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A nonempty shortest conjugacy representative is a cyclically reduced word. If $w=sv{s}^{-1}$ with first and last letters inverse, $v$ is a shorter word representing a conjugate. Consequently $w^n$ has length exactly $n|w|$ as a freely reduced word, and any segment of the periodic word of length at most $|w|$ fits in a cyclic rotation of $w$. This observation turns a Dehn shortening segment crossing copy boundaries into a shortening of a conjugacy representative.

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.

A shortest conjugacy representative is a word having the fewest written letters among all words representing elements of a fixed conjugacy class. This minimum equals the least group word length of an element in the class; a longer spelling of that element is not itself a shortest representative. Such a word exists because lengths are nonnegative integers. It is freely and cyclically reduced: free cancellation shortens the same representative, while removing mutually inverse first and last letters shortens a conjugate. Every cyclic rotation has the same minimal length.

In a finite Dehn presentation, every nonidentity finite-order element has a shortest conjugacy representative of length at most half the length of some relator. Indeed a positive power contains a Dehn segment. If the representative had length at least $k=\lfloor |r|/2\rfloor +1$, the first $k$ letters of that segment would fit into a cyclic rotation and could be replaced by fewer letters, contradicting minimality. Bounded relator lengths and a finite alphabet therefore give finitely many conjugacy classes of finite-order elements. With no relators the group is free and torsion-free, so only the identity class remains.