A Magnus theorem for some amalgamated products

TL;DR

This paper proves the Magnus property for certain amalgamated products, including the fundamental group of a non-orientable surface of genus 3, extending previous results for higher genus surfaces.

Contribution

It establishes the Magnus property for specific amalgamated products, notably including genus 3 non-orientable surface groups, answering an open question in the field.

Findings

Magnus property holds for some amalgamated products

Includes fundamental groups of non-orientable surfaces of genus 3

Extends results previously known for genus greater than 3

Abstract

A group $G$ possesses the Magnus property if for every two elements $u,v \in G$ with the same normal closure, $u$ is conjugate in $G$ to $v$ or $v^{-1}$. We prove the Magnus property for some amalgamated products including the fundamental group of a closed non-orientable surface of genus 3. This answers a question of O. Bogopolski and K. Sviridov, who obtained the analogous result for genus $g > 3$.

Equations

$$N=⟨\bigcup _{i\in \mathbb{Z}}{Y}_i|b_i{u}_i=b_{i+k}(i\in \mathbb{Z})⟩.$$

$$N_l=\dots \underset{Z_{l-k}}{\ast }{G}_{l-k}\underset{Z_l}{\ast }{G}_l\underset{Z_{l+k}}{\ast }{G}_{l+k}\underset{Z_{l+2k}}{\ast }\dots$$

$$\mathcal{B}(i)=\{b_i,b_{i+1},\dots ,b_{i+k-1}\}\cup \{y_{m,j}\mid 1⩽m⩽n,j\in \mathbb{Z}\}.$$

$$B_l:=\{b_l\}\cup \{y_{m,l+tk}|1⩽m⩽n,t\in \mathbb{Z}\}.$$

$$\begin{array}[]{llllllll}\emph{b-left basis}&\mathcal{B}^{+}(i)&:=&\{b_{i},b_{i+1},\dots,b_{i+k-1}\}&\cup&\{y_{m,j}\mid 1\leqslant m\leqslant n,j\geqslant i\}&\text{of}&G_{i,\infty}\ \ \text{and the}\\ \emph{b-right basis}&\mathcal{B}^{-}(i)&:=&\{b_{i-k+1},b_{i-k+2},\dots,b_{i}\}&\cup&\{y_{m,j}\mid 1\leqslant m\leqslant n,j\leqslant i\}&\text{of}&G_{-\infty,i}.\\ \end{array}$$

$$\{b_{i+1},b_{i+2},\dots ,b_{i+k}\}\cup \{y_{m,l}\mid 1⩽m⩽n,i⩽l\}.$$

$$G_{i,\infty }=G_{i+1,\infty }\ast ⟨y_{m,i}\mid 1⩽m⩽n⟩.$$

$$\mathcal{B}(i{)}^{\prime }=\{b_i^{\prime },b_{i+1}^{\prime },\dots ,b_{i+k-1}^{\prime }\}\cup \{y_{m,j}^{\prime }\mid 1⩽m⩽n,j\in \mathbb{Z}\}.$$

$$\begin{array}[]{llllllll}\emph{$b^{\prime}$-left basis}&\mathcal{B}^{+}(i)^{\prime}&:=&\{b_{i}^{\prime},b_{i+1}^{\prime},\dots,b_{i+k-1}^{\prime}\}&\cup&\{y_{m,j}^{\prime}\mid 1\leqslant m\leqslant n,j\geqslant i\}&\text{of}&G_{i,\infty}^{\prime}\ \text{and the}\\ \emph{$b^{\prime}$-right basis}&\mathcal{B}^{-}(i)^{\prime}&:=&\{b_{i-k+1}^{\prime},b_{i-k+2}^{\prime},\dots,b_{i}^{\prime}\}&\cup&\{y_{m,j}^{\prime}\mid 1\leqslant m\leqslant n,j\leqslant i\}&\text{of}&G_{-\infty,i}^{\prime}.\\ \end{array}$$

$$r\in ⟨b_{{\alpha }_r},b_{{\alpha }_r+1},\dots ,b_{{\alpha }_r+k-1},y_{m,{\alpha }_r},y_{m,{\alpha }_r+1},\dots ,y_{m,{\omega }_r-1}\mid 1⩽m⩽n⟩.$$

$$r\in ⟨b_{{\alpha }_r-k},\dots ,b_{{\alpha }_r-1},y_{m,{\alpha }_r-k},y_{m,{\alpha }_r-k+1},\dots ,y_{m,{\omega }_r-1}\mid 1⩽m⩽n⟩.$$

$$N/\langle\!\langle\widetilde{r}_{-1},\widetilde{r}_{0},\widetilde{r}_{1},\widetilde{r}_{2}\rangle\!\rangle\ \cong\ G_{-\infty,4}/\langle\!\langle\widetilde{r}_{-1},\widetilde{r}_{0},\widetilde{r}_{1}\rangle\!\rangle\underset{\begin{array}[]{rl}w_{1}\!\!\!\!&=\,b_{5},\vspace*{-3mm}\\ w_{2}\!\!\!\!&=\,b_{6}\vspace*{-3mm}\\ w_{3}\!\!\!\!&=\,b_{7},\vspace*{-3mm}\\ w_{4}\!\!\!\!&=\,b_{8}\end{array}}{\underset{G_{3,4}}{\ast}}G_{3,\infty}/\langle\!\langle\widetilde{r}_{2}\rangle\!\rangle.$$

$$N/\langle\!\langle\widetilde{r}_{j}\rangle\!\rangle\ \cong\ G_{-\infty,t}\underset{\begin{array}[]{ll}w_{t-k+1}\!\!&=b_{t+1},\vspace*{-3mm}\\ \dots&\vspace*{-3mm}\\ w_{t}\!\!&=b_{t+k}\\ \end{array}}{\underset{G_{s,t}}{\ast}}G_{s,\infty}/\langle\!\langle\widetilde{r}_{j}\rangle\!\rangle.$$

$$\{b_s,b_{s+1},\dots ,b_{s+k-1},y_{m,s},y_{m,s+1},\dots ,y_{m,t}\mid 1⩽m⩽n\}.$$

$$N/\langle\!\langle\widetilde{r}_{i},\widetilde{r}_{i+1}\dots,\widetilde{r}_{j+1}\rangle\!\rangle\ \cong\ G_{-\infty,t+1}/\langle\!\langle\widetilde{r}_{i},\widetilde{r}_{i+1},\dots,\widetilde{r}_{j}\rangle\!\rangle\underset{\begin{array}[]{ll}w_{t-k+2}\!\!&=b_{t+2},\vspace*{-3mm}\\ \dots&\vspace*{-3mm}\\ w_{t+1}\!\!&=b_{t+k+1}\\ \end{array}}{\underset{G_{s+1,t+1}}{\ast}}G_{s+1,\infty}/\langle\!\langle\widetilde{r}_{j+1}\rangle\!\rangle.$$

