# Hyperbolic one-relator groups

Marco Linton  University of Oxford, Oxford, OX2 6GG, UK
<EMAIL_ADDRESS>
(Date: August 27, 2024)

###### Abstract.

We introduce two families of two-generator one-relator groups called primitive
extension groups and show that a one-relator group is hyperbolic if its
primitive extension subgroups are hyperbolic. This reduces the problem of
characterising hyperbolic one-relator groups to characterising hyperbolic
primitive extension groups. These new groups moreover admit explicit
decompositions as graphs of free groups with adjoined roots. In order to
obtain this result, we characterise $2$-free one-relator groups with
exceptional intersection in terms of Christoffel words, show that hyperbolic
one-relator groups have quasi-convex Magnus subgroups and build upon the one-
relator tower machinery developed in the authors previous article.

##  1\. Introduction

Since the introduction of hyperbolic groups by Gromov in the 80’s, a wealth of
powerful tools have been developed to study them. Thus, when studying a class
of groups, a classification of those that are hyperbolic can be very useful.
We here focus on the class of one-relator groups; that is, groups of the form
$F(\Sigma)/\langle\langle w\rangle\rangle$, where $F(\Sigma)$ denotes the free
group generated by $\Sigma$. The best possible statement that one could hope
for is known as Gersten’s conjecture [Ger92b] which asserts that one-relator
groups without Baumslag–Solitar subgroups are hyperbolic. In this article, we
present some progress in this direction, building on our previous work in
[Lin22].

A _Magnus subgroup_ of a one-relator group is a subgroup generated by a subset
of the generators $\Sigma$, omitting at least one generator mentioned in the
cyclic reduction of $w$. The theory of one-relator groups originated in 1930,
when Magnus proved the Freiheitssatz [Mag30].

######  Theorem (Freiheitssatz).

Magnus subgroups of one-relator groups are free.

The proof of the Freiheitssatz makes use of a hierarchy of one-relator groups
known as the Magnus hierarchy: a one-relator group $G$ splits as a HNN-
extension with one-relator vertex group $H$ of lower ‘complexity’ and where
the edge groups are Magnus subgroups of $H$. By understanding the splittings
that arise in this way, one may use induction to make conclusions about $G$.
See [MKS66, LS01, Lin22] for various versions and applications of the Magnus
hierarchy.

Almost a century later, the Magnus hierarchy is still a powerful tool for the
study of one-relator groups. However, the splittings that arise remain
somewhat mysterious. In [Lin22], the case that $G$ is a $2$-free one-relator
group was considered in detail: there it was shown that if $G$ is a finitely
generated $2$-free one-relator group, then $G$ is hyperbolic and acts
acylindrically on any Bass-Serre tree associated with a one-relator splitting.
When $G$ has torsion, the same properties hold [New68, Wis20, Lin22]. On the
other hand, the fact that Baumslag–Solitar groups cannot act acylindrically on
a tree [MO15] implies that the general case is more complicated.

We say that a one-relator group $F(\Sigma)/\langle\langle w\rangle\rangle$ has
an _exceptional intersection_ if there are subsets $A,B\subset\Sigma$ that
generate Magnus subgroups and such that the following holds in the quotient:

$\langle A\rangle\cap\langle B\rangle\neq\langle A\cap B\rangle.$

These groups were first studied by Collins in [Col04]. Howie then obtained
generalisations in the context of one-relator products of locally indicable
groups in [How05]. Examples of one-relator groups with exceptional
intersections include torus knot groups $\langle a,b\mid a^{p}=b^{q}\rangle$
and closed orientable surface groups of genus at least two $\langle
a_{1},b_{1},...,a_{g},b_{g}\mid[a_{1},b_{1}]=[a_{2},b_{2}]...[a_{g},b_{g}]\rangle$.

In [Lin22], it is shown that if $G=H*_{\psi}$ is a one-relator splitting where
$G$ does not contain any Baumslag–Solitar subgroups, $H$ is hyperbolic and the
edge groups of the splitting are quasi-convex and do not have exceptional
intersection, then $G$ acts acylindrically on its Bass-Serre tree and so is
hyperbolic by [BF92]. In light of this (and Theorem 4.8), when attempting to
understand the hyperbolicity of one-relator groups, the case of interest lies
when $H$ has an exceptional intersection. In this article, we take a step in
this direction and characterise $2$-free one-relator groups with exceptional
intersection.

Let $G$ be a one-relator group with exceptional intersection. One of the
following holds:

  1. (1)

$G$ is $2$-free,

  2. (2)

there is a two-generator one-relator generalised Baumslag–Solitar subgroup
$H<G$ such that every non-free two-generator subgroup of $G$ is conjugate into
$H$.

It turns out that $2$-free one-relator groups with an exceptional intersection
have presentations of a particular form that can easily be described in terms
of Christoffel words. We call these primitive exceptional intersection groups;
see Section 3.1 for their description. The two-generator generalised
Baumslag–Solitar subgroup appearing in the statement is a $w$-subgroup in the
sense of [LW22].

Thanks to the dichotomy provided by Theorem 3.19, with a little work we are
able to establish quasi-convexity of Magnus subgroups of all hyperbolic one-
relator groups.

Magnus subgroups of hyperbolic one-relator groups are quasi-convex.

As a consequence, we prove Theorem 4.9, a strengthening of the main tool from
[Lin22]. Note that hyperbolic one-relator groups can have distorted free
subgroups as there are many of hyperbolic (finitely generated free)-by-cyclic
one-relator groups. See [Kap99] for an explicit example. Theorem 4.8 was
already known in the case of one-relator groups with torsion by Newman’s
Spelling Theorem [New68]. It was also known in the case of hyperbolic one-
relator groups with quasi-convex one-relator hierarchies [Lin22].

Before stating our main result, we introduce two new families of two-generator
one-relator groups. Let $p/q\in\mathbb{Q}_{>0}$ and denote by

$A_{i,j}=\\{a_{i},a_{i+1},...,a_{j}\\}\leavevmode\nobreak\ .$

The first family is the fundamental group of the following graph of groups:

${{F(A_{0,k-1})}}$${{F(A_{1,k-1})*\langle
w\rangle}}$${{F(A_{1,k})}}$$\scriptstyle{\langle A_{1,k-1},x\rangle=\langle
A_{1,k-1},w^{p}\rangle}$$\scriptstyle{\langle A_{1,k-1},w^{q}\rangle=\langle
A_{1,k-1},y\rangle}$$\scriptstyle{\langle A_{0,k-1}\rangle=\langle
A_{1,k}\rangle}$

where $\\{\langle x\rangle,\langle y\rangle\\}$ form a malnormal family in
$F(A_{0,k})$ and $x,y\notin F(A_{1,k-1})$. The upper edge homomorphism simply
shifts the generators along. The second family is the fundamental group of the
following graph of groups:

${{F(A_{0,k-1})}}$${H}$${{F(A_{1,k})}}$$\scriptstyle{\langle
A_{1,k-1},x\rangle=\langle A_{1,k-1},x\rangle}$$\scriptstyle{\langle
A_{1,k-1},y\rangle=\langle A_{1,k-1},y\rangle}$$\scriptstyle{\langle
A_{0,k-1}\rangle=\langle A_{1,k}\rangle}$

where $H$ takes the following form

${{F(A_{1,k-1})}}$${\langle w\rangle}$${{F(x,y)}}$$\scriptstyle{\langle
z\rangle=\langle w^{p}\rangle}$$\scriptstyle{\langle w^{q}\rangle=\langle
xy\rangle}$

and where $\langle z\rangle$ is malnormal in $F(A_{1,k-1})$ and $x,y\notin
F(A_{1,k-1})$. We call these groups _primitive extension groups_.

In Section 5 it is shown that primitive extension groups are two-generator
one-relator groups. Examples of primitive extension groups include all one-
relator ascending HNN-extensions of finitely generated free groups and, in
particular, $\operatorname{BS}(1,n)$. These correspond to the families where
$x=a_{0}$ and $p=1$, see Example 5.6. By [Mut21], the hyperbolicity of one-
relator groups in these subfamilies are understood. Although two-generator
one-relator hyperbolic groups have received some attention [KW99, GKL21],
there are currently no further criteria that do not rely on small
cancellation-like criteria to determine when a primitive extension group is
hyperbolic.

Our main result can now be stated as follows.

A one-relator group is hyperbolic (and virtually special) if its primitive
extension subgroups are hyperbolic (and virtually special).

It is immediate that Gersten’s conjecture [Ger92b] needs only to be proved for
primitive extension groups in order to hold for all one-relator groups.

Gersten’s conjecture is true if it is true for primitive extension groups.

### Acknowledgments

The author would like to thank Saul Schleimer for many helpful conversations,
in particular for pointing him in the direction of Christoffel words. A large
part of the work in this article also appears in the author’s PhD thesis,
completed at the University of Warwick. While carrying out part of this work,
the author also received funding from the European Research Council (ERC)
under the European Union’s Horizon 2020 research and innovation programme
(Grant agreement No. 850930).

##  2\. Preliminaries on free groups

In this section, we recall some standard material from the theory of free
groups and prove some technical results that will be of use to us for the
proof of Theorem 3.19.

If $\Sigma$ is a set, we denote by $F(\Sigma)$ the free group, freely
generated by $\Sigma$. If $\Delta$ is another set, we write
$F(\Sigma,\Delta)=F(\Sigma\sqcup\Delta)$.

###  2.1. Proper powers

We say an element $w\in F(\Sigma)$ is a _proper power_ if there is some $u\in
F(\Sigma)$ and $n\geq 2$ such that $w=u^{n}$.

######  Lemma 2.1.

Let $1\neq y^{-1}zy\in F(\Sigma)$ be freely reduced with $z$ cyclically
reduced and not equal to a proper power. Then:

  1. (1)

if $y=1$ and there is some $i\in\mathbb{Z}$, $g,h\in F(\Sigma)$ such that
$z^{i}=gzh$ is freely reduced, then $i\geq 1$ and $g,h\in\langle z\rangle$,

  2. (2)

if $y\neq 1$ and there is some $i,j\in\mathbb{Z}$, $g,h\in F(\Sigma)$ such
that $y^{-1}z^{i}y=gy^{-1}z^{j}yh$ is freely reduced, then $g,h=1$ and $i=j$.

###### Proof.

Suppose for a contradiction that the first assertion does not hold. We may
assume that $i=2$ or $i=-2$. Then
$z=z^{\prime}z^{\prime\prime}=z^{\prime\prime}z^{\prime}$ in the first case
and $z=z^{\prime}z^{\prime\prime}=(z^{\prime})^{-1}(z^{\prime\prime})^{-1}$ in
the second. But then the first case cannot happen by [LS62, Lemma 3] and the
second case cannot happen as a non-trivial element of a free group cannot be
conjugate to its own inverse. Thus, the first assertion holds.

