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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 $d(g, \gamma)$.

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, then $d(\gamma g_{t'}, 1) < d(\gamma g_{t'}, \gamma)$, a contradiction. Similarly $t'. 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 G$$gHg^{-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.