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

Advertisements