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 June 27, 2002 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 as subgroups of the group of permutations of the vertices of the solid. 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 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), 1

2 where G/H is the set of left cosets of H in G. This homomorphism is given as follows. For g G, let P 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) is the largest normal subgroup of G contained in H. If [G : H] = n, then Perm(G/H) = S n, so ϕ yields a homomorphism from G to S n. These facts can be found in Walker [2, 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. For the cube and octahedron, the symmetry group is isomorhic to a semidirect 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 a semidirect product of A 5 by Z 2. At the end of this note, we will investigate semidirect products of S 4 and A 5 by Z 2. A consequence of the result we prove will show that the cube and the octahedron have isomorphic symmetry groups, as do the dodecahedron and the icosahedron. In all cases we choose H 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. 1 The Tetrahedron 2

3 The tetrahedron T is a solid with four faces and four vertices. We view Sym(T ) as a subgroup of S 4 by viewing the symmetries of T as permutations of the four vertices. It is clear that Sym(T ) 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 Sym(T ) is isomorphic to S 4. Let a 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}): > grouporder(g); 24 From the Maple output, we see that Sym(T ) = 24. Therefore, Sym(T ) = S 4. 2 The Cube The cube C has six faces and eight vertices. We view Sym(C) as a subgroup of S 8 by viewing the symmetries of C as permutations of the eight vertices of C. We use a counting argument to get an upper bound for Sym(C). First, there are eight choices of where vertex 1 can be sent. 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, Sym(C) = 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 3

4 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}): > grouporder(g); > R := permgroup(8,{a,b}): > grouporder(r); > P := Sylow(R,3): > H := normalizer(r,p): > grouporder(h); > core(h,r); > d := mulperms(c,mulperms(a,c)); permgroup(8, {}) d := [[1, 2, 3, 4], [5, 6, 7, 8]] > mulperms(mulperms(b,c),mulperms(b,c)); > groupmember(d,r); [] true From this, we see that Sym(C) = 48 and {a, b} generates a subgroup R of 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 [2, Theorem ] is injective, and so R = S 4 since these groups have the same order. Furthermore, the Maple output shows that conjugation by c sends R to itself, but is not the identity map on R, as 4

5 one of the commands shows that (bc) 2 = 1, so cbc 1 = cbc = b 1 b. Thus, Sym(C) is a nontrivial semidirect product of S 4 and c = Z 2. 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 Sym(O), 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, Sym(O) = 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}): > grouporder(g); > R := permgroup(6,{a,b}): > grouporder(r); > P := Sylow(R,3): > H := normalizer(r,p): > grouporder(h);

6 6 > core(h,r); permgroup(6, {}) > d1 := mulperms(c,mulperms(a,c)); d1 := [[1, 2, 3, 4]] > d2 := mulperms(c,mulperms(b,c)); d2 := [[1, 2, 6], [3, 4, 5]] > groupmember(d1,r); true > groupmember(d2,r); true From the output, we get Sym(O) = 48 and that the group R generated by a and b has order 24. As in the case with the cube, we conclude that Sym(O) is a nontrivial semidirect product of S 4 and 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 Sym(D), 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, 6

7 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, Sym(D) = 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}): > grouporder(g); > R := permgroup(20,{a,b}): > grouporder(r); > P := Sylow(R,2): > H := normalizer(r,p): > grouporder(h); > core(h,r); > H2 := normalizer(g,p): > grouporder(h2); > grouporder(core(h2,g)); permgroup(20, {}) > mulperms(mulperms(a,c),mulperms(a,c)); 24 2 [] 7

8 > d:=mulperms(c,mulperms(b,c)); d := [[1, 6, 15, 14, 5], [2, 7, 20, 13, 4], [3, 8, 16, 19, 12], [9, 17, 18, 11, 10]] > groupmember(d,r); true We conclude that Sym(D) = 120 and that the subgroup R generated by a, b 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. Moreover, G is not isomorphic to S 5 since we see that if H 2 = N G (P ), then the core of H 2 in G is nontrivial. However, the only nontrivial normal subgroup of S 5 is A 5 ; we thus conclude that G S 5. However, we see that Sym(D) is a nontrivial semidirect product of A 5 and 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}): > grouporder(g); > R := permgroup(12,{a,b}): > grouporder(r); 120 8

