I'm reading the proof of Theorem 1.64 in http://www.jmilne.org/math/CourseNotes/GT.pdf
my questions are
1) Why don't we immediately start with a prime $p$ not dividing $mnr$? Using the same strategy it seems we get elements of order $m$,$n$ with product of order $r$ directly in the group $SL_2(\mathbb{F}_q)$. Why is the 2 important?
2) Why in the definition of $a$ there is an $u^{-1}$ and in the definition of $b$ a $v^{-1}$? Why don't we define $a$ to be $\begin{pmatrix} u&1\\0&u\end{pmatrix}$ and $b$ to be $\begin{pmatrix}v&0\\t&v\end{pmatrix}$ with $t$ such that $2uv+t=w+w^{-1}$?
For Question 1, the group in which we are finding elements of these orders is not ${\rm SL}(2,{\mathbb F}_q)$ but ${\rm PSL}(2,{\mathbb F}_q)$, and so you choose an element of order $2m$ in ${\rm SL}(2,{\mathbb F}_q)$ in order for its image to have order $m$ in ${\rm PSL}(2,{\mathbb F}_q)$. The argument will not work in ${\rm SL}(2,{\mathbb F}_q)$ if any of $m,n,r$ is $2$, because $-I_2$ is the only element of ${\rm SL}(2,{\mathbb F}_q)$ of order $2$, and that lies in the centre.
The answer to the second question is that the elements $a$ and $b$ must have determinant $1$ for them to lie in ${\rm SL}(2,{\mathbb F}_q)$.