Content-Type: text/shitpost

Subject: Octahedron path group
Path: you​!your-host​!walldrug​!epicac​!thermostellar-bomb-20​!twirlip​!am​!plovergw​!ploverhub​!shitpost​!mjd
Date: 2018-11-14T20:24:16
Newsgroup: talk.mjd.octahedron-path-group
Message-ID: <>
Content-Type: text/shitpost

Note to self: On the octahedron, we don't have !!xy=yx!!, but we do have !!xyx=yxy!!.

This is a consequence of !!x^2=1!! and !!(xy)^3=1!!.

Subject: Some ideas for counting paths on a dodecahedron
Path: you​!your-host​!walldrug​!prime-radiant​!skordokott​!mechanical-turk​!goatrectum​!plovergw​!plovervax​!shitpost​!mjd
Date: 2018-11-14T15:11:28
Newsgroup: talk.mjd.dodecahedron-ideas
Message-ID: <>
Content-Type: text/shitpost

Your original idea labeled the endpoints of the edges instead of the edges themselves. Putting the same label at each endpoint means that !!x^2=1!! for all !!x!!. Maybe you don't need this. Or what if we go partway in this direction and label the endpoints !!x!! and !!x^{-1}!!?

Your original idea was to think of the dodecahedron as a Cayley graph, but then you didn't really follow this up. Go back and think about Cayley graphs more carefully.

In the tetrahedron, each face is a product !!abc!! in some order, so at every starting point !!xyz=1!! whenever !!x,y,!! and !!z!! are all different. In the cube the corresponding property for faces is !!xyxy=1!!. The coloring of the dodecahedron that you're using has no such good property. Can you find one that does? This would probably require that you use five labels.

The fact that you found a labeling of the tetrahedron and cube where label order didn't matter is an expression of some fact about the symmetry of the polyhedron itself that you aren't looking at directly. What's really going on there?

It's obvious why this labeling exists for the cube (and for the !!n!!-cube generally). But why does it exist for the tetrahedron? What's going on there?