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Recall from last time:

Theorem 16: Every \gamma \in \mathrm{Isom}\ T is semisimple. If \gamma is loxodromic, then \mathrm{Min}(\gamma) is isometric to \mathbb{R} and \gamma acts on \mathrm{Min}(\gamma) as translation by |\gamma|.

In particular, for every \langle \gamma \rangle, \mathrm{Min}(\langle \gamma \rangle) is a \langle \gamma \rangle-invariant subtree of T.

Exercise 21: If \gamma, \delta \in \mathrm{Isom}\ T and \mathrm{Min}(\gamma) \cap \mathrm{Min}(\delta) = \emptyset, then \gamma \delta is loxodromic, \mathrm{Min}(\gamma \delta) \cap \mathrm{Min}(\gamma) \neq \emptyset, and \mathrm{Min}(\gamma \delta) \cap \mathrm{Min}(\delta) \neq \emptyset.

(Hint: It is enough to construct an axis for \gamma \delta.)

Lemma 20 (Helly’s Theorem for Trees): If S_1, \ldots, S_n \subseteq T are closed subtrees and S_i \cap S_j \neq \emptyset for every i,j, then \bigcap _i S_i \neq \emptyset.

Proof. Case n=3: Let x_i \in S_j \cap S_k where \{ i,j,k \} = \{1, 2, 3\}. Let o be the center of the triangle with vertices x_1, x_2, x_3. Then o \in S_1 \cap S_2 \cap S_3.

General case: Let S_i ' = S_i \cap S_n for 1\leq i \leq n-1. Then S_i' \cap S_j' = S_i \cap S_j \cap S_n \neq \emptyset by the n=3 case. Now, by induction,

\bigcap _{i=1} ^n S_i = \bigcap _{i=1}^{n-1} S_i ' \neq \emptyset .

Corollary: If a finitely generated group G acts on a tree T with no global fixed point, then there is a (unique) G-invariant subtree T_G that is minimal with respect to inclusion, among all G-invariant subtrees. Furthermore, T_G is countable.

Proof. Let

eq

For any \gamma-invariant T' \subseteq T, it is clear that \mathrm{Axis}(\gamma) \subseteq T'. Therefore T_G is minimal. Let S be a finite generating set for G. Suppose that every element of G is elliptical, so for all s, s' \in S, ss' is elliptical. Then for all s,s' \in S,

\mathrm{Fix}(s) \cap \mathrm{Fix}(s') \neq \emptyset.

Therefore by Lemma 20, \bigcap _{s\in S} \mathrm{Fix}(s) \neq \emptyset, a contradiction.

Suppose that \gamma, \delta \in G are loxodromic and \mathrm{Axis}(\gamma) \cap \mathrm{Axis}(\delta) = \emptyset. By Exercise 21, \mathrm{Axis}(\gamma \delta) intersects \mathrm{Axis}(\gamma) and \mathrm{Axis}(\delta) non-trivially. Thus, T_G is connected.

It remains to show that T_G is G-invariant. Let \gamma, \delta \in G. For any x\in T,

d(x, \delta ^{-1} \gamma \delta x) = d(\delta x, \gamma(\delta x)).

This implies that |\delta^{-1} \gamma \delta | = |\gamma |, and

\delta \mathrm{Axis}(\gamma) = \mathrm{Axis} (\delta^{-1} \gamma \delta).

So we conclude that T_G is G-invariant.

Definition: If \Gamma' \subseteq \Gamma is connected, then the graph of groups carried by \Gamma' is the graph of groups \mathcal{G}' with underlying graph \Gamma' such that the vertex v \in V(\Gamma') \subseteq V(\Gamma) is labelled by G_v, and the edge e \in E(\Gamma') \subset E(\Gamma) is labelled G_e (with obvious edge maps). There is a natural map G' \rightarrow G, where G = \pi_1 (\mathcal{G}) and G' = \pi_1(\mathcal{G}'). This map is an injection by the Normal Form Theorem.

