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As before, are and are graphs of spaces equipped with maps , , , and such that

commutes.

**Lemma 28: ** Suppose that every edge map of is an elevation. Then the map is -injective.

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

If 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 . Suppose . Then is a subgroup of . Let be the corresponding covering space of . Since contains , there is a lift of . Now replace by and repeat. After infinitely many repetitions, the result satisfies the hypothesis of Stalling’s condition, and is contained in a subgraph of spaces.

**Theorem 18:** If and are residually finite groups then is also residually finite.

**Proof:** Let , let be a point, and fix maps . This defines a graph of spaces . By the Seifert-van Kampen Theorem, . Let be the graph of spaces structure on the universal cover of , and let be a compact subset. We may assume that is connected; we may also assume that if (Bass-Serre tree) and , then . Let be the map to the underlying map of the Bass-Serre tree. Then is a finite connected subgraph, . For each , let , a compact subspace of . Because are residually finite, we have a diagram

where is a finite-sheeted covering map and embeds in . Let be defined as follows. Set ; for a vertex , the vertex space is the corresponding to . For each corresponding to , define so that the diagram

commutes. Now sum.

**Exercise 26: **If and are residually finite and is finite, prove that is residually finite.

Finally, we are in a position to prove that a hyperbolic group has no subgroup isomorphic to .

**Theorem 11.** Let with . Then .

**Proof.** By Lemma 10, we can assume that is not conjugate to any element of length by replacing with a power of itself. Suppose . We need to bound .

Replacing with for some , we may assume that . We will be done if we can bound .

Suppose . By dividing into triangles, we see that any geodesic rectangle is -slim, in the same way that triangles are -slim.

Because the rectangle with vertices is -slim, there exists such that .

If , then , a contradiction. Similarly . So . Therefore, .

But . This is a contradiction since we assumed that is not conjugate to anything so short. Therefore . Thus .

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 be a torsion-free hyperbolic group. Whenever is not a proper power, then is malnormal.

**Definition.** A subgroup of a group is *malnormal* if for all , , then .

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

**Proof of Lemma.** Suppose .

Therefore for some , .

By Lemma 10, . Therefore . Thus . Therefore .

**Exercise 17.** Prove that if where is hyperbolic and torsion-free and and and , then . That is, 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 , 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 is a finitely generated non-residually finite group, then Lemma 4 implies that the kernel of a surjection is not separable in . However, finitely generated subgroups of free groups are separable:

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

This proof is associated with Stallings.

**Proof:** As usual, let where is a rose. Let be a covering map with finitely generated. Let be compact. We need to embed in an intermediate finite-sheeted covering.

Enlarging if necessary, we may assume that is connected and that . Note that we have . By Theorem 5 (see below), the immersion extends to a covering . Then . So lifts to a map .

The main tool in the proof above is this:

**Theorem 5:** The immersion can be completed to a finite-sheeted covering into which embeds:

**Proof: ** Color and orient the edges of . Any combinatorial map of graphs corresponds uniquely to a coloring and orientation on the edges of . A combinatorial map is an immersion if and only if at every vertex of , 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 be the number of vertices of . For each color , let be the number of edges of colored . Then there are vertices of missing “arriving” edges colored , and there are vertices of missing “leaving” edges colored . Choose any bijection between these two sets and use this to glue in edges colored . When this is done for all colors, the resulting map is clearly a covering.

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

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

**Exercise 7 (Greenberg’s Theorem):** If and is finitely generated, then is of finite index in .

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