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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

fig_28_11

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:fig_28_2

If \chi' does not satisfy Stalling’s condition then one of these maps is not surjective.  Without loss of generality, suppose thatfig_28_31

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 diagramfig_28_6

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 diagramfig_28_81

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).

henrylec151

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<t'-2\delta, then d(\gamma g_{t'}, 1) < d(\gamma g_{t'}, \gamma), a contradiction. Similarly t'<t-2\delta. 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 GgHg^{-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:

diagram

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.

Completing $latex \Delta$ to a finite-sheeted covering $latex \hat{X} \rightarrow X$.

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.