Lemma 21: If \Gamma is countable and G = \pi_1 \mathcal{G} is finitely generated, then there is a finite subgraph \Gamma' \subset \Gamma such that \pi_1 \mathcal{G}' = \pi_1 \mathcal{G} (where \mathcal{G}' is the graph of groups carried by \Gamma').

Proof. Let \Gamma_0 \subseteq \Gamma_1 \subseteq \cdots \subseteq \Gamma be an exhaustion of \Gamma by finite connected subgraphs. Let \mathcal{G}_n denote the graph of groups carried by \Gamma_n and set G_n = \pi_1 (\mathcal{G}_n). Since G is finitely generated, there is an n such that G_n contains each generator of G, and since G_n \leq G, we conclude that G_n = G.

Lemma 22: If \Gamma is countable and G = \pi_1 \mathcal{G} is finitely generated then there is a finite minimal (wrt inclusion) subgraph \Gamma' that carries G.

Proof. Let T be the Bass-Serre tree of \mathcal{G}. Let \Gamma' = G \backslash T_G \subseteq G \backslash T = \Gamma. It’s an easy exercise to check that this is as required.

We will refer to \Gamma' (and \mathcal{G}', the graph of groups carried by \Gamma') as the core of \mathcal{G}.

Theorem 17: If H \leq G = \pi_1 \mathcal{G} is finitely generated, then H decomposes as \pi_1 of a finite graph of groups \mathcal{H}. The vertex groups of \mathcal{H} are conjugate into the vertex groups of \mathcal{G}. The edge groups are likewise.

Proof. Let T be the Bass-Serre tree of \mathcal{G}, and set \mathcal{H} = H \backslash T_H.

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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 \mathcal{G} 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:

BSpqgrp

For example \mathrm{BS} (1,1) \cong \mathbb{Z}^{2}.

Consider \mathrm{BS}(2,3).  We will show that this group is non-Hopfian (i.e., that there exists a homomorphism \varphi: \mathrm{BS}(2,3) \to \mathrm{BS}(2,3) that is a surjection with nontrivial kernel), hence is not RF.  Consider \psi: \mathrm{BS}(2,3) \to \mathrm{BS}(2,3) defined by \psi(t) = t and \psi(a) = a^{2}.  First, we show this is a homomorphism, then a surjection, then that it has nontrivial kernel.

Homomorphism:  \psi(ta^{2}t^{-1}) = ta^{4}t^{-1} = ta^{2}t^{-1}ta^{2}t^{-1} = a^{3}a^{3} = a^{6} = \psi(a^{3}).  Hence, the map is a homomorphism.

Epimorphism: To see that \psi is a surjection, we check that a, t \in \mathrm{Im}(\psi).  This is obvious for t.  Further a^{2} = \psi(a) \in \mathrm{Im}(\psi) and so ta^{2}t^{-1} = a^{3} \in \mathrm{Im}(\psi).  But then a = a^{3}a^{-2} \in \mathrm{Im}(\psi).

Nontrivial kernel: Consider c = ata^{-1}t^{-1}a^{-2}tat^{-1}a.  Then \psi(c) = a^{2}ta^{-2}t^{-1}a^{-4}ta^{2}t^{-1}a^{2} = a^{2}a^{-3}a^{-4}a^{3}a^{2} = 1.  So we just need to show that c is nontrivial.  We use Britton’s Lemma (Normal Form Theorem).  What Britton’s Lemma says is that there will be t's that can be removed from our word using the relation if it is trivial.  But this can’t be done with c, so we must have c \neq 1 as required.

Remark: \mathrm{BS}(2,3) can never embed in a word-hyperbolic group by a previous Lemma.

Exercise 20: Show that \mathrm{BS}(1,2) \cong \mathbb{Q}_{2} \rtimes_{2} \mathbb{Z}, where \mathbb{Q}_{2} is the dyadic rationals and we have \mathbb{Z} acting via multiplication by 2.  One can deduce that \mathrm{BS}(1,2) is linear, hence RF by Selberg’s Lemma.

Theorem 15 (Higman-Neumann-Neumann): Every countable group embeds in a 3-generator group.

