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Last time we saw that if $G$ acts on a tree, $T$ then $G/T$ has the structure of a graph of groups (remeber stabilizers). Such a $\mathcal{G}=G/\Gamma$ is called developable.

Exercise 18: Show that $SL_2\mathbb{Z}\cong(\mathbb{Z}/4\mathbb{Z})\ast_{\mathbb{Z}/2\mathbb{Z}}(\mathbb{Z}/6\mathbb{Z})$.

Theorem 13 (Scott-Wall): Let $\mathscr{X}$ be a graph of spaces. The universal cover of $X_{\mathscr{X}}$ is itself a graph of spaces. Also, each vertex space (resp. edge space) is the universal cover of a vertex space of $\mathscr{X}$ (.edge space).

Sketch of Proof: For any $v\in V(\Xi)$, let $L_v=X_v\cup(\cup_{e\in E}X_e\times [\pm 1,0])$, where $E$ is the set of edges that are the edges incident to $v$. It should be noted that $X_v$ is a deformation retract of $L_v$. Also, recall that the edge maps are $\pi_1$ injective. From covering space theory we know that given a map $f:Y\to X$ and a covering space $\hat X$ of $X$ that $f$ lifts to a map $\hat f:Y\to \hat X$ if and only if $f_\ast (\pi_1(Y))\subset \pi_1(\hat X)$. It therefore follows that $\tilde X_v \hookrightarrow \tilde L_v$. So $\tilde L_v$ is built from $\tilde X_v$ by attaching covering spaces of edge spaces. Because $\partial_{\pm}$ is $\pi_1$ injective, these covering spaces of edge spaces really are universal covers. By iteratively gluing together copies of the $\tilde L_v$ we can construct a simply connected cover of $X_{\mathscr{X}}$.

Given $X_{\mathscr{X}}$ we have constructed the universal cover $\tilde X_{\mathscr{X}}$. The underlying graph, $T$ of $\tilde X_{\mathscr{X}}$ is a tree because $\pi_1(\tilde X_{\mathscr{X}})\to\pi_1(T)$ is a surjection. We now need to check that the action of $G=\pi_1(X_{\mathscr{X}})$ on $T$ is interesting.

Let $\mathcal{G}$ be a graph of groups and let $G=\pi_1(\mathcal{G})$. Fix a base point $\ast\in X_v \subset X_{\mathcal{G}}$  and a choice of lift $\tilde\ast\in \tilde X_{\tilde v}\subset \tilde X_{\mathcal{G}}$. Let $g\in G_v=\pi_1(X_v)$.  The space $\tilde X_{\tilde v}$ is a universal cover of $X_v$ by Theorem 13, and so the lift of $g$ to the universal cover of $X_{\mathcal{G}}$ at $\tilde\ast$ is contained in $\tilde X_{\tilde v}$. Therefore the preimages of $\ast$ that are contained in $\tilde X_{\tilde v}$ correspond to the elements of $G_v$.

Now consider $g\in G\smallsetminus G_v$.  If $g$ is lifted at $\tilde \ast$ then the terminus of this lift is not in $\tilde X_{\tilde v}$, but in some other component of the preimage of $X_v$.  Call the component where the lift terminates $\tilde X_{\tilde v_1}$. If $g,h\in G$ are such that both have lifts that terminate in $\tilde X_{\tilde v_1}$ then $h^{-1}g\in G_v$. We have just proved the following lemma.

Lemma 16: Let $\mathcal{G}$ be a graph of groups and let $T$ be the underlying graph of the universal cover $\tilde X_{\mathcal{G}}$. For any vertex $v\in V(\Gamma)$ the set of vertices of $T$ lying above $v$ is in bijection with $G/G_v$ and $G$ acts by left translation.

We can also prove the following two lemmas in a similar fashion.

Lemma 17: For any $e\in E(\Gamma)$ the set of edges of $T$ that lie above $e$ is in bijection with $G/G_e$ and $G$ acts by left translation.

Lemma 18: If $e\in E(\Gamma)$ adjoins a vertex $v\in V(\Gamma)$ then for any $\tilde v\in V(T)$ lying above $v$ the set of edges of $T$ adjoining $\tilde v$ lying above $e$ is in bijection wiht $G_v/G_e$ and $G_v$ acts by left translation.

