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