Brian Bi
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## Section 6.12. Finite Subgroups of the Rotation Group

Exercise 6.12.1 To every dodecahedron there corresponds an inscribed icosahedron whose vertices are located at the centres of the dodecahedron's faces. Every symmetry of the dodecahedron is therefore a symmetry of this inscribed icosahedron. The converse is also true: to every icosahedron we can associate a dodecahedron whose vertices are located at the centres of the icosahedron's faces, implying that every symmetry of the icosahedron is a symmetry of the inscribed dodecahedron also. This implies that the symmetry groups of the dodecahedron and icosahedron are the same.

Exercise 6.12.2 The poles fall into three subsets: the vertex poles $$P_v$$ which are unit vectors directed from the centre of the octahedron to one of the vertices; the edge poles $$P_e$$ which are directed from the centre to the midpoint of one of the edges; and the face poles $$P_f$$ which are directed from the centre to the centre of one of the faces. Any symmetry of the octahedron can only map a pole to another pole of the same type, by virtue of the fact that it maps vertices to vertices, edges to edges, and faces to faces. But $$O$$ acts transitively on the vertices, on the edges, and on the faces of the octahedron, so there is only a single orbit for each of $$P_v$$, $$P_e$$, and $$P_f$$.

Exercise 6.12.3 The order of $$O$$ is 24, and $$O$$ acts transitively on $$S$$ which has size 4, since to carry one body diagonal onto another it suffices to carry one of the vertices of the source diagonal onto one of the vertices of the target diagonal. Therefore the stabilizer $$O_D$$ of a single diagonal $$D$$ has order $$|O|/|S| = 6$$. The 6 elements of $$O$$ that fix $$D$$ are the three rotations $$\rho_0 = 1, \rho_{2\pi/3}, \rho_{4\pi/3}$$ around $$D$$ and the three rotations $$\rho_0 r, \rho_{2\pi/3}r, \rho_{4\pi/3}$$ where $$r$$ is some rotation that exchanges the two vertices of $$D$$; it is easy to see that these are 6 distinct elements of $$O$$. There is an isomorphism from $$O_D$$ to $$\operatorname{Perm}(S \setminus \{D\})$$ mapping a rotation to the permutation it induces on the other three body diagonals: the rotations by 120 and 240 degrees act as 3-cycles while $$r$$ acts as a transposition so that the $$\rho r$$'s correspond to odd permutations of $$S \setminus \{D\}$$. Therefore $$O_D$$ is isomorphic to $$S_3$$.

Exercise 6.12.4 The group $$O$$ has order 24, and acts transitively on the set $$S = \{t_1, t_2\}$$ of the two inscribed tetrahedra (for example, any of the 90 degree rotations in $$O$$ will exchange the two tetrahedra). Therefore the stabilizer $$H = O_{t_1}$$ has order 12. Now consider the rotational symmetries of $$t_1$$, which may or may not be symmetries of the containing cube. There are 12 such symmetries (the order of $$T$$). In fact, any permutation of the four vertices $$A, B, C, D$$ of $$t_1$$ must also permute the four affine combinations $$A + B + C - 2D, B + C + D - 2A, C + D + A - 2B, D + A + B - 2C$$, but these are precisely the other four vertices of the cube. So $$T$$ is a subgroup of $$H$$. And $$H$$ has the same order as $$T$$, so $$H = T$$.

Exercise 6.12.5 The icosahedral group has order 60 (Example 6.9.5(a)) and acts transitively on the set of 6 long diagonals of the icosahedron (diagonals that join a vertex to the opposite vertex). To see this, observe that to carry one diagonal onto another it suffices to carry a vertex of the source diagonal onto a vertex of the target diagonal. Therefore the stabilizer subgroup of one of these six diagonals has order $$60 \div 6 = 10$$.

Exercise 6.12.6

1. Every subgroup of the tetrahedral group is a finite subgroup of $$SO(3)$$. By the classification theorem (6.12.1), such a group must be cyclic, dihedral, or one of the groups $$T, O, I$$. Obviously $$T$$ is a subgroup of $$T$$ while $$O$$ and $$I$$ are not.

All elements of $$T$$ have order 2 or 3, which rules out any cyclic groups of order 4 or greater. (A 180 degree rotation that fixes an edge has order 2, while a 120 degree rotation that fixes a vertex and the opposite face has order 3.) The elements of orders 2 and 3 generate subgroups of $$T$$ isomorphic to $$C_2$$ and $$C_3$$.

Again, any $$D_n$$ with $$n > 3$$ is ruled out since $$T$$ has no elements of order greater than 3. The group $$D_1$$ is just $$C_2$$. Now consider the set consisting of the 3 pairs of opposite edges of the tetrahedron; $$T$$ acts transitively on this set, so the stabilizer of a particular pair of edges is a subgroup of order 4. This subgroup can't be $$C_4$$, so it must be $$D_2$$. To answer the question of whether $$D_3 = S_3$$ occurs as a subgroup of $$T$$, we use the fact from Exercise 6.9.4 that $$T$$ is isomorphic to $$A_4$$, and the solution to Exercise 6.11.1, which implies that the only faithful actions of $$S_3$$ on a set of 4 elements are those that fix one element while permuting the other three, therefore their images contain odd permutations and are not subgroups of $$A_4$$.

