Let \(p\) be prime and \(n\ge1\). Determine all pairs \((n,p)\) for which \[ \SL_n(\FF_p) \] is solvable.
Solution
The solvable cases are \[ n=1\quad\text{for every prime }p, \] and \[ (n,p)=(2,2),(2,3). \]
1. If \(n=1\), then \(\SL_1(\FF_p)=1\), so it is solvable.
2. The groups \(\SL_2(\FF_2)\) and \(\SL_2(\FF_3)\) are solvable.
We have \[ \SL_2(\FF_2)\cong S_3, \] which is solvable.
For \(p=3\), the center of \(\SL_2(\FF_3)\) is \(\{\pm I\}\) and \[ \PSL_2(\FF_3)\cong A_4. \] Thus \(\SL_2(\FF_3)\) is an extension of the solvable group \(A_4\) by the abelian group \(C_2\), hence is solvable.
3. All other cases with \(n\ge2\) are nonsolvable.
The classical simplicity theorem for projective special linear groups says that \[ \PSL_n(\FF_p) \] is nonabelian simple for \(n\ge2\), except for \[ (n,p)=(2,2),(2,3). \] Since \(\PSL_n(\FF_p)\) is a quotient of \(\SL_n(\FF_p)\), solvability of \(\SL_n(\FF_p)\) would force solvability of \(\PSL_n(\FF_p)\). A nonabelian simple group is not solvable, so no other pair occurs.