Brian Bi
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Section 6.9. The Counting Formula

Exercise 6.9.1 The group $$T$$ of rotational symmetries of the tetrahedron acts transitively on the vertices of a tetrahedron, so the orbit of a vertex has size 4. The stabilizer of a vertex consists of the three rotations around the axis passing through that vertex and the centroid of the opposite face, with angles $$0, 2\pi/3, 4\pi/3$$. So in total the order of $$T$$ is $$4 \times 3 = 12$$.

The group $$O$$ of rotational symmetries of the cube acts transitively on the vertices of a cube, so the orbit of a vertex has size 8. The stabilizer of a vertex again consists of three rotations, where the axis is the body diagonal containing the vertex. So the order of $$O$$ is $$8 \times 3 = 24$$.

Exercise 6.9.2 $$G_v$$ consists of the rotations through angles of 0, 120, and 240 degrees around the body diagonal containing $$v$$. These rotations fix $$v$$ and the vertex opposite $$v$$, while $$G_v$$ acts transitively on the three vertices adjacent to $$v$$ and on the three vertices adjacent to the opposite vertex. So the formula here is $$8 = 1 + 1 + 3 + 3$$.

$$G_e$$ consists of the identity and the 180 degree rotation around the axis joining the midpoint of $$e$$ to the midpoint of the opposite edge. These two edges are fixed by $$G_e$$, while none of the other 10 edges are. So the other 10 edges can be partitioned into 5 orbits consisting of 2 edges each, where each edge is mapped to the other by the non-identity element of $$G_e$$. So the formula is $$12 = 1 + 1 + 2 + 2 + 2 + 2 + 2$$.

$$G_f$$ consists of the rotations through angles of 0, 90, 180, and 270 degrees around the axis that passes through the centre of $$f$$ and the face opposite to $$f$$. These rotations fix $$f$$ and the face opposite $$f$$ while $$G_f$$ acts transitively on the other 4 faces. So the formula here is $$6 = 1 + 1 + 4$$.

Exercise 6.9.3 The order of the group $$I$$ consisting of the orientation-preserving symmetries of a dodecahedron was computed using the counting formula in Example 6.9.5(a); it equals 60. The dodecahedron is also symmetric with respect to inversion, $$i$$, which is orientation-reversing. The map $$g \mapsto ig$$ takes orientation-preserving symmetries to orientation-reversing symmetries, is one-to-one, and is also onto since if $$h$$ is an orientation-reversing symmetry then $$i(ih) = h$$ and $$ih$$ is orientation-preserving. This implies that there are also 60 orientation-reversing symmetries, for a total of 120.

Exercise 6.9.4 Label the four vertices of the tetrahedron as $$P_1, P_2, P_3, P_4$$. Consider the scalar triple product $$\Delta = \det(P_1 - P_2 \mid P_1 - P_3 \mid P_1 - P_4)$$. Observe that any orientation-preserving symmetry of the tetrahedron will preserve $$\Delta$$ while any orientation-reversing symmery will change the sign of $$\Delta$$. Also observe that $$\Delta$$ changes sign under any transposition of two vertices. This implies that an orientation-preserving symmetry must induce an even permutation of the vertices while any orientation-reversing symmetry must induce an odd permutation of the vertices. Say $$Q_1, Q_2, Q_3, Q_4$$ is some even permutation of the points $$P_1, P_2, P_3, P_4$$; then if we perform some rotation that maps $$P_1$$ to $$Q_1$$, since this induces an even permutation of the four vertices, the remaining permutation required to bring the other three vertices to their desired positions $$Q_2, Q_3, Q_4$$ must also be even, so it is some 3-cycle, which can be effected by a rotation around the axis passing through $$Q_1$$ and the centroid of the opposite face. This implies that to each even permutation of the vertices there corresponds at least one rotational symmetry of the tetrahedron. But there are 12 even permutations and 12 rotational symmetries, so these are in bijection. If on the other hand $$Q_1, Q_2, Q_3, Q_4$$ is an odd permutation of the points $$P_1, P_2, P_3, P_4$$ then we can first perform a reflection of the tetrahedron across one of its planes of symmetry (which passes through a vertex and an altitude of the opposite face) to make the remaining permutation required even, which can then be effected by a rotation; this implies that to each of the 12 odd permutations there corresponds some orientation-reversing symmetry of the tetrahedron that induces that permutation on its vertices, and again, as there are 12 of the latter, the two sets are in bijection. This implies that the full group of symmetries of the tetrahedron is isomorphic to $$S_4$$ when a symmetry is identified with the permutation it induces on the tetrahedron's vertices, and the orientation-preserving subgroup $$T$$ is isomorphic to $$A_4$$.