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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’.

Residual finiteness

Let G be a group and g\in G with g\neq 1. Then we call G residually finite (hereafter RF) if there exists a subgroup K\subset G of finite index such that g\notin K. In other words, for every nontrivial element of G, there exists a finite index subgroup that does not contain that particular element.

Example. Finite groups are RF, since the trivial subgroup has finite index and does not contain any of the nontrivial elements of G.

Metaquestion. How general is the class of RF groups? In particular, which finitely generated/finitely presented groups are RF?

Remarks. Assume that G is finitely generated.

(i) The definition can be strengthened so that we may assume that our finite index subgroup K\subset G is normal in G. Indeed, if G is finitely generated, then there are only finitely many subgroups of a given fixed index k. (See the exercise from the second lecture.) If g_0\neq 1, and K\subset G is a subgroup of finite index that does not contain g_0, then

\textrm{core}(K)=\bigcap_{g\in G}gKg^{-1}

is a subgroup of finite index in G which excludes g_0, since gKg^{-1} and K have the same index in G for all g\in G, so this intersection is really the intersection of finitely many subgroups of finite index. Thus \textrm{core}(K) also has finite index in G.

(ii) Equivalently, G is RF if and only if for each g\in G not the identity, there exists a homomorphism \phi:G\to A, where A is a finite group, for which \phi(g) is not the identity in A. Indeed, if G is RF and K_0 is normal subgroup provided by (i), then g does not die under the natural homomorphism from G to G/K_0. Conversely, given such a homomorphism, the kernel of \phi is a subgroup of finite index that does not contain g.

(iii) Also, G is RF if and only if

\bigcap_{K\subset G,\ [G:K]<\infty}K=\{1\}.

That is, the intersection of all the subgroups of finite index in G is the trivial subgroup. Were some nonidentity element g to be contained in this intersection, then it would be contained in each subgroup of finite index in G, so this element prevents G from being RF. Conversely, if this intersection is trivial, each nonidentity element of G must be excluded from some finite index subgroup, so G is RF.

(iv) If G is RF and g_1,\dots,g_n\in G, then there exists a finite index subgroup K\subset G with g_j\notin K for all 1\leq j\leq n. Here, just take the intersection of the finite index subgroup associated with each g_j. This is again a finite index subgroup of G.

Lemma 2: Let G be a finitely generated group.

(i) If G is RF and H\subset G is a subgroup, then H is RF.

(ii) If H is RF and H\subset G with H finite index in G, then G is RF.

That is, RF passes to all subgroups and also to supergroups of finite index.

Proof. For (i), choose h\in H. Considered as an element of G, there exists a homomorphism \phi to a finite group A for which \phi(h) is not the identity. The restriction of \phi to H is also a homomorphism to a finite group for which the image of h is nontrivial. The kernel of this restricted homomorphism is a subgroup of finite index in H that does not contain h. (Note that we did not assume that H is finitely generated.)

For (ii), choose g\in G. If g\notin H, then H is a finite index subgroup not containing g, and we are done. If g\in H, then there exists a finite index subgroup K\subset H that does not contain g. However, K is also a finite index subgroup of G, so we are done. \square

Topological reformulation of RF

Now, we would like to connect RF with a topological property of a space with fundamental group G. Let M be a compact manifold with universal covering \widetilde{M} and G=\pi_1(M). Accordingly, we assume throughout that G is finitely generated. (We will see later that the manifold condition can be relaxed.)

Theorem 2: The group G is RF if and only if the following condition holds: for every compact subset C\subset\widetilde{M} there exists a finite sheeted covering M_C\to M for which C embeds homeomorphically in M_C.

Proof. Assume the topological condition holds and choose any g\in G not the identity. This corresponds to a loop (based at a point dependent only upon our choice of universal covering) in M which we also denote by g. To this loop, there also is a corresponding lift to a connected arc a in \widetilde{M} with distinct endpoints x and y.

