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 $1\trianglelefteq R\trianglelefteq D_{2n}$ 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.

Solution

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

 Solution (source code)