Small-Cancellation Disc Diagrams and the Torsion Toolkit

Let a diagram have a face. There is a face edge bordering the unbounded region of the union of faces: a generic ray from an interior point has a last crossing of that finite union. Deleting this open edge and its incident open face retracts that polygon onto its complementary boundary path, leaving a connected simply connected planar complex. At the chosen outer occurrence write the boundary as $AeB$ and the face word as $r=e{q}^{-1}$, so the new boundary is $AqB$. In the free group $AeB=(Ar{A}^{-1})(AqB)$, since the inserted $q^{-1}q$ and $A^{-1}A$ cancel. Repeat until every face has been removed; each removal contributes exactly one conjugate. The remaining connected simply connected graph is a tree, whose boundary freely cancels to the empty word by deleting end edges. Thus the original boundary is freely equal to a product of exactly $m$ conjugates when there are $m$ faces. This peeling is the algebraic unfolding of the diagram into face polygons and conjugating paths. It also treats $m=0$ directly.

Conversely suppose $w$ lies in the normal closure. First freely reduce it to $\bar{w}$. By [F2] express $\bar{w}$ as a product ${\prod }_{j=1}^m{u}_j{r}_j{u}_j^{-1}$ in the free group, using the least possible $m$. Draw $m$ disjoint polygons in planar order, joining their basepoints to a common point by separate whiskers labelled $u_j$. Its outer word is that literal product. With $m=0$, a sequence of inverse-pair insertions into the empty walk gives a tree reading any freely trivial word.

Fold consecutive outer edges whose letters cancel, identifying them with opposite traversals. If they already form an end spur, delete the spur. Otherwise the two edges bound a sector in the unbounded region and can be identified across that sector. Distinct other endpoints are merged; the sector closes to a slit, so the resulting complex remains planar and simply connected. If the other endpoints already agree, the two edges instead enclose a component. The fold would seal this component into a sphere attached at a point. Before sealing it, discard its interior and identify the two edges: its boundary is precisely the cancelling pair, so the outside boundary word undergoes the same free cancellation. Any positive-area discarded component would give a diagram with fewer than $m$ faces for the same freely reduced word; step 1.1 would give a product with fewer than $m$ factors, contrary to minimality. A component of area zero is a tree and is removed by spur deletions. This accounts for the possible spherical closure, including when the outside word is empty; then minimality already forces $m=0$.

Each fold decreases outer length by two, so the finite reduction terminates with a planar simply connected diagram reading exactly $\bar{w}$. To recover $w$, reverse its chosen free reduction: for each insertion of $a{a}^{-1}$ at a boundary occurrence attach a fresh edge labelled $a$ in that occurrence's exterior sector. This adds a spur, introduces no face or hole, and inserts exactly the required pair. Together with step 1.1 this proves both directions and the $m$-factor assertion.

The set of face counts of diagrams for $w$ is nonempty by [F2] and is a subset of the natural numbers. Its least member exists by [F3] and, being in this set, is attained by a diagram. This is a single existential choice, even for an infinite presentation.

To decompose a diagram, split at any cut vertex into its incident components together with that vertex, and repeat in each component. Each split partitions a finite nonempty set of edges into smaller sets, so it terminates. The incidence graph of resulting blocks and splitting vertices is connected. It has no cycle: such a cycle would provide a path avoiding one of the vertices that was a cut vertex at the corresponding split. Hence this incidence graph is a finite tree. A block with no cycle is a single bridge. In a block with a cycle, the outer boundary is a simple closed curve: a repeated boundary vertex would separate two successive exterior sectors and be a cut vertex. Every bounded region is filled, since an unfilled bounded region would be a hole in the original simply connected planar complex. Thus this block is a nonsingular disc.

