natural map $SL_2(\mathbb{Z})\to SL_2(\mathbb{Z}/n\mathbb{Z})$ is surjective 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.
 A: 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 $\text{SL}_n$.
A: Assume $n=p^k$, given your matrix $M=\pmatrix{u&v\\ w&z}\in SL_2(\mathbb{Z}/p^k\mathbb{Z})$, if $\gcd(u,p)\ne 1$ replace $M$ by $\pmatrix{0 & 1\\ -1 & 0}M=\pmatrix{U&V\\ W&Z}$, let $-cU+W=0$ so that  $$M=\pmatrix{1 & 0 \\ c & 1} \pmatrix{a & B \\ 0 & a^{-1}}= \pmatrix{1 & 0 \\ c & 1} \pmatrix{a & 0 \\ 0 & a^{-1}}\pmatrix{1 & b \\ 0 & 1}$$
the middle factor is easy to lift: $ad\equiv 1 \bmod p^{2k}, ad-1=p^{2k}bc, \pmatrix{a & bp^k\\ bc&d}\in SL_2(\Bbb{Z})$
Then it is the chinese remainder theorem to extend from $p^k$ to any $n$.
