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Fix a prime $p$ and integer $n>1$, along with the ring $R$ of integers in a finite extension of the field $\mathbb{Q}_p$ (for example $R = \mathbb{Z}_p$).

Is there an upper bound $C(n,p)$ on the orders of finite subgroups of $\mathrm{GL}_n(R)$? Or can finite subgroups be arbitrarily large?

Probably this question has been resolved one way or the other in the literature but I don't recall a relevant source.

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Concerning Qiaochu's question, the bound certainly has to involve something about $R$, such as its rank over ${\mathbf Z}_p$, because otherwise you could let $R$ be the integers in the degree $d$ unramified extension of ${\mathbf Q}_p$ and use the group of matrices ${\rm diag}(a,1,\dots,1)$ where $a$ runs over the group of $p^d-1$-th roots of unity in $R$. That is a group of order $p^d-1$ inside ${\rm GL}_n(R)$ and its order gets arbitrarily large as $d \rightarrow \infty$. So probably the question should have the upper bound expressed as $C(n,p,r)$, where $r ={\mathbf Z}_p$-rank of $R$. – KConrad Apr 20 '13 at 20:57
The bound is not even uniform in $[K:\mathbb{Q}_p]$ when $n = 1$, in which case we are looking at roots of unity in $p$-adic fields. – Pete L. Clark Apr 21 '13 at 1:09
Since $\overline{\mathbf{Q}_p}$ is abstractly isomorphic to $\mathbf{C}$, every finite subgroup of $\mathrm{GL}_n(\mathbf{C})$ occurs in $\mathrm{GL}_n(\mathcal{O}_K)$ for some finite extension $K/\mathbf{Q}_p$. – François Brunault Apr 21 '13 at 16:02
@Francois: True. But actually the Cassels Embedding Theorem shows more than this: every finite subgroup of $\operatorname{GL}_n(\mathbb{C})$ occurs in $\operatorname{GL}_n(\mathbb{Z}_p)$ for infinitely many primes $p$. – Pete L. Clark Apr 21 '13 at 17:49
I made it a new question :… – François Brunault Apr 22 '13 at 7:19
up vote 10 down vote accepted

Yes, for any fixed $p$-adic field $K$ the supremum of orders of finite subgroups of $\operatorname{GL}_n(K)$ is finite and can be explicitly bounded above. There is a beautiful discussion of this tucked away somewhere in Serre's Lie Algebras and Lie Groups. (What I say in the following is almost entirely derived from that, so you would do at least as well just to go back to the source.)

The first observation is that the answer is the same or $K$ as for its ring of integers $R$: a simple compactness argument shows that any finite subgroup of $\operatorname{GL}_n(K)$ can be conjugated into $\operatorname{GL}_n(R)$. Then the idea is to show that for sufficiently large $k$, there is no torsion in the kernel of the natural surjective map $\operatorname{GL}_n(R) \rightarrow \operatorname{GL}_n(R/\pi^k R)$: this comes down to analysis of torsion in formal groups, which explains why it shows up in Serre's LALG.

In fact Chapter 4 of these notes is devoted to precisely this and closely related questions, for instance including a proof of Selberg's Theorem: if $K$ is any field of characteristic zero and $G$ is a finitely generated subgroup of $\operatorname{GL}_n(K)$, then $G$ has a finite index torsionfree subgroup. But the proof here is kind of antithetical to your precise question: one reduces to the case in which $K$ itself is finitely generated and applies a theorem of Cassels: any such field can be embedded in $\mathbb{Q}_p$ for some odd $p$.

My notes give (standard) explicit upper bounds in the case of $\operatorname{GL}_n(\mathbb{Q}_p)$.

Okay, but wait: I know how to do the general case too. You want to combine the above arguments with:

Proposition: Let $K$ be a finite extension of $\mathbb{Q}_p$ with ramification index $e$, and let $R$ be its ring of integers, with maximal ideal $\mathfrak{m}$. Let $F(X,Y)$ be any formal group law (of any finite dimension; here $X$ and $Y$ are vector variables) over $R$, with associated "standard" $K$-analytic Lie group $G^1 = F(\mathfrak{m})$. Then the exponent of any finite subgroup of $G^1$ divides $p^{\gamma_p(e)}$, where for any $m \in \mathbb{Z}^+$, $\gamma_p(m) = \lfloor \log_p \left( \frac{pm}{p-1} \right) \rfloor $.

(This is Proposition 9 from this paper of myself and Xavier Xarles. It was well known to just about anyone who had worked in the area, but we couldn't find it in the literature, and in fact our paper was cited at least once for precisely this result.)

So I think that this does exactly what you want: let me know if I'm mistaken.

Finally, I find it striking that the answer is completely different for local fields of positive characteristic: it is not possible to bound torsion in formal groups in this setting -- somehow $e = \infty$ in the above setup -- and Selberg's Theorem is false. If I am not mistaken, there are indeed arbitrarily large finite subgroups of $\operatorname{GL}_n(\mathbb{F}_q((t)))$ for all prime powers $q$ and all $n \geq 2$.

