Proper isometry group/Finite subgroup/Numerical properties/Fact/Proof

Proof

For two opposite semiaxes ${}H$ and ${}-H$, we have ${}G_{H}=G_{-H}$. For two semiaxes ${}H_{1}$ and ${}H_{2}$ that do not belong to the same axis (in particular, they are different) we have the relation ${}G_{H_{1}}\cap G_{H_{2}}=\operatorname {Id}$, because an isometry with two rotation axes is the identity. Since ${}G$ is the union of all ${}G_{H}$, ${}H\in {\mathfrak {H}}(G)$, we have a union

${}G\setminus \{\operatorname {Id} \}=\bigcup _{H\in {\mathfrak {H}}(G)}{\left(G_{H}\setminus \{\operatorname {Id} \}\right)}\,,$

where every group element ${}g\neq \operatorname {Id}$ appears twice on the right-hand side. Therefore,

${}2(n-1)=\sum _{H\in {\mathfrak {H}}(G)}{\left(\operatorname {ord} \,(G_{H})-1\right)}\,.$

The classes ${}K_{i}$ contain ${}n/n_{i}$ elements. Hence,

${}2(n-1)=\sum _{H\in {\mathfrak {H}}(G)}{\left(\operatorname {ord} \,(G_{H})-1\right)}=\sum _{i=1}^{m}\sum _{H\in K_{i}}(n_{i}-1)=\sum _{i=1}^{m}{\frac {n}{n_{i}}}(n_{i}-1)\,.$

Dividing by ${}n$ yields the claim.

To fact