For those who have the book, this question is regarding p181 in Thurston's "Three Dimensional Geometry and Topology" (although I will do my best to summarize it). Basically, there's an entire paragraph where all of the implications make no sense to me.
We have a connected, simply connected manifold $X$. $G$ is a Lie group that acts transitively by diffeomorphisms on $X$, and has compact point stabilizers. We also require that $G$ is maximal among all Lie groups satisfying these conditions. The goal is to enumerate the $(G,X)$ pairs that serve as geometries for a 3-manifold, in the sense that there is a manifold with charts in $X$ such that the compatibility maps are local restrictions of elements of $G$.
The problem paragraph is as follows:
In... we will first look at the connected component of the identity of $G$ - call it $G'$. The action of $G'$ is still transitive, and the stabilizers $G_x'$ of points $x \in X$ are connected. This is because the quotients $G_x' / (G_x')_0$, where $(G_x')_0$ is the component of the identity of $G_x'$, form a covering space of $X$. Since $X$ is simply connected, the covering is trivial.
Therefore $G_x'$ is a connected closed subgroup of $SO(3)$.
I assume the main purpose of restricting to the component of the identity is to cut out orientation reversing components (as in the case of $G = O(3)$). However, I have no idea what he means by the $G_x' / (G_x')_0$ being a covering of $X$. In the example of $X = \mathbb{R}^3$ and $G$ the group of Euclidean isometries of 3 space, wouldn't we have $G'_x = (G'_x)_0 = SO(3)$, so the quotient would be trivial? So this would be saying that trivial points are covering spaces of $\mathbb{R}^3$? Even in cases where it wasn't trivial, how does it make sense for quotients of pieces of $G$ to be covering spaces for $X$ in a canonical way?
I'd greatly appreciate the help of anybody who can translate the paragraph. Not really sure what tags to use here, since it's so specific.
Here is my reading of this text:
Consider the topological space $$ \tilde X= \bigsqcup_{x\in X} G'_x/(G'_x)_0. $$ This space has the natural topology coming from the fact that each $G'_x$ is a subset (in fact, a subgroup) of a topological group $G$.
Edit: The natural topology I had in mind here is coming from the weakest topology on $\sqcup_{x\in X} G'_x$ in which the map from this space to $G'$ is continuous. Then pass to the quotient topology on $\tilde X$. But it is then unclear why this topology on $\tilde X$ is Hausdorff. See the edit below for a clean argument.
There is a natural map $$ p: \tilde X\to X $$ sending each $G'_x/(G'_x)_0$ to $x$. It is clear that this map is continuous; one then has to check that this map is also a covering map. One also needs to check that $\tilde X$ is connected (this follows from connectedness of $G'$). Since $X$ is simply-connected it then follows that $p$ is a homeomorphism.
Edit: Here is a cleaner and complete argument. Pick a base-point $z\in X$. Then the natural map $$ G'/G'_z\to X $$ induced by the orbit map $g\to gz$, is a diffeomorphism. Since $G'$ acts transitively on $X$, all subgroups $(G'_x)_0$ are conjugate to $(G'_z)_0$; moreover, $K=(G'_z)_0$ is a normal subgroup in $G'_z$ with finite quotient group $F=G'_z/(G'_z)_0$. Now, consider the space $\tilde X= G'/(G'_z)_0$. The group $F$ acts on this quotient via right multiplication, for $f\in F$ define
$$ f\cdot gK =gKf= gfK. $$ We obtain diffeomorphisms $$ \tilde X/F\cong G'/G'_z\cong X. $$ One then sees that the action of the finite group $F$ is free, hence, the quotient map $p: \tilde X\to \tilde X/F$ is a covering map. The space $\tilde X$ is connected since $G'$ is. Therefore, $p$ is a covering map. Since $X$ is simply-connected, $p$ is a diffeomorphism. Therefore, $F=1$ and $G'_z$ is connected.
Edit 2. In fact, one can eliminate the entire covering argument, by appealing to the long exact sequence of homotopy groups/sets of the fibration $$ G_x\to G\to X $$ which immediately shows that simple connectivity of $X$ and connectivity of $G$ imply connectivity of $G_x$.