Let $T$ be a $k$-regular tree for some $k>2$ and let $G=\text{Aut}(T)$ the group of automorphisms on the tree. I need to find a closed subgroup $H$ of $G$, so $H$ is a free group (and therefore discrete) that acts cocompactly on $T$ (and therefore by some thereom $H$ is a lattice).
The topology on $G$, if it helps, is generated from $$G(T_0)=\{g \in G: gx=x , \forall x\in T_0\}$$ when $T_0$ is some finite subtree. So $G$ is locally compact and totally disconnected.
One can construct a free subgroup $F_n$ of $G$ genereated by $n$ hyperbolic automorphisms of $T$ with pairwise disjoint axes, using the ping-pong lemma. But, who says this subgroup acts cocompactly on $T$?
So another approach is to find a finite subtree $K$ (which is compact using the discrete topology on the tree), and find some free subgroup $F_n$ so $K$ is a fundamental domain with the action of $F_n$ on $T$ (that is, $F_n K=T$), and therefore the action is cocompact (maybe some Cayley graph?).
Thanks!
Start with a finite connected graph $X$ of valence $k$, e.g. the complete graph on $k+1$ vertices. Let $Y\to X$ be the universal covering of $X$ and $G$ the group of covering transformations. Thus, $G$ is a free group acting freely on the tree $Y$ (of valence $k$) with the quotient space $X$. You can find details for instance in Hatcher's book "Algebraic topology", chapter 1.