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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.

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closed as too broad by YCor, Chris Godsil, Yemon Choi, abx, Stefan Kohl Jan 10 '15 at 17:33

There are either too many possible answers, or good answers would be too long for this format. Please add details to narrow the answer set or to isolate an issue that can be answered in a few paragraphs.If this question can be reworded to fit the rules in the help center, please edit the question.

The question is stated is such vague terms that several interpretations of the question are possible (how to prove that some (given?) infinite subgroup has infinite index). I voted to close the question as too broad; could be also closed as unclear. In any case I don't see any relevance in posting new answers now. – YCor Jan 10 '15 at 15:13

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.

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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 $x \in X$ 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 $f: G \rightarrow X$ such that $H$ contains the kernel of $f$. Then $f: G/\operatorname{ker}(f) \hookrightarrow X$, so $\operatorname{ker}(f)$ has finite index, so $H$, which contains $\operatorname{ker}(f)$, has finite index.

Note that both of these will, in principle, always work. In Case 1, take $X = G/H$. In Case 2, let $H' = \bigcap_{g \in G} gHg^{-1}$ 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.

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I'm not sure if you're interested in classical groups over Q_p or over Z_p; in the latter case, one can often show that a closed subgroup H of G is in fact the whole group G once you know it projects surjectively onto some explicit finite quotient of G. See, for instance, [IV.3.4, Lemma 3] of Serre's "Abelian l-adic representations and elliptic curves." I explain how this works for GL_n (this is definitely not original to me, I just wanted the paper to be self-contained) in Lemma 3 of this paper about K3 surfaces.

Of course this won't work unless you know for some reason that H is closed; for instance, two random elements of GL_2(Z_p) will (I think) almost always generate a (discrete) free subgroup of GL_2(Z_p) which is dense, i.e. which projects surjectively onto every finite quotient, but which is obviously not finite-index.

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This may be tangentially related: a subgroup H of a nilpotent group G is in fact the whole group if it projects surjectively onto the abelianisation of G. – Terry Tao Oct 26 '09 at 16:49
Good point; similarly, if H and G are finitely generated pro-p groups, all you need is for H to surject onto the finite quotient G^{ab} / p G^{ab}. – JSE Oct 26 '09 at 17:05

From a computational perspective, you could try coset enumeration (or Knuth-Bendix) over the candidate subgroup (starting with generators given in terms of a known presentation of the overgroup). If it terminates, then the subgroup has finite index.

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I often try to show that a subgroup contains another subgroup known to be of finite index. There are various refinements. One technique is to prove that your subgroup contains an intersection of a finite number of subgroups, each of which is known to have finite index. For example, suppose you know that the central quotient G/Z(G) of a group G is finite. (One way to show this is to show that G is a union of finitely many abelian subgroups.) If it is too hard to show that your subgroup H contains the centre, but you can show that H contains the intersection of Z(G) with the derived subgroup [G,G] of G, then the index of H in G is finite.

Another idea is to try to show that (e.g.) every nilpotent quotient of your group is finite, and then show that your subgroup must contain some term of the lower central series. (You could replace nilpotent with soluble and the LCS with the derived series, and so on.)

This may not count as a general method; perhaps it is more of a "trick", but I've seen it used to good effect in proving some commutativity theorems for groups and rings. If you can show that your subgroup is a union of (at most) two subgroups known to have finite index, then you are done.

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You could try computing the virtual cohomological dimension of the ambient group, and then the subgroup. If they're torsion free, you only need to know the cohomological dimensions, as a theorem of Serre tells you that a torsion free group and its finite index subgroups have the same dimension. (You'd be surprised how often this works.)

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I guess that's more of a "show it has infinite index" test, but still useful. – Richard Kent Nov 4 '09 at 0:47
On the other hand, in a lot of settings, infinite index subgroups will have cd smaller than the ambient group. – Richard Kent Nov 4 '09 at 0:53

I am not fully sure if you also include this type of groups but considering $\mathbb Q$ as an abelian group and a subgroup $\mathbb Z$, then the order $|\mathbb Z|$ is obviously infinite and so does the index $[\mathbb Q : \mathbb Z]=\infty $. This can be shown by the following :

If an abelian group $H<\mathbb Q$ is a subgroup of $\mathbb Q$ and the index $[\mathbb Q : H]=n < \infty $, then the order of the quotient group $\mathbb Q /H$ is $n$. That is, for any element $x\in \mathbb Q$ of $\mathbb Q$, $n\cdot x \in H$. This implies that $H=\mathbb Q$ as for any $x\in \mathbb Q$, $x/n \in \mathbb Q$, so that $n \cdot x/n = x \in H$. Hence all subgroups of abelian group $\mathbb Q$ are of infinite index.

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The question asks for a way to verify that an infinite group has a subgroup of finite index. It is not asking for examples. – Chris Godsil Jan 10 '15 at 14:15
@ChrisGodsil: the question is stated is such vague terms that both interpretations of the question are possible (how to prove that some (given?) infinite subgroup has infinite index). I voted to close the question as too broad, could be also closed as unclear. – YCor Jan 10 '15 at 15:12
@ChrisGodsil It seems to me that the question is implying every infinite subgroup of a group must have finite index. As it is not true, I wrote this example. In fact, I was almost believing that every infinite subgroup of a group always has finite index. – sock Jan 11 '15 at 11:14
I agree with YCor that the question was not clear. I just came to a different interpretation than you did. – Chris Godsil Jan 11 '15 at 14:54

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