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

A Dehn presentation is a finite group presentation with the following strict shortening property: every nonempty freely reduced word representing the identity contains a consecutive segment $u$ of a relator $r$ satisfyingRelators are taken as cyclically reduced words with a symmetrized relator set: their inverse words and cyclic permutations are included. This convention allows the matched segment to start anywhere on either orientation of a relator. The symmetrized set remains finite and has the same maximum relator length.

If $r=uv$, the relation replaces $u$ by $v^{-1}$, of length $|r|-|u|<|u|$. The Dehn algorithm alternates this replacement with free reduction. Every step shortens the word; it terminates at the empty word exactly for words representing the identity. The strict inequality matters: a half-perimeter match alone need not shorten anything.

Finiteness is part of the convention used here and is needed for the maximum and counting argument in the last part. With an unrestricted infinite relator set, merely imposing the shortening property would not by itself justify that conclusion.

Minimality of the conjugacy representative implies that $w$ is freely reduced; otherwise free reduction produces a shorter representative of the same element. It is also a cyclically reduced word. If $w=sv{s}^{-1}$ as a freely reduced word with its first and last letters inverse, the shorter word $v$ represents a conjugate, contradicting minimality. This is the cyclic reduction of a shortest conjugacy representative.

The word is nonempty because the element has order greater than one. Since it is cyclically reduced, no cancellation occurs between successive copies in $w^n$, and $w^n$ is a nonempty freely reduced word. It represents the identity because conjugation preserves the order. Apply the defining property of the Dehn presentation to obtainThe relator belongs to the finite symmetrized set fixed in the definition. No assumption that the group itself is torsion-free is made.

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 $w$.

Assume first that the finite relator set $R$ is nonempty and putFor every nonidentity finite-order element choose a shortest conjugacy representative. The preceding parts give a word representing its conjugacy class of length at most $M$. The identity class has the empty representative.

There are only finitely many such words because the generating alphabet is finite. If it has $q=2|A|$ formal letters and $N=\lfloor M\rfloor$, a sufficient upper bound for the number of words isDistinct classes cannot require more representatives than there are words; different words may of course represent the same class. ConsequentlyThis is the torsion conjugacy bound for a Dehn presentation. It does not claim that there are only finitely many finite-order elements.

If $R$ is empty, the maximum in the PDF is undefined. Handle this case separately: the group is a free group on the finite alphabet, and a nonempty cyclically reduced word has no freely trivial positive power. Thus there is no nonidentity torsion and only the identity conjugacy class. If the alphabet is empty, the group is trivial and the same conclusion holds.