$$P\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}$$

$$N/⟨⟨{\tilde{r}}_i,{\tilde{r}}_{i+1},\dots ,{\tilde{r}}_j⟩⟩\cong G_{-\infty ,{\omega }_{{\tilde{r}}_j}-1}/⟨⟨{\tilde{r}}_i,{\tilde{r}}_{i+1},\dots ,{\tilde{r}}_{j-1}⟩⟩\underset{A}{\ast }{G}_{{\alpha }_{{\tilde{r}}_j},\infty }/⟨⟨{\tilde{r}}_j⟩⟩$$

$${\mathcal{B}}^{-}({\omega }_r)=\{b_i|i\in I\}\cup \{y_{m,\ell }\mid 1⩽m⩽n,\ell ⩽{\omega }_r\},$$

$${\mathcal{B}}^{+}({\alpha }_r)=\{b_j|j\in J\}\cup \{y_{m,\ell }\mid 1⩽m⩽n,\ell ⩾{\alpha }_r\},$$

$$H=⟨x,b,y_1,\dots ,y_n|[x^k,b]u⟩$$

$$\tilde{H}=⟨x,b⟩\cong F_2\text{and}H=\tilde{H}\underset{[x^k,b]=u^{-1}}{\ast }⟨y_m\mid 1⩽m⩽n⟩,$$

$$H/⟨⟨\tilde{r}⟩⟩$$

Full text

1 Introduction

A group $G$ possesses the Magnus property if for every two elements $u,v\in G$ with the same normal closure, $u$ is conjugate in $G$ to $v$ or $v^{-1}$. The Magnus property was named after W. Magnus who proved the so-called Freiheitssatz (see Theorem 4.1) and the Magnus property for free groups [Mag30]. Since then, many mathematicians proved or disproved the Freiheitssatz and the Magnus property for certain classes of groups (see e.g. [Bog05], [BS08], [Edj89], [How81], [How04]).

Let $\pi_{1}(S_{g}^{+})$, respectively $\pi_{1}(S_{g}^{-})$, be the fundamental group of the compact orientable, respectively non-orientable, surface of genus $g$. The Magnus property of $\pi_{1}(S_{g}^{+})$ for all $g$ was proved independently by O. Bogopolski [Bog05] (by using algebraic methods) and by J. Howie [How04] (by using topological methods). As observed in [Bog05], there is a third, model theoretic method: two groups $G_{1}$, $G_{2}$ are called elementarily equivalent if their elementary theories coincide: $Elem(G_{1})=Elem(G_{2})$, see [CG05]. It is easy to show that elementarily equivalent groups either both possess the Magnus property, or both do not possess it. Since the groups $\pi_{1}(S_{g}^{+})$ for $g\geqslant 2$ and $\pi_{1}(S_{g}^{-})$ for $g\geqslant 4$ are elementarily equivalent to the free group on two generators, all these groups possess the Magnus property.

In [BS08], O. Bogopolski and K. Sviridov proved the following theorem:

Theorem 1.1**.**

[BS08**, Main Theorem]*** Let $G=\langle a,b,y_{1},\dots,y_{e}\mid[a,b]uv\rangle$, where $e\geqslant 2$, $u,v$ are non-trivial reduced words in letters $y_{1},\dots,y_{e}$, and $u,v$ have no common letters. Let $r,s\in G$ be two elements with the same normal closures. Then $r$ is conjugate to $s$ or $s^{-1}$.*

As a corollary of that theorem (​​[BS08, Corollary 1.3]), they showed the Magnus property of $\pi_{1}(S_{g}^{-})$ for $g\geqslant 4$. Since the Magnus property trivially holds for genus $1$ and $2$, the authors asked, whether it also holds for the fundamental group of the non-orientable surface of genus 3. With our Main Theorem, that proves the Magnus property for a slightly larger subclass of one-relator groups than in Theorem 1.1, we answer this question positively. The difficulty with genus $g=3$ is essential since it is well known that the group $\pi_{1}(S_{3}^{-})=\langle x,y,z\,|\,x^{2}y^{2}z^{2}\rangle$ is not even existentially equivalent to a free group $F_{n}$ on $n$ generators. However, in large parts, our proof follows the proof of the Main Theorem in [BS08].

Main Theorem 1.2**.**

Let $G=\langle a,b,y_{1},\dots,y_{n}\mid[a,b]u\rangle$, where $n\in\mathbb{N}$ and $u$ is a non-trivial reduced word in the letters $y_{1},\dots,y_{n}$. Then $G$ possesses the Magnus property.

Using the isomorphism $\varphi$ between $\pi_{1}(S_{3}^{-})=\langle x,y,z\,|\,x^{2}y^{2}z^{2}\rangle$ and $\widetilde{H}=\langle a,b,c\mid[a,b]c^{2}\rangle$ defined by $\varphi(x)=ca^{-1},\varphi(y)=b^{-1}c^{-1}$ and $\varphi(z)=cbcac^{-1}$, we get the following corollary.

Corollary 1.3**.**

The group $\pi_{1}(S_{3}^{-})$ possesses the Magnus property.

In the proof of Theorem 1.1, the authors of [BS08] used an automorphism of $G$ with certain convenient properties. This automorphism is in general absent for the group given in our Main Theorem. For example, it is absent for the group $\pi_{1}(S_{3}^{-})$. So we introduce an additional tool which we call $\alpha$- and $\omega$-limits (see Section 3). Together with the results of [BS08, Bog05, How04] Corollary 1.3 implies the following.

Corollary 1.4**.**

The fundamental groups of all compact surfaces possess the Magnus property.

We will start the proof of our main theorem by recapitulating the notation of [BS08, Section 3] with some small alterations. In Section 3 we introduce $\alpha$- and $\omega$-limits, $\alpha$-$\omega$-length $||\cdot||_{\alpha,\omega}$ and suitable elements. In Section 4 we give the proof for the case, where the $\alpha$-$\omega$-length of suitable elements is positive (the consideration of this case is close to that in [BS08]). In Section 5 we complete the proof in the remaining case.

2 Reduction to a new group and left/right bases

We denote the normal closure of an element $g$ in a group $G$ by $\langle\!\langle g\rangle\!\rangle_{G}$ and the exponent sum of an element $g\in G$ in a letter $x$ by $g_{x}$. Note that this sum is well-defined if all relations of $G$ considered as words of a free group have exponent sum [math] in $x$. Our main theorem can be deduced from the following proposition in the same way as in the Main Theorem of [BS08] from [BS08, Proposition 2.1]. Therefore we leave this argumentation out.

Proposition 2.1**.**

Let $H=\langle x,b,y_{1},\dots,y_{n}\ |\ [x^{k},b]u\rangle$, where $k,n\geq 1$ and $u$ is a non-trivial reduced word in the letters $y_{1},\dots,y_{n}$. Further, let $r,s\in H\backslash\{1\}$ with $r_{x}=0$. Then $\langle\!\langle r\rangle\!\rangle_{H}=\langle\!\langle s\rangle\!\rangle_{H}$ implies that $r$ is conjugate to $s$ or $s^{-1}$.

