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

$g_if_{\ast}(\pi_1(C))=hg_jf_{\ast}(\pi_1(C))$

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. $\Box$