Suppose two adjacent faces cancel. Cut them apart along any further common edges, keeping copies of those edges on the attached outside sectors. Delete their interiors and the selected common edge; pair their complementary boundary paths position by position. The paths have equal labels in the same direction because the full face words agree when based on that common edge. Glue each paired pair, carrying its outside sectors with it in their inherited planar order. This is the collapse of a folded pair of polygons to one path: it can be performed in a small planar neighbourhood of those polygons after the cuts. Components pinched off at a vertex are retained as vertex-attached components; any closed interior components are discarded. The outside boundary occurrences and their labels are preserved; a disappearing backtrack can be restored by an exterior spur. No hole is introduced, since the removed region is replaced by its paired boundary path. The result is a diagram for $w$ with at most two fewer faces. This contradicts step 1.1.

Traversing the outer boundary visits each branch of this finite tree in planar order, returning to its attachment before continuing in the parent block. Therefore an end disc block has just one possible interruption, at its attachment vertex; all boundary arcs not passing through that vertex internally occur uninterrupted in the full walk. An end bridge has a terminal vertex of degree one, whose excursion reads $a{a}^{-1}$. The one-point diagram needs no blocks, and a single disc needs no attachment. These observations prove the asserted decomposition and boundary qualifications.

Start each incident face-side reading at the initial endpoint of $a$, choosing its orientation to traverse $a$ in the given direction. The two readings have opposite planar orientations, even if they belong to the same face. Their words $r,s$ belong to the symmetrised set and both begin with the nonempty label $p$ of $a$. No free cancellation occurs inside $p$ since it is a segment of a cyclically reduced relator.

No nonempty cyclically reduced word equals a cyclic rotation of its inverse. Indeed, writing its letters $b_i$ with indices modulo its length $m$, such equality would give $b_i=b_{k-i}^{-1}$ for some integer $k$. If $m$ is odd, or if $m$ and $k$ are both even, $2i=k$ has a solution modulo $m$, forcing a letter to equal its formal inverse, impossible in $X^{\pm 1}$. If $m$ is even and $k$ is odd, take $i=(k-1)\mathrm{/}2$ modulo $m$; then $b_i=b_{i+1}^{-1}$, contrary to cyclic reduction. This includes proper-power words and uses no hypothesis on the presented group's torsion.

For distinct faces, equality $r=s$ would give a cancellable pair, excluded by [F2]. For the same face, the two opposite orientations make $s$ a cyclic rotation of $r^{-1}$, so step 1.2 excludes equality. Thus in both cases $r\mathrm{\ne }s$ and $p$ is a piece by [F1]. Under $C^{\mathrm{\prime }}(1\mathrm{/}6)$, apply its bound to $r$ and separately to $s$ to obtain $\mathrm{\mid }p\mathrm{\mid }<\mathrm{\mid }r\mathrm{\mid }\mathrm{/}6$ and $\mathrm{\mid }p\mathrm{\mid }<\mathrm{\mid }s\mathrm{\mid }\mathrm{/}6$. No injectivity of a face attaching walk has been assumed.

Every boundary edge of an interior face is internal. The arcs partition its boundary occurrences, so $\mathrm{\mid }r\mathrm{\mid }={\sum }_{j=1}^d\mathrm{\mid }{a}_j\mathrm{\mid }$. If an arc occurs twice at this face it contributes twice to this sum, and [F1] applies to each occurrence; no distinct-face or simple-boundary hypothesis is needed. The relator is nonempty, hence $\mathrm{\mid }r\mathrm{\mid }>0$ and $d\ge 1$. By [F1], each summand is strictly below $\mathrm{\mid }r\mathrm{\mid }\mathrm{/}6$.

Summing gives $\mathrm{\mid }r\mathrm{\mid }<d\mathrm{\mid }r\mathrm{\mid }\mathrm{/}6$. Dividing by the positive number $\mathrm{\mid }r\mathrm{\mid }$ gives $d>6$; because $d$ is an integer, $d\ge 7$. In particular equality at six arcs is excluded by strictness.

