Small Cancellation and Dehn Algorithms - Examples

The cyclic conjugates of $r=abc$ are $abc$, $bca$, and $cab$.

The inverse word is $r^{-1}=c^{-1}{b}^{-1}{a}^{-1}$, whose cyclic conjugates are $c^{-1}{b}^{-1}{a}^{-1}$, $b^{-1}{a}^{-1}{c}^{-1}$, and $a^{-1}{c}^{-1}{b}^{-1}$. Applying [L1] gives exactly the displayed six-element set.

The words $r_1$ and $r_2$ start with the same letter $a$ and then immediately diverge, so $a$ is a piece by [L1]. Their inverse words have cyclic conjugates starting with $a^{-1}$ and then immediately diverging, so $a^{-1}$ is also a piece. Every other letter occurs in only one cyclic position among the two relators and their inverses, so there is no other nontrivial piece and no piece of length greater than $1$.

The relevant relator lengths are $\mathrm{\mid }{r}_1\mathrm{\mid }=7$ and $\mathrm{\mid }{r}_2\mathrm{\mid }=8$, while both pieces have length $1$. Since $1<7\mathrm{/}6$ and $1<8\mathrm{/}6$, [L2] shows that the $C^{\mathrm{\prime }}(1\mathrm{/}6)$ inequalities hold for every symmetrised occurrence.

The first seven letters of $w$ form the defining relator itself, so [L1] replaces that block by the empty word. The result is the single relator $x_1{x}_2{x}_3{x}_4{x}_5{x}_6{x}_7$.

Apply the same move again to the remaining relator block. A second Dehn replacement produces the empty word.

In a one-face disc diagram, every edge of the unique face lies on the outer boundary of the whole diagram. So the distinguished face contributes exactly $\mathrm{\mid }r\mathrm{\mid }$ boundary edges to the outer boundary.

Since $\mathrm{\mid }r\mathrm{\mid }>\mathrm{\mid }r\mathrm{\mid }\mathrm{/}2$, the unique face satisfies the conclusion of [L1]. Thus this one-cell disc is a concrete Greendlinger face.

Write $r=x_1\cdots x_7$. Its seven positive cyclic conjugates have distinct first letters, as do its seven negative inverse conjugates; no positive first letter equals a negative one. Thus no distinct symmetrised relators share a nonempty prefix, so there are no pieces and strict $C^{\mathrm{\prime }}(1\mathrm{/}6)$ holds vacuously. The cyclic word is not a proper power: each generator occurs exactly once, whereas a repetition at least twice would repeat each occurring letter.

Add the inverse relator $r^{-1}$ as a redundant relator and delete $r$, since each is a consequence of the other. Now $r^{-1}=x_7^{-1}w$ with $w=(x_1\cdots x_6{)}^{-1}$, which contains no $x_7$. There are no other relators, so the legal inverse dictionary move deletes $x_7$ and $r^{-1}$. The resulting presentation is $⟨x_1,\dots ,x_6\mid \mathrm{\varnothing }⟩$. Hence [L2] identifies $G$ with the free group on these six generators.

By [L3] that free group has no nonidentity element of finite order. An isomorphism preserves powers and the identity, so $G$ is torsion-free.

The common initial segment of the two relators is the one-letter word $a$, so $a$ is a piece. Both relators have length $6$.

Hence the piece length is exactly $1=6\mathrm{/}6$. This meets the weak inequality $\mathrm{\mid }p\mathrm{\mid }\le \mathrm{\mid }r\mathrm{\mid }\mathrm{/}6$, but it does not satisfy the strict inequality required by [L1].

So the strict convention is genuinely stronger: exact one-sixth overlap fails it. This refutes the statement.

As in FALSE: a presentation with no proper-power relators is automatically torsion-free, neither relator $a{b}^2$ nor $a^{2}b$ is a proper power.

The same substitution calculation gives $a=b^{-2}$ and then $b^{-4}b=1$, so $b^3=1$. Moreover the assignment $a\mapsto t$, $b\mapsto t$ to the cyclic group $C_3=⟨t\mid t^3=1⟩$ satisfies both relators, so it induces a surjective homomorphism $G\to C_3$. Thus $G$ has nontrivial torsion.

Therefore no-proper-power relators do not by themselves prevent torsion. By [L1], the missing small-cancellation hypothesis is essential.