# Regular covering implies transitive automorphism group action

Theorem

Let $$p: (Y,y) \rightarrow (X,x)$$ be a covering, with $$Y$$ connected and $$X$$ locally path connected, and let $$p(y) = x$$. If $$p_*(\pi_1(Y,y))$$ is a normal subgroup of $$\pi_1(X,x)$$, then $$\pi_1(X,x)/p_*(\pi_1(Y,y))$$ is isomorphic to Aut(Y/X).

Question

As the situation above, T.F.A.E

(a) the covering is normal (i.e., $$p_*(\pi_1(Y,y))$$ is a normal subgroup of $$\pi_1(X,x)$$);

(b) the action of $$Aut(X/Y)$$ on $$p^{-1}(x)$$ is transitive;

(c) for every loop $$\sigma$$ at $$x$$, if one lifting of $$\sigma$$ is closed (i.e., still a loop at some $$y \in p^{-1}(x)$$), then all liftings are closed.

My approach

I already proved (b) $$\implies$$ (c) and (c) $$\implies$$ (a). So I need to prove (a) $$\implies$$ (b), and I know I need to use the theorem above.

So, is that true that for any $$y$$ and $$y' \in p^{-1}(x)$$, the there exists $$[\sigma] \in \pi_1(X,x)$$ such that $$y' = [\sigma]y$$? (If this is true, then we are done!)

Thank you very much!

Yes, your property is true : $\pi_1(X,x)$ acts transitively on $p^{-1}(x)$ :
First, assume that $Y$ is path connected : let $y,y'\in p^{-1}(x)$, you can find a path $\gamma:y\rightsquigarrow y'$. Then notice that $\gamma$ is the unique lift of $p\circ\gamma:x\rightsquigarrow x$ with starting point y. So $y'=\gamma(1)=y\cdot[p\circ\gamma]$.
Now, notice that $p$ is a local homeomorphism, so $Y$ is connected and locally path connected (since $X$ is), so $Y$ is path connected.