How many ways can you color the edges of a hexagon in two colors?

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SUMMARY

The discussion focuses on determining the number of unique ways to color the edges of a hexagon using two colors, considering symmetries from rotations and reflections. Utilizing Burnside's Lemma and the Orbit Stabilizer Lemma, the unique colorings are calculated for various numbers of colors. For two colors, the total unique colorings amount to 13. The symmetry group involved is Di6, which has an order of 12, comprising 6 rotation and 6 reflection symmetries.

PREREQUISITES
  • Understanding of Burnside's Lemma
  • Familiarity with the Orbit Stabilizer Lemma
  • Knowledge of symmetry groups, specifically Di6
  • Basic combinatorial principles related to colorings and permutations
NEXT STEPS
  • Study advanced applications of Burnside's Lemma in combinatorial enumeration
  • Explore the properties and applications of Dihedral groups in geometry
  • Learn about necklace permutations and their relation to combinatorial problems
  • Investigate other symmetry considerations in coloring problems, such as Cayley's formula
USEFUL FOR

Mathematicians, combinatorial theorists, and students studying group theory or symmetry in geometric contexts will benefit from this discussion.

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Homework Statement



How many ways can you color the edges of a hexagon in two colors? It is assumed two colorings are identical if there is a way to flip or rotate the hexagon.

Homework Equations



Orbit Stabilizer Lemma and Burnside's Lemma

The Attempt at a Solution



This, implements the Orbit Stabilizer Lemma and Burnside's Lemma (think necklace permutations) however, is there anything special or different to computing this because you are now dealing with the faces/edges rather than the vertices (or beads of a necklace)?

Thanks.
 
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To answer my own question: it does not matter.

Di6 symmetry with order 12.
6 rotation symmetries
6 reflection symmetries

By Burnside's Lemma:

ƒ(n) = \frac{1}{12}⋅(2⋅n + 2⋅n^{2} + 4⋅n^{3} + 3⋅n^{4} + n^{6})

Where
n := # of colors
ƒ(n) := # of unique colorings

n = 2
ƒ(2) = 13


n = 3
ƒ(3) = 92

n = 4
ƒ(4) = 430

n = 5
ƒ(5) = 1505


\cdots
 

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