# Groups of F-Type

Benjamin Fine, Gerhard Rosenberger and Leonard Wienke  Department of
Mathematics, Fairfield University, Fairfield, Connecticut 06430, United States
<EMAIL_ADDRESS>Department of Mathematics, University of Hamburg, Bundesstraße
55, 20146 Hamburg, Germany<EMAIL_ADDRESS>Department of
Mathematics, University of Bremen, Bibliothekstraße 5, 28359 Bremen, Germany
<EMAIL_ADDRESS>
###### Abstract.

We consider a class of groups, called groups of F-type, which includes some
known and important classes like Fuchsian groups of geometric rank $\geq 3$,
surface groups of genus $\geq 2$, cyclically pinched one-relator groups and
torus-knot groups, and discuss algebraic and geometric properties of groups of
F-type.

###### Key words and phrases:

group presentations, groups of F-type, virtually torsion-free, coherent,
conjugacy separable, subgroup separable, SQ-universal, Tits alternative,
essential representations, free products with amalgamation, Euler
characteristic.

##  1\. Introduction

A group $G$ is of F-type if it admits a presentation of the following form

$G=\langle a_{1},\dots,a_{n}\mid
a_{1}^{e_{1}}=\dots=a_{n}^{e_{n}}=U(a_{1},\dots,a_{p})V(a_{p+1},\dots,a_{n})=1\rangle$

where $n\geq 2,e_{i}=0$ or $e_{i}\geq 2$, for $i=1,\dots,n$, $1\leq p\leq
n-1$, $U(a_{1},\dots,a_{p})$ is a cyclically reduced word in the free product
on $a_{1},\dots,a_{p}$ which is of infinite order and $V(a_{p+1},\dots,a_{n})$
is a cyclically reduced word in the product on $a_{p+1},\dots,a_{n}$ which is
of infinite order. With $p$ understood we write $U$ for $U(a_{1},\dots,a_{p})$
and $V$ for $V(a_{p+1},\dots,a_{n})$.

Now, if $U=a_{1}^{\pm 1}$ then $e_{1}$ must be equal zero since we assume that
$U$ has infinite order. In this case, $G$ reduces to

$G=\langle a_{1},\dots,a_{n}\mid a_{2}^{e_{2}}=\dots=a_{n}^{e_{n}}=1\rangle$

which is a free product of cyclics. Therefore if $p=1$ (or $p=n-1$) we
restrict groups of F-type to those where $U=a_{1}^{m}$ (or $V=a_{n}^{m}$) with
$m\geq 2$. It follows then that in all cases the group $G$ decomposes as a
non-trivial free product with amalgamation:

$G=G_{1}\Asterisk_{A}G_{2}$

with $G_{1}\neq A\neq G_{2}$ where the factors are free products of the
cyclics

$G_{1}=\langle a_{1},\dots,a_{p}\mid
a_{1}^{e_{1}}=\dots=a_{p}^{e_{p}}=1\rangle,$ $G_{2}=\langle
a_{p+1},\dots,a_{n}\mid a_{p+1}^{e_{p+1}}=\dots=a_{n}^{e_{n}}=1\rangle,$

and

$A=\langle U^{-1}\rangle=\langle V\rangle.$

We first mention the observation that a group of F-type is coherent, that is,
finitely generated subgroups are finitely presented. Observe that finitely
generated subgroups of $G_{1}$ and $G_{2}$, respectively, are finitely related
and every subgroup of $A$ is finitely generated, in fact cyclic because $A$ is
cyclic. Hence, $G$ is coherent by [20]. This answers the conjecture in
affirmatively.

In what follows we make a further restriction. Suppose $UV$ omits some
generator. For instance, suppose that $UV$ does not involve $a_{1}$. Then $G$
is a free product $H_{1}\ast H_{2}$, where

$H_{1}=\langle a_{1}\mid a_{1}^{e_{1}}=1\rangle$

and

$H_{2}=\langle a_{2},\dots,a_{n}\mid
a_{2}^{e_{2}}=\dots=a_{n}^{e_{n}}=UV=1\rangle.$

This does not affect the validity of the upcoming results in the next
sections. Hence, we may assume that $UV$ involves all the generators. In the
following sections we consider essentially faithful representations in general
and for groups of F-type and discuss many consequences of this. Then we
classify the hyperbolic groups of F-type. The hyperbolic groups of F-type have
a faithful representation into $\operatorname{PSL}(2,\mathbb{R})$ which gives
interesting consequences for groups of F-type. Then we give some further
algebraic properties and finally some quotients of groups of F-type. The paper
is in large parts a survey but it contains also many new results. Many results
have appeared in many different locations but it would be convenient and
important to have the known results in just one place. As background for the
used notations, definitions, and results one may take . Groups of F-type were
originally introduced in [15] and [16].

Sadly, our coauthor and friend Ben Fine died during the final preparation of
the paper. The remaining two authors dedicate the paper to the memory of Ben
and to all the work that he inspired.

##  2\. Essential Representations and Algebraic Consequences

Let $G$ be a group. We say that a linear representation $\rho$ over a field of
characteristic 0 is an essentially faithful representation if $\rho$ is
finite-dimensional with torsion-free kernel. We first give some general
statements and show then that a group of F-type has an essentially faithful
representation into $\operatorname{PSL}(2,\mathbb{C})$.

######  Proposition 1.

Let $G$ be a finitely generated group. Then $G$ admits an essentially faithful
representation if and only if $G$ is virtually torsion-free.

######  Proof 2.1.

Suppose $G$ is finitely generated, and $\rho\colon G\to\text{Linear Group}$ is
an essentially faithful representation. Since $G$ is finitely generated
$\rho(G)$ is a finitely generated linear group. From a result of Selberg, see
[23], $\rho(G)$ is then virtually torsion-free. Let $H$ be a torsion-free
normal subgroup of $\rho(G)$ of finite index and let $H^{\ast}$ be the
pullback of $H$ in $G$. $H^{\ast}$ has finite index in $G$. If $g\neq 1$ has
finite order then $\rho(g)$ has exactly the same order since $\rho$ has
torsion-free kernel. Therefore $g$ cannot be in $H^{\ast}$ since its image
would then be an element of finite order in the torsion-free group $H$.
Therefore $H^{\ast}$ must be torsion-free and $G$ is virtually torsion-free.

