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Last time, we used the following lemma without justification, so let’s prove it now.

Lemma 30. Let \mathcal{G} be a graph of groups with \Gamma finite and G = \pi_1 \mathcal{G} finitely generated. If G_e is finitely generated for every edge e \in E ( \Gamma ) , then G_v is finitely generated for every v \in V ( \Gamma ) .

This is not completely trivial: it is certainly possible for finitely generated group to have subgroups that are not finitely generated. Indeed, recall the proof that every countable group embeds in the free group on three generators.

Pf. Let S be a finite generating set for G , and for each e \in E ( \Gamma ) let S_e be a finite generating set for the edge group G_e . By the Normal Form Theorem, every g \in S can be written in the form

g = g_0 t_1^{\pm 1} g_1 \cdots t_n^{\pm 1} g_n

where each t_i is a stable letter and each g_i \in G_{v_i} for some v_i \in V ( \Gamma ) . For a fixed v \in V ( \Gamma ) , let

\displaystyle S_v = \bigcup_{g\in S} \{ g_i : g_i \in G_v \} \cup \bigcup_{e \text{ adjoining } v} \partial_e^{\pm} ( S_e ) ,

where for each e \in E ( \Gamma ) adjoining v the plus or minus is chosen so that \partial_e^\pm : G_e to G_v . It is clear then that S_v is contained in G_v . To see that S_v is finite, note that since S is finite, the first union is finite; and since \Gamma is finite there can be only finitely many edges adjoining a given vertex, so the second union is finite. Hence it remains to prove that S_v generates G_v .

Let \gamma \in G_v . Because S generates G , we have

\gamma = \gamma_1 \cdots \gamma_m

where \gamma_j \in S . Each \gamma_j has a normal form as above, so we get an expression of the form

\gamma = g_0 t_1^{\pm 1} g_1 t_n^{\pm 1} g_n , or 1 = \gamma^{-1} g_0 t_1^{\pm 1} g_1 t_n^{\pm 1} g_n .

By the Normal Form Theorem, the expression on the right can then be simplified. After the simplification process, we have no stable letters left, and every g_i is either contained in G_v or is a product of elements of the incident edge groups, and in both cases lie in S_v . \square

Remember that Theorem 19 said D is LERF, answering our question (b). But in fact, we get more from the proof of Theorem 19.

Definition. Recall that H \subset G is a retract if the inclusion H \hookrightarrow G has a left inverse \rho : G \to H . Similarly, we call H a virtual retract if H is a retract of a finite index subgroup of G .

For instance, Marshall Hall’s Theorem implies that every finitely generated subgroup of a free group is a virtual retract.

Theorem 20. Every finitely generated subgroup of D is a virtual retract.

Pf. Consider the setup of the proof of Theorem 19. We start with a subgroup H = \pi_1 X_{\mathfrak{X}'} and end up with a finitely sheeted covering space X_{\widehat{\mathfrak{X}}} \to X_{\mathfrak{X}'} . The graph of spaces \widehat{\mathfrak{X}} is built using the “obvious” bijection between elevations to \mathfrak{X}^+ and elevations to \mathfrak{X}^- . Thus the identification X_{\mathfrak{X}^+} \to X_{\mathfrak{X}^-} extends to a topological retraction X_{\widehat{\mathfrak{X}}} \to X_{\mathfrak{X}^+} . Now, we shall build a map \pi_1 X_{\mathfrak{X}^+} \to \pi_1 X_{\mathfrak{X}'} = H . We build it at each vertex space, one at a time. From the proof of Lemma 29, we see that the core of each X_{v'} is a topological retract of the corresponding X_{v^+} . Furthermore, we can choose the retraction so that for each long loop of degree d that we added is mapped to a null-homotopic loop in X_{v'} . This allows you to piece together the map X_{v^+} \to X_{v'} into a retraction \pi_1 X_{\mathfrak{X}^+} \to \pi_1 X_{\mathfrak{X}'} = H . \square

Exercise 14 asserted that a (virtual) retract is quasi-isometrically embedded, and so Theorem 20 has the following corollary:

Corollary. Every finitely generated subgroup of D is quasi-convex.

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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