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Theorem 12 (Gromov): Let \Gamma be torsion-free \delta-hyperbolic group.  If u,v \in\Gamma such that uv\neq vu, then for all sufficiently large m,n, \langle u^m,v^n\rangle \cong F_2.

Remark: The torsion-free hypothesis is not necessary, but it allows us to avoid some technicalities.  For instance, it is a non-obvious fact that an infinite hyperbolic group contains a copy of \mathbb{Z}.

For the rest of this lecture \Gamma will be a torsion-free \delta-hyperbolic group, uv\neq vu where u,v are primitive (i.e. not proper powers).

Recall that for \Gamma torsion-free \delta-hyperbolic, u primitive implies that \langle u \rangle = C(u)= C(u^m).

If u and v do not commute we can show there is some point u^p on \langle u \rangle arbitrarily far from \langle v \rangle .
pic11Hence we have the following lemma.

Lemma 13: d_{haus}(\langle u \rangle , \langle v \rangle )=\infty

If u and v do not commute there is some point u^p on \langle u\rangle arbitrarily far from \langle v\rangle .

Proof: Suppose not. That means \exists R_0 > 0 such that \forall u^p \in \langle u \rangle \exists v^q \in \langle v \rangle such that  d(u^p,v^q) = d(1,u^{-p}v^q) < R_0.  So u^{-p}v^q is in B(1,R_0).  But the Cayley graph is locally finite so B(1,R_0) has finitely many elements.  By the Pigeonhole Principle \exists p\neq r such that u^{-p}v^q=u^{-r}v^s for some q, s.  Then \langle u \rangle = C(u) =C(u^{p-r})=C(v^{q-s})=C(v)=\langle v \rangle .  But then uv=vu. \Rightarrow\Leftarrow .

For a moment view \langle u \rangle and \langle v \rangle as the horizontal and vertical geodesics in \mathbb{H}.  For two points x on \langle u \rangle and y on \langle v \rangle , we can argue that the geodesic between them curves toward the origin.

pic2And so we have Lemma 14.

Lemma 14: There exists R > 0 such that \forall m,n, [u^m,v^n]\cap B(1,R)\neq \emptyset .


pic3Recall that \phi : \mathbb{Z}\to \Gamma by \phi (i)= u^i is a quasi-isometric embedding.  So by Theorem 6, d_{haus}({1,u,u^2,\dots, u^m},[1,u^m]) < R_1 and d_{haus}({1,v,v^2,\dots, v^n},[1,v^n]) < R_1

pic4By Lemma 13 choose u^p \in \langle u \rangle such that
d(u^p,\langle v \rangle) > 2R_1 + \delta .  Choose u_p \in [1,u^m] such that d(u_p,u^p) < R_1 .  Now, u_p must be \delta-close to [u^m,v^n] so for some point x on the geodesic between v^n and u^m, d(u_p, x) < \delta .  Then d(1,[u^m,v^n])\leq d(1,u^p) + d(u^p,u_p) + d(u_p, x) \leq l(u^p)+R_1 + \delta \Box

For a subgroup H \subseteq \Gamma, one can choose a closest point projection \Pi_H : \Gamma \to H which is H-equivariant. (Write \Gamma = \cup H{g_i}.  Choose \Pi_H(g_i)=h_i where h_i and g_i are close and declare \Pi_H to be H-equivariant.)  \Pi_{H} is typically not a group homomorphism.

We’re interested in \Pi_{\langle u\rangle} and \Pi_{\langle v\rangle}.
pic5In \mathbb{H}^2, there is some m such that \forall x\in \mathbb{H}^2 either l(\Pi_{<u>}(x)) \leq m or l(\Pi_{<v>}(x)) \leq m.


Lemma 15: \exists M such that \forall x\in Cay(\Gamma), l(\Pi_{<u>}(x)) \leq M or l(\Pi_{<v>}(x)) \leq M .


Let y\in[\Pi_{<u>}(x), \Pi_{<v>}(x)]\cap B(1,R).  WLOG, y is \delta-close to p\in[x,\Pi_{<u>}(x)] and d(1, \Pi_{<u>}(x) \leq d(1,p)+d(p,\Pi_{<u>}(x)) \leq d(1,p) +d(p,1) since \Pi_{<u>}(x) is the closest point to x (in particular compared to u^0=1).  So d(1,\Pi_{<u>}(x)) \leq 2d(1,p)\leq 2(R+\delta)\Box .

Now we can prove the theorem.

Proof of Theorem 12:

The idea is to use the Ping-Pong Lemma on the Cayley graph.

pic8Let X_1 = \Pi_{<u>}^{-1}(\lbrace u^m\mid l(u^m) > M\rbrace) and let X_2= \Pi_{<v>}^{-1}(\lbrace v^n\mid l(v^n) > M\rbrace), where M is provided by Lemma 15.  For all x_1\in X_1 we have l(\Pi_{<v>}(x_1))\leq M and likewise for all x_2\in X_2 we have l(\Pi_{<u>}(x_2))\leq M.  In particular,  X_1 \cap X_2 = \emptyset.

