The Szilassi polyhedron has seven hexagonal faces and 14 trihedral vertices. This is enough to make it a "regular toroid" in according to Szilassi.[1] However abstract polytopes are required to have a flag-transitive automorphism group, which is a more restrictive definition of regular.
So the question arises:
Is the Szilassi polyhedron abstractly regular?
Since the Szilassi polyhedron has 7 hexaongal faces it should have 84 flags, and thus, if it were regular, it would have an automorphism group of size 84. When I check the Atlas of Small Abstract Regular Polytopes there is no Abstract Regular polytope of Schläfli type {6,3} with a automorphism group of size 84.
The pages on the AoSARP are generated by a program exhaustively up to 1024 flags, so either there's a bug in that program or the Szilassi polyhedron is not abstractly regular, and I'm inclined to believe the latter over the former. (Or I have made a mistake in using the AoSARP, that is always a possibility)
But I'd like to prove it. I could construct the adjacency matrix and brute force check every pair of flags for symmetry, but this is time consuming and not very enlightening at the end.
I know a couple of easy checks that can show a polyhedron is not abstractly regular:
- Check every face has the same number of sides (Yes n=6)
- Check every vertex has the same number of faces (Yes n=3)
- Check every k-zigzag (here k=2) has the same number of edges (Yes n=14)
- Check the above for the dual. (Yes)
None of these are sufficient to be abstractly regular, but they are all necessary, and the Szilassi polyhedron seems (the last one is a bit tough to compute since the dual has such high degree vertices), to pass all of them.
Is there a nice proof that the Szilassi polyhedron is not abstractly regular?