For the proof of the second assertion, we use the first assertion and the fact
that $z$ is cyclically reduced. ∎

######  Lemma 2.2.

Let $F(\Sigma)$ be a free group with $\Sigma=A\sqcup B$ and let $b\in\langle
B\rangle$ and $z\in F(\Sigma)$ be freely reduced elements. Suppose that $z$
begins and ends with generators in $A\sqcup A^{-1}$, that $zb$ is not a proper
power and that there are elements $i\in\mathbb{Z}$, $g,h\in F(\Sigma)$ such
that $(zb)^{i}=gzh$ is freely reduced. Then $i\geq 1$ and $g\in\langle
zb\rangle$.

###### Proof.

By Lemma 2.1, we may assume that $b\neq 1$. If $i\leq-1$, since $b$ does not
mention any $A\sqcup A^{-1}$ generators, it follows that $z$ has a non-trivial
prefix equal to its own inverse. Since this is not possible, we may also
assume that $i\geq 1$.

Now suppose for a contradiction that $i\geq 1$ and $g\notin\langle zb\rangle$.
Then we have equalities of the form:

$\displaystyle z$ $\displaystyle=z_{1}z_{2}z_{3}\leavevmode\nobreak\ ,$
$\displaystyle z$ $\displaystyle=z_{3}bz_{1}\leavevmode\nobreak\ ,$

for some $z_{1},z_{2},z_{3}\in F(\Sigma)$, where the words on the right hand
side are freely reduced.

We show by induction on $\left\lvert z_{1}\right\rvert+\left\lvert
z_{3}\right\rvert$ that $z_{2}=b$. If $\left\lvert
z_{1}\right\rvert=\left\lvert z_{3}\right\rvert$, then clearly $z_{2}=b$. So
suppose that $\left\lvert z_{1}\right\rvert<\left\lvert z_{3}\right\rvert$,
the other case is symmetric. Since the first letter of $z_{3}$ is in $A\sqcup
A^{-1}$ and $b\in\langle B\rangle$, we have that $\left\lvert
z_{3}\right\rvert>\left\lvert z_{1}z_{2}\right\rvert$. Now by [LS62, Lemma 2],
we have $z_{3}=(z_{1}z_{2})^{i}z^{\prime}$ for some $i\geq 1$ and where
$z^{\prime}$ is a proper prefix of $z_{1}z_{2}$. Then we get:

$\displaystyle z$
$\displaystyle=(z_{1}z_{2})^{i+1}z^{\prime}\leavevmode\nobreak\ ,$
$\displaystyle z$
$\displaystyle=(z_{1}z_{2})^{i}z^{\prime}bz_{1}\leavevmode\nobreak\ .$

By comparing suffixes, we have:

$z_{1}z_{2}z^{\prime}=z^{\prime}bz_{1}\leavevmode\nobreak\ .$

But now we obtain equalities of the same form as before, with $z^{\prime}$
playing the role of $z_{3}$. Since $\left\lvert
z^{\prime}\right\rvert<\left\lvert z_{3}\right\rvert$, by induction, we see
that $z_{2}=b$.

If $z_{2}=b$, then $zb$ is a conjugate of itself and so must be a proper power
by [LS62, Lemma 3]. Thus we obtain the required contradiction and conclude
that $g\in\langle zb\rangle$. ∎

###  2.2. Subgroups

A _graph_ is a $1$-dimensional CW-complex. A morphism of graphs
$\Gamma\to\Delta$ is a map sending vertices to vertices and edges
homeomorphically to edges. A morphism of graphs is an _immersion_ , denoted by
$\looparrowright$, if it is locally injective. It is a fundamental observation
due to Stallings [Sta83] that subgroups of free groups can be represented by
immersions of pointed graphs $(\Gamma,x)\looparrowright(\Delta,y)$. The _core_
of a graph $\Gamma$, denoted by $\operatorname{Core}(\Gamma)$, is the union of
the images of all immersed cycles $S^{1}\looparrowright\Gamma$. Then conjugacy
classes of subgroups of free groups can be represented by immersions of core
graphs $\Gamma\looparrowright\Delta$.

######  Lemma 2.3.

Let $F(A,B,C)$ be a free group and let $x\in\langle A,B\rangle-\langle
B\rangle$ and $y\in\langle B,C\rangle-\langle B\rangle$. If $H<F(A,B,C)$ is a
subgroup of rank two, containing $\langle x,y\rangle$, then one of the
following holds:

  1. (1)

there are elements $u,v$ and non zero integers $i,j$, such that $H=\langle
u,v\rangle$ and $x=u^{i}$, $y=v^{j}$,

  2. (2)

there are elements $u,v,w$ and non zero integers $i,j$ such that $H=\langle
u^{-1}wu,u^{-1}v\rangle$ and $x=u^{-1}w^{i}u$, $y=v^{-1}w^{j}v$.

###### Proof.

Let $\Delta$ be a rose graph with one edge for each element in $A\sqcup
B\sqcup C$ so that we may identify $\pi_{1}(\Delta)$ with $F(A,B,C)$. Then we
may represent $\langle x,y\rangle$ and $H$ by graph immersions
$\Gamma\looparrowright\Delta$ and $\Lambda\looparrowright\Delta$. By
assumption, $\Gamma\looparrowright\Delta$ factors through
$\Lambda\looparrowright\Delta$. Thus, there must be loops based at the same
vertex in $\Lambda$ with labels $x$ and $y$, covering $\Lambda$. Since a path
labelled by $x$ cannot traverse any $C$-edges and a path labelled by $y$
cannot traverse any $A$-edges, it follows that there is a decomposition
$Q^{(1)}=Q_{1}\cup Q_{2}$ where $\chi(Q_{1}),\chi(Q_{2})=0$ and $Q_{1}$ only
contains $A$-edges and $B$-edges and $Q_{2}$ only contains $B$-edges and
$C$-edges. Moreover, the path labelled by $x$ is supported in $Q_{1}$ and the
path labelled by $y$ is supported in $Q_{2}$. If $Q_{1}\cap Q_{2}$ is
connected, (1) must hold. If $Q_{1}\cap Q_{2}$ is not connected, then (2) must
hold. ∎

######  Lemma 2.4.

Let $F(A,B,C)$ be a free group and let $x\in\langle A,B\rangle-\langle
B\rangle$, $y\in\langle B,C\rangle-\langle B\rangle$ and $z\in\langle
B\rangle-1$. If $H<F(A,B,C)$ is a subgroup of rank two, containing $\langle
xy,z\rangle$, then there is an element $u$ and a non zero integer $i$, such
that $H=\langle xy,u\rangle$ and $z=u^{i}$.

###### Proof.

Just as in the proof of Lemma 2.3, let $\Delta$ be a rose graph with one edge
for each element in $A\sqcup B\sqcup C$ so that we may identify
$\pi_{1}(\Delta)$ with $F(A,B,C)$. Then we may represent $\langle xy,z\rangle$
and $H$ by graph immersions $\Gamma\looparrowright\Delta$ and
$\Lambda\looparrowright\Delta$. By assumption, $\Gamma\looparrowright\Delta$
factors through $\Lambda\looparrowright\Delta$. Thus, there must be loops
based at the same vertex in $\Lambda$ with labels $xy$ and $z$, covering
$\Lambda$. Since $z$ only traverses $B$-edges, it follows that there is a
decomposition $Q^{(1)}=Q_{1}\cup Q_{2}\cup Q_{3}$ where $\chi(Q_{1})=0$,
$Q_{1}$ only contains $B$-edges, $\chi(Q_{2})=\chi(Q_{3})=1$, $Q_{2}$ only
contains $A$-edges and $B$-edges and $Q_{3}$ only contains $B$-edges and
$C$-edges. Moreover, $z$ is supported in $Q_{1}$, $x$ is supported in
$Q_{1}\cup Q_{2}$ and $y$ is supported in $Q_{1}\cup Q_{3}$. Now the result
follows. ∎

The _fibre product_ $\Gamma\times_{\Delta}\Lambda$ of two graph immersions
$f:\Gamma\looparrowright\Delta$, $g:\Lambda\looparrowright\Delta$ is defined
as the graph with vertices:

$V(\Gamma\times_{\Delta}\Lambda)=\\{(v_{\Gamma},v_{\Lambda})\in
V(\Gamma)\times V(\Lambda)\mid f(v_{\Gamma})=g(v_{\Lambda})\\}$

and edges:

$E(\Gamma\times_{\Delta}\Lambda)=\\{(e_{\Gamma},e_{\Lambda})\in
E(\Gamma)\times E(\Lambda)\mid f(e_{\Gamma})=g(e_{\Lambda}),\text{ respecting
orientations}\\}$

As elucidated in [Sta83], there is a correspondence between the double cosets
$\pi_{1}(\Lambda)h\pi_{1}(\Gamma)$ such that
$\pi_{1}(\Gamma)\cap\pi_{1}(\Lambda)^{h}\neq 1$ and components of the core of
the fibre product graph $\operatorname{Core}(\Gamma\times_{\Delta}\Lambda)$
given by the $\pi_{1}$ functor.

##  3\. Exceptional intersection groups

The interactions between Magnus subgroups of one-relator groups are well
understood. The following is [Col04, Theorem 2].

######  Theorem 3.1.

Let $F(\Sigma)/\langle\langle w\rangle\rangle$ be a one-relator group and
suppose $\Sigma=A\sqcup B\sqcup C$. If $\langle A,B\rangle$ and $\langle
B,C\rangle$ are Magnus subgroups, then one of the following holds:

  1. (1)

$\langle A,B\rangle\cap\langle B,C\rangle=\langle B\rangle$,

  2. (2)

$\langle A,B\rangle\cap\langle B,C\rangle=\langle B\rangle*\mathbb{Z}$.

We say the Magnus subgroups $\langle A,B\rangle$ and $\langle B,C\rangle$ have
_exceptional intersection_ if the latter situation occurs.

######  Definition 3.2.

A one-relator presentation $\langle\Sigma\mid w\rangle$ is an _exceptional
intersection presentation_ if there are disjoint subsets $A,B,C\subset\Sigma$
such that $\langle A,B\rangle$ and $\langle B,C\rangle$ have exceptional
intersection. A one-relator group $G$ is an _exceptional intersection group_
if it has an exceptional intersection presentation.

The following result appears as [Col04, Corollary 2.3].

######  Corollary 3.3.

Exceptional intersection groups are torsion-free.

###  3.1. Primitive exceptional intersection groups

In this section, we introduce two families of one-relator groups called
primitive exceptional intersection groups. Our aim will be to show that they
are precisely the exceptional intersection groups that are $2$-free.