Conversely, suppose $G$ is virtually torsion-free. Let $H$ be a torsion-free
subgroup of $G$ of finite index. The intersection of the conjugates of $H$ in
$G$ is a normal subgroup of finite index. Hence, $G$ must contain a torsion-
free normal subgroup $H^{\ast}$ of finite index. Choose a faithful finite-
dimensional representation $\rho^{\ast}$ of the finite group $G/H^{\ast}$. The
composition of this with the natural homomorphism from $G$ to $G/H^{\ast}$
will give the desired representation.

This has two immediate consequences.

######  Proposition 2.

  1. 1.

Let $G$ be a finitely generated group. Then $G$ is residually finite if and
only if for each $g\in G$, $g\neq 1$, there exists a non-trivial linear
representation $\rho$ of $G$ with $\rho(G)\neq\\{1\\}$.

  2. 2.

Let $G$ be a finitely generated group with a balanced representation

$G=\langle a_{1},\dots,a_{n}\mid R_{1}^{m_{1}}=\dots=R_{n}^{m_{n}}=1\rangle$

where each $R_{i}$ is a non-trivial cyclically reduced word in the free
product on $\\{a_{1},\dots,a_{n}\\}$. If at least one $m_{j}\geq 2$, then $G$
is non-trivial.

######  Proof 2.2.

For 1. there is nothing to show. Now we prove 2. It is enough to show the
claim for $n=2$. Consider the group $G^{\ast}=\langle a_{1},a_{2}\mid
R_{1}^{m_{1}}=1\rangle$, for $m_{1}\geq 1$. Its Abelianization has rank at
least 1. If we adjoin $R_{2}^{m_{2}}$ with $m_{2}\geq 2$ to the
Abelianization, the resulting Abelian group is non-trivial.

We now show that groups of F-type admit an essentially faithful representation
into $\operatorname{PSL}(2,\mathbb{C})$. More concretely, we show the
following.

######  Theorem 3.

Let $G$ be a group of F-type. Then $G$ admits a representation $\rho\colon
G\to\operatorname{PSL}(2,\mathbb{C})$ such that $\rho|_{G_{1}}$ and
$\rho|_{G_{2}}$ are faithful. Further, if neither $U$ nor $V$ is a proper
power then $G$ has a faithful representation in
$\operatorname{PSL}(2,\mathbb{C})$.

######  Proof 2.3.

Choose faithful representations

$\rho_{1}\colon G_{1}\to\operatorname{PSL}(2,\mathbb{C})\text{ and
}\rho_{2}\colon G_{2}\to\operatorname{PSL}(2,\mathbb{C})$

such that

$\rho_{1}(U^{-1})=\begin{pmatrix}t_{1}&0\\\ 0&t_{1}^{-1}\end{pmatrix}\text{
and }\rho_{2}(V)=\begin{pmatrix}t_{2}&0\\\ 0&t_{2}^{-1}\end{pmatrix}$

where $t_{1}$ and $t_{2}$ are transcendental. This may be done since $U$ and
$V$ both have infinite order and if $G_{i}$, $i=1$ or $i=2$, is non-cyclic
then the dimension of the character space as an affine variety is positive.
This, especially, implies that we may choose $\rho_{1}$ and $\rho_{2}$ such
that $t_{1}=t_{2}$. Now define $\rho\colon
G\to\operatorname{PSL}(2,\mathbb{C})$ via $\rho_{i}=\rho|_{G_{i}}$ for
$i=1,2$.

This gives the desired representation of the theorem. Further if neither $U$
nor $V$ is a proper power, the above construction leads to the existence of a
faithful representation of $G$ because $\rho(g)$, $g\in G_{1}\setminus\langle
U\rangle$, and $\rho(U)$ have no common fixed point, considered as a linear
fractional transformation, analogously for $G_{2}$ and $V$. This gives the
result inductively.

We note that if both $U$ and $V$ are proper powers then there is no faithful
representation in $\operatorname{PSL}(2,\mathbb{C})$. If $U=U_{1}^{\alpha}$,
$\alpha\geq 2$, and $V=V_{1}^{\beta}$, $\beta\geq 2$, then $\rho(U_{1})$ and
$\rho(V_{1})$ commute but $U_{1}$ and $V_{1}$ do not commute.

We now describe some straightforward algebraic consequences which we get from
the essentially faithful representation of a group of F-type into
$\operatorname{PSL}(2,\mathbb{C})$.

######  Corollary 4.

Let $G$ be a group of F-type. Then

  1. 1.

$G$ is virtually torsion-free.

  2. 2.

If neither $U$ nor $V$ is a proper power then $G$ is residually finite and
thus Hopfian.

  3. 3.

    1. (i)

If $e_{i}\geq 2$ then $a_{i}$ has order exactly $e_{i}$.

    2. (ii)

Any element of finite order in $G$ is conjugate to a power of some $a_{i}$.

    3. (iii)

Any finite subgroup is cyclic and conjugate to a subgroup of some $\langle
a_{i}\rangle$.

    4. (iv)

Any Abelian subgroup is cyclic or free Abelian of rank 2.

The next result is concerned with the Tits alternative, that is, a group
satisfies the Tits alternative if it either contains a free subgroup of rank 2
or is virtually solvable. We use for the Tits alternative some simple facts:

  1. 1.

A subgroup $H$ of $\operatorname{PSL}(2,\mathbb{C})$ is elementary if the
commutator of any two elements of infinite order has trace 2; or equivalently,
$G$ is elementary if any two elements of infinite order, regarded as linear
fractional transformations, have at least one common fixed point. A non-
elementary subgroup contains a free subgroup of rank two. Hence, if $G$ is a
group of F-type and $\rho\colon G\to\operatorname{PSL}(2,\mathbb{C})$ an
essential representation such that $\rho(G)$ is non-elementary, then $G$ has a
free subgroup of rank 2.

  2. 2.

If

$G=\langle a_{1},a_{2}\mid a_{1}^{p}=a_{2}^{q}=1\rangle$

with $2\leq p,q$ and $p+q\geq 5$ then $G$ contains a free subgroup of rank 2.

This now gives the following result.

