 Solution

All six reflections square to the identity, and the rotation $R^3$ is the half-turn. These exhaust the nonidentity involutions: $R^k$ has square one only when $2k\equiv 0(\mathrm{mod}6)$, giving $k=3$ apart from the identity. Thus the order-two subgroups areA reflection cannot have order three. The rotations of order three are $R^2$ and $R^4$, and they generate the same subgroup:All these subgroups are cyclic, because a group of prime order is cyclic; the displayed nonidentity elements supply their generators. This gives seven subgroups of order two and one of order three, consistently with involutions in an even dihedral group and the classification of subgroups of a dihedral group.