Application of Burnside's Lemma

  • Thread starter Jay121
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  • #1
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Struggling to apply Burnside's Lemma to an example for the Dihedral group (3)

i.e

We colour the vertices of an equilateral triangle with three colours: red, blue, green.
We do not distinguish colourings if they can be obtained from each other by rotations
or reflections. Use Burnside’s Lemma to count the colourings.

Any help would be great (particularly on how to identify fixed points!)

Cheers
 

Answers and Replies

  • #2
22,089
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So, Burnside's lemma tells us that if G is a finite group that acts on X, then

[tex]|X/G|=\frac{1}{|G|}\sum_{g\in G}{|X^g|}[/tex]

Where X/G is the number of orbits of G, and Xg is the elements fixed by G.

Now, we put X the set of all coloured triangles. Dih(3) acts on this. Two triangles are the same under rotation and reflection if and only if they are in the same orbit. So we need to count X/G.

So, we need to count Xg for each g. In total, there are 33 number of triangles.
  • There are 33 triangles fixed by the identity.
  • There are 32 triangles fixed under reflection, and there are 3 reflections
  • There are 3 triangles fixed under rotation, and there are 2 rotations.

Thus

[tex]|X/G|=\frac{1}{6}(3^3+3^2+2.3)[/tex]
 
  • #3
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Do you mean

[tex]
|X/G|=\frac{1}{6}(3^3+3.3^2+2.3)
[/tex]

As there are 3 reflections?

Thanks for this its helped alot :)
 
  • #4
22,089
3,291
Yes, that's what I meant :smile:
 
  • #5
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Found a trickier one =\

We colour the vertices of a regular hexagon with two different colours. We do not distinguish colourings if they can be obtained from each other by rotations or reflections. Use Burnside's Lemma to count the colourings.

So |G|=2^6=64

Any help on how I go about identifying the rotations and reflections? I know that D(6) acts on this so we have 6 rotations and 6 reflections

I think my main issue with these problems is actually defining what is meant by a fixed point in this situation

Cheers
 
  • #6
22,089
3,291
Draw a hexagon. Can you identify all the rotations and reflections in the hexagon? Turning the hexagon clockwise gives you rotations. The reflections can be described by picking two points and use that as reflection axis.

Now, for a certain transformation, a fixed point is simply a vertex that remains in the same spot before and after the transformation...

If you really understood the triangle example, then I don't think this should be a problem...
 

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