Prove that \(SL_2(\ZZ)\) is not solvable.
Solution
Reduce matrices modulo \(5\).
1. The reduction map\[ \rho:SL_2(\ZZ)\longrightarrow SL_2(\FF_5) \]is surjective.
The group \(SL_2(\FF_5)\) is generated by the elementary matrices \[ E_{12}(a)=\begin{pmatrix}1&a\\0&1\end{pmatrix}, \qquad E_{21}(a)=\begin{pmatrix}1&0\\a&1\end{pmatrix} \qquad(a\in\FF_5). \] Indeed, elementary row operations reduce any matrix of determinant \(1\) to the identity; equivalently, these elementary matrices generate \(SL_2\) over a field.
Each \(E_{ij}(a)\) is the reduction modulo \(5\) of an integral elementary matrix of determinant \(1\). Hence every generator of \(SL_2(\FF_5)\) lies in the image of \(\rho\), so \(\rho\) is surjective.
2. The group \(SL_2(\FF_5)\) has a nonsolvable quotient.
Its center is \[ Z=\{\pm I\}. \] The quotient \[ PSL_2(\FF_5)=SL_2(\FF_5)/Z \] is the classical group of order \(60\) isomorphic to \(A_5\). Since \(A_5\) is nonabelian simple, it is not solvable.
3. Therefore \(SL_2(\ZZ)\) is not solvable.
2. This is impossible.