We briefly summarise the concept of left and right bases from [BS08]. For a given group $G$ and an element $g\in G$ we denote by $g_{i}$ the element $x^{-i}gx^{i}$ for $i\in\mathbb{Z}$.

Let $H$ be as in Proposition 2.1. We consider the homomorphism $\varphi:H\rightarrow\mathbb{Z}$ which sends $x$ to $1$ and $b,y_{1},y_{2},\dots,y_{n}$ to [math]. For each $i\in\mathbb{Z}$ let $Y_{i}=\{b_{i},y_{1,i},y_{2,i},\dots,y_{n,i}\}$, where $y_{j,i}:=x^{-i}y_{j}x^{i}$. Using the rewriting process of Reidemeister-Schreier, we obtain that the kernel $N$ of $\varphi$ has the presentation

[TABLE]

The group $N$ is the free product of the free groups $G_{i}=\langle Y_{i}\ |\ \rangle\ (i\in\mathbb{Z})$ with amalgamation, where $G_{i}$ and $G_{i+k}$ are amalgamated over a cyclic group $Z_{i+k}$ that is generated by $b_{i}u_{i}$ in $G_{i}$ and by $b_{i+k}$ in $G_{i+k}$. This gives $N=N_{t}\ast\dots\ast N_{t+k-1}$ for all $t\in\mathbb{Z}$, where

[TABLE]

for $l\in\mathbb{Z}$.

Proposition 2.2**.**

For every $i\in\mathbb{Z}$, the group $N$ is free with basis

[TABLE]

Proof.

Since $N=N_{i}\ast\dots\ast N_{i+k-1}$, it suffices to show that each $N_{l}$ has the basis

[TABLE]

We have $N_{l}=\underset{p\in\mathbb{N}}{\bigcup}N_{l,p}$, where $N_{l,p}=G_{l-pk}\underset{Z_{l-(p-1)k}}{\ast}\dots\ \underset{Z_{l}}{\ast}G_{l}\underset{Z_{l+k}}{\ast}\dots\underset{Z_{l+pk}}{\ast}G_{l+pk}$. Using Tietze transformations, one can show that $N_{l,p}$ is free with basis $B_{l,p}:=\{b_{l}\}\cup\{y_{m,l+tk}\mid 1\leqslant m\leqslant n,-p\leqslant t\leqslant p\}$. Since $B_{l,p}\subset B_{l,p+1}$, the group $N_{l}$ is free with basis $B_{l}=\underset{p\in\mathbb{N}}{\bigcup}B_{l,p}$. ∎

Notation. Let $G_{i,j}=\langle G_{i},G_{i+1}\dots,G_{j}\rangle$, $G_{i,\infty}=\langle G_{l}\mid l\geqslant i\rangle$ and $G_{-\infty,i}=\langle G_{l}\mid l\leqslant i\rangle$ for all $i,j\in\mathbb{Z}$ with $i\leqslant j$. Further, we use the

[TABLE]

Remark 2.3**.**

Let $r\in N$ be written as a word in the letters $b_{j},y_{m,j}\ (1\leqslant m\leqslant n,j\in\mathbb{Z})$. For an arbitrary $i\in\mathbb{Z}$ we describe how to rewrite $r$ in the basis $\mathcal{B}^{+}(i)$. First, replace each letter $b_{j}$ of $r$ by $b_{j-k}u_{j-k}$, if $j>i+k-1$, and by $b_{j+k}u_{j}^{-1}$, if $j<i$, and reduce. If the resulting word contains $b_{j}$ that do not belong to $\mathcal{B}^{+}(i)$, we repeat this procedure. After finitely many steps, we will obtain the desired form of $r$. Analogously, we can rewrite $r$ in the basis $\mathcal{B}^{-}(i)$.

3 $\alpha$- and $\omega$-limits

We keep using the notation introduced in Section $2$. An arbitrary element $r\in N\setminus\{1\}$ can be written in many ways as a reduced word in letters $b_{i},y_{m,i}$. For example, $b_{0}u_{0}u_{1}^{-1}b_{1}^{-1}=b_{k}b_{k+1}^{-1}=b_{2k}u_{k}^{-1}u_{k+1}b_{2k+1}^{-1}$. We give an algorithm which finds a word $r^{*}$ representing $r$ such that the smallest index of letters used in $r^{*}$ is maximum possible. We call such index the $\alpha$-limit of $r$ and denote it by $\alpha_{r}$. In other words, $\alpha_{r}$ is the largest index such that $r$ is an element of $G_{\alpha_{r},\infty}$. The following algorithm rewrites an arbitrary word $r\in N\setminus\{1\}$ into the presentation $r^{*}$ of $r$ written in basis $\mathcal{B}^{+}(\alpha_{r})$ (see Lemma 3.2). For a word $r$ in the alphabet $\{b_{i}\,|\,i\in\mathbb{Z}\}\cup\{y_{m,i}\,|\,1\leqslant m\leqslant n,i\in\mathbb{Z}\}$, let $\min(r)$ denote the minimal index of letters of $r$.

Algorithm 3.1**.**

Let $r\in N\backslash\{1\}$. Suppose that $r$ is given as a finite word in letters $b_{i},b_{i+1},\dots$ and $y_{m,i},y_{m,i+1},\dots$ ($1\leqslant m\leqslant n$). In particular, $r\in G_{i,\infty}$. Let $r[0]$ be the reduced word representing $r$ in basis $\mathcal{B}^{+}(i)$. Increasing $i$ if necessary, we may assume that $i=\min(r[0])$.

• (1)

Let $r[1]$ be the word obtained from $r[0]$ by replacement of each occurrence of the letter $b_{i}$ by $b_{i+k}u_{i}^{-1}$ followed by free reduction. Then $r[1]$ presents $r$ in the following basis of $G_{i,\infty}$:

[TABLE]

1. (1a)

If $r[1]$ does not contain a letter $y_{m,i}$ ($1\leqslant m\leqslant n$), we reset $r[0]:=r[1]$, $i:=\min(r[1])$, and go back to $(1)$. Clearly, the new $i$ is larger than the old one. 2. (1b)

If $r[1]$ contains a letter $y_{m,i}$ ($1\leqslant m\leqslant n$), the algorithm ends with $r^{*}=r[0]$ and $\alpha_{r}=i$.

Lemma 3.2**.**

The output $r^{*}$ of Algorithm 3.1 coincides with the word representing $r$ in Basis $\mathcal{B}^{+}(\alpha_{r})$, where $\alpha_{r}$ is the $\alpha$-limit of $r$.

Proof.

We shall prove that the algorithm ends and that the integer assigned to $\alpha_{r}$ by the algorithm is really the $\alpha$-limit of $r$.

