Proof. Let $G = \langle a, b\mid S^n\rangle$ , $n\geq1$ maximal.

Proof. We start with (iii) $\Leftrightarrow$ (iv).

(iii) $\Rightarrow$ (iv) It is a standard result in the theory of group presentations that $\langle a, b\mid S^n\rangle\cong\langle a, b\mid \psi(S^n)\rangle$ , so the result follows by taking $T(a, y)\in F(a, y)$ such that $T(a, b^{-1}ab)=\psi(S)$ .

(iv) $\Rightarrow$ (iii) This holds as the word $S':=T(a, b^{-1}ab)\in\langle a, b^{-1}ab\rangle$ is in the ${\mathrm{Aut}}(F(a, b))$ -orbit of S [ Reference PridePri77 ], as required.

Finally we prove (v) $\Rightarrow$ (i) $\Rightarrow$ (ii) $\Rightarrow$ (iv) $\Rightarrow$ (v).

(v) $\Rightarrow$ (i) This is clear as the stated $\mathcal{Z}_{\max}$ -JSJ decomposition is non-trivial.

(i) $\Rightarrow$ (ii) This is immediate from the definitions (see Convention 2·8).

(ii) $\Rightarrow$ (iv) Suppose G has an essential $\mathbb{Z}$ -splitting. Then G has infinite outer automorphism group [ Reference LevittLev05a , Theorem 5·1], and so there exists $S'\in\langle a, b^{-1}ab\rangle$ such that $G\cong \langle a, b\mid S'\rangle$ [ Reference LoganLog16b , Lemma 5·1]. Setting T(a, y) to be the word such that $T(a, b^{-1}ab)=S'$ gives the result.

(iv) $\Rightarrow$ (v) The group G has presentation

\[\langle a, b\mid T(a, b^{-1}ab)^n\rangle\cong\langle a, b, y\mid T(a, y)^n, y=b^{-1}ab\rangle\]

which has the claimed form. This is an HNN-extension with base group $H=\langle a, y\mid T(a, y)^n\rangle$ and stable letter b as the subgroups $\langle a\rangle$ and $\langle y\rangle$ of H are isomorphic, because a, and hence its conjugate $y=b^{-1}ab$ , have infinite order in G by the B.B. Newman Spelling Theorem [ Reference Lyndon and SchuppLS77 , Theorem IV·5·5]. Indeed, G has presentations $\langle a, b\mid T(a, b^{-1}ab)^n\rangle$ and $\langle y, b\mid T(byb^{-1}, y)^n\rangle$ , whence it follows that the subgroups $\langle a\rangle$ and $\langle y\rangle$ are maximal cyclic subgroups of G [ Reference NewmanNew73 , Lemma 2·1].

Therefore, the HNN-extension describes a graph of groups decomposition ${\boldsymbol{\Gamma}}$ of G with a single vertex and a single loop edge, and where the edge groups are maximal cyclic in G. By Proposition 2·7, the vertex is rigid and ${\boldsymbol{\Gamma}}$ is the $\mathcal{Z}_{\max}$ -JSJ decomposition of G, as required.

Proof. For a conjugacy class [U] of an element $U\in F(a, b)$ , we write $|[U]|:=\min\{|V|\mid V\in [U]\}$ . So $|[U]|$ is simply the length of U after cyclic reduction. Therefore, the proposition says in particular that for $W\in\langle a, b^{-1}ab\rangle$ , there is no automorphism $\alpha\in{\mathrm{Aut}}(F(a, b))$ such that $|[\alpha(W)]|\lt |[W]|$ . Clearly the result is true if [W] contains some power of a. Therefore, assume that $[W]\cap \langle a\rangle=\emptyset$ , and suppose that there exists an automorphism $\alpha\in{\mathrm{Aut}}(F(a, b))$ such that $|[\alpha(W)]|\lt |[W]|$ . We find a contradiction.

