Can every covering space be obtained from a principal $G$-bundle with $G$ discrete? I was wondering if every covering space on a base space $X$ is the "underlying covering space" of some principal $G$-bundle on $X$ with $G$ a discrete group?
If we assume $X$ has some nice properties, e.g. admits a finite covering such that every intersection of open sets in the covering is either contractible or empty, then we can reduce the above to the following question:
Given a set $F$, and a finite collection $f_\alpha \in \mathrm{Bijections}(F)$, does there exist a group structure on $F$ such that each $f_\alpha$ equals left-multiplication by some $g_\alpha \in F$?
 A: The quick answer to your question is no, not every covering map is obtained from a principle $G$-bundle.
But there's an important key word here, namely a regular covering map, also known as a normal covering map. What's true is this:

For any covering map $p : E \to X$ such that $E,X$ are path connected and $X$ is locally path connected, $p$ comes from a principle $G$-bundle if and only if $p$ is a regular covering map.

Here's some details. Consider any covering map $p : E \to X$ satisfying the properties listed above. There is an associated deck transformation group, namely the group $\text{Deck}(p)$ of all homeomorphisms $d : E \to E$ such that $p \circ d = p$. The homeomorphisms $d \in \text{Deck}(p)$ are called the deck transformations of the covering map $p$. Note that for each $x \in X$ the group $\text{Deck}(p)$ acts on the fiber $p^{-1}(x)$. The covering map $p$ is regular if and only if the action of $\text{Deck}(p)$ on each fiber $p^{-1}(x)$ is transitive.
The relevant theorems you learn about covering spaces in a first year course on algebraic topology are the following. Let $p : E \to X$ be a covering map, where $X$ is path connected and locally path connected, and $E$ is path connected. Pick base points $e \in E$, $x \in X$ such that $p(e)=x$.

*

*The induced homomorphism $p_* : \pi_1(E,e) \to \pi_1(X,x)$ is injective.

*The covering map $p$ is regular if and only if the image subgroup $p_*(\pi_1(E,e)) < \pi_1(X,x)$ is normal in $\pi_1(X,x)$.

*Assuming that the covering map $p$ is regular, the deck transformation group is isomorphic to the quotient of $\pi_1(X,x)$ modulo the above normal subgroup:
$$\text{Deck}(p) = \pi_1(X,x) \, \bigm/ \, p_*(\pi_1(E,e))
$$
In this situation, letting $G = \text{Deck}(p)$, the covering map $p : E \to X$ is a principle $G$-bundle. The converse holds as well: if $p$ is a principle $G$-bundle then the covering map $p$ is regular and $G$ is isomorphic to $\text{Deck}(p)$. Proving these facts should be straightforward exercises once you've worked through the theory, in particular the proofs of the above theorems. One key underlying fact of the theory is a theorem saying that the action of $\text{Deck}(p)$ on $p^{-1}(x)$ is always a free action. So one can say that $p$ is regular if and only if each fiber $p^{-1}(x)$ is a torsor for the group $\text{Deck}(p)$.
This MSE answer gives examples of non-regular covering maps, hence ones that are not associated with a principle $G$-bundle.
