\section{Introduction}
In \cite{lodha2016nonamenable}, Lodha and Moore introduced the group $G_0$ consisting of piecewise projective homeomorphisms of the real projective line.
This group is a finitely presented torsion free counterexample to the von Neumann-Day problem \cite{Neumann1929, day1950means}, which asks whether every nonamenable group contains nonabelian free subgroups.
Although counterexamples of the von Neumann-Day problem are known \cite{MR586204, adian1979burnside, MR682486, ol2003non, ivanov2005embedding,  monod2013groups}, it is still an open question whether the Thompson group $F$ can be a new counterexample.

The Lodha--Moore group $G_0$ has similar properties to the Thompson group $F$.
Indeed, the Lodha--Moore group can also be defined as a finitely generated group consisting of homeomorphisms of the space of infinite binary sequences, whose generating set is obtained by adding an element to the well-known finite generating set of $F$.
Both have (small) finite presentations \cite{brown1987finiteness, lodha2016nonamenable}, normal forms with infinite presentations \cite{brown1984infinite, lodha2020nonamenable}, simple commutator subgroups \cite{cannon1996introductory, burillo2018commutators}, trivial homotopy groups at infinity \cite{brown1984infinite, zaremsky2016hnn}, no nonabelian free subgroups \cite{brin1985groups, lodha2016nonamenable}, and are of type $F_\infty$ \cite{brown1984infinite, lodha2020nonamenable}.

On the other hand, there exist various generalizations of the Thompson group $F$.
One of the most natural ones is the $n$-adic Thompson group $F(n)$, which is obtained by replacing infinite binary sequences with infinite $n$-ary sequences.
Even now, it is still being actively studied whether what is true for the group $F$ is also true for the generalized group and what interesting properties can be obtained under the generalization.

In this paper, we generalize the Lodha--Moore group similarly and study its properties.
Namely, we define the $n$-adic group $G_0(n)$ of the Lodha--Moore group $G_0$ and show that several properties which hold for $G_0$ also hold for $G_0(n)$.
We remark that $G_0(2)$ is isomorphic to $G_0$.

Let $n, m \geq 2$.
We show the following:
\begin{theorem}
\begin{enumerate}[font=\normalfont]
\item The group $G_0(n)$ admits an infinite presentation with a normal form of elements.
\item The group $G_0(n)$ is finitely presented.
\item The group $G_0(n)$ is nonamenable.
\item The group $G_0(n)$ has no free subgroups.
\item The group $G_0(n)$ is torsion free.
\item The groups $G_0(n)$ and $G_0(m)$ are isomorphic if and only if $n=m$ holds.
\item The commutator subgroup of the group $G_0(n)$ is simple.
\item The center of the group $G_0(n)$ is trivial.
\item There does not exist any nontrivial direct product decomposition of the group $G_0(n)$.
\item There does not exist any nontrivial free product decomposition of the group $G_0(n)$.
\end{enumerate}
\end{theorem}

This paper is organized as follows.
In Section \ref{section_F(n)}, we generalize Dehornoy's infinite presentation of $F$ to $F(n)$, which will be used to construct that of $G_0(n)$.
To the best of the author's knowledge, this is a new presentation of $F(n)$.
In Section \ref{section_G0(n)}, we first recall the definition of the group $G_0$ and define the group $G_0(n)$.
Then by using the infinite presentation of $F(n)$ constructed in Section \ref{section_F(n)}, we generalize Lodha's method to obtain that of $G_0(n)$ and its normal form.
Finally, in Section \ref{section_G0(n)_properties}, we study several properties of $G_0(n)$.

Let us mention some open problems which are known to hold in the case of $G_0$.
First, it is an interesting question whether $G_0(n)$ can be realized as a subgroup of the group of piecewise projective homeomorphisms of the real projective line.
The second problem is whether this group is of type $F_{\infty}$, and all homotopy groups are trivial at infinity.
If it has these two properties, then ${G_0(n)}$ is an example of an (infinite) family of groups satisfying all Geoghegan's conjectures for the Thompson group $F$.

Furthermore, we can consider some groups related to $G_0(n)$.
The first one is constructed by using another definition of the map $y$ defined in Section \ref{subsection_def_G_0(n)}.
Although we define the map so that $G_0$ is naturally a subgroup of $G_0(n)$, we can consider several different generalizations.

We can also construct groups that contain $G_0(n)$.
In \cite{lodha2016nonamenable}, the group $G$ is defined, where $G$ contains $G_0$ as a subgroup.
For our group, by adding some of the generators $y_0$, $y_{(n-1)1}$, $\dots$, $y_{(n-1)(n-2)}$, $y_{(n-1)}$, we can define not only the group $G(n)$, which corresponds to $G$ but also the groups ``between'' $G_0(n)$ and $G(n)$.
\section{The generalized Thompson group $F(n)$}\label{section_F(n)}
\subsection{Definition} \label{subsection_F(n)_definition}
Let $n \geq 2$.
There exist several ways to define the generalized Thompson group $F(n)$.
In this paper, we define it as a group of homeomorphisms on the $n$-adic Cantor set.
We use tree diagrams to represent elements of the group visually.

We define $\bm{N}$ to be the set $\{0, 1, \dots, n-1 \}$.
We endow $\bm{N}$ with the discrete topology and endow $\N^\mathbb{N}$ with the product topology.
Note that $\N^\mathbb{N}$ and the Cantor set are homeomorphic.
We also consider the set of all finite sequences on $\bm{N}$ and write $\N^{<\mathbb{N}}$ for it.
For $s \in \N^{<\mathbb{N}}$ and $t \in \N^{<\mathbb{N}}$ (or $\N^\mathbb{N}$), the concatenation is denoted by $st$.
The group $F(n)$ is a finitely generated group that is generated by the following $n$  homeomorphisms on $\N^\mathbb{N}$:
\begin{align*}
x_0(\zeta)&=
\begin{cases}
0\eta & ( \zeta=00\eta )\\
1\eta & ( \zeta=01\eta ) \\
 &\vdots \\
(n-2)\eta & (\zeta=0(n-2)\eta) \\
(n-1)0\eta & (\zeta=0(n-1)\eta) \\
(n-1)1\eta & (\zeta=1\eta) \\
&\vdots \\
(n-1)(n-1)\eta & (\zeta=(n-1)\eta),
\end{cases} \\
x_1(\zeta)&=
\begin{cases}
0\eta & (\zeta=0\eta) \\
1\eta & (\zeta=10\eta) \\
2\eta & (\zeta=11\eta) \\
&\vdots \\
(n-2)\eta & (\zeta=1(n-3)\eta) \\
(n-1)0\eta & (\zeta=1(n-2)\eta) \\
(n-1)1\eta & (\zeta=1(n-1)\eta) \\
(n-1)2\eta & (\zeta=2\eta) \\
&\vdots \\
(n-1)(n-1)\eta & (\zeta=(n-1)\eta),
\end{cases} \\
&\vdots \\
x_{n-2}(\zeta)&=
\begin{cases}
0\eta & (\zeta=0\eta) \\
&\vdots \\
(n-3)\eta & (\zeta=(n-3)\eta) \\
(n-2)\eta & (\zeta=(n-2)0\eta) \\
(n-1)0\eta & (\zeta=(n-2)1\eta) \\
&\vdots \\
(n-1)(n-2)\eta & (\zeta=(n-2)(n-1)\eta) \\
(n-1)(n-1)\eta & (\zeta=(n-1)\eta),
\end{cases}
\shortintertext{and}
\x{0}{(n-1)}(\zeta)&=
\begin{cases}
(n-1)x_0(\eta) & (\zeta=(n-1)\eta) \\
\zeta & (\zeta \neq (n-1)\eta).
\end{cases}
\end{align*}

These maps are represented by tree diagrams as in Figure \ref{generator_Fn}.
\begin{figure}[tbp]
	\centering
	\includegraphics[width=150mm]{generator_of_Fn.pdf}
	\caption{Tree diagrams of the homeomorphisms. }
	\label{generator_Fn}
\end{figure}
Here, we briefly review the definition of tree diagrams.
See \cite{burillo2001metrics} for details.
An \textit{$n$-ary tree} is a finite tree with a top vertex (\textit{root}) with $n$ edges, and all vertices except the root have degree only $1$ (\textit{leaves}) or $n+1$.
We define a \textit{caret} to be an $n$-ary tree with no vertices whose degree is $n+1$ (see Figure \ref{n-caret}).
\begin{figure}[tbp]
	\centering
	\includegraphics[width=30mm]{n-caret.pdf}
	\caption{A caret. }
	\label{n-caret}
\end{figure}
Then, each $n$-ary tree is obtained by attaching carets to a leaf of a caret.
We always assume that the root is the top and the others are descendants.

Each $n$-ary tree can be regarded as a finite subset of $\N^{<\mathbb{N}}$.
To do this, we label each edge of each caret by $0, 1, \dots, n-1$ from the left.
Since every leaf corresponds to a unique path from the root to the leaf, we can regard it as an element in $\N^{<\mathbb{N}}$.

Let $T_+$ and $T_-$ be $n$-ary trees with $m$ leaves.
Let $a_1, \dots, a_m$ be elements in $\N^{<\mathbb{N}}$ with lexicographic order corresponding to the leaves of $T_+$.
For $T_-$, define $b_1, \dots, b_m$ in the same way.
Then, for every $\zeta \in \N^\mathbb{N}$, there exists $i$ uniquely such that $\zeta=a_i\eta$ for some $\eta \in \N^\mathbb{N}$.
Thus we obtain a homeomorphism $a_i \eta \mapsto b_i \eta$.
It is known that every homeomorphism obtained from two $n$-ary trees with the same number of leaves in this way is generated by the composition of $x_0, x_1, \dots, x_{n-2}, \x{0}{(n-1)}$.
See \cite[Corollary 10.9]{meier2008groups} for the case $n=2$.

We define $\epsilon$ to be the empty word.
Let $i$ in $\{0, \dots, n-2 \}$ and $\alpha$ in $\N^{<\mathbb{N}} \cup \{\epsilon \}$.
We define the map $\x{i}{\alpha}: \N^\mathbb{N} \to \N^\mathbb{N}$ by
\begin{align*}
\x{i}{\alpha}(\zeta)&=
\begin{cases}
\alpha x_i(\eta) & (\zeta=\alpha \eta) \\
\zeta & (\zeta \neq \alpha \eta),
\end{cases}
\end{align*}
and we define
\begin{align*}
X(n)&:=
\left\{ \x{i}{\alpha} \mid i = 0, \dots, n-2, \alpha \in \N^{<\mathbb{N}} \cup \{ \epsilon \} \right\}.
\end{align*}
This set contains the well-known infinite generating set of $F(n)$, which we describe below, and is the key generating set in Section \ref{subsubsection_New_presentation_F(n)}.

Let $X^\prime(n):=\{ \x{0}{s}, \dots, \x{n-2}{s} \mid s=\epsilon, (n-1), (n-1)(n-1), \dots \}$.
We denote each element as $x_0=X_0, \dots, x_{n-2}=X_{n-2}, \x{0}{(n-1)}=X_{n-1}, \x{1}{(n-1)}=X_{n}, \dots$ only in this section for the sake of simplicity.
For $i<j$, we have $X_i^{-1}X_jX_i=X_{j+n-1}$.
This implies that every element in $F(n)$ has the following form:
\begin{align*}
\X{i_1}{r_1}\X{i_2}{r_2}\cdots \X{i_m}{r_m}\X{j_k}{-s_k}\cdots\X{j_2}{-s_2}\X{j_1}{-s_1}
\end{align*}
where $i_1<i_2<\cdots<i_m\neq j_k>\cdots>j_2>j_1$ and $r_1, \dots, r_m, s_1, \dots, s_k>0$.
We require that this form satisfies the following additional condition:
if there exist $X_i$ and $X_i^{-1}$, then there also exists one of
\begin{align*}
X_{i+1}, \X{i+1}{-1}, X_{i+2}, \X{i+2}{-1}, \dots, X_{i+n-1}, \X{i+n-1}{-1}.
\end{align*}
It is known that this form with the additional condition always uniquely exists.
Thus we call this form \textit{normal form} of elements of $F(n)$.
The proof for the case $n=2$ is in \cite[Section 1]{brown1984infinite}.
\subsection{A presentation of the generalized Thompson group}
While the group $F(n)$ has the well-known infinite and finite presentations,
we construct another presentation with respect to $X(n)$ for Section \ref{infinite_presentation}.

We list the relations of the elements in $F(n)$ as follows.
Here, $A_i$s are defined at the beginning of Section \ref{subsubsection_New_presentation_F(n)}, and the shift-map is defined in Definition \ref{Def_shiftmap}.
\begin{table}[H]
{\small
\begin{tabular}{ccccc}
$A_0A_1=\A{1}{(n-1)}A_0$, & $A_1A_2=\A{2}{(n-1)}A_1$, & $\dots$, & $A_{n-2}\A{0}{(n-1)}=\A{0}{(n-1)^2}A_{n-2}$, & $\dots$ \\
$A_0A_2=\A{2}{(n-1)}A_0$, & $A_1A_3=\A{3}{(n-1)}A_1$, & $\dots$, & $A_{n-3}\A{0}{(n-1)}=\A{0}{(n-1)^2}A_{n-3}$, & $\dots$ \\
$A_0A_3=\A{3}{(n-1)}A_0$, & $\dots$ \\
$\vdots$ & $\ddots$ \\
$A_0A_{n-2}=\A{n-2}{(n-1)}A_0$, \\
$A_0\A{0}{(n-1)}=\A{0}{(n-1)^2}A_0$, \\
$A_0\A{1}{(n-1)}=\A{1}{(n-1)^2}A_0$, \\
$\vdots$
\end{tabular}
}
\centering
\caption{Moving to a right column corresponds to the sift-map. }
\label{relations_1}
\end{table}
\begin{table}[H]
{\small
\begin{tabular}{cccc}
$A_0A_0=\A{0}{(n-1)}A_0\A{0}{0}$, & $A_1A_0=\A{0}{(n-1)}A_0\A{1}{0}$, & $\dots$, & $A_{n-2}A_0=\A{0}{(n-1)}A_0\A{n-2}{0}$, \\
$A_1A_1=\A{1}{(n-1)}A_1\A{0}{1}$, \\
$\vdots$ & $\ddots$ \\
\end{tabular}
}
\centering
\caption{Moving to a lower row corresponds to the sift-map. }
\label{relations_2}
\end{table}
\begin{table}[H]
{\small
\begin{tabular}{ccc}
$\A{k}{0\alpha}A_1=A_1\A{k}{0\alpha}$, & $\A{k}{1\alpha}A_2=A_2\A{k}{1\alpha}$, & $\dots$ \\
$\A{k}{0\alpha}A_2=A_2\A{k}{0\alpha}$, & $\A{k}{1\alpha}A_3=A_3\A{k}{1\alpha}$, & $\dots$ \\
$\A{k}{0\alpha}A_3=A_3\A{k}{0\alpha}$, \\
$\vdots$ & $\ddots$
\end{tabular}
}
\centering
\caption{For each $\alpha$ in $\N^{<\mathbb{N}} \cup \{\epsilon \}$ and $k$ in $\{0, \dots, n-2\}$, moving to a right column corresponds to the sift-map. }
\label{relations_3}
\end{table}
\begin{table}[H]
{\small
\begin{tabular}{ccc}
$\A{k}{0\alpha}A_0=A_0\A{k}{00\alpha}$, & $\A{k}{1\alpha}A_1=A_1\A{k}{10\alpha}$, & $\dots$ \\
$\A{k}{1\alpha}A_0=A_0\A{k}{01\alpha}$, & $\A{k}{2\alpha}A_1=A_1\A{k}{11\alpha}$, & $\dots$ \\
$\A{k}{2\alpha}A_0=A_0\A{k}{02\alpha}$, \\
$\vdots$ & $\ddots$ \\
$\A{k}{(n-2)\alpha}A_0=A_0\A{k}{0(n-2)\alpha}$, \\
$\A{k}{(n-1)0\alpha}A_0=A_0\A{k}{0(n-1)\alpha}$,
\end{tabular}
}
\centering
\caption{For each $\alpha$ in $\N^{<\mathbb{N}} \cup \{\epsilon \}$ and $k$ in $\{0, \dots, n-2\}$, moving to a right column corresponds to the sift-map. }
\label{relations_4}
\end{table}
\subsubsection{Dehornoy's results}
For the ``geometric'' presentation of $F(n)$, we generalize Dehornoy's method in \cite{dehornoy2005geometric}.
Section 1 of \cite{dehornoy2005geometric} gives the way to find a presentation of a group.
So we recall the general setting and the way.

\begin{definition}\cite[Section 1.1]{dehornoy2005geometric}
Let $G$ be a group (monoid) and $T$ be a set.
We define a \textit{partial} (\textit{right}) \textit{action} to be a map $\phi$ from $G$ to the set of injections $\{f: T^\prime \to T \mid T^\prime \subset T \}$ such that the following are satisfied (in the following, we write $t \cdot g$ for the image of $t$ under $\phi(g)$ if it is defined):
\begin{itemize}
\item[$({PA}_1)$] For every $t \in T$, $t \cdot e=t$ holds;
\item[$({PA}_2)$] For every $g, h \in G$, and $t \in T$, if $t \cdot g$ is defined, then $(t\cdot g)\cdot h$ is defined if and only if $t\cdot gh$ is defined, and if one of them is defined, we have $(t\cdot g)\cdot h=t \cdot gh$;
\item[$({PA}_3)$] For every finite family $g_1, \dots, g_m$, there exists $t \in T$ such that $t \cdot g_1, \dots, t \cdot g_m$ are defined.
\end{itemize}
\end{definition}
Dehornoy also introduced a stronger condition on partial actions.
\begin{definition}\cite[Section 1.3]{dehornoy2005geometric}
We assume that $G$ has a partial action on $T$.
Then we call a subset $S \subset T$ is \textit{discriminating} if:
\begin{enumerate}
\item In (${PA}_3$), we can take $t$ in the set $S \cdot G=\{s \cdot g \mid s \in S,  \mbox{$g \in G$ such that $s \cdot g$ is defined} \}$;
\item Each $G$-orbit contains at most one element of $S$;
\item For every $s \in S$, its stabilizer is trivial.
\end{enumerate}
\end{definition}
\begin{remark}
In the above setting, for every $t \in S \cdot G$, there exist the unique $s \in S$ and $g \in G$ such that $t=s \cdot g$.
\end{remark}

When $R$ is a family of relations of a group, we write $w\equiv_R z$ if one can rewrite $w$ to $z$ by using elements in $R$.
The following theorem holds.
\begin{theorem}[{\cite[Proposition 1.4]{dehornoy2005geometric}}]
\label{dehornoy_presentation}
Let $G$ be a group with a partial action on a set $T$.
Let $X$ be a subset of $G$ and $R$ be a collection of relations satisfied in $G$ by the elements of $X$.
Assume that $S$ is a discriminating subset of $T$ and that, for each $s$ in $S$ and $t$ in the $G$-orbit of $s$, a word $w_t$ on $X$ is chosen so that $t=s \cdot w_t$ holds.
Then $\langle X \mid R \rangle$ is a presentation of $G$ if and only if for all $t, t^\prime$ in $S \cdot G$ and $x$ in $X$,
\begin{align} \label{condition}
\mbox{$t^\prime=t \cdot x$ implies $w_{t^\prime} \equiv_R w_t \times x$},
\end{align}
where $\times$ denotes the concatenation of words in $X$.
\end{theorem}
\subsubsection{A presentation of $F(n)$}\label{subsubsection_New_presentation_F(n)}
In this section, we construct a partial action of $F(n)$ and give a presentation by using Theorem \ref{dehornoy_presentation}.
Except for the construction of $w_t$, the discussions are almost the same as the case $n=2$.

