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)<k 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.