Proof Decidability for $n = 1$ is discussed in Remark 3.1, while the case for $m=1$ will be covered in Remark 3.4. For $m,n \geq 2$ , we will obtain an injection $i \colon A(P_4) \longrightarrow G_{m,n}$ in Lemma 3.6. In [Reference Lohrey and Steinberg33], it is shown that $A(P_4)$ contains a fixed submonoid where membership is undecidable. So, if $m,n \geq 2$ , then the submonoid membership problem is undecidable in $G_{m,n}$ .

Proof Note that $K = K_{m,n}$ is normal in $G = G_{m,n}$ , as it is closed under conjugation by both x and y, with quotient equal to

$$\begin{align*}G/K = \operatorname{Gp}\big\langle y\:|\:y^{2n}=1 \big\rangle \cong \mathcal{C}_{2n} = \text{ cyclic group of order }2n. \end{align*}$$

We proceed to find a presentation for K using the Reidemeister-Schreier procedure (see [Reference Lyndon and Schupp34]) with

$$\begin{align*}G = \operatorname{Gp}\big\langle S\:|\:R \big\rangle, \text{ where } S = \{x,y\}, \text{ and } R = x^m y^n x^m y^{-n}. \end{align*}$$

A Schreier transversal for K in G is $T = \{\, y^i \mid 0 \leq i \leq 2n - 1 \,\}$ , giving a set of generators for K as $U = \{ts(\overline {ts})^{-1}\mid t\in T,\; s \in S,\; ts \not \in T \}$ where $\overline {w}$ is the representative of w in T.

Any $t\in T$ can be written as $y^i$ for $0 \leq i \leq 2n-1$ , so

$$\begin{align*}ts(\overline{ts})^{-1} = y^{i}s(\overline{y^{i}s})^{-1} = \begin{cases} y^{i}x(\overline{y^{i}x})^{-1} & \text{if } s = x,\\ y^{i}y(\overline{y^{i}y})^{-1} & \text{if } s = y. \end{cases} \end{align*}$$

As $x\in K$ , one has $\overline {y^{i}x} = y^i$ for all $0 \leq i \leq 2n-1$ . Also, $\overline {y^{i}y} = y^{i+1}$ for $0 \leq i \leq 2n-2$ and $\overline {y^{2n-1}y} = 1$ . One obtains $\beta = y^{2n}$ and $\alpha _i = y^ixy^{-i}$ for $0 \leq i \leq 2n-1$ as generators of K; hence,

$$\begin{align*}U = \{\, \beta = y^{2n},\, \alpha_i = y^ixy^{-i} \text{ for }0 \leq i \leq 2n-1\,\} \end{align*}$$

gives a set of generators for K.

To get relations for K, rewrite $tRt^{-1}$ for $t\in T$ and $R = x^m y^n x^m y^{-n}$ , using generators in U.

Write any $t\in T$ as $y^i$ for some $0 \leq i \leq 2n-1$ . One obtains

$$ \begin{align*} tRt^{-1} = y^i(x^m y^n x^m y^{-n})y^{-i} & = \begin{cases} (y^i x^m y^{-i}) (y^{i + n} x^m y^{-i - n}) & \text{if } 0 \leq i \leq n - 1 \\ (y^i x^m y^{-i}) y^{2n} (y^{i - n} x^m y^{-i + n}) y^{-2n} & \text{if } n \leq i \leq 2n - 1 \end{cases} \\ & = \begin{cases} \alpha_i^m \alpha_{i+n}^m & \text{if } 0 \leq i \leq n - 1, \\ \alpha_i^m \beta \alpha_{i-n}^m \beta^{-1} & \text{if } n \leq i \leq 2n - 1. \end{cases} \end{align*} $$

The first relation gives $\alpha _{i+n}^m = \alpha _i^{-m}$ for any $0 \leq i \leq n - 1$ ; substituting it in the second line, one obtains the relation $[\alpha _i^m, \beta ]$ for all $0 \leq i \leq n - 1$ .

So, $ V = \{\, \alpha _i^m \alpha _{i+n}^m,\, [\alpha _i^m, \beta ] \mid 0 \leq i \leq n - 1 \,\}$ gives the relations of K, as desired.

Proof Recall from Lemma 3.3 that the subgroup

$$\begin{align*}K_{m,n} = \operatorname{Gp}\big\langle y^{2n}, y^i x y^{-i} \text{ for }0 \leq i \leq 2n - 1 \big\rangle\end{align*}$$

of $G_{m,n}$ , has the following presentation:

$$\begin{align*}\operatorname{Gp}\big\langle \beta,\, \alpha_i \text{ for }0 \leq i \leq 2n - 1\:|\:\alpha_i^m \alpha_{i+n}^m=1,\, [\alpha_i^m, \beta]=1 \text{ for }0 \leq i \leq n - 1 \big\rangle, \end{align*}$$

where $\beta $ corresponds to $y^{2n}$ , and $\alpha _i$ corresponds to $y^i x y^{-i}$ for all $0 \leq i \leq 2n - 1$ . Let $K \cong K_{m,n} $ denote the group defined by the above presentation, let $B = \operatorname {B}(S_{n,m})$ , and consider the two maps

$$ \begin{align*} s: B \longrightarrow K\, & \text{ given by } s(d) = \beta,\, s(c_i) = \alpha_i\, \text{ for any } 0 \leq i \leq n - 1,\\ \rho: K \longrightarrow B\, & \text{ given by } \rho(\beta) = d,\, \rho(\alpha_i) = c_i,\, \rho(\alpha_{i+n}) = c_i^{-1}\, \text{ for any } 0 \leq i \leq n - 1. \end{align*} $$

Obviously, both s and $\rho $ induce well-defined homomorphisms, as they respect the relations. Moreover, one has $\rho \circ s = \text {id}_B$ , which implies that s is injective. It follows that there is an injective homomorphism $\sigma :\operatorname {B}(S_{n,m}) \hookrightarrow G_{m,n}$ that maps $d \mapsto y^{2n}$ and $c_i \mapsto y^i x y^{-i}$ for all $0 \leq i \leq n - 1$ .

