Quote:
Originally Posted by ThePerfectHacker  This is a way to approach this problem.
Let  be the group of all trasformations of the octahedron. In the other thread it was shown that  . And  set of all colorings (even the possibly similar one).
Now put these symettries into conjugacy classes which can be found here. For each conjugacy class, say 90 degree rotation, find which elements in are fixed and how many of them. This number gets multipled by six since there are 90 degree rotations. And go through each class. After you done the hard part the rest follows by Burnside's lemma - be sure to look at the example with the cube in that link, it is very similar. |
ok im not SO much confused anymore. ive made my own little octahedron and figured out the conjugacy classes:
- 1 identity
- 9 vertex rotations (90degrees)
- 6 edge rotations (180 degrees)
- 8 face rotations (4 of which are 120 degrees and 4 of which are 240 degrees)
i still dont know how to work out which and how many elements in
are fixed. i understand burnside's lemma buttttt i dont know how many of the
elements are unchanged in each conjugacy class...PLEASE HELP?!?!?!