By the “peak reduction” lemma from Whitehead’s algorithm [ Reference Lyndon and SchuppLS77 , Proposition I·4·20], there exists a Whitehead automorphism $\beta$ of the second kind (so a Whitehead automorphism which is not a permutation of $\{a, b\}^{\pm1}$ ) such that $|[\beta(W)]|\lt |[W]|$ . Writing $\gamma_b\in{\mathrm{Aut}}(F(a, b))$ for conjugation by b, the only four such Whitehead automorphisms which do not fix [W] are $\beta_{1}\colon a\mapsto ab$ , $b\mapsto b$ , $\beta_{-1}\colon a\mapsto ab^{-1}$ , $b\mapsto b$ , $\gamma_b\beta_1$ , and $\gamma_b\beta_{-1}$ .

We find a contradiction for the automorphisms $\beta_1$ and $\beta_{-1}$ ; the cases of $\gamma_b\beta_1$ and $\gamma_b\beta_{-1}$ follow immediately as $[\gamma_b\beta_1(W)]=[\beta_1(W)]$ and $[\gamma_b\beta_{-1}(W)]=[\beta_{-1}(W)]$ . Consider a cyclically reduced conjugate W ^′ of W which begins with a or $a^{-1}$ and ends with b. Then $W'\in\langle a, b^{-1}ab\rangle$ , and we write this word as a reduced word $U(a, b^{-1}ab)$ . Let X, Y denote the total number of x-terms, y-terms respectively in U(x, y), and ${\mathrm{Syl}}(X), {\mathrm{Syl}}(Y)$ the total number of x-syllables, y-syllables respectively in U(x, y). Then $|W'|=X+Y+2{\mathrm{Syl}}(Y)$ . Now, $[\beta_1(W')]=[U(ba, ab)]$ and $[\beta_{-1}(W')]=[U(ab^{-1}, b^{-1}a)]$ , and no free reduction or cyclic reduction happens when forming U(ba, ab) or $U(ab^{-1}, b^{-1}a)$ , and so $|[\beta_1(W')]|=2X+2Y=|[\beta_{-1}(W')]|$ . Now, clearly $Y\geq{\mathrm{Syl}}(Y)$ , while ${\mathrm{Syl}}(X)={\mathrm{Syl}}(Y)$ as U(x, y) starts with an x-syllable and ends with a y-syllable, by our choice of W ^′, and so $X\geq{\mathrm{Syl}}(Y)$ . Therefore, $2X+2Y\geq X+Y+2{\mathrm{Syl}}(Y)$ . However, by assumption $2X+2Y\lt X+Y+2{\mathrm{Syl}}(Y)$ , so we have our promised contradiction.

Proof. Suppose (i) holds. Then G is not Fuchsian and so, by Proposition 3·1, S is not conjugate to $[a, b]^{\pm1}$ . Then by Theorem 3·2 there exists an automorphism $\psi\in{\mathrm{Aut}}(F(a, b))$ such that $\psi(S)\in\langle a, b^{-1}ab\rangle$ . By Proposition 3·3, there exists a cyclic shift T of $\psi(S)$ which has shortest length in its ${\mathrm{Aut}}(F(a, b))$ -orbit, and $T\in\langle a, b^{-1}ab\rangle$ . Moreover, as S is not conjugate to $[a, b]^{\pm1}$ , neither is T [ Reference Magnus, Karrass and SolitarMKS04 , Theorem 3·9]. Hence, the word T satisfies (ii).

If (ii) holds then (i) follows from Theorem 3·2.