Proof From the proof of Proposition 3.5 in [Reference Nyberg-Brodda47] (see also Proposition 3.3 there), one obtains an embedding of $A(P_4)$ in $\operatorname {B}(S_{n,m})$ for $n, m \geq 2$ . One such embedding is the following:

$$\begin{align*}j: \operatorname{Gp}\big\langle A, B, C, D\:|\:AB = BA, BC = CB, CD = DC \big\rangle \hookrightarrow \operatorname{B}(S_{n,m}), \end{align*}$$

given by $j(A) = c_1^m,\, j(B) = d,\, j(C) = c_0^m,\, j(D) = c_0dc_0^{-1}$ . Indeed, to show that j is an embedding, we note that by Lemma 3.5, the group $\operatorname {B}(S_{n,m})$ injects into $G_{m,n}$ , where

$$\begin{align*}\operatorname{B}(S_{n,m}) = \operatorname{Gp}\big\langle d, c_i \text{ for }0 \leq i \leq n - 1\:|\:[c_i^m, d]=1 \text{ for }0 \leq i \leq n - 1 \big\rangle, \end{align*}$$

with $d \mapsto y^{2n}$ , and for all $0 \leq i \leq n - 1$ , one has $c_i \mapsto y^ixy^{-i}$ . In particular, the RABSAG $\operatorname {B}(S_{2,m})$ (in the sense of [Reference Nyberg-Brodda47]) defined by the graph

injects in $\operatorname {B}(S_{n,m})$ . By [Reference Nyberg-Brodda47, Proposition 3.5] and its proof, the group $\operatorname {B}(P_{3,m})$ injects into $\operatorname {B}(S_{n,m})$ , with one embedding being via the RABSAG defined by the graph

Here, we have abused notation by letting the vertices of the graph $P_{3,m}$ above be labeled by their images in $\operatorname {B}(S_{n,m})$ under the prescribed embedding. That is, inside $\operatorname {B}(S_{n,m})$ , defined as above, the elements $c_1^m, d$ , and $c_0$ generate an isomorphic copy of $\operatorname {B}(P_{3,m})$ where, in terms of the presentation given in equation (3), this isomorphism is given by mapping $w_0$ to d, $w_2$ to $c_1^m$ , and $w_1$ to $c_0$ . Continuing, and using the analogous slight abuse of notation also for RAAGs, the RAAG defined by the graph

now embeds in our chosen copy of $\operatorname {B}(P_{3,m})$ by unpacking the proof of [Reference Nyberg-Brodda47, Proposition 3.3] together with the generalization of that result explained in [Reference Nyberg-Brodda47] in the paragraph immediately after the proof of [Reference Nyberg-Brodda47, Proposition 3.3]. That is, inside $\operatorname {B}(S_{n,m})$ as above, the elements $c_1^m, d, c_0^m$ , and $c_0dc_0^{-1}$ generate a copy of $A(P_4)$ . It follows that j is indeed an embedding of $A(P_4)$ into $\operatorname {B}(S_{n,m})$ .

Now the composition $i = \sigma \circ j$ where $\sigma $ is the injection $\sigma \colon \operatorname {B}(S_{n,m}) \hookrightarrow G_{m,n}$ defined by (4), from Lemma 3.5, gives the desired embedding $i\colon A(P_4) \hookrightarrow G_{m,n}$ .

Proof Consider the relation $x^m y^n x^m y^{-n}$ . We want to change it to an equivalent positive relation, so we introduce new generators $a,b$ with $y = a$ and $x = b a ^{n}$ . Now we write the group relation in terms of a and b as

$$ \begin{align*} (b a^{n})^m a^n (b a^{n})^m a^{-n} & = (b a^{n})^m a^n (b a^{n})^{m-1} (b a^{n}) a^{-n} \\ & = (b a^{n})^m a^n (b a^{n})^{m-1} b \\ & = (b a^{n})^m (a^{n} b)^m. \end{align*} $$

This way, we obtain another equivalent presentation for $G_{m,n}$ as

(5) $$ \begin{align} G_{m,n} \cong \operatorname{Gp}\big\langle a,b\:|\:(b a^n)^m(a^n b)^m = 1 \big\rangle. \end{align} $$

Thus, $G_{m,n}$ is a positive one-relator group.

Proof The equivalence of (i) and (ii) follows from [Reference Gray14, Remark 2.3].

(i) $\Rightarrow $ (iii): By [Reference Kim and Koberda28, Theorem 1.8], if $\Gamma $ is a finite forest, then $A(\Gamma )$ embeds in $A(P_4)$ , which in turn embeds in a positive one-relator group by Lemma 3.6 and Lemma 3.7 above.

(iii) $\Rightarrow $ (v): This follows from the result of Perrin and Schupp [Reference Perrin and Schupp50] showing that any positive one-relator group admits a one-relation monoid presentation.

(v) $\Rightarrow $ (iv) is trivial.

(iv) $\Rightarrow $ (ii): If $A(\Gamma )$ embeds into $\operatorname {Mon}\big \langle A\:|\:u=v \big \rangle $ , then it must embed in a maximal subgroup of $\operatorname {Mon}\big \langle A\:|\:u=v \big \rangle $ , which by the results of [Reference Lallement30] must itself be a one-relator group. Hence, $A(\Gamma )$ embeds in a one-relator group.

Construction 4.2 Let G be a positive one-relator group and let Q be a finitely generated submonoid of G. Let $\operatorname {Mon}\big \langle A\:|\:q=1 \big \rangle \cong \operatorname {Gp}\big \langle A\:|\:q=1 \big \rangle $ be a one-relation monoid presentation for the group, which exists by [Reference Perrin and Schupp50] since G is a positive one-relator group, and let a denote the first letter of the word q. Let Q be a finitely generated submonoid of G. Let $X = \{w_1, \ldots , w_k \} \subseteq A^+$ be a set of positive words such that $Q = \operatorname {Mon}\big \langle w_1, \ldots , w_k \big \rangle \leq G$ . Such a set of positive words X exists since every element of G can be expressed by a positive word over A, as this is true in the monoid M. For any $w \in A^+$ , let $\overline {w} \in A^+$ such that $\overline {w} = w^{-1}$ in G (i.e., $w \overline {w} = \overline {w} w = 1$ in G). Let $z_i$ be word obtained from $w_i$ after replacing a with $tx$ for every letter a. Similarly, let $\overline {z_i}$ be the word obtained by replacing a with $tx$ in $\overline {w_i}$ for every occurrence of the letter a. Let r be the word obtained by replacing every occurrence of a with $tx$ in the word q. Let $B = A \setminus \{a\}$ . Then r, $z_i$ , and $\overline {z_i}$ for $1 \leq i \leq k$ are all positive words over $B \cup \{x,t\}$ , and r begins with $tx$ since the first letter of q is a. Write $r \equiv ts$ (i.e., making s the positive word obtained by deleting the first letter of r). In particular, s begins with the letter x.

