Compute #different ways of drawing a diagonal line on every face of a regular cube. 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? 
 A: 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.



A: 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 $8$.
