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16. Free Subgroups of Hyperbolic Groups (from Sam Kim)

3 March 2009 in Course notes | Tags: Free groups, Hyperbolic groups | by elandes | 2 comments

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 $\langle v \rangle$ .
Hence 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.

And so we have Lemma 14.

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

Proof:

Recall 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 $d_{haus}({1,v,v^2,\dots, v^n},[1,v^n]) < R_1$

By 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 $\Pi_{\langle v\rangle}$ .
In $\mathbb{H}^2$ , there is some $m$ such that $\forall x\in \mathbb{H}^2$ either $l(\Pi_{}(x)) \leq m$ or $l(\Pi_{<v>}(x)) \leq m$ .

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

Proof:

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

Now we can prove the theorem.

Proof of Theorem 12:

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

Let $X_1 = \Pi_{}^{-1}(\lbrace u^m\mid l(u^m) > M\rbrace)$ 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 $l(\Pi_{}(x_2))\leq M$ . In particular, $X_1 \cap X_2 = \emptyset$ .

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

$\Pi_{}(u^m x_2)=u^m\Pi_{} (x_2)$

for any $m$ . In particular,

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

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 $\langle u^m, v^n \rangle \cong \mathbb{F}_2$ .

6. Marshall Hall’s Theorem

4 February 2009 in Course notes | Tags: Free groups, Residual finiteness, Separability | by katerman | Leave a comment

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:

diagram

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 $F_r$ .

4. Separability (from Sam Kim)

28 January 2009 in Course notes | Tags: Free groups, Linear groups, Separability | by jasoncallahan | 3 comments

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:

a and b have infinite orders.
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.

Separability

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.

G is Extended RF (ERF) if any subgroup of G is separable.
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 $\hat{X}$ .