Connection between two groups related to covering How the the monodromy group and deck transformation group of a covering $p\colon Y\rightarrow X$ are related in the cases when covering is branched and covering is unbranched?
 A: I'm referring to http://en.wikipedia.org/wiki/Monodromy for the definition of monodromy group.
It seems to me, that the two coincides in the case of unbranchend Galois covering.
Let $p:\widetilde X \to X$ be a covering map, $p$ is Galois if $p_*(\pi_1(\widetilde X, \widetilde x_0))$ is a normal subgroup of $\pi_1(X,x_0)$. In this case we have the following description of the deck transformations:
$$
\mbox{Aut}(p) = \pi(X, x_0)/p_*(\pi_1(\widetilde X, \widetilde x_0))
$$
To prove it, let the fundamental group act on every fiber. This way you obtain a map $\Phi_{[\gamma]}:\widetilde X \to \widetilde X$ for every $[\gamma] \in \pi_1(X,x_0)$, which turns out to be an automorphism of the covering.
This construction defines a surjective homomorphism $\pi(X, x_0) \to \mbox{Aut}(p)$ whose kernel is exactly $p_*(\pi_1(\widetilde X, \widetilde x_0))$. According to the wikipedia article, the same quotient is isomorph to the monodromy group.
Maybe you can also prove it directly. The group of deck transformation preserves fibers, therefore you have a restriction homomorphism $\mbox{Aut}(p) \to \mbox{Mon}(p)$. You should be able to prove that this morphism is an isomorphism.
