# A free semigroup

In the appendix here, we make the claim that the semigroup

$\Gamma_A=\langle \begin{pmatrix}a & 1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix}b & 1 \\ 1 & 0 \end{pmatrix}\mid 1\leq a,b\leq A\rangle$

is free on the given generators.

Here, I will give a proof. It is a simpler version of the ping-pong argument commonly used to prove that a group is free on a given set of generators.

First a simple observation. it suffices to prove that the set

$\left\{ \begin{pmatrix}a & 1 \\ 1 & 0 \end{pmatrix}\mid 1\leq a\leq A\right \}$

generates a free semigroup.

Now, consider the action of our semigroup on the interval $(1,\infty)$ by fractional linear transformations:

$\begin{pmatrix}a & b \\ c & d \end{pmatrix}\cdot x=\frac{ax+b}{cx+d}$

In particular, note that

$\begin{pmatrix}a & 1 \\ 1 & 0 \end{pmatrix}\cdot x=a+\frac{1}{x}.$

Now suppose that

$\begin{pmatrix}y_1 & 1 \\ 1 & 0 \end{pmatrix}\cdots\begin{pmatrix}y_m & 1 \\ 1 & 0 \end{pmatrix}=\begin{pmatrix}z_1 & 1 \\ 1 & 0 \end{pmatrix}\cdots\begin{pmatrix}z_n & 1 \\ 1 & 0 \end{pmatrix}$

Apply both sides to some $x\in(1,\infty)$. The LHS lies in $(y_1,y_1+1)$ and the RHS lies in $(z_1,z_1+1)$. Therefore $y_1=z_1$ and the rest is an easy induction.