Past exam of the mathematics course of the University of Cambridge / 2002 / ia / Paper 3 / 2B (Algebra and Geometry)

A group $G$ is a cyclic group if some element $g\in G$ generates it:For a finite cyclic group, the order of a group element of a generator equals the number of elements of the group.

An example of a finite abelian group that is not a cyclic group is the Klein four-group, realized aswith componentwise addition modulo two. Addition commutes, so this is an abelian group of four elements. Each of its three nonzero elements satisfies $g+g=0$, so the cyclic subgroup it generates has just two elements; the identity generates only itself. No element generates all four elements.

Yes: every finite group of rotations about the origin in the plane is cyclic. A planar rotation is determined by an angle modulo $2\pi$, and composition adds those angles. If $G$ consists only of the identity, it is a cyclic group. Otherwise, finiteness allows us to choose the smallest positive representative angle $\alpha \in (0,2\pi )$ occurring in $G$.

Take any planar rotation in $G$ with representative angle $\theta \in [0,2\pi )$. Write $\theta =q\alpha +r$, where $q$ is an integer and $0\le r<\alpha$. The composition $R_{\theta }{R}_{\alpha }^{-q}$ belongs to $G$ and has angle $r$. Minimality of $\alpha$ forces $r=0$, so every element of $G$ is a power of $R_{\alpha }$.

Similarly, write $2\pi =m\alpha +r$ with $0\le r<\alpha$. If $r>0$, the planar rotation $R_{\alpha }^{-m}$ has representative angle $r$, a contradiction. Thus $\alpha =2\pi /m$ andThis proves finite groups of planar rotations are cyclic by an elementary angle argument.