Before defining our groups, we first need to discuss certain primitive
elements of the free group of rank two. Recall that an element of a free group
$w\in F(\Sigma)$ is _primitive_ if $w$ forms part of a free basis for
$F(\Sigma)$. Otherwise, $w$ is _imprimitive_. The primitive elements of
interest to us are known as Christoffel words and were first introduced in
[Chr73]. They are parametrised by a rational slope $p/q\in\mathbb{Q}_{>0}$.
Let $\Gamma\subset\mathbb{R}^{2}$ denote the Cayley graph for $\mathbb{Z}^{2}$
on the generating set $a=(1,0)$, $b=(0,1)$. Let $L\subset\mathbb{R}^{2}$ be
the line segment beginning at the origin and ending at the vertex $(q,p)$. Now
let $P\subset\Gamma$ be the shortest length edge-path connecting the endpoints
of $L$, remaining below $L$ and such that there are no integral points
contained in the region enclosed by $L\cup P$. See Figure 1 for an example.
The word in $a$ and $b$ traced out by $P$ is denoted by:

$\operatorname{pr}_{p/q}(a,b)\leavevmode\nobreak\ .$

By [OZ81, Theorem 1.2], every primitive element of $F(a,b)$ is conjugate into
the set

$\left\\{a^{\pm 1},b^{\pm 1},\operatorname{pr}_{p/q}\left(a^{\pm 1},b^{\pm
1}\right)\Bigm{|}\frac{p}{q}\in\mathbb{Q}_{>0}\right\\}\leavevmode\nobreak\ .$

Figure 1. $L$ is in green with slope $5/6$. The $a$ edges are in blue and the
$b$ edges are in red, so $\operatorname{pr}_{5/6}(a,b)=a^{2}babababab$.

Now consider the free group $F(A,B,C)$, freely generated by disjoint sets
$A,B,C$. Let $p/q\in\mathbb{Q}_{>0}$ and let:

$\displaystyle x$ $\displaystyle\in\langle A,B\rangle-\langle
B\rangle\leavevmode\nobreak\ ,$ $\displaystyle y$ $\displaystyle\in\langle
B,C\rangle-\langle B\rangle\leavevmode\nobreak\ .$

Then we call $\operatorname{pr}_{p/q}(x,y)$ a _primitive exceptional
intersection word of the first type_ if the following hold:

  1. (1)

$\\{\langle x\rangle,\langle y\rangle\\}$ is a malnormal family (that is,
$\langle x\rangle$ and $\langle y\rangle$ are malnormal and no conjugate of
$\langle x\rangle$ intersects $\langle y\rangle$ non-trivially),

  2. (2)

if $p/q=1$, then there is no $a\in\langle A,B\rangle-\langle B\rangle$,
$c\in\langle B,C\rangle-\langle B\rangle$ such that
$\operatorname{pr}_{1}(x,y)=\operatorname{pr}_{1}(a,c)$ and $\\{\langle
a\rangle,\langle c\rangle\\}$ is not a malnormal family.

By definition, we see that $\langle x,y\rangle$ is an infinite cyclic subgroup
of $G=F(A,B,C)/\langle\langle w\rangle\rangle$. We find that $G$ has the
following exceptional intersection:

$\langle A,B\rangle\cap\langle B,C\rangle=_{G}\langle B\rangle*\langle
x^{q}\rangle=\langle B\rangle*\langle y^{-p}\rangle\leavevmode\nobreak\ .$

The following example demonstrates why we require the second condition in the
definition.

######  Example 3.4.

Consider the word $\operatorname{pr}_{1}(a^{2}b^{-1},bc^{2})\in F(a,b,c)$.
Although the subgroups $\\{\langle a^{2}b^{-1}\rangle,\langle
bc^{2}\rangle\\}$ form a malnormal family, we have

$\operatorname{pr}_{1}(a^{2}b^{-1},bc^{2})=\operatorname{pr}_{1}(a^{2},c^{2})=a^{2}c^{2}\leavevmode\nobreak\
,$

where $\\{\langle a^{2}\rangle,\langle c^{2}\rangle\\}$ is not a malnormal
family. Hence, $\operatorname{pr}_{1}(a^{2}b^{-1},bc^{2})$ is not a primitive
exceptional intersection word of the first type.

Now let $z\in\langle B\rangle-1$. We call $\operatorname{pr}_{p/q}(xy,z)$ a
_primitive exceptional intersection word of the second type_ if the following
hold:

  1. (1)

$\langle z\rangle$ is malnormal,

  2. (2)

if $p/q=k\in\mathbb{N}$, then there is no $a\in\langle A,B\rangle-\langle
B\rangle$, $c\in\langle B,C\rangle-\langle B\rangle$ such that
$\operatorname{pr}_{k}(xy,z)=\operatorname{pr}_{1}(a,c)$ and $\\{\langle
a\rangle,\langle c\rangle\\}$ is not a malnormal family.

By definition, we see that $\langle xy,z\rangle$ is an infinite cyclic
subgroup of $G=F(A,B,C)/\langle\langle w\rangle\rangle$ and so
$(xy)^{-1}z(xy)=_{G}z$. We find that $G$ has the following exceptional
intersection:

$\langle A,B\rangle\cap\langle B,C\rangle=_{G}\langle B\rangle*\langle
x^{-1}zx\rangle=\langle B\rangle*\langle yzy^{-1}\rangle\leavevmode\nobreak\
.$

A word $w\in F(A,B,C)$ is a _primitive exceptional intersection word_ if $w$
is a primitive exceptional intersection word of the first or second type.

######  Definition 3.5.

A group $G$ is a _primitive exceptional intersection group_ if $G\cong
F(\Sigma)/\langle\langle w\rangle\rangle$ where $w$ is a primitive exceptional
intersection word.

######  Example 3.6.

Consider the word
$\operatorname{pr}_{k}(a^{2}b^{2}c^{2},b)=a^{2}b^{2}c^{2}b^{k}\in F(a,b,c)$.
Since this is equal to $\operatorname{pr}_{1}(a^{2}b^{2-k},b^{-k}c^{2}b^{k})$
where $\langle b^{-k}c^{2}b^{k}\rangle$ is not malnormal, it is not a
primitive exceptional intersection word of the second type. However, it is
also equal to the word $\operatorname{pr}_{2}(a^{2}b^{2-k},b^{-k}cb^{k})$ (or
$\operatorname{pr}_{1/2}(a,b^{2}c^{2}b^{2})$ when $k=2$) which is a primitive
exceptional intersection word of the first type. Thus

$G=F(a,b,c)/\langle\langle a^{2}b^{2}c^{2}b^{k}\rangle\rangle$

is a primitive exceptional intersection group for all $k$.

The proof of the following theorem is rather involved and will take up the
remainder of this section.

######  Theorem 3.7.

Primitive exceptional intersection groups are $2$-free.

Before proceeding with the proof, we mention some important definitions and a
result from [LW22]. Define the _primitivity rank_ of an element $w\in
F(\Sigma)$ as the following quantity:

$\pi(w)=\min{\\{\operatorname{rk}(K)\mid w\in K<F,\text{ $w$ not primitive in
$K$}\\}}\in\mathbb{N}\cup\\{\infty\\}.$

Given an element $w\in F(\Sigma)$ such that $\pi(w)<\infty$, a subgroup
$K<F(\Sigma)$ is a _$w$ -subgroup_ if the following hold:

  1. (1)

$w\in K$ and $w$ is not primitive in $K$,

  2. (2)

$\operatorname{rk}(K)=\pi(w)$,

  3. (3)

if $K<K^{\prime}$, then $\operatorname{rk}(K)<\operatorname{rk}(K^{\prime})$.

In [LW22], Louder and Wilton connect the primitivity rank $\pi(w)$ with
subgroup properties of the one-relator group $F(\Sigma)/\langle\langle
w\rangle\rangle$. The result we shall need is the following, appearing as
[LW22, Theorem 1.5].

######  Theorem 3.8.

A one-relator group $F(\Sigma)/\langle\langle w\rangle\rangle$ is $2$-free if
and only if $\pi(w)>2$.

The idea of the proof of Theorem 3.7 will then be to show that if $w\in
F(\Sigma)$ is a primitive exceptional intersection word, then there cannot be
any two-generator $w$-subgroups.

###### Proof of Theorem 3.7.

If $G$ is a primitive exceptional intersection group, then there is a free
group $F(A,B,C)$, a rational number $p/q\in\mathbb{Q}_{>0}$, and elements:

$\displaystyle x$ $\displaystyle\in\langle A,B\rangle-\langle
B\rangle\leavevmode\nobreak\ ,$ $\displaystyle y$ $\displaystyle\in\langle
B,C\rangle-\langle B\rangle\leavevmode\nobreak\ ,$ $\displaystyle z$
$\displaystyle\in\langle B\rangle-1\leavevmode\nobreak\ ,$

such that $G\cong F(A,B,C)/\langle\langle w\rangle\rangle$ where one of the
following holds:

  1. (1)

$w=\operatorname{pr}_{p/q}(x,y)$ is a primitive exceptional intersection word
of the first type,

  2. (2)

$w=\operatorname{pr}_{p/q}(xy,z)$ is a primitive exceptional intersection word
of the second type.

Let us assume for sake of contradiction that $G$ is not $2$-free. Then by
Corollary 3.3 and Theorem 3.8, we have that $\pi(w)=2$. Thus, there is some
$w$-subgroup $K<F(A,B,C)$ such that $\operatorname{rk}(K)=2$. We will handle
the two cases at the same time as they are very similar.

Let $\Delta$ be a rose graph with one edge for each element in $A\sqcup
B\sqcup C$. The edges of $\Delta$ can then be partitioned into $A$-edges,
$B$-edges and $C$-edges. Denote by $\omega:S^{1}\looparrowright\Delta$ the
cycle representing $w$. There is an immersion of core graphs
$\Lambda\looparrowright\Delta$ representing the conjugacy class of $K$ and
such that there is a lift $\lambda:S^{1}\looparrowright\Lambda$ of $\omega$.
Now let $\Gamma\looparrowright\Delta$ be the graph immersion of core graphs
representing the conjugacy class of $\langle x,y\rangle<\pi_{1}(\Delta)$ (or
$\langle xy,z\rangle$ if we are in the second case). Then there is a lift
$\gamma:S^{1}\looparrowright\Gamma$ of $\omega$.

######  Lemma 3.9.

There is some connected component
$\Theta\subset\operatorname{Core}(\Gamma\times_{\Delta}\Lambda)$ such that
$\Theta\cong S^{1}$ and $\lambda=p_{\Lambda}\mid\Theta$.

###### Proof.

By [Ken09, Theorem 1], either
$\chi(\operatorname{Core}(\Gamma\times_{\Delta}\Lambda))=0$, or
$\operatorname{rk}(\langle\pi_{1}(\Lambda)^{g},\pi_{1}(\Gamma)\rangle)=2$ for
some $g\in\pi_{1}(\Delta)$. In the first case, we must have that $\lambda$
factors through some component
$S^{1}\subset\operatorname{Core}(\Gamma\times_{\Delta}\Lambda)$. Since $G$ is
torsion-free, $w$ cannot be a proper power by [KMS60]. Therefore, $\omega$ and
$\lambda$ cannot factor through $S^{1}$ as a proper cover and we are done.