Let x_2\in X_2.  By \langle u\rangle-equivariance,

\Pi_{<u>}(u^m x_2)=u^m\Pi_{<u>} (x_2)

for any m.  In particular,

l(\Pi_{<u>}(u^m x_2))\geq l(u^m)-l(\Pi_{<u>}(x_2))\geq l(u^m)-M

by the triangle inequality.  Similarly,

l(\Pi_{<v>}(v^n x_1))\geq l(v^n)-l(\Pi_{<v>}(x_1))\geq l(v^n)-M

for all x_1\in X_1 and all n.  Because \langle u\rangle and \langle v\rangle are quasi-isometrically embedded, it follows that u^mX_2 \subset X_1 and v^n X_1\subset X_2 for m,n >>0.

Therefore, by the Ping-Pong Lemma \langle u^m, v^n \rangle \cong \mathbb{F}_2.

Fact: There exists a finitely generated non-Hopf group.  (An example is the Baumslag-Solitar group B(2,3), although we cannot prove it yet.)  So, by Lemma 5, there is a finitely generated non-residually finite group.  Thus, free groups are not ERF: if G is a finitely generated non-residually finite group, then Lemma 4 implies that the kernel of a surjection F_r \rightarrow G is not separable in F_r.  However, finitely generated subgroups of free groups are separable:

Marshall Hall’s Theorem (1949): F_r is LERF.

This proof is associated with Stallings.

Proof: As usual, let F_r = \pi_1(X) where X is a rose. Let X' \rightarrow X be a covering map with \pi_1 (X') finitely generated.  Let \Delta \subset X' be compact. We need to embed \Delta in an intermediate finite-sheeted covering.

Enlarging \Delta if necessary, we may assume that \Delta is connected and that \pi_1 (\Delta) = \pi_1(X').  Note that we have \Delta \subset X' \rightarrow X. By Theorem 5 (see below), the immersion \Delta \rightarrow X extends to a covering \hat{X} \rightarrow X. Then \pi_1 (X') = \pi_1(\Delta) \subset \pi_1 (\hat{X}). So X' \rightarrow X lifts to a map X' \rightarrow \hat{X}.

The main tool in the proof above is this:

Theorem 5: The immersion \Delta \rightarrow X can be completed to a finite-sheeted covering \hat{X} \rightarrow X into which \Delta embeds:


Proof: Color and orient the edges of X. Any combinatorial map of graphs \Delta \rightarrow X corresponds uniquely to a coloring and orientation on the edges of \Delta.  A combinatorial map is an immersion if and only if at every vertex of \Delta, we see each color arriving at most once and leaving at most once. Likewise, it’s a covering map if and only if at each vertex, we see each color arriving exactly once and leaving exactly once.

Completing $latex \Delta$ to a finite-sheeted covering $latex \hat{X} \rightarrow X$.

Let k be the number of vertices of \Delta.  For each color c, let k_c be the number of edges of \Delta colored c. Then there are k-k_c vertices of \Delta missing “arriving” edges colored c, and there are k-k_c vertices of \Delta missing “leaving” edges colored c.  Choose any bijection between these two sets and use this to glue in k-k_c edges colored c. When this is done for all colors, the resulting map \Delta \subset \hat{X} \rightarrow X is clearly a covering.

Note that the proof in fact gives us more.  For instance:

Exercise 6: If H is a finitely generated subgroup of F_r, then H is a free factor of a finite-index subgroup of F_r.

Exercise 7 (Greenberg’s Theorem): If H\triangleleft F_r and H is finitely generated, then H is of finite index in F_r.

Ping-Pong Lemma

Question. Let G be a group and a, b \in G.  When is \langle a, b \rangle \cong F_2?

Ping-Pong Lemma: Let G be a group acting on a set X and a, b \in G. Assume:

  1. a and b have infinite orders.
  2. There exist X_1, X_2 \subseteq X such that X_2 \nsubseteq X_1 and a^m X_2\subseteq X_1, b^m X_1 \subseteq X_2 for all m \neq 0.

Then \langle a, b \rangle \cong F_2.

Proof. Consider \varphi : F_2 = \langle x, y \rangle \twoheadrightarrow \langle a,b \rangle \leq G such that \varphi (x) = a and \varphi (y) = b.  Choose a reduced word w for a nontrivial element in F_2, i.e., either w = x^{m_1} y^{n_1} x^{m_2} \cdots or w = y^{m_1} x^{n_1} y^{m_2} \cdots and m_i, n_i \neq 0.

