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As before, are $\chi'$ and $\chi$ are graphs of spaces equipped with maps $\Phi \colon X_{\chi'} \to X_{\chi}$, $\varXi' \to \varXi$, $\phi_{v'} \colon X_{v'} \to X_v$, and $\phi_{e'} \colon X_{e'} \to X_e$ such that

commutes.

Lemma 28: Suppose that every edge map of $\chi'$ is an elevation.  Then the map $\Phi$ is $\pi_1$-injective.

Proof: The idea is to add extra vertex spaces to $\chi'$ so that $\chi'$ satisfies Stalling’s condition.  As before, we have inclusions:

If $\chi'$ does not satisfy Stalling’s condition then one of these maps is not surjective.  Without loss of generality, suppose that

does not arise as an edge map of $\chi'$.  Suppose $\partial_{e}^{-} \colon X_e \to X_u$.  Then $(\partial_{e}^{-} \circ \phi_{e'})_{*}(\pi_1(X_{e'}))$ is a subgroup of $\pi_1(X_u)$.  Let $X_{u'}$ be the corresponding covering space of $X_u$.  Since $\pi_1(X_{u'})$ contains $(\partial_{e}^{-} \circ \phi_{e'})_{*}(\pi_1(X_{e'}))$, there is a lift $\partial_{e'} \colon X_{e'} \to X_{u'}$ of $\partial_{e'} \circ \phi_{e'}$.  Now replace $\chi'$ by $\chi' \cup X_{u'}$ and repeat.  After infinitely many repetitions, the result $\hat{\chi}$ satisfies the hypothesis of Stalling’s condition, and $\chi'$ is contained in a subgraph of spaces.

Theorem 18: If $G_{+}$ and $G_{-}$ are residually finite groups then $G_{+} * G_{-}$ is also residually finite.

Proof: Let $X_{\pm} = K(G_{\pm}, 1)$, let $e = *$ be a point, and fix maps $\partial^{\pm} \colon e \to X_{\pm}$.  This defines a graph of spaces $\chi$.  By the Seifert-van Kampen Theorem, $\pi_1(X_{\chi}) \cong G_{+} * G_{-}$.  Let $\tilde{\chi}$ be the graph of spaces structure on the universal cover of $X_{\chi}$, and let $\Delta \subseteq X_{\tilde{\chi}}$ be a compact subset.  We may assume that $\Delta$ is connected; we may also assume that if $\tilde{e} \in \tilde{\varXi}$ (Bass-Serre tree) and $\Delta \cap \tilde{e} \ne \emptyset$, then $\tilde{e} \subseteq \Delta$.  Let $\eta \colon X_{\tilde{\chi}} \to \tilde{\varXi}$ be the map to the underlying map of the Bass-Serre tree.  Then $\eta(\Delta)$ is a finite connected subgraph, $\varXi'$.  For each $v' \subseteq v(\varXi')$, let $\Delta_{v'} = \Delta \cap X_{v'}$, a compact subspace of $X_{v'}$.  Because $G_{\pm}$ are residually finite, we have a diagram

where $X_{\hat{v}} \to X_{\pm}$ is a finite-sheeted covering map and $\Delta_{v'}$ embeds in $X_{\hat{v}}$.  Let $\hat{\chi}$ be defined as follows.  Set $\hat{\varXi} = \varXi'$; for a vertex $\hat{v} \in \hat{\varXi}$, the vertex space is the $X_{\hat{v}}$ corresponding to $v'$.  For each $\hat{e} \in E(\hat{\varXi})$ corresponding to $e' \in E(\varXi')$, define $\partial_{\hat{e}}^{\pm}$ so that the diagram

commutes.  Now sum.

Exercise 26: If $G_1$ and $G_2$ are residually finite and $H$ is finite, prove that $G_1 *_H G_2$ is residually finite.

Finally, we are in a position to prove that a hyperbolic group has no subgroup isomorphic to $\mathbb Z^2$.

Theorem 11. Let $\gamma \in \Gamma$ with $o(\gamma)=\infty$. Then $|C(\gamma):\langle \gamma \rangle|<\infty$.

Proof. By Lemma 10, we can assume that $\gamma$ is not conjugate to any element of length $\leq 4\delta$ by replacing $\gamma$ with a power of itself. Suppose $g\in C(\gamma)$. We need to bound $d(g, \gamma)$.

Replacing $g$ with $\gamma^{-r}g$ for some $r$, we may assume that $d(1,g)=d(g,\langle \gamma \rangle)$. We will be done if we can bound $l(g)$.

Suppose $l(g)>2(l(\gamma)+2\delta)$. By dividing into triangles, we see that any geodesic rectangle is $2\delta$-slim, in the same way that triangles are $\delta$-slim.

Because the rectangle with vertices $1, \gamma, g\gamma, g$ is $2\delta$-slim, there exists $g_t, g_{t'} \in [1,g]$ such that $d(g_t, \gamma g_{t'}) \leq 2\delta$.

If $t, then $d(\gamma g_{t'}, 1) < d(\gamma g_{t'}, \gamma)$, a contradiction. Similarly $t'. So $|t-t'|<2\delta$. Therefore, $d(g_t, \gamma g_t)<4\delta$.

