Definition. A group G splits freely if G acts on a tree T without global fixed point and such that every edge stailizer is trivial. If G does not split freely, then G is called freely indecomposable.

Example. \mathbb Z=\pi_1(S^1). Equivalently, \mathbb Z acts on \mathbb R without global fixed points. So \mathbb Z splits freely.

If G \ncong \mathbb Z but G splits freely, then G=G_1 \ast G_2 for G_1, G_2 neq 1.

Definition. The rank of G is the minimal r such that F_r surjects G.

It is clear that rank(G_1\ast G_2)\leq rank(G_1)+rank(G_2).

Grushko’s Lemma. Suppose \varphi:F_r \longrightarrow G is surjective and r is minimal. If G=G_1 \ast G_2, then F_r=F_1 \ast F_2 such that \varphi(F_i)=G_i for i=1,2.

Pf. Let X_i=K(G_i,1) (i=1,2) be simplicial and let \mathfrak{X} be a graph of spaces with vertex spaces X_1, X_2 and edge space a point. So G=\pi_1(X_{\mathfrak{X}}, x_0) where x_0=(*, \frac{1}{2}).

Let \Gamma be a graph so that \pi_1(\Gamma)\cong F_r and realize \varphi as a simplicial map f: \Gamma \longrightarrow X_{\mathfrak{X}}. Let y_0 \in f^{-1}(x_0). Because r is minimal, f^{-1}(x_0) is a forest, contained in \Gamma. The goal is to modify f by a homotopy to reduce the number of connected components of f^{-1}(x_0).

Let U \subseteq f^{-1}(x_0) be the component that contains y_0. Let V \subseteq f^{-1}(x_0) be some other component. Let \alpha a path in \Gamma from y_0 to V.

Look at f \circ \alpha \in \pi_1(X_{\mathfrak{X}}, x_0). Because \varphi is surjective, there exists \gamma\in \pi_1(\Gamma, y_0) such that f \circ \gamma = f \circ \alpha. Therefore if \beta= \gamma^{-1} \cdot \alpha, then f \circ \beta is null-homotopic in X_{\mathfrak{X}} and \beta gives a path from y_0 to V.

We can write \beta as a concaternation as \beta=\beta_1 \cdot \beta_2 \cdot \cdot \cdot \beta_n such that for each i, f \circ \beta_i \subseteq X_{\mathfrak{X}}\smallsetminus {x_0}. By the Normal Form Theorem, there exists i such that f\circ \beta_i is null-homotopic in X.

We can now modify f by a homotopy so that im (f\circ\beta_i)={x_0}. Therefore \beta_i \subseteq f^{-1}(x_0) and the number of components of f^{-1}(x_0) has gone down. By induction, we can choose f so that f^{-1}(x_0) is a tree. Now f factors through \Gamma'=\Gamma/ f^{-1}(x_0). Then F_r\cong \pi_1(\Gamma') and there is a unique vertex of \Gamma' that maps to x_0. So every simple loop in \Gamma' is either contained in X_1 or X_2 as required. square

An immediate consequence is that rank(G_1\ast G_2)=rank (G_1) + rank (G_2).

Grushko’s Theorem. Let G be finitely generated. Then G\cong G_1 \ast \cdots\ast G_m \ast F_r where each G_i is freely indecomposable and F_r is free. Furthermore, the integers m and r are unique and the G_i are unique up to conjugation and reordering.

Pf. Existence is an immediate corollary of the fact that rank is additive.

Suppose G=H_1\ast \cdots \ast H_n \ast F_s. Let \mathcal{G} be the graph of groups. Let T be the Bass-Serre tree of \mathcal{G}.

Consider the action of G_i on T. Because G_i is freely indecomposable, G_i stabilize a vertex of T. Therefore G_i is conjugate into some H_i.

Now consider the action of F_r on T. F_r\smallsetminus T is a graph of groups with underlying graph \Delta, say, and \pi_1(\Delta) is a free factor in F_r. But there is a covering map F_r\smallsetminus T \longrightarrow \mathcal{G} that induces a surjection \pi_1(\Delta) \longrightarrow F_s. Therefore, r\geq s. The other inequality can be obtained by switching F_r and F_s. \square

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.

