previous result by Kramer [@kramer]. From a group theoretical point of view, this implies that there exists a subgroup of $B_6$ isomorphic to the octahedral group $\mathcal{O}$ (i.e. the symmetry group of a cube, with order $48$), which contains ${\widetilde{\mathcal{T}}}$ as a subgroup.

Dihedral group ${\mathcal{D}}_{10}$ {#d10}
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The dihedral group ${\mathcal{D}}_{10}$ is generated by two elements $g_{2d}$ and $g_{5d}$ such that $g_{2d}^2 = g_{5d}^5 = (g_{2d}g_{5d})^2 = e$. Its character table is as follows [@jones]:

An explicit representation ${\widetilde{\mathcal{D}}}_{10}$ as a matrix subgroup of ${\widetilde{\mathcal{I}}}$ is given by In order to compute the Schur rotations associated with ${\mathcal{D}}_{10}$, we proceed in a similar way as in the tetrahedral case. The projection matrix $R$ in decomposes ${\widetilde{\mathcal{D}}}_{10}$ as $$\label{d10_rel}
\hat{{\mathcal{D}}}_{10}:= R^{-1} {\widetilde{\mathcal{D}}}_{10} R = D_1 \oplus D_2,$$ where $D_1$ and $D_2$ are matrix subgroups of $\rho_3$ and $\rho_3'$, respectively, that are *reducible* representations of ${\mathcal{D}}_{10}$. In particular, from its character table we have that $D_1 \simeq A_2 \oplus E_1$ and $D_2 \simeq A_2 \oplus E_2$ in $GL(3, {\mathbb{R}})$. In order to find the Schur operators for ${\mathcal{D}}_{10}$, we first reduce $D_1$ and $D_2$ into irreps, using tools from the representation theory of finite groups (see, for example, [@fulton]). In particular, 