The algorithm ends since the length of $r[0]$ decreases with each iteration of (1) which does not terminate the algorithm. This can be verified in the following way: If we arrive in (1a), then the word $r[1]$ does not contain a letter $y_{m,i}$ ($1\leqslant m\leqslant n$). Recall that $r[1]$ was obtained from $r[0]$ by replacing all $b_{i}$ with $b_{i+k}u_{i}^{-1}$ and reducing the resulting word. The fact that there is no letter $y_{m,i}$ ($1\leqslant m\leqslant n$) left in $r[1]$ means that each $b_{i}$ in $r[0]$ occurred in a subword of the form $b_{i}u_{i}$. Therefore $r[1]$ is shorter than $r[0]$.

Finally, we show that $r$ is not an element of $G_{i+1,\infty}$, where $i$ as in (1b). Observe that

[TABLE]

By (1b), $r$, written in basis ($\star$ ‣ (1)), uses a letter $y_{m,i}$ ($1\leqslant m\leqslant n$). Therefore $r\notin G_{i+1,\infty}$ and $\alpha_{r}=i$. ∎

Corollary 3.3**.**

Let $r$ be an element from $N\backslash\{1\}$ given as a word in some $b$-left basis. Starting Algorithm 3.1 with $r$, we either get the same presentation or a presentation of shorter length.

Proof.

In the proof of Lemma 3.2, we already showed that the length of $r[0]$ in Algorithm 3.1 decreases with each iteration of (1) which does not terminate the algorithm. Moreover, the last iteration of Algorithm 3.1 does not change the current presentation $r[0]$. ∎

Analogously, we can find a presentation $r_{*}$ of a word $r\in N\backslash\{1\}$ such that the largest used index of letters in $r_{*}$ is minimum possible. We call this index $\omega$-limit of $r$ and denote it by $\omega_{r}$. In other words, $\omega_{r}$ is the smallest index such that $r$ is an element of $G_{-\infty,\omega_{r}}$. The algorithm to find $r_{*}$ can be received from Algorithm 3.1 by “mirroring” this algorithm using the $b$-right basis of $G_{-\infty,i}$ and replacements of $b_{i}$ by $b_{i-k}u_{i-k}$. We define the $\alpha$-$\omega$-length of $r$ by $||r||_{\alpha,\omega}:=\omega_{r}-\alpha_{r}+1$. Note that the $\alpha$-$\omega$-length of a non-trivial element can be non-positive.

Examples.

• (i)

Let $r=b_{-2}u_{-2}y_{1,0}b_{4}u_{1}^{-1}$ and $k=3$. To determine $\alpha_{r}$, we write $r=b_{-2}u_{-2}y_{1,0}b_{4}u_{1}^{-1}=b_{1}y_{1,0}b_{4}u_{1}^{-1}$ and get $\alpha_{r}=0$. For $\omega_{r}$ we have $r=b_{-2}u_{-2}y_{1,0}b_{4}u_{1}^{-1}=b_{-2}u_{-2}y_{1,0}b_{1}u_{1}u_{1}^{-1}=b_{-2}u_{-2}y_{1,0}b_{1}$ $=b_{-2}u_{-2}y_{1,0}b_{-2}u_{-2}.$ Thus, $\omega_{r}=0$ and $||r||_{\alpha,\omega}=1$. Note that $r\notin G_{\alpha_{r},\omega_{r}}=G_{0}$.

• (ii)

Let $r=b_{5}b_{6}^{-1}$ and $k=4$. Clearly, $\alpha_{r}=5$. To determine $\omega_{r}$, we write $r=b_{5}b_{6}^{-1}=b_{5}u_{2}^{-1}b_{2}^{-1}$ $=b_{1}u_{1}u_{2}^{-1}b_{2}^{-1}=b_{1}u_{1}u_{2}^{-1}u_{-2}^{-1}b_{-2}^{-1}$. Thus, $\omega_{r}=2$ and $||r||_{\alpha,\omega}=-2$.

Lemma 3.4**.**

Let $r\in N$. Then $\alpha_{r_{i+j}}=\alpha_{r_{i}}+j$ and $\omega_{r_{i+j}}=\omega_{r_{i}}+j$ for all $i,j\in\mathbb{Z}$. In particular, $||r_{i}||_{\alpha,\omega}=||r_{j}||_{\alpha,\omega}$ for all $i,j\in\mathbb{Z}$.

Proof.

Indeed, $r_{i+j}$ can be obtained from $r_{i}$ by increasing the (second) indices of all letters by $j$. ∎

Notation. Let $r\in N$. For the following two lemmata, let $r(i)$ be the presentation of $r$ written in basis $\mathcal{B}(i)$.

Lemma 3.5**.**

Let $r\in N\backslash\{1\}$. The following statements are equivalent:

• (1)

For some $i\in\mathbb{Z}$, the word $r(i)$ begins with a positive power of a $b$-letter.

• (2)

For all $i\in\mathbb{Z}$, the word $r(i)$ begins with a positive power of a $b$-letter.

Proof.

The word $r(i+1)$ can be obtained from $r(i)$, by replacement of each occurrence of $b_{i}$ in $r(i)$ by $b_{i+k}u_{i}^{-1}$ followed by free reduction. The new letter $b_{i+k}$ does not lie in $\{b_{i},\dots,b_{i+k-1}\}$. That prevents cancellation between $b$-letters in $r(i+1)$. Therefore $r(i)$ starts with a positive exponent of a $b$-letter if and only if $r(i+1)$ starts with a positive exponent of a $b$-letter. This proves the equivalence $(1)\Leftrightarrow(2)$. ∎

Corollary 3.6**.**

For every $r\in N\backslash\{1\}$ there exists a conjugate $\widetilde{r}$ of $r$ such that $\widetilde{r}(i)$ is cyclically reduced for each $i\in\mathbb{Z}$.

Proof.

Using conjugation, we may assume that $r(0)$ is cyclically reduced. If $r(0)$ contains only $y$-letters, we are done with the element $\widetilde{r}$ represented by $r(0)$. Suppose that $r(0)$ contains a $b$-letter. Let $\widetilde{r}$ be the element represented by a cyclic permutation of $r(0)$ which either starts with a positive power of a $b$-letter, or ends with a negative power of a $b$-letter (but not both). By Lemma 3.5, this $\widetilde{r}$ has the desired property. ∎

Definition 3.7**.**

Let $r\in N\backslash\{1\}$. Any element $\widetilde{r}$ as in Lemma 3.6 is called suitable conjugate for $r$.