Proof. Let X denote the Cayley 2-complex associated to the one-relator presentation $\langle x, y\mid R_N\rangle$ of G. Since G is torsion-free, X is aspherical [ Reference LyndonLyn50 ]. As G is finitely presentable, there exists a finite subset $\mathbf{t}_N^{\prime}$ of $\mathbf{t}_N$ which also normally generates N. Therefore, G admits a finite presentation

\[\langle x, y\mid \mathbf{t}_N^{\prime}(x, y^{-1}xy), R_N\rangle,\]

and let Y be the Cayley 2-complex of this presentation. Note that X is a subcomplex of Y. Moreover, since every relator r in $\mathbf{t}_N^{\prime}(x, y^{-1}xy)$ corresponds to a closed loop $\gamma_r$ in X, we can form a 3-complex Z from Y by G-equivariantly gluing in 3-cells, one G orbit for each element in $\mathbf{t}_N^{\prime}(x, y^{-1}xy)$ , in such a way that the boundary of the 3-cell is the union of two 2-discs glued along their boundary, where the first disc fills in the relator r in Y, and the second fills in the loop $\gamma_r$ in X. It is clear that Z is a 3-dimensional cofinite G-complex which retracts onto X, and hence is aspherical.

We will now use Z, together with the fact that G satisfies the Atiyah conjecture [ Reference Jaikin-Zapirain and López-ÁlvarezJZLA20 ] and that it is $L^2$ -acyclic [ Reference Dicks and LinnellDL07 ], and compute the $L^2$ -torsion polytope $P^{(2)}(G)$ of G, as defined by Friedl–Lück [ Reference Friedl and LückFL17 ]. Note that, in general, $P^{(2)}$ is not necessarily a polytope, but a formal difference of two polytopes. Nevertheless, when G is a one-relator group then $P^{(2)}(G)$ coincides with the Friedl–Tillmann polytope by [ Reference Friedl and LückFL17 , Remark 5·5], and is a single polytope.

We start by looking at the cellular chain complex $C_\bullet$ of Z. For every n, the n-chains $C_n$ form a free $\mathbb Z G$ -module. We pick a natural cellular basis for $C_\bullet$ , namely: the vertex 1 forms the basis for $C_0$ ; the edges connecting 1 to y and x form the basis of $C_1$ ; the discs filling-in $R_N$ in X and the relators from $\mathbf{t}_N^{\prime}(x, y^{-1}xy)$ in Y form the basis for $C_2$ ; and finally the 3-cells attached to the basic discs filling-in relators from $\mathbf{t}_N^{\prime}(x, y^{-1}xy)$ form the basis of $C_3$ .

We may now identify $C_\bullet$ with

\[ \mathbb Z G^\alpha \to \mathbb Z G^{\alpha+1} \to \mathbb Z G^2 \to \mathbb Z G,\]

where $\alpha = \vert \mathbf{t}_N^{\prime}(x, y^{-1}xy) \vert$ .

Before proceeding any further, we need to deal with the special case in which the Fox derivatives ${\partial r}/{\partial x}$ are zero for all $r \in \mathbf{t}_N^{\prime}(x, y^{-1}xy)$ . The fundamental formula of Fox calculus [ Reference FoxFox53 ] tells us that

\[ \dfrac{\partial r}{\partial x}(1-x) + \dfrac{\partial r}{\partial y}(1-y) = 1-r = 0\]

in $\mathbb Z G$ . Hence $({\partial r}/{\partial y})(1-y) = 0$ , which forces ${\partial r}/{\partial y} = 0$ , as G satisfies the Atiyah conjecture and hence $\mathbb Z G$ has no non-trivial zero divisors. (A careful reader might observe that for this argument we need $y \neq 1$ in G, which is true since otherwise the abelianisation of G would not be $\mathbb Z^2$ .) In fact, we can conclude that $\mathbb Z G$ does not have non-trivial zero-divisors from an earlier work of Lewin–Lewin [ Reference Lewin and LewinLL78 ]. Now, since both Fox derivatives of r vanish, the 1-cycle corresponding to r is trivial.

