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

The rotations form a cyclic subgroup $R\cong C_n$ of index two, so it is normal. The quotient $D_{2n}/R$ has order two and is cyclic. Hence $1⊴R⊴D_{2n}$ has abelian factors, proving $\boxed{D_{2n}\text{ is solvable}}$. Its commutator subgroup is contained in the abelian rotation subgroup, so its second derived subgroup is already trivial.