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 $R = \mathbb{Z}\_p$). 



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



Upper bound on order of finite subgroups of GL_n(Z_p)?
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](http://alpha.math.uga.edu/%7Epete/8410FULL.pdf) 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:



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> 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 $.
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(This is Proposition 9 from [this paper](http://alpha.math.uga.edu/%7Epete/cx6.pdf) 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 $n \geq 2$.


