Periodic words: a relator root or Dehn-reduced powers

Assume the first alternative fails. Let $S$ be a segment longer than half a relator $R$ in a positive power of $w$. Rotate $w$ so that the start of $S$ is the start of a period $v$. If $\mathrm{\mid }S\mathrm{\mid }>2L$, write $S=v^{k}a$, with $k\ge 2$ and $a$ a proper prefix of $v$. Write $R=v^{k}a{T}^{-1}$. By [F2], either $v,R$ have a common root (excluded), or $(k-1)L<\mathrm{\mid }R\mathrm{\mid }\mathrm{/}6$. In the latter case $L\le (k-1)L<\mathrm{\mid }R\mathrm{\mid }\mathrm{/}6$ and $\mathrm{\mid }a\mathrm{\mid }<L$, so $\mathrm{\mid }S\mathrm{\mid }=(k-1)L+L+\mathrm{\mid }a\mathrm{\mid }<\mathrm{\mid }R\mathrm{\mid }\mathrm{/}2$, a contradiction. Thus every such segment is contained in the square of a rotation. If none exists, every power is freely reduced (since $w$ is cyclically reduced) and Dehn-reduced, the second alternative.

Otherwise choose such an $S$ of greatest length among all rotations and relators. This maximum exists because $1\le \mathrm{\mid }S\mathrm{\mid }\le 2L$; an infinite set of relators creates no problem for this finite set of lengths. By [F1], $\mathrm{\mid }S\mathrm{\mid }>L$. Consequently $S=pqp$, $v=pq$, where $p\mathrm{\ne }\mathrm{\varnothing }$, $q$ is possibly empty and $\mathrm{\mid }p\mathrm{\mid }\le L$. Rotate $R$ to write $R=pqp{t}^{-1}$. Since the length-$L$ initial segment $v$ is Dehn-reduced, $L\le \mathrm{\mid }R\mathrm{\mid }\mathrm{/}2$, while $\mathrm{\mid }pqp\mathrm{\mid }>\mathrm{\mid }R\mathrm{\mid }\mathrm{/}2$ gives $\mathrm{\mid }t\mathrm{\mid }<\mathrm{\mid }R\mathrm{\mid }\mathrm{/}2$.

If $q$ is empty, $R=v^2{t}^{-1}$; [F2] either gives the excluded common root or $L<\mathrm{\mid }R\mathrm{\mid }\mathrm{/}6$, contradicting $2L=\mathrm{\mid }S\mathrm{\mid }>\mathrm{\mid }R\mathrm{\mid }\mathrm{/}2$. Thus $q$ is nonempty. Also $t$ is nonempty: otherwise $2(\mathrm{\mid }p\mathrm{\mid }+\mathrm{\mid }q\mathrm{\mid })\le \mathrm{\mid }R\mathrm{\mid }=2\mathrm{\mid }p\mathrm{\mid }+\mathrm{\mid }q\mathrm{\mid }$ forces $\mathrm{\mid }q\mathrm{\mid }=0$. The joins $pq$ and $qp$ are reduced because $v$ is cyclically reduced. The joins $qt$ and $tq$ are reduced as well. Indeed cancellation in $qt$ would make the last letter of $q$ the last letter of $t^{-1}$, extending the match $pqp$ one letter to the left around the periodic word; cancellation in $tq$ extends it one letter to the right. Since $t\mathrm{\ne }\mathrm{\varnothing }$, either extension is still a segment of $R$ and is longer than half $R$. Step 1.1 then bounds its length by $2L$, contradicting maximality of $S$.

Compare $R=pqp{t}^{-1}$ with its rotation $p{t}^{-1}pq$, based at the second copy of $p$. Equality would, after cancelling its first $p$, give $qp{t}^{-1}=t^{-1}pq$, whose first letters force cancellation in $tq$ (both $q,t$ are nonempty). Hence these rotations are distinct and their common prefix $p$ is a piece. Therefore $\mathrm{\mid }p\mathrm{\mid }<\mathrm{\mid }R\mathrm{\mid }\mathrm{/}6$. Put $P=\mathrm{\mid }p\mathrm{\mid }$, $Q=\mathrm{\mid }q\mathrm{\mid }$, $T=\mathrm{\mid }t\mathrm{\mid }$, $N=\mathrm{\mid }R\mathrm{\mid }=2P+Q+T$. The inequalities $T<N\mathrm{/}2$ and $P<N\mathrm{/}6$ give $Q=N-2P-T>N\mathrm{/}6$. The bound $P+Q\le N\mathrm{/}2$ gives $T=N-2P-Q\ge N\mathrm{/}2-P>N\mathrm{/}3$, in particular $T>N\mathrm{/}6$. Also $Q\le L\le N\mathrm{/}2$.

