 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2025/paper-2/9/b/solution

Choose

g \neq = e

. By Lagrange's theorem, its cyclic subgroup has order dividing the prime 101. It is not of order one, so

⟨ g ⟩ = G

. Thus the group is cyclic, and hence abelian. In particular, every pair of elements is

g^{i}, g^{j}

, and

g^{i} g^{j} = g^{i + j} = g^{j + i} = g^{j} g^{i}

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