Lemma 29: Suppose \{f_i':C_i\longrightarrow\Gamma^{H}\} is a finite set of infinite degree elvations and \Delta \subseteq \Gamma^{H} is compact. Then for all sufficiently large d>0, there exists an intermediate covering \Gamma_d such that

(a) \Delta embeds in \Gamma_d

(b) every f'_i descends to an elevation \hat{f_i}:\hat{C_i}\longrightarrow \Gamma_d of degree d

(c) the \hat{f_i} are pairwise distinct

Proof: We claim that the images of f_i' never share an infinite ray (a ray is an isometric embedding of [0,\infty)). Neither do two ends of the same elevation f_i'. Let’s claim by passing to the universal cover of \Gamma, a tree T.

For each i, lift f_i' to a map \tilde{f_i}:\mathbb{R}\longrightarrow T. If f'_i and f'_j share an infinite ray then there exists h\in H such that \tilde{f_i} and h\tilde{f_j} overlay in an infinite ray. The point is that \tilde{f_i}, \tilde{f_j} correspond to cosets g_if_{\ast}(\pi_1(C)) and g_jf_{\ast}(\pi_1(C)). But this implies that


This implies that Hg_if_{\ast}(\pi_1(C))=Hg_jf_{\ast}(\pi_1(C)). So f'_i=f'_j. A similar argument implies that the two ends of f'_i do not overlap in an infinite ray. This proves the claim.

Let \Gamma' be the core of \Gamma^{H}. Enlarging \Delta if necessary, we can assume that

(i) \Gamma'\subseteq\Delta;

(ii) \Delta is a connected subgraph;

(iii) for each i, for some x_i\in C'_i, f'_i(x_i)\in\Delta;

(iv) for each i, |im(f'_i)\cap \delta\Delta|=2.

For each i identifying C'_i with \mathbb{R} so that C is identified with \mathbb{R}/\mathbb{Z} and x_i is identified with 0. Let

\Delta_d=\Delta\cup(\bigcup_i f'_i([-d/2,d/2]))

For all sufficiently large d,

f'_i(\pm d/2)\notin\Delta

Now, the restriction of \Delta_d \longrightarrow \Delta factors through \Delta_d/\sim\longrightarrow\Gamma, where f'_i(d/2)\sim f'_i(-d/2). This is a finite-to-one immersion, so, by theorem 5, we can complete it to a finite-sheeted covering map as required. \Box

Theorem 19: D is LERF.

Recall the set-up from the previous lecture. We built a graph of spaces \mathscr{X} for D.

Proof: Let H\subset D be finitely generated. Let X_H be the corresponding covering space of X_{\mathscr{X}} and let \Delta\subseteq X_H be compact. Because H is finitely generated, there exists a subgraph of spaces X' such that \pi_1(X') =H. We can take X' large enough so that \Delta \subseteq X'. We can enlarge \Delta so that it contains every finite-degree edge space of X'. Also enlarge \Delta so that

\partial^{\pm}_{e'}(\Delta\times (\mathrm{interval}))\subseteq\Delta

for any e'\in E(\Xi'). For each v'\in V(\Xi') let \Delta_{v'}=\Delta\cap X_{v'} and let {f'_i}=\{ incident edge map of infinite degree \}.

Applying lemma 29 to \Gamma^H=X_{v'}, for some large d, set X_{\hat{v}}=\Gamma_d. (Here we use the fact that vertex groups of \mathscr{X}' are finitely generated)

Define \mathscr{X}^+ as follows:

\bullet \Xi^+=\Xi^-

\bullet For each v^+\in V(\Xi^+), the edge space is the X_{v^+} that the lemma produced from the corresponding v'.

Now, by construction, \bigcup_{v^+}X_{v^+} can be completed to a graph of spaces \mathscr{X}^+ so that the map

X_{\mathscr{X}'}\longrightarrow X_{\mathscr{X}}

factors through X_{\mathscr{X}'}\longrightarrow X_{\mathscr{X}} and \Delta embeds. Let \mathscr{X}^- be identical to \mathscr{X}^+ except with +’s and -‘s exchanged. Clearly \mathscr{X}^+\cup\mathscr{X}^- satisfies Stallings condition, as required. \Box