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     {\huge СИБИРСКИЕ \ ЭЛЕКТРОННЫЕ} \\[2mm]
     {\huge МАТЕМАТИЧЕСКИЕ ИЗВЕСТИЯ} \\[2mm]
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     {\flushleft\it Том 19, стр. 144--144 (2022) \hspace{65mm}{\rm\small УДК \udcs}}
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\title[Mutual embeddings of RAAGs and GBS groups]{Mutual embeddings of right-angled Artin groups and generalized Baumslag-Solitar groups}
\author{{F.A. Dudkin}}%
\address{Fedor Anatol'evich Dudkin  %Контактные данные всех авторов и место работы указываются только на английском языке
\newline\hphantom{iii} Sobolev Institute of Mathematics, 4, pr. Koptyuga, Novosibirsk, 630090, Russia
\newline\hphantom{iii} Novosibirsk State University, 1, Pirogova str., Novosibirsk, 630090, Russia}%

\email{dudkinf@math.nsc.ru}%

\thanks{\sc Dudkin, F.A.,
Mutual embeddings of right-angled Artin groups and generalized Baumslag-Solitar groups}
\thanks{\copyright \ 2022 Dudkin F.A.}
\thanks{\rm The work was supported by the Russian Science Foundation, project 19-11-00039}
\thanks{\it Received August 26, 2022, published December 31, 2022}%


\semrtop \vspace{1cm}
\maketitle {\small
\begin{quote}
\noindent{\sc Abstract.} A finitely generated group $G$ acting on a tree so that all vertex and edge stabilizers are
infinite cyclic groups is called a {\it generalized Baumslag–Solitar group} ($GBS$ group). In this paper, we study when a given $GBS$ group can be embedded in a {\it right-angled Artin group} ($RAAG$) and vice versa. An exhaustive description has been obtained in both cases. If an embedding exists, then we discuss its construction.\medskip

\noindent{\bf Keywords:} right-angled Artin group, generalized Baumslag–Solitar group, embedding problem.
 \end{quote}
}


\section{Introduction}

In combinatorial group theory, one can distinguish groups whose presentations are somehow connected with some graphs built on generating elements as on vertices. One of the most well-known and studied classes of such groups is the class of Artin groups, which includes braid groups, {\it right-angled Artin group}, Coxeter groups, and some other classes of groups. Right-angled Artin groups are also called {\it free partially commutative} or {\it graph groups}. We will briefly call groups from this class $RAAG$. This is an interesting and actively studied class of groups.

In what follows, we will use letters $\Gamma, \Lambda$ to denote undirected graphs without loops and multiple edges. Let $V(\Gamma)$ and $E(\Gamma)$ denote the sets of vertices and edges of the graph $\Gamma$. Any such graph defines $RAAG$ $A(\Gamma)$ as follows:

$$ A(\Gamma)\cong\left\langle~V(\Gamma)~|~ [a,b]=1,\ \mbox{for all}\ \{a,b\}\in E(\Gamma)~\right\rangle. $$

The class $RAAG$ includes different groups from free groups to free Abelian groups. They have applications not only in various areas of mathematics, but also in computer science and robotics. These groups were studied by A. Baudisch, C. Droms, and many other authors (see, for example, the survey \cite{DuReTr20}).

Another important and interesting class of groups defined by graphs is the class of generalized Baumslag–Solitar groups ($GBS$ groups). By the Bass–Serre theorem, $G$ is presentable as
$\pi_1(\mathbb{A})$, the fundamental group of a graph of groups $\mathbb{A}$ (see \cite{Serr80}) whose vertex and edge groups are infinite cyclic groups.

To each $GBS$-group $G$ we can associate a labeled graph $\mathbb{A}$, a particular kind of a graph of groups.
This labeled graph corresponds to an action of~$G$ on a tree and defines a presentation of $G$ (see \cite{ClayFor08} for more details about labeled graphs and their properties).

As D.~Robinson \cite{StA} noted, $GBS$ groups occupy an important position in combinatorial group theory due to the following properties:
non-cyclic $GBS$ groups are exactly those finitely generated groups of cohomological dimension 2 that have a commensurate infinite
cyclic subgroup; $GBS$ groups are coherent (every finitely generated subgroup admits a finite presentation).

