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Our goal is to understand the structure of subgroups of graphs of groups. This will enable us to prove things like: is LERF.
Exercise 22. A group is coherent if every fg subgroup is fp. Prove that of a finite graph of groups with coherent vertex groups and cyclic edge group is coherent.
Recall, that if is fg then H has an induced graph of groups structure. where T is the Bass-Serre tree of .
Topologically, defines a covering space , which inherits a graph of spaces structure with underlying graph .
is a finite connected subgraph and the image of in is a connected subcomplex such that . has the structure of a graph of space with underlying graph
There is an explicit algebraic description of , which follows immediately from Lemmas 16,17,18.
Lemma 23. (a) The vertices of are in bijection with .
The vertex is labelled by: , well-defined up to conjugation in .
(b) The edges of are in bijection with . The vertex is labelled with .
(c) The edges of adjoining the vertex corresponding to are in bijection with .
Now we will try to understand this topologically. In particular, we will understand the edge maps of a covering space of a graph of spaces.
Let be a continuous map and a covering map. If makes the diagram commutes, then is a lift of .
Lemma 24. Fix basepoint , , and mapping to . There is a lift with if and only if . Furthermore, if this lift exists it is unique.
It may be impossible to lift at . But it is possible if we pass to a covering space. Eg., if is the universal cover. Intuitively, an elevation is a minimal lift.
Definition: Let be as above. A based connected covering space together with a based map such that
commutes is an elevation (of at ) if whenever the diagram
commutes and is a covering map of degree larger than 1 then the composition does not lift to at .
The unbased covering map or equivalently the conjugacy class of the subgroup is called the degree of the elevation .
Recall from last time:
Theorem 16: Every is semisimple. If is loxodromic, then is isometric to and acts on as translation by .
In particular, for every , is a -invariant subtree of .
Exercise 21: If and , then is loxodromic, , and .
(Hint: It is enough to construct an axis for .)
Lemma 20 (Helly’s Theorem for Trees): If are closed subtrees and for every , then .
Proof. Case : Let where . Let be the center of the triangle with vertices . Then .
General case: Let for . Then by the case. Now, by induction,
Corollary: If a finitely generated group acts on a tree with no global fixed point, then there is a (unique) -invariant subtree that is minimal with respect to inclusion, among all -invariant subtrees. Furthermore, is countable.
Proof. Let
For any -invariant , it is clear that . Therefore is minimal. Let be a finite generating set for . Suppose that every element of is elliptical, so for all , is elliptical. Then for all ,
Therefore by Lemma 20, a contradiction.
Suppose that are loxodromic and . By Exercise 21, intersects and non-trivially. Thus, is connected.
It remains to show that is -invariant. Let . For any ,
This implies that , and
So we conclude that is -invariant.
Definition: If is connected, then the graph of groups carried by is the graph of groups with underlying graph such that the vertex is labelled by , and the edge is labelled (with obvious edge maps). There is a natural map , where and . This map is an injection by the Normal Form Theorem.
Lemma 21: If is countable and is finitely generated, then there is a finite subgraph such that (where is the graph of groups carried by ).
Proof. Let be an exhaustion of by finite connected subgraphs. Let denote the graph of groups carried by and set . Since is finitely generated, there is an such that contains each generator of , and since , we conclude that .
Lemma 22: If is countable and is finitely generated then there is a finite minimal (wrt inclusion) subgraph that carries .
Proof. Let be the Bass-Serre tree of . Let . It’s an easy exercise to check that this is as required.
We will refer to (and , the graph of groups carried by ) as the core of .
Theorem 17: If is finitely generated, then decomposes as of a finite graph of groups . The vertex groups of are conjugate into the vertex groups of . The edge groups are likewise.
Proof. Let be the Bass-Serre tree of , and set .
In this lecture, we will use the Normal Form Theorem to understand and construct groups. In particular, we will construct a group that is not residually finite (RF). Note that in the case when is an HNN-extension, then the Normal Formal Theorem is called Britton’s Lemma.
Example: The (p,q)-Baumslag-Solitar group has the following presentation:
For example .
Consider . We will show that this group is non-Hopfian (i.e., that there exists a homomorphism that is a surjection with nontrivial kernel), hence is not RF. Consider defined by and . First, we show this is a homomorphism, then a surjection, then that it has nontrivial kernel.
Homomorphism: Hence, the map is a homomorphism.
Epimorphism: To see that is a surjection, we check that . This is obvious for . Further and so . But then .
Nontrivial kernel: Consider . Then . So we just need to show that is nontrivial. We use Britton’s Lemma (Normal Form Theorem). What Britton’s Lemma says is that there will be that can be removed from our word using the relation if it is trivial. But this can’t be done with , so we must have as required.
Remark: can never embed in a word-hyperbolic group by a previous Lemma.
Exercise 20: Show that , where is the dyadic rationals and we have acting via multiplication by 2. One can deduce that is linear, hence RF by Selberg’s Lemma.
Theorem 15 (Higman-Neumann-Neumann): Every countable group embeds in a 3-generator group.
Proof: Let . Consider . This has some nice free subgroups, unlike : Let . Let . By the Normal Form Theorem, (basically since there will always be some g’s between the s’s whenever you multiply any two elements together). Let . Again, . Since $latex\Sigma_{1}$ and are both countable, .
Consider the HNN-extension $\Gamma = (G \ast \langle s \rangle)\ast_{F_{\Sigma_{1}} \tilde F_{\Sigma_{2}}}$ and let t be the stable letter. For every , . But is generated by , , and : , so or . So by induction . for all n. But by construction is generated by .
Remark: In fact 3 can be replaced by 2.
Isometries of Trees
Before we saw that graphs of groups correspond to groups acting on trees. As such, we now turn to isometries of trees. Let be a tree and let . The translation length of is defined to be . Let .
Definition: If , is called semisimple. If is semisimple, and , is called elliptic. If , it is called loxodromic.
Theorem 16: Every is semisimple. If is loxodromic, is isometric to , and acts on as translation by .
Notation: If is elliptic, we write . If is loxodromic$, we write . Note that is connected since if we have two points fixed by any path between them must be fixed, for otherwise we would not be in a tree.
Proof of Theorem 16: Consider any and the triangle with vertices (i.e., ). Let , and let be the midpoint of .
Case 1:
In this case we have . So , and is elliptic.
Case 2:
Let . Now . Therefore is isometric to and acts as translation by . Furthermore, . Therefore unless is on the line just constructed, .
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