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

We prove the assertion by induction on $n$. Write $w=av$, where $a=s_{i_1}$ and $v=s_{i_2}\cdots s_{i_n}$; both displayed words are reduced. The first clause of the folding condition and part a imply that $\ell (vs)$ is either $n-2$ or $n$.

If $\ell (vs)=n-2$, the induction hypothesis deletes one unique letter from the reduced word for $v$. Prefixing $a$ gives the required deletion from $w$. If another deletion were possible, induction rules out another internal position, while deletion of $a$ would give $ws=v$ and hence $vs=av$, contradicting the lengths $n-2$ and $n$.

If $\ell (vs)=n$, both $av$ and $vs$ increase the length of $v$. Since $\ell (avs)=\ell (ws)=n-1$, the other alternative in the folding condition must hold:Thus $ws=avs=v$, which deletes the first letter. An additional internal deletion would give $v=a{v}^{\prime }$ for a word $v^{\prime }$ of length $n-2$; multiplying by $a$ would make the length-$n$ element $av=w$ equal to $v^{\prime }$, impossible. The deletion position is therefore unique in every case.

Solved by gpt-5.6-sol high.