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Last time we proved that if is a graph of groups, then acts on a tree , namely the underlying graph of the universal cover of . Lemmas 16, 17, and 18 describe the vertex and edge sets of and the action of . The following is immediate.
Theorem 14 (Serre): Every graph of groups is developable.
Corollary: If is a graph of groups and , then the map is injective.
The tree is called the Bass-Serre Tree of . We will usually equip with a length metric in which each edge has length 1. The approach we’ve taken is due to Scott-Wall. Here’s a sample application.
Lemma 19: Let . If is torsion, then is conjugate into or .
Proof. Let be the Bass-Serre Tree. Fix a vertex amd consider . This is a finite set.
Exercise 19: There is an that minimizes (when is torsion).
Such an is fixed by . Therefore, stabilizes a vertex. The stabilizers of the vertices are precisely the conjugates of and .
Our next goal is to understand the elements of — which are trivial and which not? We’ll start with a presentation. If is a tree, then can be thought of as a sequence of amalgamated free products. If is a rose, then can be thought of as a sequence of HNN-extensions. In general, fix a maximal tree . This determines a way of decomposing as a sequence of amalgamated products followed by a sequence of HNN-extensions. This process is sometimes called “Excision.” It follows that has a “presentation” like this:
We can write any in the following form:
where , is a loop in , and where is the terminus of and the origin of . A priori, we just know that . Suppose, say, that doesn’t originate from . Then insert stable letters corresponding to edges in that form a path from to the origin of . Now repeat.
Ping-Pong Lemma
Question. Let G be a group and . When is ?
Ping-Pong Lemma: Let G be a group acting on a set X and . Assume:
- a and b have infinite orders.
- There exist such that and , for all .
Then .
Proof. Consider such that and . Choose a reduced word w for a nontrivial element in , i.e., either or and .
Case 1: . Then and , so and , and so on until , so . Therefore .
Case 2: . Then, by Case 1, , so . Therefore .
Case 3: . (similar to above)
Case 4: . (similar to above)
Free Groups Are Linear
Theorem 3. is linear.
Proof. acts on by linear transformations. Let and . Then and . Let and . Then for all , so by the Ping-Pong Lemma.
Corollary 1. Finitely generated free groups are linear, hence residually finite.
Proof. The case of is obvious; otherwise, this follows from as proved in Exercise 1.
Separability
Definition. Let G be a group. The profinite topology on G is the coarsest topology such that every homomorphism from G to a finite group (equipped with the discrete topology) is continuous.
Definition. A subgroup H is separable in G if H is closed in the profinite topology of G.
Exercise 3. Let G be a group. is separable if and only if for all , there exists a homomorphism to a finite group such that . Note that if X = {1}, this is equivalent to: G is RF if and only if {1} is separable.
Hint. For the “if” direction, let and consider . For the other direction, use the definition of subbase and that .
Definition. Let G be a group.
- G is Extended RF (ERF) if any subgroup of G is separable.
- G is Locally ERF (LERF, subgroup separable) if any finitely generated subgroup is separable.
Lemma 3. Let G be a group. A subgroup H of G is separable if and only if for all , there exists a finite-index subgroup such that and .
Proof. In the “only if” direction, by the previous exercise, for all , there exists a homomorphism to a finite group such that . Then . Conversely, let . By hypothesis, there exists a finite-index subgroup such that and . Let . Note that this is a finite number of intersections (, to be precise). There exists a finite quotient . Then . Therefore, , i.e., , and the lemma follows by the previous exercise.
Scott’s Criterion (1978). Let X be a Hausdorff topological space and . Let be a covering and . Then H is separable in G if and only if for any compact , there exists and intermediate finite-sheeted cover such that embeds into .
Exercise 4. Let be a separable subgroup.
- If , then is separable in G’.
- If has finite index, then is separable in G’.
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