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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: $F \ast_{z} F$ is LERF.

Exercise 22. A group is coherent if every fg subgroup is fp. Prove that $\pi_1$ of a finite graph of groups with coherent vertex groups and cyclic edge group is coherent.

Recall, that if $H \subseteq \pi_1 \mathcal{G}=G$ is fg then H has an induced graph of groups structure. $\mathcal{H} =H \backslash T_H \subseteq H \backslash T$ where T is the Bass-Serre tree of $\mathcal{G}$.

Topologically, $H$ defines a covering space $X^H \rightarrow X_g$, which inherits a graph of spaces structure with underlying graph $H\backslash T$.

$\Gamma' =H \backslash T_H \subseteq H \backslash T$ is a finite connected subgraph and the image of $\Gamma'$ in $X^H$ is a connected subcomplex $X'$ such that $\pi_1 X' = \pi_1 X^H =H$. $X'$ has the structure of a graph of space with underlying graph $\Gamma'$

There is an explicit algebraic description of $H\backslash T$, which follows immediately from Lemmas 16,17,18.

Lemma 23. (a) The vertices of $H\backslash T$ are in bijection with $\coprod_{v \in V(\Gamma)} H \backslash G/G_v=\coprod_{v \in V(\Gamma)}\{HgG_v\mid g \in G\}$.

The vertex $HgG_v$ is labelled by: $H \cap gG_vg^{-1}$, well-defined up to conjugation in $H$.

(b) The edges of $H\backslash T$ are in bijection with $\coprod_{e \in E(\Gamma)} H \backslash G/G_e$. The vertex $HgG_e$ is labelled with $H \cap g G_e g ^{-1}$.

(c) The edges of $H\backslash T$ adjoining the vertex corresponding to $HgG_v$  are in bijection with $\coprod_{e~\mathrm{adjoining}~v} (g^{-1} H g \cap G_v ) \backslash G_v/G_e$.

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 $f:X \rightarrow Y$ be a continuous map and $Y' \rightarrow Y$ a covering map. If $f':X \rightarrow Y'$ makes the diagram commutes, then $f'$ is a lift of $f$.

Lemma 24. Fix basepoint $x \in X$, $y=f(x) \in Y$, and $y' \in Y'$ mapping to $y$. There is a lift $f':X \rightarrow Y'$ with $f'(x)=y'$ if and only if $f_{\ast} \pi_1(X,x) \subseteq \pi_1(Y',y')$.  Furthermore, if this lift exists it is unique.

It may be impossible to lift at $y'$. But it is possible if we pass to a covering space. Eg., if $\tilde{X} \rightarrow X$ is the universal cover.  Intuitively, an elevation is a minimal lift.

Definition: Let $X,Y,Y', x,y,y'$ be as above. A based connected covering space $(X',x') \rightarrow (X,x)$ together with a based map $f':(X',x')\to (Y',y')$ such that

commutes is an elevation (of $f$ at $y'$) if whenever the diagram

commutes and $X' \rightarrow \bar{X}$ is a covering map of degree larger than 1 then the composition $(\bar{X},\bar{x}) \rightarrow (X,x) \rightarrow (Y,y)$ does not lift to $Y'$ at $y'$.

The unbased covering map $X' \rightarrow X$ or equivalently the conjugacy class of the subgroup $\pi_1(X') \subseteq \pi_1(X)$ is called the degree of the elevation $f'$.

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:

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

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

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, $d(x, \gamma x) > d(\gamma^{-1} O, O) = |\gamma|$. $\square$

Normal form theorem for graphs of groups. Let $\mathcal{G}$ be a graph of groups and $G=\pi_1(G)$.

1. Any $g \in G$ can be written as
$g = g_0 t_{e_1}^{\epsilon_1} g_1 \cdots t_{e_n}^{\epsilon_n} g_n\quad$
as before.
2. If $g = 1$, this expression includes `backtracking’, meaning that for some $i$, $e_i = e_{i + 1}$ with $\epsilon_i = - \epsilon_{i + 1}$, and furthermore that if $\epsilon_i = \pm 1$, then $g_i \in \partial_\pm ( G_{e_i} )$.

