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 $G$.