# coloured faces of an octahedron

• Sep 9th 2008, 05:57 AM
wik_chick88
coloured faces of an octahedron
how many distinct octahedra (all of the same size) are possible if each face of each octahedron is either red, blue or green? the octahedron is a regular solid with 6 vertices, 12 edges and 8 triangular faces.

im having trouble working out how many symmetries there are of an octahedron. 8 faces and 3 colours so different ways to colour the faces would be 8 choose 3 = 56. but obviously 2 ways might have the same physical result if the octahedron was rotated on a vertice, edge or face.

• Sep 9th 2008, 07:20 AM
ThePerfectHacker
Quote:

Originally Posted by wik_chick88
how many distinct octahedra (all of the same size) are possible if each face of each octahedron is either red, blue or green? the octahedron is a regular solid with 6 vertices, 12 edges and 8 triangular faces.

im having trouble working out how many symmetries there are of an octahedron. 8 faces and 3 colours so different ways to colour the faces would be 8 choose 3 = 56. but obviously 2 ways might have the same physical result if the octahedron was rotated on a vertice, edge or face.

This is a way to approach this problem.

Let \$\displaystyle G\$ be the group of all trasformations of the octahedron. In the other thread it was shown that \$\displaystyle |G| = 48\$. And \$\displaystyle X\$ 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 \$\displaystyle X\$ 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.
• Sep 16th 2008, 01:23 AM
wik_chick88
Quote:

Originally Posted by ThePerfectHacker
This is a way to approach this problem.

Let \$\displaystyle G\$ be the group of all trasformations of the octahedron. In the other thread it was shown that \$\displaystyle |G| = 48\$. And \$\displaystyle X\$ 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 \$\displaystyle X\$ 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.

i am still very confused. any more help you can give me?
• Sep 17th 2008, 07:02 AM
wik_chick88
Quote:

Originally Posted by ThePerfectHacker
This is a way to approach this problem.

Let \$\displaystyle G\$ be the group of all trasformations of the octahedron. In the other thread it was shown that \$\displaystyle |G| = 48\$. And \$\displaystyle X\$ 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 \$\displaystyle X\$ 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 \$\displaystyle X\$ are fixed. i understand burnside's lemma buttttt i dont know how many of the \$\displaystyle 3^8\$ elements are unchanged in each conjugacy class...PLEASE HELP?!?!?!
• Sep 18th 2008, 07:18 AM
wik_chick88
please anyone i really need help understanding this. i looked at the cube example in burnside's lemma in wikipedia but im stuck on how they get \$\displaystyle 3^3\$ and \$\displaystyle 3^4\$ etc for each conjugacy class...please someone help me understand this i even made my own little cube and octahedron but i still dont understand!!(Crying)
• Sep 18th 2008, 07:30 AM
asw-88
Does 486 seem like an outrageous answer? because thats the answer that I got...
• Sep 18th 2008, 04:51 PM
awkward
Quote:

Originally Posted by wik_chick88
how many distinct octahedra (all of the same size) are possible if each face of each octahedron is either red, blue or green? the octahedron is a regular solid with 6 vertices, 12 edges and 8 triangular faces.

im having trouble working out how many symmetries there are of an octahedron. 8 faces and 3 colours so different ways to colour the faces would be 8 choose 3 = 56. but obviously 2 ways might have the same physical result if the octahedron was rotated on a vertice, edge or face.