Find the number of different significant colorings of tetrahedron in 2 colors.
Find, using the Burnside's lemma. It's not difficult to find this just exhausting all options, but it is necessary to use the Burnside's lemma. But I do not quite understand how to use it. Could you explain, please?
Solution 1:
Assuming that "different" means up to rotation (i.e. you can't turn the shape inside out), then the group has exactly twelve elements, i.e twelve "rotations" including the identity. Choose your favorite corner; it has 4 corners where it can go, and there are three orientations in which it can be placed there.
For each of these twelve operations, count how many colorings are invariant under that rotation. Example:
The identity rotation has every coloring invariant, so there are $2^4=16$ colorings.
Rotating at a corner, permutes the three other points. Hence those 3 points must all be the same color, and the corner has a color, so $2^2=4$ colorings.
Do this for all 12 rotations, then use the Burnside lemma formula, and you're done. It's actually not that bad, and it's a good way to be sure you didn't miss any coloring. Also, if your replace 2 colors with 20, it's no more difficult with Burnside, but good luck doing it on your own.
More clarity, as requested. A rotation moves the tetrahedron in a way that after the motion it occupies the exact same space as before. Corners go to corners, edges to go edges, and so on. If it weren't colored, it would look identical. If it is colored, it might still look identical if the colors are in the same places as before. A rotation of 180 degrees, through any axis, is not allowed because the tetrahedron is not in the same place as before.
One possible motion is no motion at all -- this is the identity. There are 2 colors for each of the four corners, or $2^4$ colorings, and all of them are unchanged if you don't move the shape. Now consider keeping the top corner in place, and rotating the base by 1/3. For the shape to look identical after this rotation, the colors on the bottom must be the same, or else the rotated shape would look different (a black corner suddenly turns white).
Solution 2:
There are twelve rotations of a tetrahedron:
- 1 identity rotation,
- 8 rotations around the altitude,
- 3 rotations around the axes passing through the midpoints of the opposite sides.
The identity rotation is a neutral permutation (no rotation at all with different colors). There are $2^4$=16 colorings under it.
A edges incident with the endpoint of the altitude around which we're rotating are the same color, and all the rest of the edges are another color. There are $2^2$=4 colorings under this rotation.
With such a rotation (around the axes passing through the midpoints of the opposite sides) two edges change places on one side and two on the other side. So there are $2^2$=4 colorings under it.
By Burnside's lemma, we have $$ 1/12(1*16+8*4+3*4)=5 $$ significantly different colorings of a tetrahedron with 2 colors.