Now suppose that
$\operatorname{rk}(\langle\pi_{1}(\Lambda)^{g},\pi_{1}(\Gamma)\rangle)=2$. By
definition of $w$-subgroups, we must have that
$\pi_{1}(\Gamma)<\pi_{1}(\Lambda)^{g}$. Now it follows from our assumptions on
$x,y,z$ and Lemmas 2.3 and 2.4, that $\pi_{1}(\Gamma)=\pi_{1}(\Lambda)^{g}$,
contradicting the fact that $w$ is not primitive in $K$. ∎

We will make use of the following factorisations of $x$ and $y$:

$\displaystyle x$ $\displaystyle=b_{1}\cdot x_{1}^{-1}\cdot x_{2}\cdot
x_{1}\cdot b_{2}\leavevmode\nobreak\ ,$ $\displaystyle y$
$\displaystyle=b_{3}\cdot y_{1}^{-1}\cdot y_{2}\cdot y_{1}\cdot
b_{4}\leavevmode\nobreak\ ,$

as freely reduced words, where $b_{1},b_{2},b_{3},b_{4}\in\langle B\rangle$,
$x_{1}^{-1}\cdot x_{2}\cdot x_{1}$ and $y_{1}^{-1}\cdot y_{2}\cdot y_{1}$ do
not begin or end with a $B$-letter and $x_{2}$ and $y_{2}$ are cyclically
reduced.

If $n\geq 1$, denote by $u_{n}$ the free reduction of
$x_{1}^{-1}x_{2}x_{1}(b_{2}b_{1}x_{1}^{-1}x_{2}x_{1})^{n-1}$. Similarly,
denote by $v_{n}$ the free reduction of
$y_{1}^{-1}y_{2}y_{1}(b_{4}b_{3}y_{1}^{-1}y_{2}y_{1})^{n-1}$.

######  Lemma 3.10.

Let $n\geq 1$ and let $\alpha:I\looparrowright\Lambda$ be a path labelled by
$u_{n}$ or $v_{n}$. Then $\alpha$ must traverse a vertex of degree at least
three, other than at its endpoints.

###### Proof.

We shall prove the result for $u_{n}$ as the other case is identical. Firstly,
we show that $\Gamma$ supports at most one path with label $u_{n}$. If
$\pi_{1}(\Gamma)=\langle xy,z\rangle$, then $\Gamma$ supports one path with
label $u_{1}$ and no path with label $u_{k}$ for any $k\geq 2$. Let us suppose
that $\pi_{1}(\Gamma)=\langle x,y\rangle$ and that there are two paths in
$\Gamma$ with label $u_{n}$. Since $u_{n}$ begins and ends with an $A$-letter
and does not contain any $C$-letters, it follows that:

  1. (1)

if $b_{2}b_{1}\neq 1$ or $b_{2}b_{1}=x_{1}=1$, then $u_{n}$ must be a subword
of $u_{n+1}$ that is not a prefix or a suffix, or $u_{n}$ is a subword of
$u_{n+1}^{-1}$,

  2. (2)

if $b_{2}b_{1}=1$ and $x_{1}\neq 1$, then $u_{n}$ must be a subword of
$u_{m}^{\pm 1}$ for some $m>n$.

Since $\langle x\rangle$ is malnormal by definition, $x$ is not a proper
power. The first situation cannot happen by Lemma 2.2. The second situation
cannot happen by Lemma 2.1.

Now, since there is at most one path in $\Gamma$ with label $u_{n}$, there can
be at most one lift of $\alpha$ to
$\operatorname{Core}(\Gamma\times_{\Delta}\Lambda)$ by definition of the fibre
product graph. If $\alpha$ does not traverse vertices of degree three or more,
except possibly at its endpoints, then each edge it traverses can have at most
one preimage in $\operatorname{Core}(\Gamma\times_{\Delta}\Lambda)$. So by
Lemma 3.9, since $\lambda$ is surjective, it must traverse some edge in
$\Lambda$ precisely once. But then this would imply that $\lambda$ represents
a primitive element of $\pi_{1}(\Lambda)$, contradicting the fact that $K$ was
a $w$-subgroup. Therefore, we conclude that $\alpha$ must traverse a vertex of
degree at least three, other than at its endpoints. ∎

We now use Lemma 3.10 to derive a contradiction to the definitions of
primitive exceptional intersection words. Let $\alpha:I\looparrowright\Lambda$
be a path satisfying the following:

  1. (1)

$\alpha$ factors through $\lambda$,

  2. (2)

$\alpha$ is labelled by $u_{n}$ for some $n>0$,

  3. (3)

there is no path $\alpha^{\prime}:I\looparrowright\Lambda$, strictly extending
$\alpha$ and satisfying the above.

We similarly define $\beta:I\looparrowright\Lambda$, replacing $u_{n}$ with
$v_{n}$. Such paths exist by definition of $w$.

By Lemma 3.10, $\alpha$ and $\beta$ must traverse a vertex of degree at least
three, away from their endpoints. Now the idea is to use this fact to divide
$\Lambda$ according to where $\alpha$ or $\beta$ are supported. Since $\alpha$
does not traverse any $C$-edges and $\beta$ does not traverse any $A$-edges,
they will block each other from traversing certain regions of $\Lambda$.

Since $\chi(\Lambda)=-1$, and $\Lambda$ is a core graph, we only have three
topologically distinct cases to consider:

  1. (1)

$\Lambda$ is a rose graph, see Figure 2,

  2. (2)

$\Lambda$ is a theta graph, see Figure 3,

  3. (3)

$\Lambda$ is a spectacles graph, see Figure 5.

Figures 2, 3 and 5 contain all the different cases, up to symmetry. Before
proceeding with the case analysis, we briefly explain the diagrams. The red
regions indicate sections that $\alpha$ traverses and must contain an
$A$-edge; $\beta$ cannot traverse any edge in a red region. The blue regions
indicate sections that $\beta$ traverses and must contain a $C$-edge; $\alpha$
cannot traverse any edge in a blue region. The yellow regions indicate
sections that $\alpha$ or $\beta$ or both $\alpha$ and $\beta$ traverse. In
any case, the yellow regions must only contain $B$-edges, but are allowed to
have length zero when this does not change the topology of the underlying
graph. The black regions indicate sections that are not traversed by either
$\alpha$ or $\beta$ and are also allowed to have length zero when this does
not change the topology of the underlying graph. The red vertices and blue
vertices indicate the start and endpoints of $\alpha$ and $\beta$
respectively.

Topologically, in each diagram there can be at most three edges. The path
$\alpha$ must leave one of these edges by Lemma 3.10 and re-enter another
edge, leaving enough space for $\beta$ to do the same elsewhere. Given these
constraints, the reader should check that these are indeed all the cases to
consider.

Case 1. We handle this case more in detail than the others as the arguments
are mostly identical. We have three subcases to consider, according to Figure
2.

Figure 2. Rose graph cases.

Suppose we are in the situation of the first subdiagram. When $\lambda$
traverses a red segment from a red vertex, it must then be followed by the
other red segment. Otherwise we would obtain a contradiction to Lemma 3.10.
Similarly for the blue segments. Thus, since $\lambda$ is primitive, it must
traverse each edge precisely once. Hence, $\lambda$ would not represent a
primitive element of $\pi_{1}(\Lambda)$ which is a contradiction.

Now consider the second subdiagram. Any (maximal) $u_{n}$ labelled path must
begin at one red vertex and end at the other red vertex. Similarly for the
$v_{n}$ labelled paths. But this then implies that only one power of $x$ and
one power of $y$ appears in $w$. Thus, if $w$ is of the first type, then it
must be equal to $xy$. If $w$ is of the second type, it must be equal to
$xyz^{i}$ for some $i\geq 1$. By Lemma 3.10, $\lambda$ must traverse both
loops at least twice. We may now deduce that there exist elements:

$\displaystyle a$ $\displaystyle\in\langle A,B\rangle-\langle
B\rangle\leavevmode\nobreak\ ,$ $\displaystyle b$ $\displaystyle\in\langle
B\rangle-1\leavevmode\nobreak\ ,$ $\displaystyle c$ $\displaystyle\in\langle
B,C\rangle-\langle B\rangle\leavevmode\nobreak\ ,$

such that $w=(ab^{-1})(bc)$ and that $\pi(a),\pi(c)\neq 1$. Hence, $\\{\langle
a\rangle,\langle c\rangle\\}$ is not a malnormal family, contradicting our
assumptions on $w$.

Let us move onto the third subdiagram. Similarly to the second subcase, we see
that if $w$ is of the first type, it must be equal to $xy$ and if $w$ is of
the second type, it must be equal to $xyz^{i}$ for some $i\geq 1$. From the
diagram we may now deduce that there exist elements:

$\displaystyle a$ $\displaystyle\in\langle A,B\rangle-\langle
B\rangle\leavevmode\nobreak\ ,$ $\displaystyle b$ $\displaystyle\in\langle
B\rangle\leavevmode\nobreak\ ,$ $\displaystyle c$ $\displaystyle\in\langle
B,C\rangle-\langle B\rangle\leavevmode\nobreak\ ,$

such that $w=(ab^{-1})(bc)$ and that $\langle a\rangle^{g}\cap\langle
c\rangle\neq 1$ for some $g\in F(A,B,C)$. Hence, $\\{\langle a\rangle,\langle
c\rangle\\}$ is not a malnormal family, contradicting our assumptions on $w$.
It follows that $\Lambda$ cannot be a rose graph.

Figure 3. Theta graph cases.

Case 2. Consider the first subdiagram in Figure 3. By collapsing the yellow
edge, we see that we may handle this case in the same way as the first subcase
of the rose case. Similarly, we may reduce the second and third subcases to
the second and third subcases of the rose case.

So now let us consider the new cases appearing in the fourth and fifth
subdiagrams. Now if $b_{2}b_{1}\neq 1$ and $\alpha$ was labelled by $u_{n}$
with $n\geq 2$, then we would have that $\Lambda$ would support a path
labelled by $u_{1}$ that does not traverse a vertex of degree at least three,
away from its endpoints. But this contradicts Lemma 3.10. So if
$b_{2}b_{1}\neq 1$ and $n=1$, then there can be no other $u_{m}$-labelled path
beginning or ending at a red vertex as $\Lambda\looparrowright\Delta$ is an
immersion. This would imply that there can be no other $u_{m}$-labelled paths
in $\Lambda$ for any $m\geq 1$ and so $\lambda$ traverses the red segments
only once by Lemma 3.9. Hence, $\lambda$ could not represent a primitive
element of $\pi_{1}(\Lambda)$, a contradiction. So now we may assume that
$b_{2}b_{1}=1$. By a symmetric argument, we may assume that $b_{4}b_{3}=1$.
Similarly, we must have $x_{1},y_{1}\neq 1$. As before, there must be at least
one other path $\alpha^{\prime}:I\looparrowright\Lambda$ labelled by $u_{m}$
for some $m\geq 1$. We may assume that $\alpha^{\prime}$ is maximal in the
same sense that $\alpha$ was maximal. We see that $\alpha$ and
$\alpha^{\prime}$ must traverse a common segment. We now have two subcases to
consider, up to symmetry, depending on whether $\alpha$ and $\alpha^{\prime}$
traverse a common segment in the same direction or the opposite direction. See
Figure 4. In either case, there can only be one lift of the segment with label
$x_{2}^{m}x_{1}$ to $\operatorname{Core}(\Gamma\times_{\Delta}\Lambda)$ for
the following reason: the projection of any loop in
$\operatorname{Core}(\Gamma\times_{\Delta}\Lambda)$ traversing the segment
labelled $x_{2}^{m}x_{1}$, must then traverse a blue segment labelled by
$y_{1}^{-1}y_{2}^{k}$ for some $k\geq 0$. Since there is only one path in
$\Gamma$ with this label, there can be only one lift of these segments to
$\operatorname{Core}(\Gamma\times_{\Delta}\Lambda)$. Now Lemma 3.9 tells us
that $\lambda$ does not represent a primitive element of $\pi_{1}(\Lambda)$.
It follows that $\Lambda$ cannot be a theta graph.

