Solutions 4

1. Let A be generated by r₁ : a rotation by 2p/n about a and let B be generated by r₂ : a rotation by 2p/n about b. Let g be any isometry that maps a to b. (for example T_a-b)
   Then r₂ = g ^-1r₁g (just check the fixed point.) and so B = g ^-1Ag as required.
   By definition the subgroups C_n are generated by such rotations and so are conjugate.
   To prove that any two copies of D_n are conjugate, suppose that they are generated by rotations about points a_i and reflections in lines l_i containing a_i for i = 1, 2.
   This time choose a symmetry which maps a₁ a₂ and l₁ l₂ (for example, T_a-b followed by a suitable rotation about b) and proceed as before.
   [Note that Leonardo's Theorem is really a classification of the finite subgroups of O(2) up to conjugacy by elements of O(2) .]
   Since conjugation by any element preserves the determinant, we cannot have D₁ (determinant = -1) conjugate to C₂ (determinant = +1).

2. If a subgroup of I(R) contained translations by arbitrarily small amounts it would not be discrete. Hence there is a translation T_m by some smallest number m. Then if there is a translation T_k by an amount which was not a multiple of m, we could write k = am + r with |r| < |m| and T_r would also be in the subgroup, giving a contradiction.

3. If the subgroup is generated by the smallest translation T_m then it is an infinite cyclic group C ( Z under +).
   Otherwise it will also contain a reflection and conjugating this reflection by the smallest translation will give a reflection in a different point and it may then be generated by these two reflections whose product will be the translation.

   The frieze groups (i) and (ii) are each generated by a single element of infinite order and hence are C.
   Note that although the two subgroups are isomorphic as groups, they are not conjugate subgroups in the group of all symmetries of the strip.
   The group (iii) is generated by a translation and an element of order 2 (reflection in the horizontal) which commutes with the translation and hence is C D₁.
   The group (iv) is generated by a pair of reflections (one through the centre of each symmetric motif and the other between each pair of motifs).
   The group (v) is generated by a pair of elements of order 2 (this time half turns, one about the centre of each motif and the other about a point between each pair of motifs).
   The group (vi) is generated by a pair of elements of order 2 (a reflection through the centre of each motif and a half turn about a point between each pair of motifs).
   The group (vii) is generated by the same pair of elements as in any of the previous three cases (giving the group D) together with reflection in the horizontal which commutes with everything else. Hence the group is D D₁.

4. Inversion in a is a rotatory reflection: reflection in any plane followed by rotation by p about an axis perpendicular to the plane.
   Following this with a rotation about the same axis gives a rotatory reflection. Hence any rotatory inversion is a rotatory reflection.
   The converse works similarly.

5. 1. To check the homomorphism property (AB) = (A) (B) for all A, B O(3) we must verify that det(AB) AB = det(A) A det(B) B and that det(AB) = det(A) det(B). The second of these is the standard property of the determinant while the first follows from the fact that multiplication by the scalar det(B) (which is 1) commutes with the matrix A.
      It is easy to check that is a bijection.

   2. The full symmetry group of X (which is a subgroup of O(3)) is mapped by to a subgroup of SO(3) < J >.
      [Note that (J) = (I, -1) which is why we take J as the generator of the second factor.]
      This subgroup is the group S_d(X) < J >.