in which we learn about the symmetry groups of polyhedra, and start exploring conjugacy classes in the symmetric groups .
Today we had another look at the rotations of a cube, and found that they form a group isomorphic to . We proved this by acting on the four diagonals of the cube and being able to generate all of as the image. The group of all symmetries of a cube is , because the “reflection in the centre point” symmetry commutes with all other symmetries. And by duality of regular polyhedra, this tells us the group of symmetries of an octahedron as well: the cube and the octahedron are dual in the sense that I can put each vertex of the cube onto the midpoint of a face of the octrahedron. See this picture.
We also investigated the symmetries of a tetrahedron and found that the rotations give us and the group of all symmetries is . While proving this we looked at the stabilisers of a vertex, say , and realised we can view it as either the rotations or all symmetries of the triangular face opposite to , which makes it nice and visual. Notice that the tetrahedron does not have the “reflection in mid-point” as a symmetry, because the vertices are opposite midpoints of faces rather than other vertices.
After these polyhedral symmetries we revisited symmetric groups and looked at their conjugacy classes. In , conjugacy classes are determined by cycle type. That makes them very easy to deal with.
Understanding today’s lecture
Rotations and general symmetries of regular polyhedra can be a bit confusing if you find visualising them difficult. But hopefully the Taylor-cube and a self-built tetrahedron might help with the visualisation.
The ccls in are something people usually get used to very quickly. Just be careful you don’t get too used to it, it doesn’t quite work like that in ! So never switch off your brain.
Preparing for Lecture 17
In the next lecture we will see what the ccls sizes of tell us about its normal subgroups and so its quotients. We will investigate how ccls work in . They are not quite the same there: some ccls from can split when we go to . We will look at ccls of and and use them to prove that is simple.
Going a little deeper
Having investigated the symmetry groups of cubes/octahedrons and tetrahedrons, you might wonder about the symmetry group of the dodecahedron (and icosahedron). Those are a bit more difficult, and I believe often get done in GRM in second year. But here are some thoughts so you can start trying. There are five cubes inscribed in a dodecahedron, which you can see by looking at Gareth Taylor’s picture and/or building your own from his net. So you can act with the rotations of the dodecahedron on those five cubes, and do similar arguments as we did for the cube acting on diagonals. You will find that you get as the rotational group. Then do the same argument as we did for the full symmetry group of a cube to see that the full symmetry group of a dodecahedron is . Notice it is not , as one might be tempted to think! You could also look here for some explanation by John O’Connor from St Andrews.