Earlier, embeddings of $GBS$  groups \cite{Lev15q, Dud15} and embeddings of $RAAGs$ \cite{KiKo13} were actively studied. In this paper, we will study when a $GBS$ group can be embedded in a $RAAG$ and vice versa. Our approaches to the study of embedding provide not only a criterion for the existence of an embedding, but also allow us to construct such an embedding if it exists.

\begin{theorem} If a $GBS$ group $G$ is embedded in a $RAAG$ $H$, then $G\cong\mathbb{F}_n\times\mathbb{Z}, n\geqslant 0$. The embedding of $G\cong\mathbb{F}_n\times\mathbb{Z}$ in a $RAAG$ $H$ can be constructed algorithmically.
\label{GBSembRAAG}
\end{theorem}

\begin{theorem} If a finitely generated $RAAG$ $H$ embeds in a $GBS$ group $G$, then $H$ is either free or $H\cong\mathbb{F}_n\times\mathbb{Z}, n\geqslant 0$. The embedding of a $RAAG$ $H$ in $G$ can be constructed algorithmically.
\end{theorem}

\section{Preliminaries}

To describe $GBS$ groups, it is convenient to use the class of oriented pseudographs (loops and multiple edges allowed). A {\emph{pseudograph}} $A$ is a set of vertices $V(A)$,
set of edges $E(A)$, maps $\partial_0, \partial_1\colon E(A)\to
V(A)$~--- endpoints of the edge and inversion
$\bar{\phantom{x}}\colon E(A)\to E(A)$ such that
$\partial_0(\overline{e})=\partial_1(e),\partial_1(\overline{e})=\partial_0(e),\,\bar{\bar{e}}=e,\, \bar{e}\neq
e$.

Given a $GBS$-group $G$, we can represent the corresponding graph of groups by a {\it labeled graph} $\mathbb{A}=(A,\lambda)$, where $A$ is a finite connected pseudograph and $\lambda\colon E(A)\to \mathbb{Z}\setminus\{0\}$ labels the edges of $A$. The label $\lambda(e)$ of an edge $e$ with the origin $v=\partial_0(e)$ defines the embedding $\alpha_e\colon e\to v^{\lambda(e)}$ of the cyclic edge group $\langle e \rangle$ into the cyclic vertex group $\langle v \rangle$.

{\it The fundamental group} $\pi_1(\mathbb{A})$ {\it of a labeled graph} $\mathbb{A}=(A, \lambda)$ is described by generators and defining
relations. Denote by $\overline{A}$ the undirected pseudograph obtained from $A$ by identifying $e$ and $\overline{e}$. Denote by $\beta(A)$ the number of edges outside the maximal subtree in the pseudograph $\overline{A}$. Each maximal subtree $T$ in $\overline{A}$ defines the following presentation:

\begin{equation}
\left\langle
\begin{array}{lcl}
g_v, v\in V(\overline{A}), &\vrule&  g_{\partial_0(e)}^{\lambda(e)}=g_{\partial_1(e)}^{\lambda(\overline{e})}, e\in E(T),\\
t_e, e\in E(\overline{A})\setminus E(T) &\vrule&t_e^{-1}g_{\partial_0(e)}^{\lambda(e)}t_e=g_{\partial_1(e)}^{\lambda(\overline{e})}, e\in E(\overline{A})\setminus E(T)
\end{array}
\right\rangle.
\label{GBSRep}
\end{equation}
For different maximal subtrees, corresponding presentations define isomorphic groups \cite{Serr80}.

A generalized Baumslag–Solitar group is called {\it elementary} if it is isomorphic to $\mathbb{Z}, \mathbb{Z}^2$ or the Klein bottle group $\langle~a,t~|~t^{-1 }at=a^{-1}~\rangle$.

