Rigid motions of the dodecahedron 

 Denote the group of rigid motions of the dodecahedron and its dual, the icosahedron, by Icos. 

Icos is a group of order 60 which is found as follows. 

First there are twelve pentagonal faces so six pairs of opposite faces. Each face axis gives 4 rotations of order 5, namely rotations by 2p/5, 4p/5, 6p/5 and 8p/5.

 Then there are 20 corners so ten pairs of opposite corners. Each corner axis gives two rotations of order 3, namely rotations by 2p/3 and 4p/3.

Finally, there are 30 edges so 15 pairs of opposite edges. Each edge axis gives a rotation of order 2. 

There can be no more rigid motions of the dodecahedron: there are 12 ways to choose which face is at the bottom, and when chosen, it can be in any of 5 positions. Thus there are 60 possibilities and we have found 60 different rigid motions.

 It remains to show that in fact Icos is isomorphic to A₅. To do this, we show that it has a faithful action on a set of five elements whose image is A₅. The most transparent such set is constructed as follows:

 Choose any face, and on that face choose any diagonal AB connecting two non-adjacent corners. There are two other faces adjacent to corner B, say U and V. On the face U there are two diagonals adjacent to B, and one of them, say BC is at right angles to AB. Similarly, on face V there are two diagonals adjacent to B and one of them, say BE is at right angles to both AB and BC. Now go to vertex C and repeat the procedure. There are two face diagonals at right angles to each other and to BC. Hence one of them, say CD is parallel to the original AB. Then AD is also a diagonal of a face of the dodecahedron, and you have found a square ABCD cutting off a little 'hip-roof' of the dodecahedron. Now repeat the procedure at the vertex E to get another congruent square ABEF at right angles to ABCD. The whole procedure can now be repeated at corner A, and you finish up with a regular polyhedron of type {3,4}, in other words, a cube.

 The edges of this cube are diagonals of each of the 12 faces of the dodecahedron. The cube is completely determined once the diagonal AB is chosen. Since there are 5 possible choices for this diagonal, there are five such distinct cubes embedded in the dodecahedron, and they use up all the diagonals. Now every rigid motion in Icos maps diagonals to diagonals and preserves angles and distances, so in fact it maps cubes to cubes. The identity map preserves cubes, and if a maps cube C₁ to cube C₂, and b maps cube C₂ to cube C₃, then a.b maps cube  C₁ to cube C₃. In other words, we have a group action of of Icos on the set of five cubes.

 Only the identity map fixes all cubes so the action is faithful, and hence Icos is isomorphic to a subgroup of S₅ of size 60. But there is only one such subgroup, A₅. 

To see how that the 6 types of rigid motion correspond to conjugacy types in A₅, recall as in Section 4 that conjugate permutations have the same cycle type, but only conjugation by elements of A₅ preserves conjugacy type. Thus there is only one conjugacy type of elements of order 1 or of order 2, but two of elements of order 5 and two of elements of order 3.

By the way, this trick of embedding one platonic solid in another was well-known to the Ancient Greeks and to Kepler (German astronomer, 1571-1630). For example, you can embed two tetrahedra in the cube as follows: draw all the face diagonals; start with one face diagonal and trace it from corner A to corner B. At corner B there are two faces each with a diagonal adjacent to B. Continue until you return to corner A. Since all the diagonals are the same length and there are 3 at each corner you have the skeleton of a {3,3} i.e. a tetrahedron. You get the other one by starting with the diagonal crossing AB. Now the octahedral group acts faithfully on this set of two diagonals, so it must have a kernel of order 12. What is it? In other words, which rigid motions of the cube map each tetrahedron to itself?

Simplicity of A₅. 

1. Icos has 15 elements of order 2, all conjugate. They are rotations by p  about axes joining midpoints of opposite edges, and thay are all conjugate, since the icosahedral group is transitive on such axes.

2. Icos has 20 elements of order 3, which are rotations about the ten axes joining pairs of opposite corners.

 The ten rotations by 2p/3 form one conjugacy class, and the ten by 4p/3 form another. 