Let $T(n)$ be a set consisting of $n$-ary trees and the root (the graph with a single vertex).
We first define partial actions of $n-1$ elements $A_0, \dots, A_{n-2}$ on $T(n)$ as illustrated in Figure \ref{partial_action}.
\begin{figure}[tbp]
	\centering
	\includegraphics[width=160mm]{partial_action.pdf}
	\caption{Partial actions of $A_1, \dots, A_{n-2}$. }
	\label{partial_action}
\end{figure}
We note that the actions of $A_i$ ($i=0, \dots, n-2$) on a single caret are not defined.

Let $\alpha$ in $\N^{<\mathbb{N}}$.
Then, for $n$-ary trees which contain $\alpha$ as a subpath, we define a partial action of $\A{i}{\alpha}$ to be the partial mapping obtained by applying $A_i$ to the element of $T(n)$ positioned just below $\alpha$ (if it is defined).
See Figure \ref{monoid_def} and observe that the bottom caret is the only one moved by the action.
\begin{figure}[tbp]
	\centering
	\includegraphics[width=60mm]{monoid_def_color.pdf}
	\caption{Example of the action of $\A{1}{2}$ for $n=4$. }
	\label{monoid_def}
\end{figure}
\begin{definition}
We define $\mathscr{G}_n(\mathscr{A})$ to be the monoid generated by $X(n):=\{\A{i}{\alpha} \mid i \in \{0, \dots, n-2\}, \alpha \in \N^{<\mathbb{N}} \cup \{ \epsilon \} \}$ and their inverse, where $\epsilon$ is the empty word and $\A{i}{\epsilon}:=A_i$.
\end{definition}
From the construction, the monoid $\mathscr{G}_n(\mathscr{A})$ naturally has a partial action on $T(n)$.
In order to make $\mathscr{G}_n(\mathscr{A})$ into a group, we introduce a congruence on $\mathscr{G}_n(\mathscr{A})$.
\begin{definition}
We assume that a group $G$ has a partial action on a set $T$.
If there exists $t \in T$ such that $t \cdot g$ and $t \cdot g^\prime$ are defined, and they coincide for all such $t$, then we define $g$ and $g^\prime$ to be \textit{near-equal} and write it as $g \approx g^\prime$.
\end{definition}
As in \cite[Corollary 2.4]{dehornoy2005geometric}, we can show that $\mathscr{G}_n(\mathscr{A})$ and the set of all-right trees $S(n)$ satisfy the assumptions in \cite[Lemma 2.2]{dehornoy2005geometric} (see Figure \ref{T_m} as an example of an all-right tree).
Thus, near-equality is a congruence on $\mathscr{G}_n(\mathscr{A})$, and the quotient monoid $\mathscr{G}_n(\mathscr{A}) / {\approx}$ is a group.
We write $G_n(\mathscr{A})$ for this group.
Moreover, $S(n)$ is discriminating for the induced partial action on $G_n(\mathscr{A})$.
We omit the proofs because they are the same for $n=2$, but we recall the induced action for the reader's convenience.
\begin{definition}
For $t, t^\prime$ in $T(n)$ and $x$ in $G_n(\mathscr{A})$, we define $t \cdot x=t^\prime$ if $t \cdot g =t^\prime$ holds for some $g$ in $\mathscr{G}_n(\mathscr{A})$ such that $g$ is a representative of $x$ (if there exists such $g$).
\end{definition}
By the definitions, it is easy to see that $G_n(\mathscr{A})$ and $F(n)$ are isomorphic.
We simply write $g$ for the class of $g$ in $G_n(\mathscr{A})$ for the sake of simplicity.

In order to give a presentation of $G_n(\mathscr{A})$, we first define the ``shift-map'' on $G_n(\mathscr{A})$.
\begin{definition}\label{Def_shiftmap}
Define the map $[1]$ by the following:
\begin{align*}
A_0 \overset{[1]}{\mapsto} &A_1 \overset{[1]}{\mapsto} \cdots \overset{[1]}{\mapsto} A_{n-2} \overset{[1]}{\mapsto} \A{0}{(n-1)} \overset{[1]}{\mapsto} \A{1}{(n-1)} \overset{[1]}{\mapsto} \cdots \\
&\overset{[1]}{\mapsto} \A{n-2}{(n-1)} \overset{[1]}{\mapsto} \A{0}{(n-1)(n-1)} \overset{[1]}{\mapsto} \cdots,
\end{align*}
and for each $k \in \{0, \dots, n-2\}$ and $\alpha \in \N^{<\mathbb{N}} \cup \{ \epsilon \}$,
\begin{align*}
\A{k}{0\alpha} \overset{[1]}{\mapsto} &\A{k}{1\alpha} \overset{[1]}{\mapsto} \cdots \overset{[1]}{\mapsto} \A{k}{(n-2)\alpha} \overset{[1]}{\mapsto} \A{k}{(n-1)0\alpha} \overset{[1]}{\mapsto} \A{k}{(n-1)1\alpha} \overset{[1]}{\mapsto} \cdots \\
&\overset{[1]}{\mapsto} \A{k}{(n-1)(n-2)\alpha} \overset{[1]}{\mapsto} \A{k}{(n-1)(n-1)0\alpha} \overset{[1]}{\mapsto} \cdots \overset{[1]}{\mapsto} \A{k}{(n-1)(n-1)1\alpha} \overset{[1]}{\mapsto} \cdots.
\end{align*}
For the empty word $\epsilon$, define $[1](\epsilon)=\epsilon$.
Furthermore, for an element $\A{k_1}{\alpha_1} \cdots \A{k_m}{\alpha_m}$ in $G_n(\mathscr{A})$, we define $[1](\A{k_1}{\alpha_1} \dots \A{k_m}{\alpha_m})=[1](\A{k_1}{\alpha_1}) \cdots [1](\A{k_m}{\alpha_m})$.
A map $[i]$ denotes the composition of $[1]$ $i$ times.
\end{definition}
\begin{remark}\label{shift_n-1}
By the definition, for each $\A{k}{\alpha}$ (even if $\alpha$ is empty), we have
\begin{align*}
[n-1](\A{k}{\alpha})=\A{k}{(n-1)\alpha}.
\end{align*}
This fact is useful to check the relations.
\end{remark}
Next, for an $n$-ary tree $t$, we define inductively two words $w_t$ and $w_t^\ast$ as follows:
\begin{definition}
\begin{enumerate}
\item If $t$ is a single vertex, we define
\begin{align*}
w_t=w_t^\ast=\epsilon.
\end{align*}
\item If $t$ is an $n$-ary tree, as in Figure \ref{inductive_tree}, we define
\begin{align*}
w_t&=w_{t_0}^\ast \times [1](w_{t_1}^\ast) \times [2](w_{t_2}^\ast) \times \cdots  \times [n-2](w_{t_{n-2}}^\ast) \times [n-1](w_{t_{n-1}}), \\
w_t^\ast&=w_{t_0}^\ast \times [1](w_{t_1}^\ast) \times [2](w_{t_2}^\ast) \times \cdots  \times [n-2](w_{t_{n-2}}^\ast) \times [n-1](w_{t_{n-1}}^\ast) \times A_0,
\end{align*}
where $\times$ denotes the concatenation.
\end{enumerate}
\end{definition}
\begin{figure}[tbp]
	\centering
	\includegraphics[width=30mm]{inductive_tree.pdf}
	\caption{An $n$-ary tree $t$ with subtrees (or leaves) $t_0, \dots, t_{n-1}$}
	\label{inductive_tree}
\end{figure}
Now, we show that $t=s \cdot w_t$ holds.
We note that $s$ is an $n$-ary tree with the same number of carets of $t$.
\begin{lemma} \label{lemma1_presentation}
Let $t$ in $T(n)$ with $m$ carets.
We define $T_m$ to be an all-right tree with $m$ carets if $m \geq 1$ and a single vertex if $m=0$.
Then, for every $t^\prime$ in $T(n)$, we have $T_m \cdot w_t=t$, and the equality in Figure $\ref{presentation_assumption}$ holds.
\begin{figure}[tbp]
	\centering
	\includegraphics[width=80mm]{presentation_assumption.pdf}
	\caption{The second claim in Lemma \ref{lemma1_presentation}. }
	\label{presentation_assumption}
\end{figure}
\begin{proof}
We show by induction on $m$.
For the base case, since $t, T_m$ are single vertices and, we have $w_t=w_t^\prime=\epsilon$, the equality is clear.

Let $t$ be in Figure \ref{inductive_tree} and $n_i$ be the number of carets of $t_i$ for $i=0, \dots, n-1$.
By the definitions,
\begin{align*}
w_t=w_{t_0}^\ast \times [1](w_{t_1}^\ast) \times [2](w_{t_2}^\ast) \times \cdots  \times [n-2](w_{t_{n-2}}^\ast) \times [n-1](w_{t_{n-1}})
\end{align*}
 holds and $T_m$ is in Figure \ref{T_m}.
\begin{figure}[tbp]
	\centering
	\includegraphics[width=70mm]{n1nn-1carets.pdf}
	\caption{The $n$-ary tree $T_m$. }
	\label{T_m}
\end{figure}
By the inductive hypothesis for $w_{t_0}^\ast$, it acts on $T_m$ as in Figure \ref{wt0_action}.
\begin{figure}[tbp]
	\centering
	\includegraphics[width=125mm]{wt0_action.pdf}
	\caption{The action of $w_{t_0}^\ast$. }
	\label{wt0_action}
\end{figure}
Since $w_t, w_t^\ast$ are the words on $\{A_i, \A{i}{(n-1)}, \A{i}{(n-1)(n-1)}, \dots \mid i=0, \dots, n-2 \}$, the sift-map $[1]$ shits indices just one.
Thus, by applying the inductive hypothesis repeatedly, $w_{t_0}^\ast \times [1](w_{t_1}^\ast) \times [2](w_{t_2}^\ast) \times \cdots  \times [n-2](w_{t_{n-2}}^\ast)$ acts on $T_m$ as in Figure \ref{wto_wtn-2_action}, and we write $\tilde{t}$ for the resulting tree.
\begin{figure}[tbp]
	\centering
	\includegraphics[width=160mm]{wto_wtn-2_action.pdf}
	\caption{The action of $w_{t_0}^\ast \times [1](w_{t_1}^\ast) \times [2](w_{t_2}^\ast) \times \cdots  \times [n-2](w_{t_{n-2}}^\ast)$. }
	\label{wto_wtn-2_action}
\end{figure}
Recall Remark \ref{shift_n-1}, the action of $[n-1](w_{t_{n-1}})$ on $\tilde{t}$ is obtained from that of $w_{t_{n-1}}$ on the subtree $T_{n_{n-1}}$ in $\tilde{t}$.
By the inductive hypothesis for $w_{t_{n-1}}$, we have $T_{n_{n-1}} \cdot w_{t_{n-1}}=t_{n-1}$.
This completes the proof of the first claim that $w_t$ satisfies the equality.

By the similar argument as in the case of the action of $w_t$, that of $w_t^\ast$ is calculated as shown in Figure \ref{wtast}.
\begin{figure}[tbp]
	\centering
	\includegraphics[width=160mm]{wtast.pdf}
	\caption{The action of $w_t^\ast$. }
	\label{wtast}
\end{figure}
\end{proof}
\end{lemma}
It remains to prove that the condition \eqref{condition} in Theorem \ref{dehornoy_presentation} holds for a collection of relations.
Let $R(n)$ be the set of the elements in Tables \ref{relations_1}, \ref{relations_2}, \ref{relations_3}, and \ref{relations_4}.
It is easy to see that all elements in $R(n)$ are relations of $F(n)$.
\begin{lemma}
Let $t^\prime=t \cdot \A{k}{\alpha}$.
Then we have
\begin{align*}
w_{t^\prime} &\equiv_{R(n)} w_t \cdot \A{k}{\alpha}, \\
{w_{t^\prime}}^\ast &\equiv_{R(n)} {w_t}^\ast \cdot \A{k}{0\alpha}.
\end{align*}
\begin{proof}
When we rewrite a word $w$ to $z$ by applying a relation in Table $i$ ($1 \leq i \leq 4$), we denote by $w \equiv_i z$.
We show by induction on the length of $\alpha$ as an $n$-ary sequence.
For the base case, first, we assume $k=0$.
Then we can illustrate trees $t, t^\prime$ as in Figure \ref{t_tprime}.
\begin{figure}[tbp]
	\centering
	\includegraphics[width=100mm]{t_tprime.pdf}
	\caption{The $n$-ary trees $t$ (left) and $t^\prime$ (right) for $\alpha=\epsilon$ and $k=0$. }
	\label{t_tprime}
\end{figure}
Let $\tilde{t}$ be the subtree of $t^\prime$ consisting of $t_0, \dots, t_{n-1}$ (as in Figure \ref{inductive_tree}).
Then we have
\begin{align*}
w_{t^\prime}
&=w_{\tilde{t}}^\ast \times [1](w_{t_n}^\ast) \times \dots \times [n-2](w_{t_{2n-3}}^\ast) \times [n-1](w_{t_{2n-2}}) \\
&=w_{t_0}^\ast \times [1](w_{t_1}^\ast)\times \dots \times [n-1]({w_{t_{n-1}}}^\ast) \times A_0 \\
&\; \times  [1](w_{t_n}^\ast) \times \dots \times [n-2](w_{t_{2n-3}}^\ast) \times [n-1](w_{t_{2n-2}}) \\
&\equiv_1 w_{t_0}^\ast \times [1](w_{t_1}^\ast)\times \dots \times [n-1]({w_{t_{n-1}}}^\ast) \\
&\; \times[n-1]\left([1](w_{t_n}^\ast) \times \dots \times [n-2](w_{t_{2n-3}}^\ast) \times [n-1](w_{t_{2n-2}})\right) \times A_0 \\
&= w_{t_0}^\ast \times [1](w_{t_1}^\ast)\times \cdots \\
&\; \times [n-1]\bigl(({w_{t_{n-1}}}^\ast) \times[1](w_{t_n}^\ast) \times \dots \times [n-2](w_{t_{2n-3}}^\ast) \times [n-1](w_{t_{2n-2}})\bigr) \times A_0 \\
&= w_t \cdot A_0,
\end{align*}
and
\begin{align*}
w_{t^\prime}^\ast
&=w_{t_0}^\ast \times [1](w_{t_1}^\ast)\times \dots \times [n-1]({w_{t_{n-1}}}^\ast) \times A_0 \\
&\; \times  [1](w_{t_n}^\ast) \times \dots \times [n-2](w_{t_{2n-3}}^\ast) \times [n-1](w_{t_{2n-2}}^\ast) \times A_0 \\
&\equiv_1 w_{t_0}^\ast \times [1](w_{t_1}^\ast)\times \cdots \\
&\; \times [n-1]\bigl(({w_{t_{n-1}}}^\ast) \times[1](w_{t_n}^\ast) \times \dots \times [n-2](w_{t_{2n-3}}^\ast) \times [n-1](w_{t_{2n-2}}^\ast)\bigr) \times A_0A_0 \\
&\equiv_2 w_{t_0}^\ast \times [1](w_{t_1}^\ast)\times \cdots \\
&\; \times [n-1]\bigl(({w_{t_{n-1}}}^\ast) \times[1](w_{t_n}^\ast) \times \dots \times [n-2](w_{t_{2n-3}}^\ast) \times [n-1](w_{t_{2n-2}}^\ast)\bigr) \\
&\; \times \A{0}{(n-1)}A_0\A{0}{0} \\
&=w_{t_0}^\ast \times [1](w_{t_1}^\ast)\times \cdots \\
&\; \times [n-1]\bigl(({w_{t_{n-1}}}^\ast) \times[1](w_{t_n}^\ast) \times \dots \times [n-2](w_{t_{2n-3}}^\ast) \times [n-1](w_{t_{2n-2}}^\ast)\times A_0 \bigr) \\
&\; \times A_0\A{0}{0} \\
&=w_t^\ast \times \A{0}{0}.
\end{align*}
The case $k\geq1$ can be proved in the same way.

Let $\alpha=0\beta$.
That is, we consider the case when the condition $t_0^\prime=t_0 \cdot \A{k}{\beta}$ holds for $t$ and $t^\prime$, as shown in Figure \ref{inductive_t_tprime}.
\begin{figure}[tbp]
	\centering
	\includegraphics[width=80mm]{inductive_t_tprime.pdf}
	\caption{The $n$-ary trees $t$ (left) and $t^\prime$ (right) for $\alpha=0\beta$. }
	\label{inductive_t_tprime}
\end{figure}
When we rewrite a word by applying the inductive hypothesis, we use $\equiv_I$ to denote its equality.
We note that
\begin{align*}
[1](w_{t_1}^\ast) \times \dots \times [n-2](w_{t_{n-2}}^\ast) \times [n-1](w_{t_{n-1}})
\shortintertext{and}
 [1](w_{t_1}^\ast) \times \dots \times [n-2](w_{t_{n-2}}^\ast) \times [n-1](w_{t_{n-1}}^\ast)
 \end{align*}
are words on the set
\begin{align*}
\{ A_1, \dots, A_{n-2}, \A{i}{\alpha} \mid i= 0, \dots, n-2, \alpha=(n-1), (n-1)^2, \dots \}.
\end{align*}
Then we have
\begin{align*}
w_{t^\prime}&=w_{t_0^\prime}^\ast \times [1](w_{t_1}^\ast) \times \dots \times [n-2](w_{t_{n-2}}^\ast) \times [n-1](w_{t_{n-1}}) \\
&\equiv_I w_{t_0}^\ast \times \A{k}{0\beta} \times [1](w_{t_1}^\ast) \times \dots \times [n-2](w_{t_{n-2}}^\ast) \times [n-1](w_{t_{n-1}}) \\
&\equiv_3 w_{t_0}^\ast \times [1](w_{t_1}^\ast) \times \dots \times [n-2](w_{t_{n-2}}^\ast) \times [n-1](w_{t_{n-1}}) \times \A{k}{0\beta} \\
&=w_t \times \A{k}{\alpha},
\shortintertext{and}
w_{t^\prime}^\ast&=w_{t_0^\prime}^\ast \times [1](w_{t_1}^\ast) \times \dots \times [n-2](w_{t_{n-2}}^\ast) \times [n-1](w_{t_{n-1}}^\ast) \times A_0 \\
&\equiv_I w_{t_0}^\ast \times \A{k}{0\beta} \times [1](w_{t_1}^\ast) \times \dots \times [n-2](w_{t_{n-2}}^\ast) \times [n-1](w_{t_{n-1}}^\ast) \times A_0 \\
&\equiv_3 w_{t_0}^\ast \times [1](w_{t_1}^\ast) \times \dots \times [n-2](w_{t_{n-2}}^\ast) \times [n-1](w_{t_{n-1}}^\ast) \times \A{k}{0\beta}A_0 \\
&\equiv_4 w_{t_0}^\ast \times [1](w_{t_1}^\ast) \times \dots \times [n-2](w_{t_{n-2}}^\ast) \times [n-1](w_{t_{n-1}}^\ast) \times A_0 \A{k}{00\beta} \\
&=w_{t}^\ast \times \A{k}{0\alpha}.
\end{align*}
Since the collection of relations is closed under the shift-map, we can apply the inductive hypothesis when considering any other $i \beta$ ($1 \leq i \leq n-1$).
Although the only case $w_t$ for $i=n-1$ is rewritten slightly differently (since $w_{t_{n-1}}$ appears in the word $w_t$ instead of $w_{t_{n-1}}^\ast$), all can be shown similarly.
This completes the proof.
\end{proof}
\end{lemma}
\section{The Lodha--Moore group and its generalization}\label{section_G0(n)}
\subsection{The Lodha--Moore group}
In this section, we briefly review the original Lodha--Moore group $G_0$.
In some papers, this group is denoted by $G$.
Let $x_0$, $\x{0}{1}$ be the maps defined in Section \ref{subsection_F(n)_definition} for $n=2$.
The group $G_0$ is generated by these two maps and one more generator called $y_{10}$.
To define this generator, we first define the homeomorphism called $y$.
\begin{definition}\label{definiton_y_2}
The map $y$ and its inverse $y^{-1}$ is defined recursively based on the following rule:
\begin{align*}
&y: 2^\mathbb{N} \to 2^\mathbb{N} & &y^{-1}: 2^\mathbb{N} \to 2^\mathbb{N} \\
&y(00\zeta)=0y(\zeta) & &y^{-1}(0\zeta)=00y^{-1}(\zeta) \\
&y(01\zeta)=10y^{-1}(\zeta) & &y^{-1}(10\zeta)=01y(\zeta) \\
&y(1\zeta)=11y(\zeta), & &y^{-1}(11\zeta)=1y^{-1}(\zeta).
\end{align*}
\end{definition}
For each $s$ in $2^{<\mathbb{N}}$, we also define the map $y_s$ by setting
\begin{align*}
y_s(\xi)&=
 \left \{
\begin{array}{cc}
s y(\eta), & \xi=s\eta \\
\xi, & \mbox{otherwise}.
\end{array}
\right.
\end{align*}
We give an example of a calculation of the $y_{001}$ on $00101101 \cdots$:
\begin{align*}
y_{001}(00101101\cdots)&=001y(01101\cdots)\\
&=00110y^{-1}(101\cdots)\\
&=0011001y(1\cdots).
\end{align*}
\begin{definition}
The group $G_0$ is a group generated by $x_0$, $\x{0}{1}$, and $y_{10}$.
\end{definition}
The group $G_0$ is also realized as a group of piecewise projective homeomorphisms.
\begin{proposition}[{\cite[Proposition 3.1]{lodha2016nonamenable}}]\label{proposition_piecewiseprojective}
The group $G_0$ is isomorphic to the group generated by the following three maps of $\mathbb{R}$:
\begin{align*}
a(t)&=t+1,
&
b(t) &= \left \{
\begin{array}{cc}
t & \mbox{if $t \leq 0$} \\
\frac{t}{1-t} & \mbox{if $0 \leq t \leq \frac{1}{2}$} \\
3-\frac{1}{t} & \mbox{if $\frac{1}{2} \leq t \leq 1$} \\
t+1 & \mbox{if $1 \leq t$},
\end{array}
\right.
&
c(t) &= \left \{
\begin{array}{cc}
\frac{2t}{1+t} & \mbox{if $0 \leq t \leq 1$} \\
t & \mbox{otherwise}.
\end{array}
\right.
\end{align*}
\end{proposition}

