# Group action and covering spaces

Let $X$ be a path-connected and locally path-connected topological space. The action of a topolgical group $G$ on $X$ is a covering space action. For any subgroup $H < G$, we have a composition of covering space $X \rightarrow X/H \rightarrow X/G$.

1. Prove that any covering space of $X/G$ between $X$ and $X/G$ is isomorphic to $X/H$, for a subgroup $H$ of $G$.
2. $X/H_1 \rightarrow X/G$ and $X/H_2 \rightarrow X/G$ are isomorphic as covering spaces if and only if $H_1$ and $H_2$ are conjugate subgroups of $G$.
3. The covering space $X/H \rightarrow X/G$ is normal if and only if $H$ is a normal subgroup of $G$, in which case there is an automorphism $$\text{Aut} (X/H \rightarrow X/G) \cong G/H.$$

I have no idea as to (1). The sufficiency parts of (2) and (3) seems to be easy. As to the necessity part of (2), supposet that $\phi: X/H_1 \rightarrow X/H_2$ is an isomorphism of $X$-covering spaces, I can prove that for any $x \in X$, there exists $g_x \in G$, such that $\phi(x) = g_x(x)$. But can I find a universal $g$ such that $\phi(x)=g(x)$ for all $x \in X$? Similar problem occurs in the proof of (3). For the isomorphism part of (3), I can construct a map $\psi: G \rightarrow \text{Aut}(X/H \rightarrow X/G)$, which sends $g \in G$ to an automorphism of the covering space $X/H$ over $X/G$ sending $x \mapsto g(x)$. The kernel of this map is $H$, but I don't know how to show it is surjective.

Thank you very much.

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If you are still interested in this, you can indeed show that $\phi$ is given by multiplication by a universal $g$. To see this, show that you can lift $\phi$ to a deck transformation. – Andrew Hanlon Jan 6 at 4:25

We consider $X \overset{p}\longrightarrow C \overset{q}\longrightarrow X/G$. We may assume that $p$ and $q$ are surjective. Consider $\text{Aut}(p) \leqslant \text{Aut}(qp)\simeq G$ (the canonical isomorphism given by the hypothesis on $X$). Then we have $X \twoheadrightarrow X/\text{Aut}(p)$ respects the relation of $X \twoheadrightarrow C$, so as quotients, they are isomorphic, which gives a desired isomorphism with $H = \text{Aut}(p)$.
Comment. I am not so sure if this is enough since I have not found $H$ as a subgroup of $G$ but an isomorphic copy of a subgroup.