Ianagol on Nostr: I hadn’t seen this before. Presumably these are the orbits of Z_10 x Z_10 subgroups ...

I hadn’t seen this before. Presumably these are the orbits of Z_10 x Z_10 subgroups of the 14400 - order symmetry group. The orientation preserving subgroup is generated by the product of binary icosahedral groups acting on the right and left on the unit quaternions (with center acting trivially). This is then extended by an orientation-reversing element. In the icosahedral group (which is A_5), there is a cyclic element of order 5. The preimage of this element in the binary icosahedral group is order 10. So one gets (Z_10xZ_10)/(-1,-1). Maybe one can adjoin an orientation-reversing element to get the full Z_10 x Z_10? Or maybe it’s some sort of Z/2-extension? In any case, this suggests that on the Clifford torus, the tetrahedra come in two mirror image groups of 50. This seems to be the case looking at the visualization.