Our goal is to understand the abelian subgroups of a hyperbolic group; e.g., can a hyperbolic group contain $\mathbb{Z}^2$ as a subgroup?  We already know that it cannot be a quasiconvex subgroup, but it may be possible for $\mathbb{Z}^2$ to be “twisted” in some manner.

Theorem 10. The intersection of two quasiconvex subgroups is quasiconvex.

Pf. As usual, we work in a $\delta$-hyperbolic $\text{Cay}_S ( \Gamma )$.  Let $H , K \subset \Gamma$ be quasiconvex subgroups each with the same corresponding constant $\kappa$.  Let $\gamma \in H \cap K$, and let $g_0 \in [ 1 , \gamma ] \cap \Gamma$.  Our goal is therefore to show that $g_0$ is in a bounded neighborhood of $H \cap K$.  Let $g_D$ be the (or more precisely, a particular) closest element of $H \cap K$ to $g_0$, and suppose that $d ( g_0 , g_D ) = D$.  Let $h_0 \in H$ be such that $d ( g_0 , h_0 ) \leq \kappa$ and let $k_0 \in K$ be such that $d ( g_0 , k_0 ) \leq \kappa$; such elements exist since $H$ and $K$ are quasiconvex.  Let $g_t \in [ g_0 , g_D ]$ be such that $d ( g_0 , g_t ) = t$.  We sketch the situation below. Consider the geodesic triangle with vertices $g_0$, $h_0$, and $g_D$.  Because this triangle is $\delta$-slim, for each $t$ there exists some $h_t \in H$ such that $d ( h_t , g_t ) \leq \delta + \kappa$. Likewise, for each $t$ there exists $k_t \in K$ such that $d ( g_t , k_t ) \leq \delta + \kappa$.  Next, let $u_t = g_t^{-1} h_t$ and $v_t = g_t^{-1} k_t$.  Then $\ell ( u_t ) , \ell ( v_t ) \leq \delta + \kappa$.

Suppose $D > ( \# B ( 1 , \delta + \kappa ) )^2$.  Then by the Pigeonhole Principle there are integers $s$ and $t$ with $s > t$ such that $u_s = u_t$ and $v_s = v_t$.  Now, we can use this information to find a closer element of $H \cap K$. Consider $g_t g_s^{-1} g_D$.  By the triangle inequality, $d ( g_0 , g_t g_s^{-1} g_D ) \leq d ( g_0 , g_t ) + d ( g_t , g_t g_s^{-1} g_D ) = t + D - s < D$.

All that remains is to prove that $g_t g_s^{-1} g_D \in H \cap K$.  But since $u_t = u_s$, $g_t g_s^{-1} g_D = g_t u_t u_t^{-1} g_s^{-1} g_D = ( g_t u_t ) ( g_s u_s )^{-1} g_D = h_t h_s^{-1} g_D \in H$.

Playing this same game with $v_t = v_s$, we get $g_t g_s^{-1} g_D = k_t k_s^{-1} g_D \in K$, and hence we have found our contradiction. $\square$

For a group $G$, recall that $Z ( G ) = \{ g \in G : g g' = g' g \text{ for all } g' \in G \}$ is the center of $G$.

Corollary. For any $\gamma \in \Gamma$ (where $\Gamma$ is $\delta$-hyperbolic) of infinite order, the subgroup generated by $\gamma$ is quasiconvex.  Equivalently, the map $c : \mathbb{Z} \to \text{Cay}_S ( \Gamma )$ sending $n \mapsto \gamma^n$ is a quasigeodesic (which is a sensible statement to make given that $\mathbb{Z}$ is quasi-isometric to $\mathbb{R}$).

Pf. By Theorem 9, we deduce that $C ( \gamma )$ is quasiconvex, and so is finitely generated.  Let $T$ be a finite generating set for $C ( \gamma )$.  Notice $Z ( C ( \gamma ) ) = \displaystyle \bigcap_{t\in T} C_{C(\gamma)} ( t )$

where $C_{C(\gamma)} ( t )$ is the centralizer in $C ( \gamma )$ of $t$, so $Z ( C ( \gamma ) )$ is quasiconvex by Theorem 10 and thus $Z ( C ( \gamma ) )$ is a finitely generated abelian group containing $\langle \gamma \rangle$.  By Exercise 16 (below), we deduce that $\langle \gamma \rangle$ is quasi-isometrically embedded in $Z ( C ( \gamma ) )$, and hence is quasiconvex in $\Gamma$. $\square$

Exercise 16. Cyclic subgroups of finitely generated abelian groups are quasi-isometrically embedded.

Lemma 10. Suppose the order of $\gamma \in \Gamma$ is infinite.  If $\gamma^p$ is conjugate to $\gamma^q$, then $| p | = | q |$.

Pf. Suppose $t \gamma^p t^{-1} = \gamma^q$.  An easy induction on $n$ shows $t^n \gamma^{(p^n)} t^{-n} = \gamma^{(q^n)}$.  Therefore, applying the triangle inequality, $\ell ( \gamma^{(q^n)} ) \leq \ell ( \gamma^{(p^n)} ) + 2 \ell ( t^n ) \leq | p |^n \ell ( \gamma ) + 2 n \ell ( t )$.

But $\langle \gamma \rangle \hookrightarrow \Gamma$ is a $( \lambda , \epsilon )$-quasi-isometric embedding, so $| p |^n \ell ( \gamma ) + 2 n \ell ( t ) \geq \ell ( \gamma^{(q^n)} ) \geq \lambda^{-1} | q |^n - \epsilon$.

This is impossible unless $| q | \leq | p |$ (eventually the exponential growth dominates).  Similarly, reversing the roles of $p$ and $q$ in the above argument implies that $| p | \leq | q |$, so $| p | = | q |$. $\square$