Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 133 / 4 / c / Solution

Put $m=|w|$ and $\ell =|r|$. In fact minimality yields the stronger bound $m\le \lfloor \ell /2\rfloor$. Suppose instead thatBecause $|u|>\ell /2$, the first $k$ letters of the matched segment $u$ form a relator segment $v$ of length $k$. They also form a segment of the periodic word $w^n$. Since $k\le m$, that segment is contained in one cyclic permutation $w^{\prime }$ of $w$, starting at the same position in the period. A cyclic permutation represents a conjugate and has length $m$.

Rotate the symmetrized relator to write it as $vs$. Replacing the prefix $v$ of $w^{\prime }$ by $s^{-1}$ preserves the represented element and gives a word of length at mostFree reduction can only shorten further. This contradicts the choice of $w$. ThusThe periodic-segment argument treats matches that cross a boundary between copies of $w$; it would be incorrect simply to assume that the entire long segment $u$ lies in the original written copy of $w$.