Small Cancellation and Dehn Algorithms

Let $N=⟨\text{\hspace{0.17em}⁣}⟨R⟩\text{\hspace{0.17em}⁣}⟩$. Because $N$ is normal by [F1], it contains $r^{-1}$ whenever it contains $r$, and it contains $ur{u}^{-1}$ for every $u\in F(X)$. Hence [L1] implies that every element of $R^{\mathrm{sym}}$ already lies in $N$. Therefore $⟨\text{\hspace{0.17em}⁣}⟨R^{\mathrm{sym}}⟩\text{\hspace{0.17em}⁣}⟩\subseteq N$.

Every relator of $R$ belongs to $R^{\mathrm{sym}}$ by definition, so the normal closure of $R^{\mathrm{sym}}$ contains $R$. By the minimality clause of [F1], $N\subseteq ⟨\text{\hspace{0.17em}⁣}⟨R^{\mathrm{sym}}⟩\text{\hspace{0.17em}⁣}⟩$.

The two containments from steps 1.1 and 1.2 are equalities, so the normal closures agree.

Let $r=p_1\cdots p_n$ be a factorisation of a relator into pieces. Applying [L1] to each $p_i$ gives $\mathrm{\mid }{p}_i\mathrm{\mid }<\lambda \mathrm{\mid }r\mathrm{\mid }$ for every $i$. Summing these inequalities yields $\mathrm{\mid }r\mathrm{\mid }={\sum }_{i=1}^n\mathrm{\mid }{p}_i\mathrm{\mid }<n\lambda \mathrm{\mid }r\mathrm{\mid }\mathrm{.}$

Because every relator is nonempty, $\mathrm{\mid }r\mathrm{\mid }>0$. Thus step 1.1 implies $1<n\lambda$, hence $n>1\mathrm{/}\lambda$. This is exactly the $C(\lambda )$ condition from [L1].

If $D$ has area $0$, then [L1] forces $D$ to be the degenerate one-vertex diagram. Its boundary label is the empty word, so it represents the identity in the free group and therefore in the quotient group of [F1].

Assume that $D$ has positive area. Choose a base vertex on the outer boundary, a spanning tree in the $1$-skeleton, and a spanning tree in the dual graph rooted at the exterior face. Reading the 2-cells in an order compatible with the rooted dual tree gives the standard disc-shelling identity $\mathrm{Lab}(\mathrm{\partial }D)={\prod }_{k=1}^m{u}_k{r}_k^{{\epsilon }_k}{u}_k^{-1}$ in the free group, where each $r_k\in R$, each ${\epsilon }_k\in \{\pm 1\}$, and the words $u_k$ are labels of paths from the base vertex to the corresponding cells. Interior edges cancel in opposite orientations, leaving exactly the outer boundary label.

Every factor in step 1.2 lies in the normal closure of $R$. Hence the boundary label lies in that normal closure and represents the identity in the quotient group of [F1]. Together with step 1.1 this proves the claim in all cases.

If $w$ is the boundary label of a finite van Kampen diagram, then [L1] says that $w$ represents the identity in $G$.

Conversely, suppose that $w$ represents the identity in $G$. Then $w$ lies in the normal closure of $R$, so [F1] gives a factorisation $w={\prod }_{k=1}^m{u}_k{r}_k^{{\epsilon }_k}{u}_k^{-1}$ with $r_k\in R$ and ${\epsilon }_k\in \{\pm 1\}$.

For each factor $u_k{r}_k^{{\epsilon }_k}{u}_k^{-1}$, take one 2-cell with boundary word $r_k^{{\epsilon }_k}$ and attach to its boundary a whisker labelled $u_k$ from a common basepoint. Gluing these $m$ discs along the whiskers produces a finite planar diagram whose outer boundary label is exactly the product in step 1.2, namely $w$.

Steps 1.1 and 2.1 prove both directions of the equivalence.

Let $D$ be any van Kampen diagram for $w$ with $m$ faces. Reading the faces one by one as in the proof of The boundary label of a van Kampen diagram is trivial in the presented group expresses $w$ as a product of $m$ conjugates of relators and their inverses. Hence the algebraic relator area of $w$ is at most $m$.

