January 24th
Today I learned the Ping-pong lemma: if a group $G$ generated by $a$ and $b$ acts on a set $X$ has subsets $X_1$ and $X_2$ which do not contain each other but $a^\bullet X_1\subseteq X_2$ and $b^\bullet X_2\subseteq X_1$ (for nonzero $\bullet$), then $G=\langle a,b\rangle$ is free. This is not terribly interesting. Essentially, we need to show that any nontrivial word $g\in G$ is not the identity with the given action; visually, we are watching $X_1$ or $X_2$ "bounce'' back and forth with each power of $a$ or $b.$
We can write any element $g\in G$ as $a^\bullet b^\bullet\cdots.$ Let's say that our element is\[g_k=a^{n_1}b^{n_2}\cdots b^{n_{k-1}}a^{n_k}\]for now; here $n_\bullet\ne0$ and $k$ is odd. In other words, $g$ alternates starting and ending with $a.$ We watch where this takes $X_1$ because that's all of the information we have now, so I guess we claim $g_kX_1$ by induction. If $k=1,$ this is the hypothesis. Otherwise, we note that\[g_{k+2}X_1=g_kb^{n_{k+1}}a^{n_{k+2}}X_1\subseteq g_kb^{n_{k+1}}X_2\subseteq g_kX_1,\]which is a subset of $X_2$ by the inductive hypothesis.
Now, having established the group action, we can quickly conclude that all of the $g_k$ aren't the identity. Indeed, this would imply\[X_1=g_kX_1\subseteq X_2,\]which is false by hypothesis.
The rest of the proof requires a trick that I had to look up. For any other element $g\in G,$ there aren't that many more forms for us to deal with, but they can all be reduced to the above by sufficient conjugation by $a.$ (Conjugation is the trick I missed.) For example, if $g$ looks like\[g=b^{n_1}a^{n_2}\cdots a^{n_{k-1}}b^{n_k}\]starting and ending with $b,$ then $aga^{-1}$ will start and end with $a$ and is therefore not the identity; it follows $g$ is also not the identity. Similarly, if $g$ looks like\[g=a^{n_1}b^{n_2}\cdots a^{n_{k-1}}b^{n_k}\]starting with $a$ and ending with $b,$ then $a^{n_1}ga^{-n_1}$ will start and end with $a$ and is therefore not the identity, so $g$ is again not the identity. Finally, if $g$ looks like\[g=b^{n_1}a^{n_2}\cdots b^{n_{k-1}}a^{n_k}\]starting with $b$ and ending with $a,$ then $g^{-1}$ starts with $a$ and ends with $b,$ so $g^{-1}$ and therefore $g$ are not the identity. This completes the proof that $G$ is freely generated by $a$ and $b.$
As an example, we show that the subgroup of $\op{SL}_2(\ZZ)$ generated by\[A=\begin{bmatrix}1&2\\0&1\end{bmatrix},\qquad\text{and}\qquad B=\begin{bmatrix}1&0\\2&1\end{bmatrix}\]is free. For this, we let $A$ and $B$ act on the upper-half plane $\mathcal H$ of $\CC$ in the usual way, and we let\[X_1=\{z\in\mathcal H:|z| \lt 1\},\qquad\text{and}\qquad X_2=\{z\in\mathcal H:|z| \gt 1\}.\]These sets are disjoint and therefore do not contain each other. It remains to check that $A^nX_1\subseteq X_2$ and $B^n\subseteq X_1$ for nonzero $n.$ To start, we can inductively show that\[A^n=\begin{bmatrix}1&2n\\0&1\end{bmatrix},\qquad\text{and}\qquad B=\begin{bmatrix}1&0\\2n&1\end{bmatrix}.\]So for $z\in X_1,$ we see\[A^nz=\frac{z+2n}1=z+2n.\]Namely, $A$ shifts $z$ right by $2n,$ and because the unit disk has diameter $2,$ every element in $X_1$ is displaced outside into $X_2.$ On the other hand, for $z\in X_2,$ we see\[B^nz=\frac z{2nz+1}=\frac1{2n+1/z}.\]Because $z\in X_2$ has magnitude larger than $1,$ $1/z$ has magnitude smaller than $1$ and will live in the unit disk. So by the same shifting argument with $A,$ $2n+1/z$ is back in $X_2,$ but then $B^nz$ forces us into $X_1$ again. This is the last condition we had to check, so we are done here.