Solution
= Solution
Since $a^4=I$, multiplication adds exponents modulo four: $a^ja^k=a^{j+k\bmod4}$. Hence the <Cayley table> of the <cyclic group of order four>, with the left factor indexing rows, is
$$
\boxed{\begin{array}{c|cccc}
\cdot&I&a&a^2&a^3\\\hline
I&I&a&a^2&a^3\\
a&a&a^2&a^3&I\\
a^2&a^2&a^3&I&a\\
a^3&a^3&I&a&a^2
\end{array}.}
$$
The table is symmetric because the group is Abelian. Its row for each element permutes all four members, as cancellation requires.