# Burnside's lemma - show that there are just five necklaces

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!

• Do you have any thoughts? Have you tried anything? – blue May 26 '14 at 18:25

## 2 Answers

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?

• What do they mean with "Any choice of positions for the three black beads leaves three gaps of total length 5. These cannot be of equal size, so no rotation can ﬁx them" ? – Jane May 26 '14 at 19:36
• @Jane They mean exactly what they just said right in the quote. What part don't you understand? Do you know what they mean by gaps? Do you know why the gaps' total length is five? Do you know why they can't be equal size? Anyway, the hint I have in my answer is different from that hint (which, as it happens, you failed to mention in your question). – blue May 26 '14 at 20:10
• I did find the hint later. Yes i know what they mean by gaps but i dont know why they can't be equal size. But, i know that we have 7 rotations, 8 reflections where 4 of them passes through the opposite vertices and 3 of them passes their midpont. A conﬁguration ﬁxed by a reﬂection of this type are 2 (for each reflection) ? – Jane May 26 '14 at 21:57
• I did know unterstand the other (see the comment on the answer below :)). But i still have problem with the rotation. Would appreciate if someone could help me with it :) – Jane May 26 '14 at 22:34
• @Jane the gaps comprise the 5 white beads. Can the number 5 be split into 3 equal parts? – blue May 27 '14 at 1:11

$$5=\frac{{8 \choose 3}+6+6+6+6}{16}$$

• hey. sorry for late answer. But you did get 4*6 like this, right? 4 och de 8 reflections are the one passing through the opposite vertices. If we let one of the black beads be on one of the 2 vertices on the axis, and the other 2 black beads occupies one of the 3 other axis so that that the black beads are on the opposite verticex. If we do this, we will get 2*3=6 configurations for each reflection of this kind. @you-sir-33433 – Jane May 26 '14 at 22:30
• if we look on the line symmetries passing through the midpoint of opposite sides then none of the beads will be on the axis. and since we have 3 black beads we can't let the black beads occupies in the opposite vertices (one of the 3 will have an white bead as opposite and therefor it will not be fixed) – Jane May 26 '14 at 22:33
• @Jane That's exactly right. For the rotations, consider the orbit of a single bead. For a 90 degree rotation to fix one of the necklaces, for example, it must be that every bead rotates into one of the same color. This cannot happen due to parity. – Slade May 26 '14 at 22:38
• Thanks. Now i did understand. I made an different exercise with 15 white beads and 3 black beads arranged on the edge on a triangle with 7 beads on each edge. Any chose of the arrangemen of the three black beads will give us gaps with total length 5. Then i did check if any on the length will give equal length and saw that we could get the equal length 5. total we have 6 of these arrangement so this gives us 6+6=12 configurations (for the 2 possible rotations). And for the reflections i did almost as same argument in the comment above. But the rotations in squares with $nxn$ grid are harder:/ – Jane May 27 '14 at 1:24