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Normal form theorem for graphs of groups. Let \mathcal{G} be a graph of groups and G=\pi_1(G).

  1. Any g \in G can be written as
    g = g_0 t_{e_1}^{\epsilon_1} g_1 \cdots t_{e_n}^{\epsilon_n} g_n\quad
    as before.
  2. If g = 1, this expression includes `backtracking’, meaning that for some i, e_i = e_{i + 1} with \epsilon_i = - \epsilon_{i + 1}, and furthermore that if \epsilon_i = \pm 1, then g_i \in \partial_\pm ( G_{e_i} ).

Similar to the case of the free group, the proof boils down to the fact that the Bass–Serre tree T is a tree.

Proof. To simplify notation, set t_i = t_{e_i}^{\epsilon_i}, so

g = g_0 t_1 g_1 \cdots t_n g_n.

Fix base points *_i in the vertex spaces X_{v_i}, which are chosen to coincide when the vertices do. Then g_i is a loop in X_{v_i} based at *_i, and t_i is a path, crossing the corresponding edge space, from *_i to *_{i + 1}. This allows us to consider g as a loop in X_{\mathcal{G}} based at *_0. (We may assume v_0 = v_n by adding letters from a maximal tree.)

Consider the universal covering \widetilde{X}_{\mathcal{G}} and fix a base point \widetilde{*}_0 over *_0 in \widetilde{X}_{\widetilde{v}_0}. Let \widetilde{g} be the lift of g based at \widetilde{*}_0 and \gamma its image in the Bass–Serre tree T. We now analyze \widetilde{g} and \gamma closely.

Choose \widetilde{e}_i adjoining \widetilde{X}_{\widetilde{v}_0} and \widetilde{X}_{\widetilde{v}_1} so that the edge traversed by t_i when lifted at \widetilde{*}_{i - 1} corresponds to the coset 1 \cdot G_{e_i} \subseteq G_{v_i} / G_{e_i}.

Then g_0 lifts to a path in \widetilde{X}_{\widetilde{v}_0} which terminates at g_0 \widetilde{*}_0. Similarly, t_1 lifts at *_0 to a path across the edge \widetilde{e}_1 to the vertex space t_1\widetilde{X}_{\widetilde{v}_1} terminating at t_1 \widetilde{*}_1. Therefore, g_0 t_1 lifts at \widetilde{*}_0 to a path which crosses the edge space g_0 \widetilde{e}_1 and ends at g_0 t_1 \widetilde{*}_1.

Then, g_1 lifts at \widetilde{*}_1 to a path in \widetilde{X}_{\widetilde{v}_1} ending at \widetilde{*}_1, and t_2 lifts at \widetilde{*}_1 to a path across the edge \widetilde{e}_2 into the vertex space \widetilde{X}_{\widetilde{v}_2}, and terminating at t_2 \widetilde{*}_2. Thus g_0 t_1 g_1 t_2 lifts at \widetilde{*}_0 to a path which crosses g_0 \widetilde{e}_1, through g_0 \widetilde{X}_{\widetilde{v}_1}, across g_0 t_1 g_1 \widetilde{e}_2, and ending at

g_o t_1 g_1 t_2 \widetilde{*}_2 \in g_0 t_1 g_1 t_2 \widetilde{X}_{\widetilde{v}_2}.


We continue this process until we have explicitly constructed \widetilde{g}. By hypothesis, g = 1, so \widetilde{g} and \gamma are both loops in \widetilde{X}_{\mathcal{G}} and T, respectively. Since T is a tree, \gamma must backtrack.


This implies that \widetilde{e}_i=\widetilde{e}_{i + 1} and that \epsilon_i = -\epsilon_{i + 1}. That is, by Lemma 18,

g_{i + 1} \in \partial_\pm^{e_i} ( G_{e_i} ).

Therefore, we have found a backtracking, and can accordingly shorten g. This proves the theorem. \square

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.