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 (and more generally any limit group) is a condensation group using small cancellation groups.

For precise references, we may mention:

- About marked groups:
*Limit groups as limits of free groups*by C. Champetier and V. Guirardel. - About Cantor-Bendixson rank: Our previous note.
- About small cancellation groups:
*Combinatorial Group Theory*, R. Lyndon and P. Schupp, Springer. - About a proof of theorem 1 using HNN extensions and products with amalgation, see our note Amalgamated products and HNN extensions (I): A theorem of B.H. Neumann.

**Definition:** A marked group is a finitely generated group endowed with a finite generator set . A morphism between two marked groups and is a morphism sending to for all .

Often, when is a presentation of a group , we identify with the marked group . For example, the free group may be identified with .

Let denote the set of marked groups whose marking has cardinality up to isomorphism of marked groups. If , we suppose that using the canonical injection given by .

The space of marked groups is .

For any marked group , we can build the Cayley graph of with respect to ; here, is oriented and edge-labelled by (each edge of corresponding to the generator is labelled by ). Notice that two marked groups and are isomorphic (as marked groups) if, and only if, and are isomorphic (as oriented and labelled graphs).

Let us make a metric space by defining a distance in the following way: if , we set where is the greatest integer such that the closed balls and (centered at the neutral element) are isomorphic (as oriented and labelled graphs). More generally:

**Property:** is a compact ultrametric space. Therefore, with the final topology associated to the inclusions , the space is locally compact.

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

**Property:** Let . Then where is the greatest interger such that in if, and only if, in for all word of length .

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 be a topological space. We define by transfinite induction:

- ,
- is the set of non-isolated points of , for any ordinal ,
- for any limit ordinal .

If and if there exists an ordinal such that , we say that the Cantor-Bendixson rank of is . Otherwise, we say that is a condensation point.

**Property:** Let be a locally compact topological space. Then is a condensation point if, and only if, every neighborhood of is uncountable. In fact, its cardinality turns out to be at least .

A remarkable fact is that the Cantor-Bendixson rank of a marked group as a point of 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 be a sequence of marked groups converging to some . If is another marking of , then there exist markings such that converges to .

**Proof.** For any , there exists a word such that . Let . We can show that works.

Let be a word such that (resp. ) in . But and so (resp. ) for large enough.

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 is a condensation group:

**Property:** Let where is finitely presented. There exists a neighborhood of in such that is a quotient of for any .

**Proof.** Let be a finite presentation of . If , let be the open ball of radius centered at . Then if the relations hold in : Consequently is a quotient of .

In order to prove that is a condensation group, recall some facts about small cancellation groups:

**Definition:** Let be a presentation of group, where is finite and where each element of is freely and cyclically reduced; let denote the closure under taking cyclic permutations and inverses of .

A nontrivial freely reduced word over is called a piece if there exist such that is the maximal initial segment common to and .

Let . We say that satisfies the condition if for any containing a piece , . When , we say that is a small cancellation group.

Our main result about small cancellation groups is:

**Theorem:** *(Greendlinger)* Let be a presentation of group satisfying the condition with . Let be a nontrivial freely reduced word over such that in . Then has a subword in common with a relator such that .

**Theorem:** is a condensation group.

**Proof.** Suppose that there exists a small cancellation group such that . For any , let . Notice that is again a small cancellation group.

Let be two subsets of . Without loss of generality, take ; then in . By contradiction, suppose that in . According to Greendlinger theorem, there exist and a subword common to and such that . But is a small cancellation group so .

Hence ie. whereas by assumption. Therefore, in , and and are distinct marked groups. Moreover, a group has only countably many markings so the family contains uncountably many non-isomorphic small cancellation groups.

Notice that where is the smallest length of a nontrivial relation of . According to Greendlinger theorem, . Because , for any neighborhood of in there exists such that ; in particular, is uncountable.

To conclude the proof, it is sufficient to show that is such a family.

First, . Then for , and have no subword in common; for , a piece between two cyclic permutations of has the form , whose length is ; for , a piece between two cyclic permutations of and has the forme , whose length is .

Because if , with this choice is a small cancellation group and clearly .

Noticing that the set of condensation groups is closed in (see *Cantor-Bendixson theorem*) we deduce that:

**Corollary:** A limit group is a condensation group.

**Proof of theorem 1.** Let be a neighborhood of in such that is a quotient of if ; in particular, is a two-generator group. Because is a condensation group, is uncountable. Because any group has only countably many markings, we deduce that contains uncountably many non-isomorphic two-generator groups.