Exercise 1.7.12 (Kernel of the action of $D_{2n}$ on the set of pairs of opposite vertices)

Proof. The vertices are labelled $1,2,\dots ,n$ where $n$ is even.

We must treat the case of the square separately ( $n=4$). Let $r$ be the rotation about the center through $\pi ∕2$ radian, and $s$ the reflection through the $x$-axis. The pairs of opposite vertices are $\{1,3\}$ and $\{2,4\}$. The array of Exercise 12 shows that $r^2$ (the rotation of $\pi$ radians) fixes these two pairs, and also $\mathit{rs}$ and $r^{3}s$ (reflections through the two diagonals), but no other non identity symmetry. Therefore the kernel $K$ of the action on the pairs of opposite vertices is

Suppose now that the number $n$ of vertices satisfies $n\ge 6$.

The set of the pairs of opposite vertices has $n∕2\ge 3$ elements. Since a reflection fixes at most $2$ pairs, no reflection can fix all the pairs, so no reflection $r^{h}s$ is in the kernel $K$. No rotation $\rho$ of $𝜃=2\mathit{k\pi }∕n$ radians fixes these pairs, unless $𝜃\equiv 0$ or $\pi (\mathit{mod}2\pi )$. In these cases $\rho =e$ or $\rho =r^{n∕2}$ fixes all the pairs . So

In either case, the rotation of $\pi$ radian $r^{n∕2}$ is in the kernel, so the action of $D_{2n}$ on the set of unordered pairs of opposite vertices is never faithful. □