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

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An involution is a group element $s$ satisfying $s^2=1$. Some conventions require $s\ne1$.


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 Incoming links: Involution