######  Corollary 5.

Let $G$ be a group of F-type. Then either $G$ has a free subgroup of rank 2 or
is solvable and isomorphic to a group with one of the following
representations:

  1. (1)

$H_{1}=\langle a,b\mid a^{2}b^{2}=1\rangle$,

  2. (2)

$H_{2}=\langle a,b,c\mid a^{2}=b^{2}=abc^{2}=1\rangle$, and

  3. (3)

$H_{3}=\langle a,b,c,d\mid a^{2}=b^{2}=c^{2}=d^{2}=abcd=1\rangle$.

From our previous discussion we get a close tie between the existence of non-
Abelian free subgroups and SQ-universality. Recall that a group $H$ is SQ-
universal if every countable group can be embedded isomorphically as a
subgroup of a quotient of $H$.

######  Theorem 6.

Let $G$ be a group of F-type. If $G$ is not solvable then $G$ has a subgroup
of finite index which maps onto a free group of rank 2. In particular, a group
of F-type is either SQ-universal or solvable.

######  Proof 2.4.

Suppose $G$ is not solvable. First assume that $n\geq 5$ or $n=4$ and at least
one $e_{i}\neq 2$. We may assume without loss of generality that each
$e_{i}\geq 2$, passing to an epimorphic image if necessary. Let $\rho\colon
G\to\operatorname{PSL}(2,\mathbb{C})$ be an essential representation, and let
$N$ be a normal torsion-free subgroup of finite order in $\rho(G)$. Let $\pi$
be the canonical epimorphism from $\rho(G)$ onto the finite group $\rho(G)/N$.
Now consider

$X=\langle a_{1},\dots,a_{n}\mid a_{1}^{e_{1}}=\dots=a_{n}^{e_{n}}=1\rangle.$

There is a canonical epimorphism $\varepsilon\colon X\to G$. Consider the
sequence

$\textstyle{X\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\varepsilon}$$\textstyle{\ignorespaces\ignorespaces\ignorespaces\ignorespaces
G}$$\scriptstyle{\rho}$$\textstyle{\rho(G)\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\pi}$$\textstyle{\rho(G)/N.}$

This yields an epimorphism $\psi\colon X\to\rho(G)/N$. Let $Y=\ker(\psi)$.
Then $Y$ is a normal subgroup of finite index $j$ in $X$, and $Y\cap\langle
a_{i}\rangle=\\{1\\}$ for $i=1,\dots,n$. Then by the Kurosh theorem $Y$ is a
free group of finite index rank $r$. The finitely generated free product of
cyclic groups $Y$ may be considered as a Fuchsian group of finite hyperbolic
area. From the Riemann-Hurwitz formula we have

$j\left(n-1-\left(\frac{1}{e_{1}}+\dots+\frac{1}{e_{n}}\right)\right)=r-1.$

Therefore

$r=1-j\left(\left(\frac{1}{e_{1}}+\dots+\frac{1}{e_{n}}\right)-n+2\right).$

The group $G$ is obtained from $X$ by adjoining the additional relation $UV=1$
and thus $G=X/K$ where $K$ is the normal closure of $UV$ in $X$. Since
$K\subset Y$ the factor group $Y/K$ may be regarded as a subgroup of finite
index $j$ in $G$, and $Y/K$ can be defined on $r$ generators and $j$
relations. Then the deficiancy of this presentation is given by

$\displaystyle d$
$\displaystyle=r-j=1-\left(\left(\frac{1}{e_{1}}+\dots+\frac{1}{e_{n}}\right)-n+2\right)$
$\displaystyle=1+j\left(n-2-\left(\frac{1}{e_{1}}+\dots+\frac{1}{e_{n}}\right)\right).$

If $n\geq 5$ or $n=4$ and at least one $e_{i}\neq 2$, then $d\geq 2$. It
follows then from [4] that $G$ contains a subgroup of finite index which maps
onto a free group of rank 2. Next suppose $n=4$ and all $e_{i}=2$. Then
necessarily $p=2$ and $U=(a_{1}a_{2})^{s},V=(a_{3}a_{4})^{t}$ with
$|s|,|t|\geq 1$. Since $G$ is non-solvable, then without loss of generality we
may assume that $s\geq 2$ and $t\geq 1$. Then $G$ has as a factor group the
free product

$\overline{G}=\langle a_{1},a_{2},a_{3}\mid
a_{1}^{2}=a_{2}^{2}=(a_{1}a_{2})^{s}=a_{3}^{2}=1\rangle.$

The group $\overline{G}$ has as a normal subgroup of index 2 a group that is
isomorphic to $\langle x,y,z\mid x^{s}=y^{2}=z^{2}=1\rangle$. Therefore,
$\overline{G}$ and hence $G$ also has a subgroup of finite index mapping onto
a free group of rank 2. If $n=3$ then the result follows by a similar case-by-
case consideration, see [17].

One of the most powerful techniques in the study of Fuchsian groups is the
Riemann-Hurwitz formula relating the Euler characteristic of the whole group
to that of a subgroup of finite index.

The concept of a rational Euler characteristic is extended to more general
finitely presented groups. Further, these general rational Euler
characteristics satisfy the Riemann-Hurwitz formula. For the general
development of group homology and Euler characteristic we refer to [6].

######  Theorem 7.

Let $G$ be a group of F-type. Then $G$ has a rational Euler characteristic
$\chi(G)$ given by

$\chi(G)=2+\sum_{i=1}^{n}a_{i},$

where $a_{i}=-1$ if $e_{i}=0$ and $a_{i}=-1+\frac{1}{e_{i}}$ if $e_{i}\geq 2$.
If $|G:H|<\infty$ then $\chi(G)$ is defined and $\chi(H)=|G:H|\cdot\chi(G)$,
the Riemann-Hurwitz formula. In addition, if $G$ is of finite homological type
WFL, that is, $G$ is virtually torsion-free and for every torsion free
subgroup of finite index $\mathbb{Z}$ admits a finite free resolution over the
group ring $\mathbb{Z}G$, and $G$ has virtual cohomological dimension
$\operatorname{vcd}(G)\leq 2$.