Remark 3.8**.**

(dual structure of $N$) Denote $b^{\prime}_{i}:=b_{-i}u_{-i}$, $y_{m,i}^{\prime}:=y_{m,-i}^{-1}$ ($1\leqslant m\leqslant n$) and $G_{i}^{\prime}:=G_{-i}$. We call the elements $b_{i}^{\prime},y_{m,i}^{\prime}$ ($1\leqslant m\leqslant n$) dual to $b_{i},y_{m,i}$ ($1\leqslant m\leqslant n$) and the subgroup $G_{i}^{\prime}$ dual to $G_{i}$. Expressing $b$-letters and $y$-letters via their dual, we obtain $b_{i}=b_{-i}^{\prime}u_{-i}^{\prime}$, $y_{m,i}=y_{m,-i}^{\prime-1}$ ($1\leqslant m\leqslant n$), where $u^{\prime}_{-j}$ ($j\in\mathbb{Z}$) is the word obtained from $u_{j}$ by replacing each letter $y_{m,j}$ with $y_{m,-j}^{\prime-1}$. That justifies the terminology.

Observe that the old relations $b_{i}u_{i}=b_{i+k}$ $(i\in\mathbb{Z})$ can be rewritten in dual letters as $b_{i}^{\prime}u_{i}^{\prime}=b_{i+k}^{\prime}$ $(i\in\mathbb{Z})$. Thus, the relations preserve their form. Moreover, we have $G_{i}^{\prime}=\langle b_{i}^{\prime},y_{m,i}^{\prime}\mid 1\leqslant m\leqslant n\rangle$ that repeats the form $G_{i}=\langle b_{i},y_{m,i}\mid 1\leqslant m\leqslant n\rangle$. Other dual objects can be defined analogously: For example, $G_{i,\infty}^{\prime}:=\langle G_{i}^{\prime},G_{i+1}^{\prime}\dots\rangle=G_{-\infty,-i}$ and $G_{-\infty,i}^{\prime}:=\langle\dots,G_{i-1}^{\prime},G_{i}^{\prime}\rangle=G_{-i,\infty}$. We use the following bases of $N$  ($i\in\mathbb{Z}$):

[TABLE]

along with the

[TABLE]

For $r\in N\backslash\,\{1\}$, let $\alpha^{\prime}_{r}$ be the largest index $i$ such that $r$ is an element of $G^{\prime}_{i,\infty}$ and let $\omega^{\prime}_{r}$ be the smallest index $i$ such that $r$ is an element of $G^{\prime}_{-\infty,i}$. We denote $||r||_{\alpha^{\prime},\omega^{\prime}}:=\omega_{r}^{\prime}-\alpha_{r}^{\prime}+1$.

Lemma 3.9**.**

For any non-trivial $r\in N$ the following statements are valid.

1. (1)

We have $\alpha_{r}^{\prime}=-\omega_{r}$, $\omega_{r}^{\prime}=-\alpha_{r}$, and $||r||_{\alpha^{\prime},\omega^{\prime}}=||r||_{\alpha,\omega}$. 2. (2)

Suppose that $r$ written in a basis $\mathcal{B}(i)$ is cyclically reduced and begins with a positive power of a $b$-letter. Then $r$ written in the basis $\mathcal{B}(-i-k+1)^{\prime}$ is cyclically reduced and begins with a positive power of a $b^{\prime}$-letter.

Proof.

Both statements can be verified straightforward by using the relations $b_{i}=b_{-i}^{\prime}u_{-i}^{\prime}$, $y_{m,i}=y_{m,-i}^{\prime-1}$ ($1\leqslant m\leqslant n$). ∎

4 Proof of Proposition 2.1 for $\widetilde{r}$ with positive $\alpha$-$\omega$-length

4.1 Properties of $\widetilde{r}$ with positive $\alpha$-$\omega$-length

We use the following version of Magnus’ Freiheitssatz:

Theorem 4.1** (Magnus’ Freiheitssatz (cf. [Mag30])).**

Let $F$ be a free group on a basis $X$, and let $g$ be a cyclically reduced word in $F$ with respect to $X$, containing a letter $x\in X$. Then the subgroup generated by $X\backslash\{x\}$ is naturally embedded into the group $F/\langle\!\langle g\rangle\!\rangle_{F}$.

By abuse of notation, we write $A/\langle\!\langle a\rangle\!\rangle$ instead of $A/\langle\!\langle a\rangle\!\rangle_{A}$, where $A$ is a group and $a\in A$.

Corollary 4.2**.**

Let $\widetilde{r}\in N$ be a suitable element. Then we get the embeddings $G_{\alpha_{\widetilde{r}}+1,\infty}\hookrightarrow N/\langle\!\langle\widetilde{r}\rangle\!\rangle$ and $G_{-\infty,\omega_{\widetilde{r}}-1}\hookrightarrow N/\langle\!\langle\widetilde{r}\rangle\!\rangle$.

Proof.

By the definition of $\alpha$-limits, $\widetilde{r}$ written in basis $\mathcal{B}^{+}(\alpha_{\widetilde{r}})$ contains at least one letter in $Y_{\alpha_{\widetilde{r}}}$, and by Definition 3.7, the suitable element $\widetilde{r}$ is cyclically reduced in this basis. We extend $\mathcal{B}^{+}(\alpha_{\widetilde{r}})$ to a free basis of $N$ by adding the letters $\{y_{m,i}\mid 1\leqslant m\leqslant n,i<\alpha_{\widetilde{r}}\}$. The presentation of $\widetilde{r}$ will not change by that. Now, $G_{\alpha_{\widetilde{r}}+1,\infty}\hookrightarrow N/\langle\!\langle\widetilde{r}\rangle\!\rangle$ follows immediately from Theorem 4.1. The other embedding follows analogously. ∎

Lemma 4.3**.**

Suppose that $r\in N$ satisfies $||r||_{\alpha,\omega}\geqslant 1$. Then $r$, written in the basis $\mathcal{B}^{+}(\alpha_{r})$ of $G_{\alpha_{r},\infty}$, contains at least one letter $y_{m,l}$ with $l\geqslant\omega_{r}$.

Proof.

By assumption, $\alpha_{r}\leqslant\omega_{r}$. To the contrary, suppose that

[TABLE]

Then, using relations $b_{i}=b_{i-k}u_{i-k}$, we obtain

[TABLE]

Hence $r\in G_{-\infty,\omega_{r}-1}$. A contradiction. ∎

4.2 The structure of some quotients of $N$

This section and the next one are very similar to [BS08, Section 4 and 5], but due to some important changes we cannot skip them. Our aim in this subsection is to present $N/\langle\!\langle\widetilde{r_{i}},\widetilde{r}_{i+1},\dots,\widetilde{r}_{j}\rangle\!\rangle$ as an amalgamated product. We denote $w_{i}=b_{i}u_{i}$.

Lemma 4.4**.**

Let $\widetilde{r}\in N$ be a suitable element with $||\widetilde{r}||_{\alpha,\omega}\geqslant 1$, and let $i,j$ be two integers with $i\leqslant j$. We denote $s=\alpha_{\widetilde{r_{j}}}$ and $t=\omega_{\widetilde{r}_{j}}-1$. Then we have:

• (1)

N/\langle\!\langle\widetilde{r_{i}},\widetilde{r}_{i+1},\dots,\widetilde{r}_{j}\rangle\!\rangle\ \cong\ G_{-\infty,t}/\langle\!\langle\widetilde{r}_{i},\widetilde{r}_{i+1},\dots,\widetilde{r}_{j-1}\rangle\!\rangle\underset{\begin{array}[]{ll}w_{t-k+1}\!\!&=b_{t+1},\vspace*{-3mm}\\ \dots&\vspace*{-3mm}\\ w_{t}\!\!&=b_{t+k}\\ \end{array}}{\underset{G_{s,t}}{\ast}}G_{s,\infty}/\langle\!\langle\widetilde{r}_{j}\rangle\!\rangle.

