Proper symmetry group/Finite subgroup/Three classes of semiaxes/2/2/Dihedral/Fact/Proof

Proof

There are three classes of semiaxes, say ${}K_{1}$, ${}K_{2}$, ${}K_{3}$. Two of them have order ${}2$ (each class with ${}k$ semiaxes), and one has order ${}k$ and two semiaxes (the numbers of the semiaxes follow with ${}n_{1}=n_{2}=2$ from fact). For ${}k\geq 3$, the two semiaxes from the third class form a line, because the opposite semiaxis has the same ordwe. For ${}k=2$, every semiaxis is equivalent to its opposite semiaxis. We denote the axis of ${}K_{3}$ by ${}A_{3}$. Every group element with another rotation axis must transform the two semiaxes from ${}K_{3}$ into each other, so that all other axes are orthogonal to ${}A_{3}$. Let ${}g$ denote a generating rotation around ${}A_{3}$. For a semiaxis ${}H_{1}$ from ${}K_{1}$, the

$g^{i}(H_{1}),i=0,\ldots ,k-1,$

are exactly all semiaxes from ${}K_{1}$. These (or rather the points on these axes of norm ${}1$) form a regular ${}k$-gon in the plane orthogonal to ${}A_{3}$. The same holds for ${}g^{i}(H_{2})$ with ${}H_{2}\in K_{2}$. Every rotation about ${}180$ degree around an axis from ${}K_{1}$ permutes the semiaxes from ${}K_{2}$. Therefore, the semiaxes from ${}K_{2}$ yield a "bisection“ of the ${}k$-gon. Hence, the group is the (improper) symmetry group of a regular ${}k$-gon, or the proper symmetry group of a bipyramid over a regular ${}k$-gon, that is, it is a dihedral group ${}D_{k}$.

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