Theorem 8: Let $\Gamma$ be a $\delta$-hyperbolic group with respect to $S$. If $u,v \in \Gamma$ are conjugate then there exists $\gamma\in\Gamma$ such that

$\gamma u\gamma^{-1}=v\\l(\gamma)\leq M(l(u),l(v))$

where $M$ depends only on $\Gamma$.

Proof: We work in $Cay_S(\Gamma)$. Let $\gamma\in \Gamma$ be such that $\gamma u\gamma^{-1}=v$.  Let $\gamma_t \in [1,\gamma]$ be such that $d(1,\gamma_t)=t$. We want to find a bound on $d(\gamma_t, v\gamma_t)$.

Let $c=[1,\gamma u]$. By Lemma 9,

$d(\gamma_t,c(t))\leq 2(\delta+l(v))\\ d(v\gamma_t,c(l(\gamma u)-(l(\gamma)-t)))\leq 2(\delta+l(v))$

Also

$d(c(t),c(l(\gamma u)-(l(\gamma)-t)))=l(\gamma u)-l(\gamma)+t-t\\=l(\gamma u)-l(\gamma)\leq l(u)$

So $d(\gamma_t,v\gamma_t)\leq 4(\delta +l(u)+l(v))= R(l(u),l(v))$. Thus $l(\gamma_t^{-1}v\gamma_t)\leq R$. Suppose that $l(\gamma)> \#B(1,R)$. By the Pigeonhole Principle there exist integers $s>t$ such that $\gamma_t^{-1}v\gamma_t=\gamma_s^{-1}v\gamma_s$. It follows that one can find a shorter conjugating element by cutting out the section of $\gamma$ between $\gamma_t$ and $\gamma_s$.

Recall, for $\gamma \in \Gamma$, $C(\gamma)=\{g\in\Gamma: g\gamma=\gamma g\}$ is the centralizer of $\gamma$.

Theorem 9: If $\Gamma$ is $\delta$-hyperbolic with respect to $S$ and $\gamma\in\Gamma$, then $C(\gamma)$ is quasi-convex in $\Gamma$.

Proof: Again we work in $Cay_S(\Gamma)$. Let $g\in C(\gamma)$, $h\in [1,g]$. We need to prove that $H$ is in a bounded neighborhood $C(\gamma)$.

Just as in the proof of Theorem 8,

$l(h^{-1}\gamma h)=d(h, \gamma h)\leq 4(\delta+2l(\gamma))$

Well, $g$ and $h^{-1}\gamma h$ are conjugate. By Theorem 8 there exists $k\in \Gamma$ such that

$k^{-1}\gamma k=h^{-1}\gamma h \\ l(k)\leq M(l(\gamma), l(h^{-1}\gamma h))\\ \leq M(l(\gamma),4(\delta+2l(\gamma)))$

But $h^{-1}k\gamma=\gamma hk^{-1}$ so that $hk^{-1}\in C(\gamma)$ and $d(h,hk^{-1})=d(1,k^{-1})=l(k^{-1})=l(k)\leq M$.

Exercise 15: Prove that

$\Gamma_A=\mathbb{Z}^2\rtimes_A\mathbb{Z}$

is not hyperbolic for any Anosov $A$.