Conversely, let $w={\prod }_{k=1}^m{u}_k{r}_k^{{\epsilon }_k}{u}_k^{-1}$ be an algebraic expression with $m$ minimal as in [F1]. The converse construction in A word is trivial in a presented group exactly when it bounds a finite van Kampen diagram produces a van Kampen diagram with exactly $m$ faces and boundary word $w$. Therefore the minimal diagram area is at most the algebraic relator area.

Step 1.1 gives one inequality between the two minima and step 1.2 gives the reverse inequality. Therefore the two minimal areas are equal.

Suppose $D$ were not reduced. Then by [L1] there would be two adjacent 2-cells whose common edge can be cancelled. Delete those two faces and glue together the remaining boundary arcs.

The surgery of step 1.1 does not change the outer boundary word, but it removes exactly two 2-cells. By [F1], the new diagram therefore has strictly smaller area than $D$.

This contradicts the assumed minimality of $D$. Hence $D$ is reduced.

If $D$ has exactly one $2$-cell, that face has empty inner boundary and is therefore a shell with zero internal arcs. Hence assume that $D$ has at least two faces.

Collapse every maximal arc of $D$---boundary arcs as well as internal arcs---by suppressing its valence-$2$ internal vertices. The resulting diagram $\bar{D}$ is an arc-reduced combinatorial disc with the same faces and face incidences. Removing subdivisions neither creates a cancellable face pair nor changes which face-boundary portions are internal or external.

Every internal edge of $\bar{D}$ represents a maximal internal arc of $D$. By reducedness, the two incident face occurrences do not cancel, so [F2] identifies the arc label as a piece. If $f$ is an interior face, these piece-arcs cover $\mathrm{\partial }f$; [F1] makes each one shorter than $\mathrm{\mid }\mathrm{\partial }f\mathrm{\mid }\mathrm{/}6$, so $f$ has at least seven sides. This is exactly the arc-reduced $C^{\mathrm{\prime }}(1\mathrm{/}6)$ setup preceding the proposition cited in [F2].

Apply the Touikan proposition cited in [F2] to $\bar{D}$. It gives a boundary face whose inner boundary consists of $i$ internal arcs for some $1\le i\le 3$. Expanding the suppressed valence-$2$ vertices turns those $i$ edges back into the same $i$ maximal internal arcs of $D$, without changing the face or its outer boundary. Together with the one-face case in step 1.1, this proves the claim.

By [L1], some boundary face $f$ of $D$ is a shell whose inner boundary is a concatenation $q_1\cdots q_i$ of maximal internal arcs with $0\le i\le 3$. (For $i=0$, this is the empty concatenation.) Let $p$ be the complementary outer arc of $\mathrm{\partial }f$ lying on $\mathrm{\partial }D$.

If $i=0$, the sum of the inner-arc lengths is $0<\mathrm{\mid }\mathrm{\partial }f\mathrm{\mid }\mathrm{/}2$. If $1\le i\le 3$, each internal arc $q_j$ is shared with a distinct neighbouring face, so reducedness makes its label a piece. Because the presentation satisfies $C^{\mathrm{\prime }}(1\mathrm{/}6)$, every such arc satisfies $\mathrm{\mid }{q}_j\mathrm{\mid }<\mathrm{\mid }\mathrm{\partial }f\mathrm{\mid }\mathrm{/}6$, and hence $\mathrm{\mid }{q}_1\mathrm{\mid }+\cdots +\mathrm{\mid }{q}_i\mathrm{\mid }<i\frac{\mathrm{\mid }\mathrm{\partial }f\mathrm{\mid }}{6}\le \frac{\mathrm{\mid }\mathrm{\partial }f\mathrm{\mid }}{2}\mathrm{.}$ Thus in every case the total inner-arc length is less than half of $\mathrm{\mid }\mathrm{\partial }f\mathrm{\mid }$.

