[Math] Burnside’s lemma – show that there are just five necklaces

abstract-algebradiscrete mathematicsgroup-theorypermutationssymmetric-groups

Show that there are just five different necklaces which can be
constructed from five white beads and three black beads. Sketch them.

The lemma tells us that

The number of orbits of G on X is $$\frac{1}{|G|} \sum _{ g \in
G}{|F(g)|}$$

The only thing i know is

  • |G|=16 since we have 8 corners (this gives us 7 rotations), and 8 line symmetry gives us 8 reflections and then we have the identity.

I would appreciate if you could tell something about $F(g)$ and how i calculate them. For an example, how do i know that there is none fixed configurations of the 7 rotations etc ? IF there are already same questions here in math.stack, then i would appreciate if you could link them. Thanks!

Best Answer

A necklace fixed by a reflection across a line containing no beads (how many of these are there?) would have an even number of black and white beads. Can you prove this?

Same goes for any nontrivial rotation $g$: partition the set of bead positions into orbits under $g$.

What kind of dihedral symmetries are left to account for? Can you count their fixed necklaces?

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