# Burnside's Lemma and Necklaces

I have the following question that I am stuck on.

A jeweler is making necklaces containing 12 beads of two colors that can be worn on either side (that is, we consider necklaces up to both circular and flip symmetry). How many distinct necklaces can be created?

I know that two necklaces are distinct if one can not be rotated or flipped to form the other. I also know that we need Burnside's Lemma for this question. Burnisde's Lemma tells us that for a finite group $G$ acting on a set $X$, the number $|\mathcal{O}|$ of orbits is $$|\mathcal{O}|=\frac{1}{|G|}\sum_{g\in G}\mathrm{Fix}(g).$$

So in this case, the set $X$ is the collection of all configurations of necklaces, so $X$ is a set with cardinality $2^{12}$. Because we are considering both circular and flip rotations, the group acting on $X$ should be $D_{24}$, the set of symmetries of the regular 12-gon, right?

From here, I think that all we need to do is examine the elements of $D_{24}$ and find their fixed points. However, I'm not sure how to actually find the set of fixed points for all 24 elements in the group. Thanks in advance for any help!

• Have you tried to find the fixed point set of any element of the group at all, other than the identity? You will notice some useful patterns. – Qiaochu Yuan Oct 29 '17 at 22:52
• Yes, I have. So do I basically have to do that for all 24 elements in the group $D_{24}$? – Sir_Math_Cat Oct 29 '17 at 22:53
• That depends on how many patterns you noticed. One of them relates to when the answer will be the same for two different elements of the group. – Qiaochu Yuan Oct 29 '17 at 22:58
• I'm sorry, but I don't think that I follow. – Sir_Math_Cat Oct 29 '17 at 22:59
• You need to find the size of the fixed point set for every $g \in D_{24}$. But the size only depends on the conjugacy class of $g$ and that should cut the work down enough to make the problem manageable. – Rob Arthan Oct 29 '17 at 23:55