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Theorem 12 (Gromov): Let be torsion-free -hyperbolic group. If such that , then for all sufficiently large , .
Remark: The torsion-free hypothesis is not necessary, but it allows us to avoid some technicalities. For instance, it is a non-obvious fact that an infinite hyperbolic group contains a copy of .
For the rest of this lecture will be a torsion-free -hyperbolic group, where are primitive (i.e. not proper powers).
Recall that for torsion-free -hyperbolic, primitive implies that .
If and do not commute we can show there is some point on arbitrarily far from .
Hence we have the following lemma.
Lemma 13:
If and do not commute there is some point on arbitrarily far from .
Proof: Suppose not. That means such that such that . So is in . But the Cayley graph is locally finite so has finitely many elements. By the Pigeonhole Principle such that for some . Then . But then . .
For a moment view and as the horizontal and vertical geodesics in . For two points on and on , we can argue that the geodesic between them curves toward the origin.
And so we have Lemma 14.
Lemma 14: There exists such that , .
Proof:
Recall that by is a quasi-isometric embedding. So by Theorem 6, and
By Lemma 13 choose such that
. Choose such that . Now, must be -close to so for some point on the geodesic between and , . Then .
For a subgroup , one can choose a closest point projection which is -equivariant. (Write . Choose where and are close and declare to be -equivariant.) is typically not a group homomorphism.
We’re interested in and .
In , there is some such that either or .
Lemma 15: such that , or .
Proof:
Let . WLOG, is -close to and since is the closest point to (in particular compared to ). So . .
Now we can prove the theorem.
Proof of Theorem 12:
The idea is to use the Ping-Pong Lemma on the Cayley graph.
Let and let , where is provided by Lemma 15. For all we have and likewise for all we have . In particular, .
Let . By -equivariance,
for any . In particular,
by the triangle inequality. Similarly,
for all and all . Because and are quasi-isometrically embedded, it follows that and for .
Therefore, by the Ping-Pong Lemma .
Fact: There exists a finitely generated non-Hopf group. (An example is the Baumslag-Solitar group , although we cannot prove it yet.) So, by Lemma 5, there is a finitely generated non-residually finite group. Thus, free groups are not ERF: if is a finitely generated non-residually finite group, then Lemma 4 implies that the kernel of a surjection is not separable in . However, finitely generated subgroups of free groups are separable:
Marshall Hall’s Theorem (1949): is LERF.
This proof is associated with Stallings.
Proof: As usual, let where is a rose. Let be a covering map with finitely generated. Let be compact. We need to embed in an intermediate finite-sheeted covering.
Enlarging if necessary, we may assume that is connected and that . Note that we have . By Theorem 5 (see below), the immersion extends to a covering . Then . So lifts to a map .
The main tool in the proof above is this:
Theorem 5: The immersion can be completed to a finite-sheeted covering into which embeds:
Proof: Color and orient the edges of . Any combinatorial map of graphs corresponds uniquely to a coloring and orientation on the edges of . A combinatorial map is an immersion if and only if at every vertex of , we see each color arriving at most once and leaving at most once. Likewise, it’s a covering map if and only if at each vertex, we see each color arriving exactly once and leaving exactly once.
Let be the number of vertices of . For each color , let be the number of edges of colored . Then there are vertices of missing “arriving” edges colored , and there are vertices of missing “leaving” edges colored . Choose any bijection between these two sets and use this to glue in edges colored . When this is done for all colors, the resulting map is clearly a covering.
Note that the proof in fact gives us more. For instance:
Exercise 6: If is a finitely generated subgroup of , then is a free factor of a finite-index subgroup of .
Exercise 7 (Greenberg’s Theorem): If and is finitely generated, then is of finite index in .
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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