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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’.
Residual finiteness
Let be a group and with . Then we call residually finite (hereafter RF) if there exists a subgroup of finite index such that . In other words, for every nontrivial element of , there exists a finite index subgroup that does not contain that particular element.
Example. Finite groups are RF, since the trivial subgroup has finite index and does not contain any of the nontrivial elements of .
Metaquestion. How general is the class of RF groups? In particular, which finitely generated/finitely presented groups are RF?
Remarks. Assume that is finitely generated.
(i) The definition can be strengthened so that we may assume that our finite index subgroup is normal in . Indeed, if is finitely generated, then there are only finitely many subgroups of a given fixed index . (See the exercise from the second lecture.) If , and is a subgroup of finite index that does not contain , then
is a subgroup of finite index in which excludes , since and have the same index in for all , so this intersection is really the intersection of finitely many subgroups of finite index. Thus also has finite index in .
(ii) Equivalently, is RF if and only if for each not the identity, there exists a homomorphism , where is a finite group, for which is not the identity in . Indeed, if is RF and is normal subgroup provided by (i), then does not die under the natural homomorphism from to . Conversely, given such a homomorphism, the kernel of is a subgroup of finite index that does not contain .
(iii) Also, is RF if and only if
That is, the intersection of all the subgroups of finite index in is the trivial subgroup. Were some nonidentity element to be contained in this intersection, then it would be contained in each subgroup of finite index in , so this element prevents from being RF. Conversely, if this intersection is trivial, each nonidentity element of must be excluded from some finite index subgroup, so is RF.
(iv) If is RF and , then there exists a finite index subgroup with for all . Here, just take the intersection of the finite index subgroup associated with each . This is again a finite index subgroup of .
Lemma 2: Let be a finitely generated group.
(i) If is RF and is a subgroup, then is RF.
(ii) If is RF and with finite index in , then is RF.
That is, RF passes to all subgroups and also to supergroups of finite index.
Proof. For (i), choose . Considered as an element of , there exists a homomorphism to a finite group for which is not the identity. The restriction of to is also a homomorphism to a finite group for which the image of is nontrivial. The kernel of this restricted homomorphism is a subgroup of finite index in that does not contain . (Note that we did not assume that is finitely generated.)
For (ii), choose . If , then is a finite index subgroup not containing , and we are done. If , then there exists a finite index subgroup that does not contain . However, is also a finite index subgroup of , so we are done.
Topological reformulation of RF
Now, we would like to connect RF with a topological property of a space with fundamental group . Let be a compact manifold with universal covering and . Accordingly, we assume throughout that is finitely generated. (We will see later that the manifold condition can be relaxed.)
Theorem 2: The group is RF if and only if the following condition holds: for every compact subset there exists a finite sheeted covering for which embeds homeomorphically in .
Proof. Assume the topological condition holds and choose any not the identity. This corresponds to a loop (based at a point dependent only upon our choice of universal covering) in which we also denote by . To this loop, there also is a corresponding lift to a connected arc in with distinct endpoints and .
Let be the compact set . Then, there exists a finite sheeted covering for which and are distinct points of . By the lifting property of covering spaces, this implies that if is the subgroup of finite index corresponding to the covering , then . Therefore is RF.
Conversely, suppose that is RF and choose any subset . Then, since acts freely and properly discontinuous on , the set of those (not the identity) for which intersects nontrivially is finite. Now, choose of finite index containing none of the elements of . If is the corresponding finite sheeted covering, we have that for all . That is, embeds homeomorphically in .
Remark. We never really used here that was a manifold, only that its fundamental group acted properly discontinuous on the universal covering. (The action need not be free either, since this would only add another finite number of elements to our set .) Thus, it suffices to assume that is Hausdorff and locally compact.
Examples.
(1) Finitely generated abelian groups are RF. (Exercise.)
(2) Selberg’s Lemma (Malcev-Selberg): If is a finitely generated linear group, that is, for some , then is RF.
Proof. We begin with the case . Since RF passes to arbitrary subgroups, it suffices to prove that is RF.
