$S_n$ is solvable if and only if $n\leq 4$

For \(n\leq4\), \(S_n\) is isomorphic to a subgroup of \(S_4\), and subgroups of solvable groups are solvable. The series \(\theset{e}\normal V_4\normal A_4\normal S_4\), with \(V_4=\theset{e,(12)(34),(13)(24),(14)(23)}\), has abelian quotients \(V_4\), \(C_3\), and \(C_2\), so \(S_4\) is solvable.

For \(n\geq5\), \(A_n\) is simple and nonabelian. Its commutator subgroup \([A_n,A_n]\) is normal and nontrivial, so \([A_n,A_n]=A_n\), and the derived series of \(A_n\) is constant at \(A_n\) and never reaches \(\theset e\). Hence \(A_n\) is not solvable, and neither is \(S_n\), which contains it.