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# Question about symmetry groups watch

1. http://www2.imperial.ac.uk/~mwl/m2p2/M2P210SH6.PDF

Question 6 is an application of Burnside's lemma. For 6(a) why are we using the rotation group of the tetrahedron, rather than the rotation and reflection group?
2. Because if you can't get from one colouring to another by rotation, then they're distinguishable. (Unless you have some magical way of physically reflecting a tetrahedron).
3. (Original post by DFranklin)
Because if you can't get from one colouring to another by rotation, then they're distinguishable. (Unless you have some magical way of physically reflecting a tetrahedron).
I don't quite follow. Then, why, when we use 2D shapes, do we take rotations and reflections? Is it because, in real life, we can't physically do a reflection on a tetrahedron?
4. Yes. (or at least, it depends on what the question says. If you color edges, you can reasonably talk about reflecting (turning the shape over). If you only color the faces on one side, you can't turn the shape over, so you can't get any reflections).

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Updated: April 8, 2011
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