In this note, we deal with a construction on finite graphs due to Stallings, exposed in his article Topology of finite graphs, admitting some exciting recent generalizations to some graphs of groups (see Henry Wilton’s articles Elementary free groups are subgroup separable and Hall’s theorem for limit groups) or cube complexes (see Haglund and Wise’s article A combination theorem for special cube complexes). A next note will be dedicated to Wise’s construction for Salvetti complexes, and more generally for special cube complexes, but our main result here is:
Main Theorem: Finitely generated groups are subgroup separable.
Definition: Let be a group and be a subgroup. We say that is separable if, for every , there exists a finite-index subgroup containing and such that .
Moreover, if the trivial subgroup is separable, is said residually finite or RF; if every subgroups of are separable, is said ERF (Extended Residually Finite); if every finitely-generated subgroups of are separable, is said subgroupe separable or LERF (Locally Extended Residually Finite).
Theses definitions have topological interpretations:
Definition: Let be a group. Defining the set of finite-index subgroups as a fundamental system of neighborhoods of the identity element, we get the profinite topology. Clearly, endowed with its profinite topology, is a topological group. Moreover, a morphism between two groups endowed with their profinite topologies is automatically continuous.
Nota Bene: Because a finite-index subgroup always contains a normal finite-index subgroup (namely ), finite-index subgroups in the definitions of profinite topology and separable subgroup may be supposed normal.
Lemma: Let be a group and be a subgroup. Then is separable if and only if it is closed for the profinite topology.
Proof. Suppose that is separable and let . By hypothesis, there exists a finite-index subgroup containing but not containing . Then is an open set disjoint from : if there exists , then there exists such that , hence , a contradiction. Therefore, is closed.
Conversely, suppose that is closed and let . By hypothesis, there exists an open subset, say for some normal finite-index subgroup and , containing and disjoint from . Because we chose normal, is a subgroup of ; clearly, it contains and it is a finite-index subgroup of . Furthermore, it does not contain : indeed, implies , hence , a contradiction. Therefore, is separable.
Corollary: A group is residually finite if and only if its profinite topology is Hausdorff.
Genarally, it is not easy to prove that a given subgroup is separable. A useful tool for that is:
Definition: Let be a group and be a subgroup. We say that is a retract if there exists a morphism (a retraction) such that . We say that is a virtual retract if it is a retract in some finite-index subgroup of .
Lemma: Let be a residually finite group and be a subgroup. If is a virtual retract then it is separable.
Sketch of proof. It is sufficient to notice that the image of a (topological) retraction in a Hausdorff topological space is closed. So let be a Hausdorff space and be a retraction. Here, we only work with finitely-generated groups so that profinite topologies are second-countable, therefore it is sufficient to show that is sequentially closed; for the more general case, the same argument works with ultrafilters.
Let be a sequence converging to some . Then, converges to . Because is Hausdorff, we deduce that . Therefore, is sequentially closed.
The construction introduced by Stallings, is the following (recall that an immersion is a locally injective simplicial map):
Theorem: Let be a finite bouquet of circles, be a finite graph and be an immersion. Just by adding edges to , it is possible to construct a covering and a retraction . Moreover, we get a commutative diagram:
We say that is the canonical completion and the canonical retraction.
Now, let us see how to conclude thanks to Stallings’ construction:
Proof of the main theorem: Let be a finitely generated free groups. We first prove that is residually finite in order to apply the previous lemma.
Let be a bouquet of circles satisfying ; let denote its universal covering. Now, thinking of as a non trivial loop in , let be a finite subgraph containing a lift of . Then, the fundamental group of the canonical completion is a finite-index subgroup of not containing . We just proved that is residually finite.
Now let be a finitely generated subgroup of . Because is residually finite, it is sufficient to prove that is a virtual retract to conclude that is separable.
Again, let us see as the fundamental group of a bouquet of circles and let denote the covering associated to the subgroup . If is a maximal subtree, because is finitely generated, has only finitely many edges; let be a ball containing all theses edges. Then the canonical retraction gives a retraction
so that be a retract in , and therefore a virtual rectract in .
During the proof above, the construction of an immersion with can be made directly. For example, if and :
Notice that the main theorem cannot be extended to the separability of all subgroups, ie. every non-abelian (finitely-generated) free group is not ERF: Suppose by contradiction that there exists an ERF non-abelian free group. Because any quotient of an ERF group is RF, we deduce that any two-generator group is RF, and a fortiori Hopfian. However, we already proved in a previous note that the Baumslag-Solitar group is not Hopfian.
We now turn on the proof of Stallings’ construction. The figure below illustrates an example; the reader is encouraged to keep in mind this particular case in order to visualize how the argument is simple.
Proof of Stallings’ construction: Because is an immersion, for any vertex and any label , there is at most one edge labelled starting from or ending at . We want to add edges to so that exactly one edge of any label start from and end at any vertex; such a graph will be a covering of .
Step 1: Let be an edge labelled and let denote the path of oriented edges labelled containing of maximal length (such a path is uniquely defined). Two cases happens: either is a loop, and there is nothing to do, or is not a loop, and we add an edge labelled between the extremities of .
Doing the same thing for every edges of and for every label, we get a graph satisfying: for any vertex and any label , either there is no edge labelled adjacent to , or there are two such edges, one starting from and the other ending at .
Step 2: Let be a vertex and a label. If there is no edge labelled adjacent to , then add a loop labelled based at . Keep going for every every vertex and for every label.
Let denote our new graph. It is a covering extending by construction the immersion , so we have the commutative diagram.
To conclude the proof, let us define a retraction . Of course, we only have to define , where is an edge added to . Two cases happen: either is a loop based at a vertex , and we define , or completes a path to a loop, and we send on (first choose a subdivision of ) so that .
Nota Bene: In general, the retraction defined above is not simplicial.
See also Graphs and Free Groups.