Limitations on orientations of a character on the surfaces of a cube When looking at a cube you see either two surfaces across an edge or three
surfaces around an apex. Is it possible to arrange a non-symmetrical character (e.g."R") at  0,90,180 or 270 degrees rotation on each surface of a cube such that no two characters viewed across an edge appear the same way up and no three characters viewed from an apex appear as if they rotate around the apex (e.g. had all the top left corners of the character's square touching at the apex) ?
It is possibly obvious that I am not a mathematician which means two things :
1 responses need to be couched in "layman's" terms
2 my efforts so far have been experimental rather than mathematical:
So far I can produce arrangements that fulfil the first (edge)
criteria but can't get rid of one set of characters that rotate around an apex.
When I destroy a rotation at one apex it pops up around another apex ......
I'd like a mathematician to tell me my target either can or can't be achieved so I know whether to keep trying or not.
Regards
Ken
 A: Sorry for posting misleading comments before, I didn't read very carefully. It's still an example of group theory, but requires a lot less abstraction. Since a cube looks the same around every vertex, the solution lies in determining the allowed starting arrangements and what other vertex arrangements have to be used after a starting arrangement is selected.
Before ruling out any arrangements, you want to find out exactly which arrangements are distinct around a vertex. Notice you can rotate any of the combinations to get an equivalent picture. Looking at three faces prepared to rotate, label one square's edges, rotate the cube, and label the edges of the other squares to match where the labeled square lands.
Now the orientations of the R can be stored as just the label for its top edge, and rotating the arrangement of 3 faces will just shift the list to the right or to the left. (a,b,c)*rot=(b,c,a), and so on. If you play with it you can see how to list the distinct vertex arrangements without skipping or relisting any. I counted 25 arrangements. After using your two laws, I count 8.
EDIT: And last night I found there can be no solution.