Figure 4. Extra theta cases. Figure 5. Spectacles graph cases.

Case 3. The first subdiagram in Figure 5 is analogous to the first subdiagram
of the rose case. The second, third and fourth subdiagrams are analogous to
the second subdiagram of the rose case. The fifth subdiagram is analogous to
the third subdiagram of the rose case. Hence, $\Lambda$ cannot be a spectacles
graph.

Now we may conclude that there can be no $w$-subgroups of rank two and hence,
$G$ must must be $2$-free. ∎

###  3.2. The general case

In this section, we consider the general case and show that primitive
exceptional intersection groups are the only exceptional intersection groups
which are $2$-free.

Theorem 3.1 was generalised to one-relator products in [How05]. By
specialising [How05, Theorem C] to the case of one-relator groups, we may
obtain the following result.

######  Theorem 3.11.

Let $F(\Sigma)/\langle\langle w\rangle\rangle$ be a one-relator group and
suppose $\Sigma=A\sqcup B\sqcup C$. If $\langle A,B\rangle$ and $\langle
B,C\rangle$ are Magnus subgroups with exceptional intersection, then there is
a monomorphism of free groups

$\iota:F(a,c)\hookrightarrow F(\Sigma)\leavevmode\nobreak\ ,$

with the following properties. There is some $r\in F(a,c)$ with $\iota(r)$
conjugate to $w$ and some $m,n\neq 0$, such that one of the following holds:

  1. (1)

$a^{m}=c^{n}$ in $F(a,c)/\langle\langle r\rangle\rangle$ with $\iota(a)=x$,
$\iota(c)=y$ and

$\displaystyle x\in\langle A,B\rangle-\langle B\rangle\leavevmode\nobreak\ ,$
$\displaystyle y\in\langle B,C\rangle-\langle B\rangle\leavevmode\nobreak\ .$

  2. (2)

$ac^{m}a^{-1}=c^{n}$ in $F(a,c)/\langle\langle r\rangle\rangle$ with
$\iota(a)=xy$, $\iota(c)=z$ and

$\displaystyle x$ $\displaystyle\in\langle A,B\rangle-\langle
B\rangle\leavevmode\nobreak\ ,$ $\displaystyle y$ $\displaystyle\in\langle
B,C\rangle-\langle B\rangle\leavevmode\nobreak\ ,$ $\displaystyle z$
$\displaystyle\in\langle B\rangle-1\leavevmode\nobreak\ .$

Using this result and the algorithm to compute the centre of a one-relator
group from [BT68], Howie also showed that a generating set for the
intersection of given Magnus subgroups is computable [How05, Theorem E].

In the discussion following [Col04, Theorem 4], Collins points out the
following.

######  Corollary 3.12.

Assume the notation of Theorem 3.11 and suppose that we are in the first case.
Denote by

$H=\langle B\rangle*F(a,c)/\langle\langle r\rangle\rangle\leavevmode\nobreak\
.$

Then we have:

$G\cong\langle A,B\rangle\underset{\langle B,\iota(a)\rangle=\langle
B,a\rangle}{*}H\underset{\langle B,c\rangle=\langle
B,\iota(c)\rangle}{*}\langle B,C\rangle\leavevmode\nobreak\ .$

######  Corollary 3.13.

Assume the notation of Theorem 3.11 and suppose that we are in the second
case. Let $\iota(a)=x\cdot y$ where $x\in\langle A,B\rangle-\langle B\rangle$
and $y\in\langle B,C\rangle-\langle B\rangle$ and denote by

$H=\langle B\rangle\underset{\langle\iota(c)\rangle=\langle
c\rangle}{*}F(a,c)/\langle\langle r\rangle\rangle\underset{\langle
a\rangle=\langle de\rangle}{*}F(d,e)\leavevmode\nobreak\ .$

Then we have:

$G\cong\langle A,B\rangle\underset{\langle B,x\rangle=\langle
B,d\rangle}{*}H\underset{\langle B,e\rangle=\langle B,y\rangle}{*}\langle
B,C\rangle\leavevmode\nobreak\ .$

######  Remark 3.14.

There is a minor typographical error in the splitting provided by Collins for
the second case. Corollary 3.13 is the corrected version.

The following follows directly from Corollaries 3.12 and 3.13.

######  Corollary 3.15.

Assume the notation of Theorem 3.11. The monomorphism $\iota$ descends to a
monomorphism of one-relator groups:

$\bar{\iota}:F(a,c)/\langle\langle r\rangle\rangle\hookrightarrow
F(\Sigma)/\langle\langle w\rangle\rangle\leavevmode\nobreak\ .$

Theorem 3.11, coupled with Corollary 3.15, finds us two-generator one-relator
subgroups of exceptional intersection groups. We now characterise precisely
what these subgroups can be.

######  Lemma 3.16.

Let $H=F(a,c)/\langle\langle r\rangle\rangle$ be torsion-free such that
$a^{m}=c^{n}$ in $H$ for some $m,n\neq 0$. Then one of the following hold:

  1. (1)

$r\in F(a,c)$ is primitive and so $H\cong\mathbb{Z}$,

  2. (2)

$H$ is non-cyclic with non-trivial centre.

###### Proof.

Note that $H$ has non-trivial centre as $\langle a^{m}\rangle$ is an infinite
subgroup contained in the centre. By [LS01, Chapter II Proposition 5.11], $H$
is cyclic if and only if $r$ is primitive. ∎

######  Lemma 3.17.

Let $H=F(a,c)/\langle\langle r\rangle\rangle$ be torsion-free such that
$a^{-1}c^{m}a=c^{n}$ in $H$ for some $m,n\neq 0$. Then one of the following
hold:

  1. (1)

$r\in F(a,c)$ is primitive and so $H\cong\mathbb{Z}$,

  2. (2)

$H$ is non-cyclic with non-trivial centre,

  3. (3)

$r$ or $r^{-1}$ is conjugate to $\operatorname{pr}_{p/q}(c^{-1},a^{-1}ca)$ for
some $p/q\in\mathbb{Q}_{>0}$ and so $H\cong BS(p,q)$.

  4. (4)

$r$ or $r^{-1}$ is conjugate in $F(a,c)$ to an element $r^{\prime}\in\langle
c,a^{-1}ca\rangle$ such that $\langle c,a^{-1}ca\rangle\cong
F(c,a^{-1}ca)/\langle\langle r^{\prime}\rangle\rangle$ is non-cyclic with non-
trivial centre.

###### Proof.

Suppose that $\left\lvert m\right\rvert$ is smallest possible. If $n=m=\pm 1$,
then either $r$ is primitive or $H\cong\mathbb{Z}^{2}$. If $n=m\neq\pm 1$,
then $H$ has non-trivial centre generated by $c^{n}$. So if $n=m$, we have
obtained conclusion (1) or (2). Now suppose that $n\neq m$. Therefore, the
exponent sum of $c$ in $r$ must be non-zero. There is a single epimorphism, up
to sign change, $\phi:F(a,c)\to\mathbb{Z}$ such that $\phi(r)=0$, given by

$\displaystyle a$
$\displaystyle\to\frac{-\sigma_{c}(r)}{\gcd(\sigma_{a}(r),\sigma_{c}(r))}\leavevmode\nobreak\
,$ $\displaystyle c$
$\displaystyle\to\frac{\sigma_{a}(r)}{\gcd(\sigma_{a}(r),\sigma_{c}(r))}\leavevmode\nobreak\
,$

where $\sigma_{a}(r),\sigma_{c}(r)$ denote the exponent sums of $a$ and $c$ in
$r$ respectively. But now $\phi(a^{-1}c^{m}ac^{-n})=0$ from which it follows
that

$0=\phi(c^{m})-\phi(c^{n})=(m-n)\sigma_{a}(r)\leavevmode\nobreak\ .$

By Magnus rewriting, since $\sigma_{a}(r)=0$, we have that

$H=\langle c_{0},...,c_{k},a\mid
ac_{0}a^{-1}=c_{1},...,ac_{k-1}a^{-1}=c_{k},r^{\prime}\rangle,$

where $r^{\prime}$ is the rewriting of $r$ in terms of $c=c_{0},...,c_{k}$. By
the Freiheitssatz (see [MKS66, Theorem 4.10]), it follows that $k=1$. If
$\langle c,aca^{-1}\rangle\cong\mathbb{Z}$, then we have obtained conclusion
(3). If not, then we have obtained conclusion (4). ∎

Recall that a _generalised Baumslag–Solitar group_ is group that splits as a
graph of groups where each vertex and edge group is infinite cyclic.

######  Proposition 3.18.

Let $G=\langle a,c\mid r\rangle$ be a one-relator group in which either
$a^{m}=c^{n}$ or $ac^{m}a^{-1}=c^{n}$ holds. Then $G$ is a generalised
Baumslag–Solitar group.

###### Proof.

The first case follows from Lemma 3.16 and [Pie74, Theorems 1 & 3]. Similarly,
in the second case we may conclude that $G$ is a generalised Baumslag–Solitar
group by Lemma 3.17, unless $n\neq m$ and $r$ or $r^{-1}$ is conjugate in
$F(a,c)$ to an element $w\in\langle c,a^{-1}ca\rangle$ such that $H=\langle
c,a^{-1}ca\rangle\cong F(c,a^{-1}ca)/\langle\langle w\rangle\rangle$ is non-
cyclic with non-trivial centre. So let us suppose that we are in the latter
case. Since $ac^{m}a^{-1}=c^{n}$ holds in $H$, the rank of the abelianisation
of $H$ must be one. Then by [Pie74, Theorem 1], we have:

$H\cong\langle a_{1},...,.a_{m}\mid
a_{1}^{p_{1}}=a_{2}^{q_{1}},...,a_{m-1}^{p_{m-1}}=a_{m}^{q_{m-1}}\rangle$

where $p_{i},q_{i}\geq 2$ and $\gcd(p_{i},q_{j})=1$ for all $i>j$. Then by
[Pie74, Lemma 3], $c$ and $a^{-1}ca$ are both conjugate within $H$ to some
subgroup $\langle a_{i}\rangle<H$. Suppose that $c^{g}=a_{i}^{k}$ and
$(a^{-1}ca)^{h}=a_{j}^{l}$. Then we have that:

$G\cong\langle a_{1},...,a_{m},b\mid
a_{1}^{p_{1}}=a_{2}^{q_{1}},...,a_{m-1}^{p_{m-1}}=a_{m}^{q_{m-1}},b^{-1}a_{i}^{k}b=a_{j}^{l}\rangle$

where $b=g^{-1}ah$. Thus, we may conclude that $G$ is a generalised
Baumslag–Solitar group. ∎

We are now ready to prove the main result of this section.

