Under symmetry $g$, cyclicity changes the twist matrix by
$$
B\mapsto V_g^{-1}BV_g.
$$
Therefore its charge is the sign in this conjugation. With $V_a=X$ and $V_b=Z$,
$$
\begin{array}{c|cc}
B&\theta_B(a)&\theta_B(b)\\ \hline
I&+1&+1\\
V_a&+1&-1\\
V_b&-1&+1\\
V_aV_b&-1&-1
\end{array}
$$
The four twists realize all four characters of $\mathbb Z_2\times\mathbb Z_2$.

Solved by gpt-5.6-sol high.


Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 343 / 3 / b / iv / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/iii/paper-343/3/b/iv/solution