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I have a bunch of polyhedroa. In each of them all of the vertices are painted with different colors. For each polyhedron, I start the process with an initial permutation $\mathcal{P}_0$ and I want to generate all other permutations, excluding those that correspond to rigid rotations of $\mathcal{P}_0$.

I thought about doing that with reflections through symmetry planes. Now, these planes can be known beforehand due to the symmetry groups that these reflections belong to, when acting on each polyhedron. And, when you know how they form a group with the rigid rotations, you can also know beforehand which combinations of reflections generate rotations.

So I imagined that I could know beforehand which reflections and combinations of reflections could generate permutations that did not correspond to rigid rotations right from the start, by construction.

But then it hit me that this would be doomed to failure if it were not possible to generate all possible permutations by the application of reflections alone. So here is my question:

Is it possible to generate all of the permutations of $\mathcal{P}_0$ that do not correspond to rigid rotations of $\mathcal{P}_0$ just by using reflections through planes containing the center of the polyhedron?

urquiza
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    Are these all regular polyhedra? – aschepler Aug 11 '23 at 13:24
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    Are you aware of the fact that in $\Bbb{R}^3$ (or any $\Bbb{R}^n$ for that matter) any combination of two reflections is a rigid rotation (about the line of intersection of the two planes with respect to you reflect). This generalizes. A combination of an even number of reflections is a rigid rotation, but an odd number of reflections isn't). – Jyrki Lahtonen Aug 11 '23 at 13:28
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    See, for example, the animation in an old answer of mine. – Jyrki Lahtonen Aug 11 '23 at 13:29
  • But, for example with a cube you do get combinations of three reflections that is neither a reflection (w.r.t. a plane) nor a rigid rotation. Such as the symmetry that maps each vertex of the cube to the diagonally opposite one. – Jyrki Lahtonen Aug 11 '23 at 13:32
  • Irregular polyhedra do have rotational symmetries that cannot be obtained as combinations of reflectional symmetries. Think swastika (with a 4-fold rotational symmetry) for a simple to grasp example. So I'm seconding @aschepler's query. – Jyrki Lahtonen Aug 11 '23 at 13:34
  • Given a particular vertex permutation of (say) a cube, let's call it P0=[1 2 3 4 5 6 7 8]. I want to know whether it is possible to hit every permutation of vertices except those that correspond to rigid rotations of the cube in the P0 permutation, only using the symmetry reflections of the cube. I'm guessing that it is possible because I can know beforehand which combination of reflections correspond to a rotation, but there is that nagging feeling that I am missing something... I'm not sure if these are regular polyhedra, I don't think all of them are. – urquiza Aug 14 '23 at 21:07

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