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I couldn't find a demonstration of this theorem:

Given $A \in SO_2(\mathbb{Z}[{1 \over q_1},\dots,{1 \over q_k}])$ and $p$ prime $\notin \{q_1,\dots,q_k\}$

$\exists n \in \mathbb{N} : A^n=Id$ and $A\equiv_p Id \implies n = p^{\space \alpha}$

In other words if $A$ is of finite order and

$ A = \begin{pmatrix} a & -b \\ b & a \end{pmatrix} $

$a \equiv_p 1$ and $b \equiv_p 0$

then the order of $A$ is a power of $p$

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    $\begingroup$ I think that a conceptual way to see this would be to embed the global group $G(Z[1/q_1,...,1/q_k])$ (with $G=SO_2$) into the local group $G(Z_p)$, with $Z_p$ the $p$-adic integers. The matrices in $G(Z_p)$ congruent to the identity form a pro-$p$ group (because this group is filtered by matrices congruent to the identity mod $p^n$ as $n$ goes to infinity), so any element of finite order in this (much larger) group must have $p$-power order. This strategy works in far more generality and is a calculation-free way of arguing. $\endgroup$
    – znt
    Commented Sep 13, 2016 at 8:04

2 Answers 2

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Much stronger statements are true ( and well-known to experts): let $\mathbb{Z}_{p}$ denote the (incomplete) localization at $p$ in $\mathbb{Q}$ ( that is, the rational numbers with denominators prime to $p$ ( together with $0$)).Then when $p$ is odd, only the identity element of ${\rm GL}(n,\mathbb{Z}_{p})$ has $p$-power order and is congruent to the identity (mod $p$). When $p = 2$, any such element has order one or two.

In the case $p$ is odd, we only need to exclude elements of order $p$ which are congruent to the identity (mod $p$). Suppose that $x$ is such an element, and suppose that every entry of $x-I$ is divisible by $p^{j},$ but some element of $x-I$ is not divisible by $p^{j+1}$. Then $x^{p}-I = (x-I)( I + x + \ldots + x^{p-1})$ has every entry divisible by $p^{j+1}$, but $$I + x + x^{2}+ \ldots + x^{p-1} = pI + (x-I)(I + (x+I) + \ldots + ( I + x + \ldots + x^{p-2}))$$ is congruent to $pI$ ( mod $p^{j+1}$) since $(I + (x+I) + \ldots + ( I + x + \ldots + x^{p-2})$ is conguent (mod $p^{j}$) to $I ( 1 + 2 + \ldots + (p-1)),$ which is congruent to the zero matrix (mod $p$).

Hence some entry of $(x^{p}-I)$ is divisible by $p^{j+1}$, but not by $p^{j+2}$, and certainly $x^{p} \neq I.$ I omit the similar proof when $p=2.$

A similar (well-known) argument proves that if $x \equiv I$ (mod $p^{j})$ ( but the congruence fails (mod $p^{j+1}$), and $n$ is not divisible by $p$, then $x^{n}-I \equiv n(x-I)$ ( mod $p^{j+1}$), so $x^{n} \neq I,$ which is sufficient to prove that each element of finite order in $\rm{GL}(n,\mathbb{Z}_{p})$ which is congruent to the identity (mod $p$) has $p$-power order ( so, as seen above, has order $1$ if $p$ is odd, and order dividing $2$ when $p=2$).

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  • $\begingroup$ Thank you very much, that's exactly what I was looking for! $\endgroup$
    – DarioDF
    Commented Sep 17, 2016 at 9:34
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Actually, in this case $A^p=Id$. This assertion is a rather special case of Theorem 6.2 on p. 292 of ``Variations on a a theme of Minkowski and Serre", JPAA 111 (1996), 285--302.

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