Past exam of the mathematics course of the University of Oxford / 2015 / Prelims / Mathematics I / 6 / a / ii / Solution

If $G=⟨g⟩$, any two elements are $g^a,g^b$, andThus every cyclic group is an abelian group.

For the converse, the Klein four-group $C_2\times C_2$ is an abelian group, with componentwise multiplication. Each nonidentity element has order of a group element two, so none generates its four elements. It is therefore not a cyclic group.