Solution
= Solution
By (d), every group of <group order> three is a <cyclic group>. Writing its elements as $e,a,a^2$, every product is determined by $a^3=e$; the <multiplication table> is
$$
\begin{array}{c|ccc}
\cdot&e&a&a^2\\\hline
e&e&a&a^2\\
a&a&a^2&e\\
a^2&a^2&e&a
\end{array}
$$
The map $a^j\mapsto j\pmod3$ is a bijective <group homomorphism> to addition modulo three. Therefore
$$
\boxed{G\cong\mathbb Z_3,}
$$
so there is exactly one such <group> \b[up to isomorphism].