Since $R_N$ lies in the normal closure of $\mathbf{t}_N^{\prime}(x, y^{-1}xy)$ , the 1-cycle given by $R_N$ must also be trivial. Hence, the differential $C_2 \to C_1$ is zero. We know that $C_\bullet$ is $L^2$ -acyclic. For torsion-free groups satisfying the Atiyah conjecture, like G, this amounts to saying that $\mathcal D(G) \otimes_{\mathbb Z G} C_\bullet$ is acyclic, where $\mathcal{D}(G)$ is the Linnell skew-field, a skew-field which contains $\mathbb Z G$ . In our case, if the differential $C_2 \to C_1$ is trivial, then tensoring $C_\bullet$ with $\mathcal D(G)$ yields the chain

\[ \mathcal D(G)^\alpha \longrightarrow \mathcal D(G)^{\alpha+1} \overset{0}{\longrightarrow} \mathcal D(G)^2 \longrightarrow \mathcal D(G)\]

which cannot be acyclic for dimension reasons. This is a contradiction.

Let $r \in \mathbf{t}_N^{\prime}(x, y^{-1}xy)$ be such that the Fox derivative ${\partial r}/{\partial x}$ is not zero. Let d denote the 2-chain given by the disc filling-in r in Y; note that d is a basis element of $C_2$ . We now look at the commutative diagram with exact columns

(†)

where $\langle d \rangle$ denotes the $\mathbb Z G$ -span of d. Note that (†) is really a short exact sequence of chain complexes.

By assumption, the middle row of (†) is $L^2$ -acyclic. We claim that so is the second row, which we will denote by $B_\bullet$ . To prove the claim, we need to look at the chain complex $B_\bullet$ in more detail:

\begin{equation*}\begin{aligned} \langle d \rangle \xrightarrow{\Big( \dfrac{\partial r}{\partial y} \ \dfrac{\partial r}{\partial x} \Big)} & & \mathbb Z G^2 \xrightarrow{\left( \begin{array}{c} 1-y \\ 1-x \end{array} \right) } & & \mathbb Z G,\end{aligned}\end{equation*}

Since G is not cyclic, we have $1-y \neq 0$ ; we also have ${\partial r}/{\partial x} \neq 0$ by assumption. Now $B_\bullet$ fits into the exact sequence of chain complexes

(‡)

where the middle column represents the inclusion of the first coordinate into $\mathbb Z G^2$ and then the projection onto the second coordinate. Since both horizontal differentials labelled in the diagram are multiplications by non-zero elements of $\mathbb Z G$ , they are invertible over $\mathcal{D}(G)$ , and hence the second and fourth rows of this commutative diagram are exact upon tensoring with $\mathcal{D}(G)$ . By [ Reference Friedl and LückFL17 , Lemma 2·9], the middle row also becomes exact upon tensoring with $\mathcal{D}(G)$ . This proves the claim.

We are now back to examining (†). We have just shown that the second row is $L^2$ -acyclic, and hence [ Reference Friedl and LückFL17 , Lemma 2·9] tells us that so is the fourth row, and that

\[ P^{(2)}(C_\bullet) = P^{(2)}(B_\bullet) + P^{(2)}(D_\bullet),\]

where $D_\bullet$ denotes the fourth row of (†). We can split $P^{(2)}(B_\bullet)$ further using diagram (‡) and [ Reference Friedl and LückFL17 , Lemma 1·9], and obtain

\[ P^{(2)}(B_\bullet) = P - Q,\]

where $-P$ is the $L^2$ -torsion polytope of

\[ \mathbb Z G \overset{1-y}{\longrightarrow} \mathbb Z G\]

and $-Q$ of

\[ \mathbb Z G \overset{\frac{\partial r}{\partial x}}{\longrightarrow} \mathbb Z G,\]

Now we need to recall how the polytope $P^{(2)}$ is actually constructed, at least in the above (simple) cases. When considering a chain complex of the form

\[ \mathbb Z G \overset{z}{\longrightarrow} \mathbb Z G\]

with $z \in \mathbb Z G \smallsetminus \{0\}$ , the polytope is obtained by first taking the support $\mathrm{supp}(z) \subseteq G$ , then taking the image of this set in $H_1(G;\;\mathbb R)$ , and then taking the convex hull. At the end the polytope is given a sign, negative in this situation, and in general depending on whether the single non-trivial differential starts in an even or an odd dimension.

