Finite cover, homotopy group $\pi_1$ and group extension Can one show the following two statements are true or false:


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*Given a Lie group $L$, the group extension of $L$ to $G$ by its normal subgroup $N$, then we say $G/N=L$. Can one show that finding such a $G$ for a finite $N$ for a given $L$ is equivalent to the finite covering of $L$? Namely

Finite extension of a Lie group is its finite covering.


*Finite extension of a Lie group exists, if and only if the $\pi_1(L)\neq0$. Thus $L$ must not be simply-connected Lie group.

*The if and only if statement above may be wrong. What may be the additional conditions to include? For example, Lie group needs to be compact? Not solvable? etc.
Any comments and partial answers are welcome. They will be still upvoted as long as you answer is correct.
 A: What is missing in your statements are assumptions on connectedness. For part 1, there is a standard result in Lie theory which says that a homomorphism from a Lie group to a connected Lie group which induces an isomorphism between the Lie algebras is a covering map. This easily extends to the case of a surjective homomorphism between arbitrary Lie groups, which induces an isomorphism between the Lie algebras. So an extension by a Lie group by a discrete group is a covering. (You may want to assume that the discrete group is at most countable to avoid problems with separability.) 
For part 2, connectedness really becomes important. You can always form trivial extension like $L\times\mathbb Z_2$ without problems. But if you assume that $L$ and $G$ are connected, then $G\to L$ indeed is a connected covering of a connected space, so if $L$ is simply connected, it is an isomorphism. Conversely, it is another standard result that the universal covering of a Lie group can be given a unique Lie group structure, so if $L$ is not simply connected, you get a non-trivial connected extension. (Again you can extend this to non-connected cases, a natural assumption under which this works would be that the extension uses an isomorphism between the component-groups of $G$ and $L$.)   
Edit (inview of the comments): As explained above, "having a finite extension" in its own right is not really an issue. Given any Lie group $L$ and any finite group $N$, the product $G:=L\times N$ is a finite extension of $L$. To make the question interesting, you probably want the extension to be non-trivial, either from an algebraic or from a topological point of view. The simplest way to express non-triviality from a topological point of view is to require both $L$ and $G$ to be connected (which clearly excludes products). In this case, the existence of a connected finite extension of $L$ can be read off the fundamental group of $L$ as described above. If you go to the case that $L$ is not connected, then you cannot require $G$ to be connected, since there cannot be a surjective continuous map from a connected space to $L$. As briefly indicated in the end of the last paragraph, my suggestion for a condition to impose in this case is that the projection $G\to L$ induces an isomorphism of the component groups. (This means that $G$ has "as many connected components" as $L$). To analyze extensions in this sense, you can first look at the connected components of the identities. By assumption $G_0\to L_0$ is a finite extension so these are understood in terms of the fundamental group of $L_0$ which coincides with the fundamental group of $L$. I don't know a general description for the last step, so one has to check directly whether for a given extension $G_0\to L_0$ one can add further connected components (which all have to be diffeomorphic to $G_0$) to cook up an extension $G$ of $L$. (For examples like Pin, this should be rather easy, since only $SO(n)$, $O(n)$ and $\mathbb Z_2$ are involved.)   