Since $\mathrm{\partial }f$ is the disjoint union of the outer arc $p$ and the inner arcs $q_1,\dots ,q_i$, step 2.1 gives $\mathrm{\mid }p\mathrm{\mid }=\mathrm{\mid }\mathrm{\partial }f\mathrm{\mid }-(\mathrm{\mid }{q}_1\mathrm{\mid }+\cdots +\mathrm{\mid }{q}_i\mathrm{\mid })>\frac{\mathrm{\mid }\mathrm{\partial }f\mathrm{\mid }}{2}\mathrm{.}$ Thus $f$ contributes more than half of its boundary to the outer boundary of $D$.

By [L1], the Dehn replacement sends $w=ausb$ to $w^{\mathrm{\prime }}=a{v}^{-1}sb$ before free reduction. Because $\mathrm{\mid }u\mathrm{\mid }>\mathrm{\mid }v\mathrm{\mid }$, one has $\mathrm{\mid }{w}^{\mathrm{\prime }}\mathrm{\mid }=\mathrm{\mid }w\mathrm{\mid }-\mathrm{\mid }u\mathrm{\mid }+\mathrm{\mid }v\mathrm{\mid }<\mathrm{\mid }w\mathrm{\mid }$.

Free reduction can only delete inverse pairs, never add letters. So the freely reduced form of $w^{\mathrm{\prime }}$ is no longer than $w^{\mathrm{\prime }}$, hence still strictly shorter than $w$.

Start by freely reducing $w$. Whenever the current freely reduced word contains a relator subword longer than half of a defining relator, perform the corresponding Dehn replacement and freely reduce again. By [L1], each such cycle strictly decreases word length, so no infinite run is possible. Thus the algorithm terminates.

Each replacement uses a relation $uv=1$ in the group and swaps $u$ for $v^{-1}$, so every step preserves the group element represented by the current word. Therefore if the algorithm reaches the empty word, the original input represented the identity.

Conversely, suppose the input represents the identity. After each iteration the current word is freely reduced and still represents the identity by step 2.1. If the current word were nonempty when the algorithm stopped, it would be Dehn-reduced by construction, contradicting the defining property of a Dehn presentation. Therefore the algorithm cannot stop before reaching the empty word, and termination from step 1.1 forces the final output to be empty.

Steps 2.1 and 3.1 prove correctness, and step 1.1 proves termination. Therefore Dehn's algorithm decides the word problem for the presentation.

By [L1], the given finite $C^{\mathrm{\prime }}(1\mathrm{/}6)$ presentation is a Dehn presentation: every nonempty freely reduced trivial word contains the required long relator subword.

Apply [L2] to that Dehn presentation. The resulting Dehn algorithm decides triviality of words.

Freely reduce $w$ to a word $u$. Because free reduction does not change the represented group element, $u$ is still null, and $\mathrm{\mid }u\mathrm{\mid }\le \mathrm{\mid }w\mathrm{\mid }$. If $u$ is the empty word, then the null diagram with no faces has area $0$, so the claim is immediate. Otherwise let $D$ be a minimal-area van Kampen diagram for $u$. By [L2], the diagram $D$ is reduced, so [L1] applies.

By [L1], some face $f$ of $D$ contributes an outer boundary arc $p$ with $\mathrm{\mid }p\mathrm{\mid }>\mathrm{\mid }\mathrm{\partial }f\mathrm{\mid }\mathrm{/}2$. Let $q$ be the complementary boundary arc of $f$, so $\mathrm{\mid }q\mathrm{\mid }<\mathrm{\mid }p\mathrm{\mid }$. Replacing $p$ by $q^{-1}$ and freely reducing gives a null word $u^{\mathrm{\prime }}$ with $\mathrm{\mid }{u}^{\mathrm{\prime }}\mathrm{\mid }\le \mathrm{\mid }u\mathrm{\mid }-1$. Conversely, attach one $f$-cell along the occurrence of $q^{-1}$ in any minimal diagram for the unreduced replacement word and add the free-cancellation strips. This constructs a diagram for $u$ with one more face, so $\mathrm{Area}(u)\le \mathrm{Area}(u^{\mathrm{\prime }})+1.$

Induct on the freely reduced boundary length. Step 1.1 gives the base case $\mathrm{\mid }u\mathrm{\mid }=0$. For $\mathrm{\mid }u\mathrm{\mid }>0$, step 2.1 yields a shorter freely reduced null word $u^{\mathrm{\prime }}$. By the induction hypothesis, $\mathrm{Area}(u)\le \mathrm{Area}(u^{\mathrm{\prime }})+1\le \mathrm{\mid }{u}^{\mathrm{\prime }}\mathrm{\mid }+1\le \mathrm{\mid }u\mathrm{\mid }\mathrm{.}$ Because $\mathrm{\mid }u\mathrm{\mid }\le \mathrm{\mid }w\mathrm{\mid }$, this is a linear isoperimetric inequality.

