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 sides). Also λμ = μλ = (1 8)(2 7)(3 6)(4 5) is the sides). Also λμ = μλ = (1 8)(2 7)(3 6)(4 5) is the rotation about a vertical axis. The group of motions is {ε, λ, μ, λμ} ≅ K₄. However, there are symmetries which are not motions. We have