But $l(g_t^{-1}\gamma g_t)=d(g_t, \gamma g_t)<4\delta$. This is a contradiction since we assumed that $\gamma$ is not conjugate to anything so short. Therefore $l(g)\leq2(l(\gamma)+2\delta)$. Thus $|C(\gamma):\langle \gamma \rangle|<\infty$.

An element of a group is torsion if its order is finite.

A group is torsion if every element is torsion.

A group is  torsion-free if no nontrivial elements are torsion.

Corollary. Every non-trivial abelian subgroup of a hyperbolic group is virtually cyclic.

Lemma 11. Let $\Gamma$ be a torsion-free hyperbolic group. Whenever $\gamma \in \Gamma$ is not a proper power, then $\langle \gamma \rangle$ is malnormal.

Definition. A subgroup $H$ of a group $G$ is malnormal if for all $g\in G$$gHg^{-1} \cap H \neq 1$ , then $g\in H$.

Remark. By Theorem 11, if $\Gamma$ is hyperbolic and torsion-free, centralizers are cyclic.

Proof of Lemma. Suppose $g \langle \gamma \rangle g^{-1}\cap \langle \gamma \rangle \neq 1$.

Therefore for some $p, q \neq 0$, $g\gamma^{p}g^{-1}=\gamma^q$.

By Lemma 10, $|p|=|q|$. Therefore $g^2\gamma^pg^{-2}=\gamma^p$. Thus $g^2\in C(\gamma^p)= \langle \gamma \rangle$. Therefore $g \in \langle \gamma \rangle$.

Exercise 17. Prove that if $x, y, z \in \Gamma$ where $\Gamma$ is hyperbolic and torsion-free and $xy=yx$ and $yz=zy$ and $y\neq 1$, then $xz=zx$. That is, $\Gamma$ is commutative transitive.

We now turn briefly to a fundamental open question about hyperbolic groups.  This question is a theme of the course.

Question. Is every word-hyperbolic group residually finite?

The fundamental groups of hyperbolic manifolds are linear, so residually finite by Selberg’s Lemma.

What about for negative curved manifolds?

Evidence for:

Theorem (Sela). Every torsion-free hyperbolic group is Hopfian.

Theorem (I. Kapovich-Wise). If every nontrivial hyperbolic group has a proper finite-index subgroup, then every hyperbolic group is residually finite.

Evidence against:

Theorem (Agol-Groves-Manning). If every hyperbolic group is residually finite, then every quasi-convex subgroup of every hyperbolic group is separable.

Fact: There exists a finitely generated non-Hopf group.  (An example is the Baumslag-Solitar group $B(2,3)$, although we cannot prove it yet.)  So, by Lemma 5, there is a finitely generated non-residually finite group.  Thus, free groups are not ERF: if $G$ is a finitely generated non-residually finite group, then Lemma 4 implies that the kernel of a surjection $F_r \rightarrow G$ is not separable in $F_r$.  However, finitely generated subgroups of free groups are separable:

Marshall Hall’s Theorem (1949): $F_r$ is LERF.

This proof is associated with Stallings.

Proof: As usual, let $F_r = \pi_1(X)$ where $X$ is a rose. Let $X' \rightarrow X$ be a covering map with $\pi_1 (X')$ finitely generated.  Let $\Delta \subset X'$ be compact. We need to embed $\Delta$ in an intermediate finite-sheeted covering.

Enlarging $\Delta$ if necessary, we may assume that $\Delta$ is connected and that $\pi_1 (\Delta) = \pi_1(X')$.  Note that we have $\Delta \subset X' \rightarrow X$. By Theorem 5 (see below), the immersion $\Delta \rightarrow X$ extends to a covering $\hat{X} \rightarrow X$. Then $\pi_1 (X') = \pi_1(\Delta) \subset \pi_1 (\hat{X})$. So $X' \rightarrow X$ lifts to a map $X' \rightarrow \hat{X}$.

The main tool in the proof above is this:

Theorem 5: The immersion $\Delta \rightarrow X$ can be completed to a finite-sheeted covering $\hat{X} \rightarrow X$ into which $\Delta$ embeds:

Proof: Color and orient the edges of $X$. Any combinatorial map of graphs $\Delta \rightarrow X$ corresponds uniquely to a coloring and orientation on the edges of $\Delta$.  A combinatorial map is an immersion if and only if at every vertex of $\Delta$, we see each color arriving at most once and leaving at most once. Likewise, it’s a covering map if and only if at each vertex, we see each color arriving exactly once and leaving exactly once.

Let $k$ be the number of vertices of $\Delta$.  For each color $c$, let $k_c$ be the number of edges of $\Delta$ colored $c$. Then there are $k-k_c$ vertices of $\Delta$ missing “arriving” edges colored $c$, and there are $k-k_c$ vertices of $\Delta$ missing “leaving” edges colored $c$.  Choose any bijection between these two sets and use this to glue in $k-k_c$ edges colored $c$. When this is done for all colors, the resulting map $\Delta \subset \hat{X} \rightarrow X$ is clearly a covering.

Note that the proof in fact gives us more.  For instance:

Exercise 6: If $H$ is a finitely generated subgroup of $F_r$, then $H$ is a free factor of a finite-index subgroup of $F_r$.

Exercise 7 (Greenberg’s Theorem): If $H\triangleleft F_r$ and $H$ is finitely generated, then $H$ is of finite index in $F_r$.