Course MT3818 Topics in Geometry

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## Symmetry groups of Platonic solids Some of the finite subgroups of I(R3) arise from these solids.

Definition
A convex regular solid in R3 is called a Platonic solid.

Remarks

1. A polyhedron is a region bounded by planes in R3. It has two-dimensional faces which meet in one-dimensional edges which meet in vertices.

2. A polyhedron is regular if all its faces, edges and vertices are equal.
That is all the faces meet at the same angle and that the same number of edges meet at the same angles at each vertex. This implies that all the faces are the same regular polygon.

3. These solids were first classified by Plato in about 400 BC. There are numerous other semi-regular solids studied later by among others Archimedes and Kepler.
Here are the numbers for the five Platonic solids.
 Number of faces Number of edges Number of vertices Edges per face Dual Tetrahedron 4 6 4 3 Tetrahedron Cube 6 12 8 4 Octahedron Octahedron 8 12 6 3 Cube Dodecahedron 12 30 20 5 Icosahedron Icosahedron 20 30 12 3 Dodecahedron

Note that all these satisfy Euler's theorem: V - E + F = 2.
The entry in the last column is the dual solid.
One gets the dual by joining the midpoints of adjacent faces and then filling in the solid.
This explains the symmetries between the entries for F and V in the table.

Theorem
The above five solids are the only regular solids.

Proof
Suppose we have r faces (each a regular n-gon) meeting at every vertex. (Such a solid is said to have Schafli symbol {n, r}.)
The angle at the corner of the n-gon is (n - 2)π/n and since the polyhedron is convex we must have r × (n - 2)π/n < 2π ⇒ (r - 2)(n - 2) < 4.
The only positive integers satisfying this are (n, r) = (3, 3), (4, 3), (3, 4), (5, 3), (3, 5) corresponding to the above five. We now investigate the symmetry groups of these solids.

1. The symmetry group of the tetrahedron S(T).

To calculate the order of the group, oberve that a given vertex can be moved to one of four positions. There is a choice of three for a second and two for a third. Hence |S(T)| = 24.
Any symmetry determines a permutation of the four vertices so we get a map θ : S(T) S4 to the Symmetric group which is easily seen to be an isomorphism.
Since a transposition (swapping a pair of vertices) corresponds to a reflection (an opposite symmetry), the subgroup Sd(T) of direct symmetries corresponds to the Alternating subgroup A4.

All the other Platonic solids are symmetric about their centres and so (See Exercises 4 Question 5) the full group of symmetries S(X) is isomorphic to the direct product Sd(X) × < J > where Sd(X) is the subgroup of direct symmetries and < J > is the subgroup generated by the map x ↦ -x.

1. The symmetry group of the cube or octahedron S(C).

Because these two solids are dual to each other they have the same symmetry group.
Arguing as in the last case, the order of the group of direct symmetries (all rotations) is |Sd(C)| = 8 × 3 = 24.
The elements are:
3 rotations (by ±π/2 or π) about centres of 3 pairs of opposite faces. 
1 rotation (by π) about centres of 6 pairs of opposite edges. 
2 rotations (by ±2π/3) about 4 pairs of opposite vertices (diagonals). 
Together with the identity this accounts for all 24 elements.

The orders of these elements suggests the SLd(C) ≅ S4. In fact every rotation determines a permutation of the four diagonals and this defines the isomorphism.

Hence Sd(C) ≅ S4 and S(C) ≅ S4 × < J > with order 48.

2. The symmetry group of the dodecahedron or icosahedron S(D).

Because these two solids are dual to each other they have the same symmetry group.
Arguing as before, the order of the group of direct symmetries (all rotations) is |Sd(D)| = 20 × 3 = 60.
The elements are:
4 rotations (by multiples of 2π/5) about centres of 6 pairs of opposite faces. 
1 rotation (by π) about centres of 15 pairs of opposite edges. 
2 rotations (by ±2π/3) about 10 pairs of opposite vertices. 
Together with the identity this accounts for all 60 elements.

This suggests that Sd(D) ≅ A5 which has 24 5-cycles, 20 3-cycles and 15 permutations of the shape (..)(..).

In fact one can embed five cubes in the dodecahedron which are permuted by each rotation. Alternatively, one may embed five tetrahedra (partitioning the 20 vertices) and these are permuted also.

Hence Sd(D) ≅ A5 and S(D) ≅ A5 × < J > with order 120.

Remark

It is tempting to believe that the full symmetry group S(D) is actually isomorphic to S5 but one can check that the reflections in S(D) lead to even permutations of the tetrahedra and so the full symmetry group is not S5.

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JOC March 2003