純粋・応用数学（含むガロア理論）6

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https://mathoverflow.net/questions/2630/infinite-subgroups-with-finite-index
mathoverflow
Infinite subgroups with finite index
Is there a general method to prove that an infinite subgroup of a group has finite index? Or, in other words, to prove that the quotient group is finite? I am particularly interested in classical groups, such as GL(n), SL(n), etc, over a nonarchimedean local field but I am looking for a general method, if there exists one.
asked Oct 26 '09 at 15:53
Sergio Mendes

7 Answers
14
This is a somewhat tautological answer, but: if you can show that the subgroup contains the kernel of a finite representation (i.e. a homomorphism to a finite group), you're done. Intuitively: "I only need a finite number of things to go my way in order to belong to this subgroup."
If the group (or some representation of that group) is compact in some topology, and the subgroup contains the connected component of the identity (or an open neighbourhood of the identity), you're also done.
answered Oct 26 '09 at 16:08
Terry Tao

For this type of problem, "general methods" tend to be quite general indeed, but here are two ideas:
1) Find an action of G on a finite set X and an element x∈X such that H is the stabilizer of x. Then, by the Orbit-Stabilizer Theorem, G/H is isomorphic to the orbit space Gx, so is finite.
2) Find a finite group X and a homomorphism f:G→X such that H contains the kernel of f. Then f:G/ker(f)?X, so ker(f) has finite index, so H, which contains ker(f), has finite index.
Note that both of these will, in principle, always work. In Case 1, take X=G/H. In Case 2, let H′=?g∈GgHg?1 be the normal core of H. It is easy to show that (since H has finite index), H′ is a finite index normal subgroup of G which is contained in H. Take X=G/H′ and f to be the quotient map.
answered Oct 26 '09 at 16:15
Pete L. Clark

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