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**Definition.** A group **splits freely** if acts on a tree without global fixed point and such that every edge stailizer is trivial. If does not split freely, then is called **freely indecomposable**.

Example. . Equivalently, acts on without global fixed points. So splits freely.

If but splits freely, then for .

**Definition. **The **rank** of is the minimal such that surjects .

It is clear that .

**Grushko’s Lemma. **Suppose is surjective and is minimal. If , then such that for .

**Pf**. Let be simplicial and let be a graph of spaces with vertex spaces and edge space a point. So where .

Let be a graph so that and realize as a simplicial map . Let . Because is minimal, is a forest, contained in . The goal is to modify by a homotopy to reduce the number of connected components of .

Let be the component that contains . Let be some other component. Let a path in from to .

Look at . Because is surjective, there exists such that . Therefore if , then is null-homotopic in and gives a path from to .

We can write as a concaternation as such that for each , . By the Normal Form Theorem, there exists such that is null-homotopic in .

We can now modify by a homotopy so that . Therefore and the number of components of has gone down. By induction, we can choose so that is a tree. Now factors through . Then and there is a unique vertex of that maps to . So every simple loop in is either contained in or as required.

An immediate consequence is that .

**Grushko’s Theorem. **Let be finitely generated. Then where each is freely indecomposable and is free. Furthermore, the integers and are unique and the are unique up to conjugation and reordering.

**Pf.** Existence is an immediate corollary of the fact that rank is additive.

Suppose . Let be the graph of groups. Let be the Bass-Serre tree of .

Consider the action of on . Because is freely indecomposable, stabilize a vertex of . Therefore is conjugate into some .

Now consider the action of on . is a graph of groups with underlying graph , say, and is a free factor in . But there is a covering map that induces a surjection . Therefore, . The other inequality can be obtained by switching and .

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.

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