• (2)

$G_{s+1,\infty}$* naturally embeds into $N/\langle\!\langle\widetilde{r}_{i},\widetilde{r}_{i+1},\dots,\widetilde{r}_{j}\rangle\!\rangle$.*

Before we give a formal proof, we consider an illustrated example. This will help to visualise a lot of technical details in the formulation of lemma.

Example. Let $k=4$. We consider the element $\widetilde{r}=\widetilde{r}_{0}=b_{4}y_{2,1}y_{1,3}b_{0}u_{0}$ and the integers $i=-1$ and $j=2$.

Algorithm 3.1 applied to $\widetilde{r}$ gives $\alpha_{\widetilde{r}_{0}}=1$. Its “mirrored” version gives $\omega_{\widetilde{r}_{0}}=3$. In particular, $||\widetilde{r}||_{\alpha,\omega}=3$. Furthermore, we have $s=\alpha_{\widetilde{r}_{2}}=3$ and $t=\omega_{\widetilde{r}_{2}}-1=4$. Then Lemma 4.4 (1) says that

[TABLE]

In Figure 1, the word $\widetilde{r}_{0}=\widetilde{r}$ is pictured by the (partially dashed) line crossing the blocks $G_{0},\dots,G_{4}$ since $\widetilde{r}$ uses letters with indices from the segment $[0,4]$. We can represent $\widetilde{r}$ by the word $b_{4}y_{2,1}y_{1,3}b_{4}$ that uses letters with indices from the segment $[1,4]$, and we can represent $\widetilde{r}$ by the word $b_{0}u_{0}y_{2,1}y_{1,3}b_{0}u_{0}$ that uses letters with indices from the segment $[0,3]$. To visualise the fact that $\alpha_{\widetilde{r}_{0}}=1$ and $\omega_{\widetilde{r}_{0}}=3$, we draw a continuous line crossing the blocks $G_{1},G_{2},G_{3}$.

Proof.

First, we prove that (1) implies (2). By Corollary 4.2, we have $G_{s+1,\infty}\hookrightarrow G_{s,\infty}/\langle\!\langle\widetilde{r}_{j}\rangle\!\rangle$, and if (1) holds, then we have $G_{s,\infty}/\langle\!\langle\widetilde{r}_{j}\rangle\!\rangle\hookrightarrow N/\langle\!\langle\widetilde{r}_{i},\widetilde{r}_{i+1},\dots,\widetilde{r}_{j}\rangle\!\rangle$. The composition of these two embeddings gives (2). Now we prove (1) for fixed $i$ by induction on $j$.

Base of induction. For $j=i$ we shall show

[TABLE]

It suffices to show the following claim.

Claim. Let $P$ be the subgroup of $N$ generated by $G_{s,t}\cup\{b_{t+1},\dots,b_{t+k}\}$. Then

1. (a)

$P$ embeds into $G_{-\infty,t}$ and $G_{s,\infty}/\langle\!\langle\widetilde{r}_{j}\rangle\!\rangle$. 2. (b)

The abstract amalgamated product in the right side of (4.1) is canonically isomorphic to $N/\langle\!\langle\widetilde{r}_{j}\rangle\!\rangle$.

Proof of the claim. (a) Clearly, $P$ embeds into $G_{-\infty,t}$. So, we show that $P$ embeds into $G_{s,\infty}/\langle\!\langle\widetilde{r}_{j}\rangle\!\rangle$. Note that $P=\langle Y_{s}\cup\dots\cup Y_{t}\cup\{b_{t+1},\dots,b_{t+k}\}\rangle=\langle b_{s},b_{s+1},\dots,b_{s+k-1},y_{m,s},y_{m,s+1},\dots,y_{m,t}\mid 1\leqslant m\leqslant n\rangle$. Thus, the group $P$ is generated by the set

[TABLE]

This set is a part of the free basis $\mathcal{B}^{+}(s)$ of $G_{s,\infty}$. By Definition 3.7, $\widetilde{r}_{j}$ written in $\mathcal{B}^{+}(s)$ is cyclically reduced. Further, by statement (1) of Lemma 4.3, the element $\widetilde{r}_{j}$ written in $\mathcal{B}^{+}(s)$ contains at least one letter $y_{m,\ell}$ with $1\leqslant m\leqslant n$ and $\ell\geqslant t+1$. In particular, this letter does not lie in the set (4.2). By Magnus’ Freiheitssatz (Theorem 4.1), $P$ embeds into $G_{s,\infty}/\langle\!\langle\widetilde{r}_{j}\rangle\!\rangle$.

Now, the amalgamated product in (4.1) is well defined. It is easy to check that the groups written in (4.1) are isomorphic by finding a common presentation. This completes the base of induction.

Inductive step $i,j\ \rightarrow i,j+1$. We need to show the formula, where $s=\alpha_{\widetilde{r_{j}}}$ and $t=\omega_{\widetilde{r}_{j}}-1$:

[TABLE]

Let $P$ be the subgroup of $N$ generated by $G_{s+1,t+1}$ and the set $\{b_{t+2},b_{t+3},\dots,b_{t+k+1}\}$. First, we prove that $P$ canonically embeds into both factors. As above $P$ embeds into $G_{s+1,\infty}/\langle\!\langle\widetilde{r}_{j+1}\rangle\!\rangle$. So we show that $P$ embeds into $G_{-\infty,t+1}/\langle\!\langle\widetilde{r}_{i},\widetilde{r}_{i+1},\dots,\widetilde{r}_{j}\rangle\!\rangle$ using the following commutative diagram:

[TABLE]

Let $\varphi$ be the composition of the canonical embedding of the subgroup $P$ into $G_{-\infty,t+1}$ and the canonical homomorphism from $G_{-\infty,t+1}$ to the factor group $G_{-\infty,t+1}/\langle\!\langle\widetilde{r}_{i},\widetilde{r}_{i+1},\dots,\widetilde{r}_{j}\rangle\!\rangle$. It remains to prove that $\varphi$ is an embedding. Considering $P$ as a subgroup of $G_{s+1,\infty}$ and using statement (2) for $i,j$ (recall that (1) implies (2)), we have an embedding of $P$ into $N/\langle\!\langle\widetilde{r}_{i},\widetilde{r}_{i+1},\dots,\widetilde{r}_{j}\rangle\!\rangle$. Since the diagram is commutative, $\varphi:P\rightarrow G_{-\infty,t+1}/\langle\!\langle\widetilde{r}_{i},\widetilde{r}_{i+1},\dots,\widetilde{r}_{j}\rangle\!\rangle$ is an embedding. Again, it is easy to check that the groups in (4.3) are isomorphic. ∎

