Because $C(t')=R(t)$,
$$
e(t')=\sum_{r\in R(t)}\operatorname{sgn}(r)\{rt'\}.
$$
Applying $\theta$ and using $r e(u)=\operatorname{sgn}(r)e(u)$ gives
$$
\theta(e(t'))=\left(\sum_{r\in R(t)}r e(t)\right)\otimes e(u).
$$
Thus $m=\sum_{r\in R(t)}r e(t)$. Every $r$ fixes the tabloid $\{t\}$, while $e(t)$ has coefficient one at $\{t\}$. Invariance of the <tabloid bilinear form> now yields
$$
\langle m,\{t\}\rangle
=\sum_{r\in R(t)}\langle r e(t),\{t\}\rangle
=\sum_{r\in R(t)}1
=\boxed{|R(t)|}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 160 / 1 / c / ii / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/iii/paper-160/1/c/ii/solution