Using group actions and counting, we now classify the finite groups of rotations of . Recall from Chapter 9 that is the group of orthogonal operators on (with the standard inner product) of determinant 1.
11.3.1 Rotations and Symmetry Groups
Proposition 11.3.1 (Elements of are rotations). Let with . The vectors fixed by form a line , the axis of , and maps the plane to itself and acts on it as a rotation by an angle . In particular fixes exactly two unit vectors , where ; we call them the poles of .
Proof. By the structure theorem for orthogonal operators (Theorem 9.5.19), is an orthogonal direct sum , where and acts on each plane by a rotation through an angle . A rotation of a plane has determinant 1, so , and is even. Since and , there are two possibilities: ,,, where acts on as the rotation by ; or ,,, where acts on as a rotation by . In both cases the fixed vectors of are those of the line .∎
Given a subset , its rotation group is a subgroup of . It may be infinite: if is a circle centered at the origin, it contains every rotation about the axis perpendicular to the plane of the circle. For the following shapes it is finite.
Example 11.3.2 (Cyclic and dihedral rotation groups). Let be a regular -gon () in the -plane, centered at the origin. Its rotation group consists of the rotations about the -axis by the angles , and the half-turns (rotations by ) about the lines in the -plane through the origin and a vertex or the midpoint of an edge. Each of these rotations maps the -plane to itself, and restricting to that plane is an isomorphism onto the symmetry group of the polygon: the rotations about the -axis restrict to the rotations of the polygon and the half-turns to its reflections, and an element is recovered from its restriction, since it acts on the -axis by . So the rotation group of is isomorphic to the dihedral group (Definition 2.2.14).
If instead is a pyramid built on the -gon, with apex on the positive -axis, the half-turns are lost (they move the apex), and the rotation group is the cyclic group of rotations about the -axis by the angles . For every , these rotations form a cyclic subgroup of of order .
Example 11.3.3 (Regular polyhedra). The rotation groups of the regular polyhedra centered at the origin are finite. The cube and the octahedron have the same rotation group: the centers of the faces of a cube are the vertices of an octahedron, and conversely, so a rotation preserving one preserves the other. In the same way the dodecahedron (12 pentagonal faces) and the icosahedron (20 triangular faces) have the same rotation group. The orders follow from the Orbit–Stabilizer Theorem applied to the faces: the rotation group acts transitively on the faces, and the stabilizer of a face with sides consists of the rotations about the axis through its center. This gives for the tetrahedron, for the cube and for the dodecahedron. We call these three groups the tetrahedral, octahedral and icosahedral groups.
11.3.2 Poles and the Classification
Let be a finite subgroup, and let be the set of poles of the elements of other than : If is a pole of and , then and ; so the poles of are the images under of the poles of . In particular , and acts on the finite set .
Lemma 11.3.4 (Stabilizers of poles). Let . Then consists of the identity and the elements of with axis ; in particular . Moreover is cyclic, of some order .
Proof. An element fixes exactly when is one of its poles, that is, when its axis is ; this condition is the same for . Each maps the plane to itself (for an orthogonal operator, the orthogonal complement of an invariant subspace is invariant, by Proposition 9.5.4), and is determined by its restriction to , since . In an orthonormal basis of whose first vector is , the matrix of is block diagonal with blocks and the matrix of , so . Restriction is therefore an injective homomorphism from to , and is isomorphic to the circle group (Proposition 9.5.17). Let . By Lagrange’s theorem every in the image satisfies , and the polynomial has at most roots, namely the th roots of unity . So the image is the group of th roots of unity, which is cyclic, generated by . Finally , because is a pole of some .∎
For we write . By Proposition 11.1.1, poles in the same orbit have the same .
Proposition 11.3.5 (The orbits of poles). Let be a finite subgroup of . The action of on its poles has two or three orbits.
If there are two orbits, each consists of a single pole, the two poles are , and is a cyclic group of rotations about the axis .
If there are three orbits , numbered so that the corresponding values of are increasing, then and are one of In the case ,, where for we write for the group .
In the cases and , all elements of of order greater than 2 are rotations about a single axis.
Proof. The set is finite and nonempty, since each has two poles. Count the set in two ways. Each has exactly two poles, so . Each is fixed by the elements of , so . Let be the orbits and the common value of on ; by the Orbit–Stabilizer Theorem . Hence Since , the left side lies in , and since , each term on the right lies in . So , that is, ; and , since then the right side would be less than 1.
Two orbits. Here reads . Each is the order of a subgroup of , so and ; equality in the sum forces . Then each orbit has element, so with both poles fixed by all of . Every has its two poles in , so its poles are and , and . So , which is cyclic by Lemma 11.3.4, and consists of rotations about .
Three orbits. Here reads If the left side is at most , so . If the left side is at most , so .
Suppose . Then , so , and . The orbit has elements, say and . The group permutes and fixes , so it fixes ; a non-identity element of has only the poles , so . By Lemma 11.3.4, is cyclic of order , and it has index 2 in . Let . Since preserves and moves ,, and then . The poles of are not , because moves and ; they are fixed by . If , its only poles would be ; hence . So every element of is a half-turn. Fix such an element . Then , with ; and since , also , that is, . These relations determine the product of any two elements of (using ), and the rotation and a reflection of the dihedral group satisfy the same relations; so is an isomorphism . (For the relations say that , the group of order 4 consisting of the identity and three half-turns.)
Suppose . Then , so , and gives respectively.
Finally, in case every element of is a rotation about , and in the case every element outside has order 2. ∎
It remains to identify the groups in the last three cases. We use that a cycle of odd length is a product of transpositions, , hence an even permutation.