Finally, we need a “mirrored” version of Lemma 4.4:

Lemma 4.5**.**

Let $\widetilde{r}\in N$ be a suitable element with $||\widetilde{r}||_{\alpha,\omega}\geqslant 1$, and let $i,j$ be two integers with $i\leqslant j$. We denote $s=\alpha_{\widetilde{r}_{i}}+1$ and $t=\omega_{\widetilde{r}_{i}}$. Then we have:

• (1)

N/\langle\!\langle\widetilde{r_{i}},\widetilde{r}_{i+1},\dots,\widetilde{r}_{j}\rangle\!\rangle\ \cong\ G_{-\infty,t}/\langle\!\langle\widetilde{r}_{i}\rangle\!\rangle\underset{\begin{array}[]{ll}w_{s-k}\!\!&=b_{s},\vspace*{-3mm}\\ \dots&\vspace*{-3mm}\\ w_{s-1}\!\!&=b_{s+k-1}\\ \end{array}}{\underset{G_{s,t}}{\ast}}G_{s,\infty}/\langle\!\langle\widetilde{r}_{i+1},\widetilde{r}_{i+1},\dots,\widetilde{r}_{j}\rangle\!\rangle.

• (2)

$G_{-\infty,t-1}$* naturally embeds into $N/\langle\!\langle\widetilde{r}_{i},\widetilde{r}_{i+1},\dots,\widetilde{r}_{j}\rangle\!\rangle$.*

Proof.

Due to the dual structure noticed in Remark 3.8, the statements of Corollary 4.2, Lemma 4.3 and Lemma 4.4 also hold for the dual objects. By rewriting the dual version of Lemma 4.4 with the help of Lemma 3.9, we get the desired statement. ∎

4.3 Conclusion of the proof for $\widetilde{r}$ with positive $\alpha$-$\omega$-length

By assumption of Proposition 2.1, we have two elements $r,s\in H$ with the same normal closure and $r_{x}=0$. Thus, $r,s$ are elements of $N$. By Lemma 3.6, we can choose conjugates $\widetilde{r},\widetilde{s}$ of $r,s$ such that $\widetilde{r}$ and $\widetilde{s}$ are suitable elements. Normal closures are invariant under conjugation. So, without loss off generality, we can replace $r,s$ by $\widetilde{r},\widetilde{s}$. In this section we assume $||\widetilde{r}||_{\alpha,\omega}\geqslant 1$. Since the normal closures of $\widetilde{r}$ and $\widetilde{s}$ in $H$ are equal, the normal closures of $\mathcal{R}:=\{\widetilde{r}_{i}\ |\ i\in\mathbb{Z}\}$ and $\mathcal{S}:=\{\widetilde{s}_{i}\ |\ i\in\mathbb{Z}\}$ in $N$ are equal. In particular, $\widetilde{s}_{0}$ is trivial in $N/\langle\!\langle\mathcal{R}\rangle\!\rangle$. Thus, there are indices $i,j\in\mathbb{Z}$ such that $\widetilde{s}_{0}$ is trivial in $N/\langle\!\langle\widetilde{r}_{i},\widetilde{r}_{i+1},\dots,\widetilde{r}_{j}\rangle\!\rangle$. We choose a pair $i,j$ with this property and $j-i$ minimal. By Lemma 4.4 (1), we have

[TABLE]

for some subgroup $A$.

Lemma 4.6**.**

Suppose that $||\widetilde{r}||_{\alpha,\omega}\geqslant 1$. Then

1. (1)

$\omega_{\widetilde{s}_{0}}\geqslant\omega_{\widetilde{r}_{j}}\geqslant\omega_{\widetilde{r}_{i}}$, 2. (2)

$\alpha_{\widetilde{s}_{0}}\leqslant\alpha_{\widetilde{r}_{i}}\leqslant\alpha_{\widetilde{r}_{j}}$, 3. (3)

$||\widetilde{s}||_{\alpha,\omega}\geqslant||\widetilde{r}||_{\alpha,\omega}.$**

Proof.

(1) We prove $\omega_{\widetilde{s}_{0}}\geqslant\omega_{\widetilde{r}_{j}}$. Suppose the contrary. Then $\widetilde{s}_{0}\in G_{-\infty,\omega_{\widetilde{r}_{j}}-1}$. Since $\widetilde{s}_{0}$ is trivial in $N/\langle\!\langle\widetilde{r}_{i},\widetilde{r}_{i+1},\dots,\widetilde{r}_{j}\rangle\!\rangle$, it is also trivial in the left factor of the amalgamated product (4.4). Therefore, $\widetilde{s}_{0}$ is trivial in $N/\langle\!\langle\widetilde{r}_{i},\widetilde{r}_{i+1},\dots,\widetilde{r}_{j-1}\rangle\!\rangle$. This contradicts the minimality of $j-i$. The inequality $\omega_{\widetilde{r}_{j}}\geqslant\omega_{\widetilde{r}_{i}}$ follows by Lemma 3.4 since $j\geqslant i$. Inequalities (2) can be proved analogously with the help of Lemma 4.5 (1). Inequality (3) follows straightforward from (1) and (2). ∎

Since $||\widetilde{s}||_{\alpha,\omega}\geqslant||\widetilde{r}||_{\alpha,\omega}\geqslant 1$, we have $||\widetilde{r}||_{\alpha,\omega}\geqslant||\widetilde{s}||_{\alpha,\omega}$ by symmetry and therefore $||\widetilde{s}||_{\alpha,\omega}=||\widetilde{r}||_{\alpha,\omega}$. By Lemma 3.4, this is equivalent to $||\widetilde{s}_{0}||_{\alpha,\omega}=||\widetilde{r}_{i}||_{\alpha,\omega}$. This and the first two statements of Lemma 4.6 imply that $\alpha_{\widetilde{s}_{0}}=\alpha_{\widetilde{r}_{i}}=\alpha_{\widetilde{r}_{j}}$, and $\omega_{\widetilde{r}_{j}}=\omega_{\widetilde{r}_{i}}=\omega_{\widetilde{s}_{0}}$, in particular, $j=i$. Therefore, $\widetilde{s}_{0}$ is trivial in $N/\langle\!\langle\widetilde{r}_{i}\rangle\!\rangle$, and the index $i$ is determined by $\alpha_{\widetilde{s}_{0}}=\alpha_{\widetilde{r}_{i}}$. One can prove in the same way that $\widetilde{r}_{i}$ is trivial in $N/\langle\!\langle\widetilde{s}_{0}\rangle\!\rangle$. Thus, $\langle\!\langle\widetilde{s}_{0}\rangle\!\rangle_{N}=\langle\!\langle\widetilde{r}_{i}\rangle\!\rangle_{N}$. From Section 2 we know that $N$ is a free group. So by Theorem 4.1, $\widetilde{s}_{0}$ is conjugate to $\widetilde{r}_{i}^{\pm 1}$ in $N$. Finally, $\widetilde{r}$ is conjugate to $\widetilde{s}^{\pm 1}$ in $H$, and the proof of Proposition 2.1 in the case $||\widetilde{r}||_{\alpha,\omega}\geqslant 1$ is completed. ∎

5 Proof of Proposition 2.1 for $\widetilde{r}$ with non-positive $\alpha$-$\omega$-length

5.1 Properties of $\widetilde{r}$ with non-positive $\alpha$-$\omega$-length

Lemma 5.1**.**

Let $r$ be an element in $N$ with $||r||_{\alpha,\omega}<1$. Then the reduced word representing $r$ in basis $\mathcal{B}^{+}(\alpha_{r})$ contains only $b_{i}$’s.