This proposition is shown by identifying $2^\mathbb{N}$ with $\mathbb{R}$ by the following maps:
\begin{align*}
&\varphi: 2^\mathbb{N} \to [0, \infty] &
&\Phi: 2^\mathbb{N} \to \mathbb{R} \cup \{ \infty \} \\
&\varphi(0\xi)=\frac{1}{1+\frac{1}{\varphi(\xi)}} &
&\Phi(0\xi)=-\varphi(\tilde{\xi}) \\
&\varphi(1\xi)=1+\varphi(\xi) &
&\Phi(1\xi)=\varphi(\xi),
\end{align*}
where $\mathbb{R} \cup \{ \infty \}$ denotes the real projective line.
We note that for every $x \in \mathbb{R} \cup \{ \infty \}$, the inverse image $\Phi^{-1}(\{x\})$ is either a one-point set or a two-point set.
\subsection{$n$-adic Lodha--Moore group} \label{subsection_def_G_0(n)}
In order to define a generalization of the Lodha--Moore group, we first define a homeomorphism on $\N^\mathbb{N}={\{0, \dots, n-1 \}}^\mathbb{N}$ corresponding to $y$ for the case of $n=2$.
We fix $n \geq 2$ and also denote this map by $y$ as in the case of $n=2$.
\begin{definition}
The map $y$ and its inverse map $y^{-1}$ is defined recursively based on the following rule:
\begin{align*}
y: \N^\mathbb{N} &\to \N^\mathbb{N} & y^{-1}: \N^\mathbb{N} &\to \N^\mathbb{N} \\
y(00\zeta)&=0y(\zeta) & y^{-1}(0\zeta)&=00y^{-1}(\zeta) \\
y(01\zeta)&=1\zeta & y^{-1}(1\zeta)&=01\zeta \\
&\;\vdots & &\;\vdots \\
y(0(n-2)\zeta)&=(n-2)\zeta & y^{-1}((n-2)\zeta)&=0(n-2)\zeta \\
y(0(n-1)\zeta)&=(n-1)0y^{-1}(\zeta) & y^{-1}((n-1)0\zeta)&=0(n-1)y(\zeta) \\
y(1\zeta)&=(n-1)1\zeta & y^{-1}((n-1)1\zeta)&=1\zeta \\
&\;\vdots & &\;\vdots \\
y((n-2)\zeta)&=(n-1)(n-2)\zeta & y^{-1}((n-1)(n-2)\zeta)&=(n-2)\zeta \\
y((n-1)\zeta)&=(n-1)(n-1)y(\zeta) & y^{-1}((n-1)(n-1)\zeta)&=(n-1)y^{-1}(\zeta)
\end{align*}
\end{definition}
\begin{remark}
Note that if we restrict the domain to $\{0, n-1\}^\mathbb{N}$, $y$ is exactly the map defined in Definition \ref{definiton_y_2} under the identification of $n-1$ with $1$.
We will use this fact to reduce the discussion to the case of $n=2$.
\end{remark}
For each $s$ in $\N^\mathbb{N}$, define the map $y_s$ by setting
\begin{align} \label{n-adicy_definition}
y_s(\xi)&=
 \left \{
\begin{array}{cc}
s y(\eta), & \xi=s\eta \\
\xi, & \mbox{otherwise}.
\end{array}
\right.
\end{align}
\begin{definition}
We define $G_0(n)$ to be the group generated by the $n+1$ elements $x_0, \dots, x_{n-2}, {x_0}_{[n-1]}$, and $y_{(n-1)0}$.
We call this group the \textit{$n$-adic Lodha--Moore group}.
\end{definition}
For the infinite presentation of $G_0(n)$ described in Section \ref{infinite_presentation}, we introduce an infinite generating set of this group.
Let $i$ in $\{0, \dots, n-2 \}$ and $\alpha$ in $\N^{<\mathbb{N}} \cup \{\epsilon \}$.
We recall that the map $\x{i}{\alpha}$ is defined as follows:
\begin{align*}
\x{i}{\alpha}(\zeta)&=
\begin{cases}
\alpha x_i(\eta) & (\zeta=\alpha \eta) \\
\zeta & (\zeta \neq \alpha \eta).
\end{cases}
\end{align*}
We also recall that the group $F(n)$ is generated by the following infinite set:
\begin{align*}
X(n)=
\left\{ \x{i}{\alpha} \mid i = 0, \dots, n-2, \alpha \in \N^{<\mathbb{N}} \cup \{ \epsilon \} \right\}.
\end{align*}
In addition, let
\begin{align*}
Y(n):=
\left\{ y_\alpha \;\middle|\;
\begin{array}{l} \alpha \in \N^{<\mathbb{N}}, \\
\alpha \neq 0\cdots0, (n-1)\cdots(n-1), \epsilon, \\
\mbox{the sum of each number in  $\alpha$ is equal to $0 \bmod {n-1}$}
\end{array}
\right\}.
\end{align*}
We remark that for $\alpha=\alpha_1 \cdots \alpha_m \in \N^{<\mathbb{N}}$, the actions of $x_0, \dots, x_{n-2}, \x{0}{n-1}$, and $y$ preserve the value of $\alpha_1+\cdots+\alpha_m \bmod {n-1}$.
Then the set $Z(n):=X(n) \cup Y(n)$ also generates the group $G_0(n)$.

\subsection{Infinite presentation and normal form}\label{infinite_presentation}
In this section, we give the unique word with ``good properties'' for each element of $G_0(n)$ (Definition \ref{normal_form}).
In this process, we also give an infinite presentation of $G_0(n)$ (Corollary \ref{G_0(n)_presentation}).
Although almost all result in the rest of this section follows along the lines of the arguments in \cite{lodha2020nonamenable, lodha2016nonamenable}, we write them down for the convenience of the reader.

Let $s $ in $\N^{<\mathbb{N}}$ and $t$ in $\N^{<\mathbb{N}}$ or $\N^\mathbb{N}$.
We write $s \subset t$ if $s$ is a proper prefix of $t$ and write $s \subseteq t$ if $s \subset t$ or $s=t$.
For $s$, $t$, we say that $s$ and $t$ are independent if one of the following holds:
\begin{itemize}
\item $s, t \in \N^{<\mathbb{N}}$ and neither $s \subseteq t$ and $t \subseteq s$ holds.
\item $s \in \N^{<\mathbb{N}}$, $t \in \N^\mathbb{N}$ and $s$ is not any prefixes of $t$.
\item $s, t \in \N^\mathbb{N}$ and $s\neq t$.
\end{itemize}
In all cases, we write $s \perp t$.

Let $s(i), t(i)$ denote the $i$-th number of $s, t$, respectively.
Then we say $s<t$ if one of the following is true:
\begin{itemize}
\item[(a)] $t \subset s$;
\item[(b)] $s \perp t$ and $s(i)<t(i)$, where $i$ is the smallest integer such that $s(i) \neq t(i)$.
\end{itemize}
We note that this order is transitive.
For elements in $\N^\mathbb{N}$, we use the same symbol to denote the lexicographical order.

We claim that the following collection of relations gives a presentation of $G_0(n)$ (Corollary \ref{G_0(n)_presentation}):
\begin{enumerate}
\item the relations of $F(n)$ in Tables \ref{relations_1}, \ref{relations_2}, \ref{relations_3}, and \ref{relations_4};
\item $y_t\x{i}{s}=\x{i}{s}y_{\x{i}{s}(t)}$ for all $i$ and $s, t \in \N^{<\mathbb{N}}$ such that $y_t \in Y(n)$ and $\x{i}{s}(t)$ is defined;
\item $y_s y_t =y_t y_s$ for all $s, t \in \N^{<\mathbb{N}}$ such that $y_s, y_t \in Y(n)$ and $s\perp t$;
\item $y_s=\x{0}{s} y_{s0} y_{s(n-1)0}^{-1}y_{s(n-1)(n-1)}$ for all $s \in \N^{<\mathbb{N}}$ such that $y_s \in Y(n)$.
\end{enumerate}
We note that $\x{0}{n-1}((n-1)0)$ is not defined, for example.
All relations can be verified directly.
We write $R(n)$ for the collection of these relations.

We first define a form, which is easy to compute the composition of maps.
\begin{definition}
A word $\Omega$ on $Z(n)$ is in \textit{standard form} if $\Omega$ is a word such as $f\y{s_1}{t_1}\cdots\y{s_m}{t_m}$ where $f$ is a word on $X(n)$ and $\y{s_1}{t_1}\cdots\y{s_m}{t_m}$ is a word on $Y(n)$ with the condition that $s_i<s_j$ if $i<j$.
\end{definition}
In some cases, it is helpful to use the following weaker form.
\begin{definition}
A word $\Omega$ on $Z(n)$ is in \textit{weak standard form} if $\Omega$ is a word such as $f\y{s_1}{t_1}\cdots\y{s_m}{t_m}$ where $f$ is a word on $X(n)$ and $\y{s_1}{t_1}\cdots\y{s_m}{t_m}$ is a word on $Y(n)$ with the condition that if $s_j \subset s_i$, then $i<j$.
\end{definition}
We can always make a word in weak standard form into one in standard form.
\begin{lemma}[{\cite[Lemma 3.11]{lodha2020nonamenable}} for $n$=2]\label{weakstandard_standard}
We can rewrite a weak standard form into a standard form with the same length by just switching the letters $($i.e., relation $(3)$$)$ finitely many times.
\begin{proof}
Let $f\y{s_1}{t_1}\cdots\y{s_m}{t_m}$ be a word in weak standard form.
We show this by induction on $m$.
It is obvious if $m=0, 1$.
For $m \geq 2$, by the induction hypothesis, we get a word $f\y{s_1^\prime}{t_1^\prime}\cdots\y{s_{m-1}^\prime}{t_{m-1}^\prime}\y{s_m}{t_m}$ where $f\y{s_1^\prime}{t_1^\prime}\cdots\y{s_{m-1}^\prime}{t_{m-1}^\prime}$ is in standard form.
Then one of the following holds:
\begin{enumerate}
\item $s_{m-1}^\prime \supset s_m$;
\item $s_{m-1}^\prime \perp s_m$ and $s_{m-1}^\prime(i)<s_m(i)$, where $i$ is the smallest number such that $s_{m-1}^\prime (i) \neq s_m(i)$;
\item $s_{m-1}^\prime \perp s_m$ and $s_{m-1}^\prime(i)>s_m(i)$, where $i$ is the smallest number such that $s_{m-1}^\prime(i)\neq s_m(i)$.
\end{enumerate}
If (1) or (2), $f\y{s_1^\prime}{t_1^\prime}\cdots\y{s_{m-1}^\prime}{t_{m-1}^\prime}\y{s_m}{t_m}$ is also in standard form.
If (3), by applying the relation (3) to $\y{s_{m-1}^\prime}{t_{m-1}^\prime}\y{s_m}{t_m}$ and using the induction hypothesis for the first $m-1$ characters again, we get a word in standard form.
\end{proof}
\end{lemma}
For an infinite $n$-ary word $w$ in $\N^\mathbb{N}$ and a word $f\y{s_1}{t_1}\cdots\y{s_m}{t_m}$ in weak standard form, we define their \textit{calculation} as follows:
First, we apply $f$ to $w$. Then apply $\y{s_1}{t_1}, \dots, \y{s_m}{t_m}$ to $f(w)$ in this order, where the latter term ``apply'' means to rewrite each $y_{s_i}$ by using its definition in equation \eqref{n-adicy_definition}, and no rewriting can be done by the definition of the map $y$.
\begin{example}
Let $n=4$. For $w=3002\cdots$ and $y_{300}^{-1}y_{30}y_1$, we apply as follows:
\begin{align*}
3002\cdots \xrightarrow{y_{300}^{-1}}300y^{-1}(2\cdots) \xrightarrow{y_{30}} 30(y(0(y^{-1}(2\cdots)))) \xrightarrow{y_1} 30(y(0(y^{-1}(2\cdots)))).
\end{align*}
Therefore the calculation of $w=3002\cdots$ and $y_{300}^{-1}y_{30}y_1$ is $30(y(0(y^{-1}(2\cdots))))$.
\end{example}
We write such the element as $30y0y^{-1}2\cdots$ and sometimes regard it as a word on $\bm{N} \cup \{y, y^{-1}\}$.
We also define calculations for finite words in $\N^{<\mathbb{N}}$ and weak standard forms in the same way, although finite words are not in the domain of $y$.
We note that, unlike infinite words, not all calculations of finite words can be defined.
For example, in $n=4$, the calculation of $w=3002$ and $y_{300}^{-1}y_{30}y_1$ is $30y0y^{-1}2$.
However, that of $w=3$ and $y_{300}^{-1}y_{30}y_1$ is not defined.

We call the operation of rewriting a calculation by using the definition of $y$ once as \textit{substitution} and call the element in $\N^\mathbb{N}$ obtained by repeating the substitution \textit{output string}.
\begin{example}
For $w=0^4(n-1)0^4(n-1)\cdots$ and $y^2$, we have the following substitutions:
\begin{align*}
y^20^4(n-1)0^4(n-1)\cdots \to y 0 y 0^2(n-1)0^4(n-1)\cdots \to y0^2 y(n-1)0^4(n-1)\cdots.
\end{align*}
Their output string is $0(n-1)^40(n-1)^4 \cdots$.
\end{example}
As described in \cite[Section 3.6]{lodha2020nonamenable}, the map $y$ can be expressed as a finite state transducer.
For the sake of simplicity, assume $n\geq3$.
Then our transducer is illustrated in Figure \ref{transducer_y}.
\begin{figure}[tbp]
	\centering
	\includegraphics[width=120mm]{transducer_y.pdf}
	\caption{The transducer of $y$, where $i=1, \dots, n-2$. }
	\label{transducer_y}
\end{figure}
In this setting, each edge is labeled by a form $\sigma \mid \tau$, where $\sigma$ and $\tau$ are in $\N^{<\mathbb{N}}$.
The element $\sigma$ represents the input string, and $\tau$ represents the output string.
The difference from the case of $n=2$ is that there exists an accepting state (double circle mark).
This difference corresponds to the fact that $y$ vanishes in $y \sigma$ by substitutions if $\sigma$ is not in $\{0, n-1 \}^{\mathbb{N}}$ or $\{0, n-1 \}^{<\mathbb{N}}$.

\begin{definition}
A calculation contains a \textit{potential cancellation} if there exists a subword of the form
\begin{align*}
y^{t_1}\sigma y^{t_2}, \sigma \in \N^{<\mathbb{N}}, t_i \in \{1, -1 \}
\end{align*}
such that we get the word $\sigma^\prime y^{-t_2}$ by substituting $y^{t_1}\sigma$ finitely many times.
\end{definition}
\begin{example}
The subwords $y0(n-1)y$ and $y00y^{-1}$ are potential cancellations.
The subword $y0y$ is not a potential cancellation since we can not apply any substitutions.
The subword $y^{-1}(n-1)00(n-1)y$ is also not a potential cancellation since we have $y^{-1}(n-1)00(n-1)y \to 0(n-1)y0(n-1)y \to 0(n-1)(n-1)0y^{-1}y$.
\end{example}
\begin{remark} \label{remark_PCA}
If a calculation contains a potential cancellation, then $\sigma$ is in $\{0, n-1\}^{<\mathbb{N}}$.
Indeed, if not, $y$ vanishes.
\end{remark}
The following holds.
\begin{lemma}[{\cite[Lemma 5.9]{lodha2016nonamenable}} for $n=2$] \label{substitution_NPCA}
Let $\Lambda$ be a calculation that contains no potential cancellations.
Then the word $\Lambda^{\prime}$ obtained by substitution at any $y^{\pm1}$ again contains no potential cancellations.
\begin{proof}
By Remark \ref{remark_PCA}, it is sufficient to consider the case $\sigma \in \{0, n-1\}^{<\mathbb{N}}$.
Then we can identify $\{0, n-1 \}$ with $\{0, 1\}$.
By \cite[Lemma 5.9]{lodha2016nonamenable}, we have the desired result.
\end{proof}
\end{lemma}
We generalize the notion of potential cancellation to the case of weak standard forms.
\begin{definition}
A weak standard form $f\y{s_1}{t_1}\cdots\y{s_m}{t_m}$ has a \textit{potential cancellation} if there exists $w$ in $\N^\mathbb{N}$ such that the calculation of $w$ by $f\y{s_1}{t_1}\cdots\y{s_m}{t_m}$ contains a potential cancellation.
\end{definition}
To construct unique words, we define the moves obtained from the relations of $G_0(n)$.
\begin{definition} \label{definition_five_moves}
We assume that every map in the following is defined.
\begin{description}
\item[Rearranging move]
$\y{t}{i}\x{j}{s}^{\pm1} \to \x{j}{s}^{\pm1}\y{\x{j}{s}^{\pm1}(t)}{i}$;
\item[Expansion move]
$y_s \to \x{0}{s}y_{s0}\y{s(n-1)0}{-1}y_{s(n-1)(n-1)}$ and $y_s^{-1} \to \x{0}{s}^{-1}\y{s00}{-1}y_{s0(n-1)}\y{s(n-1)}{-1}$;
\item[Commuting move] $y_uy_v\leftrightarrow y_v y_u$ (if $u \perp v$);
\item[Cancellation move] $\y{s}{\pm i}\y{s}{\mp i} \to \epsilon$;
\item[ER moves]
\begin{enumerate}
\item $f(\y{s_1}{t_1}\cdots\y{s_k}{t_k})y_u \to f\x{0}{u}(\y{\x{0}{u}(s_1)}{t_1}\cdots \y{\x{0}{u}(s_k)}{t_k})(y_{u0}\y{u(n-1)0}{-1}y_{u(n-1)(n-1)})$
\item $f(\y{s_1}{t_1}\cdots\y{s_k}{t_k})\y{u}{-1} \to f\x{0}{u}^{-1}(\y{\x{0}{u}^{-1}(s_1)}{t_1}\cdots \y{\x{0}{u}^{-1}(s_k)}{t_k})(\y{u00}{-1}y_{u0(n-1)}\y{u(n-1)}{-1})$.
\end{enumerate}
\end{description}
ER moves are a combination of an expansion move and rearranging moves.
\end{definition}
By the definition of potential cancellations, we note that if we apply ER move to a standard form that contains a potential cancellation, then either the resulting word also contains a potential cancellation, or we can apply a cancellation move to the resulting word.
Like the Thompson group $F$, each element of $G_0(n)$ can be represented by tree diagrams.
Then the expansion move for $y_s$ is the replacement from the left diagram to the right diagram in Figure \ref{image_expansion_move}.
\begin{figure}[tbp]
	\centering
	\includegraphics[width=110mm]{image_expansion_move.pdf}
	\caption{The black and white circles represent that we apply $y$ and $y^{-1}$ to the words corresponding to the edges below them, respectively. }
	\label{image_expansion_move}
\end{figure}
See \cite[Section 4]{lodha2016nonamenable} for details of the case $n=2$.