######  Proof 2.5.

Since $G$ is virtually finite we can apply the techniques of Brown, see [6],
to get an Euler characteristic for $G$ by the formula

$\chi(G)=\chi(G_{1})+\chi(G_{2})-\chi(A).$

Recall that $\chi(H)$ is defined if $H$ is a free product of cyclics. Since
$A$ is infinite cyclic we have $\chi(A)=0$. Therefore
$\chi(G)=\chi(G_{1})+\chi(G_{2})$. $G_{1}$ and $G_{2}$ are free products of
cyclic groups, so we can apply the computation rules for the free products,
and we get

$\chi(G)=2+\sum_{i=1}^{n}\alpha_{i}$

where $\alpha_{i}=-1$ if $e_{i}=0$ and $\alpha_{i}=-1+\frac{1}{e_{i}}$ if
$e_{i}\geq 2$. We note that the Euler characteristic of a group of F-type can
be zero. In fact, for instance, $\psi(G)=0$ if $n=2$. Now, from Section 7.6 in
[6], we get that $G$ is of finite homological type WFL with
$\operatorname{vcd}(G)\leq 2.$

##  3\. Additional Algebraic Results for Groups of F-Type

The first results follow by a straightforward application of the Nielsen
cancellation method in free products with amalgamation, see [9] and .

######  Theorem 8 (Freiheitssatz for groups of F-type).

Let $G$ be a group of F-type. Suppose that $UV$ involves all the generators.
Then

  1. 1.

Any subset of $(n-2)$-many of the given generators generates a free product of
cyclics of the obvious orders.

  2. 2.

If both $U$ and $V$ are proper powers in the respective factors $G_{1}$ and
$G_{2}$, then any subset of $(n-1)$-many of the given generators generates a
free product of cyclics of the obvious orders.

######  Theorem 9.

Let $G$ be a group of F-type and let $H$ be a non-cyclic two-generator
subgroup of $G$. Then $H$ is conjugate in $G$ to a subgroup $\langle
x,y\rangle$ satisfying one of the following conditions:

  1. (1)

$\langle x,y\rangle$ is a free product of cyclic groups.

  2. (2)

$x^{t}$ is in $\langle U\rangle=\langle V\rangle$ for some natural number $t$
and $y^{-1}x^{t}y$ is in $\langle a_{1},\dots,a_{p}\rangle$ or $y^{-1}x^{t}y$
is in $\langle a_{p+1},\dots,a_{n}\rangle$.

Recall that a subgroup $K$ in a group $H$ is malnormal in $H$ provided

$gKg^{-1}\cap K=\\{1\\}$

unless $g\in H$. If $G$ is a group of F-type then $\langle U\rangle=\langle
V\rangle$ is malnormal in $G$ if neither $U$ nor $V$ is a proper power or is
conjugate to a word of the form $xy$ for elements $x,y$ of order 2. Since non-
cyclic two-generator subgroups of the free products of cyclic groups are free
products of two cyclic groups we get the following corollary.

######  Corollary 10.

Let $G$ be a group of F-type. Suppose further that $\langle U\rangle=\langle
V\rangle$ is malnormal in $G$. Then any two-generator subgroup of $G$ is a
free product of cyclics, and $\operatorname{rank}(G)\geq 3$.

######  Remark 11.

In Subsection 3.5.4 of we just gave a list of properties of groups of F-type.
Unfortunately, the last property is not correct as stated at this place. We
just forgot to add that $\langle U\rangle=\langle V\rangle$ is malnormal in
$G$. The correct statement is given in Corollary 1.5.20 of .

Indeed, the malnormality of $\langle U\rangle=\langle V\rangle$ in $G$ is
important. For instance, the following groups of F-type

$\langle a_{1},a_{2},a_{3},a_{4}\mid
a_{1}^{2}=a_{2}^{2}=a_{3}^{2}=a_{4}^{3}=a_{1}a_{2}a_{3}a_{4}=1\rangle$

and

$\langle a_{1},a_{2},a_{3}\mid
a_{1}^{2}=a_{2}^{3}=a_{3}^{2}=a_{1}a_{2}a_{3}^{2}=1\rangle$

are two-generator groups. These examples are considered in .

We may extend Corollary 10 easily to the following.

######  Corollary 12.

Let $G$ be a group of F-type and $UV$ involves all the generators. Suppose
further that $p\geq 3$, and $n-p\geq 3$, and that $\langle U\rangle=\langle
V\rangle$ is malnormal in $G$. Then any three-generator subgroup of $G$ is a
free product of cyclics.

######  Proof 3.1.

We apply the Nielsen cancellation method for free products with amalgamation.
Let $x,y,z\in G$. If $\langle x,y,z\rangle$ is already of rank two there is
nothing to show. Hence, let $\operatorname{rank}(\langle x,y,z\rangle)=3$.
Without loss of generality, we may assume that one of the following two cases
holds:

  1. a)

$x,y,z\in\langle a_{1},\dots,a_{p}\rangle=G_{1}$,

  2. b)

$x,y\in\langle a_{1},\dots,a_{p}\rangle=G_{1}$ with $z\notin G_{2}$.

In case a) we are done. Now, let $x,y\in G_{1}$ and $z\notin G_{2}$. We
consider the subgroup $\langle G_{1},z\rangle$. We may assume that $z$ has the
normal form

$z=\mu_{1}\mu_{2}\dots\mu_{k},$

$k\geq 1$ and $\mu_{1},\mu_{k}\in G_{2}\setminus\langle V\rangle$. But then
$\langle y,x,z\rangle$ must be a free product of cyclics because $n-p\geq 3$.

If neither $U$ nor $V$ is a proper power, then from Theorem 3, $G$ is both
residually finite and Hopfian. This holds in general for groups of F-type.
Recall that a group $H$ is conjugacy seperable if given any non-trivial
$g,h\in G$ that are not conjugate then there exists a finite quotient
$H^{\ast}$ of $H$ where images of $g$ and $h$ are still not conjugate.
Conjugacy separability implies residual finiteness.

From work of Allenby in [3] we get that a group of F-type is conjugacy
separable. Allenby actually proves the following.

Let

$G=\langle a_{1},\dots,a_{n}\mid
a_{1}^{e_{1}}=\dots=a_{n}^{e_{n}}=(UV)^{m}=1\rangle,$

where $n\geq 2$, $e_{i}=0$ or $e_{i}\geq 2$, $1\leq p\leq n-1$,
$U=U(a_{1},\dots,a_{p})$ a non-trivial cyclically reduced word in the free
product on $a_{1},\dots,a_{p}$, $V=V(a_{p+1},\dots,a_{n})$ a non-trivial
cyclically reduced word in the free product on $a_{p+1},\dots,a_{n}$, and
$m\geq 2$. Then $G$ is conjugacy separable.