The total number of vertices in all links is $2E$, one per edge germ, including two for a loop. The total number of link edges is ${\sum }_{f}d(f)$, one per corner. Hence ${\sum }_v\chi (\mathrm{lk}(v))=2E-{\sum }_{f}d(f)$. Each angle occurs once in the vertex sums and once in the face sums, with opposite signs. Substitution into [F1] therefore gives ${\sum }_{v}k(v)+{\sum }_{f}k(f)=2V-2E+{\sum }_{f}d(f)-{\sum }_f(d(f)-2)=2(V-E+F)$.

If a planar diagram has a face, some edge of a face borders the unbounded region of the union of faces: take a ray from an interior point in a generic direction and its last crossing of this finite union. Such an edge has only one incident face. Remove that open edge and the open face. A polygon retracts to the complementary boundary path, with all other cells fixed; thus the remainder is connected and simply connected and still planar. This operation removes one edge and one face and preserves $V-E+F$. Repeating removes every face. The remaining graph is connected and has no cycle, since a cycle in a planar graph without faces is a hole.

The remaining finite tree, if not a point, has an end vertex: a longest simple path cannot extend at either end. Removing an end vertex and its edge preserves $V-E$ and leaves a tree. It ends at one vertex, where $V-E+F=1$. Reversing all these operations yields $V-E+F=1$ for $D$. Arc suppression also removes one edge and one vertex at each degree-two suppression, preserving this value; the whole-circle convention avoids deleting the final marked vertex.

In a tree there are no angles or faces and $k(v)=2-\mathrm{deg}(v)$, so the total is $2V-2E=2$. A deleted spur tip contributes one; at its neighbour the link loses one isolated vertex, raising that neighbour's curvature by one. Thus retaining or removing the spur preserves the total, rather than silently assigning its tip zero curvature. The one-point case contributes two. Combining step 1.1 with step 2.1 gives the identity in every case.

First consider a nonsingular multi-face disc and suppress its degree-two vertices. There is no degree-one interior tip: a face walk around such a tip would traverse its incident edge and immediately its reverse, contradicting cyclic reduction of the relator. Boundary vertices have degree at least two since the disc boundary is a circle. Thus after suppression all degrees are at least three; the all-degree-two circle case would have just one face. Assign $2\mathrm{/}d$ to each corner occurrence at an interior vertex of degree $d$, and $1\mathrm{/}(d-1)$ at a boundary vertex of degree $d$. Their links are respectively a circle with $d$ corner occurrences and an interval with $d-1$ corner occurrences, so every vertex has curvature zero. Consequently interior angles are at most $2\mathrm{/}3$ and boundary angles at most $1\mathrm{/}2$. A face visiting the same vertex several times contributes several corners, not one.

Let $n$ count the corners in the cyclic attaching walk of a face and let $b$ count those occurrences based at boundary vertices. Then $k(f)\le 2-n\mathrm{/}3-b\mathrm{/}6$. For an interior face, [F2] gives $n\ge 7$, including self-adjacent arc occurrences, so its curvature is negative even if it touches the boundary at a vertex. Distinct exterior arcs cannot be consecutive in this cyclic walk: at their common boundary vertex the corner would join the two boundary germs with no intervening internal germ in the interval link, forcing degree two, which was suppressed. Hence if the face has at least two exterior arcs, each has two endpoint corner occurrences separated from the others by an internal arc occurrence. There are at least four such boundary corner occurrences and $n\ge 4$, giving $k(f)\le 2-4\mathrm{/}3-4\mathrm{/}6=0$. Their underlying vertices may repeat; only the occurrences must be distinct. No simplicity of a face attaching walk is used.

A shell with $i$ internal arcs has $n=i+1$ and at least two boundary corners, hence $k(f)\le (4-i)\mathrm{/}3$. A multi-face disc has no zero-shell, since a face with its entire boundary exterior would be the whole disc. Thus positive-curvature faces are precisely among the shells with $1\le i\le 3$, and each contributes at most one. The total is two by [F3]; all other contributions are nonpositive. There must therefore be at least two distinct positive-curvature shells. This also proves existence without specifying an attachment.