We introduce the following notion, which makes it easy to check whether the moves are defined.
\begin{definition}
For $s$ in $\N^{<\mathbb{N}}$, we write $\| s\|$ for the length of $s$.
We define the \textit{depth} of a weak standard form $f\y{s_1}{t_1}\cdots\y{s_m}{t_m}$ to be the integer $\min_{1 \leq i \leq m} \|s_i \|$.
If a weak standard form is on $X(n)$, we define its depth to be $\infty$.
\end{definition}
We show that an arbitrary word on $Z(n)$ can be rewritten in standard form.
In order to do this, we prepare two lemmas.
\begin{lemma}[{\cite[Lemma 5.2]{lodha2016nonamenable}} for $n=2$] \label{standard_base_case}
Let $y_s$ in $Y(n)$ and $l$ in $\mathbb{N}\cup\{0\}$.
Then there exists a standard form $\Omega$ obtained by applying expansion move and rearranging move on $y_s^{\pm1}$ finitely many times that satisfies the following:
\begin{enumerate}[font=\normalfont]
\item If there exists $\x{i}{u}^t$ in $\Omega$, then $s \subset u$ holds.
\item If there exists $y_u$ or $\y{u}{-1}$, then it is the only one, and $s \subseteq u$ and $\|u \| \geq l$ hold.
\item If there exist $y_u, y_v$ with $u \neq v$, then $u \perp v$ holds.
\end{enumerate}
\begin{proof}
We show this by induction on $l-\|s\|$.
If $l-\|s\| \leq 0$, then $\Omega=y_s^{\pm1}$ satisfies all the conditions.

We assume that $l-\|s\|>0$ holds.
Because we can discuss $y_s^{-1}$ in almost the same way, we consider only $y_s$.
We apply expansion move on $y_s$ and get the word $\x{0}{s}y_{s0}\y{s(n-1)0}{-1}y_{s(n-1)(n-1)}$.
Since we have
\begin{align*}
\max \{ l-\|s0\|,~ l-\|s(n-1)0\|,~ l-\|s(n-1)(n-1)\| \} \leq l-\|s\|,
\end{align*}
by the induction hypothesis, there exists three standard forms $\Omega_{s0}, \Omega_{s(n-1)0}, \Omega_{s(n-1)(n-1)}$ for $y_{s0}$, $y_{s(n-1)}^{-1}$, $y_{s(n-1)(n-1)}$, respectively, all of which satisfy the conditions.
Thus the word is rewritten from $\x{0}{s}y_{s0}\y{s(n-1)0}{-1}y_{s(n-1)(n-1)}$ to $\x{0}{s}\Omega_{s0}\Omega_{s(n-1)0}\Omega_{s(n-1)(n-1)}$.

If there exists $\x{i}{u}^t$ in $\Omega_{s(n-1)0}$, then $u \supset s(n-1)0$ holds.
If there exists $\y{v}{\pm1}$ in $\Omega_{s0}$, then $v \supseteq s0$ holds.
Since $u \perp v$, $\x{i}{u}^t(v)$ is defined and equals to $v$.
Thus, we can apply rearranging moves on this $\x{i}{u}^t$.
A similar process can be done for the $\x{i}{u}^t$ in $\Omega_{s(n-1)(n-1)}$.
This completes the proof.
\end{proof}
\end{lemma}
\begin{lemma}[{\cite[Lemma 5.3]{lodha2016nonamenable}} for $n=2$] \label{standard_depth_lemma}
Let $\Lambda$ be a word on $X(n)$, and $k$ be the length of $\Lambda$.
Then there exists $l_0$ in $\mathbb{N}$ such that the following holds:
If $\Omega$ is a standard form with depth $l \geq l_0$, then we have a standard form $\Omega^\prime$ obtained from $\Omega \Lambda$ by rearranging moves, with the depth of $\Omega^\prime$ being at least $l-k$.
\begin{proof}
We show this by induction on the length of $\Lambda$.
For the base case, let $\Lambda=\x{i}{s}^{\pm1}$.
If $\x{i}{s}^{\pm1}(t)$ is not defined, then $t=si$ holds.
This means that if $t \in \N^{<\mathbb{N}}$ with $\|t \| \geq \|s\|+2$, then $\x{i}{s}(t)$ is defined.
Hence let $l_0=\|s\|+2$.
We note that
\begin{align*}
\|t\|-1 \leq \|\x{i}{s}^{\pm1}(t)\| \leq \| t\|+1
\end{align*}
holds.
Then we can apply rearranging moves on $\x{i}{s}^{\pm1}$ with every $y_t$ in $\Omega$.
The depth of the resulting standard form $\Omega^\prime$ is at least $l-1$.

We assume that the claim holds for $k-1$.
Let $\Lambda=\Lambda^\prime\x{i}{s}^{\pm1}$.
By the induction hypothesis, there exists $l_0^\prime$ for $\Lambda^\prime$ such that we can apply moves $\Omega\Lambda^\prime\x{i}{s}^{\pm1}\to\Omega^{\prime \prime}\x{i}{s}^{\pm1}$ where the depth of $\Omega$ is $l \geq l_0^\prime$ and that of $\Omega^{\prime \prime}$ is at least $l-(k-1)$.
Then, let $l_0=\max\{l_0^\prime, k+\|s\|+1\}$ and assume that the depth of $\Omega$ is $l \geq l_0$.
Since we have $l-(k-1) \geq \|s\|+2$, for the same reason as in the base case, we can apply rearranging moves and obtain a standard form $\Omega^\prime$ whose depth is at least $l-(k-1)-1=l-k$.
\end{proof}
\end{lemma}
Now we make arbitrary words into standard forms.
\begin{proposition}[{\cite[Lemma 5.4]{lodha2016nonamenable}} for $n=2$] \label{Prop_standard_forms_from_words}
Let $l$ be a natural number and $w$ be a word on $Z(n)$.
Then we can rewrite $w$ into a standard form whose depth is at least $l$ by moves.
\begin{proof}
If $w$ is in $X(n)$, there is nothing to do since the depth of $w$ is $\infty$.
Hence we assume that $w$ is not in $X(n)$.

We show the claim by induction on the length $m$ of $w$.
For the base case, since $w$ is not in $X(n)$, we have $w=\y{s}{\pm1}$.
By Lemma \ref{standard_base_case}, the claim holds.

Let $m>1$.
By dividing $a^{\pm(k+1)}$ in $w$ as $a^{\pm k}a^{\pm1}$ if necessary, we can decompose $w$ into two positive length words $\Omega_0\Omega_1$.
We assume that $\Omega_1$ is not on $X(n)$.
By the induction hypothesis, we obtain a standard form $\Lambda\Upsilon$ from $\Omega_1$, where $\Lambda$ is on $X(n)$, and $\Upsilon$ is on $Y(n)$, with the depth being at least $l$.

Let $k$ be the length of $\Lambda$ and $r=\max\{\|u\| \mid \mbox{$\y{u}{i}$ belongs to $\Upsilon$} \}+1$.
By the induction hypothesis, there exists a standard form $\Omega_0^\prime$ obtained from $\Omega_0$, with the depth being at least $\max\{l_0, r+k\}$, where $l_0$ is a natural number for $\Lambda$ in Lemma \ref{standard_depth_lemma}.
Then we rewrite $\Omega_0^\prime\Lambda \to \Omega_0^{\prime\prime}$ where $\Omega_0^{\prime\prime}$ is in standard form with the depth being at least $r$.

The form $\Omega_0^{\prime\prime}\Upsilon$ obtained through the previous transformations $w \to \Omega_0\Omega_1 \to \Omega_0^\prime\Lambda\Upsilon \to \Omega_0^{\prime\prime}\Upsilon$ is in weak standard form.
Indeed, since $\Omega_0^{\prime\prime}$ and $\Upsilon$ (on $Y(n)$) are in standard form, by the definition of $r$, the claim holds.
By Lemma \ref{weakstandard_standard}, we get the standard form without changing the depth.

If $\Omega_1$ is in $X(n)$, then we set $\Lambda=\Omega$, $\Upsilon=\epsilon$, and $r=l$.
Then we can apply the above argument.
\end{proof}
\end{proposition}
Next, we list three lemmas about ER moves.
These are useful for getting words in standard form without potential cancellations.
\begin{lemma}[{\cite[Lemma 3.21]{lodha2020nonamenable}} for $n$=2]\label{ER_move_preserve_wsf}
Let $f(\y{s_1}{t_1}\cdots\y{s_k}{t_k})y_u^{\pm1}(\y{p_1}{q_1}\cdots\y{p_m}{q_m})$ be in weak standard form.
Then the word obtained from $f(\y{s_1}{t_1}\cdots\y{s_k}{t_k})y_u^{\pm1}(\y{p_1}{q_1}\cdots\y{p_m}{q_m})$ by the ER move on $y_u^{\pm1}$ is also in weak standard form.
\begin{sproof}
We show that only the case $y_u$.
Let
\begin{align*}
f\x{0}{u}(\y{\x{0}{u}(s_1)}{t_1}\cdots \y{\x{0}{u}(s_k)}{t_k})(y_{u0}\y{u(n-1)0}{-1}y_{u(n-1)(n-1)})(\y{p_1}{q_1}\cdots\y{p_m}{q_m})
\end{align*}
be the word obtained from the given word by ER move.
We show this by considering whether each $s_i$ is independent with $u$ or not.
If $s_i \perp u$ holds, then $\x{0}{u}(s_i)=s_i$, and there is nothing to do.
If $s_i \supset u$ holds, since $\x{0}{u}(s_i)$ is defined, $\x{0}{u}(s_i) \supset u$ holds, and it can never be a proper prefix of $u0$, $u(n-1)0$, or $u(n-1)(n-1)$. Since $p_j$ satisfies either $p_j\perp u$ or $p_j \subseteq u$, the claim holds in both cases.
\end{sproof}
\end{lemma}
\begin{lemma}[{\cite[Lemma 3.22]{lodha2020nonamenable}} for $n$=2]\label{lem_ERmove_preserve_NPCA}
Let $f \lambda_1$ be in weak standard form with no potential cancellations and
$g\lambda_2$ be a word obtained from $f \lambda_1$ by an ER move.
Then $g\lambda_2$ is in weak standard form with no potential cancellations.
\begin{proof}
Let $\lambda_1=\y{s_1}{t_1}\cdots\y{s_m}{t_m}$ with every $t_i$ in $\{\pm1\}$.
We apply ER move on $\y{u_i}{t_i}$.
Since $f$ does not affect the existence of potential cancellations, we assume that $f$ is the empty word.
We only consider the case $t_i=1$ since the cases $t_i=1$ and $t_i=-1$ are shown similarly.
Let $s_j^{\prime}:=\x{0}{s_i}(s_j)$ $(1 \leq j \leq i-1)$.
By the definition of ER move, we have
\begin{align*}
g\lambda_2=\x{0}{s_i}(\y{s_1^\prime}{t_1}\cdots\y{s_{i-1}^\prime}{t_{i-1}})(y_{s_i0}\y{s_i(n-1)0}{-1}y_{s_i(n-1)(n-1)})(\y{s_{i+1}}{t_{i+1}}\cdots\y{s_m}{t_m}).
\end{align*}

Since it is clear by Lemma \ref{ER_move_preserve_wsf} that $g\lambda_2$ is in weak standard form, suppose that $g\lambda_2$ has a potential cancellation.
Let $\tau \in \N^\mathbb{N}$ be an element such that the calculation $\Lambda$ of $\tau$ by $(\y{s_1^\prime}{t_1}\cdots\y{s_{i-1}^\prime}{t_{i-1}})(y_{s_i0}\y{s_i(n-1)0}{-1}y_{s_i(n-1)(n-1)})(\y{s_{i+1}}{t_{i+1}}\cdots\y{s_m}{t_m})$ contains a potential cancellation.
Then, $s_i$ is a prefix of $\tau$.
Indeed, we have the following:
\begin{enumerate}
\item if $s_j \perp s_i$, then $\y{s_j^\prime}{t_j}=\y{s_j}{t_j}$;
\item if $s_j \perp s_i$, then $\y{s_j}{t_j}(\tau^\prime)\perp s_i$ where $\tau^\prime$ is in $\N^\mathbb{N}$ with $\tau^\prime \perp s_i$;
\item if $s_j \supset s_i$, since $\x{0}{s_i}(s_j) \supset s_i$, we have $\y{s_j^\prime}{t_j}(\tau^\prime)=\y{\x{0}{s_i}(s_j)}{t_j}(\tau^\prime)=\tau^\prime$ and $y_{s_j}(\tau^\prime)=\tau^\prime$ where $\tau^\prime$ is in $\N^\mathbb{N}$ with $\tau^\prime \perp s_i$;
\item $y_{s_i}$, $y_{s_i0}$, $y_{s_i(n-1)0}^{-1}$, and $y_{s_i(n-1)(n-1)}$ all fix $\tau^\prime$ where $\tau^\prime$ is in $\N^\mathbb{N}$ with $\tau^\prime \perp s_i$.
\end{enumerate}
Since ER move is defined, each $y_j$ satisfies either $s_j\perp s_i$ or $s_j \supset s_i$.
If $s_i$ is not a prefix of $\tau$, $\tau \perp s_i$ holds.
Then the calculations of $\tau$ with $f\lambda_1$ and with $g\lambda_2$ are the same.
This contradicts the assumption that $f\lambda_1$ does not have a potential cancellation. Hence $s_i$ is a prefix of $\tau$.

Let $\Lambda^\prime$ be the calculation of $\x{0}{s_i}^{-1}(\tau)$ by $\y{s_1}{t_1}\cdots\y{s_m}{t_m}$.
By the assumption, this calculation does not contain a potential cancellation.
On the other hand, $\Lambda$ is obtained from $\Lambda^\prime$ by substituting once.
However, this also contradicts Lemma \ref{substitution_NPCA}.
\end{proof}
\end{lemma}
\begin{lemma} \label{lem_depth_ER_moves}
Let $l$ be an arbitrary natural number.
Then, by applying ER moves to the weak standard form $f\y{s_1}{t_1}\cdots\y{s_m}{t_m}$ finitely many times, we obtain a weak standard form whose depth is at least $l$.
\begin{sproof}
If $\|s_i\|< l$ holds, we apply ER moves on $\y{s_i}{t_i}$ (if $t_i \neq \pm1$, we apply on the last one).
We note that the move may not be defined.
In that case, before we do that, we apply ER moves to $y_{s_j}$ $(j<i)$ which is the cause that the ER move on $y_{s_i}$ is not defined.
If it is also not defined, repeat the process.
Since we can always apply ER moves on $fy_{s_1}$, this process is finished.
We do this process repeatedly until the depth of the obtained word is at least $l$.
By Lemma \ref{ER_move_preserve_wsf}, it is also in weak standard form.
\end{sproof}
\end{lemma}

Since subwords play an essential role in the notion of potential cancellations, we introduce the following definitions for simplicity.
\begin{definition}
Let $f\y{s_1}{t_1}\cdots\y{s_m}{t_m}$ be in weak standard form.
We say the pair $(\y{s_j}{t_j}, \y{s_i}{t_i})$ is \textit{adjacent} if the following two conditions hold:
\begin{enumerate}
\item $s_i \subset s_j$,
\item if $u$ in $\N^{<\mathbb{N}}$ satisfies $s_i \subset u \subset s_j$, then $u \notin \{ s_1, \dots, s_m\}$.
\end{enumerate}
\end{definition}
We define an adjacent pair $(\y{s_j}{t_j}, \y{s_i}{t_i})$ is a \textit{potential cancellation} if $y^{t_i}\sigma y^{t_j}$ is a potential cancellation, where $\sigma$ is the word satisfying $s_i\sigma=s_j$.
It is clear from the definition that a weak standard form contains a potential cancellation if and only if there exists an adjacent pair such that it is a potential cancellation.

By the definition, for example, if $(\y{j}{t}, y_s)$ is a potential cancellation, then either $\y{j}{t}=\y{s00}{-1}$, $y_{0(n-1)}$, or $\y{s(n-1)}{-1}$ holds, or $(\y{\x{0}{s}(j)}{t}, y_{s0})$, $(\y{\x{0}{s}(j)}{t}, \y{s(n-1)0}{-1})$, or $(\y{\x{0}{s}(j)}{t}, y_{s(n-1)(n-1)})$ is also a potential cancellation.
As mentioned in Remark \ref{remark_PCA}, we note that if an adjacent pair is a potential cancellation, then $\sigma$ is in $\{0, n-1\}^{<\mathbb{N}}$.

We introduce the moves which are the ``inverse'' of ER moves.
First, we define the conditions where the moves are defined.
\begin{definition}\label{def_potential_contraction}
We say that a weak standard form contains a \textit{potential contraction} if it satisfies either of the following:
\begin{enumerate}
\item there exists a subword $y_{s0}\y{s(n-1)0}{-1}y_{s(n-1)(n-1)}$, but there does not exist $\y{s(n-1)}{\pm1}$;
\item there exists a subword $\y{s00}{-1}y_{s0(n-1)}\y{s(n-1)}{-1}$, but there does not exist $\y{s0}{\pm1}$.
\end{enumerate}
We also call the same when the condition above is satisfied by commuting moves.
\end{definition}
\begin{example} \label{example_potential_contraction}
Let $n=4$. A weak standard form $y_{300}y_{3030}^{-1}y_{3031}y_{3033}y_{30}$ contains a potential contraction.
\end{example}
We now define moves.
\begin{definition}\label{def_contraction_move}
Let $f\lambda$ be in standard form.
We assume that this form contains a potential contraction in the sense of (1).
We apply commuting move to all $y_u^{v}$ except the subword in the standard form satisfying $s0<u \leq s(n-1)(n-1)$ to the left of $y_{s0}$ while preserving the order, and we replace $y_{s0}\y{s(n-1)0}{-1}y_{s(n-1)(n-1)}$ with $\x{0}{s}^{-1}y_s$.
Then, move $\x{0}{s}^{-1}$ to just next to $f$ by rearranging moves.
Finally, we apply cancellation moves or moves $x^ax^b \to x^{a+b}$ if necessary.
The rearranging moves are defined, and the word obtained by the above sequence of moves is in standard form.
We call this sequence of moves a \textit{contraction move}.
We also define the contraction move for (2) in the same way.
\end{definition}
The moves and the claim in this definition are justified by the following lemma.
We describe only case (1), but the similarity holds for case (2).
\begin{lemma}\label{lem_contraction_move}
In Definition $\ref{def_contraction_move}$, the rearranging moves are defined, and the resulting word is again in standard form.
\begin{proof}
By the definition of standard form, for $y_u^v$, we have either $s0\perp u$ or $s0 \subseteq u$.
In the former case, we have either $s \perp u$ or $s \subset u$.
In the latter case, we have $s \subset u$.
By the definition of potential contraction, there does not exist $\y{s(n-1)}{\pm1}$.
Hence $\x{0}{s}^{-1}(u)$ is defined, and the rearranging moves can be done.

