I have a pyramid in the shape of a tetrahedron which is made out of 6 pegs and I want to use Burnside's Theorem to count the number of distinct designs if there are:

2 each of red,white and blue pegs.

I am having trouble determining the $|fix(g)|$.

One thing I think I know is that $|fix(identity)| = 6!$

But then any other rotation of Tetra only fixes edges not faces, so does this mean:

$orbits = 6!/12 = 60$ ?


Suppose $SABC$ is a tetrahedron. Let $M$ and $N$ be midpoints of $AS$ and $BC$ respectively. Conceder a rotation about $MN$ on $180^\circ$. Then $AS$ and $BC$ are fixed. Also we obtain: $AB↔SC$ and $AC↔SB$. And we have permutation on edges of the type $(.)(.)(..)(..)$.

There are 12 rotations of a tetrahedron: the identity, 8 rotations about the line connecting a vertex and the center of the opposite face (2 rotations on: $120^\circ$ and $240^\circ$ for each vertex), and 3 rotations (on $180^\circ$) about the line connecting midpoints of opposite edges. In terms of permutations on edges they are: 1 of type $(.)(.)(.)(.)(.)(.)$; 8 of type $(...)(...)$; and 3 of type $(.)(.)(..)(..)$.

Let $G$ be the group of rotations of a tetrahedron. Then we are looking for fixed points of elements of $G$ acting on the set of words of length 6 over 3-letter alphabet ($R$ - red, $W$ - white and $B$ - blue, and we're allowed to use two $B$'s, two $W$'s and two $R$'s). So, we have a fixed point if all cycles of a permutation are 'painted' in one color. For example, $\textrm{WWRRBB}$ is a fixed point of permutation $(1)(2)(34)(56)$.

Hint: In this situation the identity has the following cyclic structure $id = (.)(.)(.)(.)(.)(.)$. So, you need to 'paint' each cycles in one color. You can choose 2 red cycles in $C_6^2=\frac{6\cdot 5}{2}$ ways, then choose another 2 blue cycles in $C_4^2=\frac{4\cdot 3}{2}$ ways, the rest of cycles are white. So, we obtain $C_6^2 \cdot C_4^2=90$ fixed points for the identity. Fixed points for other rotations can be calculated similarly.

For example: A rotation of the type $(.)(.)(..)(..)$ has $3!$ possible ways to paint each cycle in one color (with restrictions given above). So, it has $3!=6$ fixed points.

  • $\begingroup$ I'm still lost, but thank you. $\endgroup$ – user96137 Mar 20 '14 at 16:57
  • $\begingroup$ Maybe I could help. What part is not clear? $\endgroup$ – user35603 Mar 20 '14 at 17:01
  • $\begingroup$ There are no rotations of Tetra that fix edges besides the identity? $\endgroup$ – user96137 Mar 20 '14 at 17:59
  • $\begingroup$ @user96137: I've updated my post. $\endgroup$ – user35603 Mar 20 '14 at 18:49

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