######  Theorem 3.19.

Let $G$ be an exceptional intersection group. Then one of the following holds:

  1. (1)

$G$ is a primitive exceptional intersection group and so is $2$-free,

  2. (2)

there is a two-generator one-relator generalised Baumslag–Solitar subgroup
$H<G$ such that every non-free two-generator subgroup of $G$ is conjugate into
$H$.

###### Proof.

Assume that $G$ is not a primitive exceptional intersection group. Let
$H<G\cong F(A,B,C)/\langle\langle w\rangle\rangle$ be the two-generator
subgroup from Theorem 3.11. We may assume that $H$ is maximal in the sense
that there is no subgroup properly containing $H$ and that is of the same form
as the two-generator subgroup from Theorem 3.11. By Proposition 3.18, $H$ is a
generalised Baumslag–Solitar group. Clearly $H$ has rank at most two, so we
now consider the two possible cases.

Let us first assume that $H$ has rank one. Thus, $G$ has one of the following
presentations:

  1. (1)

$G\cong\langle\Sigma\mid\operatorname{pr}_{p/q}(x,y)\rangle$ for some
$x\in\langle A,B\rangle-\langle B\rangle$ and $y\in\langle B,C\rangle-\langle
B\rangle$,

  2. (2)

$G\cong\langle\Sigma\mid\operatorname{pr}_{p/q}(xy,z)\rangle$ for some
$x\in\langle A,B\rangle-\langle B\rangle$, $y\in\langle B,C\rangle-\langle
B\rangle$ and $z\in\langle B\rangle$.

where $H=\langle x,y\rangle$ in the first case, and $H=\langle xy,z\rangle$ in
the second case.

Suppose that we are in the first situation. By Definition 3.5, we may assume
that either $\\{\langle x\rangle,\langle y\rangle\\}$ is not a malnormal
family (in $F(A,B,C)$), or that $p/q=1$ and that there exist elements
$a\in\langle A,B\rangle-\langle B\rangle$ and $c\in\langle B,C\rangle-\langle
B\rangle$ such that $\operatorname{pr}_{1}(x,y)=\operatorname{pr}_{1}(a,c)$
and $\\{\langle a\rangle,\langle c\rangle\\}$ is not a malnormal family (in
$F(A,B,C)$). As the two cases are identical, it suffices to assume that
$\\{\langle x\rangle,\langle y\rangle\\}$ is not a malnormal family. Now, if
$\\{\langle x\rangle,\langle y\rangle\\}$ is not a malnormal family, then
either $x$ is a proper power, $y$ is a proper power, or a conjugate of
$\langle x\rangle$ intersects $\langle y\rangle$ non-trivially. If either $x$
or $y$ is a proper power, by adjoining a root of $x$ or $y$ to $H$, we obtain
a contradiction to maximality of $H$. If $\langle x\rangle^{f}\cap\langle
y\rangle\neq 1$ for some $f\in F(\Sigma)$, it follows that there must be
elements $g\in\langle A,B\rangle-\langle B\rangle$, $h\in\langle
B,C\rangle-\langle B\rangle$, $d\in\langle B\rangle$ such that $\langle
x\rangle^{g}\cap\langle y\rangle^{h^{-1}}<\langle d\rangle<\langle B\rangle$.
Now $H$ would be properly contained in $\langle gh,h^{-1}dh\rangle$. However,
since $(gh)(h^{-1}dh)^{n}(gh)^{-1}=(h^{-1}dh)^{m}$ holds for some $m,n\neq 0$,
we obtain a contradiction to maximality of $H$ again.

Now suppose that we are in the second situation. As before, we may assume that
$\langle z\rangle$ is not malnormal by Definition 3.5. Then $z$ is a proper
power and we contradict maximality of $H$ once again.

We may conclude from the above that $H$ must have rank two. Since $H$ cannot
be free, we have that $\pi(w)=2$ and $G$ has a two-generator $w$-subgroup $K$
into which $H$ is conjugate by [LW22, Corollary 1.10]. If $H\neq K$, we get a
contradiction to maximality of $H$ by the definition of $w$-subgroups and
Lemmas 2.3 and 2.4. Thus, (2) holds by [LW22, Corollary 1.10]. ∎

######  Remark 3.20.

The $w$-subgroup from Theorem 3.19 is always a subgroup of the form given in
Theorem 3.11.

It follows from [Lin22, Theorem 8.2] and Theorem 3.19 that an exceptional
intersection group is hyperbolic if and only if it contains no
Baumslag–Solitar subgroup. Note that this can also be derived using
Corollaries 3.12 and 3.13 and the combination theorem [BF92]. However, in
order to prove our main results, we need the stronger dichotomy established in
Theorem 3.19. This dichotomy is harder to establish using Corollaries 3.12 and
3.13 as the splittings do not satisfy the hypothesis of existing $2$-free
combination theorems (see [Bau68] for example).

######  Example 3.21.

We give two examples of groups with exceptional intersection that are not
$2$-free. Let $p/q\in\mathbb{Q}_{>0}$ and $n,m\neq 0$. Consider the group with
presentation:

$G\cong\langle
c_{0},c_{1}\mid\operatorname{pr}_{p/q}(c_{0}^{m},c_{1}^{-n})\rangle\leavevmode\nobreak\
.$

The relation $c_{0}^{qm}=c_{1}^{pn}$ holds in $G$ and so it has an exceptional
intersection of the first type. In [MPS73], this group was shown to be
isomorphic to a non-cyclic generalised Baumslag–Solitar group with
presentation:

$\langle c_{0},b,c_{1}\mid
c_{0}^{m}=b^{p},b^{q}=c_{1}^{n}\rangle\leavevmode\nobreak\ ,$

and so is not $2$-free.

Now consider the HNN-extension:

$\displaystyle\langle c_{0},b,c_{1},a\mid
c_{0}^{m}=b^{p},b^{q}=c_{1}^{n},ac_{0}a^{-1}=c_{1}\rangle$
$\displaystyle\cong\langle c_{0},c_{1},a\mid
ac_{0}a^{-1}=c_{1},\operatorname{pr}_{p/q}(c_{0}^{m},c_{1}^{-n})\rangle$
$\displaystyle\cong\langle
a,c\mid\operatorname{pr}_{p/q}(c^{m},a^{-1}c^{-n}a)\rangle$

The relation $a^{-1}c^{mp}a=c^{nq}$ holds in this group and so

$\langle
a_{0},a_{1},c\mid\operatorname{pr}_{p/q}(c^{m},(a_{0}a_{1})^{-1}c^{-n}(a_{0}a_{1})\rangle$

has an exceptional intersection of the second type. This group also contains a
non-cyclic generalised Baumslag–Solitar subgroup and so is not $2$-free.

##  4\. One-relator towers

A _one-relator complex_ is a combinatorial $2$-complex with a single $2$-cell.
If $X$ is a one-relator complex, we write $X=(\Gamma,\lambda)$ where $\Gamma$
is a $1$-complex and $\lambda:S^{1}\looparrowright\Gamma$ is the attaching
map.

If $p:Y\looparrowright X$ is an infinite cyclic cover of a CW-complex, a _tree
domain_ for $p$ is a subcomplex $Z\subset Y$ satisfying the following:

  1. (1)

$\mathbb{Z}\cdot Z=Y$,

  2. (2)

for all $k\in\mathbb{N}$ and every cell $c\subset Z$, if $k\cdot c\subset Z$,
then $i\cdot c\subset Z$ for all $0\leq i\leq k$,

  3. (3)

$Z\cap 1\cdot Z$ is connected and non-empty.

By [Lin22, Proposition 3.10], if the map $\pi_{1}(Z\cap 1\cdot
Z)\to\pi_{1}(Z)$ induced by inclusion is injective, then we obtain a splitting

$\pi_{1}(X)\cong\pi_{1}(Z)*_{\pi_{1}(Z\cap 1\cdot Z)}$

By [Lin22, Proposition 4.7], if $p:Y\looparrowright X$ is an infinite cyclic
cover of a one-relator complex, then there always exists a one-relator tree
domain $Z$ for $p$. Moreover, by the Freiheitssatz, the maps $\pi_{1}(Z\cap
1\cdot Z)\to\pi_{1}(Z)$ are always injective.

A _one-relator tower_ is a sequence of immersions of one-relator complexes

$X_{N}\looparrowright...\looparrowright X_{1}\looparrowright X_{0}=X,$

where $X_{i+1}\looparrowright X_{i}$ factors as

$X_{i+1}\overset{\iota}{\hookrightarrow}Y\overset{p}{\looparrowright}X_{i}$

where $p$ is an infinite cyclic covering map and $\iota$ is an inclusion of a
tree domain for $p$. A one-relator tower is _maximal_ if $\chi(X_{N})=1$.

By the above, it follows that one-relator hierarchies correspond to iterated
HNN-extensions over one-relator groups.

The following is [Lin22, Proposition 5.1] and can be interpreted as a modern
version of the well-known Magnus–Moldavanskii hierarchy.

######  Proposition 4.1.

Let $X$ be a finite one-relator complex. Then $X$ has a maximal one-relator
tower.

If $X=(\Gamma,\lambda)$ is a one-relator complex, then for each $w$-subgroup
$K<\pi_{1}(\Gamma)$, where $w=[\lambda]$, there is an immersion of one-relator
complexes $Q=(\Omega,\omega)\looparrowright X$ where
$\Omega\looparrowright\Gamma$ is the core graph immersion representing $K$. We
say that $Q\looparrowright X$ represents a $w$-subgroup. The following is
[Lin22, Theorem 5.15].

######  Theorem 4.2.

Let $X$ be a one-relator complex and let $Q\looparrowright X$ be an immersion
of a one-relator complex, representing some $w$-subgroup. There exists a one-
relator tower

$Q=X_{K}\looparrowright...\looparrowright X_{1}\looparrowright X_{0}=X.$

Equipped with Theroem 4.2, we make the following definition.

######  Definition 4.3.

We say a one-relator tower $X_{N}\looparrowright...\looparrowright X_{0}=X$ is
_factored_ if for every immersion $Q\looparrowright X_{i}$ representing a
$w$-subgroup of $\pi_{1}(X_{i})$ with $\chi(Q)=0$, either there is some $j\geq
0$ such that $X_{i+j}=Q\looparrowright X_{i}$, or $Q\looparrowright X_{i}$
factors through $X_{N}\looparrowright X_{i}$.

The proof of the following is essentially identical to that of Proposition
4.1, but we include it for completeness.

######  Proposition 4.4.

Every one-relator complex has a maximal factored one-relator tower.