For an end multi-face disc block with attachment $v$, apply steps 1.1–3.1 to the block by itself, temporarily suppressing $v$ if it has degree two there. At most one of the distinct shells can have $v$ in the interior of its exterior arc: an interior point of an exterior edge is incident to just one face. Choose another positive shell. If $v$ was not suppressed, it is an arc endpoint and cannot lie internally on a single exterior arc. In either event the chosen shell's exterior arc avoids the attachment internally and transfers by [F1]. This supplies the attachment qualification directly; it does not assume that a shell crossing the attachment is a contiguous global segment.

A one-face end block has no internal arcs and its complete boundary is an excursion beginning and ending at the attachment, so it is a zero-shell. If the finite block tree has an end bridge, its terminal endpoint is a spur by [F1]. Otherwise an end block is a disc and step 4.1 or the one-face argument applies. A diagram with no faces is a nontrivial finite tree and has an end bridge. These alternatives cover every non-point diagram.

The diagram exists and is reduced by [F3]. Root its finite block tree at the boundary basepoint (or the block containing that point). A terminal bridge away from the root gives a spur excursion wholly inside the linear word, hence consecutive inverse letters, impossible since $w$ is freely reduced. If there are no faces, the diagram is a tree; any nontrivial finite rooted tree has such a terminal spur away from the root. Since $w$ is nonempty, the diagram is not a point. There is therefore a terminal disc block, with no attachments away from its parent attachment. When the root block is the only block, it is a disc and the chosen basepoint is the sole point to avoid inside the exterior arc.

For a multi-face terminal block, choose one of the two shells supplied by [F1] whose exterior arc does not contain the attachment in its interior. If this is the root disc, avoid the basepoint instead. At most one of the two distinct exterior arcs can contain that specified point internally. The chosen arc therefore appears contiguously in the original outer boundary walk. For a one-face terminal block, the entire face boundary is a contiguous excursion from the attachment back to itself; when it is the root disc, start the full face reading at the given basepoint, so it is the original linear word. This also handles conjugating bridges leading from the basepoint to a single disc.

For a shell with $i\in \{1,2,3\}$ internal arcs, their total length $t$ satisfies $t<i\mathrm{\mid }r\mathrm{\mid }\mathrm{/}6\le \mathrm{\mid }r\mathrm{\mid }\mathrm{/}2$ by [F2]. Its exterior arc has length $\mathrm{\mid }r\mathrm{\mid }-t>\mathrm{\mid }r\mathrm{\mid }\mathrm{/}2$. For a zero-shell it has length $\mathrm{\mid }r\mathrm{\mid }>\mathrm{\mid }r\mathrm{\mid }\mathrm{/}2$, since relators are nonempty. Step 2.1 places this segment in the original linear word. Choose the orientation and cyclic conjugate of the face relator that starts with this segment; symmetrisation ensures that this is again a defining relator. This proves the stated initial-segment conclusion.

Prove the claim by recursion on the positive integer $\mathrm{\mid }x\mathrm{\mid }+\mathrm{\mid }y\mathrm{\mid }$. When $\mathrm{\mid }x\mathrm{\mid }=\mathrm{\mid }y\mathrm{\mid }$, the prefixes of that length in the equality $xy=yx$ give $x=y$; take $c=x$ and $a=b=1$. This includes the smallest possible total length two.