Finally, [F1] identifies van Kampen area with algebraic relator area, so the same linear bound holds in the algebraic formulation.

Because $g$ is a nontrivial torsion element, the hypotheses of [F2] apply directly. Therefore $g$ is conjugate to a power of a root of some defining relator.

Powers are interpreted as in [F1], so step 1.1 is exactly the claimed conclusion.

Let $g$ be a torsion element. By [L1], $g$ is conjugate to $v^k$, where some defining relator has the form $v^m$.

The no-proper-power hypothesis forces $m=1$. Thus the root word $v$ is itself a defining relator and represents the identity in the presented group, so every power $v^k$ is trivial by [F1]. Hence $g=1$.

Since every torsion element is trivial, the group is torsion-free.

The reduced subword $ab$ appears twice inside $r$, namely in positions $1$ through $2$ and $3$ through $4$. So it is certainly a repeated subword of one relator.

The cyclic conjugates of $r$ are only $abab$ and $baba$, while the cyclic conjugates of $r^{-1}$ start with inverse letters. Among these symmetrised occurrences, $ab$ is an initial segment only of $abab$, and in that word the continuation is always the same suffix $ab$. Therefore there is no second distinct ordered pair $(s,v)$ with $s=abv$, so [L1] says that $ab$ is not a piece.

So a repeated interior subword need not be a piece. The statement is false.

In the displayed one-relator presentation, the symmetrised relator set has no nontrivial piece: distinct cyclic conjugates begin with different letters, and the inverse cyclic conjugates do as well. Hence the $C^{\mathrm{\prime }}(1\mathrm{/}6)$ condition holds vacuously by [L1].

The unique defining relator has length $7$, which is strictly greater than $6$.

So a $C^{\mathrm{\prime }}(1\mathrm{/}6)$ presentation can have relators longer than $6$. The statement is false.

The relator $ab{a}^{-1}{b}^{-1}$ and its cyclic conjugates have long overlaps, so the presentation does not satisfy the $C^{\mathrm{\prime }}(1\mathrm{/}6)$ hypothesis required by [L1].

In the square grid van Kampen diagrams for commutator powers in ${\mathbb{Z}}^2$, every face can meet the outer boundary in exactly two of its four edges, never in more than half. Thus the characteristic Greendlinger conclusion fails for these null words.

Therefore Greendlinger's lemma does not extend to arbitrary finite presentations. The statement is false.

The word $x_1{x}_2{x}_3{x}_4{x}_5{x}_6{x}_7$ is freely reduced, because no adjacent inverse letters occur.

Nevertheless [L1] applies to the whole word: it is itself a relator and is longer than half of that relator, so one Dehn move replaces it by the empty word.

Thus a word can admit a Dehn reduction while admitting no free reduction. The two notions are different, so the statement is false.

Neither relator $a{b}^2$ nor $a^{2}b$ is a proper power: each is cyclically reduced of length $3$ and is not a repetition of a shorter cyclic word.

From $a{b}^2=1$ one gets $a=b^{-2}$, and substituting this into $a^{2}b=1$ gives $b^{-4}b=1$, hence $b^3=1$ in the sense of [F1]. Moreover, if $C_3=⟨t\mid t^3=1⟩$, then the assignment $a\mapsto t$, $b\mapsto t$ satisfies both relators, so it induces a surjective homomorphism $G\to C_3$. Therefore the image of $b$ is nontrivial and $G$ contains a nontrivial torsion element.

Therefore the absence of proper-power relators alone does not force torsion-freeness. By [L1], the missing small-cancellation hypothesis is load-bearing.