Suppose that we have finished applying commuting move.
Let $s_j^\prime:=\x{0}{s}^{-1}(s_j)$, and $\x{0}{s}^{-1}(\y{s_1^\prime}{t_1}\cdots\y{s_k^\prime}{t_k})y_s(\y{p_1}{q_1}\cdots\y{p_m}{q_m})$ be the resulting word.
If $s_k^\prime=s_k$, then $s_k\perp s$ holds. Since the original word is in standard form, we have $s_k^\prime <s$.

If $s_k^\prime\neq s_k$, then $s_k^\prime \supset s$.
By the transitivity of this order, $\x{0}{s}^{-1}(\y{s_1^\prime}{t_1}\cdots\y{s_k^\prime}{t_k})y_s$ is in standard form.

We show that $y_s(\y{p_1}{q_1}\cdots\y{p_m}{q_m})$ is in standard form.
We note that $s(n-1)(n-1)<p_1$ holds.
If $s(n-1)(n-1) \supset p_1$, either $s=p_1$ or $s \supset p_1$ holds.
In the former case, by $y_s\y{p_1}{t_1}\to\y{s}{t_1+1}$ or $\to \epsilon$, we get the standard form.
In the latter case, we have $s<p_1$.
If $s(n-1)(n-1) \perp p_1$, since $(n-1)$ is the largest number, we have $s<p_1$.
By the transitivity, $\x{0}{s}^{-1}(\y{s_1^\prime}{t_1}\cdots\y{s_k^\prime}{t_k})y_s(\y{p_1}{q_1}\cdots\y{p_m}{q_m})$ is in standard form.
\end{proof}
\begin{example}
In example \ref{example_potential_contraction}, we obtain $\x{0}{30}^{-1}y_{301}\y{30}{2}$ by applying a contraction move.
\end{example}
\end{lemma}
Contraction moves have the following property.
\begin{lemma}\label{lem_contraction_move_preserve_NPCA}
Let $f\y{s_1}{t_1}\cdots\y{s_m}{t_m}$ be in standard form which contains a potential contraction and no potential cancellations.
Then the word obtained by contraction move contains no potential cancellations.
\begin{proof}
We show only the case of potential contraction condition (1).
We assume that we have produced a potential cancellation after applying a contraction move.
If there exists such an adjacent pair, the pair must be $(\y{s}{k}, \y{s^\prime}{k^\prime})$, where $s^\prime$ satisfies $s^\prime \subset s$.
Indeed, if not, each adjacent pair is either $(\y{u_1}{j_1}, \y{u_2}{j_2})$ where $u_1, u_2 \neq s$, or $(y_{s^{\prime\prime}}^{k^{\prime\prime}}, y_s^k)$ where $s^{\prime\prime}$ satisfies $s^{\prime\prime} \supset s$.
In the former case, since either $\x{0}{s}(u_i)\neq u_i$ ($i=1, 2$) or $\x{0}{s}(u_i)= u_i$ ($i=1, 2$) holds, this contradicts the assumption that $f\y{s_1}{t_1}\cdots\y{s_m}{t_m}$ contains no potential cancellations.
In the latter case, we apply ER move on $\y{s}{k}$.
Since the contraction move is the inverse of the ER move, if $y_{s^{\prime\prime}}^{k^{\prime\prime}}$ is $\y{s0}{-1}$, $y_{s(n-1)0}$, or $\y{s(n-1)(n-1)}{-1}$, then it contradicts the assumption that $f\y{s_1}{t_1}\cdots\y{s_m}{t_m}$ is in standard form.
If $y_{s^{\prime\prime}}^{k^{\prime\prime}}$ is none of them, then it contradicts that $f\y{s_1}{t_1}\cdots\y{s_m}{t_m}$ contains no potential cancellations.

There does not exist $\y{s}{k_1}$ in $f\y{s_1}{t_1}\cdots\y{s_m}{t_m}$.
Indeed, if there exists $y_s^{k_1}$ ($k_1<0$), since there exists $y_{s0}$ by the assumption of potential contraction, $f\y{s_1}{t_1}\cdots\y{s_m}{t_m}$ contains a potential cancellation.
If there exists $\y{s}{k_1}$ ($k_1>0$), since $(\y{s}{k}, \y{s^\prime}{k^\prime})$ is a potential cancellation in the word after applying the contraction move, the adjacent pair $(y_s^{k_1}, \y{s^\prime}{k^\prime})$ is a potential cancellation in $f\y{s_1}{t_1}\cdots\y{s_m}{t_m}$.
This contradicts the assumption about no potential cancellations.

Now we consider the adjacent pair $(y_s, y_{s^\prime}^{k^\prime})$.
Let $\sigma$ be the finite word such that $s^\prime\sigma=s$ holds.
By the assumption, $y\sigma y$ or $y^{-1}\sigma y$ is potential cancellation.
In the former case, by substitutions, we have $y\sigma=\sigma^\prime y^{-1}$.
Then, we have $y\sigma 0y=\sigma^\prime y^{-1}0y=\sigma^\prime00y^{-1}y$.
This contradicts the assumption that the standard form contains no potential cancellations.
Similarly, in the latter case, we have $y^{-1}\sigma=\sigma^\prime y^{-1}$, which also contradicts the assumption.
\end{proof}
\end{lemma}
Our goal in the rest of this section is to give the unique word for each element in $G_0(n)$, which satisfies the following:
\begin{definition} \label{normal_form}
Let $f\y{s_1}{t_1}\cdots\y{s_m}{t_m}$ be in standard form with no potential cancellations, no potential contractions, and $f$ is in the normal form in the sense of $F(n)$.
Then we say that $f\y{s_1}{t_1}\cdots\y{s_m}{t_m}$ is in the \textit{normal form}.
\end{definition}
See Section \ref{subsection_F(n)_definition} for the definition of the normal form of elements in $F(n)$.

We obtain a normal form from an arbitrary word on $Z(n)$ through the following four steps:
\begin{description}
\item[Step 1] Convert an arbitrary word into a standard form (Proposition \ref{Prop_standard_forms_from_words});
\item[Step 2] Convert a standard form into a standard form that contains no potential cancellations;
\item[Step 3]Convert a standard form that contains no potential cancellations into a standard form $f\y{s_1}{t_1}\cdots\y{s_m}{t_m}$ which contains no potential cancellations and no potential contractions;
\item[Step 4]Convert $f$ into $g$, where $g$ is the unique normal form in $F(n)$ (Section \ref{subsection_F(n)_definition}).
\end{description}
The remaining steps are only 2 and 3.
\begin{lemma}[{\cite[Lemma 4.5]{lodha2020nonamenable}} for $n$=2]\label{lem_for_step2}
Let $(\y{s_1}{t_1}\cdots\y{s_m}{t_m})\y{s}{t}$ be in standard form such that the following hold:
\begin{enumerate}[font=\normalfont]
\item $t_1, \dots, t_m, t \in \{1, -1\}$;
\item any two $s_i, s_j$ are independent;
\item for any $\y{s_i}{t_i}$, $\y{s_i}{t_i}$ and $\y{s}{t}$ is an adjacent pair with being potential cancellation.
\end{enumerate}
Then by applying ER moves and cancellation moves, we obtain a standard form $f\y{u_1}{v_1}\cdots\y{u_k}{v_k}$ such that any two $u_i, u_j$ are independent and $v_1, \dots, v_k \in \{1, -1\}$.
\begin{proof}
We consider only the case $t=1$.
We claim that $\x{0}{s}(s_i)$ is defined for every $i$.
Indeed, for $s_i$ such that $s \subset s_i$, the only case where $\x{0}{s}(s_i)$ is not defined is the case $s_i=s0$.
Then since the adjacent pair $(\y{s0}{t_i}, y_s)$ is not potential cancellation whether $t_i$ is $1$ or $-1$, this contradicts the assumption $(3)$.

Let $s_i^\prime=\x{0}{s}(s_i)$.
By applying ER move to $y_s$, we have
\begin{align*}
(\y{s_1}{t_1}\cdots\y{s_m}{t_m})y_s=\x{0}{s}(\y{s_1^\prime}{t_1}\cdots\y{s_m^\prime}{t_m})y_{s0}\y{s(n-1)0}{-1}y_{s(n-1)(n-1)}.
\end{align*}
We note that $s0$, $s(n-1)0$, and $s(n-1)(n-1)$ are independent of each other.
For every $\y{s_i^\prime}{t_i}$, one of $y_{s0}$, $\y{s(n-1)0}{-1}$, or $y_{s(n-1)(n-1)}$ corresponds to it, either as the inverse or as an adjacent pair.
Let $\y{\sigma}{\tau}$ be the corresponding one.
If it is the inverse, we apply a cancellation move.
If it forms an adjacent pair, the distance in an $n$-ary tree between $s_i^\prime$ and $\sigma$ is smaller than the distance between $s_i$ and $s$.
Thus, by iterating this process, we obtain the desired result.
\end{proof}
\end{lemma}
The following lemma completes step 2.
\begin{lemma}[{\cite[Lemma 4.6]{lodha2020nonamenable}} for $n$=2]\label{lem_step2}
By applying moves, any weak standard form can be rewritten into a standard form that contains no potential cancellations.
\begin{proof}
Let $f\y{s_1}{t_1}\cdots\y{s_m}{t_m}$ $(t_i \in \{-1, 1 \})$ be a weak standard form.
We show by induction on $m$.
If $m\leq1$, since there exists no adjacent pair, it is clear.

We assume that $m>1$ holds.
By applying the inductive hypothesis, Lemmas \ref{lem_ERmove_preserve_NPCA}, and \ref{lem_depth_ER_moves} to $f\y{s_1}{t_1}\cdots\y{s_{m-1}}{t_{m-1}}$, we obtain a weak standard form $g\y{p_1}{q_1}\cdots \y{p_k}{q_k}$ with at least depth $\|s_m\|+1$ and without potential cancellations.
Since $p_i \subset s_m$ does not hold by its depth, $g\y{p_1}{q_1}\cdots \y{p_k}{q_k}\y{s_m}{t_m}$ is in weak standard form.

If there exists an adjacent pair that is a potential cancellation, it is only of the form $(\y{p_i}{q_i}, \y{s_m}{t_m})$.
We record all such $\y{p_i}{q_i}$.
By applying commuting moves, we obtain a weak standard form $h(\y{v_1}{k_1}\cdots\y{v_l}{k_l})(\y{u_1}{v_1}\cdots\y{u_o}{v_o})\y{s_m}{t_m}$ which satisfies the following:
\begin{enumerate}
\item $v_1, \dots, v_o \in \{ -1, 1\}$;
\item For $j=1, \dots, o$, $\y{u_j}{v_j}$ and $\y{s_m}{t_m}$ is an adjacent pair which is a potential cancellation;
\item All other adjacent pairs are not potential cancellations.
\end{enumerate}
Indeed, since $s_m$ is the shortest word, each adjacent element is the ``second shortest.''
Hence we can apply commuting moves to get a word.

We again note that Lemmas \ref{lem_ERmove_preserve_NPCA} and \ref{lem_depth_ER_moves}, i.e., we can increase the depth by ER moves while preserving in weak standard form without potential cancellations.
We apply ER moves to $h(\y{v_1}{k_1}\cdots\y{v_l}{k_l})(\y{u_1}{v_1}\cdots\y{u_o}{v_o})\y{s_m}{t_m}$ in two phases.
First, we apply ER moves to the part of $h(\y{v_1}{k_1}\cdots\y{v_l}{k_l})$ so that we can apply ER moves to the word on $X(n)$ that appears when we apply Lemma \ref{lem_for_step2} to $(\y{u_1}{v_1}\cdots\y{u_o}{v_o})\y{s_m}{t_m}$.
Secondly, we apply Lemma \ref{lem_for_step2} to $(\y{u_1}{v_1}\cdots\y{u_o}{v_o})\y{s_m}{t_m}$.
By the same argument in Lemma \ref{lem_ERmove_preserve_NPCA}, no new potential cancellations are produced in this process.

Finally, by Lemma \ref{weakstandard_standard}, we have the desired result.
\end{proof}
\end{lemma}
The following lemma completes step 3.
\begin{lemma}[{\cite[Lemma 4.7]{lodha2020nonamenable}} for $n$=2]\label{lem_step3}
Any standard form which contains no potential cancellations can be rewritten into a standard form that contains no potential cancellations and no potential contractions.
\begin{proof}
We apply contraction moves repeatedly.
By Lemmas \ref{lem_contraction_move} and \ref{lem_contraction_move_preserve_NPCA}, the applied word is again in standard form and contains no potential cancellations.
Since this move makes the word on $Y(n)$ in the standard form strictly shorter, the process finishes.
\end{proof}
\end{lemma}
The only thing that remains to be proved is the uniqueness of the normal form.
We will show this by contradiction.
In order to do this, we define an ``invariant'' of the forms.
\begin{definition}
A calculation that contains no potential cancellations has \textit{exponent $m$} if $m$ is the number of $y^{\pm1}$ satisfying the condition that the number appearing after it is only $0$, $(n-1)$, or $y^{\pm1}$.
\end{definition}
\begin{example}
Let $n=4$.
The string $y0y02y^{-1}(n-1)y00\cdots$ has exponent $2$.
The string $y100\cdots$ has exponent $0$.
\end{example}
For the proof of uniqueness, we prepare two lemmas.
\begin{lemma}[{\cite[Lemma 5.10]{lodha2016nonamenable}} for $n=2$]\label{lem_exponent_1}
Let $\Lambda$ be a finite word on $\bm{N} \cup \{y, y^{-1}\}$ that contains no potential cancellations.
Let $m$ be the exponent of $\Lambda$.
Then there exists a finite word $u$ on $\{0, n-1\}$ and $v$ in $\N^{<\mathbb{N}}$ such that $\Lambda u$ can be rewritten into $vy^m$ by substitutions.
\begin{proof}
By substitutions, we rewrite $\Lambda$ into $v^\prime\Lambda^\prime$, where $v^\prime$ is in $\N^{<\mathbb{N}}$ and $\Lambda^\prime$ is in $\{0, (n-1), y, y^{-1}\}$.
By the definition of exponent, $\Lambda^\prime$ also has exponent $m$, and by Lemma \ref{substitution_NPCA}, this also contains no potential cancellations.
Hence, by identifying $\{0, n-1\}$ with $\{0, 1\}$, the claim comes down to the case $n=2$.
\end{proof}
\end{lemma}
For $u, s$ in $\N^{<\mathbb{N}}$, we say that \textit{$u$ dominates $s$} if the condition $u \perp s$ or $u \supset s$ is satisfied.
\begin{lemma}[{\cite[Lemma 4.8]{lodha2020nonamenable}} for $n=2$]\label{lem_exponent_2}
Let $f\y{s_1}{t_1}\cdots\y{s_l}{t_l}$ and $g\y{p_1}{q_1}\cdots\y{p_m}{q_m}$ be in standard form which represent the same element in $G_0(n)$.
Let $u$ be in $\N^{<\mathbb{N}}$ such that the following hold:
\begin{enumerate}[font=\normalfont]
\item $f(u)=:u_1$ and $g(u)=:u_2$ are defined;
\item $u_1$ dominates $s_1, \dots, s_l$;
\item $u_2$ dominates $p_1, \dots, p_m$.
\end{enumerate}
Let $\Theta$ be the calculation of $u$ and $f\y{s_1}{t_1}\cdots\y{s_l}{t_l}$, and let $\Lambda$ be the calculation of $u$ and $g\y{p_1}{q_1}\cdots\y{p_m}{q_m}$.
Assume that these calculations contain no potential cancellations.
Then two exponents of the calculations are the same.
\begin{proof}
Let $e(\Lambda)$ and $e(\Theta)$ be the two corresponding exponents.
We show this by contradiction.
We can assume that $e(\Lambda)>e(\Theta)$ without loss of generality.
Since exponents are non-negative integer, let $k:=e(\Lambda)>0$.
By Lemma \ref{lem_exponent_1}, we have $wy^k$ obtained from $\Lambda v$ by substitutions, where $v \in \N^{<\mathbb{N}}$.
Since $f\y{s_1}{t_1}\cdots\y{s_l}{t_l}$ and $g\y{p_1}{q_1}\cdots\y{p_m}{q_m}$ are equal as elements of $G_0(n)$, the output strings of
\begin{align*}
\Theta v 0^{2^k}(n-1)0^{2^k}(n-1)\cdots
\shortintertext{and}
\Lambda v 0^{2^k}(n-1)0^{2^k}(n-1)\cdots
\end{align*}
are equal as elements of $\N^\mathbb{N}$.
Then the latter is
\begin{align*}
\Lambda v 0^{2^k}(n-1)0^{2^k}(n-1)\cdots = w y^k 0^{2^k}(n-1)0^{2^k}(n-1)\cdots=w0(n-1)^{2^k}0(n-1)^{2^k}\cdots.
\end{align*}
Since $e(\Lambda)>e(\Theta)$ holds, by the definition of $y$, this implies a contradiction.
\end{proof}
\end{lemma}
\begin{corollary} \label{G_0(n)_presentation}
The group obtained from the presentation $\langle Z(n) \mid R(n)\rangle$ is the $n$-adic Lodha--Moore group $G_0(n)$.
\begin{proof}
We show that any standard form which contains no potential cancellations and represents an element of $F(n)$ is always a word on $X(n)$.
Let $f\y{s_1}{t_1}\cdots\y{s_m}{t_m}=g$ be standard forms, where $f, g \in F(n)$, and $m \geq 1$.
Assume that they contain no potential cancellations.
Let $u=s_100\cdots$ in $\N^\mathbb{N}$.
Then the exponent of the calculation of $f^{-1}(u)$ and $f\y{s_1}{t_1}\cdots\y{s_m}{t_m}$ is strictly greater than $0$, and that of the calculation of $f^{-1}(u)$ and $g$ is $0$.
By Lemma \ref{lem_exponent_2}, this is a contradiction.

In particular, for a word on $Z(n)$ representing the identity element of $G_0(n)$, we can rewrite it into a standard form that contains no potential cancellations by Proposition \ref{Prop_standard_forms_from_words} and Lemma \ref{lem_step2}.
Since this standard form is a word on $X(n)$, as shown above, we can reduce it to the empty word by using the relations of $F(n)$.
\end{proof}
\end{corollary}
\begin{remark}
From the above argument, the uniqueness of the normal form of the elements of $F(n)$ in $G_0(n)$ follows.
\end{remark}\label{remark_uniqueF(n)}
The following completes the proof of the uniqueness.
\begin{theorem}[{\cite[Theorem 4.4]{lodha2020nonamenable}} for $n=2$]\label{thm_normal_form_uniqueness}
For each element in $G_0(n)$, its normal form is unique.
\begin{proof}
By Remark \ref{remark_uniqueF(n)}, we can assume that every normal form in the following argument is not in $F(n)$.

We show this by contradiction.
Let $f\y{s_1}{t_1}\cdots\y{s_l}{t_l}$ and $g\y{p_1}{q_1}\cdots\y{p_m}{q_m}$ be different normal forms representing the same element.
We can assume that $s_l \leq p_m$ without loss of generality.
One of the following three holds:
\begin{enumerate}
\item $s_l=p_m$ and $t_l=q_m$;
\item $s_l=p_m$ and $t_l \neq q_m$;
\item $s_l<p_m$.
\end{enumerate}
First, we show that it is sufficient to consider only case $(3)$.

