 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2024/paper-2/8/e/solution

e, a, a^{2}

, every product is determined by

a^{3} = e

\cdot e a a^{2} e e a a^{2} a a a^{2} e a^{2} a^{2} e a

a^{j} \mapsto j (mod 3)

is a bijective group homomorphism to addition modulo three. Therefore

G ≅ Z_{3},

so there is exactly one such group up to isomorphism.

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