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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