9 60 > P := Sylow(R,2): > H := normalizer(r,p): > grouporder(h); 12 > core(h,r); permgroup(12, {}) > mulperms(mulperms(a,c),mulperms(a,c)); [] > mulperms(mulperms(b,c),mulperms(b,c)); [] From the output we have the same conclusion as for the dodecahedron; the group R is isomorphic to A 5, and Sym(I) is a semidirect product of A 5 and Z 2. 6 Semidirect Products of N by Z 2 We have described symmetry groups of the Platonic solids in terms of semidirect products. It would appear that we have not completely determined them. For example, the symmetry group of the cube and octahedron are both semidirect products of S 4 by Z 2. Recall that if ϕ : H Aut(N) is a group homomorphism, then the semidirect product N ϕ H of N by H (determined by ϕ) is the set N H with multiplication given by (n, h)(n, h ) = (nϕ(h)(n ), hh ). We note that the isomorphism class of N ϕ H depends on the map ϕ. As an example, consider S n, which contains the normal subgroup A 5 of order 2. View Z 2 = [1, 2] S n. By the relation between internal and external semidirect products, we see that S n = An [1,2] Z 2. To simplify notation, if ϕ : Z 2 Aut(G), and if σ = ϕ(1), we write G σ Z 2 for G ϕ Z 2. Moreover, if σ = Int(u), we further simplify notation and write G u Z 2. Note that if σ = Int(u), then σ 2 = ϕ(1 + 1) = id, so u 2 Z(G). Since Z(S 4 ) and Z(A 5 ) are both trivial, any such u satisfies u 2 = e. We claim that, for any two nontrivial homomorphisms ϕ 1, ϕ 2 : Z 2 Aut(S 4 ), we have S 4 ϕ1 Z 2 = S4 ϕ2 Z 2. To verify this, we first recall that Sym(S 4 ) = S 4 ; this follows since every automorphism of S 4 is inner and Z(S 4 ) = e. Furthermore, every automorphism of A 5 9

10 is of the form Int(u) for some u S 5. These facts about the automorphism groups can be found in Section 3.2 of [1]. If u S 5 and σ = Int(u) A5, we write A 5 u Z 2 for A 5 σ Z 2. We note that if u A 5, then Int(u) A5 is an inner automorphism of A 5, while if u S 5 A 5, then it is not inner. Proposition If u, v S 4 have order 2, then S 4 u Z 2 = S4 v Z 2. Therefore, any two nontrivial semidirect products of S 4 by Z 2 are isomorphic, and are isomorphic to the symmetry groups of the cube and of the octahedron. 2. If u, v A 5 have order 2, then A 5 u Z 2 = A5 v Z 2 = Sym(D) = Sym(I). If u, v S 5 A 5, then A 5 u Z 2 = A5 v Z 2 = S5, and S 5 is not isomorphic to these symmetry groups. Proof. To prove (1), a short calculation shows that (e, 1)(g, 0)(e, 1) 1 = (e, 1)(g, 0)(e, 1) = (ugu, 0) in S 4 u Z 2. By some experimentation, one finds that (vu, 1)(g, 0)(vu, 1) 1 = (vu, 1)(g, 0)(vu, 1) = (vgv, 0). We define a map f : S 4 v Z 2 S 4 u Z 2 by f(g, 0) = (g, 0) and f(e, 1) = (vu, 1). In order for this to be a group homomorphism, we note that (g, 1) = (g, 0)(e, 1). Thus, we set f(g, 1) = (gvu, 1). A short check shows that f is indeed a bijective group homomorphism, which proves (1). A similar calculation shows that A 5 u Z 2 = A5 v Z 2 when u, v A 5 or u, v S 5 A 5. Note that in either of these cases, vu A 5, so the argument for (1) works in this case. Thus, there are at most two nonisomorphic (nontrivial) semidirect products of A 5 by Z 2 since [S 5 : A 5 ] = 2. We know that S 5 is one, and that S 5 = A 5 [1,2] Z 2. By our Maple calculations, Sym(D) and Sym(I) both contain a normal subgroup of order 2. Thus, they must both be isomorphic to A 5 u Z 2 for u A 5. By the way, if u A 5, we can see directly that A 5 u Z 2 has a normal subgroup of order 2, since if N = (u, 1), then (g, 0)(u, 1)(g, 0) 1 = (g, 0)(u, 1)(g 1, 0) = (gu, 1)(g 1, 0) = (gu ug 1 u, 1) = (u, 1) and (e, 1)(u, 1)(e, 1) 1 = (e, 1)(u, 1)(e, 1) = (u, 0)(e, 1) = (u, 1). Thus, N is indeed a normal subgroup of A 5 u Z 2 of order 2. References [1] M. Suzuki, Group Theory I, Springer-Verlag, Berlin, [2] E. Walker, Introduction to Abstract Algebra. 10