In case (1), since $\y{s_l}{t_l}=\y{p_m}{q_m}$, we start from $f\y{s_1}{t_1}\cdots\y{s_{l-1}}{t_{l-1}}$ and $g\y{p_1}{q_1}\cdots\y{p_{m-1}}{q_{m-1}}$.
They are two different standard forms representing the same element.
They contain no potential cancellations but possibly contain a potential contraction.
Since we only cancel $\y{s_l}{t_l}=\y{p_m}{q_m}$, the only case that contains a potential cancellation at this step is when $\y{s_l}{t_l}$ or $\y{p_m}{q_m}$ or both play the role of $\y{s(n-1)}{\pm1}$ for some $s$ (see Definition \ref{def_potential_contraction} (1)).
By the assumption of the standard form, the case with $\y{s0}{\pm 1}$ (Definition \ref{def_potential_contraction} (2)) does not occur.
We further divide case (1) into the following four subcases:
\begin{description}
\item[\rm (1-1)] Both $f\y{s_1}{t_1}\cdots\y{s_{l-1}}{t_{l-1}}$ and $g\y{p_1}{q_1}\cdots\y{p_{m-1}}{q_{m-1}}$ contain a potential contraction, and $t_{l-1}=q_{m-1}$;
\item[\rm (1-2)] Both $f\y{s_1}{t_1}\cdots\y{s_{l-1}}{t_{l-1}}$ and $g\y{p_1}{q_1}\cdots\y{p_{m-1}}{q_{m-1}}$ contain a potential contraction, and $t_{l-1}\neq q_{m-1}$;
\item[\rm (1-3)] $f\y{s_1}{t_1}\cdots\y{s_{l-1}}{t_{l-1}}$ contains a potential contraction, and $g\y{p_1}{q_1}\cdots\y{p_{m-1}}{q_{m-1}}$ contains no potential contractions, or vice versa;
\item[\rm (1-4)] $f\y{s_1}{t_1}\cdots\y{s_{l-1}}{t_{l-1}}$ and $g\y{p_1}{q_1}\cdots\y{p_{m-1}}{q_{m-1}}$ contain no potential contractions.
\end{description}

In case (1-1), we consider $f\y{s_1}{t_1}\cdots\y{s_{l-2}}{t_{l-2}}$ and $g\y{p_1}{q_1}\cdots\y{p_{m-2}}{q_{m-2}}$ instead of the original words in order to eliminate the potential contraction part.
Since the assumption of containing a potential contraction and being in standard form, $s_l=p_m=s(n-1)$, $s_{l-1}=p_{m-1}=s(n-1)(n-1)$ and $t_{l-1}=q_{m-1}>0$ hold.
Then $f\y{s_1}{t_1}\cdots\y{s_{l-2}}{t_{l-2}}$ and $g\y{p_1}{q_1}\cdots\y{p_{m-2}}{q_{m-2}}$ are different normal forms.
Indeed, the only case where a potential contraction is generated  by canceling $\y{s_{l-1}}{t_{l-1}}=\y{p_{m-1}}{q_{m-1}}$ is when $\y{s_{l-1}}{t_{l-1}}=\y{p_{m-1}}{q_{m-1}}$ plays the role of $y_{s^\prime(n-1)}$ for $s^\prime=s(n-1)$, but this does not occur since $\y{s^\prime0}{k_1}=\y{s(n-1)0}{k_1}$ and $\y{s^\prime0}{k_2}=\y{s(n-1)0}{k_2}$ exist in two forms respectively and $k_1, k_2<0$ holds by the assumption of potential contractions.
Thus, we get ``shorter'' normal forms that satisfy all the assumptions.

In case (1-2), we can assume that $0<t_{l-1}<q_{m-1}$ without loss of generality.
We consider $f\y{s_1}{t_1}\cdots\y{s_{l-2}}{t_{l-2}}$ and $g\y{p_1}{q_1}\cdots\y{p_{m-1}}{q_{m-1}-t_{l-1}}$ instead.
For the same reason as for case (1-1), the former is in normal form.
By the assumption of a potential contraction, $\y{s_{l-2}}{t_{l-2}}=\y{s(n-1)i\sigma}{t_{l-2}}$, where $i$ is in $\bm{N}$ and $\sigma$ is a word in $\N^{<\mathbb{N}} \cup \{\epsilon\}$.
Note that $s(n-1)0 \leq s(n-1)i \sigma$ holds.
For the latter, by performing contraction moves as in step 3, the last letter is $\y{u}{k}$ where $u \subseteq s$.
This is the situation in case (3).

In case (1-3), we only consider the former situation.
By the assumptions, $s_l=p_m=s(n-1)$ holds for some $s$.
By applying contraction moves to $f\y{s_1}{t_1}\cdots\y{s_{l-1}}{t_{l-1}}$ as in step3, the last latter is $\y{u}{k}$ where $u \subseteq s \subset s_l=p_m$.
Since $g\y{p_1}{q_1}\cdots\y{p_{m-1}}{q_{m-1}}$ is in normal form, and in particular in standard form, we have $y_{p_{m-1}}\neq y_u$.
This is the situation in case (3).

In case (1-4), both are ``shorter'' normal forms that satisfy all the assumptions.
Therefore, any subcase of case (1) yields shorter words or in case (3).

In case (2), we can assume that neither $0<q_m<t_l$ nor $t_l<q_m<0$ without loss of generality.
Consider $f\y{s_1}{t_1}\cdots\y{s_{l-1}}{t_{l-1}}$ and $g\y{p_1}{q_1}\cdots\y{p_{m}}{q_{m}-t_{l}}$ instead.
The latter is in normal form, but as in case (1), the former may contain a potential contraction.
Similarly, we divide case (2) into two subcases:
\begin{description}
\item[\rm (2-1)] $f\y{s_1}{t_1}\cdots\y{s_{l-1}}{t_{l-1}}$ contains a potential contraction;
\item[\rm (2-2)] $f\y{s_1}{t_1}\cdots\y{s_{l-1}}{t_{l-1}}$ contains no potential contractions.
\end{description}
In case (2-1), the same argument as in (1-3) can be applied.
In case (2-2), since $s_{l-1}\neq p_m$, this is the situation in case (3).

We note that case (1-1) or (1-4) happens only finitely many times due to the uniqueness of the normal form of $F(n)$.
Thus, we consider case (3).

We only consider the case $q_m>0$.
We note that $f\y{s_1}{t_1}\cdots\y{s_l}{t_l}\y{p_m}{-1}$ is in standard form since $s_l < p_m$.
One of the following holds:
\begin{enumerate}[label=(\roman*)]
\item there exists an infinite word $\sigma$ on $\{0, n-1\}$ such that for any finite word $\sigma_1 \subset \sigma$, $p_m\sigma_1$ is not in $\{s_1, \dots, s_l \}$;
\item there exists an adjacent pair of the form $p_m, s_i$ where $p_mu=s_i$ for $u$ on $\{0, n-1\}$ which is not a potential cancellation.
\end{enumerate}
Indeed, if both are false, it contradicts that $f\y{s_1}{t_1}\cdots\y{s_l}{t_l}$ contains no potential contractions.
Then, in either case, there exists a finite word $w$ on $\{0, n-1\}$ such that the calculation $\Lambda$ of $f\y{s_1}{t_1}\cdots\y{s_l}{t_l}\y{p_m}{-1}$ and $f^{-1}(p_mw)$ contains no potential cancellations.
By expanding $w$ if necessary, we can consider the following three calculations:
\begin{enumerate}[label=(\alph*)]
\item $\Theta$ is the calculation of $g\y{p_1}{q_1}\cdots\y{p_m}{q_m-1}$ and $f^{-1}(p_mw)$;
\item $\Lambda^\prime$ is the calculation of $f\y{s_1}{t_1}\cdots\y{s_l}{t_l}$ and $f^{-1}(p_mw)$;
\item $\Theta^\prime$ is the calculation of $g\y{p_1}{q_1}\cdots\y{p_m}{q_m}$ and $f^{-1}(p_mw)$.
\end{enumerate}

All calculations contain no potential cancellations.
Indeed, for $\Lambda^\prime$ and $\Theta^\prime$, it is clear from the assumption of normal form.
For $\Theta$, either $q_m-1$ is zero or strictly greater than zero since $q_m>0$.
In both cases, it is clear from the definitions and the assumption that $g\y{p_1}{q_1}\cdots\y{p_m}{q_m}$ contains no potential cancellations.

Let $e(\Theta)$, $e(\Theta^\prime)$, $e(\Lambda)$, and $e(\Lambda^\prime)$ be the its exponent, respectively.
Since we have
\begin{align*}
f\y{s_1}{t_1}\cdots\y{s_l}{t_l}\y{p_m}{-1}=g\y{p_1}{q_1}\cdots\y{p_m}{q_m-1}
\end{align*}
as elements of $G_0(n)$, by Lemma \ref{lem_exponent_2}, $e(\Lambda)=e(\Theta)$ holds.
Similarly, we have $e(\Theta^\prime)=e(\Lambda^\prime)$.
By the construction of $\Lambda$ and $\Lambda^\prime$, we have $e(\Lambda)>e(\Lambda^\prime)$.
Similarly, we have $e(\Theta)\leq e(\Theta^\prime)$ since $q_m>0$.
Combining them, we obtain
\begin{align*}
e(\Theta)\leq e(\Theta^\prime)=e(\Lambda^\prime)<e(\Lambda),
\end{align*}
which is a contradiction.

If $q_m<0$, we can prove the same for $f\y{s_1}{t_1}\cdots\y{s_l}{t_l}y_{p_m}$ and $g\y{p_1}{q_1}\cdots\y{p_m}{q_m+1}$.
\end{proof}
\end{theorem}

\section{Several properties of $G_0(n)$}\label{section_G0(n)_properties}
We show that $G_0(n)$ has ``expected'' properties.
\subsection{The finite presentation of $G_0(n)$}
Using the infinite presentation given in Section \ref{infinite_presentation}, we construct a finite presentation of $G_0(n)$ with respect to the generating set $\{x_0, \dots, x_{n-2}, {x_0}_{[n-1]}, y_{(n-1)0}\}$.
As in \cite[Theorem 3.3]{lodha2016nonamenable}, we use two properties of $F(n)$.
Since $G_0=G_0(2)$ is already known to be finitely presented, we assume that $n>2$.

For $s$ in $\N^{<\mathbb{N}}$ that is neither $0\cdots0$ nor $(n-1)\cdots(n-1)$, we define a map $f_s$ in $F(n)$ as follows:
Let $a$ be the sum of each number in $s$, and let $d, k$ be the integers such that $a=(n-1)d+k$ holds, where $0\leq d$ and $0\leq k<n-1$.
Then $f_s$ is defined as in Figure \ref{definition_fs}.
\begin{figure}[tbp]
	\centering
	\includegraphics[width=100mm]{definition_fs.pdf}
	\caption{The definition of $f_s$. The rightmost carets are for adjustment. The triangle labeled $s$ denotes the minimal $n$-ary tree containing $s$. }
	\label{definition_fs}
\end{figure}
We fix a word on $\{x_0, \dots, x_{n-2}, \x{0}{(n-1)}\}$ representing $f_s$.
Since the leaf corresponding to $s$ in the minimal $n$-ary tree containing $s$ is the $a$-th leaf counting from the left (from $0$), we note that $f_s((n-1)k)=s$ holds.

\begin{lemma}\label{lem_finitely_presented1}
For the above $f_s$, we have $\x{i}{(n-1)k}f_s=f_s \x{i}{s}$ and $y_{(n-1)0}f_s=f_sy_s$.
\begin{sproof}
We note that we have $f_s^{-1}\x{i}{(n-1)k}f_s=\x{i}{f_s((n-1)k)}=\x{i}{s}$.
Indeed, if $w=sw^\prime \in \N^\mathbb{N}$, the left hand side sends $w$ to $sx_i(w^\prime)$ via $(n-1)kw^\prime$ and $(n-1)kx_i(w^\prime)$.
If $w \neq sw^\prime$, since $f_s^{-1}(w)$ does not contains $(n-1)k$ as a prefix, $\x{i}{(n-1)k}(f_s^{-1}(w))=f_s^{-1}(w)$.

Another statement can be shown in the same way.
\end{sproof}
\end{lemma}
\begin{lemma}\label{lem_finitely_presented2}
Let $u, v$ be in $\N^{<\mathbb{N}}$ such that $u<v$, $u \perp v$, and $y_u, y_v$ are in $Y(n)$.
Then there exists $g$ in $F(n)$ such that $g(0(n-1))=u$ and $g((n-1)0)=v$ hold.
\begin{sproof}
Like the construction of $f_s$, by adding some carets, we can obtain two trees (may no be tree diagram) such that $0(n-1) \mapsto u$.
Since $n>2$, we can add some carets between two leaves corresponding to $0(n-1)$ and $(n-1)0$ so that $(n-1)0 \mapsto v$.
Finally, we add some carets to the rightmost if necessary, to make it a tree diagram.
See Figure \ref{construction_g_uv} for the sketch of the construction of $g$.
\end{sproof}
\end{lemma}
\begin{figure}[tbp]
	\centering
	\includegraphics[width=80mm]{construction_g_uv.pdf}
	\caption{The construction of $g$ in Lemma \ref{lem_finitely_presented2}. }
	\label{construction_g_uv}
\end{figure}
\begin{theorem}[{\cite[Theorem 3.3]{lodha2016nonamenable}} for $n=2$]\label{Thm_finitely_presented_G0(n)}
$G_0(n)$ admits a finite presentation.
\begin{proof}
Since $F(n)$ is finitely presented \cite[Theorem 4.17]{brown1987finiteness},
the relation (1) in $R(n)$ is expressed by finite relations.
Thus we only consider the other three relations (2), (3), and (4).

For the relation $y_t\x{i}{s}=\x{i}{s}y_{\x{i}{s}(t)}$,
we can rewrite as follows:
\begin{align*}
y_t\x{i}{s}&=f_t^{-1}y_{(n-1)0}f_t \x{i}{s}, \\
\x{i}{s}y_{\x{i}{s}(t)}&=\x{i}{s}f^{-1}_{\x{i}{s}(t)}y_{(n-1)0}f_{\x{i}{s}(t)}.
\end{align*}
Thus, it is sufficient to show that
\begin{align}
y_{(n-1)0}f_t \x{i}{s}f^{-1}_{\x{i}{s}(t)}=f_t \x{i}{s}f^{-1}_{\x{i}{s}(t)}y_{(n-1)0},  \label{relation2_into_finite}
\end{align}
by using a finite number of relations.
We note that $f_t \x{i}{s}f^{-1}_{\x{i}{s}(t)}$ maps $(n-1)0$ to itself.
Therefore, $f_t \x{i}{s}f^{-1}_{\x{i}{s}(t)} \in F(n)$ is rewritten into a word on the finite set
\begin{align*}
X_{(n-1)0}(n):=
&\{x_0x_i^{-1}, \x{i}{(n-1)} \mid i=1, \dots, n-2 \} \\
&\cup \{\x{i}{0}, \x{i}{1}, \dots, \x{i}{n-2}, \x{i}{(n-1)1}, \dots, \x{i}{(n-1)(n-1)}\mid i=0, \dots, n-2 \} \\
&\cup \{ \x{0}{0(n-1)}, \dots, \x{0}{(n-2)(n-1)}, \x{0}{(n-1)1(n-1)}, \dots, \x{0}{(n-1)(n-1)(n-1)}\},
\end{align*}
by finite relations of $F(n)$.
Since each element is commute with $y_{(n-1)0}$, the collection of the finite relations $hy_{(n-1)0}=y_{(n-1)0}h$ where $h$ is in $X_{(n-1)0}(n)$ implies equation \eqref{relation2_into_finite}.

For the relation $y_sy_t=y_ty_s$ where $s \perp t$, we note that we have $y_{(n-1)0}y_{0(n-1)}=y_{0(n-1)}y_{(n-1)0}$.
By using this one and relation (2), we have
\begin{align*}
y_sy_t=f^{-1}_sy_{(n-1)0}f_sf^{-1}_ty_{(n-1)0}f_t=g^{-1}y_{0(n-1)}gg^{-1}y_{(n-1)0}g=g^{-1}y_{(n-1)0}gg^{-1}y_{0(n-1)}g=y_ty_s,
\end{align*}
where $g$ is the element such that $g(0(n-1))=s$ and $g((n-1)0)=t$.

Finally, for the relation $y_s=\x{0}{s}y_{s0}\y{s(n-1)0}{-1}y_{s(n-1)(n-1)}$, we note that we have $y_{(n-1)0}=\x{0}{(n-1)0}y_{(n-1)00}\y{(n-1)0(n-1)0}{-1}y_{(n-1)0(n-1)(n-1)}$.
Since $f_s((n-1)0w)=sw$ for $w \in \N^{<\mathbb{N}}$, by using relation (2), we have
\begin{align*}
y_s&=f^{-1}_sy_{(n-1)0}f_s \\
&=f^{-1}_s\x{0}{(n-1)0}y_{(n-1)00}\y{(n-1)0(n-1)0}{-1}y_{(n-1)0(n-1)(n-1)}f_s \\
&=f^{-1}_s\x{0}{(n-1)0}f_sf^{-1}_sy_{(n-1)00}f_sf^{-1}_s\y{(n-1)0(n-1)0}{-1}f_sf^{-1}_sy_{(n-1)0(n-1)(n-1)}f_s \\
&=\x{0}{s}y_{s0}\y{s(n-1)0}{-1}y_{s(n-1)(n-1)}.
\end{align*}
This completes the proof.
\end{proof}
\end{theorem}
\subsection{Nonamenability of $G_0(n)$}
In this section, we discuss embeddings of the groups $G_0(n)$.
For nonamenability, we only use the fact that a subgroup of an amenable group is also amenable \cite[Theorem 18.29 (1)]{dructu2018geometric}.
The idea for the following proposition comes from \cite{burillo2001metrics}.
\begin{theorem}\label{prop_embedding_G_0(p)G_0(q)}
Let $p, q \geq 2$ and assume that there exists $d$ in $\mathbb{N}$ such that $q-1=d(p-1)$ holds.
Then there exists an embedding $I_{p, q}: G_0(p) \to G_0(q)$.
Moreover, the equality $I_{p,r}=I_{p, q}I_{q, r}$ holds for the maps $I_{p, q}$, $I_{q, r}$ and $I_{p, r}$ defined for $r\geq q \geq p \geq2$ such that any two satisfy the condition.
\begin{proof}
For the sake of clarity, we label elements of $G_0(p)$ with ``tildes'' and elements of $G_0(q)$ with ``hats.''
We first recall the definition of the embedding $F(p) \to F(q)$ given in \cite[Section 3, example (3)]{burillo2001metrics}.
This homomorphism is defined by
\begin{align*}
&\tilde{x}_0\mapsto\hat{x}_0 & &\tilde{x}_1 \mapsto \hat{x}_d & &\cdots & &\tilde{x}_{p-2} \mapsto \hat{x}_{d(p-2)}& &\tx{0}{(p-1)} \mapsto \hx{0}{(q-1)},
\end{align*}
and extended to the (quasi-isometric) embedding \cite[Theorem 6]{burillo2001metrics}.
By considering $F(p)$ and $F(q)$ as pairs of $p$-ary trees and $q$-ary trees, this embedding is regarded as a ``caret replacement.''
Indeed, for every pair of $p$-ary trees, inserting $d-1$ edges between every pair of adjacent edges of each $p$-caret corresponds to the embedding.
See Figure \ref{embedding_F3F7}, for example.
\begin{figure}[tbp]
	\centering
	\includegraphics[width=120mm]{embedding_F3F7.pdf}
	\caption{Example of embedding for $p=3$ and $q=7$. }
	\label{embedding_F3F7}
\end{figure}

Define the map $i_{p, q}: \Pset^{<\mathbb{N}}=\{0, \dots, p-1\}^{<\mathbb{N}} \to \Qset^{<\mathbb{N}}=\{0, \dots, q-1\}^{\mathbb{N}}$ by setting
\begin{align*}
w_1 w_2\cdots w_k \mapsto (dw_1)(dw_2)\cdots (dw_k),
\end{align*}
namely, we multiply each number of a given word by $d$.
In addition, define the map $Y(p)\to Y(q)$ by setting $\tilde{y}_s \mapsto \hat{y}_{i_{p, q}(s)}$.
By the definition  of $i_{p, q}$, we have that $\hat{y}_{i_{p, q}(s)}$ is in $Y(q)$.

By combining the two maps $F(p) \to F(q)$ and $Y(p) \to Y(q)$, we obtain the map $I_{p, q}: F(p)\cup Y(p) \to F(q)\cup Y(q)$.
This map can be extended to the homomorphism $I_{p, q}: G_0(p) \to G_0(q)$.
Indeed, since $i_{p, q}(\tx{i}{s}(t))=\hx{di}{i_{p, q}(s)}(i_{p, q}(t))$ and $I_{p, q}(\tilde{y}_s)=\hat{y}_{i_{p, q}(s)}$ hold, we can verify that the relations of the infinite presentation (Corollary \ref{G_0(n)_presentation}) are preserved under $I_{p, q}$ by directly calculations.

