1. Label the figure as shown. Clearly (1 3) and (2 4) are motions, as is their product. Hence the group of motions is {
ε, (13), (24), (13)(24)}, isomorphic to the Klein group
K4.

3. Label the figure as shown. Then (123) and (132) are motions (rotations of 120

and 240

about a line through vertex 4 and the center of the triangle base). Clearly every motion (indeed every symmetry) must fix vertex 4. Hence the group of motions is
G = {
ε, (123), (132)}. However (12), (13) and (23) are all symmetries (which are not motions), so the group of symmetries if
S3.

5. Label the figure as shown. Clearly (12)(34) and (14)(23) are such symmetries, and hence their product is (13)(24). The rest of the symmetries of the square do not preserve blue edges, so the group is {
ε, (12)(34), (13)(24), (14)(23)} ≅
K4.

7. Label the vertices as shown. Let λ = (1 3)(2 4)(5 7)(6 8) and μ = (1 6)(2 5)(3 8)(4 7). These are motions (rotations of π radians about axes through the ...