Given $n$ find an element of that order in $PGL_{2}$ over rationals I have the following question,
Suppose I consider the projective general linear group, $PGL_{2}(\mathbb{Q})$.
Now given a positive integer $n$, I want to find an element say $M\in PGL_{2}(\mathbb{Q})$ such that order of $M$ is $n$.
My attempt: 
**n=3 **
\begin{pmatrix}0 & 1 \\ -1 & 1 \end{pmatrix}
n=4
\begin{pmatrix}1 & 1 \\ -1 & 1 \end{pmatrix}
But what about other cases, 
UPDATE
Over Complex numbers it is easy to do 
\begin{pmatrix} exp(2\pi i/n) & 0 \\ 0 & 1 \end{pmatrix}
 A: Edit 1. We search the integers $n\geq 2$ s.t. there is $a\in\mathbb{Q}^*,X\in M_2(\mathbb{Q})$ satisfying $X^n=aI_2$. Note that $X$ is diagonalizable over $\mathbb{C}$. For every positive rational $u$, $order(X)=order(uX)$; then we may assume that $X\in M_2(\mathbb{Z})$ and $a\in \mathbb{Z}^*$. If the eigenvalues of $X\not=I$ are real, then they are opposite and $n=2$ works. Otherwise $spectrum(X)=e^{\pm ki\pi/n}b^{1/n}$ with $b=\pm a>0$. Thus $tr(X)=2\cos(k\pi/n)b^{1/n}$ and $\det(X)=b^{2/n}$ are integers. Then $\cos^2(k\pi/n)$ is a rational and, consequently, $\cos(2k\pi/n)$ too.
EDIT 2. It is known that if $(\theta,\cos(\theta \pi))\in \mathbb{Q}^2$, then $\theta\pi\in U=\{\pi/2,3\pi/2,\pi/3,2\pi/3,4\pi/3,5\pi/3\}$. Thus $2k\pi/n\in U$ and $k\pi/n\in \{\pi/4,\pi/6,5\pi/6,\pi/3,2\pi/3\}$.
Finally, we obtain the orders $n=2,3,4,6$ (as in $GL_2(\mathbb{Q}))$.
$n=2$ for $A=\begin{pmatrix}1&0\\0&-1\end{pmatrix}$.
$n=3$ for $A=\begin{pmatrix}0&1\\-1&1\end{pmatrix}$ ($k\pi/n=\pi/3,b=1$; same order when $k\pi/n=2\pi/3$). Note that $order(A)=6$ in $GL_2$.
$n=4$ for $A=\begin{pmatrix}1&1\\-1&1\end{pmatrix}$ ($k\pi/n=\pi/4,b=4$).
$n=6$ for $A=\begin{pmatrix}1&1\\-1&2\end{pmatrix}$ ($k\pi/n=\pi/6,b=27$; same order when $k\pi/n=5\pi/6$).
