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’.
3 comments
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1 February 2009 at 2.57 pm
Henry Wilton
Thanks Jason (and Sam!). I’ve just added some numbering, as in the comment on the last post.
1 February 2009 at 5.39 pm
Henry Wilton
To be absolutely precise, in Scott’s Criterion we should really assume that is locally path-connected and semi-locally simply connected, in order to guarantee that has a covering corresponding to .
2 February 2009 at 8.43 am
Sam Kim
Oh, right. I should’ve mentioned that $X$ has a universal cover.