###### Proof.

Let $X=(\Gamma,\lambda)$ be a one-relator complex. Denote by $\deg(\lambda)$
the largest degree covering map $S^{1}\looparrowright S^{1}$ that $\lambda$
factors through and denote by $X_{\lambda}$ the smallest one-relator
subcomplex of $X$. Then define the quantity

$c(X)=\left(\frac{\left\lvert\lambda\right\rvert}{\deg(\lambda)}-\left\lvert
X_{\lambda}^{(0)}\right\rvert,-\chi(X)\right).$

The proof proceeds by induction on $c(X)$.

If $\pi_{1}(X)$ is $2$-free or has torsion, then any maximal one-relator tower
is a factored tower by definition and so the result follows from Proposition
4.1. If $Q\looparrowright X$ represents a $w$-subgroup, then it is clear that
we must have $c(Q)\leq c(X)$ with equality if and only if $Q=X$. Thus, by
Theorem 4.2, we may assume that $\chi(X)=0$. Then as $\chi(X)\neq 1$,
$\pi_{1}(X)$ is indicable and there is some infinite cyclic cover
$p:Y\looparrowright X$. By [Lin22, Proposition 4.9], there is some one-relator
tree domain $Z$ for $p$ such that $c(Z)<c(X)$. Hence, by induction, the proof
is complete. ∎

###  4.1. Acylindrical, quasi-convex and stable one-relator towers

Let $X_{K}\looparrowright...\looparrowright X_{1}\looparrowright X_{0}$ be a
one-relator tower. Denote by $T_{i}$ the Bass-Serre tree associated with the
splitting $\pi_{1}(X_{i})\cong\pi_{1}(X_{i+1})*_{\psi_{i}}$. We call this an
_acylindrical one-relator tower_ if there is some constant $k\geq 0$ such that
the stabilisers of segments of length $k$ in $T_{i}$ are finite. We call it a
_quasi-convex one-relator tower_ if the inclusions
$\tilde{A}_{i+1},\tilde{B}_{i+1}\hookrightarrow\tilde{X_{i}}$ are quasi-
isometric embeddings, where $A_{i+1}=X_{i+1}\cap 1\cdot X_{i+1}$ and
$B_{i+1}=-1\cdot X_{i+1}\cap X_{i+1}$. This last definition is due to Wise
[Wis21], adapted to the one-relator case. In [Lin22], a _stable one-relator
tower_ is defined in terms of the identifying homomorphisms $\psi_{i}$. Since
we shall not need this definition, we instead record the following, which is a
reformulation of [Lin22, Lemma 7.4].

######  Lemma 4.5.

A one-relator tower $X_{N}\looparrowright...\looparrowright
X_{1}\looparrowright X_{0}$ is stable if and only if there is some $k\geq 0$
such that the pointwise stabilisers of segments of length $k$ in $T_{i}$ have
rank at most one.

In [Lin22], the author established a connection between these three types of
one-relator towers. The aim of this section is to improve on that result.

###  4.2. Primitive extension complexes

A one-relator complex $X$ is a _primitive extension complex_ if $\chi(X)=0$
and there is a one-relator tower $Z\looparrowright X$ where $\pi_{1}(Z)$ is
$2$-free and such that

$\pi_{1}(Z\cap 1\cdot Z)\cap\pi_{1}(Z\cap-1\cdot Z)\neq\pi_{1}(-1\cdot Z\cap
Z\cap 1\cdot Z),$

after possibly adding edges to $Z$ and extending the $\mathbb{Z}$-action so
that $-1\cdot Z\cap Z\cap 1\cdot Z$ is connected. By Theorem 3.19, we can see
that $\pi_{1}(Z)*F$ is a primitive exceptional intersection group for some
finitely generated free group $F$.

Let $X=(\Gamma,\lambda)$ be a one-relator complex with $\chi(X)=0$. Let
$T\subset\Gamma$ be a spanning tree and let $\langle a,b\mid w\rangle$ be the
induced one-relator presentation. If $a^{m}=b^{n}$ for some
$m,n\in\mathbb{Z}-\\{0\\}$ and $\pi_{1}(X)$ is not cyclic, then we call $X$ a
_powered one-relator complex_. By Proposition 3.18, $\pi_{1}(X)$ is a
generalised Baumslag–Solitar group. Moreover, we have the following.

######  Lemma 4.6.

Powered one-relator complexes are primitive extension complexes.

###### Proof.

Let $X=(\Gamma,\lambda)$ be a powered one-relator complex. Then there is a
spanning tree in $\Gamma$ such that $a^{m}=b^{n}$ where $\langle a,b\mid
w\rangle$ is the induced presentation for $\pi_{1}(X)$. By virtue of the fact
that $a^{m}=b^{n}$, there is only a single epimorphism
$\phi:\pi_{1}(X)\to\mathbb{Z}$ and by [Mur64], we have that $\ker(\phi)$ is
finitely generated and free. Now let $p:Y\looparrowright X$ be the induced
cyclic cover and $Z\subset Y$ a one-relator tree domain. Since $\pi_{1}(Y)$ is
finitely generated, this in particular implies that

$\pi_{1}(Z\cap 1\cdot Z)=\pi_{1}(Z)=\pi_{1}(-1\cdot Z\cap Z)$

and so that $X$ is a primitive extension complex. ∎

It turns out that primitive extension complexes obstruct stable one-relator
towers.

######  Proposition 4.7.

Let $X$ be a one-relator complex and $X_{N}\looparrowright...\looparrowright
X_{1}\looparrowright X_{0}=X$ a factored one-relator tower. Then one of the
following holds:

  1. (1)

$X_{N}\looparrowright...\looparrowright X_{1}\looparrowright X_{0}=X$ is
stable,

  2. (2)

$X_{K}\looparrowright...\looparrowright X_{1}\looparrowright X_{0}=X$ is
stable and $X_{K}$ is a primitive extension complex for some $K\leq N$,

  3. (3)

$X_{K}$ is a powered one-relator complex for some $K\leq N$.

###### Proof.

Let $p:Y\looparrowright X$ be the cyclic cover such that $X_{1}$ is a tree
domain for $p$. Let $A=X_{1}\cap-1\cdot X_{1}$ and $B=X_{1}\cap 1\cdot X_{1}$,
where the $\mathbb{Z}$ action is the covering action. Up to possibly adding
finitely many edges to $X$ (and so to $X_{1}$), we may assume that $A\cap B$
is connected. Now by [Lin22, Lemma 7.8], either $X_{1}\looparrowright X_{0}=X$
is stable, or $\pi_{1}(A)\cap\pi_{1}(B)\neq\pi_{1}(A\cap B)$. If
$X_{1}\looparrowright X_{0}$ is stable, we proceed by induction. So suppose
that it is not. By Theorem 3.1, either $\pi_{1}(X_{1})$ is $2$-free, or there
is an immersion $Q\looparrowright X_{1}$ representing a $w$-subgroup such that
$\chi(Q)=0$ and $\pi_{1}(Q)$ is a generalised Baumslag–Solitar group. Let us
first consider the latter case. By definition of factored one-relator towers,
there is some $i$ such that $X_{i}=Q$. Moreover, by Lemmas 3.16 and 3.17,
either $X_{i}$ is a powered one-relator complex, or $X_{i+1}$ is. So now let
us consider the case that $\pi_{1}(X_{1})$ is $2$-free. If $\chi(X)<0$, then
by definition, we have that $X$ is $2$-free. Thus, by [Lin22, Theorems 7.8 &
7.9], we have that $X_{N}\looparrowright...\looparrowright X_{0}=X$ is a
stable tower. If $\chi(X)=0$, then $X$ is a primitive extension complex. ∎

###  4.3. Improved one-relator towers

Recall that a _Magnus subgroup_ of a one-relator group $\langle\Sigma\mid
w\rangle$ is a subgroup generated by a subset of the generators
$A\subset\Sigma$ such that $A$ omits a generator that appears in the cyclic
reduction of $w$. Using Proposition 4.7 we may now prove that Magnus subgroups
of hyperbolic one-relator groups are quasi-convex. This was previously known
in the case of one-relator groups with torsion by Newman’s spelling theorem
and in the case of $2$-free one-relator groups by [Lin22, Theorems 8.1 & 8.2].

######  Theorem 4.8.

Magnus subgroups of hyperbolic one-relator groups are quasi-convex.

###### Proof.

Let $X$ be a one-relator complex with $\pi_{1}(X)$ hyperbolic. By Proposition
4.4, there is a maximal factored one-relator tower
$X_{N}\looparrowright...\looparrowright X_{1}\looparrowright X_{0}=X$. By
[Ger96, Corollary 7.8], $\pi_{1}(X_{i})$ is hyperbolic for all $i\geq 1$. By
Proposition 4.7, either this tower is stable, or there is some $1\leq L<N$
such that $X_{L}\looparrowright...\looparrowright X_{1}$ is stable and $X_{L}$
is a primitive extension complex.

In the first case, since hyperbolic groups cannot contain Baumslag–Solitar
subgroups, the result follows by [Lin22, Theorems 8.1 & 8.2]. So now let us
assume that we are in the second case. The proof proceeds by induction on
tower length. For the base case $L=1$, since $\chi(X_{L})=0$, all Magnus
subgroups of $\pi_{1}(X_{L})$ must by quasi-convex as they are all cyclic. So
now we assume the inductive hypothesis. In other words, that all Magnus
subgroups of $\pi_{1}(X_{1})$ are quasi-convex. Then by [Lin22, Theorems 3.6 &
6.10], all Magnus subgroups of $\pi_{1}(X)$ are quasi-convex and the proof is
complete. ∎

As a consequence, we may improve on the main tool developed in [Lin22].

######  Theorem 4.9.

Let $X$ be a one-relator complex and let $Z\looparrowright X$ be a one-relator
tower. If $\pi_{1}(Z)$ is hyperbolic (and virtually special), then the
following are equivalent:

  1. (1)

$Z\looparrowright X$ is a quasi-convex tower and $\pi_{1}(X)$ is hyperbolic
(and virtually special),

  2. (2)

$Z\looparrowright X$ is an acylindrical tower,

  3. (3)

$Z\looparrowright X$ is a stable tower and $\pi_{1}(X)$ contains no
Baumslag–Solitar subgroups.

Moreover, if any of the above is satisfied, then $\pi_{1}(Z)<\pi_{1}(X)$ is
quasi-convex.

###### Proof.

By Theorem 4.8, Magnus subgroups of $\pi_{1}(Z)$ are quasi-convex. By [Lin22,
Theorem 7.16] and [Wis21, Theorem 13.1], we may establish the equivalence
between (2) and (3). Similarly, by [Lin22, Theorem 3.6] and [Wis21, Theorem
13.1], we may establish the equivalence between (2) and (1). ∎

Combining Proposition 4.7 and Theorem 4.9 and using induction, we obtain the
following corollary.