We claim that $I_{p, q}: G_0(p) \to G_0(q)$ is injective.
We show this by contradiction.
Assume that $\operatorname{Ker}(I_{p, q})$ is not trivial.
By the construction, the restriction $I_{p, q}\restr{F(p)}$ coincides with the embedding $F(p) \to F(q)$ mentioned above.
Hence if there exists $x$ in $\operatorname{Ker}(I_{p, q})$ that is not identity, then $x$ is not in $F(p)$.
We note that the map $I_{p, q}$ preserves being in normal form.
In particular, if $x$ is not in $F(p)$, then $I_{p, q}(x)$ is not in $F(q)$ by the uniqueness (Theorem \ref{thm_normal_form_uniqueness}).
This implies that $I_{p, q}(x)$ is not identity, as desired.

Let $d_1$, $d_2$ and $d_3$ be natural numbers such that $q-1=d_1(p-1)$, $r-1=d_2(q-1)$ and $r-1=d_3(p-1)$ holds respectively.
We note that $d_3=d_1d_2$ holds.
By considering the definitions of the map $i_{p, q}$ and the homomorphism from $F(p)$ to $F(q)$, the equality of $I_{p,r}$ and $I_{p, q}I_{q, r}$ follow immediately.
\end{proof}
\end{theorem}
\begin{corollary}
Let $n\geq2$. Then $G_0(n)$ is nonamenable.
\begin{proof}
Since we have $n-1=(n-1)(2-1)$ for every $n$, by Theorem \ref{prop_embedding_G_0(p)G_0(q)}, we have an embedding from $G_0(2)=G_0$ into $G_0(n)$.
This implies that $G_0(n)$ has a subgroup that is isomorphic to $G_0$.
Since $G_0$ is nonamenable \cite[Theorem 1.1]{lodha2016nonamenable}, $G_0(n)$ is also nonamenable.
\end{proof}
\end{corollary}
\begin{corollary}
Let $s_i:=2^{i-1}+1$.
Then the sequence $G(s_1), G(s_2), \dots$ forms an inductive system of groups.
\end{corollary}
\subsection{$G_0(n)$ has no free subgroups}
We assume $n \geq 3$.
In order to show that $G_0(n)$ does not contain the free groups, we will follow \cite[Section 3]{brin1985groups}.
The difference in this paper is that the domain and range set of the maps are $\N^\mathbb{N}$ instead of $\mathbb{R}$.

For an element $g$ in $G_0(n)$, we write the set-theoretic support $\supp{g}$ for the set $\{\xi \in \N^\mathbb{N} \mid g(\xi)\neq\xi\}$.
Although we equip $\N^\mathbb{N}$ with the topology that is homomorphic to Cantor set, which is totally disconnected, we can consider ``connected components'' of $\supp{g}$, using the total order of $\N^\mathbb{N}$.

For $a, b \in \N^\mathbb{N}$ with $a<b$, set $(a, b):=\{\xi \in \N^\mathbb{N} \mid a<\xi<b\}$.
\begin{lemma}\label{lem_supp_finite_component}
For $g \in G_0(n)$, there exists a sequence $a_1<a_2 \leq a_3<a_4\leq \cdots < a_{2m}$ in $\N^\mathbb{N}$ such that we have
\begin{align*}
\supp{g}=(a_1, a_2) \sqcup (a_3, a_4) \sqcup \cdots \sqcup (a_{2m-1}, a_{2m}).
\end{align*}
We call each $(a_{2i}, a_{2i+1})$ a \textit{connected component} of $\supp{g}$.
\begin{proof}
Let $f\y{s_1}{t_1}\cdots \y{s_m}{t_m}$ be in normal form of $g$.
We decompose $\N^\mathbb{N}$ into disjoint sets $\{n_1\eta \mid \eta \in \N^\mathbb{N} \}, \dots, \{n_p\eta \mid \eta \in \N^\mathbb{N}\}$ by using $n_1, \dots, n_p$ in $\N^{<\mathbb{N}}$ such that the following holds:
\begin{enumerate}
\item For each $n_l$, $f(n_l)$ is defined;
\item For each $s_l$, there exists $n_{l^\prime}$ such that $s_l \leq f(n_{l^\prime})$ holds.
\end{enumerate}

We show the claim in the lemma by contradiction.
Assume that there exist $a_1<b_1<\cdots$ such that each $a_i$ is a fixed point, and each $b_i$ is in $\supp{g}$.
Since $\N^\mathbb{N}=\{n_1\eta \mid \eta \in \N^\mathbb{N} \}\cup \dots \cup \{n_p\eta \mid \eta \in \N^\mathbb{N}\}$ holds, we can assume that each $a_i$ and $b_i$ is in some set $\{n_j\eta \mid \eta \in \N^\mathbb{N} \}$ without loss of generality.
Since $a_{i}<b_i<a_{i+1}$ holds, we can write each $a_i$ and $b_i$ as $n_ja_i^\prime$ and $n_jb_i^\prime$, respectively.

We claim that each $a_i^\prime$ and $b_i^\prime$ can be replaced by words in $\{0, n-1\}^{\mathbb{N}}$, respectively.
Indeed, if $a_i^\prime$ is not in $\{0, n-1\}^{\mathbb{N}}$, then there exist $\hat{a}_i \in \{0, n-1\}^{<\mathbb{N}}\cup \{\epsilon\}$, $k \in \{1, \dots, n-2\}$, and $a_i^{\prime \prime} \in \N^\mathbb{N}$ such that $a_i^\prime=\hat{a}_i k a_i^{\prime \prime}$ holds.
Then, by the definition of $y$, $g(n_j\hat{a}_i k)$ is defined and equals to $n_j\hat{a}_i k$ since $a_i=n_j \hat{a}_i k a_i^{\prime \prime}$ is a fixed point of $g$.
Moreover, for any other $k^\prime \in \{1, \dots, n-2\}$, we have $g(n_j\hat{a}_i k^\prime)=n_j\hat{a}_i k^\prime$.
This implies that we have $g(n_j\hat{a}_i 0 \overline{(n-1)})=n_j\hat{a}_i 0 \overline{(n-1)}$, where $\overline{(n-1)}$ denotes the element $(n-1)(n-1)(n-1)\cdots $ in $\N^\mathbb{N}$.
By a similar argument for $b_i^\prime$, we have $g(n_j\hat{b}_i k)=n_j\tilde{b_i}k$, where $\hat{b}_i \neq \tilde{b}_i$.
Since we have $g(n_j\hat{b}_i k^\prime)\neq n_j\hat{b}_i k^\prime$ for any other $k^\prime$, $n_j\hat{b}_i 0\overline{(n-1)}$ is in $\supp{g}$.
Even if each $a_i$ or $b_i$ is replaced by the above one, the order $a_1<b_1<\cdots$ is also preserved.

Let $w_1, \dots, w_t$ be in $\{0, \dots, n-1\}$ such that $f(n_j)=w_1 \cdots w_t$ holds.
For $n_j \zeta \in \N^\mathbb{N}$ and $f\y{s_1}{t_1}\cdots \y{s_m}{t_m}$, let $w_1y^{l_1}w_2 y^{l_2}\cdots w_t y^{l_t} \zeta$ be their calculation.
Then some $l_q$ may be zero.
Assume that $f(n_j)=w_1 \cdots w_t$ is not in $\{0, n-1\}^{<\mathbb{N}}$ and let $z$ be the maximal number such that $w_z$ is in $\{1, \dots, n-2\}$.
By applying finitely many substitutions to all $y^{l_{z^\prime}}$ ($z^\prime<z$), we obtain a word
$w y^{l_1^\prime} w_2^\prime y^{l_2^\prime} \cdots w_{t^\prime}^\prime y^{l_{t^\prime}^\prime}\zeta$,
where $w_j^\prime \in \{0,  n-1\}$.
Since $n_j a_i$ is the output string of $w y^{l_1^\prime} w_2^\prime y^{l_2^\prime} \cdots w_{t^\prime}^\prime y^{l_{t^\prime}^\prime} a_i$, we have $w \leq n_j$.
Indeed, it is clear that $w \perp n_j$ does not hold, and if $w > n_j$ holds, then it contradicts the facts that $a_i$ is in $\{0, n-1\}^\mathbb{N}$, and the last number of $w$ is $w_j$.
Let $n_j=w n_j^\prime$.
Then we note that $n_j^\prime$ is in $\{0, n-1\}^{<\mathbb{N}} \cup \{\epsilon\}$ since $y^{l_1^\prime} w_2^\prime y^{l_2^\prime} \cdots w_{t^\prime}^\prime y^{l_{t^\prime}^\prime}a_i$ is in $\{0, n-1\}^\mathbb{N}$ and its output string equals to $n_j^\prime a_i$.

We recall that $G_0(n)$ has a subgroup that is isomorphic to $G_0$ (Theorem \ref{prop_embedding_G_0(p)G_0(q)}).
We construct an element $g^\prime$ which is in the subgroup and satisfies the assumption of $g$.
Let $g^\prime$ be the element represented by
\begin{align*}
f^\prime \y{(n-1)0w_2^\prime \cdots w_{t^\prime}^\prime}{l_{t^\prime}^\prime} \y{(n-1)0w_2^\prime \cdots w_{t^\prime-1}^\prime}{l_{t^\prime-1}^\prime} \cdots \y{(n-1)0w_2^\prime}{j_2^\prime} \y{(n-1)0}{j_1^\prime},
\end{align*}
where $f^\prime$ is a word on $\{x_0, \x{0}{(n-1)}\}$ satisfying $f^\prime((n-1)0n_j^\prime)=(n-1)0w_2^\prime \cdots w_{t^\prime}^\prime$.
Since $n_j^\prime$ and $w_2^\prime \cdots w_{t^\prime}^\prime$ are in $\{0, n-1\}^{<\mathbb{N}}$, there must exist such an element $f^\prime$.
By the construction, $g^\prime$ is in the subgroup that is isomorphic to $G_0$.
In addition, each $(n-1)0n_j^\prime a_i$ is a fixed point of $g^\prime$, and $(n-1)0n_j^\prime b_i$ is in $\supp{g^\prime}$.
By the construction of the map $I_{2, n}$ in Theorem \ref{prop_embedding_G_0(p)G_0(q)}, if we restrict the domain set of $g^\prime$ to $\{0, n-1\}^\mathbb{N}$ and identify $n-1$ with $1$, then we obtain the element of $G_0$ which satisfies all the assumptions about fixed points and elements of support.
However, since this does not happen by Proposition \ref{proposition_piecewiseprojective}, this is a contradiction.

Finally, assume that $f(n_j)=w_1 \cdots w_t$ is in $\{0, n-1\}^{<\mathbb{N}}$.
Since $w_1y^{l_1}w_2 y^{l_2}\cdots w_t y^{l_t} a_i$ is in $\{0, n-1\}^{\mathbb{N}}$, $n_j$ is also in $\{0, n-1\}^{<\mathbb{N}}$.
Therefore, $f$ is represented by a word on $\{x_0, \x{0}{(n-1)}\}$.
We again note that some $l_q$ may be zero.
Then $g^\prime=f\y{w_1\cdots w_t}{l_t} \cdots \y{w_1w_2}{l_2} \y{w_1}{l_1}$ is in $G_0(n)$, and an element of $G_0$ can be similarly constructed from $g^\prime$.
\end{proof}
\end{lemma}
\begin{remark}
By the construction, the sequence $a_1, \dots, a_{2m}$ in Lemma \ref{lem_supp_finite_component} is uniquely determined.
\end{remark}
For a group $G$, $[G, G]$ denotes its commutator subgroup.
For two elements of $x, y \in G$, $[x, y]$ denotes the commutator $xyx^{-1}y^{-1}$.
\begin{theorem}\label{Theorem_no_free_group}
The group $G_0(n)$ has no free subgroups.
\begin{proof}
Similarly to \cite{brin1985groups}, we will show that if $G_1$ is a subgroup of $[G_0(n), G_0(n)]$, then either $\mathbb{Z}^2$ is a copy of a subgroup of $G_1$ or $[G_1, G_1]$ is trivial.
Indeed, in that case, if $G$ is a subgroup of $G_0(n)$, then either $\mathbb{Z}^2$ is a copy of a subgroup of $G$ or $[G, G]$ is abelian, and the free group $F_2$ does not have such a property.

Let $G_1$ be a subgroup of $[G_0(n), G_0(n)]$, and assume that $[G_1, G_1]$ is not trivial.
Then there exist elements $f$ and $g$ in $G_1$ such that $z=fgf^{-1}g^{-1}\neq \mathrm{Id}$.
For $f$ and $g$, let $a_1, \dots, a_{2s}$ and $b_1, \dots, b_{2t}$ be sequences given in Lemma \ref{lem_supp_finite_component}, respectively.
By choosing elements appropriately from $a_1, \dots, a_{2s}, b_1, \dots, b_{2t}$, we obtain $c_1, \dots, c_{2m}$ such that
\begin{align*}
\supp{f} \cup \supp{g}=(c_1, c_2) \sqcup (c_3, c_4) \sqcup \cdots \sqcup (c_{2m-1}, c_{2m}).
\end{align*}
Let $d_1, \dots, d_{2l}$ be a sequence given by Lemma \ref{lem_supp_finite_component} for $[f, g]$.
Then we have that for each $x \in \supp{[f, g]}$, there exist $i$ and $i^\prime$ such that
\begin{align}
c_{2i^\prime-1}<d_{2i-1}<x<d_{2i}<c_{2i^\prime} \label{support_condition_commutator}
\end{align}
holds.
Indeed, it is sufficient to show that for each common fixed point $x$ of $f$ and $g$,  there exists a prefix $u$ such that $[f, g]$ fixes all the element of $\{u\xi \mid \xi \in \N^\mathbb{N}\}$.
If $x$ is not in $\{0, n-1\}^{\mathbb{N}}$, then $y$ ``vanishes.''
Thus by regarding $f$ and $g$ as piecewise linear map of $F(n)$, it is clear.
If $x$ is in $\{0, n-1\}^{\mathbb{N}}$, by identifying $\{0, n-1\}$ with $\{0, 1\}$, it is also clear by Proposition \ref{proposition_piecewiseprojective}.

We write $\supp{[f, g]} \subsetneq \supp{f} \cup \supp{g}$ for the ``proper'' inclusion of each connected component of the support described above.
Let
\begin{align*}
W:=\{h \in \langle f, g \rangle \mid \supp{h} \subsetneq \supp{f} \cup \supp{g}, h \neq \mathrm{Id} \},
\end{align*}
where $\langle f, g \rangle$ is the subgroup generated by $f$ and $g$.
The set $W$ is not empty since $z$ is in $W$.

Define the function $\kappa: W \to \mathbb{Z}_{\geq 0}$ by setting
\begin{align*}
\kappa(h)= \# \left\{ i \in \{ 1, 3, \dots, 2m-1\} \mid \supp{h}\cap (c_i, c_{i+1}) \neq \emptyset \right\}.
\end{align*}
Here $\#$ denotes the cardinality of the set.
Let $z^\prime$ be a minimizer of $\kappa$ and let $e_1, \dots, e_{2p}$ denote a sequence given by Lemma \ref{lem_supp_finite_component} for $z^\prime$.
Since $z^\prime$ is not identity map, there exist $i$ and $i^\prime$ such that $c_{2i^\prime-1}<e_{2i-1}<e_{2i}<c_{2i^\prime}$ holds.
We note in the following that each element of $G_0(n)$ preserves the order of $\N^\mathbb{N}$.
Since there may exist more than one such $i$, we denote the smallest one by $i_-$ and the largest one by $i_+$.
Then there exists $w \in \langle f, g \rangle$ such that $w(e_{2{i_-} -1})>e_{2i_+}$ holds.
By construction of $w$, we have
\begin{align*}
\supp{z^\prime} \cap \supp{w^{-1}z^\prime w} \cap (c_{2i^\prime-1}, c_{2i^\prime})=\emptyset.
\end{align*}
This implies that we have
\begin{align*}
\supp{[z^\prime, w^{-1}z^\prime w]} \cap (c_{2i^\prime-1}, c_{2i^\prime})=\emptyset.
\end{align*}
Indeed, if $x$ is in $\supp{z^\prime}$, then $z^\prime(x)$ is in $\supp{z^\prime}$, thus $z^\prime(x)$ is not in $\supp{w^{-1}z^\prime w}$ and we have $[z^\prime, w^{-1}z^\prime w](x)=x$.
If $x$ is in $\supp{w^{-1}z^\prime w}$, then the image of $x$ by $w^{-1}z^\prime w$ is also in $\supp{w^{-1}z^\prime w}$. Since $z^\prime(x)=x$ holds, we have $[z^\prime, w^{-1}z^\prime w](x)=x$.

We note that
\begin{align*}
\supp{[z^\prime, w^{-1}z^\prime w]} \subsetneq \supp{z^\prime} \cup \supp{w^{-1}z^\prime w} \subsetneq \supp{f} \cup \supp{g}.
\end{align*}
However, $[z^\prime, w^{-1}z^\prime w]$ is not in $W$.
Indeed, if $\supp{z^\prime} \cap (c_{2j-1}, c_{2j}) =\emptyset$ for any other $j$, then we have
\begin{align*}
\supp{[z^\prime, w^{-1}z^\prime w]}\cap  (c_{2j-1}, c_{2j})=\emptyset,
\end{align*}
since if $x$ is in $(c_{2j-1}, c_{2j})$ then $w(x)$ is also in $(c_{2j-1}, c_{2j})$.
Thus if $[z^\prime, w^{-1}z^\prime w]$ is in $W$, it contradicts the minimality of $\kappa(z^\prime)$.
This implies that we have
\begin{align*}
z^\prime(w^{-1}z^\prime w)=(w^{-1}z^\prime w)z^\prime.
\end{align*}
Therefore, these two maps $z^\prime$ and $(w^{-1}z^\prime w)$ generate the group which is isomorphic to $\mathbb{Z}^2$.
We have the desired result.
\end{proof}
\end{theorem}
\begin{remark} \label{remark_torsion_free}
From this proof, we can see that $G_0(n)$ is torsion free.
Indeed, let $g$ $(g \neq \textrm{Id})$ be in $G_0(n)$ and assume that $g^m= \textrm{Id}$ holds.
Since $g$ preserves the order of $\N^\mathbb{N}$, for $x$ in $\supp{g}$, we have either
\begin{align*}
x < g(x)<g^2(x)< \cdots < g^m(x)=x
\shortintertext{or}
x>g(x)>g^2(x)<\cdots <g^m(x)=x.
\end{align*}
This is a contradiction.
\end{remark}
\begin{remark}
If there exists an embedding from $G_0(n)$ into either $G_0$ or the Monod group $H$ \cite{monod2013groups}, the theorem follows immediately because $G_0$ and $H$ contain no free subgroups.
\end{remark}
\subsection{The abelianization of $G_0(n)$ and simplicity of the commutator subgroup}
The idea for the theorems in this section comes from \cite{brown1987finiteness} and \cite{burillo2018commutators}.

We note that the group $F(n)$ is called $F_{n, 1}$ in \cite[section 4D]{brown1987finiteness}.
Thus there exists a surjective homomorphism $\phi: F(n) \to \mathbb{Z}^n$.
We briefly recall its definition to compute the abelianization of $G_0(n)$.