Using the data above, we define a two-relator group presentation

$$\begin{align*}H_{G,X} = \operatorname{Gp}\big\langle B,x,t\:|\: r = 1, t z_1 s t\overline{z_1} s \ldots s t z_k s t\overline{z_k} s = 1 \big\rangle. \end{align*}$$

For future reference, we also use $M_{G,X}$ to denote the corresponding inverse monoid presentation

$$\begin{align*}M_{G,X} = \operatorname{Inv}\big\langle B,x,t\:|\: r = 1, t z_1 s t\overline{z_1} s \ldots s t z_k s t\overline{z_k} s = 1 \big\rangle. \end{align*}$$

Proof Let

$$\begin{align*}H_{G,X} = \operatorname{Gp}\big\langle B,x,t\:|\: r = 1, t z_1 s t\overline{z_1} s \ldots s t z_k s t\overline{z_k} s = 1 \big\rangle \end{align*}$$

be the positive two-relator group given by Construction 4.2. Set $H = H_{G,X}$ , $Y = B \cup \{x,t\}$ , $u=r$ and $v = t z_1 s t\overline {z_1} s \ldots s t z_k s t\overline {z_k} s$ . It then follows from Theorem 7.5 and its proof that the membership problem for Q in G reduces to the prefix membership problem for H, that the identity map on Y induces an isomorphism $\operatorname {Gp}\big \langle Y\:|\:u=1, v=1 \big \rangle \rightarrow \operatorname {Gp}\big \langle Y\:|\:u=1 \big \rangle $ , and that $H \cong G \ast \mathbb {Z}$ . It follows that the identity map on Y induces an isomorphism

$$\begin{align*}\operatorname{Gp}\big\langle Y\:|\:u=1,v=1 \big\rangle \rightarrow \operatorname{Gp}\big\langle Y\:|\:vuv^{-1}=1 \big\rangle, \end{align*}$$

and this group is isomorphic to $G \ast \mathbb {Z}$ . Now since $v=1$ in this group, the prefix monoids of $\operatorname {Gp}\big \langle Y\:|\:u=1,v=1 \big \rangle $ and $\operatorname {Gp}\big \langle Y\:|\:vuv^{-1}=1 \big \rangle $ are both generated by $\operatorname {Pref}( u ) \cup \operatorname {Pref}( v )$ . Hence, the isomorphism induced by the identity map on Y maps the prefix monoid of $\operatorname {Gp}\big \langle Y\:|\:u=1,v=1 \big \rangle $ bijectively to the prefix monoid of $\operatorname {Gp}\big \langle Y\:|\:vuv^{-1}=1 \big \rangle $ . Therefore, if $\operatorname {Gp}\big \langle Y\:|\:vuv^{-1}=1 \big \rangle $ has decidable prefix membership problem, then so does $\operatorname {Gp}\big \langle Y\:|\:u=1,v=1 \big \rangle $ , which in turn by Theorem 7.5 implies that the membership problem for Q in G is decidable. The result then follows by taking $G'$ to be the one-relator group with generating set Y and defining relator the reduced form of $vuv^{-1}$ .

Proof of Theorem 4.1

By Theorem 3.8, there exists a positive one-relator group G with a fixed finitely generated submonoid Q such that the membership problem for Q in G is undecidable. Hence, by Theorem 4.3, there is a quasi-positive one-relator group $G'$ such that the membership problem for Q in G reduces to the prefix membership problem for $G'$ . Hence, $G'$ is a quasi-positive one-relator group with undecidable prefix membership problem.

Proof Set $r \equiv ba^{i_1}ba^{i_2} \ldots ba^{i_k} b a$ . Since r begins and ends with $ba$ , we can write $r \equiv \alpha ba \equiv ba \gamma $ . As $a=r=ba\gamma $ in M and $\gamma $ ends in the letter a, it follows that $a \gamma \mathcal {L} a$ . In M, we have $ a \gamma = \alpha ba \gamma = \alpha a $ ; hence, $\alpha a \mathcal {L} a$ . Since $r \equiv \alpha ba$ and $i_k \geq 1$ , it follows that the last letter of $\alpha $ equals a, so we can write $\alpha \equiv \alpha ' a$ . But now, $\alpha a \mathcal {L} a$ implies $\alpha ' a^2 \mathcal {L} a$ so $\delta \alpha ' a^2 = a$ for some word $\delta $ . It follows that $a^2 \mathcal {L} a$ in M.

Next, observe that $ba \mathcal {L} a$ since $ba$ is a suffix of r, from which it follows that $ba(ba) \mathcal {L} a(ba)$ and also $ba(a) \mathcal {L} a(a)$ . Also $aba \mathcal {L} a$ since $aba$ is a suffix of r. We have shown $aa \mathcal {L} a$ , $a(ba) \mathcal {L} a$ , $(ba)(ba) \mathcal {L} aba \mathcal {L} a$ , and $(ba)a \mathcal {L} aa \mathcal {L} a$ ; that is, all two-factor products of the words $\{a,ba\}$ are $\mathcal {L}$ -related to a. Now the result follows by induction since setting $w_1 \equiv a$ , $w_2 \equiv ba$ , for any product $ w_{i_1} w_{i_2} \ldots w_{i_k} $ with $k \geq 3$ from above, we have $w_{i_1} w_{i_2} \mathcal {L} a$ from which, since $\mathcal {L}$ is a right congruence, it follows that $ w_{i_1} w_{i_2} w_{i_3} \ldots w_{i_k} \mathcal {L} a w_{i_3} \ldots w_{i_k} $ where by induction, $a w_{i_3} \ldots w_{i_k} \equiv w_1 w_{i_3} \ldots w_{i_k} \mathcal {L} a$ .