With the construction in mind, it is now immediate that, since r is a word in x and $y^{-1}xy$ , the polytope P is contained within the strip $[-1,0] \times \mathbb R$ . The polytope Q is obtained from $1-y$ , and hence it is the segment $[0,1]\times\{0\}$ .

Since $D_\bullet$ is concentrated around a single differential from odd to even degree chains, $P^{(2)}(D_\bullet) = -R$ where R is some polytope. Hence

\[ P^{(2)}(G) =P^{(2)}(C_\bullet) = P - Q - R,\]

Since G is a one-relator group, $P^{(2)}(G)$ is actually a single polytope. Hence, in particular, $P-Q$ must be a single polytope, and therefore it must lie on the line $\{-1\} \times \mathbb R$ . Subtracting a further polytope from such a polytope does not alter this fact, and hence we have finished the proof.

Next, we prove the converse to Theorem 4·1. This is applied to prove that (iii) implies (ii) in the proof of Theorem 5·4.

Proof. Suppose that the Friedl–Tillmann polytope is a line. By applying a suitable Nielsen automorphism, we may assume that G is given by the presentation $ \langle x,y \mid S \rangle$ and that the polytope lies on the line $\{0\} \times \mathbb R$ .

To construct the Friedl–Tillmann polytope, we first define an auxiliary polytope P by tracing the closed loop the word S gives in the Cayley graph of the free part of the abelianisation of G taken with respect to the image of $\{x,y\}$ as a generating set, and then taking the convex hull of the image of the loop in $H_1(G;\;\mathbb R)$ . Then we observe that P has to be the Minkowski sum of another polytope, say P ^′, and the square $[-1,0] \times [-1,0]$ . The polytope P ^′ is precisely the Friedl–Tillmann polytope. Note that the polytope is only well-defined up to translation, so it does not matter which square we choose. (When G is torsion-free, the polytope P ^′ coincides with the $L^2$ -torsion polytope constructed above.)

Now, we know that P ^′ is a line, which forces P to lie inside the strip $[-1,0] \times \mathbb R$ . Hence, the loop given by S must also lie in this strip. Since $S \in F(x,y)'$ , it can be written as a product of right conjugates of x by powers of y. We now see that the only conjugates of x appearing can be x and $y^{-1}xy$ . Therefore, there exists a word T ^′ in the ${\mathrm{Aut}}(F(a, b))$ -orbit of R such that $T'\in\langle a, b^{-1}ab\rangle$ , and so by Proposition 3·3 there exists a word T of shortest length in the ${\mathrm{Aut}}(F(a, b))$ -orbit of R with $T\in\langle a, b^{-1}ab\rangle$ , as required.

Now suppose that there exists a word T of shortest length in the ${\mathrm{Aut}}(F(a, b))$ -orbit of R such that $T\in\langle a, b^{-1}ab\rangle$ . Then we may assume that G is given by a presentation $\langle a, b \mid T' \rangle$ with $T'\in\langle a, bab^{-1}\rangle$ . It follows that the loop given by T must lie inside the strip $[-1,0] \times \mathbb R$ , and so P also lies inside this strip. As P is the Minkowski sum of P ^′ and the square $[-1,0] \times [-1,0]$ , we see that P ^′ is a straight segment.

Proof. Suppose the Friedl–Tillmann polytope P ^′ of $\mathcal{P}$ is a point. As in the proof of Lemma 4·2, we may assume that G is given by the presentation $\langle x,y \mid S \rangle$ and that the polytope lies on the line $\{0\} \times \mathbb R$ . As P ^′ is a point, the polytope P is the square $[0,1] \times [0, 1]$ (up to translation). This means precisely that $S = [a, b]^k$ for some non-zero $k\in\mathbb{Z}$ .

Now suppose that R is conjugate to $[a, b]^k$ for some $k\in\mathbb{Z}$ . Then we may assume that G is given by the presentation $ \langle a, b \mid (aba^{-1}b^{-1})^k \rangle$ , and this presentation is easily seen to have Friedl–Tillmann polytope a point.