However, in this proof $m\geq 2$ is only used in the case where either $U$ or
$V$ has finite order in the respective free product on the generators which
they involve. In a group of F-type $U$ and $V$ are assumed to have infinite
order so the result goes through. Hence, we have the following.

######  Theorem 13.

A group of F-type is conjugacy separable and, hence, residually finite and
Hopfian.

A detailed proof can be found in [17]. Recall that a group $H$ is subgroup
separable or LERF (locally extended residually finite) if given any subgroup
$K$ of $H$ and any element $g$ not in $K$ there exists a subgroup $K^{\ast}$
of finite index in $H$ such that $K$ is in $K^{\ast}$ and $g$ is not in
$K^{\ast}$. From work of Aab and Rosenberger, see [1], we can deduce that
groups of F-type are subgroup separable. Details can be found in .

######  Theorem 14.

A group of F-type is subgroup separable.

Tang, see [24], considered a class of groups which contains the groups of
F-type.

######  Theorem 15.

A group of F-type has solvable generalized word problem, and hence, solvable
word problem.

Recall that the generalized word problem is the following. Let $H=\langle
X\mid R\rangle$ be a finite presentation and $W$ a finite set of words
$\\{w_{1},\dots,w_{m}\\}$ in $X$. Let $K$ be the subgroup of $G$ generated by
$W$. Given a word $v$ in $X$, can one decide whether or not $v$ is in $K$?
From Theorem 4.6 in [25] we see that groups of F-type have solvable conjugacy
problem. The applied technique also answers the power conjugacy problem for
groups of F-type, that is, given two elements to determine if a power of one
is conjugate to a power of the other. Hence, we have the following.

######  Theorem 16.

Groups of F-type have both solvable conjugacy problem and solvable power
conjugacy problem.

##  4\. Hyperbolic Groups of F-Type

In the following we present the results by A. Juhász and G. Rosenberger, see
[19], and consider the combinatorial curvature of a class of one-relator
products which generalize groups of F-type. Recall from [18, p.134] that a
presentation $\langle X\mid R\rangle$ has non-positive (negative)
combinatorial curvature if the hyperbolic area of every inner face of every
van Kampen diagram over $R$ has nonnegative (positive) hyperbolic area. It is
known that groups with a presentation having negative curvature are hyperbolic
in the sense of Gromov. We give a more detailed explanation below. The main
result is the following

######  Theorem 17.

Let $G^{(1)}=\Asterisk_{\alpha\in\mathcal{T}_{1}}G_{\alpha}^{(1)}$,
$G^{(2)}=\Asterisk_{\beta\in\mathcal{T}_{2}}G_{\beta}^{(2)}$, and let
$G^{(0)}=G^{(1)}\ast G^{(2)}$. Let $U\in G^{(1)}$ and $V\in G^{(2)}$ be
cyclically reduced words of infinite order in $G^{(0)}$. Assume that if $U\in
G_{\alpha}^{(1)}$, then $G_{\alpha}^{(1)}$ is free and $U$ is cyclically
reduced in $G_{\alpha}^{(1)}$. Assume the analogue for $V$. Let $G$ be the
quotient of $G^{(0)}$ by the normal closure of $UV$ in $G^{(0)}$. Then

  1. 1.

$G$ has non-positive combinatorial curvature as a quotient of ${G}^{(0)}$.

  2. 2.

$G$ is hyperbolic in the sense of Gromov if and only if $G^{(i)}$ is
hyperbolic for $i=1,2$, and the following property holds:

    1. ($\dagger$)

At least one of $U$ or $V$ is neither a proper power nor a product of two
elements of order 2.

Since groups of F-type are special cases of the groups mentioned in the
theorem and cyclic groups are hyperbolic, we get the following corollary.

######  Corollary 18.

Groups of F-type are hyperbolic unless $U$ is a proper power or a product of
two elements of order $2$ and $V$ also is a proper power or a product of two
elements of order $2$. In the last case they have non-positive combinatorial
curvature. In particular, they satisfy a quadratic isoperimetric inequality.

We consider van Kampen diagrams. All the unexplained terms concerning them can
be found in [21].

######  Remark 19.

  1. 1.

Let $M$ be a diagram over a free group or free product $F$. We shall denote by
$|\partial M|$ the length of a cyclically reduced label of $M$ over $F$.

  2. 2.

Assume $A$ is given by a presentation $\langle X\mid R\rangle$. Then we shall
always assume that $R$ is cyclically reduced. Also we shall assume that our
van Kampen diagram contains a minimal number of regions for a simply connected
given boundary label.

  3. 3.

Recall from [18] that a presentation $\langle X\mid R\rangle$ has non-positive
(negative) combinatorial curvature if every inner region $D$ of every van
Kampen diagram over $R$ satisfies that the excess
$\kappa(D)=2\pi+\sum_{i=1}^{n}\left(\theta_{i}-\pi\right)$ is non-positive
(negative). Here, $\partial D=v_{1}e_{1}v_{2}e_{2}\cdots v_{n}e_{n}v_{1}$,
$v_{i}$ vertices on $\partial D$, $e_{i}$ edges on $\partial D$, and
$\theta_{i}$ are the inner angles of the polygon $D$ at $v_{i}$. In general,
one takes $\theta_{i}=\frac{2\pi}{d(v_{i})}$ where $d\left(v_{i}\right)$ is
the valency of $v_{i},$ that is, distribute the curvature equally among the
regions containing $v_{i}$. However, from the point of view of the theory of
groups with non-positive curvature, see [5] and the theory of hyperbolic
groups, see [18], it is immaterial how the angles around $v_{i}$ are
distributed, as long as they sum up to $2\pi$ and
$\theta_{i}-\frac{2\pi}{d(v_{i})}<\frac{2\pi}{6}.$ In the following we shall
make use of this remark.

The following lemma is an immediate consequence of the construction of
diagrams over free products. We omit its proof.

######  Lemma 20.

Let $A=\langle X\mid R\rangle$, $B=\langle Y\mid S\rangle$ be finitely
generated, and let $G=A\ast B/D,$ where $D$ is the normal closure of $T\subset
A\ast B$. Let $Q=R\cup S\cup T$. Thus $Q\subset F(X\cup Y)$. Assume that there
are constants $C$ and $a$ such that the following isoperimetric inequalities
hold for $R$-diagrams $M$, $S$-diagrams $N$ and $T$-diagrams $H$ respectively:

$\mathrm{Vol}(M)\leq C|\partial M|^{a},\mathrm{Vol}(N)\leq C|\partial
N|^{a}\text{ and }\mathrm{Vol}(H)\leq C|\partial H|^{a}$

where for a $P$-diagram $Z$, $\mathrm{Vol}(Z)$ is the number of regions of
$Z$.

