A shortest finite-order representative shares a word root with a relator

The word $z$ is nonempty and cyclically reduced by [F2]. Every cyclic rotation represents a conjugate of the original element. If such a rotation contained $s$ with $r=st$ and $\mathrm{\mid }s\mathrm{\mid }>\mathrm{\mid }r\mathrm{\mid }\mathrm{/}2$, replace $s$ by $t^{-1}$ in that rotation. The quotient element is unchanged since $st=1$, while length strictly decreases since $\mathrm{\mid }t\mathrm{\mid }<\mathrm{\mid }s\mathrm{\mid }$; subsequent free reduction cannot increase length. This contradicts shortest conjugacy length. Hence $z$ is cyclically Dehn-reduced.

Any nonempty word $a$ all of whose positive powers are Dehn-reduced has infinite order. Indeed if $a^n=1$, its nonempty freely reduced literal word has a long cyclic relator segment by [F1]. Every cyclic segment of $a^n$ of length at most $n\mathrm{\mid }a\mathrm{\mid }$ appears as a literal segment of $a^{2n}$ by taking two copies of that cyclic word. This contradicts Dehn reduction of $a^{2n}$. This argument also covers n=1 and a segment crossing the chosen basepoint.

Apply [F3] using step 1.1. Its all-powers alternative for $z$ contradicts finite order by step 1.2. Its square alternative gives $z^2$ conjugate to a nonempty $y$ of infinite order by step 1.2. But if $z^n=1$ then $(z^2{)}^n=1$, and conjugation preserves this equation; hence $y$ has finite order, a contradiction, even when $z^2=1$. Only the common-root alternative remains. The inverse of a defining relator has the inverse root, so the conclusion may equally be expressed using cyclic conjugates of the original oriented defining words and integer powers of their roots.