Burnside's Lemma: Lighting Arrangements

[Elzair]

Homework Statement

A ring of ten lights is hanged around the inside of a window in the following fashion.

1   2   3
10      4
9       5
8   7   6

Each light may be either on or off. Use Burnside's Lemma to find the number of unique arrangements of lights if the window can be be observed from either side (but obviously not rotated).

Homework Equations

Given a group G acting on a set X,
the number of distinct orbits = [tex] \frac{\sum_{g \epsilon G} \left| X_{g} \right| }{ \left| G \right| } [/tex]
where [tex] X_{g} = \left\{ x \epsilon X | g * x = x \right\} [/tex]

The Attempt at a Solution

If each light can be either on or of, the number of colorings = 2
If the window can only be flipped, then [tex] G = \left\{ \mu_{0} , \mu_{1} \right\} [/tex], so [tex] \left| G \right| = 2[/tex]

When the window is flipped, the 2 & 7 lights remain the same:

3   2   1
4       10
5       9
6   7   8

[tex]\mu_{0} =[/tex]
( 1 2 3 4 5 6 7 8 9 10 )
( 1 2 3 4 5 6 7 8 9 10 )

[tex]\mu_{1} =[/tex]
( 1 2 3 4 5 6 7 8 9 10 )
( 3 2 1 10 9 7 8 6 5 4 )
[tex]\mu_{1} = (1 3)(4 10)(5 9)(6 7)[/tex]

Alright, how do I proceed?

 

[lippyka]

X_{\mu_{0}} = \left\{ (1, 2, 3, 4, 5, 6, 7, 8, 9, 10) \right\}X_{\mu_{1}} = \left\{ (1, 3), (4, 10), (5, 9), (6, 7) \right\}Therefore, the number of distinct orbits = \frac{\left| X_{\mu_{0}} \right| + \left| X_{\mu_{1}} \right|}{\left| G \right|} = \frac{1 + 4}{2} = 2.5