Group isomorphism for deck transformation in covering space.

When reading Lee's book, I encountered the following exercise:

Let $\mathcal{P}\colon M\rightarrow G\backslash M$ be the covering arising from a free and proper discrete group action of $G$ on $M$ and suppose $M$ is connected. Let $\Gamma_G:=\{l_g\in \text{Diff}(M):g\in G\}$, then $G$ is isomorphic to $\Gamma_G$ by the obvious map $g\mapsto l_g$ and furthermore $\Gamma_G= \text{Deck}(\mathcal{P})$

I have two questions:

(1) Why do we need $M$ to be connected? I think the conclusion is obvious and does not involve connectedness of $M$.

(2) There is no assumption that the discrete group action is smooth, so isn't $\Gamma_G$ the empty set? Or we can deduce smoothness of the action from the connectedness of $M$?

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There is an implicit assumption that the group acts smoothly. – Mariano Suárez-Alvarez Oct 23 '12 at 5:29

Suppose $M=\mathbb R\sqcup\mathbb R\sqcup\mathbb R$ is the disjoint union of three copies of $\mathbb R$, and that $G$ is a cyclic group of order $3$ acting which permutes the copies transitively. Then $M/G=\mathbb R$. Can you see what the group of covering transformations $M\to M/G$ is?
It is $S^3$? So $\Gamma_G$ and $\text{Deck}(\mathcal{P})$ are not isomorphic.. But I can't come up with an rigorous argument why connectness is important here, any hint?. – hxhxhx88 Oct 23 '12 at 5:58
Sorry, I found my previous thought was wrong. Consider following condition: let $M$ be $S^1$, and $G$ be 3-order cyclic group of rotation 1, $2\pi/3,4\pi/3$. Then Diff($M$) is $S_3$, which is not isomorphic to $G$...but $M$ is connected. any problem? – hxhxhx88 Oct 23 '12 at 6:35