Proof.

We have $\omega_{r}<\alpha_{r}$. By definition, $r\in G_{-\infty,\omega_{r}}$ and $r\in G_{\alpha_{r},\infty}$.

$\bullet$ Let $r^{-}$ be the reduced word representing $r$ in the basis $\mathcal{B}^{-}(\omega_{r})$ of $G_{-\infty,\omega_{r}}$. Recall that

[TABLE]

where $I=\{\omega_{r}-k+1,\dots,\omega_{r}\}$.

$\bullet$ Let $r^{+}$ be the reduced word representing $r$ in the basis $\mathcal{B}^{+}(\alpha_{r})$ of $G_{\alpha_{r},\infty}$. Recall that

[TABLE]

where $J=\{\alpha_{r},\dots,\alpha_{r}+k-1\}$.

For each $i\in I$, there is a unique $j\in J$ such that $j\equiv i\mod k$. Then $r^{+}$ can be obtained from $r^{-}$ by replacements of all letters $b_{i}$ by $b_{j}u_{j-k}^{-1}u_{j-2k}^{-1}\dots u_{i}^{-1}$ followed by reduction (see Remark 2.3). The second indices of $y$-letters appearing in these replacements lie in the interval $(-\infty,\max J-k]=(-\infty,\alpha_{r}-1]$. The second indices of $y$-letters of $r^{-}$ lie in the interval $(-\infty,\omega_{r}]\subseteq(-\infty,\alpha_{r}-1]$. Therefore, the second indices of $y$-letters of $r^{+}$ (if exist) lie in $(-\infty,\alpha_{r}-1]$. Hence, $r^{+}$ does not have $y$-letters. ∎

5.2 Conclusion of the proof for $\widetilde{r}$ with non-positive $\alpha$-$\omega$-length

As in Section 4.3, we can replace the elements $r,s\in N$ from Proposition 2.1 by suitable elements $\widetilde{r},\widetilde{s}$. In this section, we consider the case $||\widetilde{r}||_{\alpha,\omega}<1$. Assuming $||\widetilde{s}||_{\alpha,\omega}\geqslant 1$, we immediately get a contradiction by applying Lemma 4.6 with reversed roles of $\widetilde{r}$ and $\widetilde{s}$. Thus, we have $||\widetilde{s}||_{\alpha,\omega}<1$.

By Lemma 5.1, there are presentations of $\widetilde{r}$ and $\widetilde{s}$ in $N$ which use only $b_{i}$, $i\in\mathbb{Z}$. So by means of the relations $b_{i}=x^{-i}bx^{i}$, we get presentations of $\widetilde{r}$ and $\widetilde{s}$ as elements in

[TABLE]

that only use $x$ and $b$.

Thus, $\widetilde{r},\widetilde{s}\in\widetilde{H}$, where $\widetilde{H}$ is the subgroup of $H$ generated by $x$ and $b$. We have the presentations

[TABLE]

where $F_{2}$ is the free group in two generators and the isomorphism to $F_{2}$ follows from the Magnus’ Freiheitssatz.

Convention. For $h\in\widetilde{H}$, let $(\!(h)\!)$ be the normal closure of $h$ in $\widetilde{H}$ and let $\langle\!\langle h\rangle\!\rangle$ be the normal closure of $h$ in $H$.

Consider the canonical homomorphism $\iota:\widetilde{H}\rightarrow\widetilde{H}/(\!(\widetilde{r})\!)$. There are two cases:

Case 1. $\ [x^{k},b]^{p}\notin(\!(\widetilde{r})\!)$ for all $p\geqslant 1$.

In this case $\iota\mid_{\langle[x^{k},b]\rangle}:\widetilde{H}\rightarrow\widetilde{H}/(\!(\widetilde{r})\!)$ is an embedding. We identify the subgroup $\langle[x^{k},b]\rangle$ of $\widetilde{H}$ with its image in $\widetilde{H}/(\!(\widetilde{r})\!)$. Then

[TABLE]

Case 2. $[x^{k},b]^{p}\in(\!(\widetilde{r})\!)$ for some $p\geqslant 1$.

We choose the smallest $p$ with this property. Using Tietze transformations, we write:

[TABLE]

Thus, in both cases, we have the embedding $\widetilde{H}/(\!(\widetilde{r})\!)\hookrightarrow H/\langle\!\langle\widetilde{r}\rangle\!\rangle$.

We already know that $\widetilde{s}$ is an element of the subgroup $\widetilde{H}$ of $H$. Further, $\widetilde{s}$ is trivial in $H/\langle\!\langle\widetilde{r}\rangle\!\rangle$ since $\langle\!\langle\widetilde{r}\rangle\!\rangle=\langle\!\langle\widetilde{s}\rangle\!\rangle$. Now, we use the embedding $\widetilde{H}/(\!(\widetilde{r})\!)\hookrightarrow H/\langle\!\langle\widetilde{r}\rangle\!\rangle$ to conclude that $\widetilde{s}$ is trivial in $\widetilde{H}/(\!(\widetilde{r})\!)$. By symmetry, $\widetilde{r}$ is trivial in $\widetilde{H}/(\!(\widetilde{s})\!)$, and we get $(\!(\widetilde{r})\!)=(\!(\widetilde{s})\!)$.

As we have seen in (5.1), the group $\widetilde{H}$ is isomorphic to the free group of rank 2, and this group possesses the Magnus property (cf. [Mag30]). Hence, $\widetilde{r}$ is conjugate to $\widetilde{s}^{\pm 1}$ in $\widetilde{H}$, and since $\widetilde{H}\hookrightarrow H$, the element $\widetilde{r}$ is conjugate to $\widetilde{s}^{\pm 1}$ in $H$. ∎

Acknowledgements. This work is based on my M.Sc. thesis, Heinrich Heine Universität Düsseldorf, 2015. I would like to express my special thanks to my supervisor Professor Oleg Bogopolski for the subject proposal and the support during and after the preparation of my M.Sc. thesis. I also want to thank Professor Benjamin Klopsch for his editorial help.