 Solution

The Burnside lemma states that the number of orbits of a finite group action isFor $m=2n-1\ge 3$, the dihedral group consists of $m$ rotations $r^j$ by $2\pi j/m$, and $m$ reflections $r^{j}s$, each across the line through one vertex and the midpoint of the opposite edge. Thus $|G|=2m=4n-2$.

For the pentagon, the identity fixes $v^5{e}^5$ colourings. Each of the four nonidentity rotations has a single cycle on vertices and a single cycle on edges, fixing $ve$ colourings. Each of the five reflections has one fixed vertex and two vertex pairs, and one fixed edge and two edge pairs, fixing $v^3{e}^3$ colourings. Applying the Burnside lemma to simultaneous vertex and edge colourings givesVertex colours and edge colours may be chosen independently; each symmetry must preserve both assignments.