Then there is a constant $C^{\prime}\geq C$ depending on $Q$ such that for
every word $W\in F(X\cup Y)$ which represents $1$ in $G$ there is a
$Q$-diagram $U$ with boundary label $W$ such that $\mathrm{Vol}(U)\leq
C^{\prime}|W|^{a}$.

We also mention the following corollary.

######  Corollary 21.

  1. 1.

If $A$ and $B$ are hyperbolic and the presentation of $G$ as a quotient of the
free product $A\ast B$ satisfies a linear isoperimetric inequality, then $G$
is hyperbolic.

  2. 2.

If $A$ and $B$ satisfy a polynomial isoperimetric inequality of degree $k$ and
$G$ has nonpositive combinatorial curvature as a quotient of $A\ast B$, then
$G$ satisfies a polynomial isoperimetric inequality of degree $\max(k,2)$.

Let $R$ be the set of all the cyclically reduced cyclic conjugates of $UV$ and
$(UV)^{-1}$. We now describe the structure of the van Kampen diagrams for the
groups given in Theorem 17. Let $M$ be a reduced $R$-diagram, and let $D$ be
an inner region of $M$. Then $D$ has a boundary cycle $v_{0}\mu v_{1}\nu
v_{0}$ such that the label of $\mu$ is $U$ and the label of $v$ is $V$. From
now on, we shall not distinguish between a path in $M$ and its label in
$G^{(0)}$. Since the vertices $v_{0}$ and $v_{1}$ separate $\mu$ from $v$ we
shall call them separating vertices, see Figure 1.

••$D$$v_{0}$$v_{1}$$\mu$$\nu$ Figure 1.

######  Theorem 22.

Separating vertices of D have valency at least $4$.

######  Proof 4.1.

Assume $v_{0}$ has valency $3$. Then there are regions $D_{1}$ and $D_{2}$
which contain $v_{0}$ on their boundary. Let $x$ be the first letter on the
edge common to $D_{1}$ and $D_{2}$ which contains $v_{0}$. If $x\in G^{(1)}$,
then $v_{0}$ is a separating vertex of $D_{2}$, see Figure 2 (A). Consequently
a non-trivial tail of $v$ is a common edge of $D$ and $D_{2}.$ But then, since
$U$ is cyclically reduced and $U\neq U^{-1}$, $D$ completely cancels $D_{n}$.
This violates the assumption that $M$ is reduced.

•$D_{1}$$D_{2}$$v_{0}$$x$$D$ (a)

•$D_{1}$$D_{2}$$v_{0}$$x$$D$ (b)

Figure 2.

Similarly, if $x\in G^{(2)}$, then $v_{0}$ is a separating vertex of $D_{1}$,
see Figure 2 (B), leading to the same contradiction. Thus $v_{0}$ has valency
$\neq 3$. Finally assume $v_{0}$ has valency $2$. Then, since $U$ and $V$ are
cyclically reduced, $v_{0}$ is a separating vertex of $D_{1}$ and $D_{1}$
cancels $D$ completely as in Figure 3.

$u$$u$$D$$D_{2}$ Figure 3.

######  Corollary 23.

If $M$ satisfies the small cancellation condition $C(5)$, then $M$ has a non-
positive combinatorial curvature.

The inner region $D$ always has two vertices with valency $\geq 4$ (namely
$v_{0}$ and $v_{1}$).

######  Remark 24.

Neither $U$ nor $V$ can be a piece, for then as we have seen, the diagram is
not reduced. Consequently $U$ and $V$ are each the product of at least two
pieces. Since by the theorem no region $D_{1}$ can be a neighbor of $D$ with a
common edge, see Figure 4, which contains $v_{0}$ as an inner vertex, $D$ has
at least $4$ neighbors.

•$v_{0}$$D_{1}$$D$ Figure 4.

This proves the following.

######  Corollary 25.

  1. 1.

$M$ satisfies the small cancellation condition $C(4)$.

  2. 2.

An inner region $D$ has $4$ neighbors if and only if its boundary cycle is
subdivided into $4$ edges in such a way that the label on the first two edges
is $U$ and the label on the second two edges is $V$. In particular, if such a
region exists, then $U$ and $V$ are each the product of two pieces.

Thus, in what follows, we shall study the situation when $U$ is the product of
two pieces. Denote by $\mathcal{D}_{0}$ the set of all the inner regions in
$M$ which have a boundary path with label in $\\{U^{\pm 1},V^{\pm 1}\\}$ and
have a decomposition to two pieces by a vertex which is not a separating
vertex.

Denote the set of all the inner vertices of $M$ which are not separating
vertices and divide a boundary path of a region $D$ with label $U^{\pm 1}$ or
label $V^{\pm 1}$ to two parts by $\mathcal{V}_{0}$. For $i\geq 3$ denote by
$\mathcal{V}_{i}$ the set of all the vertices of $M$ not in $V_{0}$ which have
valency $i$. Denote by $\kappa(U)$ the contribution of all the vertices of an
inner region $D$ which are inner vertices of the boundary path $\mu$ with
label $U$ to the excess of $D$, and let $\kappa(V)$ be defined analogously.
Then

$\displaystyle\kappa(D)=2\pi+\kappa(U)+\kappa(V)+(\theta_{0}-\pi)+(\theta_{t}-\pi)=\kappa(U)+\kappa(V)+\theta_{0}+\theta_{t}$
(1)

where $\theta_{0}$ and $\theta_{t}$ denote the angles of the two separating
vertices $v_{0}$ and $v_{t}$, compare Figure 1.

  1. 1.

A corner consists of a pair $(e,e^{1})$ of edges which have the same initial
vertex and are such that the path $e^{-1}e^{1}$ is a subpath of a boundary
cycle of a region associated with the corner.

  2. 2.

Given a vertex $u\in\mathcal{V}_{0},$ a bad corner at $u$ is a corner
$(e,e^{1})$ such that the path $e^{-1}e^{1}$ has label one of $U^{\pm
1},V^{\pm 1}$. A good corner is a corner which is not bad.

