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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

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As before, are \chi' and \chi are graphs of spaces equipped with maps \Phi \colon X_{\chi'} \to X_{\chi}, \varXi' \to \varXi, \phi_{v'} \colon X_{v'} \to X_v, and \phi_{e'} \colon X_{e'} \to X_e such that

fig_28_11

commutes.

Lemma 28: Suppose that every edge map of \chi' is an elevation.  Then the map \Phi is \pi_1-injective.

Proof: The idea is to add extra vertex spaces to \chi' so that \chi' satisfies Stalling’s condition.  As before, we have inclusions:fig_28_2

If \chi' does not satisfy Stalling’s condition then one of these maps is not surjective.  Without loss of generality, suppose thatfig_28_31

does not arise as an edge map of \chi'.  Suppose \partial_{e}^{-} \colon X_e \to X_u.  Then (\partial_{e}^{-} \circ \phi_{e'})_{*}(\pi_1(X_{e'})) is a subgroup of \pi_1(X_u).  Let X_{u'} be the corresponding covering space of X_u.  Since \pi_1(X_{u'}) contains (\partial_{e}^{-} \circ \phi_{e'})_{*}(\pi_1(X_{e'})), there is a lift \partial_{e'} \colon X_{e'} \to X_{u'} of \partial_{e'} \circ \phi_{e'}.  Now replace \chi' by \chi' \cup X_{u'} and repeat.  After infinitely many repetitions, the result \hat{\chi} satisfies the hypothesis of Stalling’s condition, and \chi' is contained in a subgraph of spaces.

Theorem 18: If G_{+} and G_{-} are residually finite groups then G_{+} * G_{-} is also residually finite.

Proof: Let X_{\pm} = K(G_{\pm}, 1), let e = * be a point, and fix maps \partial^{\pm} \colon e \to X_{\pm}.  This defines a graph of spaces \chi.  By the Seifert-van Kampen Theorem, \pi_1(X_{\chi}) \cong G_{+} * G_{-}.  Let \tilde{\chi} be the graph of spaces structure on the universal cover of X_{\chi}, and let \Delta \subseteq X_{\tilde{\chi}} be a compact subset.  We may assume that \Delta is connected; we may also assume that if \tilde{e} \in \tilde{\varXi} (Bass-Serre tree) and \Delta \cap \tilde{e} \ne \emptyset, then \tilde{e} \subseteq \Delta.  Let \eta \colon X_{\tilde{\chi}} \to \tilde{\varXi} be the map to the underlying map of the Bass-Serre tree.  Then \eta(\Delta) is a finite connected subgraph, \varXi'.  For each v' \subseteq v(\varXi'), let \Delta_{v'} = \Delta \cap X_{v'}, a compact subspace of X_{v'}.  Because G_{\pm} are residually finite, we have a diagramfig_28_6

where X_{\hat{v}} \to X_{\pm} is a finite-sheeted covering map and \Delta_{v'} embeds in X_{\hat{v}}.  Let \hat{\chi} be defined as follows.  Set \hat{\varXi} = \varXi'; for a vertex \hat{v} \in \hat{\varXi}, the vertex space is the X_{\hat{v}} corresponding to v'.  For each \hat{e} \in E(\hat{\varXi}) corresponding to e' \in E(\varXi'), define \partial_{\hat{e}}^{\pm} so that the diagramfig_28_81

commutes.  Now sum.

Exercise 26: If G_1 and G_2 are residually finite and H is finite, prove that G_1 *_H G_2 is residually finite.

Recall that our goal is to determine, for a given map of graphs of spaces such as the one shown below, whether the map \Phi can be extended to a covering map \widehat{\Phi}.

Figure 1

Figure 1

Let \mathcal{X}, \mathcal{X}' be graphs of spaces equipped with maps \Xi' \to \Xi, \phi_{v'} : X_{v'} \to X_v and \phi_{e'} : X_{e'} \to X_e as before.  Recall that in Stallings’ proof of Hall’s Theorem, we completed an immersion to a covering map by gluing in edges.  We will aim to do the same thing with graphs of spaces.

