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 $G$ be a group and $H$ be a subgroup. We say that $H$ is separable if, for every $g \notin H$, there exists a finite-index subgroup $K$ containing $H$ and such that $g \notin K$.

Moreover, if the trivial subgroup is separable, $G$ is said residually finite or RF; if every subgroups of $G$ are separable, $G$ is said ERF (Extended Residually Finite); if every finitely-generated subgroups of $G$ are separable, $G$ is said subgroupe separable or LERF (Locally Extended Residually Finite).

Theses definitions have topological interpretations:

Definition: Let $G$ 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, $G$ 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 $H$ always contains a normal finite-index subgroup (namely $\bigcap\limits_{g \in G} gHg^{-1}$), finite-index subgroups in the definitions of profinite topology and separable subgroup may be supposed normal.

Lemma: Let $G$ be a group and $H$ be a subgroup. Then $H$ is separable if and only if it is closed for the profinite topology.

Proof. Suppose that $H$ is separable and let $g \notin H$. By hypothesis, there exists a finite-index subgroup $K$ containing $H$ but not containing $g$. Then $gK$ is an open set disjoint from $H$: if there exists $x \in H \cap gK$, then there exists $k \in K$ such that $x=gk$, hence $g=xk^{-1} \in K$, a contradiction. Therefore, $H$ is closed.

Conversely, suppose that $H$ is closed and let $g \notin H$. By hypothesis, there exists an open subset, say $pK$ for some normal finite-index subgroup $G$ and $p \in G$, containing $g$ and disjoint from $H$. Because we chose $K$ normal, $HK$ is a subgroup of $G$; clearly, it contains $H$ and it is a finite-index subgroup of $G$. Furthermore, it does not contain $g$: indeed, $g \in HK$ implies $pK \cap HK \neq \emptyset$, hence $pK \cap H \neq \emptyset$, a contradiction. Therefore, $H$ is separable. $\square$

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 $G$ be a group and $H$ be a subgroup. We say that $H$ is a retract if there exists a morphism (a retraction) $r : G \to H$ such that $r_{|H}= \mathrm{Id}_H$. We say that $H$ is a virtual retract if it is a retract in some finite-index subgroup of $G$.

Lemma: Let $G$ be a residually finite group and $H$ be a subgroup. If $H$ 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 $X$ be a Hausdorff space and $r : X \to Y \subset X$ 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 $Y$ is sequentially closed; for the more general case, the same argument works with ultrafilters.

Let $(x_n) \subset Y$ be a sequence converging to some $\in X$. Then, $(r(x_n))=(x_n)_n$ converges to $r(x)$. Because $X$ is Hausdorff, we deduce that $x=r(x) \in Y$. Therefore, $Y$ is sequentially closed. $\square$

The construction introduced by Stallings, is the following (recall that an immersion is a locally injective simplicial map):

Theorem: Let $B$ be a finite bouquet of circles, $X$ be a finite graph and $X \to B$ be an immersion. Just by adding edges to $X$, it is possible to construct a covering $C(X,B) \to B$ and a retraction $r : C(X,B) \to X$. Moreover, we get a commutative diagram:

We say that $C(X,B)$ is the canonical completion and $r$ the canonical retraction.

Now, let us see  how to conclude thanks to Stallings’ construction:

Proof of the main theorem: Let $F$ be a finitely generated free groups. We first prove that $F$ is residually finite in order to apply the previous lemma.

Let $B$ be a bouquet of circles satisfying $\pi_1(B) \simeq F$; let $\widehat{B}$ denote its universal covering. Now, thinking of $g \in F \backslash \{1 \}$ as a non trivial loop in $B$, let $X \subset \widehat{B}$ be a finite subgraph containing a lift of $g$. Then, the fundamental group of the canonical completion $C(X,B)$ is a finite-index subgroup of $\pi_1(B) \simeq F$ not containing $g$. We just proved that $F$ is residually finite.

Now let $H$ be a finitely generated subgroup of $F$. Because $F$ is residually finite, it is sufficient to prove that $H$ is a virtual retract to conclude that $H$ is separable.

Again, let us see $F$ as the fundamental group of a bouquet of circles $B$ and let $Y \to B$ denote the covering associated to the subgroup $H$. If $T \subset Y$ is a maximal subtree, because $H$ is finitely generated, $Y \backslash T$ has only finitely many edges; let $X \subset Y$ be a ball containing all theses edges. Then the canonical retraction $r: C(X,B) \to X$ gives a retraction

$r_* : K:= \pi_1(C(X,B)) \to H \simeq \pi_1(X)$

so that $H$ be a retract in $K$, and therefore a virtual rectract in $F$. $\square$

During the proof above, the construction of an immersion $X \to B$ with $\pi_1(X) \simeq H$ can be made directly. For example, if $F = \langle a,b \mid \ \rangle$ and $H = \langle ab^{-2}a^{-1}, abab^2 \rangle$:

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 $BS(1,2)$ 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 $X \to B$ is an immersion, for any vertex $v \in V$ and any label $x_i$, there is at most one edge labelled $x_i$ starting from $v$ or ending at $v$. We want to add edges to $X$ so that exactly one edge of any label start from and end at any vertex; such a graph will be a covering of $B$.

Step 1: Let $e \in X$ be an edge labelled $x_i$ and let $p = (e_0, \dots, e_r)$ denote the path of oriented edges labelled $x_i$ containing $e$ of maximal length (such a path is uniquely defined). Two cases happens: either $p$ is a loop, and there is nothing to do, or $p$ is not a loop, and we add an edge labelled $x_i$ between the extremities of $p$.

Doing the same thing for every edges of $X$ and for every label, we get a graph $X'$ satisfying: for any vertex $v \in X'$ and any label $x_i$, either there is no edge labelled $x_i$ adjacent to $x_i$, or there are two such edges, one starting from $v$ and the other ending at $v$.

Step 2: Let $v \in X$ be a vertex and $x_i$ a label. If there is no edge labelled $x_i$ adjacent to $v$, then add a loop labelled $x_i$ based at $v$. Keep going for every every vertex and for every label.

Let $C(X,B)$ denote our new graph. It is a covering $C(X,B) \to B$ extending by construction the immersion $X \to B$, so we have the commutative diagram.

To conclude the proof, let us define a retraction $r : C(X,B) \to X$. Of course, we only have to define $r(e)$, where $e$ is an edge added to $X$. Two cases happen: either $e$ is a loop based at a vertex $v \in X$, and we define $r(e)=v$, or $e$ completes a path $p\subset X$ to a loop, and we send $e$ on $p$ (first choose a subdivision of $e$) so that $r(e)=p$. $\square$

Nota Bene: In general, the retraction defined above is not simplicial.