Set $y=tq$. It is nonempty and cyclically reduced by step 3.1. In the quotient, $pqp=t$, hence $v^2=pqpq=tq=y$. Since $v$ is a cyclic rotation of $w$, this makes $y$ conjugate to $w^2$. To prove all powers of $y$ Dehn-reduced, suppose a segment $U$ in some $y^n$ is longer than half a relator $R^{\ast }$. Regard $y^n$ as part of the bi-infinite alternating string of nonempty blocks $t,q$. Whenever two cyclically read relators share a segment at least one sixth of either relator, they must be identical when based at that segment; otherwise this is a forbidden piece by [F3]. This observation permits comparisons both before and after the matched segment, because equal based relators are equal cyclic strings.

If $U$ contains a whole $t$ block, compare $R^{\ast }$ based there with $R^{-1}$ based at $t$, which reads $t{p}^{-1}{q}^{-1}{p}^{-1}$. Since $\mathrm{\mid }t\mathrm{\mid }>N\mathrm{/}6$, these based relators must agree. If $U$ continues past $t$, its next letter is the first of $q$, whereas the relator prescribes the first of $p^{-1}$; equality contradicts reducedness of $pq$. If $U$ begins before $t$, the preceding letter is the last of $q$, whereas the relator prescribes the last of $p^{-1}$; equality contradicts reducedness of $qp$. If neither occurs, $U=t$ and $\mathrm{\mid }U\mathrm{\mid }=T<N\mathrm{/}2=\mathrm{\mid }{R}^{\ast }\mathrm{\mid }\mathrm{/}2$, again impossible.

If $U$ contains a whole $q$ block, compare $R^{\ast }$ there with $R$ based at $q$, namely $qp{t}^{-1}p$. The bound $\mathrm{\mid }q\mathrm{\mid }>N\mathrm{/}6$ forces equality. An extension after $q$ equates the first letters of $t$ and $p$, contradicting the cyclically reduced join $t^{-1}p$ in $R$. An extension before $q$ equates the last letters of $t$ and $p$, contradicting the reduced join $p{t}^{-1}$ in $R$. With no extension, $U=q$ has length at most $N\mathrm{/}2$, impossible.

Suppose $U$ lies inside a single $t$ or $q$ block. Its length exceeds $\mathrm{\mid }{R}^{\ast }\mathrm{\mid }\mathrm{/}2$, so comparison with that occurrence in $R^{-1}$ or $R$ forces equality of the based relators. Then $\mathrm{\mid }{R}^{\ast }\mathrm{\mid }=N$, whereas both block lengths are at most $N\mathrm{/}2$, a contradiction. This includes a segment touching either endpoint of just one block.

Otherwise $U$ contains no whole block and crosses one join. For a $tq$ join write $U=t_s{q}_p$ with nonempty proper suffix $t_s$ of $t$ and nonempty proper prefix $q_p$ of $q$. At least one part has length greater than $\mathrm{\mid }{R}^{\ast }\mathrm{\mid }\mathrm{/}4$, hence greater than $\mathrm{\mid }{R}^{\ast }\mathrm{\mid }\mathrm{/}6$. If it is $t_s$, comparison with its occurrence in $R^{-1}$ forces equal cyclic relators; their next letters are respectively the first of $q$ and the first of $p^{-1}$, contrary to reducedness of $pq$. If it is $q_p$, comparison with $R$ forces equality; immediately preceding $q$ the letters would be the last of $t$ and the last of $p$, contrary to reducedness of $p{t}^{-1}$.

At a $qt$ join write $U=q_s{t}_p$. If the suffix $q_s$ exceeds $\mathrm{\mid }{R}^{\ast }\mathrm{\mid }\mathrm{/}4$, align it with $R$; the following letters would equate the first of $t$ and the first of $p$, contrary to reducedness of $t^{-1}p$. If the prefix $t_p$ exceeds $\mathrm{\mid }{R}^{\ast }\mathrm{\mid }\mathrm{/}4$, align it with $R^{-1}$; the preceding letters would equate the last of $q$ and the last of $p^{-1}$, contrary to reducedness of $qp$. At least one inequality holds because $\mathrm{\mid }U\mathrm{\mid }>\mathrm{\mid }{R}^{\ast }\mathrm{\mid }\mathrm{/}2$. These are all possibilities: crossing two joins would include a whole intervening block, already excluded. Thus no $y^n$ contains such $U$, proving the third alternative. Combined with step 1.1 and step 2.1 and step 3.1, this proves the asserted alternative without using any infinite-order hypothesis or finiteness of $R$.