Similar to the case of the free group, the proof boils down to the fact that the Bass–Serre tree $T$ is a tree.

Proof. To simplify notation, set $t_i = t_{e_i}^{\epsilon_i}$, so

$g = g_0 t_1 g_1 \cdots t_n g_n$.

Fix base points $*_i$ in the vertex spaces $X_{v_i}$, which are chosen to coincide when the vertices do. Then $g_i$ is a loop in $X_{v_i}$ based at $*_i$, and $t_i$ is a path, crossing the corresponding edge space, from $*_i$ to $*_{i + 1}$. This allows us to consider $g$ as a loop in $X_{\mathcal{G}}$ based at $*_0$. (We may assume $v_0 = v_n$ by adding letters from a maximal tree.)

Consider the universal covering $\widetilde{X}_{\mathcal{G}}$ and fix a base point $\widetilde{*}_0$ over $*_0$ in $\widetilde{X}_{\widetilde{v}_0}$. Let $\widetilde{g}$ be the lift of $g$ based at $\widetilde{*}_0$ and $\gamma$ its image in the Bass–Serre tree $T$. We now analyze $\widetilde{g}$ and $\gamma$ closely.

Choose $\widetilde{e}_i$ adjoining $\widetilde{X}_{\widetilde{v}_0}$ and $\widetilde{X}_{\widetilde{v}_1}$ so that the edge traversed by $t_i$ when lifted at $\widetilde{*}_{i - 1}$ corresponds to the coset $1 \cdot G_{e_i} \subseteq G_{v_i} / G_{e_i}$.

Then $g_0$ lifts to a path in $\widetilde{X}_{\widetilde{v}_0}$ which terminates at $g_0 \widetilde{*}_0$. Similarly, $t_1$ lifts at $*_0$ to a path across the edge $\widetilde{e}_1$ to the vertex space $t_1\widetilde{X}_{\widetilde{v}_1}$ terminating at $t_1 \widetilde{*}_1$. Therefore, $g_0 t_1$ lifts at $\widetilde{*}_0$ to a path which crosses the edge space $g_0 \widetilde{e}_1$ and ends at $g_0 t_1 \widetilde{*}_1$.

Then, $g_1$ lifts at $\widetilde{*}_1$ to a path in $\widetilde{X}_{\widetilde{v}_1}$ ending at $\widetilde{*}_1$, and $t_2$ lifts at $\widetilde{*}_1$ to a path across the edge $\widetilde{e}_2$ into the vertex space $\widetilde{X}_{\widetilde{v}_2}$, and terminating at $t_2 \widetilde{*}_2$. Thus $g_0 t_1 g_1 t_2$ lifts at $\widetilde{*}_0$ to a path which crosses $g_0 \widetilde{e}_1$, through $g_0 \widetilde{X}_{\widetilde{v}_1}$, across $g_0 t_1 g_1 \widetilde{e}_2$, and ending at

$g_o t_1 g_1 t_2 \widetilde{*}_2 \in g_0 t_1 g_1 t_2 \widetilde{X}_{\widetilde{v}_2}$.

We continue this process until we have explicitly constructed $\widetilde{g}$. By hypothesis, $g = 1$, so $\widetilde{g}$ and $\gamma$ are both loops in $\widetilde{X}_{\mathcal{G}}$ and $T$, respectively. Since $T$ is a tree, $\gamma$ must backtrack.

This implies that $\widetilde{e}_i=\widetilde{e}_{i + 1}$ and that $\epsilon_i = -\epsilon_{i + 1}$. That is, by Lemma 18,

$g_{i + 1} \in \partial_\pm^{e_i} ( G_{e_i} )$.

Therefore, we have found a backtracking, and can accordingly shorten $g$. This proves the theorem. $\square$