In conclusion, all the subgroups of $$T$$ are $$C_1, C_2, C_3, D_2$$, and $$T$$ itself.

2. Again we turn to (6.12.1). Consider first the case of cyclic groups. $$I$$ has elements of orders 1, 2, 3, and 5, and no other orders, so the cyclic groups $$C_1, C_2, C_3, C_5$$ are the only cyclic subgroups of $$I$$.

Similarly, as $$D_n$$ contains elements of order $$n$$, any dihedral group other than $$D_1, D_2, D_3, D_5$$ can't be a subgroup of $$I$$. Again, $$D_1$$ is just $$C_2$$, and again $$D_2$$ arises as the stabilizer group of a pair of opposite edges of an icosahedron. Now $$I$$ acts transitively on the set of the 10 pairs of opposite faces of an icosahedron, so the stabilizer group $$H$$ of one of these 10 pairs must have order 6. $$H$$ consists of three rotations by 0, 120, and 240 degrees, call them $$1, \rho, \rho^2$$, around the axis joining the centres of the two faces, plus the three rotations consisting of the composition of one of these three aforementioned rotations with some fixed rotation $$r$$ that exchanges the two faces. By labelling the three vertices of one of the faces, we can see that $$r\rho r^{-1} = \rho^{-1}$$, so $$H$$ is isomorphic to $$D_3$$. $$I$$ also acts transitively on the set of the 6 pairs of opposite faces of a dodecahedron, and by similar reasoning we find that a stabilizer subgroup of order 10 that is isomorphic to $$D_5$$.

We now turn our attention to the last three groups, $$T, O, I$$. $$I$$ is of course a subgroup of itself. $$O$$ contains elements of order 4, so it isn't a subgroup of $$I$$. We claim that $$T$$ is a subgroup of $$I$$. This argument is due to Steven Stadnicki. Let $$t$$ be one of the tetrahedra inscribed in a dodecahedron. Now $$T$$ is the symmetry group of $$t$$ and it is generated by the 120 degree rotations around an axis passing through a vertex of $$t$$ and the centre of the opposite face. But such an axis also passes through the opposite vertex on the dodecahedron, and the dodecahedron has 3-fold rotational symmetry around that axis. This implies that each rotational symmetry of $$t$$ is also a rotational symmetry of the icosahedron in which $$t$$ is inscribed. The 12 elements of $$T$$ therefore form a subgroup of $$I$$.

In conclusion, all the subgroups of $$I$$ are given by $$C_1, C_2, C_3, C_5, D_2, D_3, D_5, T$$, and the entire group $$I$$.

Exercise 6.12.7 Let $$P = (1, \alpha, 0).$$ If $$\alpha > 1,$$ then of the points $$(\pm 1, \pm \alpha, 0),$$ the one closest to $$P$$ other than itself is $$(-1, \alpha, 0),$$ which is 2 units away. The two points $$(0, -1, \pm \alpha)$$ are more than 2 units away from $$P$$, as are $$(-\alpha, 0, \pm 1).$$ Therefore the five points adjacent to $$P$$ must be $$(-1, \alpha, 0), (0, 1, \pm \alpha), (\alpha, 0, \pm 1).$$ In order for $$(1, \alpha, 0)$$ to be 2 units from $$(0, 1, \alpha),$$ we must have $$1 + (\alpha - 1)^2 + \alpha^2 = 4,$$ which has a single positive root $$\alpha = \frac{1}{2}(1 + \sqrt{5}).$$ With this choice of $$\alpha,$$ we can easily verify that the following five sets of points: \begin{array}{ccc} (1, \alpha, 0) & (-1, \alpha, 0) & (0, 1, \alpha) \\ (1, \alpha, 0) & (0, 1, \alpha) & (\alpha, 0, 1) \\ (1, \alpha, 0) & (\alpha, 0, 1) & (\alpha, 0, -1) \\ (1, \alpha, 0) & (\alpha, 0, -1) & (0, 1, -\alpha) \\ (1, \alpha, 0) & (0, 1, -\alpha) & (-1, \alpha, 0) \end{array} are equilateral triangles with side length 2, meeting at the vertex $$(1, \alpha, 0).$$ By applying rotations and reflections to $$(1, \alpha, 0)$$ we can similarly write down such a set of 5 equilateral triangles of side length 2 meeting at any of the other 12 given points. (For example, to get the five triangles meeting at $$(-1, \alpha, 0),$$ just reverse the sign of each $$\pm 1$$ entry in the table above.) Each triangle is counted 3 times, one for each of its vertices, so there are $$12 \times 5 \div 3 = 20$$ triangles in all, which therefore form the faces of a regular icosahedron.