Let C be the compact set \{x,y\}. Then, there exists a finite sheeted covering M_C\to M for which x and y are distinct points of M_C. By the lifting property of covering spaces, this implies that if K_C\subset G is the subgroup of finite index corresponding to the covering M_C\to M, then g\notin K_C. Therefore G is RF.

Conversely, suppose that G is RF and choose any subset C\subset\widetilde{M}. Then, since G acts freely and properly discontinuous on \widetilde{M}, the set T_C of those g\in G (not the identity) for which g C intersects C nontrivially is finite. Now, choose K_C\subset G of finite index containing none of the elements of T_C. If M_C is the corresponding finite sheeted covering, we have that hC\cap C=\emptyset for all h\in\pi_1(M_C). That is, C embeds homeomorphically in M_C. \square

Remark. We never really used here that M was a manifold, only that its fundamental group acted properly discontinuous on the universal covering. (The action need not be free either, since this would only add another finite number of elements to our set T_C.) Thus, it suffices to assume that M is Hausdorff and locally compact.


(1) Finitely generated abelian groups are RF. (Exercise.)

(2) Selberg’s Lemma (Malcev-Selberg): If G is a finitely generated linear group, that is, G\subset\mathrm{GL}_N(\mathbb{C}) for some N, then G is RF.

Proof. We begin with the case G\subset\mathrm{GL}_N(\mathbb{Z}). Since RF passes to arbitrary subgroups, it suffices to prove that \mathrm{GL}_N(\mathbb{Z}) is RF.

Choose any g\neq 1. This means that the matrix g-1 has some nonzero entry, say x. Since x is an integer, we can find some large prime p that does not divide x. Now, consider the homomorphism


given by reducing the entries of a given matrix modulo p. This is a homomorphism because matrix multiplication is linear in the entries. (Exercise – make this precise.) Since \mathbb{Z}/p\mathbb{Z} is a finite field, its general linear group is a finite group, i.e. the kernel K_p of \phi_p is a finite index subgroup of \mathrm{GL}_N(\mathbb{Z}), and it does not contain g by construction. Thus \mathrm{GL}_N(\mathbb{Z}) is RF, and thus any subgroup thereof is also RF.

More generally, if G\subset\mathrm{GL}_N(\mathbb{C}) is finitely generated, we can arrange that G\subset\mathrm{GL}_N(R), where R\subset\mathbb{C} is the subring generated by the entries of a finite collection of generators for G. In particular, this is an integral domain, since it is a finitely generated subring of \mathbb{C}. Thus, given x a nonzero element of g-1 as above, we can find a prime ideal \mathcal{P} of R that does not divide x, i.e. x\notin\mathcal{P}.

Now, R/\mathcal{P} is a finite integral domain, that is, a finite field, and we can build a reduction homomorphism \phi_{\mathcal{P}} analogous to the situation over the integers. The the image of the reduction homomorphism is the general linear group of a finite field, so it is a finite group. We now proceed exactly as above.  \square

This theorem is often referred to as Selberg’s Lemma, even though Malcev supposedly proved it first.

(3) Free groups F_r of finite rank are linear, so they are RF. Notice that we have seen, in the exercise from the first lecture, that F_r is a subgroup of F_2 for all r\geq 0, so it suffices to prove that F_2 is linear. One such representation the subgroup of \mathrm{SL}_2(\mathbb{Z}) generated by the two matrices

a=\left(\begin{matrix} 1 & 2 \\ 0 & 1\end{matrix}\right)

b=\left(\begin{matrix} 1 & 0 \\ 2 & 1\end{matrix}\right).

There are several ways to prove that this group is free of rank two. First, one use a bit of hyperbolic geometry and the action of \mathrm{SL}_2(\mathbb{R}) on the hyperbolic plane to prove that this group is the fundamental group of a hyperbolic manifold that is deformation equivalent to the rose with two petals. Also, one can use the so-called Ping-Pong Lemma, which says that a finite collection of homeomorphisms of a space that satisfy a certain set of conditions necessarily generate a free subgroup of the homeomorphism group.

