This is a problem in my past Qual exam
"Let $n\geq 2$ be an integer. Prove that the natural map $SL_2(\mathbb{Z})\to SL_2(\mathbb{Z}/n\mathbb{Z})$ is surjective."
I approach the problem naturally. Take a matrix $\begin{pmatrix} \bar{a}&\bar{b} \\\bar{c}&\bar{d} \end{pmatrix}\in SL_2(\mathbb{Z}/n\mathbb{Z})$. Then the preimage must be $\begin{pmatrix} a+xn&b+yn\\c+zn&d+tn \end{pmatrix}$. We need to prove there exists $x,y,z,t$ s.t. the determinant is $1$, i.e.
$$(xt-yz)n^2+(at+dx-cy-bz)n+(ad-bc)=1$$
Clealy I just complicate the problem, making it into finding 4 values that satisfy an equation. I do not how to proceed. Is my approach right? Other ways would be awesome, too.
Best Answer
One way to approach this is to prove that $\text{SL}_2(\Bbb Z/n\Bbb Z)$ is generated by the matrices $S=\pmatrix{1&1\\0&1}$ and $T=\pmatrix{1&0\\1&1}$. These can be lifted to $\text{SL}_2(\Bbb Z)$ and hence so can any element of $\text{SL}_2(\Bbb Z/n\Bbb Z)$.
Let $A=\pmatrix{a&b\\c&d}\in\text{SL}_2(\Bbb Z/n\Bbb Z)$. Use the Chinese remainder theorem to prove that there is some $j\in\Bbb Z$ such that $ja+c$ is a unit in $\Bbb Z/n\Bbb Z$. Then $T^jA=\pmatrix{a&b\\c_1&d_1}$ where $c_1$ is a unit in $\Bbb Z/n\Bbb Z$. Then there is $k\in\Bbb Z$ with $S^kT^jA=\pmatrix{1&b_2\\c_1&d_1}=T^{c_1}S^{b_2}$. Therefore $A\in\left<S,T\right>$.
This method can be adapted to $\text{SL}_n$.