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Also, although I certainly don't claim that any of these bounds are sharp, I believe, in line with the above comments, that easy examples will show that if you fix any two of $e$ (the ramification index of $K/\mathbb{Q}_p$), $f$ (the residual degree of $K/\mathbb{Q}_p$) and $n$ and let the third one grow, then there will be no uniform bound. – Pete L. Clark Apr 21 '13 at 1:11
@Pete: This looks very helpful, though I'll need to sort references out more carefully. I've tracked down my ancient copy of Serre's 1964 Harvard notes, where I found to my surprise that I had once underlined with red pencil some passages in the LG part including Theorem 1 in Appendix 3, page LG 4.35, based on earlier theorems. For some reason I looked at Bourbaki's Chapter 3 without getting any help there. I've always found Serre's notes hard to navigate, so didn't think to look there. – Jim Humphreys Apr 21 '13 at 13:54
@Pete: This works for a fixed $\mathbb{K}$ or $R.$ But I am not sure that the bound only depends on n and p, as Jim sought. I do not think there is a bound which only depends on n and p, as you can't a priori bound the number of p′-roots of unity in R just in terms of n and p. – Geoff Robinson Apr 21 '13 at 17:26
@Geoff: I'm sorry, evidently I haven't been clear enough on this point: the bound does not only depend on $n$ and $p$. It fully depends on $n$, $p$, the ramification index $e$ and the inertial degree $f$: if you fix any three of them and let the fourth go to infinity, the supremum will be infinite. – Pete L. Clark Apr 21 '13 at 17:46

There may be a section in the old Curtis and Reiner "Theorems of Blichfeld, Burnside and Frobenius" which answers this to a large extent, although they don't think about $p$-adics. Thanks to Jordan's theorem, there is an Abelian normal subgroup of (fixed) bounded index. So the question reduces to limiting the size of Abelian (normal) subgroups. But for the rings $R$ you discuss, there are only finitely many roots of unity in any such $R,$ and for any given $R,$ there is an explicit bound on the number of roots of unity in $R$, and an explicit bound on the size of Abelian subgroups (in fact, for a primitive absolutely irreducible group, the largest normal Abelian subgroup consists of scalar matrices).

An alternative approach is to note that the kernel of any reduction (mod $p$) (strictly, reduction (mod the appropriate prime ideal containing $p$) of a finite subgroup is a finite normal $p$-group, and the image group is a subgroup of a finite ${\rm GL}(n,q),$ ($q$ a fixed -in terms of $R$-power of $p$). There are many ways to obtain explicit bounds on the size of the normal $p$-subgroup- over some extension field, it is a monomial group, etc.

In view of questions below, let me expand a little. It is commonplace in modular representation theory to work with a $p$-modular system, which is a triple $(\mathbb{K},R,F)$ such that $R$ is a complete discrete valuation ring of characteristic $0$ such that $R$ has field of fractions $\mathbb{K},$ and $F$ is the residue field $R/J(R)$. This triple is usually taken to be large enough for $\mathbb{K}$ to contain a splitting field for a finite group $G$ and its subgroups ( for example, by assuming that $R$ contains a primitive $|G|$-th roots of unity, which we now do). It is also commonplace to identify $\mathbb{C}$-valued characters of $G$ with characters afforded by finite dimensonal $\mathbb{K}G$-modules. Details are often glossed over, but this is all perfectly permissible.

Any character of a finite dimensional $\mathbb{C}G$-module is afforded by some $\mathbb{Q}[\omega]G$-module, where $\omega$ is a primitive $|G|$-th complex root of unity. This module is determined uniquely up to isomorphism by its character. Since $\mathbb{Q}[\omega]$ is isomorphic to a subfield of $\mathbb{K}$ under current assumptions, every $\mathbb{C}G$-module ``comes from" a $\mathbb{K}G$-module. Conversely, every character afforded by a $\mathbb{K}G$-module may be decomposed uniquely using the standard inner product on the character ring into a sum of complex irreducible characters, each of which may be afforded by a $\mathbb{Q}[\omega]G$-module. So for most purposes, there is little difference between studying $\mathbb{C}G$-modules and $\mathbb{K}G$-modules, and, in particular, the maximum index of an Abelian normal subgroup of a finite subgroup of ${\rm GL}(n,\mathbb{K})$ can be no larger than the corresponding bound for ${\rm GL}(n,\mathbb{C}).$ As mentioned in comments, if one works with primitive irreducible groups, all Abelian normal subgroups are central. But large groups of scalar matrices can't be finessed, though Jordan's theorem in the primitive case shows that this is the only real obstacle to an absolute bound.

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@Geoff: In the final sentence, did you mean "finite index" instead of "finite"? Yet the kernel can have nontrivial torsion when $R$ is ramified or $p = 2$. But via $p$-adic logarithm, the kernel of reduction mod $pR$ (not just mod the maximal ideal of $R$, in case of ramification) is a torsion-free subgroup for odd $p$, and likewise mod $4R$ when $p = 2$, so the finite subgroup is a subgroup of ${\rm{GL}}_n(R/pR)$ for odd $p$, and ${\rm{GL}}_n(R/4R)$ for $p = 2$. So that yields a crude bound $C(n,q,e)$ where $q$ is residue field size and $e = e(R|\mathbf{Z}_p)$ is ramification degree. – user30180 Apr 20 '13 at 22:17
@ayanta: I meant finite, though it is also finite index. I gree with the rest of what you say, but left it unsaid. – Geoff Robinson Apr 20 '13 at 22:42
@Geoff: I found it tempting at first to look for some version of Jordan's bound, but classically it's been focused on working over the complex numbers. Is there a natural way to adapt to a p-adic field? – Jim Humphreys Apr 21 '13 at 14:00
@Jim: I have re-edited my answer to give a bit more detail. – Geoff Robinson Apr 21 '13 at 16:31
@Geoff: Your suggestions are quite helpful, and I will ponder them along with Pete's. It still surprises me a bit not to find the p-adic case made more explicit in the older literature. – Jim Humphreys Apr 21 '13 at 20:51

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