We begin with a lemma that among other things demonstrates that most free groups are non-trivial

Lemma 1: If T\subset S are sets then the map \iota: F_T\to F_S, induced by inclusion has a left inverse. Consequently, \iota is injective.

Proof: Let X_T and X_S be the roses with petals indexed by T and S respectively. In particular we have that F_T=\pi_1(X_T) and F_S=\pi(X_S). If we view X_T as a subspace of X_S in the obvious way then there exists p:X_S\to X_T such that p|_{X_T}=Id (One such map is obtained by crushing the petals in X_S\smallsetminus X_T). This map p induces a function p_\ast :\pi_1(X_S)\to \pi_1(X_T) and this map can easily be seen to be the desired left inverse. QED

The next thing that we would like to do is examine exactly what do elements of free groups “look like.” Informally, elements look like words in the elements of S where only the obvious cancellations by inverses are allowed. More formally, we will identify the elements of S with their images under the map \iota: S\to F_S (This is not the same \iota from Lemma 1). Next we consider products of the form w=s_1^{\epsilon_1}s_2^{\epsilon_2}\ldots s_k^{\epsilon_k}, where s_i\in S and \epsilon_i=\pm 1. Next, we prove a theorem that formalizes the previous intuition about free groups

Normal Form Theorem: Given a free group F_S the following 2 statements are true

  1. Every element of F_S is of the form w.
  2. If w=1 then w is either the empty word or there exists 1\leq j\leq k-1 such that s_j=s_{j+1} and \epsilon_j=-\epsilon_{j+1}

Proof: We will prove the theorem when S=\{a,b\}, but the proof generalizes to sets of all cardinality. Let X_S be the rose with 2 petals and wedge point v and so we have that F_S=\pi_1(X_S,v). Next, let T_S be the universal cover of X_S based at t_0. Because F_S acts freely on T_S and X_S has only one vertex we see that the vertices of T_S are in bijection with the elements of F_S. We next observe 2 facts. First, from lemma 1 we see that since T_S is simply connected that it must be a tree. Second, since each vertex of T_S looks locally like v we see that every vertex of T_S is 4-valent. From covering space theory we know that if g\in \pi_1(X_S,v) and t is a vertex of T_S then the lift of g to T_S that starts at t ends at g\cdot t.

We proceed by induction on d(t_0,g\cdot t_0), which is the minimum number of edges between t_0 and g\cdot t_0.  Suppose that d(t_0,g\cdot t_0)=k. Since T_S is a tree with all 4-valent vertices we see that there exists h\in \pi_1(X_S,v) such that d(t_0,h\cdot t_0)<k and d(h\cdot t_0, g\cdot t_0)=1. Next, observe that there exists c\in \{a,b,a^{-1},b^{-1}\} such that when c is lifted at g\cdot t_0 ends at h\cdot t_0. Therefore we see that cg\cdot t_0=h\cdot t_0 and since the group action is free this implies that h=ag which proves 1.

For the second proposition, suppose that w=1 and that the second proposition is not true. From covering space theory we know that the lift of w at v is a loop based at v. First, since w is not the empty word we see that it will not lift to the constant path. Finally, since w contains no obvious cancellations we see that T_S must contain a loop based at v that is not the constant loop, which contradicts the fact that T_S is a tree. Thus 2 must hold. QED

As a consequence of the previous theorem we see that it is possible to solve the word problem of whether it is algorithmically possible to decide if a group element is the identity.

Definition: A group G is finitely generated if and only if there exists a surjection from F_k to G for some finite k.

Exercise 2:

  1. Show that if r<\infty then F_r has finitely many subgroups of index d<\infty.
  2. Deduce that the same holds for any finitely generated group G.