3. Icos has 24 elements of order 5, which are rotations about 6 axes joining the centroids of opposite pentagonal faces.

 The 6 rotations by 2p/5 form one conjugacy class, the 6 by 4p/5 another, the 6 by 6p/5 another, and the 6 by 8p/5 another. 

4. Every element of Icos is a product of elements of order 5.

1. We first show that some rotation by 2p/3 is a product of two elements of order 5. So let ABCDE and BCFGH be two pentagonal faces sharing the edge BC, so BA, BC and BH are edges meeting at B. Let r be anticlockwise rotation by 2p/3 about the axis through the centroid of ABCDE, so r maps A->B->C->D->E->A. Let s be anticlockwise rotation by 2p/3 about the axis therough the centroid of BCFGH, so s maps B->C->F->G->H->B. Thus r.s is a rotation which fixes B and maps A to H, C to A and H to C. It does a similar thing to the vertex opposite B and the three edges attached to it, and hence r.s is an anti-clockwise rotation by 2p/3.
2. Next, note that we could have started with any vertex in place of B, so every anticlockwise rotation by 2p/3 is a product of 2 rotations by 2p/5.
3. Taking the square of this map, we see that every anticlockwise rotation by 4p/3 is a product of four rotations by 2p/5.
4. Next, we return to our original pair of pentagons and show that rotation by p about the axis passing through the midpoint of the edge AB is a product of three rotations by 2p/5. For r maps the (directed) edge AB to BC and r.s maps BC to BA. Hence r.r.s reverses the edge AB, and does the same thing to the corresponding edge on the opposite side of the icosahedron, so is the rotation by p about an axis joining the midpoints of a pair of opposite edges.
5. Since we could have started with any edge in place of AB, we see that every element of A₅ of order 2 is a product of 3 rotations by 2p/5.

 

 

Thus we can conclude that A₅ is generated by elements of order 5, and furthermore, every element of order 5 is a conjugate of a multiple of any fixed element of order 5.

5. Every element of Icos is a product of elements of order 3.

 Let A and D be two adjacent corners, and let B, C and D be the 3 corners adjacent to A and A, E and F the 3 corners adjacent to D. Let G be the fifth corner of the pentagonal face having vertices CADF. Let a be anti-clockwise rotation by 2p/3 about the axis through A and b be anticlockwise rotation by 2p/3 about the axis through D. Thus a maps B->C->D->B and b maps A->F->E->A. Then the product a.b maps A->F->C->D->G->A, and so acts on the pentagonal face ACGFD as anti-clockwise rotation by 4p/5 and on the opposite face in a similar way. Taking the square, we see that rotation of this face by 2p/5 is a product of 4 rotations by 2p/3. Since we could have started at any face, we see that every rotation of order 5 is a product of rotations of order 3.

 Combining this result with the previous paragraph, we see that every element of A₅ is a product of elements of order 3. Furthermore, every element of order 3 is a conjugate of a power of a fixed element of order 3.
 
 

6. Every element of Icos is a product of elements of order 2.

 Just as in the two paragraphs above show that g.d acts as anti-clockwise rotation of the face ABCDE by 4p/5, so anticlockwise rotation by 2p/5 is the product of 4 rotations of order 2.

 Since we could have started with any face, every rotation of order 5 is a product of rotations of order 2, so by the first paragraph, so is every rotation of order 3. Thus A₅ is generated by elements of order 2, and every element of order 2 is conjugate to a fixed element of order 2. 

7. Icos, and hence A₅, is simple.

 With the machinery we have developed it is easy to prove this: for let G be a normal subgroup of Icos different from {1}.

 If G contains an element x of order 2, it contains all the conjugates of x, so all elements of order 2, and hence all products of elements of order 2. By the previous paragraph, G = Icos. 

If G contains an element y of order 3, it contains y² and all conjugates of y and y². Hence G contains all elements of order 3, and hence Icos.

 Finally, if G contains an element z of order 5, it contains z², z³ and z⁴ and each of their conjugates. Since Icos is generated by its elements of order 5, G contains Icos.
 

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 Last update Nov 24 1999

Author: Phill Schultz, schultz@maths.uwa.edu.au