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Can you give me an example of a Lie group acting on a compact metric connected space transitively so that it has a closed finite index subgroup which does not act transitively?

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Can you give me an example of a connected Lie group that has a closed finite index subgroup? –  zroslav May 19 '11 at 16:36
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Is this really the question you meant to ask? Consider the orbit of a point under the action of the identity component of the group. –  S. Carnahan May 19 '11 at 17:14
    
@zroslav: I think such an example cannot exist -- if a subgroup is closed and finite index, then its complement is a finite union of its cosets, hence also closed; then our subgroup is open as well. –  JGordon May 20 '11 at 2:41
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The ‘number-theory’ tag is intriguing. What is its relevance? –  L Spice May 21 '11 at 6:14
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I wanted to make this a comment, but it turned out to be too long, so have to submit it as (partial) answer... sorry if I misunderstood the question.

Let $G$ be the Lie group. If such an example exists, it would have to be somewhat exotic -- by my previous comment, any closed subgroup of finite index has to be a union of connected components of $G$. On the other hand, if a Lie group acts on a differentiable manifold $X$ transitively, then its identity component also acts transitively (see e.g. Onischik and Vinberg, "Lie groups and algebraic groups", Chapter 1 section 3). Hence, to find such an example, one would have to look for $X$ that is not a manifold. On the other hand, if the stabilizer of a point is Lie subgroup, then $X$ would have a unique structure of a manifold with respect to which the action is differentiable. Hence, if you really want such an example, one has to look for some action that "cannot be made smooth" by putting a differentiable manifold structure on $X$, i.e., where the stabilizer of a point is not a submanifold of $G$ (I think traditionally in the "standard" theory of Lie groups such actions are not considered...)

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