Proposition 11.3.6 (The exceptional cases). Let be a finite subgroup of in one of the cases ,, of Proposition 11.3.5. Then ,,, respectively.
Proof. Recall that , so and have the same . When are distinct, each orbit is therefore closed under , and consists of axes.
Case . Here and . An element fixes only two unit vectors, so the action of on the four points of is faithful, and is isomorphic to a subgroup of of order 12. Since has index 2, it contains the eight 3-cycles of by Lemma 11.2.6. These are even, so is a subgroup with at least 9 elements whose order divides 12; hence .
Case . Here , and consists of 8 poles forming a set of four axes. The group acts on , since maps the line to ; let be this action. Let and let generate . The only line that maps to itself is : if for a unit vector , then is fixed by or by , both of which are non-identity elements with axis , so . Hence fixes the axis and permutes the other three axes without fixed points; since , it permutes them in a 3-cycle. So for each axis , the image contains a 3-cycle fixing and its inverse, and contains all eight 3-cycles of . Thus , so and . Suppose . As is normal, for every , so each maps the axis of to itself. Taking as above shows that the axis of is , for every ; this is impossible, since contains four different axes. So is injective, and .
Case . Here , and consist of poles, forming axes. Every has exactly one axis , and the elements with that axis are the non-identity elements of . Hence consists of the identity, 15 half-turns (one about each axis of ), 20 elements of order 3 (two about each axis of ) and 24 elements of order 5 (four about each axis of ); in particular has no element of order 4.
For , let be the half-turn with axis , so that . For , is a non-identity element fixing , so . Since is an orbit, the 15 half-turns form one conjugacy class, and the centralizer has order . Its elements other than have order 2 (there are no elements of order 4), so consists of , and two other half-turns, and it is abelian, because in a group in which for all we have . If is a half-turn in , then every element of commutes with , so , and the two groups are equal since both have order 4. So the 15 half-turns fall into five disjoint triples, and we obtain five subgroups , each of which is the centralizer of each of its three half-turns.
Since commutes with if and only if commutes with , we have . So acts on by conjugation, transitively because all half-turns are conjugate. Let be the kernel of this action; it is a normal subgroup of contained in the stabilizer of , which has order , so divides 12. If contained a half-turn, it would contain all 15 half-turns, since is normal; this is too many. By Lagrange’s theorem contains no element of order 5. If contained an element of order 3, with axis for some , then for every it would contain , a non-identity element of , and hence both elements of order 3 about the axis . As runs over , this gives all 20 elements of order 3, again too many. So , and is isomorphic to a subgroup of of order 60. Since has index 2, it contains the twenty 3-cycles and the twenty-four 5-cycles of by Lemma 11.2.6. These are even, so is a subgroup with at least 45 elements whose order divides 60; hence .∎
Theorem 11.3.7 (Finite subgroups of ). Every finite subgroup of is isomorphic to one of the following: a cyclic group (), a dihedral group (),,, or . Each of them occurs; in particular , and are the tetrahedral, octahedral and icosahedral groups.
Proof. The first statement follows from Propositions 11.3.5 and 11.3.6. The groups and for occur by Example 11.3.2, and occurs as the group of the identity and the half-turns about the three coordinate axes. The tetrahedral group has order 12 (Example 11.3.3) and contains rotations of order 3 about four different axes, those through the vertices. By the last statement of Proposition 11.3.5, it is neither of type nor of type ; since its order is 12, it is of type , and it is isomorphic to . In the same way the octahedral group, of order 24, contains rotations of order 4 about the three axes through the centers of opposite faces of the cube, so it is of type and isomorphic to ; and the icosahedral group, of order 60, contains rotations of order 5 about the six axes through the centers of opposite faces of the dodecahedron, so it is of type and isomorphic to .∎
Note. For a regular polyhedron the poles are the vertices, the centers of the faces and the midpoints of the edges (scaled to unit length). Poles at midpoints of edges have ; poles at vertices have equal to the number of faces meeting at a vertex; and poles at centers of faces have equal to the number of edges of a face. This gives for the tetrahedron (whose vertices and face centers form two different orbits with ), for the cube and the octahedron, and for the dodecahedron and the icosahedron, as in the figure.
The proof of Proposition 11.3.6 becomes concrete for the cube and the dodecahedron.
Example 11.3.8 (The cube and its four diagonals). Place the cube with vertices . Its vertices are the poles with , so in the proof of Proposition 11.3.6 the four axes in are the four long diagonals of the cube, and the rotation group of the cube is acting on its diagonals. One can also see directly that only the identity maps each diagonal to itself. The diagonals are spanned by ,, and . If a rotation maps each line to itself, then with , because preserves lengths. Since , and as are linearly independent, . So , and forces .
Example 11.3.9 (The icosahedral group and five tetrahedra). For the dodecahedron, the poles with are the centers of the 12 faces, those with are the 20 vertices, and those with are the midpoints of the 30 edges. The five subgroups of the proof of Proposition 11.3.6 have a geometric description, which the proof does not need. The 20 vertices of the dodecahedron split into five sets of four, each the vertex set of a regular tetrahedron inscribed in the dodecahedron (one of them is marked in the figure), and the rotations of the dodecahedron permute these five tetrahedra. Each consists of the identity and the three half-turns about the axes through the midpoints of opposite edges of one of the five tetrahedra. Since an isomorphism preserves the orders of elements, and the elements of of orders 5, 3 and 2 are the 5-cycles, the 3-cycles and the products of two disjoint transpositions, the isomorphism with matches
the 24 rotations about centers of faces with the 24 five-cycles,
the 20 rotations about vertices with the 20 three-cycles,
the 15 half-turns about midpoints of edges with the 15 elements .