If two labeled graphs $\mathbb{A}$ and $\mathbb{B}$ define isomorphic $GBS$ groups $\pi_1(\mathbb{A})\cong\pi_1(\mathbb{B})$, and group $\pi_1(\mathbb{A})$
is not elementary, there is a finite sequence of {\it expansions} and {\it collapses} (see Fig. \ref{ris ExpCol}) connecting $\mathbb{A}$ and $\mathbb{B}$ \cite{ For02}. A labeled graph is called {\it reduced} if it does not allow collapses (the label $\pm 1$ can appear only on a loop).

\begin{figure}[h]
\centering{\includegraphics[width=90mm]{Gr02.eps}}
\caption{Collapse and expansion}
\label{ris ExpCol}
\end{figure}

We need some definitions and statements from \cite{KiKo13}. For a given graph $\Gamma$, we construct its {\it extension} --- the graph $\Gamma^e$ \cite{KiKo13}. Vertices of the graph $\Gamma^e$ are words in the group $A(\Gamma)$ conjugate to the vertices $\Gamma$. Two vertices in $A(\Gamma^e)$ commute if and only if they commute as words in $A(\Gamma)$.

\begin{lemma} \cite[Corollary 8]{KiKo13} Let $\Lambda$ and $\Gamma$ be two finite graphs and $\Lambda$ be a forest. Then $A(\Lambda)$ embeds in $A(\Gamma)$ if and only if $\Lambda$ embeds in $\Gamma^e$.
\label{embRAAGs}
\end{lemma}

\section{Auxiliary statements}

\begin{lemma} \label{F2embGBS} A free group $\mathbb{F}_2$ of rank 2 does not embed into a $GBS$ group $G$ if and only if the group $G$ is a solvable Baumslag–Solitar group $$BS(1,n)\cong\langle~a,t~|~t^{-1}at=a^n~\rangle, n\in\mathbb{Z}.$$
\end{lemma}

\selectlanguage{english}
\begin{proof}
\selectlanguage{russian}

Note that $BS(1,0)\cong\mathbb{Z}, BS(1,1)\cong\mathbb{Z}^2$, and $BS(1,-1)$ is isomorphic to the Klein bottle group. Hence, all elementary groups are solvable Baumslag–Solitar groups. Moreover, the group $BS(1,n)$ is metabelian for $n\neq 0$. Therefore the group $BS(1,n)$ cannot contain a free group of rank 2 as a subgroup.

Assume that the $GBS$ group $G$ is given by a reduced labeled graph $\mathbb{A}$ and does not contain subgroups isomorphic to $\mathbb{F}_2$. If $\beta(A)\geqslant 2$, then $G$ contains a free subgroup of rank 2 generated by generators of the second type $t_e, e\in E(\overline{A})\setminus E(T)$, see (\ref{GBSRep}). So $\beta(A)\leqslant 1$.

If $\beta(A)=0$ and the group $G$ is not elementary, then the center of $G$ is isomorphic to $\mathbb{Z}$. In this case $G$ contains $\mathbb{F}_2\times\mathbb{Z}, n\geqslant 2$ \cite{Lev15}. You can understand how embedding works from \cite{Serr80, Lev15}.

It remains to analyze the case $\beta(A)=1$. If, moreover, $G$ is not a solvable Baumslag–Solitar group $BS(1,n)$, then either $G\cong BS(m,n)$ and $m,n\not\in\{-1, 0,1\}$; or the labeled graph contains at least two vertices.

In the first case, the group $G$ contains a subgroup
$$\left\langle~a, b, c~\mid~a^m=b^n, b^m=c^n~\right\rangle$$
and, arguing as if $\beta(A)=0$, we get an embedding.

In the second case, the group $G$ contains the subgroup
$$\left\langle~a, b, c~\mid~a^m=b^n, b^k=c^l~\right\rangle$$
and again a free group of rank 2 embeds in $G$.
\end{proof}

\begin{lemma} Let $n\geqslant 2$ be an integer and $RAAG$ $H$ be given by a finite graph $\Gamma$. A $RAAG$ $G=A(\Gamma_n)\cong\mathbb{F}_n\times\mathbb{Z}$ embeds in a $RAAG$ $H$ if and only if the graph $\Gamma$ has a subgraph $\Gamma_2$ (see Fig. \ref{ris 2}).
\label{FnZembRAAG}
\end{lemma}

\selectlanguage{english}
\begin{proof}
\selectlanguage{russian}
\begin{figure}[h]
\centering{\includegraphics[width=40mm]{Gr2.eps}}
\caption{Graph $\Gamma_n$}
\label{ris 2}
\end{figure}

Indeed, if $\Gamma_2$ is a subgraph of the graph $\Gamma$, then $\mathbb{F}_2\times\mathbb{Z}$ is embedded in $H$. Group $\mathbb{F}_n\times\mathbb{Z}, n\geqslant 2$ is embedded in the $\mathbb{F}_2\times\mathbb{Z}$ group.

