 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-oxford/2015/prelims/mathematics-i/6/a/ii/solution

G = ⟨ g ⟩

, any two elements are

g^{a}, g^{b}

, and

g^{a} g^{b} = g^{a + b} = g^{b + a} = g^{b} g^{a} .

Thus 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.

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