Let $A_{n+1}$ be the free abelian group generated by $e_-$, $e_+$, and $e_i$ ($i \in \mathbb{Z}$) and satisfying the relations $e_i =e_j$ if $i \equiv j \pmod {(n-1)}$.
Then the rank of $A_{n+1}$ is $n+1$.
From an $n$-ary tree $Y$ and an $n$-ary tree Z which contains $Y$ as a rooted subtree, we define the element $\delta(Z, Y)$ in $A_{n+1}$ as follows:
\begin{enumerate}
\item Label the leftmost leaf of Y as $-$, the rightmost as $+$, and the other leaves as
$1, 2, \dots $ from left to right.
\item To construct $Z$ from $Y$, add a caret to some leaf of $Y$.
Then record the label of the leaf.
\item Regard the obtained tree as $Y$ again.
\item Repeat the process (1) to (3) until the tree $Z$ is obtained.
\item Add up all the $e_i$ that have the recorded labels as indices.
\end{enumerate}
Since $e_i=e_j$ if $i \equiv j$ mod $(n-1)$, we can add carets in any order in process (2).
For example, for the trees $Y$ and $Z$ in Figure \ref{definition_Y_and_Z}, by adding carets from left to right, $\delta(Y, Z)=e_-+e_7+e_{10}+e_+=e_-+2e_1+e_+$.
\begin{figure}[tbp]
	\centering
	\includegraphics[width=80mm]{definition_Y_and_Z.pdf}
	\caption{The trees $Y$ and $Z$. }
	\label{definition_Y_and_Z}
\end{figure}
The process of this example is in Figure \ref{calculation_delta_Y_Z}.
\begin{figure}[tbp]
	\centering
	\includegraphics[width=150mm]{calculation_delta_Y_Z.pdf}
	\caption{A process of the calculation of $\delta(Y, Z)$. }
	\label{calculation_delta_Y_Z}
\end{figure}

For the group $F(n)$, we define the homomorphism $\phi: F(n) \to A_{n+1}$ as follows:
Let $x$ be in $F(n)$ and $(T_+, T_-)$ be a tree diagram that represents $x$.
Then we set
\begin{align*}
\phi(x):=\delta(W, T_-)-\delta(W, T_+),
\end{align*}
where $W$ is an $n$-ary tree that contains both $T_+$ and $T_-$.

From the construction, this map is independent of the choice of $W$ and $(T_+, T_-)$, and this is a homomorphism.
By calculating $\phi(x_0), \dots, \phi(x_{n-2})$, and $\phi(\x{o}{(n-1)})$, we have that $\operatorname{Im} \phi =\{\Sigma\lambda_ie_i \mid \Sigma \lambda_i=0 \} \cong \mathbb{Z}^n$.
Thus we obtain a surjective homomorphism $F(n) \to \mathbb{Z}^n$.

\begin{theorem}[{\cite[Lemma 2.1]{burillo2018commutators}} for $n=2$] \label{Th_abelianization_G0(n)}
The abelianization of $G_0(n)$ is isomorphic to $\mathbb{Z}^{n+1}$.
\begin{proof}
We define the map $\pi: Z(n) \to \mathbb{Z}^n \bigoplus \mathbb{Z}$ by
\begin{align*}
&\x{i}{s} \mapsto (\phi(\x{i}{s}), 0), & &y_s \mapsto (\bm{0}, 1).
\end{align*}
Since $G_0(n)$ is $(n+1)$-generated group, its abelianization is a quotient of $\mathbb{Z}^{n+1}$.
So if we obtain a surjective homomorphism $G_0(n) \to \mathbb{Z}^{n+1}$, then this map must be its abelianization map.
Thus it is sufficient to show that $\pi$ extends to a homomorphism $G_0(n) \to \mathbb{Z}^{n+1}$.
In order to do this, we only need to check that the relations in $R(n)$ are satisfied, which is clear except for (4).
By the definition of $\phi$, we have $\phi(\x{0}{s})=\bm{0}$ for each $\x{0}{s}$, where $s$ is in $\N^{<\mathbb{N}}$ such that $y_s$ is in $Y(n)$.
Indeed, since $s$ is neither $0\cdots0$ nor $(n-1)\cdots(n-1)$, $\phi(\x{0}{s})$ is calculated as in Figure \ref{phi_x0s=0}.
\begin{figure}[tbp]
	\centering
	\includegraphics[width=140mm]{phi_x0s=0.pdf}
	\caption{Note that $e_{s}=e_{s+(n-1)}$. }
	\label{phi_x0s=0}
\end{figure}
Thus the relation (4) is also satisfied.
\end{proof}
\end{theorem}
\begin{corollary}
Let $n, m \geq 2$.
Then the groups $G_0(n)$ and $G_0(m)$ are isomorphic if and only if $n=m$ holds.
\end{corollary}
In the following, we show that the commutator subgroup $G_0(n)^\prime=[G_0(n), G_0(n)]$ is simple.
As in the case of $n=2$, first we show that the second derived subgroup $G_0(n)^{\prime \prime}=[G_0(n)^\prime, G_0(n)^\prime]$ is simple by using Higman's Theorem, and
then we show that $G_0(n)^{\prime \prime}=G_0(n)^{\prime}$.

Let $\Gamma$ be a group of bijections of a set $E$.
For $\alpha \in \Gamma$, we write the set-theoretic support $\supp{\alpha}$ for the set $\{x \in E \mid \alpha(x) \neq x\}$.
\begin{theorem}[{\cite[Theorem 1]{MR72136}}] \label{Theorem_Higman_simple}
Suppose that for every $\alpha, \beta, \gamma \in \Gamma \setminus \{ 1_\Gamma \}$, there exists $\rho$ such that
\begin{align*}
\gamma \Bigl( \rho \bigl(\supp{\alpha}\cup \supp{\beta}) \Bigr) \cap \rho \bigl(\supp{\alpha}\cup \supp{\beta}\bigr) = \emptyset
\end{align*}
holds.
Then the commutator subgroup $\Gamma^\prime$ is simple.
\end{theorem}
\begin{theorem}[{\cite[Theorem 2]{burillo2018commutators}} for $n=2$] \label{theorem_commutator_simple}
The commutator subgroup of the group $G_0(n)$ is simple.
\end{theorem}
The following two lemmas complete the proof.
\begin{lemma}
The second derived subgroup $G_0(n)^{\prime \prime}$ is simple.
\begin{proof}
Let $\alpha, \beta, \gamma \in G_0(n)^\prime$.
Choose $x \in \supp{\gamma}$.
If $\gamma(x)>x$, then let $I$ be the set $\{x^\prime \in \N^\mathbb{N} \mid x<x^\prime<\gamma(x)\}$.
If $\gamma(x)<x$, then let $I$ be the set $\{x^\prime \in \N^\mathbb{N} \mid \gamma(x)<x^\prime<x\}$.
Then $\gamma(I) \cap I = \emptyset$ holds.

We note that $\supp{\alpha}$ and $\supp{\beta}$ are not $\N^\mathbb{N} \setminus \{00\cdots, (n-1)(n-1) \cdots\}$, by the argument in Theorem \ref{Theorem_no_free_group} (in particular, see the order \eqref{support_condition_commutator}).
Therefore there exists $\rho \in F(n)$ such that
\begin{align*}
\rho \bigl(\supp{\alpha} \cup \supp{\beta}\bigr) \subset I
\end{align*}
holds.
Then we have
\begin{align*}
\rho \bigl(\supp{\alpha} \cup \supp{\beta} \bigr) \cap \gamma \Bigl(\rho \bigl(\supp{\alpha} \cup \supp{\beta} \bigr) \Bigr) \subset I \cap \gamma(I)=\emptyset.
\end{align*}
By Theorem \ref{Theorem_Higman_simple}, $G_0(n)^{\prime \prime}$ is simple.
\end{proof}
\end{lemma}
\begin{lemma}[{\cite[Proposition 2.5]{burillo2018commutators}} for $n=2$]
$G_0(n)^\prime=G_0(n)^{\prime \prime}$.
\begin{proof}
We assume that $n \geq 3$.

Since $G_0(n)^{\prime \prime} \subset G_0(n)^\prime$ holds, we show $G_0(n)^\prime \subset G_0(n)^{\prime \prime}$.
Let $g \in G_0(n)^\prime$ and $f \y{s_1}{t_1}\cdots \y{s_m}{t_m}$ be its normal form.
By the definition of the map $\pi$ in Theorem \ref{Th_abelianization_G0(n)}, $t_1+\cdots +t_m=0$ holds and $f$ is in $F(n)^\prime$.
By \cite[Theorem 4.13]{brown1987finiteness}, we have $F(n)^\prime = F(n)^{\prime \prime} \subset G_0(n)^{\prime \prime}$.
Therefore, it is sufficient to show that $\y{s_1}{t_1}\cdots \y{s_m}{t_m}$ is in $G_0(n)^{\prime \prime}$.

We show this by induction on $k=|t_1|+\cdots + |t_m|$ for words $\y{s_1}{t_1}\cdots \y{s_m}{t_m}$ that satisfy $t_1+\cdots t_m=0$ (which may not be in normal).
We note that $k$ is an even number.
Since the base case $k=0$ is clear, we assume $k>0$.
Then, since $G_0(n)^{\prime \prime}$ is a normal subgroup of $G_0$, by cyclically conjugating, we can assume that the word starts with a subword of the form $y_sy_t^{-1}$.
For example, we can do the following:
\begin{align*}
y_{s_1}y_{s_2}\y{s_3}{-1}\y{s_4}{-1} \mapsto \y{s_1}{-1}(y_{s_1}y_{s_2}\y{s_3}{-1}\y{s_4}{-1})y_{s_1}=y_{s_2}\y{s_3}{-1}\y{s_4}{-1}y_{s_1}.
\end{align*}
By the inductive hypothesis, it is sufficient to show that $y_s \y{t}{-1}$ is in $G_0(n)^{\prime \prime}$.
We divide the proof into two cases:

Case (1): $s \perp t$.

Since $(y_s y_t^{-1})^{-1}=y_t y_s^{-1}$ holds, we can assume that $s<t$ without loss of generality.
By the definition of $Y(n)$, $s$ and $t$ are neither $000\cdots$ nor $(n-1)(n-1)(n-1)\cdots$.
Then since $n \geq 3$, there exists an element $h^\prime$ in $F(n)$ such that
\begin{align*}
h^\prime((n-1)000)=s
\shortintertext{and}
h^\prime((n-1)00(n-1)0)=t
\end{align*}
hold.
Indeed, we can construct it similarly as in Figure \ref{construction_g_uv}.
Then we have
\begin{align*}
y_s y_t^{-1}=({h^\prime}^{-1}y_{(n-1)000} h^\prime)({h^\prime}^{-1}\y{(n-1)00(n-1)0}{-1} h^\prime)={h^\prime}^{-1}y_{(n-1)000} \y{(n-1)00(n-1)0}{-1} h^\prime.
\end{align*}
Thus it is sufficient to show that $y_{(n-1)000}\y{(n-1)00(n-1)0}{-1}$ is in $G_0(n)^{\prime \prime}$.

Let $w=y_{(n-1)00}\y{(n-1)0(n-1)}{-1} \in G_0(n)^\prime$.
We note that there exists an element $h$ in $F(n)^\prime$ such that
\begin{align*}
h\bigl((n-1)00(n-1)(n-1)\bigr)=(n-1)00
\shortintertext{and}
h\bigl((n-1)0(n-1)\bigr)=(n-1)0(n-1)
\end{align*}
hold.
Indeed, let $h$ be as in Figure \ref{definition_h_in_simple}.
Then since $\phi(h)=\bm{0}$ holds where $\phi$ is the abelianization map of $F(n)$,  the element $h$ is in $F(n)^\prime$.
\begin{figure}[tbp]
	\centering
	\includegraphics[width=80mm]{definition_h_in_simple.pdf}
	\caption{The element $h$. }
	\label{definition_h_in_simple}
\end{figure}

Since $w$ is in $G_0(n)^\prime$ and $h$ is in $F(n)^\prime \subset G_0(n)^\prime$, the commutator $[w, h]=whw^{-1} h^{-1}$ is in $G_0(n)^{\prime \prime}$.
By the construction of $h$, we have $h w h^{-1}=y_{(n-1)00(n-1)(n-1)}\y{(n-1)0(n-1)}{-1}$.
Thus we have
\begin{align*}
[w, h]&=w (hwh^{-1})^{-1} \\
&=y_{(n-1)00}\y{(n-1)0(n-1)}{-1} (y_{(n-1)00(n-1)(n-1)}\y{(n-1)0(n-1)}{-1})^{-1}\\
&=y_{(n-1)00} \y{(n-1)00(n-1)(n-1)}{-1}.
\end{align*}
By applying expansion move (in Definition \ref{definition_five_moves}) to $y_{(n-1)00}$, we have
\begin{align*}
y_{(n-1)00} \y{(n-1)00(n-1)(n-1)}{-1}
&=\x{0}{(n-1)00}y_{(n-1)000}\y{(n-1)00(n-1)0}{-1}y_{(n-1)00(n-1)(n-1)}\y{(n-1)00(n-1)(n-1)}{-1} \\
&=\x{0}{(n-1)00}y_{(n-1)000}\y{(n-1)00(n-1)0}{-1}.
\end{align*}
Since $\x{0}{(n-1)00}$ is in $F(n)^\prime$ (see Figure \ref{phi_x0s=0}), we note that this element is in $F(n)^{\prime \prime} \subset G_0(n)^{\prime \prime}$.
Therefore we have
\begin{align*}
y_{(n-1)000}\y{(n-1)00(n-1)0}{-1}=\x{0}{(n-1)00}^{-1} [w, h] \in G_0(n)^{\prime \prime},
\end{align*}
as required.

Case (2): $s$ is a prefix of $t$ or vice versa.

Since $(y_s y_t^{-1})^{-1}=y_t y_s^{-1}$ holds, we can assume that $s$ is a prefix of $t$ without loss of generality.
Let $t=su$.
By expansion move, we have $y_s \y{su}{-1}=\x{0}{s}y_{s0}\y{s(n-1)0}{-1}y_{s(n-1)(n-1)}\y{su}{-1}$.
Then since $\x{0}{s}$ is in $F(n)^\prime=F(n)^{\prime \prime} \subset G_0(n)^{\prime \prime}$ (see Figure \ref{phi_x0s=0}), it is enough to show that $y_{s0}\y{s(n-1)0}{-1}y_{s(n-1)(n-1)}\y{su}{-1}$ is in $G_0(n)^{\prime \prime}$.

We further divide the proof of case (2) into two subcases.

Case (2-1): $0$ is a prefix of $u$.

We note that $s0 \perp s(n-1)0$ and $s(n-1)(n-1)\perp su$ hold.
Thus $y_{s0}\y{s(n-1)0}{-1}$ and $y_{s(n-1)(n-1)}\y{su}{-1}$ are in $G_0(n)^{\prime \prime}$, by case (1).

Case (2-2): $0$ is not a prefix of $u$.

By cyclically conjugating, it is sufficient to show that $\y{s(n-1)0}{-1}y_{s(n-1)(n-1)}\y{su}{-1}y_{s0}$ is in $G_0(n)^{\prime \prime}$.
Since $su \perp s0$ holds, $\y{s(n-1)0}{-1}y_{s(n-1)(n-1)}$ and $\y{su}{-1}y_{s0}$ are in $G_0(n)^{\prime \prime}$, by case (1).
\end{proof}
\end{lemma}
\subsection{The center of $G_0(n)$}\label{subsec_center_G0(n)}
In this section, we show that the center of the group $G_0(n)$ is trivial.
The idea for the theorem comes from \cite[Section 4]{cannon1996introductory}.
Let $D(n):=\{s \overline{0} \mid s \in \N^{<\mathbb{N}}\}$, where $\overline{0}$ denotes the element $000\cdots $ in $\N^\mathbb{N}$.
Then the following holds.
\begin{lemma} \label{lemma_rich_Thompson_F(n)}
For any $s\overline{0}$ in $D(n)$, there exists $x$ in $F(n)$ such that
\begin{align*}
\supp{x}=(s\overline{0}, \overline{(n-1)})=\{\xi \in \N^\mathbb{N} \mid s\overline{0}<\xi< \overline{(n-1)}\}
\end{align*}
holds.
\begin{proof}
If $s=0, \dots, (n-2)$, then $x_0, \dots, x_{n-2}$ satisfy the claim, respectively.
By regarding $(n-1)\overline{0}$ as $(n-1)0\overline{0}$, we can assume that the length of $s$ is greater than or equal to $2$.

Let $s=s^\prime i \overline{0}$ ($i \in \{0, \dots, n-2 \}$).
If $s^\prime=(n-1)\cdots(n-1)$, then $\x{i}{s^\prime}$ satisfies the claim.
If $s^\prime \neq (n-1)\cdots(n-1)$, by using $\x{i}{s^\prime}$, we can define an element in $F(n)$ as in Figure \ref{rich_elements_Fn}, which satisfies the claim
\begin{figure}[tbp]
	\centering
	\includegraphics[width=90mm]{rich_elements_Fn.pdf}
	\caption{The construction of an element of $F(n)$ from $\x{i}{s^\prime}$. We add a caret to each of the leaf $s(n-1)$ of  the domain tree and the rightmost leaf of the range tree. }
	\label{rich_elements_Fn}
\end{figure}
\end{proof}
\end{lemma}
We note that $D(n)$ is a dense subset of $\N^\mathbb{N}$.
\begin{theorem}[{\cite[Proposition 2.7]{burillo2018commutators}} for $n=2$] \label{theorem_center_trivial}
The center of $G_0(n)$ is trivial.
\begin{proof}
Let $f$ be an element of the center of $G_0(n)$.
For $g \in G_0(n)$, assume that $supp(g)=(b_1, \overline{(n-1)})$ holds.
Then we have $f(b_1)=b_1$.
Indeed, if not, either $f(b_1)>b_1$ or $f^{-1}(b_1)>b_1$ holds.
Since $g(b_1)=b_1$, in both cases, this contradicts that $fg=gf$ holds.

By Lemma \ref{lemma_rich_Thompson_F(n)}, for every $s\overline{0} \in D(n)$, there exists $g \in F(n) \subset G_0(n)$ such that $b_1=s \overline{0}$ holds.
Thus we have $f(s\overline{0})=s\overline{0}$ for every $s\overline{0} \in D(n)$.
Since $D(n)$ is a dense subset of $\N^\mathbb{N}$, we conclude that $f$ is the identity map.
\end{proof}
\end{theorem}
\subsection{Indecomposability with respect to direct products and free products}
In this section, we show that there exist no nontrivial ``decompositions'' using theorems of $G_0(n)$.

\begin{theorem}
There exists neither nontrivial direct product decompositions nor nontrivial free product decompositions.
\begin{proof}
Suppose that $G_0(n)$ is isomorphic to $K \times H$ for some groups $K$ and $H$.
We first assume that $H$ (or $K$) is abelian.
Then the center of $G_0(n)$ contains $\{1\} \times H$.
Since the center of $G_0(n)$ is trivial (Theorem \ref{theorem_center_trivial}), $H$ must be the trivial group.

We assume that $K$ and $H$ are not abelian.
We note that the commutator subgroup of $G_0(n)=K \times H$ is $[K, K] \times [H, H]$.
Since $[K, K]$ and $[H, H]$ are not trivial, the group $\{1\} \times [H, H]$ is a nontrivial normal subgroup of $[K, K] \times [H, H]$.
However, this contradicts that $[G_0(n), G_0(n)]$ is simple (Theorem \ref{theorem_commutator_simple}).

Finally, we assume that $G_0(n)=K \star H$ for nontrivial groups $K$ and $H$.
Let $k \in K \setminus \{1\}$ and $h \in H \setminus \{1\}$.
Since $G_0(n)$ is torsion free (Remark \ref{remark_torsion_free}), both $h$ and $k$ generate infinite cyclic groups $\langle k \rangle$ and $\langle h \rangle$, respectively.
This implies that $G_0(n)$ has a subgroup
\begin{align*}
\langle k \rangle \star \langle h \rangle \cong \mathbb{Z} \star \mathbb{Z}=F_2.
\end{align*}
By Theorem \ref{Theorem_no_free_group}, this is a contradiction.
\end{proof}
\end{theorem}

\subsection*{Acknowledgments}
I would like to appreciate Professor Motoko Kato and Professor Shin-ichi Oguni for several comments and suggestions.
I would also like to thank my supervisor, Professor Tomohiro Fukaya, for his comments and careful reading of the paper.
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