Proof: Let G = \{ 1 = g_{0}, g_{1}, \ldots\}.  Consider G \ast \mathbb{Z} = G \ast \langle s \rangle.  This has some nice free subgroups, unlike G:  Let s_{n} = g_{n}s^{n}.  Let \Sigma_{1} = \{ s_{n} \vert n \geq 1\}.  By the Normal Form Theorem, \langle \Sigma_{1} \rangle \cong F_{\Sigma_{1}} (basically since there will always be some g’s between the s’s whenever you multiply any two elements together).  Let \Sigma_{2} = \{ s_{n} \vert n \geq 2\}.  Again, \langle \Sigma_{2} \rangle \cong F_{\Sigma_{2}}.  Since $latex\Sigma_{1}$ and \Sigma_{2} are both countable, F_{\Sigma_{1}} \cong F_{\Sigma_{2}}.

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 n \geq 1, ts_{n}t^{-1} = s_{n+1}.  But \Gamma is generated by g_{1}, s, and t: s_{n+1} = ts_{n}t^{-1}, so g_{n+1}s^{n+1} = tg_{n}s^{n}t^{-1} or g_{n+1} = tg_{n}s^{n}t^{-1}s^{-n-1}.  So by induction g_{n} \in \langle g_{1}, s, t \rangle. for all n.  But by construction \Gamma is generated by \Sigma_{1} \cup \{ s\} \cup \{ t \}. \square

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 T be a tree and let \gamma \in \mathrm{Isom}(T).  The translation length of \gamma is defined to be |\gamma| := \inf_{x \in T} d(x, \gamma x).  Let \mathrm{Min}(\gamma) = \{ x \in T \vert d(x, \gamma x) = |\gamma| \}.

Definition: If \mathrm{Min}(\gamma) \neq \emptyset, \gamma is called semisimple.  If \gamma is semisimple, and |\gamma| = 0, \gamma is called elliptic.  If |\gamma| > 0, it is called loxodromic.

Theorem 16: Every \gamma \in \mathrm{Isom}(T) is semisimple.  If \gamma is loxodromic, \mathrm{Min}(\gamma) is isometric to \mathbb{R}, and \gamma acts on \mathrm{Min}(\gamma) as translation by |\gamma|.

Notation: If \gamma is elliptic, we write \mathrm{Min}(\gamma) = \mathrm{Fix}(\gamma).  If \gamma is loxodromic$, we write \mathrm{Min}(\gamma) = \mathrm{Axis}(\gamma).  Note that \mathrm{Fix}(\gamma) is connected since if we have two points fixed by \gamma any path between them must be fixed, for otherwise we would not be in a tree.

Case 1 of Theorem 16

Case 1 of Theorem 16

Proof of Theorem 16: Consider any x \in T and the triangle with vertices x, \gamma x, \gamma^{2} x (i.e., [x, \gamma x] \cup \gamma [x, \gamma x]).  Let [ x, \gamma x] \cap [ \gamma x, \gamma^{2} x ] \cap [ \gamma^{2} x, x ] = \{ O \}, and let M be the midpoint of [ x, \gamma x ].

Case 1: d(M, \gamma x) \leq d(O,\gamma x)

In this case we have d(M, \gamma x) = d(M, x) = d(\gamma M, \gamma x).  So \gamma M = M, and \gamma is elliptic.

Case 2: d(M, \gamma x) > d(O, \gamma x)

Case 2 of Theorem 16

Case 2 of Theorem 16

Let I = [\gamma^{-1} O, O] \subseteq [x, \gamma x].  Now I \cap \gamma I = \{ O \}.  Therefore \bigcup\limits_{n \in \mathbb{Z}} \gamma^{n} I is isometric to \mathbb{R} and \gamma acts as translation by [\gamma^{-1}O, O]. Furthermore, d(x, \gamma x) = d(\gamma^{-1}O, O) + 2 d(O, \gamma x).  Therefore unless x is on the line just constructed, d(x, \gamma x) > d(\gamma^{-1} O, O) = |\gamma|. \square