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

Example. \mathbb Z=\pi_1(S^1). Equivalently, \mathbb Z acts on \mathbb R without global fixed points. So \mathbb Z splits freely.

If G \ncong \mathbb Z but G splits freely, then G=G_1 \ast G_2 for G_1, G_2 neq 1.

Definition. The rank of G is the minimal r such that F_r surjects G.

It is clear that rank(G_1\ast G_2)\leq rank(G_1)+rank(G_2).

Grushko’s Lemma. Suppose \varphi:F_r \longrightarrow G is surjective and r is minimal. If G=G_1 \ast G_2, then F_r=F_1 \ast F_2 such that \varphi(F_i)=G_i for i=1,2.

Pf. Let X_i=K(G_i,1) (i=1,2) be simplicial and let \mathfrak{X} be a graph of spaces with vertex spaces X_1, X_2 and edge space a point. So G=\pi_1(X_{\mathfrak{X}}, x_0) where x_0=(*, \frac{1}{2}).

Let \Gamma be a graph so that \pi_1(\Gamma)\cong F_r and realize \varphi as a simplicial map f: \Gamma \longrightarrow X_{\mathfrak{X}}. Let y_0 \in f^{-1}(x_0). Because r is minimal, f^{-1}(x_0) is a forest, contained in \Gamma. The goal is to modify f by a homotopy to reduce the number of connected components of f^{-1}(x_0).

Let U \subseteq f^{-1}(x_0) be the component that contains y_0. Let V \subseteq f^{-1}(x_0) be some other component. Let \alpha a path in \Gamma from y_0 to V.

Look at f \circ \alpha \in \pi_1(X_{\mathfrak{X}}, x_0). Because \varphi is surjective, there exists \gamma\in \pi_1(\Gamma, y_0) such that f \circ \gamma = f \circ \alpha. Therefore if \beta= \gamma^{-1} \cdot \alpha, then f \circ \beta is null-homotopic in X_{\mathfrak{X}} and \beta gives a path from y_0 to V.

We can write \beta as a concaternation as \beta=\beta_1 \cdot \beta_2 \cdot \cdot \cdot \beta_n such that for each i, f \circ \beta_i \subseteq X_{\mathfrak{X}}\smallsetminus {x_0}. By the Normal Form Theorem, there exists i such that f\circ \beta_i is null-homotopic in X.

We can now modify f by a homotopy so that im (f\circ\beta_i)={x_0}. Therefore \beta_i \subseteq f^{-1}(x_0) and the number of components of f^{-1}(x_0) has gone down. By induction, we can choose f so that f^{-1}(x_0) is a tree. Now f factors through \Gamma'=\Gamma/ f^{-1}(x_0). Then F_r\cong \pi_1(\Gamma') and there is a unique vertex of \Gamma' that maps to x_0. So every simple loop in \Gamma' is either contained in X_1 or X_2 as required. square

An immediate consequence is that rank(G_1\ast G_2)=rank (G_1) + rank (G_2).

Grushko’s Theorem. Let G be finitely generated. Then G\cong G_1 \ast \cdots\ast G_m \ast F_r where each G_i is freely indecomposable and F_r is free. Furthermore, the integers m and r are unique and the G_i are unique up to conjugation and reordering.

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

Suppose G=H_1\ast \cdots \ast H_n \ast F_s. Let \mathcal{G} be the graph of groups. Let T be the Bass-Serre tree of \mathcal{G}.

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

Now consider the action of F_r on T. F_r\smallsetminus T is a graph of groups with underlying graph \Delta, say, and \pi_1(\Delta) is a free factor in F_r. But there is a covering map F_r\smallsetminus T \longrightarrow \mathcal{G} that induces a surjection \pi_1(\Delta) \longrightarrow F_s. Therefore, r\geq s. The other inequality can be obtained by switching F_r and F_s. \square

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.

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