# Burnside lemma application

The problem is to find rotationally indistinguishable necklaces formed from beads of 2 colors with exactly 3 beads of each color i.e, each necklace is of length 6. If $X$ is the set of all possible permutations of these 6 beads, then $|X| = 6!/(3!)^2 = 20$. I tried to apply Burnside Lemma to this by considering the cyclic group $C_6$ of rotations and get $\frac{1}{6}(20+2)$: 6 is the order of the cyclic group, 20 for the identity element, $2!=2$ if you rotate the necklace by 2 beads. I dont think that there are any elements of X fixed for any other rotations. But this answer is obviously wrong, as it is not an integer.

Can you please help explain where I went wrong and how I can apply Burnside Lemma properly (or why I can't use it here if that is the case)?

• You have to add up Fix(g) for all 6 elements of the group – Count Iblis Jun 13 '18 at 11:20
• Thanks! I thought I checked all the elements of the group and that the only cases where there were fixed elements were for identity and rotation by 2 beads. For the 2 bead rotation the fixed elements are (1, 2, 1, 2, 1, 2), (2, 1, 2, 1, 2, 1) [1 and 2 are colors]. Looking at them now, it looks like they are fixed for even rotation by 4 beads which adds another 2 to the summation. I also wrote a small program to check if there are any other fixed elements for other rotations and there are none. That makes the answer to be 4 instead of a fraction which matches what my program generates. – user625 Jun 13 '18 at 11:38

## 1 Answer

You just forgot the rotation of $4$ beads (the inverse of the rotation of $2$ beads) which adds another $2$ at the numerator and we finally obtain $$\frac{20+2+2}{6}=4$$ that is $(1,1,1,2,2,2)$, $(1,1,2,1,2,2)$, $(1,1,2,2,1,2)$, $(1,2,1,2,1,2)$.

• Yes, got it now. Being a novice in group theory I started doubting whether I made some other basic error. Accepting your answer. – user625 Jun 13 '18 at 11:41