Definition: Let \mathcal{X} be a graph of spaces, and let \eta : X_{\mathcal{X}} \to \Xi be the map to the underlying graph.  If \Delta \subseteq \Xi is a subgraph, then \eta^{-1}(\Delta) \subseteq X_{\mathcal{X}} has a graph-of-spaces structure \mathcal{Y} with underlying graph \Delta.  Call \mathcal{Y} a subgraph of spaces of \mathcal{X}.

We’re seeking a condition on \mathcal{X}' such that \mathcal{X}' is realized as a subgraph of spaces of some \widehat{\mathcal{X}} with a covering map \widehat{\Phi} : X_{\widehat{\mathcal{X}}} \to X_{\mathcal{X}} such that the following diagram commutes:

diagram1Definition: For each edge map \partial_e^{\pm} : X_e \to X_v of \mathcal{X}, and each v' \mapsto v a vertex of \mathcal{X}', let

\mathcal{E}^{\pm}(e) = \bigcup_{v' \mapsto v} \mathcal{E}^{\pm}(e, v').

For each possible degree \mathcal{D}, let \mathcal{E}_{\mathcal{D}}^{\pm}(e) \subseteq \mathcal{E}^{\pm}(e) be the set of elevations of degree \mathcal{D}.  We will say \mathcal{X}' satisfies Stallings’ condition if and only if the following two things hold:

(a) Every edge map of \mathcal{X}' is an elevation of the appropriate edge map of \mathcal{X}.
(b) For each e \in E(\Xi) and \mathcal{D}, there is a bijection \mathcal{E}_{\mathcal{D}}^+(e) \leftrightarrow \mathcal{E}_{\mathcal{D}}^-(e).

So in Figure 1, the graph of spaces \mathcal{X}' is something you might be able to turn into a covering.  In the picture, \mathcal{E}_{\mathcal{D}}^+(e) is represented by the blue circles, and \mathcal{E}_{\mathcal{D}}^-(e) is represented by the green circles.  Observe that the blue circles are in bijection with the green circles.

Corollary: \mathcal{X}' satisfies Stallings’ condition if and only if \mathcal{X}' can be realized as a subgraph of spaces of some \widehat{\mathcal{X}} such that

(a) V(\Xi ') = V(\widehat{\Xi}), and
(b) there is a covering map \widehat{\Phi} : X_{\widehat{\mathcal{X}}} \to X_{\mathcal{X}} such that the following diagram commutes:

diagram1

Proof of Corollary. First we’ll show that if \Phi can be extended to a covering map as described above, then \mathcal{X}' satisfies Stallings’ condition.  By Theorem 17, every edge map of \widehat{\mathcal{X}} is an elevation.  So there are inclusions

diagram2

Furthermore, these maps are surjective, and clearly degree-preserving.

Now assume that \mathcal{X}' satisfies Stallings’ condition.  Then we build \widehat{\mathcal{X}} as follows.  Let V(\widehat{\Xi}) = V(\Xi ').  As above, we have degree-preserving inclusions (this time, not surjections)

diagram3

Extend these inclusions to bijections \mathcal{E}_{\mathcal{D}}^+(e) \leftrightarrow \mathcal{E}_{\mathcal{D}}^-(e).  Now we set E(\widehat{\Xi}) = \bigcup_{e \in E(\Xi)} \mathcal{E}^+(e).  Each of these \widehat{e} is an elevation

diagram4

This defines an edge space X_{\widehat{e}} and an edge map \partial_{\widehat{e}}^+.  Consider the corresponding elevation in \bigcup_{e \in E(\Xi)} \mathcal{E}^-(e):

diagram5

Because \partial_{\widehat{e}}^+ and \partial_{\widehat{e}}^- are of the same degree, we have a covering transformation X_{\widehat{e}} ' \to X_{\widehat{e}}.  So we can identify them, and use \partial_e^- as the other edge map.  By construction, \widehat{\mathcal{X}} satisfies the conditions of Theorem 17, so there is a suitable covering map \widehat{\Phi} : X_{\widehat{\mathcal{X}}} \to X_{\mathcal{X}}.

Exercise 25: (This will be easier later, but we have the tools necessary to do this now.)  Prove that if G_1 and G_2 are LERF groups, then so is G_1 * G_2.