######  Corollary 4.10.

Let $X$ be a one-relator complex and suppose that $\pi_{1}(X)$ contains no
Baumslag–Solitar subgroups. If $X_{N}\looparrowright...\looparrowright
X_{1}\looparrowright X_{0}=X$ is a maximal factored one-relator tower, then
one of the following holds:

  1. (1)

$\pi_{1}(X)$ is hyperbolic and $X_{N}\looparrowright...\looparrowright
X_{1}\looparrowright X_{0}=X$ is an acylindrical tower,

  2. (2)

$X_{K}$ is a primitive extension complex for some $K\geq 0$ and, if
$\pi_{1}(X_{K})$ is hyperbolic, then $\pi_{1}(X)$ is hyperbolic and
$X_{K}\looparrowright...\looparrowright X_{1}\looparrowright X_{0}=X$ is an
acylindrical tower.

##  5\. Primitive extension groups

Let $i<j\in\mathbb{Z}$ and denote by

$A_{i,j}=\\{a^{t^{i}},a^{t^{i+1}},...,a^{t^{j}}\\}\subset
F(a,t)\leavevmode\nobreak\ .$

For each $p/q\in\mathbb{Q}_{>0}$, we define two new families of one-relator
groups. The first family is parametrised by two words

$\displaystyle x$ $\displaystyle\in\langle A_{0,k-1}\rangle-\langle
A_{1,k-1}\rangle\leavevmode\nobreak\ ,$ $\displaystyle y$
$\displaystyle\in\langle A_{1,k}\rangle-\langle
A_{1,k-1}\rangle\leavevmode\nobreak\ ,$

such that $\operatorname{pr}_{p/q}(x,y)$ is a primitive exceptional
intersection word of the first type (as an element of $F(A_{0,k})$). We then
define:

(1) $E_{p/q}(x,y)=\langle
a,t\mid\operatorname{pr}_{p/q}(x,y)\rangle\leavevmode\nobreak\ .$

We call this a _primitive extension group of the first type_.

The second family is parametrised by three words

$\displaystyle x$ $\displaystyle\in\langle A_{0,k-1}\rangle-\langle
A_{1,k-1}\rangle\leavevmode\nobreak\ ,$ $\displaystyle y$
$\displaystyle\in\langle A_{1,k}\rangle-\langle
A_{1,k-1}\rangle\leavevmode\nobreak\ ,$ $\displaystyle z$
$\displaystyle\in\langle A_{1,k-1}\rangle-1\leavevmode\nobreak\ ,$

such that $\operatorname{pr}_{p/q}(xy,z)$ is a primitive exceptional
intersection word of the second type (as an element of $F(A_{0,k})$). We then
define:

(2) $F_{p/q}(x,y,z)=\langle
a,t\mid\operatorname{pr}_{p/q}(xy,z)\rangle\leavevmode\nobreak\ .$

We call this a _primitive extension group of the second type_.

######  Definition 5.1.

A group $G$ is a _primitive extension group_ if it is a primitive extension
group of the first or second type.

It follows from Corollary 3.12 that $E_{p/q}(x,y)$ has the following graph of
groups decomposition:

${{F(A_{0,k-1})}}$${{F(A_{1,k-1})*\langle
w\rangle}}$${{F(A_{1,k})}}$$\scriptstyle{\langle A_{1,k-1},x\rangle=\langle
A_{1,k-1},w^{p}\rangle}$$\scriptstyle{\langle A_{1,k-1},w^{q}\rangle=\langle
A_{1,k-1},y\rangle}$$\scriptstyle{\langle A_{0,k-1}\rangle=\langle
A_{1,k}\rangle}$

Similarly, it follows from Corollary 3.13 that $F_{p/q}(x,y,z)$ has the
following graph of groups decomposition:

${{F(A_{0,k-1})}}$${H}$${{F(A_{1,k})}}$$\scriptstyle{\langle
A_{1,k-1},x\rangle=\langle A_{1,k-1},x\rangle}$$\scriptstyle{\langle
A_{1,k-1},y\rangle=\langle A_{1,k-1},y\rangle}$$\scriptstyle{\langle
A_{0,k-1}\rangle=\langle A_{1,k}\rangle}$

where $H$ takes the following form:

${{F(A_{1,k-1})}}$${\langle w\rangle}$${{F(x,y)}}$$\scriptstyle{\langle
z\rangle=\langle w^{p}\rangle}$$\scriptstyle{\langle w^{q}\rangle=\langle
xy\rangle}$

Note that since $xy$ is primitive in $F(x,y)$, we see that

$H\cong F(A_{1,k-1},x)\underset{\langle z\rangle=\langle
w^{p}\rangle}{\ast}\langle w\rangle.$

If $G$ is a finitely presented group, we denote by $\delta_{G}$ its Dehn
function. Although the class of one-relator groups containing groups with Dehn
function not bounded by any finite tower of exponentials [Ger92a], the Dehn
function of primitive extension groups cannot be worse than exponential.

######  Lemma 5.2.

If $G$ is a primitive extension group, then $\delta_{G}(n)\preceq\exp(n)$.

###### Proof.

Since primitive extension groups split as graphs of hyperbolic groups with
finitely generated undistorted edge groups, the upper bound follows from
[Ber94, Theorem 2]. ∎

We say two one-relator complexes $X=(\Gamma,\lambda)$ and $Q=(\Delta,\omega)$
are _Nielsen equivalent_ if there is a homotopy equivalence
$f:\Gamma\to\Lambda$ and a homeomorphism $s:S^{1}\to S^{1}$ such that
$f\circ\lambda\simeq\omega\circ s$. This is a strong version of homotopy
equivalence for one-relator complexes, introduced in [LW22].

######  Lemma 5.3.

If $X$ is a primitive extension complex, then $X$ is Nielsen equivalent to a
presentation complex for (1) or (2).

###### Proof.

Denote by $X=(\Gamma,\lambda)$. Without loss, we may assume that $\Gamma$ is a
rose graph. Denote by $a$ and $b$ the two edges in $\Gamma$. This then yields
a one-relator presentation $\langle a,b\mid w\rangle$ for $\pi_{1}(X)$.

By definition, there is a cyclic cover $p:Y\looparrowright X$ and a one-
relator tree domain $Z\subset Y$ such that

$\pi_{1}(Z\cap 1\cdot Z)\cap\pi_{1}(Z\cap-1\cdot Z)\neq\pi_{1}(-1\cdot Z\cap
Z\cap 1\cdot Z),$

after possibly adding edges to $Z$ and extending the $\mathbb{Z}$-action so
that $-1\cdot Z\cap Z\cap 1\cdot Z$ is connected.

Now the epimorphism $\pi_{1}(X)\to\mathbb{Z}$ is induced by an epimorphism
$\pi_{1}(\Gamma)\to\mathbb{Z}$. Suppose that $a$ maps to zero and $b$ maps to
$\pm 1$ (or vice versa) under this homomorphism. Then we see that $\pi_{1}(Z)$
is conjugate to $\langle a,b^{-1}ab,...,b^{-k}ab^{k}\rangle$ for some $k\geq
0$. Moreover, that $-1\cdot Z\cap Z\cap 1\cdot Z$ is connected and that
$\langle a,b\mid w\rangle$ is thus a primitive extension presentation.

Suppose instead that $a$ maps to $p$ and $b$ maps to $-q$ under the
homomorphism to $\mathbb{Z}$, with $p,q>0$. Let
$X^{\prime}=(\Gamma^{\prime},\lambda^{\prime})$ be the one-relator complex
where $\Gamma^{\prime}$ has two edges, $x$ and $y$, and $\lambda^{\prime}$ is
pulled back via the homotopy equivalence $f:\Gamma^{\prime}\to\Gamma$ defined
by mapping $x$ to $\operatorname{pr}_{p/q}(a,b)$ and $y$ to
$\operatorname{pr}_{c/d}(a,b)$, where $c,d$ are integers such that $cp-dq=1$.
The two one-relator complexes are Nielsen equivalent by construction. Now the
induced epimorphism $\pi_{1}(X^{\prime})\to\mathbb{Z}$ maps $x$ to zero and
$y$ to one and so, as before, there is a one-relator tree domain
$Z^{\prime}\subset Y^{\prime}\looparrowright X^{\prime}$ such that $-1\cdot
Z^{\prime}\cap Z^{\prime}\cap 1\cdot Z^{\prime}$ is connected. There is also
an induced homotopy equivalence between $Z^{\prime}$ and a one-relator
subcomplex of $Y$. In particular, we see that we must have

$\pi_{1}(Z^{\prime}\cap 1\cdot Z^{\prime})\cap\pi_{1}(Z^{\prime}\cap-1\cdot
Z^{\prime})\neq\pi_{1}(-1\cdot Z^{\prime}\cap Z^{\prime}\cap 1\cdot
Z^{\prime}).$

Therefore, this implies that $X^{\prime}$ is a presentation complex of a
primitive extension group of the form (1) or (2). ∎

Before proving our main theorem, we first mention some examples of one-relator
groups that are primitive extension groups.

######  Example 5.4.

If $p,q>0$ are coprime integers, then:

$E_{p/q}(a,t^{-1}a^{\pm 1}t)\cong\operatorname{BS}(p,q^{\pm 1})$

######  Example 5.5.

A one-relator group $\langle a,b\mid w\rangle$ in which $\langle
a\rangle\cap\langle b\rangle\neq 1$ is a primitive extension group by Lemma
4.6. More concretely, for all non-zero integers $m,n$, the one-relator group
$\langle a,b\mid a^{m}=b^{n}\rangle$ is a primitive extension group.

######  Example 5.6.

Any one-relator group that splits as an ascending HNN-extension of a finitely
generated free group (or in other words, has non-trivial BNS invariant) is a
primitive extension group for the following reason: if $X$ is a one-relator
complex, then by Brown’s criterion [Bro87], $\pi_{1}(X)$ splits as an
ascending HNN-extension of a free group if and only if there is a one-relator
tower $Z\looparrowright X$ such that $\pi_{1}(Z)=\pi_{1}(Z\cap 1\cdot Z)$ or
$\pi_{1}(Z)=\pi_{1}(-1\cdot Z\cap Z)$ (see also [Lin22, Section 5.3]). Thus,
when $\pi_{1}(X)$ has non-trivial BNS invariant, $\pi_{1}(Z)$ has an
exceptional intersection and is free, so $X$ is a primitive extension complex.
In fact, all such one-relator groups correspond to the subfamily $E_{p}(a,y)$.

######  Theorem 5.7.

A one-relator group is hyperbolic (and virtually special) if its primitive
extension subgroups are hyperbolic (and virtually special).

###### Proof.

Let $G$ be a one-relator group. Suppose that all primitive extension subgroups
of $G$ are hyperbolic (and virtually special). Since $\operatorname{BS}(1,n)$
is a primitive extension group and all Baumslag–Solitar groups contain some
$\operatorname{BS}(1,n)$ as a subgroup, $G$ contains no Baumslag–Solitar
subgroup. Now by Corollary 4.10 and Lemma 5.3, it follows that $G$ is
hyperbolic (and virtually special). ∎

It is now immediate that in order to characterise hyperbolic one-relator
groups, one only needs to characterise hyperbolic primitive extension groups.

######  Corollary 5.8.

Gersten’s conjecture is true if it is true for primitive extension groups.

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