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Disclaimer: Not a homework question it just only for learning

I know i'm supposed to use the formula: $\frac{1}{ \left |G \right |} \sum_{g \in G} \left | Fix(g) \right |$

And using the Burnside theorem, but I have problems running. I appreciate any explanation and help.

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  • $\begingroup$ Is the necklace the same if you put it on "backwards" or is there only one right way up? $\endgroup$ – mathreadler Jul 11 '17 at 18:16
  • $\begingroup$ This is a nice practice problem for the Burnside theorem, but it's actually easier to solve just by drawing pictures. $\endgroup$ – Barry Cipra Jul 11 '17 at 18:40
  • $\begingroup$ So the first thing you'll want to do is identify $G$. Since the necklace is a circle with six relevant "points" (equally spaced WLOG), we have $G \cong D_6$, the symmetries of a hexagon. So what is $|G|$? What do elements $g \in G$ look like? $\endgroup$ – Kaj Hansen Jul 11 '17 at 18:42
  • $\begingroup$ @Kaj "Necklace" usually means up to rotation and "bracelet" (or "turnover necklace") usually means up to rotation and reflection. $\endgroup$ – Trevor Gunn Jul 11 '17 at 20:47
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Here $G=D_6$ that is the group of the $12$ symmetries of the regular hexagon. $G$ is made of $6$ rotations and $6$ reflections. Now we evaluate the cardinality of the fixed point set for each $g\in G$:

  • $g$ is the identity then $|\mbox{Fix}(g)|=\binom{6}{3}=20$ (why?);

  • $g$ is the $\pm60^\circ$-clockwise rotation then $|\mbox{Fix}(g)|=0$ (why?);

  • $g$ is the $\pm120^\circ$-clockwise rotation then $|\mbox{Fix}(g)|=2$ (why?);

  • $g$ is the $180^\circ$-clockwise rotation then $|\mbox{Fix}(g)|=0$ (why?);

  • $g$ is a reflection across a line through two opposite vertices then $|\mbox{Fix}(g)|=4$ (why?);

  • $g$ is a riflection across a line through the midpoints of two opposite sides then $|\mbox{Fix}(g)|=0$ (why?);

Hence, by Burnside's formula, $$\frac{1}{ \left |G \right |} \sum_{g \in G} \left | \mbox{Fix}(g) \right |=\frac{20+2+2+4+4+4}{12}=3.$$

In fact, we have 1 necklace with three adjacent red beads, 1 necklace with exactly two adjacent red beads, and 1 necklace where beads are alternately colored.

P.S. If reflections are not allowed then $G=R_6$ that is the group of the six rotations of the hexagon, and, by Burnside's formula, the number of different necklaces is $$\frac{1}{ \left |G \right |} \sum_{g \in G} \left | \mbox{Fix}(g) \right |=\frac{20+2+2}{6}=4.$$

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  • $\begingroup$ By trial and error: $RRRBBB, RBRBRB, RRBBRB, BBRRBR$. Or are the last two equivalent by rotating? $\endgroup$ – farruhota Jul 11 '17 at 18:46
  • $\begingroup$ @Farrukh Ataev The last two are equivalent by reflection. Note that if we do not consider reflections then Burnside gives $\frac{20+2+2}{6}=4$. $\endgroup$ – Robert Z Jul 11 '17 at 18:50
  • $\begingroup$ "Necklace" usually means up to rotation and "bracelet" (or "turnover necklace") usually means up to rotation and reflection. $\endgroup$ – Trevor Gunn Jul 11 '17 at 20:47
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Since you can't embed images in comments I am posting this as an answer.

You can also simply draw these:

(from the Wikipedia page for necklaces, created by Wikipedia user Watchduck (Tilman Piesk)).

There are 3 bracelets (dihedral symmetry) and 4 necklaces (rotational symmetry). The fourth necklace is obtained as the mirror image of the middle one in the above picture.

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