  3. 3.

Call two corners adjacent if they have an edge in common.

######  Theorem 26.

Let the edge pair $\left(\alpha^{-1},\beta\right)$ define a bad corner at the
vertex $v$ of the region $D$. Then

  1. 1.

The two corners at $v$ which are adjacent to the given corner are good
corners.

  2. 2.

The vertex $v$ is preceded or followed (in the boundary cycle of $D$) by a
vertex of valency 4 which is a separating vertex of $D$.

•••••••$D$$D_{1}$$D_{2}$$v_{0}$$v_{1}$$w_{0}$$w_{1}$$w_{2}$$w_{3}$$\nu_{1}$$\nu_{2}$$\mu_{1}$$\mu_{2}$$\alpha$$\beta$$v$
Figure 5.

######  Proof 4.2.

Let $w_{0}$ and $w^{\prime}$ be the separating vertices of $F$, see Figure 5.
Then $w_{1}\neq v$ since $v$ is not a separating vertex by assumption and
$w_{0}\neq v_{0}$, otherwise either $D_{1}$ cancels $D$ (if $\mu$ has label
$U$) or $U$ is not cyclically reduced (if $\mu_{1}$ has label $U^{-1}$).

Now define the angles in $M$. If $v\in\mathcal{V}_{i}$, $i\geq 3$, then assign
the value $\frac{2\pi}{i}$ to every angle having $v$ as its vertex. If
$v\in\mathcal{V}_{0}$ and $d=d(v)\geq 5$ then assign to each angle having $v$
as its vertex the value $\frac{2\pi}{d}$. Let
$\varepsilon=\frac{2\pi}{8m^{2}}$, $m=2(|U|+|V|)$. If $v\in\mathcal{V}_{0}$
and $d(v)=4$, then by Theorem 26 there are at most two regions which contain
$v$ on their boundary and belong to $\mathcal{D}_{0}$ ($D_{1}$ and $D_{3}$ in
Figure 6 (A) and (B)).

If $D_{1}$ and $D_{3}$ are in $\mathcal{D}_{0},$ then assign
$\frac{2\pi}{4}-\varepsilon$ to $\gamma_{1}$ and $\gamma_{3}$ and assign
$\frac{2\pi}{4}+\varepsilon$ to $\gamma_{2}$ and $\gamma_{4}$. If
$D_{1}\in\mathcal{D}_{0}$ and $D_{3}\notin\mathcal{D}_{0},$ then assign
$\frac{2\pi}{4}-\varepsilon$ to $\gamma_{1}$,
$\frac{2\pi}{4}+\frac{\varepsilon}{2}$ to $\gamma_{2}$ and $\gamma_{4}$ and
$\frac{2\pi}{4}$ to $\gamma_{3}$. Finally, if $d(v)=3$, then by Theorem 26
exactly one of the regions containing $v$, say $D_{1}$, belongs to
$\mathcal{D}_{0}$, see Figure 6 (B). Assign $\frac{2\pi}{4}-\varepsilon$ to
$\gamma_{1}$ and assign $\frac{2\pi}{3}+\frac{2\pi}{24}+\frac{\varepsilon}{2}$
to $\gamma_{2}$ and to $\gamma_{3}$.

•$D_{2}$$D_{3}$$D_{4}$$D_{1}$$\gamma_{4}$$\gamma_{2}$$v$$\gamma_{1}$$\gamma_{3}$
(a)

•$D_{2}$$D_{3}$$D_{1}$$\gamma_{3}$$\gamma_{2}$$v$$\gamma_{1}$ (b)

Figure 6.

######  Theorem 27.

Let $D$ be an inner region of $M$. If $D$ has a vertex $v\in\mathcal{V}_{0}$
on its boundary, then $\kappa(D)<0$.

######  Proof 4.3.

Denote by $a_{i}$ the number of inner vertices of $\mu$ (see Figure 1) in
$\mathcal{V}_{i}$, $i\geq 3$, and let $a_{4}^{*}=\sum_{i\geq 4}a_{i}$. For
$i\geq 5$, denote by $b_{i}$ the number of inner vertices of $\mu$ in $V_{0}$
with valency $i$. Define numbers $b_{4}^{\prime},b_{4}^{\prime\prime},c_{4}$
and $c_{3}$ as follows:

  * •

$b_{4}^{\prime}$ is the number of inner vertices of $\mu$ in $\mathcal{V}_{0}$
which are of degree $4$ and such that

    1. a)

the corner associated with $D$ is a good corner;

    2. b)

exactly one of the two corners adjacent to the corner associated with $D$ is a
bad corner;

  * •

$b_{4}^{\prime\prime}$ is the number of inner vertices of $\mu$ in
$\mathcal{V}_{0}$ which are of degree $4$ and such that

    1. a)

the corner associated with $D$ is a good corner;

    2. b)

both the two corners adjacent to the corner associated with $D$ are bad
corners;

  * •

$c_{4}$ is the number of inner vertices of $\mu$ in $\mathcal{V}_{0}$ which
are of degree $4$ and such that

    1. a)

the corner associated with $D$ is a good corner;

    2. b)

neither of the two corners adjacent to the corner associated with $D$ are bad
corners (so that the remaining corner which is ’opposite’ to the corner
associated with $D$, must be a bad corner);

  * •

$c_{3}$ is the number of inner vertices of $\mu$ in $\mathcal{V}_{0}$ of
degree $3$ such that the corner associated with $D$ is a good corner (in which
case exactly one of the other two corners adjacent to the corner associated
with $D$ is a bad corner).

