Find thee number of necklaces given 3 beads that can take 4 colours, two necklaces are considered the same if the colours can be matched by rotation or flipping them.
My approach is by using Burnside's lemma, and so I consider the group $D_3$. To calculate the identity $|F(r^0)|$ we have 4 colours in 3 possible spot, $|F(r^0)|$=4^3=64. I imagine it as a string with 3 spots before it's combined to form a necklace.
Burnsides lemma: $|X/G|=1/|G|*\sum_{g\in G}|F(g)|$, where $G=$$D_3$, $|X|=|F(r^0)|=64$, $|G|=6$
If $g \in G$ is a flip by one of the three flip-symmetries, two of the beads with $4^2$ choices together will force the color of the third bead. And there are 3 flips i.e $|F(g)|=3*4^2$ for the flips.
If $g \in G$ is a rotation (not identity) there are only 4*2, since both of these rotations only count when all beds are of the same colour.
And so I get: $|X/G|=1/6(4^3+3*4^2+2*4)=20$ distinct necklaces, do you agree?
Thank you in advance!
Yes, that's right. But you have so few beads that you can do without Burnside's lemma: the number of necklaces of beads of one color is 4; the number of two-color necklaces is 12; the number of three-color necklaces is 4. The total is 20.