Solutions 6

1. The direct symmetries or rotations of the tetrahedron map to the elements of A₄ × {I} S₄ × < J >.
   The opposite symmetries of the tetrahedron map to symmetries of the cube of the form P × J where P is an odd permutation in S₄.
   The kernel of the map is trivial.

2. All the opposite symmetries of the dodecahedron are rotatory inversions. That is, they are of the form J R with R a rotation.
   Hence by Exercises 4 question 4 they are all rotatory reflections. The only ones which are actually reflections are those of the form J H with H a half turn and the only half-turns which are rotatory symmetries of the dodecahedron have axes joining the centres of pairs of opposite edges.
   There are 15 such pairs and the reflections are then in the planes through the mid-points perpendicular to these pairs of edges.

3. The groups C₂ , C₁ × < J > , C₂ C₁ are all cyclic groups of order 2.
   However, C₂ is generated by a half turn H, C₁ × < J > is generated by central inversion J , while C₂ C₁ is the mixed group generated by J H which is reflection in a plane.
   Hence these three are not conjugate in I(R³).

4. The direct symmetries of P_n are the rotational symmetries of the regular n-gon (or dihedron, if you like) half-way down the prism. This is the same as the full symmetry group D_n of the regular n-gon in R² (where the elements of order 2 are reflections rather than rotations by π).
   If n is even, then P_n is centrally symmetric (and has x -x as a symmetry) and so the full symmetry group is D_n × < J > . If n is odd, then, as in the last question, the full symmetry group is the mixed group D_2nD_n .

   The rotational group of the square antiprism consists of 4 rotations about the line joining the centres of the squares (including the identity!) and 4 rotations by π about lines joining the centres of the edges joining the two squares.
   The group S_d D₄ .
   (If you cut the antiprism by a plane halfway up you get a regular octagon and some of the symmetries of this are symmetries of the solid.)
   The square antiprism is not centrally symmetric and so its full symmetry group is a mixed group which must be D₈D₄ .

5. To get a figure with rotation group C_n and no opposite symmetries paint a symbol like (but with n "arms") with only rotational symmetry on the bottom face of a pyramid or a regular n-gon.
   Painting the same figure on the top and bottom faces of an n-prism will give a figure with rotation group D_n and no opposite symmetries.
   Paint a (suitable situated) 3-armed version of this figure on each face of a tetrahedron, octahedron or icosahedron to get figures with the requisite rotation groups but no opposite symmetries.

   For figures with full symmetry group D_n × < J > take either a prism (n even) or an anti-prism (n odd).
   Paint the above symbol on the top and J(this symbol) on the bottom to get a figure with full symmetry group C_n × < J >.
   The stella-octangula, cube and dodecahedron give examples of centrally symmetric figures with direct symmetry groups of the Platonic figures.

6. The mixed group D_n C_n consists of n rotations (by multiples of 2π/n about the same axis) and n opposite symmetries of the form J H with H a half turn, which are reflections in planes containing this axis.
   Suitable figures include a right pyramid on a regular n-gonal base or an n-prism with its top surface painted a different colour.

   Choose a figure with full symmetry group D_n × < J > . (Either a prism or anti-prism depending on the parity of n.) Then mark a symbol like with only rotational symmetry on the top face and J(this symbol) on the bottom face.
   This figure will then have C_n × < J > as its full symmetry group.

   Similarly, when the full symmetry group of the prism/anti-prism is the mixed group D_2n D_n marking such symbols on the upper and lower n-gons will give a figure whose full symmetry group is C_2n C_n.

7. The picture shows that if one has an array of octahedra (four are seen from above in the diagram on the right) then one may add four tetrahedra between them (the dotted lines are the top edges) to form a "well" into which another layer of octahedra can be fitted.

   If you think about the portion shown in the diagram: you should notice that when you put in the bottom half of the next layer of octahedra, the bit you see consists of ¹/₂ an octahdron, four ¹/₈'s of an octahedron and four ¹/₂ tetrahedra. So you need twice as many tetrahedra as octahedra.

8. 1. Position a corner to see that for the direct symmetry group we have |S_d| = 4.
      The direct symmetries are the same as the direct symmetries of the rectangle where the tiles meet and so S_d = D₂ .
      Since the figure has central symmetry, the full symmetry group is S_d × < J > = D₂ × < J >.
      
   2. Again |S_d| = 4 and S_d = D₂ (the identity and rotation by π about three mutually perpendicular axes). This time the figure is not centrally symmetric but it does have some opposite symmetries. So the full symmetry group is a mixed group and the only possibility is D₄D₂ .
      
   3. As before S_d = D₂ but this time there are no opposite symmetries and so the full symmetry group is also D₂ .
      
   
   Colouring the tiles removes some of the possible symmetries. We get :
   
   1. S_d = C₂ and S = D₂C₂ since there is now no central symmetry.
      
   2. S_d = C₂ and S = D₂C₂
      
   3. S_d = C₂ and S = C₂