Case 1: w = x^{m_1} y^{n_1} \cdots x^{m_k}.  Then \varphi (w) X_2 = a^{m_1} b^{n_1} \cdots a^{m_k} X_2 and a^{m_k} X_2 \subseteq X_1, so \varphi (w) X_2 \subseteq a^{m_1} b^{n_1} \cdots b^{n_{k-1}} X_1 and b^{n_{k-1}} X_1 \subseteq X_2, and so on until \varphi (w) X_2 \subseteq a^{m_1} X_2 \subseteq X_1, so \varphi (w) X_2 \neq X_2.  Therefore \varphi (w) \neq 1.

Case 2: w = y^{m_1} x^{n_1} \cdots y^{m_k}.  Then, by Case 1, \varphi (xwx^{-1}) \neq 1, so \varphi (x) \varphi (w) \varphi (x^{-1}) \neq 1.  Therefore \varphi (w) \neq 1.

Case 3: w = x^{m_1} y^{n_1} \cdots y^{n_k}.  (similar to above)

Case 4: w = y^{m_1} x^{n_1} \cdots x^{n_k}.  (similar to above)

Free Groups Are Linear

Theorem 3. F_2 is linear.

Proof. \mathrm{SL}_2\mathbb{R} acts on \mathbb{R}^2 by linear transformations.  Let a = \left(\begin{array}{cc} 1&2 \\ 0&1 \\ \end{array}\right) and b = \left(\begin{array}{cc} 1&0 \\ 2&1 \\ \end{array}\right).  Then a^m = \left(\begin{array}{cc} 1&2m \\ 0&1 \\ \end{array}\right) and b^m = \left(\begin{array}{cc} 1&0 \\ 2m&1 \\ \end{array}\right). Let X_1 = \left\{\left(\begin{array}{c} x \\ y \\ \end{array}\right) \mid |x| > |y|\right\} and X_2 = \left\{\left(\begin{array}{c} x \\ y \\ \end{array}\right) \mid |x| < |y|\right\}.  Then a^m X_2 \subseteq X_1, b^m X_1 \subseteq X_2 for all m \neq 0, so \langle a, b \rangle \cong F_2 by the Ping-Pong Lemma.

Corollary 1. Finitely generated free groups are linear, hence residually finite.

Proof. The case of \mathbb{Z} is obvious; otherwise, this follows from F_n \hookrightarrow F_2 as proved in Exercise 1.


Definition. Let G be a group.  The profinite topology on G is the coarsest topology such that every homomorphism from G to a finite group (equipped with the discrete topology) is continuous.

Definition. A subgroup H is separable in G if H is closed in the profinite topology of G.

Exercise 3. Let G be a group. X \subseteq G is separable if and only if for all g \in G \smallsetminus X, there exists a homomorphism to a  finite group q : G \rightarrow Q such that q(g) \not \in q(X).  Note that if X = {1}, this is equivalent to: G is RF if and only if {1} is separable.

Hint. For the “if” direction, let g \in G and consider q^{-1} \circ q(g).  For the other direction, use the definition of subbase and that p : G \rightarrow P, q: G \rightarrow Q \Rightarrow \text{ker}(p \times q) = \text{ker}(p) \cap \text{ker}(q).

Definition. Let G be a group.

  1. G is Extended RF (ERF) if any subgroup of G is separable.
  2. G is Locally ERF (LERF, subgroup separable) if any finitely generated subgroup is separable.

Lemma 3. Let G be a group. A subgroup H of G is separable if and only if for all g \in G \smallsetminus H, there exists a finite-index subgroup K \leq G such that H \leq K \leq G and g \not \in K.

Proof. In the “only if” direction, by the previous exercise, for all g \in G \smallsetminus H, there exists a homomorphism to a  finite group q : G \rightarrow Q such that q(g) \not \in q(H). Then g \not \in  q^{-1} \circ q(H) = K.  Conversely, let g \in G \smallsetminus H. By hypothesis, there exists a finite-index subgroup K \leq G such that H \leq K \leq G and g \not \in K.  Let L = \cap_{h \in G} K^h. Note that this is a finite number of intersections (|G/N_G (K)|, to be precise). There exists a finite quotient q : G \rightarrow G/L.  Then  q^{-1} \circ q(H) = H \cdot L.  Therefore, g \not \in  q^{-1} \circ q(H), i.e., q(g) \not \in q(H), and the lemma follows by the previous exercise.

Scott’s Criterion (1978). Let X be a Hausdorff topological space and G = \pi_1 (X,x).  Let X' \rightarrow X be a covering and H = \pi_1 (X',x'). Then H is separable in G if and only if for any compact \Delta \subseteq X', there exists and intermediate finite-sheeted cover \hat{X} \rightarrow X such that X' \rightarrow \hat{X} embeds \Delta into \hat{X}.

Exercise 4. Let H \leq G be a separable subgroup.

  1. If G' \leq G, then G' \cap H is separable in G’.
  2. If G \leq G' has finite index, then H is separable in G’.