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 $F_s$. $\square$