Computing the Symmetry Groups of the Platonic Solids with the Help of Maple

Transcription

1 Computing the Symmetry Groups of the Platonic Solids with the Help of Maple In this note we will determine the symmetry groups of the Platonic solids. We will use Maple to help us do this. The five Platonic solids are the tetrahedron, the cube, the octahedron, the dodecahedron, and the icosahedron. We will view the symmetry groups of these solids in two ways. By definition, the symmetry group of a solid is the set of isometries of R 3 that stabilize the set. In each of these five cases we may view the solid as being centered at the origin. The center of each must be fixed by any symmetry. Thus, the symmetry group is then a subgroup of the group of linear isometries, which is the orthogonal group O 3 (R) of linear transformations that preserve dot products. A second interpretation of the symmetry groups is that the group is a subgroup of the group of permutations of the vertices of the solid. We may view O 3 (R) as the group of matrices { A Gl 3 (R) :A T A = I }. Then SO 3 (R) ={A O 3 (R) :det(a) =1} is a subgroup of index 2, and it coincides with the group of rotations of R 3. If z : R 3 R 3 is defined by z(x) = x for all x R 3,then z is a reflection, z is central in O 3 (R), and O 3 (R) =SO 3 (R) z = SO 3 (R) Z 2. The element z is a symmetry of all the Platonic solids except for the Tetrahedron. Thus, it will be sufficient for the remaining four solids to determine the rotation subgroup. For each of the Platonic solids, we will use a counting argument to find an upper bound for the order of the symmetry group. We will then, by intelligent trial and error, find three elements of the symmetry group and use Maple to show that the group generated by them has order exactly equal to this upper bound. By doing so, we can conclude that the upper bound is in fact equal to the order of the group, and that the group is generated by the three symmetries. In each case we will have the symmetry group G represented as G = a, b, c with a, b rotations and c a reflection. To help our counting argument, we point out that any symmetry is determined by its action on three vertices; to see this, if we center the solid at the origin, then three vertices represent three linearly independent vectors, or a basis of R 3. A symmetry fixes the center, so is a linear transformation, which is then determined by its 1

2 action on a basis. Alternatively, we can use the fact that an isometry of R n is determined by its action on n + 1 points. Thus, an isometry of a Platonic solid is determined by its action on 4 points. Since the center must be sent to itself, an isometry is determined by what it does to three vertices. We will use this fact without further comment for each of the five solids. In order to determine the symmetry groups, we use a fact from the theory of group actions. If G is a group and H a subgroup, then there is a homomorphism ϕ : G Perm(G/H), where G/H is the set of left cosets of H in G. This homomorphism is given as follows. For g G, letp g be the permutation G/H G/H given by P g (xh) =gxh. The map ϕ is then defined by ϕ(g) =P g. It is a group homomorphism and ker(ϕ) isthelargest normal subgroup of G contained in H. If [G : H] =n, then Perm(G/H) = S n,soϕyields a homomorphism from G to S n. These facts can be found in Walker [1, Theorem ]. Maple can calculate ker(ϕ); this normal subgroup is called the core of H in G. To help understand our choice of Maple commands, we summarize our findings. We will see, without much trouble, that the symmetry group of the tetrahedron is S 4.Forthecube and octahedron, the symmetry group is isomorphic to the direct product S 4 Z 2. The group S 4 will be obtained as the rotation subgroup R of G, but more directly it will be generated by the rotations a, b, as mentioned above. We will show that R = S 4 by producing a subgroup H of R whose core in R is 1 and with [H : R] = 4. We then get an injective map R S 4. We show that R = 24, and this then yields R = S 4. With similar methods we show that the symmetry groups of the dodecahedron and icosahedron are both isomorphic to the direct product A 5 Z 2.InallcaseswechooseH to be the normalizer in R of a p-sylow subgroup of R, for an appropriate prime p. It is no coincidence that the symmetry group of the octahedron and the symmetry group of the cube are isomorphic, as are the groups for the dodecahedron and icosahedron. There is a notion of duality of Platonic solids. If you take a Platonic solid, put a point in the center of each face, and connect all these dots, you get the skeleton of another Platonic solid. The resulting solid is called the dual of the first solid. For instance, the dual of the octahedron is the cube, and the dual of the cube is the octahedron. By viewing the dual solid as being built from a solid in this way, any symmetry of the solid will yield a symmetry of the dual, and vice-versa. From this we get that the symmetry groups of a Platonic solid and its dual are isomorphic. To fix notation, in each case we write G for the symmetry group of the given solid and R for the subgroup of rotations. 2

3 1 The Tetrahedron The tetrahedron T is a solid with four faces and four vertices. We view G as a subgroup of S 4 by viewing the symmetries of T as permutations of the four vertices. It is clear that G S 4 = 24; this is the one case in which we don t use a counting argument to get an upper bound. We now show that G is isomorphic to S 4.Leta be the counterclockwise rotation of 120 that fixes vertex 4 and let b be the counterclockwise rotation of 120 that fixes vertex 1, and let c be the reflection that fixes vertices 1 and 4 and interchanges vertices 2 and 3. We now have Maple determine the group generated by a, b, c. > with(group): > a := [[1,2,3]]: > b := [[4,2,3]]: > c := [[1,2]]: > G := permgroup(4,{a,b,c}): 24 From the Maple output, we see that G = 24. Therefore, G = S 4. 2 The Cube The cube C has six faces and eight vertices. We view G as a subgroup of S 8 by viewing the symmetries of C as permutations of the eight vertices of C. Weuseacountingargument to get an upper bound for G. First, there are eight choices of where vertex 1 can be sent. 3

