Covering spaces of wedge sum of circles

Let's suppose I'd like to characterise all connected covering spaces of wedge sum of circles.When it comes to a torus, a sphere or $S^{1} \vee S^{2}$ it is easy to point out exactly how do all covers look like.All I know is how to construct universal covering of $S^{1} \vee S^{1}$ and that every subgroup $H< Z * Z$ is a free group. So is it a non-trivial thing to do ( involving graphs etc.)or is there some quite easy way to enhance my results? Thank you for all your answers.

• Basically, any group with $2$ generators is the group of some cover of $S^1\vee S^1$, so yes it's bound to be a little complicated. – Captain Lama Apr 11 '16 at 20:23

In general, the universal cover of $X$ the wedge product of $p$ copies of $S^1$ is the $p$-regular tree. That is the connected graph without cycle such that the cardinal of the arcs through a node is $p$. The covering of $X$ has $p$-regular graphs.
• I haven't done any calculations, but that should be the $(p+1)$-regular tree, I think :) – Eric Stucky Dec 4 '16 at 13:55
As usual, the connected covers of $X=S^1\vee S^1$ are in bijective correspondance with the subgroups of $\pi_1(X) = F_2$ (the free group with two elements), the universal cover being a tree $Y$ which is the Cayley graph of $F_2$. Of course those subgroups are (very) complicated, for instance $F_\infty$ (the free group on a countable set) is such a subgroup.
Suppose you are interested in the Galois (or normal) covers of $X$ : they correspond to normal subgroups $K\subset F_2$. Then write $G = F_2/K$, equipped with its canonical generating elements coming from $F_2$. The Cayley graph $\Gamma_G$ of $G$ with respect to this generating set is then the corresponding cover of $X$.