Proof Set

$$\begin{align*}\alpha =a(ba^n)^{m-1} b a^{n-1}, \; \beta =a^{2n}, \; \gamma =(ba^n)^m, \; \mu =ba^n a^{2n} (ba^n)^{m-1}. \end{align*}$$

Let $H = \operatorname {Gp}\big \langle a(ba^n)^ma^{-1}, a^{2n}, (ba^n)^m, ba^{2n}b^{-1} \big \rangle \leqslant M_{m,n}$ , and denote by $A, B, C, D$ its $4$ generators, in the given order, recalling that $M_{m,n} = \operatorname {Gp}\big \langle a, b\:|\:(b a^n)^m(a^n b)^ma = a \big \rangle $ . In Section 3, Remark 3.6 was applied to show that $H \cong A(P_4)$ with A, B, C, and D corresponding to the vertices in the path $P_4$ in this order. It then follows from a sequence of easy Tietze transformations that $\operatorname {Gp}\big \langle \alpha ,\beta ,\gamma ,\mu \big \rangle \leq M_{m,n}$ is also isomorphic to $A(P_4)$ since $\alpha = a(ba^n)^{m-1} b a^{n-1}$ is equal to A in $M_{m,n}$ , $\beta = B$ , $\gamma = C$ , and

$$\begin{align*}\mu = ba^n a^{2n} (ba^n)^{m-1} = ba^{2n}a^n (ba^n)^{m-1} = ba^{2n}b^{-1}(ba^n)^m = DC \end{align*}$$

is equal to $DC$ in $M_{m,n}$ . Since any trace monoid naturally embeds in its corresponding right-angled Artin group, it follows that $\operatorname {Mon}\big \langle \alpha ,\beta ,\gamma ,\mu \big \rangle \leq M_{m,n}$ is isomorphic to the trace monoid $T(P_4)$ , where $\alpha $ , $\beta $ , $\gamma $ , $\mu $ correspond to the vertices on the path $P_4$ in this order. Let $\phi $ be the homomorphism $\phi : R_{m,n} \rightarrow M_{m,n}$ induced by the identity map on $\{a,b\}$ . Then

$$\begin{align*}\phi(T_4) = \phi(\operatorname{Mon}\big\langle X \big\rangle) = \operatorname{Mon}\big\langle \phi(X) \big\rangle = \operatorname{Mon}\big\langle \alpha,\beta,\gamma,\mu \big\rangle \leq M_{m,n} \end{align*}$$

is isomorphic to the trace monoid $T(P_4)$ , where $T_4 = \operatorname {Mon}\big \langle X \big \rangle = \operatorname {Mon}\big \langle \alpha ,\beta ,\gamma ,\mu \big \rangle \leq R_{m,n}$ . So $\phi $ induces a surjective homomorphism from $T_4$ onto $\phi (T_4) \leq M_{m,n}$ . To show that this defines an isomorphism between $T_4$ and $\phi (T_4)$ , we need to show that $\phi $ is injective on the set $T_4$ which, since $\phi (T_4)$ is isomorphic to $T(P_4)$ , means we need to show that the defining relations of the trace monoid hold between the generators of $T_4$ in the monoid $R_{m,n}$ . Hence, to complete the proof, we just need to show that $\alpha \beta = \beta \alpha $ , $\beta \gamma = \gamma \beta $ and $\gamma \mu = \mu \gamma $ all hold in $R_{m,n}$ .

Using the rewriting rule (ii) for $i = m$ from Lemma 5.4, we obtain

(6) $$ \begin{align} (ba^n)^m a^n a = a^n (a^n b)^{m} a. \end{align} $$

Multiplying Equation (6) by $a^{n-1}$ on the right, we obtain $(ba^n)^m a^{2n} = a^{2n} (ba^n)^{m}$ ; that is, $\gamma \beta = \beta \gamma $ in $R_{m,n}$ . Multiplying Equation (6) by a on the left and $a^{n-2}$ on the right (note that $n \geq 2$ ), and writing $m = (m-1) + 1$ , we obtain

$$\begin{align*}[a(ba^n)^{m-1} ba^{n-1}]a^{2n}= a^{2n} [a(ba^n)^{m} ba^{n-1}], \end{align*}$$

which means that $\alpha \beta = \beta \alpha $ in $R_{m,n}$ . Lastly, note that $\gamma = (ba^n)^m$ commutes with all the words $ba^{n}$ , $a^{2n}$ , and $(ba^n)^{m-1}$ . So we obtain

$$\begin{align*}\mu \gamma = [ba^n a^{2n} (ba^n)^{m-1}] \gamma = \gamma [ba^n a^{2n} (ba^n)^{m-1}] = \gamma \mu, \end{align*}$$

as required.

Proof of Theorem 6.1

Let $m,n \geq 2$ , and set $R_{m,n}= \operatorname {Mon}\big \langle a, b\:|\:(b a^n)^m(a^n b)^ma = a \big \rangle $ . By Lemma 6.6, the submonoid $T_4 = \operatorname {Mon}\big \langle X \big \rangle $ of $R_{m,n}$ generated by

$$\begin{align*}X = \{ a(ba^n)^{m-1} b a^{n-1}, \; a^{2n}, \; (ba^n)^m, \; ba^n a^{2n} (ba^n)^{m-1} \} \end{align*}$$

is isomorphic to the trace monoid $T(P_4)$ , and since every word in X belongs to $\{a,ba\}^+$ , it follows from Lemma 6.5 that all nonempty products of elements of X are $\mathcal {L}$ -related to each other, since they are all $\mathcal {L}$ -related to a. Since M is left cancellative by Adian [Reference Adian1, Theorem 3], the result now follows by applying Theorem 6.2.

Proof The language $\overline {L}K \subseteq A^*$ is clearly rational as both $\overline {L}$ and K are.

Let $i \in \mathbb {N}$ . Since M is left cancellative, it follows that for any $(l,k) \in L \times K$ , we have $lw^i = k$ in M if and only if $\overline {l}lw^i = \overline {l}k$ . Applying condition (i) gives $\overline {l}lw^i = \overline {l}lww^{i-1} = w^i$ ; hence, $lw^i = k$ in M if and only if $w^i = \overline {l}k$ .

It follows that for a given $i \in \mathbb {N}$ , there exists $(l,k) \in L \times K$ satisfying $lw^i = k$ if and only if there exists $(l,k) \in L \times K$ satisfying $w^i = \overline {l}k$ which is true if and only if $w^i \in \overline {L} K$ , since $\alpha $ being surjective implies that $\overline {L} = \{\overline {l} : l \in L \}$ . But the former problem is undecidable by assumption (ii), and hence, the latter problem must also be undecidable (i.e., membership in the rational subset $\overline {L}K$ of M is undecidable).

