2^{i+1}}{2^{i-1} \cdot 3 \cdot 2^{i-1}}= \frac{1}{3 \cdot 2^{i-3}} \leq \frac{1}{6}$ if $i \geq 4$, with this choice $G$ is a small cancellation group and clearly $\mathrm{lg}(w_i(x,y)) \underset{i \to + \infty}{\longrightarrow} + \infty$. $\square$

Noticing that the set of condensation groups is closed in $\mathcal{G}$ (see Cantor-Bendixson theorem) we deduce that:

Corollary: A limit group is a condensation group.

Proof of theorem 1. Let $V$ be a neighborhood of $\mathbb{F}_2= \langle x,y \mid \ \rangle$ in $\mathcal{G}$ such that $H$ is a quotient of $\mathbb{F}_2$ if $(H,R) \in V$; in particular, $H$ is a two-generator group. Because $\mathbb{F}_2$ is a condensation group, $V$ is uncountable. Because any group has only countably many markings, we deduce that $V$ contains uncountably many non-isomorphic two-generator groups. $\square$