Choose any . This means that the matrix has some nonzero entry, say . Since is an integer, we can find some large prime that does not divide . Now, consider the homomorphism
given by reducing the entries of a given matrix modulo . This is a homomorphism because matrix multiplication is linear in the entries. (Exercise – make this precise.) Since is a finite field, its general linear group is a finite group, i.e. the kernel of is a finite index subgroup of , and it does not contain by construction. Thus is RF, and thus any subgroup thereof is also RF.
More generally, if is finitely generated, we can arrange that , where is the subring generated by the entries of a finite collection of generators for . In particular, this is an integral domain, since it is a finitely generated subring of . Thus, given a nonzero element of as above, we can find a prime ideal of that does not divide , i.e. .
Now, is a finite integral domain, that is, a finite field, and we can build a reduction homomorphism analogous to the situation over the integers. The the image of the reduction homomorphism is the general linear group of a finite field, so it is a finite group. We now proceed exactly as above.
This theorem is often referred to as Selberg’s Lemma, even though Malcev supposedly proved it first.
(3) Free groups of finite rank are linear, so they are RF. Notice that we have seen, in the exercise from the first lecture, that is a subgroup of for all , so it suffices to prove that is linear. One such representation the subgroup of generated by the two matrices
There are several ways to prove that this group is free of rank two. First, one use a bit of hyperbolic geometry and the action of on the hyperbolic plane to prove that this group is the fundamental group of a hyperbolic manifold that is deformation equivalent to the rose with two petals. Also, one can use the so-called Ping-Pong Lemma, which says that a finite collection of homeomorphisms of a space that satisfy a certain set of conditions necessarily generate a free subgroup of the homeomorphism group.
We begin with a lemma that among other things demonstrates that most free groups are non-trivial
Lemma 1: If are sets then the map , induced by inclusion has a left inverse. Consequently, is injective.
Proof: Let and be the roses with petals indexed by and respectively. In particular we have that and . If we view as a subspace of in the obvious way then there exists such that (One such map is obtained by crushing the petals in ). This map induces a function and this map can easily be seen to be the desired left inverse. QED
The next thing that we would like to do is examine exactly what do elements of free groups “look like.” Informally, elements look like words in the elements of where only the obvious cancellations by inverses are allowed. More formally, we will identify the elements of with their images under the map (This is not the same from Lemma 1). Next we consider products of the form , where and . Next, we prove a theorem that formalizes the previous intuition about free groups
Normal Form Theorem: Given a free group the following 2 statements are true
- Every element of is of the form .
- If then is either the empty word or there exists such that and
Proof: We will prove the theorem when , but the proof generalizes to sets of all cardinality. Let be the rose with 2 petals and wedge point and so we have that . Next, let be the universal cover of based at . Because acts freely on and has only one vertex we see that the vertices of are in bijection with the elements of . We next observe 2 facts. First, from lemma 1 we see that since is simply connected that it must be a tree. Second, since each vertex of looks locally like we see that every vertex of is 4-valent. From covering space theory we know that if and is a vertex of then the lift of to that starts at ends at .
We proceed by induction on , which is the minimum number of edges between and . Suppose that . Since is a tree with all 4-valent vertices we see that there exists such that and . Next, observe that there exists such that when is lifted at ends at . Therefore we see that and since the group action is free this implies that which proves 1.
For the second proposition, suppose that and that the second proposition is not true. From covering space theory we know that the lift of at is a loop based at . First, since is not the empty word we see that it will not lift to the constant path. Finally, since contains no obvious cancellations we see that must contain a loop based at that is not the constant loop, which contradicts the fact that is a tree. Thus 2 must hold. QED
As a consequence of the previous theorem we see that it is possible to solve the word problem of whether it is algorithmically possible to decide if a group element is the identity.
Definition: A group is finitely generated if and only if there exists a surjection from to for some finite .
Exercise 2:
- Show that if then has finitely many subgroups of index .
- Deduce that the same holds for any finitely generated group .
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