Every countable group acts properly discontinuous on a 4 regular graph I asked before a similar question but this one is more relevant to me.
Let $G$ be a group. We say that an action of $G$ on a topological space $X$ is properly discontinuous if for every $x \in X$ there exist a neighborhood $U$ of $x$ such that $\forall g\in G, g\ne1: (g.U)\cap U = \emptyset $.
Now I ask the following question: Can any countable group act properly discontinuous on a (possibly infinite) 4-regular graph?
I have no direction, the only 4 regular graph in topology that I know of is the universal cover of $\Bbb S^1\lor \Bbb S^1$.
 A: Enumerate the elements of $G$ as $G=\{g_i \mid i \in \mathbb Z\}$. Since $G$ is obviously generated by its elements, there is some presentation of the form
$$G = \langle g_i \mid r_j \rangle
$$
Now construct a 2 dimensional CW complex $X$ with fundamental group $G$.
The ordinary construction produces a 1-skeleton $X^{(1)}$ which is a rose graph having a single vertex, and having one edge for each generator. If we did that with the group $G$ and its generating set $\{g_i\}$ then we would get a vertex of countably infinite valence.
Instead, here's a closely related construction which produces a $1$-skeleton $X^{(1)}$ that is a 4-regular graph. The 0-skeleton $X^{(0)}$, i.e. the vertex set, is $X^{(0)} = \{x_i \mid i \in \mathbb Z\}$. The 1-skeleton $X^{(1)}$ is obtained from $X^{(0)}$ by attaching edges as follows. First, for each $i \in \mathbb Z$ attach an edge $A_i$ with endpoints $x_{i-1}$, $x_i$. The union $\bigcup_i A_i$ is a graph isomorphic to the real line with vertex set $\mathbb Z$. To complete the description of $X^{(1)}$, attach another edge $B_i$ for each $i \in \mathbb Z$, with both endpoints attached to $x_i$. Clearly the subgraph $\bigcup_i A_i$ is the maximal tree of $X^{(1)}$, and it follows that the fundamental group of $X^{(1)}$ is clearly a countable rank free group, whose generators are represented by the paths $\gamma_i = \alpha_i B_i \bar \alpha_i$ in one-to-one correspondence with the set $\{g_i\}$, where $\alpha_i$ is the path in $\bigcup_i A_i$ from $x_0$ to $x_i$.
Now attach 2-cells to $X^{(1)}$ in one-to-one correspondence with the relators $r_j$, in the usual fashion, to obtain the CW complex $X$.
The group $G$ acts properly discontinuously on the universal cover $\widetilde X$, and by restriction $G$ acts properly discontinuously on the 1-skeleton $\widetilde X^{(1)}$. The universal covering map $\widetilde X \mapsto X$ restricts to a covering map $\widetilde X^{(1)} \mapsto X^{(1)}$, and it follows that $\widetilde X^{(1)}$ is a 4-regular graph.