We begin with a lemma that among other things demonstrates that most free groups are non-trivial

Lemma 1: If $T\subset S$ are sets then the map $\iota: F_T\to F_S$, induced by inclusion has a left inverse. Consequently, $\iota$ is injective.

Proof: Let $X_T$ and $X_S$ be the roses with petals indexed by $T$ and $S$ respectively. In particular we have that $F_T=\pi_1(X_T)$ and $F_S=\pi(X_S)$. If we view $X_T$ as a subspace of $X_S$ in the obvious way then there exists $p:X_S\to X_T$ such that $p|_{X_T}=Id$ (One such map is obtained by crushing the petals in $X_S\smallsetminus X_T$). This map $p$ induces a function $p_\ast :\pi_1(X_S)\to \pi_1(X_T)$ and this map can easily be seen to be the desired left inverse. QED

The next thing that we would like to do is examine exactly what do elements of free groups “look like.” Informally, elements look like words in the elements of $S$ where only the obvious cancellations by inverses are allowed. More formally, we will identify the elements of $S$ with their images under the map $\iota: S\to F_S$ (This is not the same $\iota$ from Lemma 1). Next we consider products of the form $w=s_1^{\epsilon_1}s_2^{\epsilon_2}\ldots s_k^{\epsilon_k}$, where $s_i\in S$ and $\epsilon_i=\pm 1$. Next, we prove a theorem that formalizes the previous intuition about free groups

Normal Form Theorem: Given a free group $F_S$ the following 2 statements are true

1. Every element of $F_S$ is of the form $w$.
2. If $w=1$ then $w$ is either the empty word or there exists $1\leq j\leq k-1$ such that $s_j=s_{j+1}$ and $\epsilon_j=-\epsilon_{j+1}$

Proof: We will prove the theorem when $S=\{a,b\}$, but the proof generalizes to sets of all cardinality. Let $X_S$ be the rose with 2 petals and wedge point $v$ and so we have that $F_S=\pi_1(X_S,v)$. Next, let $T_S$ be the universal cover of $X_S$ based at $t_0$. Because $F_S$ acts freely on $T_S$ and $X_S$ has only one vertex we see that the vertices of $T_S$ are in bijection with the elements of $F_S$. We next observe 2 facts. First, from lemma 1 we see that since $T_S$ is simply connected that it must be a tree. Second, since each vertex of $T_S$ looks locally like $v$ we see that every vertex of $T_S$ is 4-valent. From covering space theory we know that if $g\in \pi_1(X_S,v)$ and $t$ is a vertex of $T_S$ then the lift of $g$ to $T_S$ that starts at $t$ ends at $g\cdot t$.

We proceed by induction on $d(t_0,g\cdot t_0)$, which is the minimum number of edges between $t_0$ and $g\cdot t_0$.  Suppose that $d(t_0,g\cdot t_0)=k$. Since $T_S$ is a tree with all 4-valent vertices we see that there exists $h\in \pi_1(X_S,v)$ such that $d(t_0,h\cdot t_0) and $d(h\cdot t_0, g\cdot t_0)=1$. Next, observe that there exists $c\in \{a,b,a^{-1},b^{-1}\}$ such that when $c$ is lifted at $g\cdot t_0$ ends at $h\cdot t_0$. Therefore we see that $cg\cdot t_0=h\cdot t_0$ and since the group action is free this implies that $h=ag$ which proves 1.

For the second proposition, suppose that $w=1$ and that the second proposition is not true. From covering space theory we know that the lift of $w$ at $v$ is a loop based at $v$. First, since $w$ is not the empty word we see that it will not lift to the constant path. Finally, since $w$ contains no obvious cancellations we see that $T_S$ must contain a loop based at $v$ that is not the constant loop, which contradicts the fact that $T_S$ is a tree. Thus 2 must hold. QED

As a consequence of the previous theorem we see that it is possible to solve the word problem of whether it is algorithmically possible to decide if a group element is the identity.

Definition: A group $G$ is finitely generated if and only if there exists a surjection from $F_k$ to $G$ for some finite $k$.

Exercise 2:

1. Show that if $r<\infty$ then $F_r$ has finitely many subgroups of index $d<\infty$.
2. Deduce that the same holds for any finitely generated group $G$.