Proof. Suppose that G admits a one-relator presentation $\langle a, b \mid R \rangle$ .

The subgroups $\langle p\rangle$ and $\langle q\rangle$ are “conjugacy separated” in H, that is, for every $h\in H$ we have $h^{-1}\langle p\rangle h\cap\langle q\rangle=1$ . To see this, suppose otherwise, so $h^{-1}p^ih=q^j$ for $|i|, |j|\gt0$ . It follows that $h^{-1}p^ih=t^{-1}p^{nj}t$ , and as G is CSA $_{\mathbb{Q}}$ we have that $th^{-1}$ and p share a common power, a contradiction by Britton’s Lemma.

As $\langle p\rangle$ and $\langle q\rangle$ are conjugacy separated in H, there exists a Nielsen transformation $\psi$ of F(a, b) and some $g\in G$ such that $\psi(a)=_Gg^{-1}p^ig$ and $\psi(b)=_Gg^{-1}thg$ for some $i\in \mathbb{Z}$ and some $h\in H$ [ Reference Kapovich and WeidmannKW99b , Corollary 3·1]. Therefore, writing $x:=p$ and $y:=th$ , the pair (x, y) generates G. But also the pair $(x^i,y)$ generates G, and as G has abelianisation $\mathbb{Z}^2$ we have that $|i|=1$ . Note that as the generating pair (a, b) of G admits a one-relator presentation, and as $\psi$ is a Nielsen transformation, we have that the pair (x, y) also admits a one-relator presentation $\langle x, y\mid R_N\rangle$ . Hence, points (i) and (ii) of the theorem hold for some subgroup $N\leq F(x, y)$ .

We now establish (iii) for $N\leq F(x, y)$ . Firstly, note that $H=\langle p, h^{-1}qh\rangle$ by [ Reference Gardam, Kielak and LoganGKL21 , Proposition 8·3], and so H has presentation $\langle x, z\mid\mathbf{s}(x, z)\rangle$ where $x:=p$ (as above) and $z:=h^{-1}qh$ . Therefore, G has presentation $\langle x, z, t\mid \mathbf{s}(x, z), t^{-1}x^nt=w(x, z)zw(x, z)^{-1}\rangle$ where w(x, z) represents the element $h\in H$ . As remarked above the theorem, because G is finitely presentable we may assume for our purposes with G (but not H) that the set $\mathbf{s}(x, z)$ is finite. As $y=th$ , we can apply Tietze transformations as follows, where the generators x and y correspond precisely to the generators x and y in the previous paragraph:

\begin{align*}&\langle x, z, t\mid \mathbf{s}(x, z), t^{-1}x^nt=w(x, z)zw^{-1}(x, z)\rangle\\&\cong\langle x, y, z, t\mid \mathbf{s}(x, z), t^{-1}x^nt=w(x, z)zw^{-1}(x, z), y=tw(x, z)\rangle\\&\cong\langle x, y, z, t\mid \mathbf{s}(x, z), y^{-1}x^ny=z, y=tw(x, z)\rangle\\&\cong\langle x, y, z\mid \mathbf{s}(x, z), y^{-1}x^ny=z\rangle\\&\cong\langle x, y\mid \mathbf{s}(x, y^{-1}x^ny)\rangle\\&\cong\langle x, y\mid \mathbf{t}_N\rangle\end{align*}

for some finite subset $\mathbf{t}_N = \mathbf{s}(x, y^{-1}x^ny)$ of $\langle x, y^{-1}xy\rangle$ . As the symbols x and y in this presentation of G and the one-relator presentation $\langle x, y\mid R_N\rangle$ of G both represent the same elements of G, it follows that N is the normal closure of the set $\mathbf{t}_N$ in F(x, y), as required.

Proof. Note that G is one-ended by Proposition 3·1, and is a CSA $_{\mathbb{Q}}$ group. Therefore, Theorem 2·9 is applicable.

Items (ii) and (iii) are equivalent by Lemma 4·2.