4 Once vertex 1 has been sent somewhere, there are three choices for where vertex 2 can be sent because there are three vertices connected to any given vertex. Finally, there are two choices for where vertex 4 can be sent since it must be sent to a vertex connected to the image of vertex 1 but not to the image of vertex 2. Thus, G = 48. Let a be the counterclockwise rotation by 90 that fixes the top and bottom faces, let b the rotation of 90 that sends the top face to the front face, and let c be the reflection about the plane parallel to the top and bottom faces that passes through the center of the square. We then use Maple to get the following output. > a := [[1,2,3,4],[5,6,7,8]]: > b := [[1,4,8,5],[2,3,7,6]]: > c := [[1,5],[2,6],[3,7],[4,8]]: > G := permgroup(8,{a,b,c}): > R := permgroup(8,{a,b}): > grouporder(r); > P := Sylow(R,3): > H := normalizer(r,p): > grouporder(h); > core(h,r); > center(g); permgroup(8, {}) permgroup(12, {[[1, 9], [2, 10], [3, 11], [4, 7], [5, 8], [6, 12]]}) From this, we see that G =48and{a, b} generates the rotation subgroup R, which then has order 24. Furthermore, since H is a subgroup of R of order 6, and since the core of H in R is trivial, we see that the homomorphism R Perm(R/H) = S 4 obtained from [1, Theorem ] is injective, and so R = S 4 since these groups have the same order. We pointed out in the introduction that if R is the rotation subgroup of the symmetry group G, then G = R Z 2.Thus,G = S 4 Z 2. 4

5 3 The Octahedron The octahedron is a solid with eight faces and six vertices. We view the symmetry group of the octahedron O as a subgroup of S 6 by viewing the symmetries of O as permutations of the six vertices. To get an upper bound for G, we note that there are six choices for where vertex 1 is sent, and once it is determined, there are four choices for where vertex 2 goes, since each vertex is connected to four other vertices. Finally, there are two choices for where vertex 5 is sent, since it must be send to a vertex connected both to the image of 1 and to the image of 2. Thus, G = 48. Let a be the rotation of 90 that fixes the top and bottom vertices, b be the rotation of 120 that fixes the front face, and let c be the reflection about the plane containing vertices 1, 2,3, 4. We have the following Maple output. > a := [[1,2,3,4]]: > b := [[5,1,2],[6,3,4]]: > c := [[5,6]]: > G := permgroup(6,{a,b,c}): > R := permgroup(6,{a,b}): > grouporder(r); > P := Sylow(R,3): > H := normalizer(r,p): > grouporder(h); > core(h,r); permgroup(6, {}) 5

6 > center(g); permgroup(6, {[[1, 3], [2, 4], [5, 6]]}) From the output, we get G = 48 and that the rotation subgroup R is generated by a and b and has order 24. As in the case with the cube, we conclude that G is the direct product S 4 Z 2. 4 The Dodecahedron The dodecahedron D has twelve faces and twenty vertices. We view the symmetries of D as permutations of the twenty vertices. To get an upper bound for G, we note that vertex 1 can be sent to any of the twenty vertices. Once vertex 1 is determined, vertex 2 must be sent to one of three vertices since each vertex is connected to three others. Finally, there are two choices for where vertex 6 is sent since it must go to a vertex connected to the image of vertex 1. Thus, G = 120. Let a be the rotation of 72 that fixes the top and bottom faces, let b the rotation of 72 sends vertex 1 to 6 and vertex 6 to vertex 7. Finally, let c be the reflection about the plane containing vertices 1, 6, 11, and 18. We have the following Maple output. > a := [[1,2,3,4,5],[6,8,10,12,14],[7,9,11,13,15],[16,17,18,19,20]]: > b := [[1,6,7,8,2],[3,5,15,16,9],[4,14,20,17,10],[12,13,19,18,11]]: > c := [[2,5],[3,4],[7,15],[8,14],[9,13],[10,12],[16,20],[17,19]]: > G := permgroup(20,{a,b,c}): > R := permgroup(20,{a,b}): > grouporder(r);

7 > P := Sylow(R,2): > H := normalizer(r,p): > grouporder(h); 12 > core(h,r); permgroup(20, {}) > grouporder(center(g)); 2 We conclude that G = 120 and that the rotation subgroup R is generated by a, b and has order 60. Since H is a subgroup of R of order 12, and the core of H in R is trivial, we have an injective homomorphism R S 5. The only subgroup of S 5 of order 60 is A 5.Thus, R = A 5. As we mentioned earlier, G = R Z 2,soG = A 5 Z 2. 5 The Icosahedron The icosahedron I has twenty faces and twelve vertices. Let a be the rotation of 72 that fixes the top and bottom vertices, and let b be the rotation that permutes vertices 1, 2, 6. Also, let c be the reflection across the plane containing vertices 6 and 12, along with 4 and 7. > a := [[1,2,3,4,5],[7,8,9,10,11]]: > b := [[6,1,2],[3,5,7],[4,11,8],[9,10,12]]: > c := [[1,2],[3,5],[8,11],[9,10]]: > G := permgroup(12,{a,b,c}): 120 7

8 > R := permgroup(12,{a,b}): > grouporder(r); 60 > P := Sylow(R,2): > H := normalizer(r,p): > grouporder(h); 12 > core(h,r); permgroup(12, {}) > grouporder(center(g)); 2 From the output we have the same conclusion as for the dodecahedron; the group R is isomorphic to A 5,andG is the direct product A 5 Z 2. References [1] E. Walker, Introduction to Abstract Algebra. 8