# Burnsides Theorem Application - Necklace

Having a lot of trouble understanding this question: How many different necklaces can be made from eight black beads and four white beads?

What I know:

Burnsides Theorem is given by: $$\frac{1}{|G|} \sum_{g \in G} |\operatorname{fix}_{\Omega} (g)|$$

8+4=12 and this corresponds to the Dihedral group of order 12 (I use the notation $$D_{2n})$$ so I will note this as $$D_{24}$$, also known as the symmetric 12-gon.

So we know that $$|G|=24$$ and the total number of necklace configurations is given by $$\binom{12}{4}=495$$ so I'm guessing we have to $$Fix(e)=495$$? But I'm unsure how to apply the rest of Burnsides theorem from here and work out the configurations of rotations and reflections.

Any help would be great, thanks.

• Think about the rotations and reflections of the dodecagon, and consider how each acts on its vertices. – Lord Shark the Unknown Jan 1 at 7:17
• That's what I'm having trouble with visualising, this is known as Fix(g) right? @LordSharktheUnknown – Reety Jan 1 at 7:25
• Make/buy a small dodecahedron. I did for, e.g., my module on the group theory of virology when I was an undergraduate; it helped considerably. – Shaun Jan 1 at 11:41

Given that we have dihedral symmetry we suppose we are working with bracelets (naming convention by OEIS). This requires the cycle index $$Z(D_{12})$$ of the dihedral group $$D_{12}.$$ We have for the cyclic group that

$$Z(C_{12}) = \sum_{d|12} \varphi(d) a_d^{12/d} = \frac{1}{12} (a_1^{12} + a_2^6 + 2 a_3^4 + 2 a_4^3 + 2 a_6^2 + 4 a_{12}).$$

We get twelve more permutations corresponding to flips about an axis passing through opposite slots or opposite edges, for a result of

$$Z(D_{12}) = \frac{1}{24} (a_1^{12} + a_2^6 + 2 a_3^4 + 2 a_4^3 + 2 a_6^2 + 4 a_{12}) + \frac{1}{24} ( 6 a_1^2 a_2^5 + 6 a_2^6).$$

By the Polya Enumeration Theorem (PET) we are interested in the quantity

$$[B^8 W^4] Z(D_{12}; B+W).$$

Working through the terms we find

• $$[B^8 W^4] (B+W)^{12} = {12\choose 8},$$
• $$[B^8 W^4] (B^2+W^2)^{6} = [B^4 W^2] (B+W)^{6} = {6\choose 4},$$
• $$[B^8 W^4] 2 (B^3+W^3)^{4} = 0$$
• $$[B^8 W^4] 2 (B^4+W^4)^{3} = [B^2 W] 2 (B+W)^{3} = 2 {3\choose 2}$$
• $$[B^8 W^4] 2 (B^6+W^6)^{2} = 0$$
• $$[B^8 W^4] 4 (B^{12}+W^{12}) = 0$$

We get from the reflections

• $$[B^8 W^4] 6 (B+W)^2 (B^2+W^2)^5 \\ = [B^6 W^4] 6 (B^2+W^2)^5 + [B^7 W^3] 12 (B^2+W^2)^5 + [B^8 W^2] 6 (B^2+W^2)^5 \\ = [B^3 W^2] 6 (B+W)^5 + [B^4 W] 6 (B+W)^5 \\ = 6 {5\choose 3} + 6 {5\choose 4}$$
• $$[B^8 W^4] 6 (B^2+W^2)^{6} = [B^4 W^2] 6 (B+W)^{6} = 6 {6\choose 4}.$$

Collecting everything we find for our result that it is

$$\frac{1}{24} \left({12\choose 8} + {6\choose 4} + 2 {3\choose 2} + 6{5\choose 3} + 6{5\choose 4} + 6 {6\choose 4}\right).$$

which yields

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