Proof In the paper [Reference Lohrey and Steinberg33, Proof of Theorem 2], the authors take $\Sigma = \{a,b,c,d\}$ and work in the trace monoid $S = \operatorname {Mon}\big \langle \Sigma \:|\:ab=ba, bc=cb, cd=dc \big \rangle $ . They show that there is a fixed rational language $L \subseteq \Sigma ^*$ and a family of languages $K_{m,n}$ with $m, n \in \mathbb {N}$ such that it is undecidable whether $K_{m,n} \cap L \neq \varnothing $ for given $m, n \in \mathbb {N}$ . For our purposes, the important thing is that $K_{m,n} = b^{2^m3^n} \Omega $ , where $\Omega \subseteq \Sigma ^*$ is a certain fixed rational language with the property that $\Omega $ does not contain the empty word.Footnote ² Their argument shows that it is undecidable whether $b^{2^m3^n} \Omega \cap L \neq \varnothing $ for given $m, n \in \mathbb {N}$ . It follows from the defining relations in the trace monoid S that for any two words $u, v \in \Sigma ^*$ , we have $u=v$ in S if and only if $\mathrm {rev}(u) = \mathrm {rev}(v)$ in S where $\mathrm {rev}$ denotes the reverse of a word. Hence, if we set $\Omega ' = \{ \mathrm {rev}(u) : u \in \Omega \}$ and $L' = \{ \mathrm {rev}(l) : l \in L \}$ , then $\Omega '$ and $L'$ are rational subsets of $\Sigma ^*$ , since word reversal clearly preserves the property of being a regular language. Furthermore, $\Omega '$ and $L'$ have the property that it is undecidable whether $\Omega ' b^{2^m3^n} \cap L' \neq \varnothing $ for given $m, n \in \mathbb {N}$ . In particular, this means that there are fixed rational subsets $\Omega ', L' \subseteq \Sigma ^*$ such that it is undecidable whether $\Omega ' b^{i} \cap L' \neq \varnothing $ for given $i \in \mathbb {N}$ . Finally, if we translate this into the notation of the trace monoid as defined in the statement of the lemma by substituting $a \mapsto u_1$ , $b \mapsto u_2$ , $c \mapsto u_3$ , and $d \mapsto u_4$ , we obtain the result, where Q does not contain the empty word since $\Omega $ does not.

Proof Let $L \subseteq A^*$ be a regular language, and suppose that every word in L has length at least two. Let $\sigma :A^* \rightarrow A^*$ be the homomorphism defined by $a_i \mapsto a_i^2$ and set $L_1 = \sigma (L)$ . Since the class of regular languages is closed under taking homomorphisms, it follows that $L_1$ is a regular language. Since every word in L has length at least two, it follows that every word in $L_1$ has length at least four. Now let $L_2$ be the language obtained by taking each word from $L_1$ and deleting the first letter and the last letter. Note that every word in $L_2$ has even length and has length at least two. Since the left or right quotient of a regular language is again regular, and deletion of the last resp. first letter is an example of taking a left resp. right quotient of a language by a regular language, it follows that $L_2$ is a regular language. Let $g:L_1 \rightarrow L_2$ be the map that deletes the first and last letter of the input word, and $L_3$ the reversal of the language $L_2$ (i.e., the language obtained by reversing every word).

Now define a new alphabet $C = \{ c_{i,j} : 1 \leq i,j \leq n \}$ and a homomorphism $\theta :C^* \rightarrow A^*$ defined by $c_{i,j} \mapsto a_i a_j$ . Note that the homomorphism $\theta $ is clearly injective with image the set of all word in $A^*$ of even length. Since regularity is preserved under taking inverse images of homomorphisms, it follows that for any regular language $W \subseteq A^*$ , the language $\theta ^{-1}(W) \subseteq C^*$ is regular. Let $\rho :C^* \rightarrow C^*$ be the word reversing map which also preserves regularity.

Now, given any word $w = a_{i_1} a_{i_2} \ldots a_{i_{k-1}} a_{i_k}\in A^*$ of length at least two, we have

$$\begin{align*}g(\sigma(w)) = a_{i_1} a_{i_2} a_{i_2} a_{i_3} a_{i_3} \ldots a_{i_{k-2}} a_{i_{k-2}} a_{i_{k-1}} a_{i_{k-1}} a_{i_k} \end{align*}$$

is a nonempty word of even length and then

$$\begin{align*}(\rho \circ \theta^{-1} \circ g \circ \sigma)(w) = c_{i_{k-1},i_k} c_{i_{k-2},i_{k-1}} \ldots c_{i_2,i_3} c_{i_1,i_2}. \end{align*}$$

Finally, let $\gamma :C^* \rightarrow A^*$ be the homomorphism defined by $c_{i,j} \mapsto b_{i,j}$ , so that

$$\begin{align*}(\gamma \circ \rho \circ \theta^{-1} \circ g \circ \sigma) (a_{i_1} a_{i_2} \ldots a_{i_{k-1}} a_{i_k}) = b_{i_{k-1},i_k} b_{i_{k-2},i_{k-1}} \ldots b_{i_2,i_3} b_{i_1,i_2}. \end{align*}$$

It follows that the mapping $\phi $ in the statement of the lemma satisfies $\phi = \gamma \circ \rho \circ \theta ^{-1} \circ g \circ \sigma $ , and since as explained above, all of the maps in this composition preserve regularity of follows that $\phi (L)$ is a regular language.

Proof of Theorem 6.2

Let M be a finitely generated left-cancellative monoid and let $U \subseteq M$ such that $u v \mathcal {L} v$ for all $u, v \in U$ , and $\operatorname {Mon}\big \langle U \big \rangle $ is isomorphic to the trace monoid $T(P_4)$ . Since every generating set of $T(P_4)$ must contain the four standard generators of the monoid that correspond to the four vertices of the path $P_4$ (this follows from the fact that the relations in the presentation of $T(P_4)$ are length preserving so one cannot recover the letters as products of words of greater length), it follows that U contains a subset $Y = \{u_1, u_2, u_3, u_4\}$ corresponding to the standard generators of the trace monoid. Since the rational subsets of a monoid do not depend on the choice of finite generating sets, without loss of generality, we can take the generating set A for M in the statement of the theorem to be a finite generating set for M so that it contains the subset $Y \subseteq A$ . So $Y \subseteq A$ with $Y = \{u_1, u_2, u_3, u_4\}$ , where $T = \operatorname {Mon}\big \langle Y \big \rangle $ is isomorphic to the trace monoid $T(P_4)$ with the generators $u_1, u_2, u_3, u_4$ corresponding to the vertices in the path $P_4$ with $u_i$ adjacent to $u_{i+1}$ for all i, and by the assumptions in the statement of the theorem $u_i u_j \mathcal {L} u_j$ for all $u_i, u_j \in Y$ .

