 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2014/paper-2/10/ii/solution

Since

a^{4} = I

a^{j} a^{k} = a^{j + k mod 4}

. Hence the Cayley table of the cyclic group of order four, with the left factor indexing rows, is

\cdot I a a^{2} a^{3} I I a a^{2} a^{3} a a a^{2} a^{3} I a^{2} a^{2} a^{3} I a a^{3} a^{3} I a a^{2} .

The table is symmetric because the group is Abelian. Its row for each element permutes all four members, as cancellation requires.

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