(ii) $\Rightarrow$ (i) The word T gives a presentation for G:

\[\langle a, b\mid T\rangle=\langle a, b\mid T_0(a, b^{-1}ab)\rangle.\]

Then (i) follows by Theorem 2·9.

(i) $\Rightarrow$ (iii) By Theorem 2·9 the conditions of Lemma 5·1 are satisfied, so we may apply Theorem 4·1, and (iii) follows.

Proof. Suppose that $\alpha, \beta \in {\mathrm{Aut}}(F(a, b))$ are such that $\alpha(S)$ and $\beta(S^n)$ are shortest-length in their respective ${\mathrm{Aut}}(F(a, b))$ -orbits. In particular, they are both cyclically reduced and since $\beta(S^n) = \beta(S)^n$ is cyclically reduced if and only if $\beta(S)$ is cyclically reduced, we see that $|\alpha(S)^n| = n |\alpha(S)|$ and $|\beta(S)^n| = n |\beta(S)|$ . Now by the shortest-length assumptions we have

\[ |\alpha(S^n)| \geq |\beta(S^n)| = n |\beta(S)| \geq n |\alpha(S)| = |\alpha(S^n)|\]

so equality holds. The conclusion follows by taking $\alpha$ or $\beta$ to be the identity as appropriate.

Proof. If there exists a word T in the ${\mathrm{Aut}}(F(a, b))$ -orbit of R which is conjugate to $[a, b]^{k}$ for some $k\in\mathbb{Z}$ , then G is either abelian (when $|k|=1$ ) or contains torsion (when $k\gt1$ ). Applying this to Lemma 4·3, we also have that if $\mathcal{P}$ is a single point then G is either abelian or contains torsion. Therefore, if G is torsion-free then the result follows by Theorem 5·2.

Suppose G contains torsion, so $R=S^n$ in F(a, b) for some $n\gt1$ maximal.

(ii) $\Leftrightarrow$ (iii) This follows from Lemmas 4·2 and 4·3.

(i) $\Rightarrow$ (ii) If G has non-trivial $\mathcal{Z}_{\max}$ -JSJ decomposition then, by Theorem B, there exists a word T ^′ of shortest length in the ${\mathrm{Aut}}(F(a, b))$ -orbit of S such that $T'\in\langle a, b^{-1}ab\rangle$ but T ^′ is not conjugate to $[a, b]^{\pm 1}$ . Then the word $T:=(T')^n$ has shortest length in its ${\mathrm{Aut}}(F(a, b))$ -orbit, by Lemma 5·3, and hence in the ${\mathrm{Aut}}(F(a, b))$ -orbit of R, is contained in $\langle a, b^{-1}ab\rangle$ , and is not conjugate to $[a, b]^{\pm n}$ (we are using uniqueness of roots in free groups here). As n is maximal, T is not conjugate to any other power of $[a, b]^{\pm1}$ , and so (ii) holds.

(ii) $\Rightarrow$ (i) For T as in (ii), write $\psi\in{\mathrm{Aut}}(F(a, b))$ for the automorphism such that $\psi(R)=T$ , and set $T':=\psi(S)$ . Then T ^′ has shortest length in the ${\mathrm{Aut}}(F(a, b))$ -orbit of S, by Lemma 5·3, is in $\langle a, b^{-1}ab\rangle$ and is not conjugate to $[a, b]^{\pm 1}$ . Therefore, by Theorem B, (i) holds.

Proof. By (ii) of Theorem A or B, as appropriate, G admits a presentation of the form $\langle a, b\mid T_0^n(a, b^{-1}ab)\rangle$ , $n\geq1$ , where $T_0^n(a, b^{-1}ab)$ and $T^n(a, b)$ are the same words. Hence, G has a presentation

\[\langle a, b, y\mid T_0^n(a, y), b^{-1}ab=y\rangle,\]

which is an HNN-extension with stable letter b, and corresponds to a graph of groups decomposition ${\boldsymbol{\Gamma}}$ as in the statement of the corollary. This decomposition corresponds to the $\mathcal{Z}_{\max}$ -JSJ decomposition of G and has rigid vertex by Theorem 2·9, if G is as in Theorem A, and by Theorem 3·2 otherwise.

