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 $p$



Order of a matrix congruent to the identity modulo p
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 $p=2$).