Now suppose that the group $G\cong\mathbb{F}_n\times\mathbb{Z}$ is embedded in a $RAAG$ $H$. If the graph $\Gamma$ does not have a subgraph $\Gamma_2$, then any three points generate one of the graphs shown in the Fig. \ref{ris 3}.

\begin{figure}[h]
\centering{\includegraphics[width=90mm]{Gr3.eps}}
\caption{Subgraphs on three points}
\label{ris 3}
\end{figure}

In this case, the graph $\Gamma$ decomposes into a union of cliques. Therefore the group $H$ is a free product of free abelian groups.

Note that the graph $\Gamma^e$ (not necessarily finite) is also a union of cliques. Therefore, the graph $\Gamma_2$ does not embed into $\Gamma^e$. Now it follows from Lemma \ref{embRAAGs} that $A(\Gamma_2)\cong \mathbb{F}_2\times\mathbb{Z}$ is not embedded in $H$. Therefore, $G$ does not embed into $H$ either. This contradiction completes the proof of Lemma.
\end{proof}

\section{Proofs of embedding theorems}

\noindent{\it Proof of Theorem 1.} Let $RAAG$ $H$ be given by the graph $\Gamma$. It is known that any $RAAG$ is residually torsion-free nilpotent group (see \cite{DuKr92, Tim22}). Since every torsion-free nilpotent group is a residually $\mathcal{F}_p$-group for every prime $p$, every $RAAG$ is a residually $\mathcal{F}_p$-group for every prime $p$. Therefore the group $G$ is a residually $\mathcal{F}_p$-group for every prime $p$.

In the paper \cite{Dud20} a criterion for the $\mathcal{F}_\pi$-residuality of $GBS$ groups is given for an arbitrary set of primes $\pi$. Using this criterion we can see that the group $G$ is represented by a reduced labeled graph $\mathbb{A}$ such that all labels on its edges are equal to $1$. Such a labeled graph can only be a bouquet of $n$ circles with labels $1$ (see Fig. \ref{ris 1}). So $G\cong\mathbb{F}_n\times\mathbb{Z}$.

\begin{figure}[h]
\centering{\includegraphics[width=60mm]{Gr1.eps}}
\caption{Bouquet of $n$ circles with labels $1$}
\label{ris 1}
\end{figure}

If $n=0$, then $G\cong\mathbb{Z}$. The embedding of $G$ in $H$ is obvious.

If $n=1$, then $G\cong\mathbb{Z}\times\mathbb{Z}$. Such a group is embedded in a $RAAG$ $H$ if and only if the corresponding graph has at least one edge. The embedding in this case is obvious.

If $n\geqslant 2$, then Lemma \ref{FnZembRAAG} gives an embedding criterion.
\hfill$\square$

\noindent{\it Proof of Theorem 2.} If a $RAAG$ $H$ is embedded in a $GBS$ group $G$, then the image of the embedding is a finitely generated subgroup of $G$. It acts on the same tree as $G$. Moreover, the vertex stabilizers of this action are subgroups of the vertex stabilizers of the action $G$. Hence the stabilizers are either trivial or cyclic.

If vertex stabilizers are trivial, then $H$ acts freely on the tree, so it is itself free. If $H\cong\mathbb{Z}$ then $H$ is always embedded in $G$. If $H\cong\mathbb{F}_n, n\geqslant 2$, then $H$ embeds in $G$ if and only if the group $G$ is not solvable Baumslag-Solitar group by Lemma \ref{F2embGBS}. Embeddings are discussed in the proof of Lemma \ref{F2embGBS}.

If vertex stabilizers are cyclic, then $H$ is a $GBS$ group. Arguing as in Theorem \ref{GBSembRAAG}, we get that $H\cong\mathbb{F}_n\times\mathbb{Z}, n\geqslant 0$. Hence $H$ is represented by the only reduced labeled graph. In this case, the problem of embedding $H$ in $G$ is algorithmically decidable \cite{Dud15}.
\hfill$\square$

\bigskip
\begin{thebibliography}{1} % Ссылки на литературу приводятся только на английском языке (даже если у источника нет англоязычной версии).


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