Characterizing the product-structure on cosets $O_{h}/C_{2v}$

Question

I consider the set of cosets $Q=G/H$ where $G=O_{h}\simeq S_{4}\times Z_2$ is the full octahedral symmetry group and $H=C_{2v}\simeq Z_2 \times Z_2$, the symmetry group of a body with a $C_2$-axis in a mirror-plane. Identifying $Z_{2}=\{1,-1\}$, the elements of $G$ can be written as $\omega$ or $-\omega$, where $\omega \in O \simeq S_4$, with the minus-sign denoting spatial inversion. By inspection I find that for every $q\in Q$, also $-q\in Q$. As $C_{2v}$ is not normal in $O_h$, $Q$ is not a group. Yet, defining the direct product on $Q\times Q$ as $q\otimes q'=\{gg'|g \in q,g' \in q'\}$, I observe the following. Either $q\otimes q'=H$ or $q\otimes q'=q'' \cup -q''$. Moreover for every $q$ there is a $q^{-1}$ such that $q\otimes q^{-1}=H$ or $q\otimes q^{-1}=H \cup -H$. This seems close to an almost group-like structure, but I am at a loss to properly characterize it.

Answer 1

Consider that $G=O_h=O\times I$, where $I=\{e,\iota\}$, with $\iota$ spatial inversion and $O$ the orientation-preserving octahedral symmetry group. Now $N=C_{2v}\times I=D_{2h}\simeq Z_2\times Z_2 \times Z_2$ is normal in $O_h$, therefore $(G/H)/I=G/(H\times I)=G/N$ is a group, explaining the observation that $G/H$ is "almost" a group.