Finally, as T is not a power of a primitive we have that $|T_0|\lt|T|$ , and so $|T_0|\lt|T|\leq|S|$ , as required.

Proof. Write R as $S^n$ with $n\geq1$ maximal. Write $\mathcal{O}_S$ for the set of shortest elements in the ${\mathrm{Aut}}(F(a, b))$ -orbit of S.

Firstly, use Whitehead’s algorithm to compute the set $\mathcal{O}_S$ ; this takes $O(|S|^2)\leq O(|R|^2)$ steps [ Reference KhanKha04 ]. Then compute $\mathcal{T}:=\mathcal{O}_S\cap\langle a, b^{-1}ab\rangle$ ; Khan gave a linear bound on the cardinality of $\mathcal{O}_S$ [ Reference KhanKha04 ] and we can (constructively) test membership of any fixed finitely generated subgroup of a free group in time linear in the length of the input word, so this takes $O(|S|^2)\leq O(|R|^2)$ steps. If the intersection $\mathcal{T}$ is empty then, by Theorem A or B as appropriate, the $\mathcal{Z}_{\max}$ -JSJ decomposition of G is trivial. Therefore, output as the $\mathcal{Z}_{\max}$ -JSJ decomposition of G the graph of groups consisting of a single vertex, with vertex group G, and no edges.

If $\mathcal{T}$ is non-empty then take some $T\in\mathcal{T}$ . Rewrite this word T as a word $T_0(a, b^{-1}ab)$ ; this takes $O(|T|)\leq O(|R|)$ steps. By Corollary C, the $\mathcal{Z}_{\max}$ -JSJ decomposition of G is the graph of groups with a single vertex, a single loop edge, vertex group $\langle a, y\mid T_0^n(a, y)\rangle$ and the attaching map $a=byb^{-1}$ , and so output this as the $\mathcal{Z}_{\max}$ -JSJ decomposition of G.

Proof. We describe an algorithm to decide in $O(|R|)$ steps whether or not the Friedl–Tillmann polytope is a line; the result then follows from Theorem A. We need to avoid computing the entire convex hull, because that cannot in general be done in linear time. Suppose that the Friedl–Tillmann polytope P is a line, namely from $(x_0, y_0)$ to $(x_1, y_1)$ with $x_0 \leq x_1$ . Then the Minkowski sum P ^′ of P and $[0, 1] \times [0, 1]$ will be the convex hull of 6 points (with only 4 needed in the vertical or horizontal case). If $y_0 \leq y_1$ , the points are $(x_0+1, y_0), (x_0, y_0), (x_0, y_0+1)$ in the bottom left corner and $(x_1, y_1+1), (x_1+1, y_1+1), (x_1+1, y_1)$ in the top right corner, and the minimal bounding box (axis-parallel rectangle) has opposite corners $(x_0, y_0)$ and $(x_1+1, y_1+1)$ . If $y_0 \gt y_1$ , the minimal bounding box instead has opposite corners $(x_0, y_0+1)$ and $(x_1+1, y_1)$

The algorithm proceeds as follows. If necessary, we put the word into reduced form in linear time using a stack (for each letter, if it is inverse to the top of the stack then discard it and pop from the stack, else push it onto the stack). We then determine two candidate polytopes $P_0'$ and $P_1'$ , each on 6 points, by computing the minimal bounding box of (the convex hull of) the path $\gamma$ by walking the path and keeping track of the minimal and maximal x and y coordinates visited, then applying the two possibilities for P ^′ in terms of the bounding box discussed in the previous paragraph. It remains to check if the convex hull of $\gamma$ is contained in and contains some candidate. For each candidate, we now walk the path again and check that we stay inside the polytope (a constant time check, as it is on 6 points) and visit the 6 defining points. If either candidate matches, we have determined that the Friedl–Tillman polytope is a line, otherwise it is not.