# Find matrix representation for rotating/reflecting hexagon

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1. Mar 11, 2015

### c3po

1. The problem statement, all variables and given/known data
Consider the set of operations in the plane that includes rotations by an angle about the origin and reflections about an axis through the origin. Find a matrix representation in terms of 2x2 matrices of the group of transformations (rotations plus reflections) that leaves the snowflake invariant. Is the group Abelian or non-Abelian?

2. Relevant equations
R = [cosθ sinθ]
[-sinθ cosθ]

Reflection so that x → -x:
[-1 0][x] = [-x]
[0 1][y] [y]

Reflection so that y → -y:
[1 0][x] = [x]
[0 -1][y] [-y]

Abelian: A.B = B.A

3. The attempt at a solution
I know that I will be rotating the regular hexagon by increments of 60° so that each time I rotate, the hexagon is invariant. I know that the regular hexagon has 2n = 12 different symmetries, so I will have 12 2x2 matrices in total. I began by constructing my rotated matrices (0°, 60°, 120°, 180°, 240°, and 300°). I know that I will have 6 axes of symmetry to reflect by . . . 3 that stretch from vertex to vertex and 3 that stretch from the midpoint on one side of the hexagon to the opposite.

I think I am finding that this group will be non-Abelian, but I am not sure. I have calculated the 6 matrices for the rotations but am unsure of what to do next. If I reflect each rotation so that the +x values become -x I get 6 matrices, but those are only for the symmetries along the axes from vertex to vertex. This cannot be correct since I am leaving out the symmetries that exist from midpoint to midpoint. How do I correct my approach so that it includes these symmetries?

I have a midterm exam in the class this material appears in tomorrow morning and I know for certain that a problem very similar to this one will be on my exam. I went to my professor for help and he explained the general method for solving it, but I need some extra guidance! I am stuck. Please help!

2. Mar 11, 2015

### RUber

What would you need to rotate by to reflect parallel to the edges? Maybe 1/2 of the angle to get from vertex to vertex?
Reflecting each rotation is not the same as reflecting over a line of symmetry...you need to rotate back when you are done. In that way, the matrices corresponding to opposing vertices will be redundant, so you will only have 3 (as expected) for the vertex-vertex reflections.

3. Mar 11, 2015

### Dick

I'm not exactly sure what you are asking. But you say you know there are 12 elements in the group. The 6 rotations form an abelian subgroup. If you multiply those by ANY reflection $r$ then you will get all 12 group elements. The other reflections are products of a rotation and $r$. To determine if the group is abelian just check if the rotations commute with $r$.

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