Let $\alpha : \mathbb{C}\cup\{\infty\} \to \mathbb{C}\cup\{\infty\}$ with $\alpha(x)=x+2$ and $\beta:\mathbb{C}\cup\{\infty\} \to \mathbb{C}\cup\{\infty\}$ with $ \beta(x)=x/(2x+1)$. Show that the group generated by $\alpha$ and $\beta$ is free on the set $\{\alpha , \beta\}$
Attempt: We know that $\alpha$ and $\beta$ are bijective. We know also that $\alpha^n\neq 1$ for all $n\in \mathbb{Z}-\{0\}$ and $ \beta^m(1/x)= 1/(x+2m)$ for all $m\in\mathbb{Z}$. And we need to show that $\alpha^{n_1}\beta^{n_2}\alpha^{n_3}\beta^{n_4}...\alpha^{n_{k-1}}\beta^{n_k}\neq1$ for all $n_1,n_2,...,n_k \in \mathbb{Z}$ with $k\in \mathbb{N}$. I was trying to apply by induction on $k$. But I didn't succeed it. Thanks for your helps!
The matrices corresponding to these maps are $\left(\begin{array}{cc}1&2\\0&1\end{array}\right)$ and $\left(\begin{array}{cc}1&0\\2&1\end{array}\right)$.
Proving that they generate a free group is a well-known application of the Ping-pong lemma: see the Wikipedia article, where this is given as an example.
So the two matrices generate a free subgroup of ${\rm GL}(2,{\mathbb Z}) < {\rm GL}(2,{\mathbb C})$. Since free groups have a trivial centre, the image of this subgroup in ${\rm PGL}(2,{\mathbb C})$, which is the given group, is also free.