The group $(\mathbb Z/7\mathbb Z)^\times$ is cyclic of order six, generated by $3$. Put $\omega=e^{2\pi i/6}$. The six characters are
$$
\chi_j(3^r)=\omega^{jr},
\qquad 0\leq j\leq5,
$$
and vanish on multiples of $7$. Since
$$
(3^0,3^1,\ldots,3^5)\equiv(1,3,2,6,4,5)\pmod7,
$$
their values are explicitly
$$
\begin{array}{c|rrrrrr}
a&1&2&3&4&5&6\\ \hline
\chi_j(a)&1&\omega^{2j}&\omega^j&\omega^{4j}&\omega^{5j}&\omega^{3j}.
\end{array}
$$
Taking $j=0,\ldots,5$ lists all six characters.


Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 123 / 3 / d / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-123/3/d/solution