Math 402 Assignment 7. Due Wednesday, November 28, 2012

Chapter 6.

Suppose that a group G permutes the elements of a set S, where the permutations are described by a homomorphism φ : G → P erm(S).

The stabilizer of an element t ∈ S is the subgroup Gt = {g ∈ G|φ(g)(t) = t}, the elements of G that give permutations of S that fix the particular element t. The orbit of the element t is the subset {phi(g)(t)|g ∈ G} of S. Denote the orbit of t by G t ⊂ S.

We have the following counting formula: ◦(G)= ◦(Gt) ◦(G t), i.e., the number of elements in the orbit of t is the ratio of the order of G to the order of the stabilizer of t.

That formula is established directly from the coset theory of stabilizer subgroup.

1(7.1) Let D4 be the group of symmetries of the square. D4 permutes the set S of the four vertices of the square.

(a). Find the stabilizer in D4 of a vertex of the square.

Solution. Using the rotation subgroup of D4, we see that the orbit of a vertex v is the whole set V of vertices. Hence, by the counting formula, the stabilizer of the vertex v consists of two elements. One element is the identity element, and the other is the reflection in the diagonal through v.

(b). Find the stabilizer in D4 of a side of the square.

Solution. Using the rotation subgroup of D4, we see that the orbit of a side s is the whole set S of sides. Hence, by the counting formula, the stabilizer of the side s consists of two elements. One element is the identity element, and the other is the reflection in the line through the origin that is perpendicular to the side s.

(c). Find the stabilizer in D4 of a diagonal of the square.

Solution. The two diagonals form a set M of two elements. Let d be a diagonal. Since one diagonal can be rotated to the other diagonal, the orbit of d is the whole set M. By

1 the counting formula, the stabilizer of d is a subgroup of 4 elements. The stabilizer is C2 × C2 = × where a is the 180-degree rotation and b is the reflection in the diagonal d.

2(7.5) Find the group G of symmetries of the cube.

Solution. There are three planes that are perpendicular to the cube and bisect a face in the line that goes from the center of one edge of the face to the center of the opposite edge of the face. The symmetries of the cube permute those planes, which form a single orbit. Hence, the order of the symmetry group of the cube equals 3 times the order of the stabilizer of a plane. The stabilizer equals the product of D4, the symmetry group of a face parallel to the plane, and the group {I, R} where R is the reflection in the plane. Hence, the stabilizer has 3(16)=48 elements. We would like to find the structure of the symmetry group G of the cube. Half of the 48 elements of the group will be rotations, which form a subgroup H of G. The symmetries I and −I form a central subgroup of G. Since −I is not a rotation (it does not equal any of the rotations as described above), G = H ∪ (−I)H = H{I, −I}. To analyze the rotation subgroup H, the elements of H permute the set T consisting of the 4 pairs of opposite vertices of the cube, producing a homomorphism φ : H → P erm{T }. One can check that the kernel of φ is just {I} ∈ H, which shows that φ is injective. Since H and P erm{T } have the same number of elements, φ is then bijective.

Therefore, G = S4 × C2 as groups.

3(sec.7) Find the group symmetries of the tetrahedron.

Solution. By problem 5 below, there are 24 symmetries of the tetrahedron. In fact, the four vertices form an orbit for the symmetry group G, and the stabilizer is the symmetry group S3 of the triangular face opposite to the vertex. Those calculations give a symmetry group of order 4(6) = 24. We would like to show that the symmetry group G of the tetrahedron is abstractly the group S4. The homomorphism φ : G → S4, that takes each symmetry to a permutation of the 4 vertices that the symmetry determines, is bijective. One can check that it is injective by checking that its kernel is {I}; hence, it is bijective since the groups have the same number of elements.

4(8.2) Let H be a subgroup of a group G and let G operate on the set of left H cosets G/H by left composition: g(kH)=(gk)H, for g ∈ G and kH ∈ G/H. What is the stabilizer of kH in G?

2 Solution. We easily image that the stabilizer subgroup of kH is kHk−1. Let’s show that that is so. The stabilizer equation g(kH) = kH can be reformed to give the equation (k−1gk)H = H, which is the same as requiring that k−1gk be an element of H, i.e., that g ∈ kHk−1. Therefore, the stabilizer of kH is the subgroup kHk−1, which is conjugate to H (That subgroup could be H). The stabilizer of the coset H is H itself.

5(9.1) Let G operate on a set S. The counting formula tells us that

◦(G)= ◦(G s) ◦ (Gs), for s ∈ S. Use the counting formula to compute the number of symmetries of the cube and the number of symmetries of the tetrahedron.

Solution. The symmetry group G of the tetrahedron permutes the four vertices, and an element of the stabilizer of a vertex v must transform the opposite triangle into itself. Hence, the stabilizer has no more than 6 elements, by counting the symmetries of the triangle. Of course, we can find 6 symmetries of the tetrahedron that fix v. Hence, the stabilizer has 6 elements. The orbit of v has the four vertices as its elements since we can find a symmetry of the tetrahedron that takes v to any specified vertex. By the formula, G has 4 6 = 24 elements.

The symmetry group G of the cube permutes the set of 6 sides. The stabilizer of a side is forced to stabilize the opposite side. Then each element of the stabizer give a symmetry of the opposite side, and so, the stabilizer can have no more than 8 elements. Of course, one can easily find 8 elements that stabilize a side and the opposite side, and so, the stabilizer has 8 elements. Again, the orbit of a side is the full set of the 6 sides. Hence, G has 48 elements. (We gave a different way to compute the order of the symmetry group of the cube in problem 2.)

6. Find the order of the symmetry group G of the dodecahedron.

Solution. The 12 sides form a single orbit. Again, an element in the stabilizer of a side also stabilizes the opposite side, which has 10 symmetries each realizable by a unique symmetry of the dodecahedron. To see the uniqueness of the symmetry, we note that there is only one symmetry permuting the 5 vertices of a side in a specified way, since the composite of one with the inverse of the other would fix the five vertices of the side, and so it would fix the vertex adjacent to a vertex of the side. Continuing in this way, we find that all the vertices are fixed.

3 The order of the symmetry group is then 12(10) = 120, by the counting formula.

For an alternative calculation, in class we looked at the fifteen planes that bisect the dodec- ahedron and contain two opposite edges of the dodecahedron. They form a single orbit, and one can check that the stabilizer of the plane contains 8 elements. By the counting formula, the symmetry group has 15(8) = 120 elements.

Let us look for the structure of the symmetry group G. As for the cube, G has a central subgroup {I, −I}, and a 60 element rotation subgroup H. Then G = H ×{I, −I}. We look for the structure of H. There are 30 edges to the dodecahedron. Pass a plane through each edge and the center of the dodecaheron. That gives 15 distinct planes since opposite edges determine the same plane. Looking at the dodecahedron, we see that the planes split up into 5 subsets consisting of 3 perpendicular planes. Let T be the set of the 5 subsets. Since symmetries of the dodecahedron permute the edges and fix the center, they permute the 15 planes, and because symmetries preserve angles, they permute the 5 subsets of per- pendicular planes, i.e., the elements of the set T .

Let φ : H → P erm{T } = S5 be the homomorphism giving those permutations of T coming from the rotation subgroup of G. One can check that kernel of φ is {I}, and conclude that the image of φ is a 60 element subgroup of S5; hence, the image is A5, the only index 2 subgroup of S5. Thus, φ : H → A5 is bijective. Therefore, the symmetry group of the dodecahedron is A5 × C2.

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