Let $X=\{$ways of drawing a diagonal line on every face of a regular cube$\}$. Group $G$ is the rotation group of regular cube. $|G|=24$.
We want to compute #different orbits of $X$ under group action $G$.

I tried to equate one specific way of drawing a diagonal line in a particular face with coloring that face in red and the other way with coloring in blue. Thus we can consider coloring the regular cube with 2 colors. (I don't know if this way of thinking is accurate.) By invoking Pólya enumeration theorem, I find #orbits=10.
However, I can only find 8 ways by drawing. Is there something wrong with my method?

  • 2
    $\begingroup$ I think $8$ is the correct answer. I didn't try drawing them, but that's what I get using Burnside's lemma. Drawing a diagonal on each face is not the same as coloring each face red or blue. Consider a 90 degree rotation about the axis through the centers of the top and bottom face: the color of the top face does not change, but the direction of a diagonal line drawn on the top face does change. $\endgroup$ – bof Oct 19 '17 at 5:28
  • $\begingroup$ My calculation: $$(1\cdot2^6+6\cdot0+3\cdot2^4+6\cdot2^3+8\cdot2^2)/24=8.$$ For coloring faces red or blue, it's the same except that the $6\cdot0$ term becomes $6\cdot2^3$ so the final result is $10$ instead of $8.$ $\endgroup$ – bof Oct 19 '17 at 5:34

That method failed because diagonals are reversed by 90 degree rotations but colours are not.

The correct answer is 8. They are:

  • Four triangles forming a tetrahedron on the surface of the cube.
  • Two triangles sharing an edge, plus a separate line.
  • Two triangles not sharing an edge.
  • One triangle, plus a separate V.
  • Two separate tripods centred at opposite corners.
  • Two separate S shapes.
  • Two separate Z shapes.
  • A quadrilateral around the cube, plus two separate edges on opposing faces.

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Consider the graph whose vertices are the vertices of the cube and whose edges are the diagonals of the facets of the cube. This graph is the disjoint union of a red $K_4$ and a blue $K_4$. Choosing a diagonal on each facet means choosing on each facet either the red or the blue diagonal, among them $r\in[0,3]$ red diagonals.

If $r=0$ only blue diagonals are chosen, forming the blue $K_4$.

If $r=1$ one edge of the blue $K_4$ is replaced by the red diagonal on the same facet.

If $r=2$ two edges of the blue $K_4$ are replaced by the red diagonals on the respective facets. These two red diagonals either form a path of length $2$ on the red $K_4$, or they are opposite edges of the red $K_4$.

If $r=3$ the three chosen red diagonals form a star, a triangle, or a path of length $3$ on the red $K_4$. The three remaining edges of the blue $K_4$ then automatically form a figure of the same kind.

All figures produced in these ways are mirror symmetric, apart from the pairs of paths of length $3$ . Both of them either form an $S$, or both of them form a $Z$.

The total number of different figures therefore is $8$.


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