Next, for every pair $i, j \in \{1,2,3,4\}$ , we fix a word $v_{i,j} \in A^*$ such that $v_{i,j}u_iu_j = u_j$ . This is possible since $u_iu_j \mathcal {L} u_j$ . Then for any nonempty word $w = u_{i_1}u_{i_2} \ldots u_{i_n} \in Y^*$ , we define

$$\begin{align*}\overline{w} = v_{i_n,2} v_{i_{n-1},i_{n}} v_{i_{n-2},i_{n-1}} \ldots v_{i_3,i_4} v_{i_2,i_3} v_{i_1,i_2} \end{align*}$$

and observe that

$$\begin{align*}\overline{w} w u_2 = v_{i_n,2} v_{i_{n-1},i_{n}} v_{i_{n-2},i_{n-1}} \ldots v_{i_3,i_4} v_{i_2,i_3} v_{i_1,i_2} u_{i_1} u_{i_2} \ldots u_{i_n} u_2 = v_{i_n,2} u_{i_n} u_2 = u_2 \end{align*}$$

in M.

By Theorem 6.8, there are two fixed rational subsets $Q, R \subseteq Y^*$ , with Q not containing the empty word, such that it is undecidable whether there exists a pair $(x,y) \in Q \times R$ satisfying $x(u_2)^i = y$ in M, for a given $i \in \mathbb {N}$ . Since $Y \subseteq A$ , it follows that $Q, R$ are also rational subsets of $A^*$ satisfying this property.

Set $\overline {Q} = \{ \overline {w} : w \in Q \}$ . We claim that $\overline {Q}$ is a rational subset of $A^*$ . To see this, note that Q is a rational subset of $A^*$ not containing the empty word, from which it follows that $Qu_2$ is a rational subset of $A^*$ in which every word has length at least two. For every word $w = u_{i_1} u_{i_2} \ldots u_{i_{k-1}} u_{i_k}\in A^*$ of length at least two, define

$$\begin{align*}\phi(u_{i_1} u_{i_2} \ldots u_{i_{k-1}} u_{i_k}) = v_{i_{k-1},i_k} v_{i_{k-2},i_{k-1}} \ldots v_{i_2,i_3} v_{i_1,i_2}. \end{align*}$$

Since $Qu_2$ is a rational subset of $A^*$ , it follows from Lemma 6.9 that $\phi (Qu_2)$ is a rational subset of $A^*$ . But $\overline {Q} = \phi (Qu_2)$ , and hence, $\overline {Q}$ is a rational subset of $A^*$ , completing the proof of the claim.

Hence, we have a left-cancellative monoid M with finite generating set A, rational languages $Q, R, \overline {Q} \subseteq A^*$ , a surjective map $\alpha : Q \longrightarrow \overline {Q}$ with $\alpha (x) = \overline {x}$ , and a fixed word $u_2 \in A^*$ , satisfying the following properties:

1. (i) $\overline {x} x u_2 = u_2$ in M, for all $x \in Q$ ,

2. (ii) it is undecidable whether there exists a pair $(x,y) \in Q \times R$ satisfying $x(u_2)^i = y$ in M, for a given $i \in \mathbb {N}$ .

Then by Theorem 6.7, the rational language $\overline {Q}R \subseteq A^*$ defines a fixed rational subset of M in which membership is undecidable.

Question 6.10 If $m=1$ or $n=1$ , then does $R_{m,n}=\operatorname {Mon}\big \langle a, b\:|\:(b a^n)^m(a^n b)^ma = a \big \rangle $ have decidable rational subset membership problem?

We know from above that the group with the same presentation does have decidable rational subset membership problem when $m=1$ or $n=1$ .

Proof Let M be the inverse monoid defined by the two-relator positive presentation

$$\begin{align*}M = M_{G,X} = \operatorname{Inv}\big\langle B,x,t\:|\: r = 1, t z_1 s t\overline{z_1} s \ldots s t z_k s t\overline{z_k} s = 1 \big\rangle \end{align*}$$

given by Construction 4.2. We claim that if M has decidable word problem, then the membership problem for Q within G is decidable. Furthermore we shall show that M is E-unitary and has maximal group image isomorphic to $G \ast \mathbb {Z}$ .

To establish this, we will perform a series of Tietze transformations on the presentation. This will come in two stages. First, we will apply a set of Tietze transformations to prove that $M \cong T$ where

$$\begin{align*}T = \operatorname{Inv}\big\langle B,x,t\:|\: r= 1, (t z_1 t^{-1}) (t z_1^{-1} t^{-1}) = 1, \ldots, (t z_k t^{-1}) (t z_k^{-1} t^{-1}) = 1 \big\rangle. \end{align*}$$

Since $\operatorname {Mon}\big \langle A\:|\:q=1 \big \rangle $ is a group, it follows that every letter from a is invertible in this monoid. From this, it follows that in the monoid $\operatorname {Mon}\big \langle B,x,t\:|\:r=1 \big \rangle $ , where r is the word defined above obtained from q by replacing a by $tx$ everywhere it appears, the element $tx$ is invertible. Indeed, the map $a \mapsto tx$ and identity on every other letter defines a homomorphism from $\operatorname {Mon}\big \langle A\:|\:q=1 \big \rangle $ to $\operatorname {Mon}\big \langle B,x,t\:|\:r=1 \big \rangle $ which must map units to units, and since a is a unit in the former monoid, $tx$ is a unit in the latter monoid.

Since $tx$ is a unit in $\operatorname {Mon}\big \langle B,x,t\:|\:r=1 \big \rangle $ , it follows that $tx$ is a unit in the inverse monoid M, since M has $r=1$ as a defining relation. The fact that $tx$ is invertible in M implies $(tx)^{-1} = x^{-1} t^{-1}$ is also invertible in the inverse monoid M. The first letter of s is x, so $t z_1 x$ is right-invertible in M since it is a prefix of the second defining relator; hence, the product $t z_1 xx^{-1} t^{-1} $ of two right-invertible elements is right-invertible, and hence, by Lemma 7.3, it is equal in M to $t z_1 t^{-1} $ . We conclude that $t z_1 t^{-1} $ is right-invertible in M.