If $\mathrm{\mid }x\mathrm{\mid }>\mathrm{\mid }y\mathrm{\mid }$, prefix comparison gives $x=yu$ with $u$ nonempty. Substitute into $xy=yx$ to obtain $yuy=yyu$, and cancel the first literal copy of $y$, yielding $uy=yu$. The pair $(u,y)$ has smaller total length, so the recursive assertion gives $u=c^a$ and $y=c^b$ with $a,b>0$. Then $x=yu=c^{a+b}$. If $\mathrm{\mid }y\mathrm{\mid }>\mathrm{\mid }x\mathrm{\mid }$, interchange $x,y$ and the identical reduction gives the assertion. Each reduction strictly lowers a positive integer, so it terminates at the equal-length case. All cancellations are prefix cancellations of strings, not group reductions.

Moving the first copy of $x$ to the end gives another symmetrised word $s=x^{m-1}ax$. The equality $r=s$ holds exactly when, after cancelling the common prefix $x^{m-1}$, $xa=ax$. Thus when $xa\mathrm{\ne }ax$, the distinct words $r,s$ have the nonempty common prefix $x^{m-1}$, a piece by [F1].

When $xa=ax$ and $a\mathrm{\ne }\mathrm{\varnothing }$, [F2] gives $x=c^b$, $a=c^d$ for positive integers $b,d$. Substitution gives $r=c^{mb+d}$. If $a$ is empty then $r=x^m$ directly. This exhausts the commuting and noncommuting possibilities without mistaking equal rotations for distinct relators.

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 $R$.

The word $z$ is nonempty and cyclically reduced by [F2]. Every cyclic rotation represents a conjugate of the original element. If such a rotation contained $s$ with $r=st$ and $\mathrm{\mid }s\mathrm{\mid }>\mathrm{\mid }r\mathrm{\mid }\mathrm{/}2$, replace $s$ by $t^{-1}$ in that rotation. The quotient element is unchanged since $st=1$, while length strictly decreases since $\mathrm{\mid }t\mathrm{\mid }<\mathrm{\mid }s\mathrm{\mid }$; subsequent free reduction cannot increase length. This contradicts shortest conjugacy length. Hence $z$ is cyclically Dehn-reduced.

Any nonempty word $a$ all of whose positive powers are Dehn-reduced has infinite order. Indeed if $a^n=1$, its nonempty freely reduced literal word has a long cyclic relator segment by [F1]. Every cyclic segment of $a^n$ of length at most $n\mathrm{\mid }a\mathrm{\mid }$ appears as a literal segment of $a^{2n}$ by taking two copies of that cyclic word. This contradicts Dehn reduction of $a^{2n}$. This argument also covers n=1 and a segment crossing the chosen basepoint.

Apply [F3] using step 1.1. Its all-powers alternative for $z$ contradicts finite order by step 1.2. Its square alternative gives $z^2$ conjugate to a nonempty $y$ of infinite order by step 1.2. But if $z^n=1$ then $(z^2{)}^n=1$, and conjugation preserves this equation; hence $y$ has finite order, a contradiction, even when $z^2=1$. Only the common-root alternative remains. The inverse of a defining relator has the inverse root, so the conclusion may equally be expressed using cyclic conjugates of the original oriented defining words and integer powers of their roots.

Choose a shortest conjugacy representative using [F2]. By [F1], a rotation $v$ of this representative and a relator $r$ satisfy $v=c^a$, $r=c^b$ for a nonempty word $c$ and positive integers $a,b$. Rotation is conjugation in the free group, since $pq$ rotates to $qp=p^{-1}(pq)p$. Therefore $g$ is conjugate in the quotient to the image of $c^a$.

Among words $d$ with $c=d^k$ literally for some positive $k$, choose one of shortest length. The finite set of candidate lengths is nonempty because $d=c,k=1$ is allowed. If $d=e^j$ with $j\ge 2$, then $c=e^{jk}$, contradicting the shorter length of $e$. Hence $d$ is not a proper power, $r=d^{kb}$, and $v=d^{ka}$. Thus $d$ is a root in [F2] and $g$ is conjugate to a power of its image, as claimed. If $kb=1$, the relator itself is the root and is trivial in the quotient, which would contradict $g\mathrm{\ne }1$; this endpoint simply cannot occur for the given element.