How Many Orbits Are Formed by Dihedral Group Actions on Colored Squares?

Ted123
Messages
428
Reaction score
0

Homework Statement



Let G=D_4 (the group of symmetries (reflections/rotations) of a square) and let X=\{ \text{colourings of the edges of a square using the colours red or blue} \} so a typical element of X is:
fyg7b4.jpg


What is the size of X?

Let G act on X in the obvious way. You are given that G has 6 orbits on X. Find a representative for each G-orbit, and its size.

The Attempt at a Solution



Obviously there are going to be various ways of colouring the edges of a square, but how can I be sure that I have them all or is there a quicker way to find the size?

How do I find a representative for each G-orbit?
 
Physics news on Phys.org
Size of X: Each side has a choice of two colours, and there are 4 sides.

Orbits of G - Some properties of X will not be affected by the symmetry operations. This will naturally lead to separate orbits.
 
Joffan said:
Size of X: Each side has a choice of two colours, and there are 4 sides.

Orbits of G - Some properties of X will not be affected by the symmetry operations. This will naturally lead to separate orbits.

The definition of orbit is \text{orb}_G(x)= \{ gx : g\in G \}
Saying that G has 6 orbits on X: does this mean there are 6 x's?

So by saying 'find a representative' does it mean find 6 different x's?
 
It means (as I read it) there are six categories of x that will be mapped only onto their own category by the symmetries in G.

For example: The x that consists of all blue edges will be mapped to itself by any G. It is an orbit of size one.
 
Joffan said:
It means (as I read it) there are six categories of x that will be mapped only onto their own category by the symmetries in G.

For example: The x that consists of all blue edges will be mapped to itself by any G. It is an orbit of size one.

Is the size of X 16?
 
That's what I get. It's a conveniently small-enough number that you can actually draw them all out, too, if you need to.
 
Last edited:
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top