Now $w_1 \overline {w_1} \kern1.3pt{=}\kern1.3pt \overline {w_1} w_1 \kern1.3pt{=}\kern1.3pt 1$ in $\operatorname {Mon}\big \langle A\:|\:q\kern1.3pt{=}\kern1.3pt1 \big \rangle $ implies $z_1 \overline {z_1} \kern1.3pt{=}\kern1.3pt \overline {z_1} z_1 \kern1.3pt{=}\kern1.3pt 1$ in ${\operatorname {Mon}\big \langle B,x,t\:|\:r\kern1.3pt{=}\kern1.3pt1 \big \rangle }$ since the map between these monoids given by $a \mapsto xt$ , and identity on all other generators, defines a homomorphism with $z_1$ the image of $w_1$ under this homomorphism, and $\overline {z_1}$ is the image of $\overline {w_1}$ . Since $r=1$ in M, it then follows that $z_1 \overline {z_1} = \overline {z_1} z_1 = 1$ in M; hence $z_1^{-1} = \overline {z_1}$ in M. Since $t z_1 t^{-1} $ is right-invertible in M, we have $ t z_1 t^{-1} t z_1^{-1} t^{-1} = 1 $ in M. Then since $z_1^{-1} = \overline {z_1}$ in M, and $r=1$ in M, it follows that $ t z_1 t^{-1} r t \overline {z_1} t^{-1} r = 1 $ in M. Since this word equals $1$ , it is right-invertible, and hence, by Lemma 7.3, it is equal in M to the word obtained by canceling the $t^{-1}$ with the first letters of r. It follows that

$$\begin{align*}t z_1 s t\overline{z_1} s = t z_1 t^{-1} r t \overline{z_1} t^{-1} r = 1 \end{align*}$$

in M. Hence,

$$\begin{align*}t z_2 s t\overline{z_2} s \ldots s t z_k s t\overline{z_k} s = t z_1 s t\overline{z_1} s \ldots s t z_k s t\overline{z_k} s = 1 \end{align*}$$

in M. It follows that $t z_2 x$ is right-invertible since s begins with x, and we can repeat the argument above to deduce that $t z_2 t^{-1} $ is right-invertible in M and

$$\begin{align*}t z_2 s t\overline{z_2} s = t z_2 t^{-1} r t \overline{z_2} t^{-1} r = 1 \end{align*}$$

in M. Repeating this for all j, we can prove that $t z_j t^{-1} $ is right-invertible in M, and we have

$$\begin{align*}t z_j s t\overline{z_j} s = t z_j t^{-1} r t \overline{z_j} t^{-1} r = 1 \end{align*}$$

in M. In particular, since $t z_j t^{-1} $ is right-invertible in M, this means that the defining relations $ (t z_j t^{-1}) (t z_j^{-1} t^{-1}) = 1 $ from the presentation for T all hold in M for $1 \leq j \leq k$ .

Conversely, since $z_j \overline {z_j} = \overline {z_j} z_j = 1$ in $\operatorname {Mon}\big \langle B,x,t\:|\:r=1 \big \rangle $ , and $r=1$ in T, it follows that $z_j^{-1} = \overline {z_j}$ in T for all $1 \leq j \leq k$ . It follows that

$$\begin{align*}(t z_1 t^{-1}) (t\overline{z_1} t^{-1}) \ldots (t z_k t^{-1}) (t\overline{z_k} t^{-1}) = (t z_1 t^{-1}) (tz_1^{-1} t^{-1}) \ldots (t z_k t^{-1}) (tz_k^{-1} t^{-1}) = 1 \end{align*}$$

in T which, since $r=1$ , implies that

$$\begin{align*}(t z_1 t^{-1}) r (t\overline{z_1} t^{-1}) r \ldots r (t z_k t^{-1}) r (t\overline{z_k} t^{-1}) r = 1 \end{align*}$$

holds in T. Since the word equals one and so right is invertible by Lemma 7.3, we can reduce canceling the $t^{-1} t$ in each subword $t^{-1}r$ and deduce that the following relation holds in T:

$$\begin{align*}t z_1 s t\overline{z_1} s \ldots s t z_k s t\overline{z_k} s = 1. \end{align*}$$

We have proved the second defining relator of M holds in T, and the second defining relator of T holds in M. Since the first defining relators are the same, and the generating sets are the same, we conclude that $M \cong T$ . In fact, we have proved that these are equivalent presentations in the sense that the identity map on $B \cup \{x,t\}$ induces an isomorphism between them.

To complete the proof, we now prove a second sequence of Tietze transformations to the presentation for T to obtain a presentation to which the main construction from [Reference Gray14] can then be applied to finish the proof. To do this, we first we add a redundant generator a and relation $a=tx$ to obtain

$$\begin{align*}\operatorname{Inv}\big\langle B,a,x,t\:|\: r= 1, (t z_1 t^{-1}) (t z_1^{-1} t^{-1}) = 1, \; \ldots, \; (t z_k t^{-1}) (t z_k^{-1} t^{-1}) = 1, \; a=tx \big\rangle. \end{align*}$$

Next, in every $z_i$ and in r, we can replace every occurrence of $xt$ by the letter a obtaining, since $A = B \cup \{a\}$ , the presentation

$$\begin{align*}\operatorname{Inv}\big\langle A,x,t\:|\: q= 1, \; (t w_1 t^{-1}) (t w_1^{-1} t^{-1}) = 1, \; \ldots, \; (t w_k t^{-1}) (t w_k^{-1} t^{-1}) = 1, \; a=tx \big\rangle. \end{align*}$$

By assumption, the relations $aa^{-1}=1$ and $a^{-1}a=1$ are both consequences of $q=1$ ; hence, from $a=tx$ , we can deduce that $1 = a^{-1}a = a^{-1}tx$ in this inverse monoid. Hence, we can deduce $x = t^{-1}a$ in this inverse monoid, giving the following presentation:

$$\begin{align*}\big\langle A,x,t \:|\: q= 1, (t w_1 t^{-1}) (t w_1^{-1} t^{-1}) = 1, \ldots, (t w_k t^{-1}) (t w_k^{-1} t^{-1}) = 1,a=tx, x = t^{-1}a \big\rangle. \end{align*}$$

