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Given a group $G$ and subgroups $U\le V\le G$, there are canonical quotient maps $G/U\to G/V$ that send cosets of $U$ to the cosets of $V$ containing them. Even when $U,V$ aren't normal, this map is a valid morphism in the category of $G$-sets (sets equipped with a $G$-action, the morphisms comprised of bijections that intertwine with the action), so we may form an inverse system of quotients of $G$ using the quotient maps. Since the system contains $G/1$ at the top, the limit is simply $G$, but if we restrict our attention to infinite $G$ and subgroups of finite index, the situation is not so trivial.

Every $G$-set decomposes as a disjoint union of its orbits, orbits are always transitive $G$-sets in their own right, and transitive $G$-sets (leave the empty set out of this) are always isomorphic to quotients of $G$, so it is sensible to ask for the number of orbits of a $G$-set isomorphic to a particular quotient of the group $G$. (This motivates the Burnside ring, I think.) Is there a straightforward way to find this number for the inverse limit of the aforementioned system (or, similarly, of the system formed from quotients by infinite subgroups, which includes finite-index ones but is generally bigger)?

To a first approximation, it might be nice to know when the number is infinite or not, based on whether the $U$ in the quotient $G/U$ is finite or infinite, or the same of its index, or its algebraic relationship to the whole group, etc., and then if this number is ever finite, what exactly it is.

I get the feeling this should be just as simple as the finite case, but nothing really definitive comes to mind, other than perhaps to investigate $\hom(-,G/U)$'s...

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The inverse limit you have constructed is called the profinite completion of the group $G$, denoted $G^\land$. Here are some facts about the profinite completion:

  • It's not just a $G$-set---it actually has the structure of a group. This is because the finite-index normal subgroups are cofinal in the directed set of all finite-index subgroups. (That is, every finite-index subgroup contains a normal subgroup of finite index.)

  • Let $N_\infty$ be the intersection of all finite-index normal subgroups of $G$. Then every orbit of $G$ on $G^\land$ is isomorphic (as a $G$-set) to $G/N_\infty$.

  • The normal subgroup $N_\infty$ is trivial if and only if the group $G$ is residually finite. In this case, $G$ can be thought of as a subgroup of $G^\land$.

  • The number of orbits is finite if and only if $G$ has finitely many subgroups of finite index. Otherwise the number of orbits is uncountable.

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Thanks! It will take some time to flesh out the reasons and explanations behind the info in these bullet points. –  blue Jun 22 '12 at 3:25
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I have revisited this answer and got most of it. The first one is Poincare's theorem, the second follows because an orbit is iso to $G$'s quotient by the stabilizer and the stabilizer of all of the finite-index normals is the intersection of the normals, the third because a homomorphic image of $G$ is finite implies the morphism's kernel contains $N_\infty$ and we can take quotient maps induced by f.i. normals of $G$ for the converse. Can you give me a hint about the only if and uncountable bit in bullet #4? –  blue Aug 25 '12 at 16:37

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