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I am trying to prove that the the natural map from the special linear group $\operatorname{SL}_2(\mathbb{Z})$ to the special linear group $\operatorname{SL}_2(\mathbb{Z}/n\mathbb{Z})$ is surjective. Clearly, the problem lies in finding an inverse with determinant one. If have following outline of a proof :

Pick a matrix $\big(\begin{smallmatrix}a &b \\c& d\end{smallmatrix}\big)\in \operatorname{SL}_2(\mathbb{Z}/n\mathbb{Z})$. Use the Chinese remainder theorem to find b'=b \text{ mod } n such that $a$ and $b$ are relatively prime... Once I have this I can find an inverse.

The problem is that I cannot figure out how the Chinese remainder theorem could be of use, since there is no reason to assume $a$ and $n$ (capital $N$ in proof) have ggd=1.

If so one could say there is a solution of b'=b \bmod N and b'= 1 \bmod a.

But I fail to see how they do it in general.

Many thanks!

2 Answers 2

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Define the subsets $S$ and $T$ of $\text{M}_2(\mathbb Z)$ by

$\bullet\ A\in S\iff A\equiv g\bmod n$ for some $g$ in $\text{SL}_2(\mathbb Z)$ depending on $A$,

$\bullet\ A\in T\iff\det(A)\equiv1\bmod n$.

In particular $ G:=\text{SL}_2(\mathbb Z)\subset S\subset T\subset\text{M}_2(\mathbb Z). $ We must show $S=T$.

Clearly

$\bullet\ G\times G$ act on $S$ and $T$ by $ (g,h)A=gAh^{-1}, $ $\bullet$ for each $A$ in $T$ there are $g,h\in G$ such that $gAh^{-1}$ is diagonal.

Thus, it suffices to show that any $\text{diag}(a,d)\in T\ $ belongs to $S$.

It is straightforward to check that, as Alex Youcis noticed, there are integers c' and d' satisfying S\ni \begin{pmatrix}a&n\\c'&d'\end{pmatrix}\equiv \begin{pmatrix}a&0\\0&d\end{pmatrix}\bmod n. The general lemma is:

If $a,b,n$ are integers such that $\gcd(a,b)=1$, then any solution mod $n$ to $ ax+by=1\tag1 $ lifts to an exact solution. That is, if $x_0$ and $y_0$ satisfy $ ax_0+by_0\equiv1\bmod n, $ then there is a solution to $(1)$ such that $ x\equiv x_0,\quad y\equiv y_0\quad\bmod n. $

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    Dear @MrOperator: I made a few changes (and Dylan corrected a typo).2011-12-30
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You're basically right. Take an arbitrary $\left(\begin{smallmatrix}a & b\\ c & d\end{smallmatrix}\right)\in\text{SL}_2(\mathbb{Z}_n)$. Choose then b'\equiv b\text{ mod }n such that (b',a)=1 (this is where you use CRT). Since (b',a)=1 Bezout's lemma tells you that there exists $x,y\in\mathbb{Z}$ with ax-b'y=1 let then c'=c+y(1-(ad-b'c)) and d'=d+x(1-(ad-b'c)). Then, \left(\begin{smallmatrix}a & b'\\ c' & d'\end{smallmatrix}\right)\in\text{SL}_2(\mathbb{Z}) and easily verified to have image under the canonical map equal to what we want.

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    Dear @AlexYoucis: I posted another answer.2011-12-30