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.

Figure 1

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 .

fig2

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 .

fig3

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

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