Reversing a word for $w$ gives a word of the same length for $w^{-1}$ because every generator is an <involution>. Applying the same argument to $w^{-1}$ proves
$$
\ell_S(w^{-1})=\ell_S(w).
$$
A shortest word for $w$ followed by $s_i$ gives $\ell_S(ws_i)\leq\ell_S(w)+1$. Conversely, $w=(ws_i)s_i$ gives $\ell_S(w)\leq\ell_S(ws_i)+1$. Hence
$$
\ell_S(ws_i)\in\{\ell_S(w)-1,\ell_S(w),\ell_S(w)+1\}.
$$

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 111 / 4 / a / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/iii/paper-111/4/a/solution