Now the relation $(t w_1 t^{-1}) (t w_1^{-1} t^{-1}) = 1$ implies that $tt^{-1}=1$ in this inverse monoid, and so from $x = t^{-1}a$ , we can deduce the relation $tx = tt^{-1}a = a$ . Since this relation is a consequence of the others, we can now remove it obtaining the presentation

$$\begin{align*}\operatorname{Inv}\big\langle A,x,t\:|\: q= 1, \; (t w_1 t^{-1}) (t w_1^{-1} t^{-1}) = 1, \; \ldots, (t w_k t^{-1}) (t w_k^{-1} t^{-1}) = 1, \; x = t^{-1}a \big\rangle \end{align*}$$

for the same monoid. Now x is a redundant generator which can be removed. This has no impact on the other relations since neither x nor $x^{-1}$ appears in any of those words. Hence, we arrive at the following presentation

$$\begin{align*}\operatorname{Inv}\big\langle A,t\:|\: q= 1, (t w_1 t^{-1}) (t w_1^{-1} t^{-1}) = 1, \ldots, (t w_k t^{-1}) (t w_k^{-1} t^{-1}) = 1 \big\rangle \end{align*}$$

for T. Finally, by assumption, the relations $cc^{-1}=1$ and $c^{-1}c=1$ for $c \in A$ are all consequences of $q=1$ , so we can add them to obtain

$$\begin{align*}&\big\langle A,t \:|\: q= 1, \; cc^{-1}=1, \ c^{-1}c=1 \; (c \in A), (t w_1 t^{-1}) (t w_1^{-1} t^{-1}) = 1, \\&\qquad \ldots, (t w_k t^{-1}) (t w_k^{-1} t^{-1}) = 1 \big\rangle. \end{align*}$$

Now it follows from [Reference Gray14, Theorem 3.8] that $M \cong T$ is an E-unitary inverse monoid, and if $M \cong T$ has decidable word problem, then the membership problem for $Q = \operatorname {Mon}\big \langle w_1, \ldots , w_k \big \rangle \leq G$ within G is decidable. The fact that [Reference Gray14, Theorem 3.8] applies to the presentation above follows from [Reference Gray14, Lemma 3.3] (see line three of the proof of Theorem 3.8 in [Reference Gray14]). The maximal group image of T is isomorphic to

$$\begin{align*}&\big\langle A,t \:|\: q= 1, \; cc^{-1}=1, \ c^{-1}c=1 \; (c \in A), (t w_1 t^{-1}) (t w_1^{-1} t^{-1}) = 1, \\&\qquad\ldots, (t w_k t^{-1}) (t w_k^{-1} t^{-1}) = 1 \big\rangle, \end{align*}$$

which is isomorphic to $\operatorname {Gp}\big \langle A\:|\:q=1 \big \rangle \ast FG(t) \cong G \ast \mathbb {Z}$ , completing the proof of the theorem.

Proof Let K be the group defined by the two-relator positive presentation

$$\begin{align*}K = H_{G,X} = \operatorname{Gp}\big\langle B,x,t\:|\: r = 1, t z_1 s t\overline{z_1} s \ldots s t z_k s t\overline{z_k} s = 1 \big\rangle \end{align*}$$

given by Construction 4.2. We claim if the prefix membership problem for the positive two-relator group K is decidable, then the membership problem for Q in G is decidable. To see this, note that it is shown in the proof of Theorem 7.4 that

$$\begin{align*}M = \operatorname{Inv}\big\langle B,x,t\:|\: r = 1, t z_1 s t\overline{z_1} s \ldots s t z_k s t\overline{z_k} s = 1 \big\rangle \end{align*}$$

is E-unitary, and if M has decidable word problem, then the membership problem for Q in G is decidable. It is also shown there that the inverse monoid M has maximal group image isomorphic to $G \ast \mathbb {Z}$ which is a one-relator group and thus has decidable word problem by Magnus’ Theorem. It then follows from [Reference Ivanov, Margolis and Meakin22, Theorem 3.3] that M has decidable word problem. But in Theorem 7.4, it is proved that decidability of the word problem for M implies decidability of the membership problem for Q in G; hence, the membership problem for Q in G is decidable, as claimed.

In the proof of Theorem 7.4, it is shown that the identity map on $B\cup \{x,t\}$ induces an isomorphism between the inverse monoids

$$\begin{align*}M = \operatorname{Inv}\big\langle B,x,t\:|\: r = 1, t z_1 s t\overline{z_1} s \ldots s t z_k s t\overline{z_k} s = 1 \big\rangle \end{align*}$$

and

$$\begin{align*}T = \operatorname{Inv}\big\langle B,x,t\:|\: r= 1, (t z_1 t^{-1}) (t z_1^{-1} t^{-1}) = 1, \ldots, (t z_k t^{-1}) (t z_k^{-1} t^{-1}) = 1 \big\rangle. \end{align*}$$

It follows that the identity map induces an isomorphism between the maximal group images of these inverse monoids

$$\begin{align*}\operatorname{Gp}\big\langle B,x,t\:|\: r = 1, t z_1 s t\overline{z_1} s \ldots s t z_k s t\overline{z_k} s = 1 \big\rangle \rightarrow \operatorname{Gp}\big\langle B,x,t\:|\:r=1 \big\rangle. \end{align*}$$

(In general, if the identity map induces an isomorphism $\operatorname {Inv}\big \langle A\:|\:R \big \rangle \rightarrow \operatorname {Inv}\big \langle A\:|\:S \big \rangle $ , then for any additional set of relations T, the identity map will also induce an isomorphism between $\operatorname {Inv}\big \langle A\:|\:R \cup T \big \rangle \rightarrow \operatorname {Inv}\big \langle A\:|\:S \cup T \big \rangle $ since S and R are consequence of each other, and hence, the same is true of $S \cup T$ and $R \cup T$ . And so in particular, it is true when T is the set of relations $aa^{-1}= 1 = aa^{-1}$ added to get the maximal group image.) Now set $u \equiv r$ and $v \equiv t z_1 s t\overline {z_1} s \ldots s t z_k s t\overline {z_k} s $ and $Y = B \cup \{x,t\}$ . Then the identity map on Y induces isomorphisms

$$\begin{align*}\operatorname{Gp}\big\langle Y\:|\:u=1,v=1 \big\rangle \rightarrow \operatorname{Gp}\big\langle Y\:|\:u=1 \big\rangle. \end{align*}$$

It follows from Theorem 7.4 that the maximal group image of M is isomorphic to $G \ast \mathbb {Z}$ . But the group $\operatorname {Gp}\big \langle Y\:|\:vuv^{-1}=1 \big \rangle $ we have constructed is isomorphic to the maximal group image $\operatorname {Gp}\big \langle Y\:|\:u=1,v=1 \big \rangle $ of M, so this completes the proof of the theorem.

Proof This follows from Theorem 7.5 together with Theorem 3.8 that shows there are positive one-relator groups containing finitely generated submonoids in which membership is undecidable.