Past exam of the number theory course of the University of Oxford / 2009 / A / 3 / b / Solution

Let $V_4=\{1,(12)(34),(13)(24),(14)(23)\}$. This is an abelian subgroup, and conjugation preserves its nonidentity cycle type, so it is normal in $S_4$ and hence in $A_4$. The serieshas factors $V_4\cong C_2\times C_2$, $A_4/V_4\cong C_3$, and $S_4/A_4\cong C_2$, all abelian. Thus $\boxed{S_4\text{ is solvable}}$.