I am trying to figure out the following problem using Burnside's lemma/formula:

How many different necklaces can we make using 12 equally spaced stones if we have 4 red, 5 green and 3 blue beads?

I am stuck when trying to find the fixed points under each action in D_{24}

I have so far found that if we let $D_{24} = <e,x,y| x^2 = 1, x^12 = 1, (xy)^2 = 1$ where x is a flip and y is a rotation of the necklace then

$|Fix(e)| = \frac{12!}{3!4!5!}$

$|Fix(y^j)| = 0 \forall j$

where $Fix(\cdot)$ is the set of elements in the set of necklaces that are fixed under the group action of an element in $D_{24}$

I am stuck trying to find $Fix(x^j)_{j \leq 2}$ and $|Fix(xy^j)|_{j \leq 12}$

any hints on how to approach this problem would be apprecaited


You've worked out the number of fixed points for every rotation. The other elements of $D_{24}$ are flips.

There are two kinds of flips: half are around an axis that pass through no beads, and half are around an axis that passes through two beads. The former are easy, the latter need some counting.


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