Assume $D\notin\mathcal{D}_{0}$ and evaluate $\kappa(U)$. Thus,

$\displaystyle\kappa(U)$ $\displaystyle=\sum_{i\geq
3}a_{i}\left(\frac{2\pi}{i}-\pi\right)+\sum_{i\geq
5}b_{i}\left(\frac{2\pi}{i}-\pi\right)+b_{4}^{\prime}\left(\frac{2\pi}{4}+\varepsilon-\pi\right)$
$\displaystyle+b_{4}^{\prime\prime}\left(\frac{2\pi}{4}+\frac{\varepsilon}{2}-\pi\right)+c_{4}\left(\frac{2\pi}{4}-\pi\right)+c_{3}\left(\frac{2\pi}{3}+\frac{2\pi}{24}-\pi+\frac{\varepsilon}{2}\right)$
$\displaystyle\leq-\frac{\pi}{4}c_{3}-\frac{\pi}{2}a_{4}^{*}-\frac{\pi}{2}\left(b_{4}^{\prime}+b_{4}^{\prime\prime}\right)+\frac{\varepsilon}{2}\left(2b_{4}^{\prime}+b_{4}^{\prime}+c_{3}\right)$
$\displaystyle\leq-\frac{\pi}{4}c_{3}-\frac{\pi}{2}a_{4}^{*}-\frac{\pi}{2}\left(b_{4}^{\prime}+b_{4}^{\prime\prime}\right)+\frac{\pi}{8m^{2}}(2m).$

Assume $c_{3}>0.$ Then, by Theorem 26, $a_{4}^{*}\geq\frac{1}{2}c_{3}$, hence
$-\frac{\pi}{2}\left(a_{4}^{*}+\frac{1}{2}c_{3}\right)\leq-\frac{\pi}{2}-\frac{\pi}{4},$
and

$\kappa(U)\leq-\left(\frac{\pi}{4}+\frac{\pi}{2}\right)+\frac{\pi}{8m^{2}}(2m)=\frac{3}{4}\pi+\frac{\pi}{4m}<-\frac{3}{4}\pi+\frac{\pi}{4}=-\frac{\pi}{2}$

as $m>1.$ Thus

$\displaystyle\text{if }D\notin D_{0}\text{ and }c_{3}>0\text{, then
}\kappa(U)<-\frac{\pi}{2}.$ (2)

Assume $c_{3}=0.$ Then $b_{4}^{\prime}+b_{4}^{\prime\prime}\neq 0,$ hence, by
Theorem 26, $a_{4}>0$. Consequently,

$k(U)\leq-\frac{\pi}{2}a_{4}-\frac{\pi}{2}\left(b^{\prime}+b^{\prime\prime}\right)+\frac{\pi}{4m}\leq-\frac{\pi}{2}-\frac{\pi}{2}+\frac{\pi}{4m}<\frac{\pi}{2}-\frac{\pi}{2}+\frac{\pi}{4}<\frac{\pi}{2}.$

$\displaystyle\text{If }D\notin\mathcal{D}_{0}\text{ and }c_{3}=0\text{, then
}\kappa(U)<-\frac{\pi}{2}.$ (3)

Assume now that $D\in\mathcal{D}_{0}$. Then, by the choice of $\gamma_{1}$,
$\kappa(U)=-\frac{\pi}{2}-\varepsilon$.

$\displaystyle\text{If }D\in\mathcal{D}_{0}\text{, then
}\kappa(U)<-\frac{\pi}{2}.$ (4)

Evaluate now $\kappa(D)$. By (4.1), (4.2), (4.3), and (4.4) we have

$\kappa(D)=\kappa(U)+\kappa(V)+\frac{\pi}{2}+\frac{\pi}{2}<-\frac{\pi}{2}-\frac{\pi}{2}+\frac{\pi}{2}+\frac{\pi}{2}=0$

which yields the theorem.

######  Corollary 28.

$D$ has non-positive combinatorial curvature. $D$ has zero curvature if and
only if $D$ has four neighbors and every boundary vertex is a separating
vertex with valency $4$.

We now prove Theorem 17.

######  Proof 4.4 (Proof of Theorem 17).

We remark that we already have the first statement in Theorem 17 by the
previous results. We now consider the relation ($\dagger$). Let the notation
be as in Figures 7 (A) and (B). Then either $v$ is a separating vertex for
$D_{1}$ or $v$ is a separating vertex for $D_{2}$.

•$D$$A$$v$$B$$D_{2}$$D_{1}$$D_{3}$$V$$U$ (a)

•$D$$A$$v$$B$$V$$V$ (b)

Figure 7.

If $v$ is a separating vertex for only one of $D_{1}$ and $D_{2},$ then it is
a separating vertex for $D_{3}$ and hence for two neighboring regions of $D$
($D_{1}$ and $D_{3}$ or $D_{2}$ and $D_{3}$). But the boundary labels of such
neighbors cancel each other, contradicting our assumption that $M$ contains a
minimal number of regions with the given boundary label. Consequently, $v$ is
a separating vertex for both $D_{1}$ and $D_{2}$, see Figure 7 (B).

Let $A$ be the label of the piece common to $D$ and $D_{1},$ and let $B$ be
the label of the piece common to $D$ and $D_{2}$. Then one of the following
equations holds: $(AB)(AB)=U(AB)V$ (see Figure 8 (A)) or $A=A^{-1}$ and
$B=B^{-1}$, see Figure 8 (B). In the first case $A=D^{\alpha}$ and
$B=D^{\beta}$ and in the second case $A$ and $B$ have order $2$ as required.

•••$A$$B$$A$$B$$A$$B$ (a)

•••$B$$A$$B$$A$ (b)

Figure 8.

Assume ($\dagger$) holds. Then by Theorem 27 the combinatorial curvature at
each inner region of a derived van Kampen diagram over $R$ is strictly
negative. Therefore $G$ is hyperbolic.

Assume now that ($\dagger$) does not hold. Then $UV$ (or $VU$) has one of the
following forms.

  1. a)

$A^{n}B^{m}$ for $n,m\geq 2$;

  2. b)

$ABCD,A=A^{-1}$, $B=B^{-1}$, $C=^{-1}$, $D=D^{-1}$;

  3. c)

$ABC^{n}$, $n\geq 2$, $A=A^{-1}$, $B=B^{-1}$.

In case a) let $H=\langle A,B\rangle$. Then it follows from the fact that
every van Kampen diagram over the symmetric closure of $UV$ has a non-positive
combinatorial curvature that $H\cong\langle a,b\mid a^{n}=b^{-m}\rangle$.

Consequently $G$ is not hyperbolic.

A similar argument shows that the subgroup of G generated by $A,B,C$ and $D$
in case b) is isomorphic to $\langle a,b,c,d\mid
a^{2},b^{2},c^{2},d^{2},abcd\rangle$ which contains a normal free Abelian
subgroup generated by $ab$ and $bc$.

Finally in case c) let $H=\langle B,BC^{n}\rangle$. Then, as above,
$H\cong\langle a,b\mid a^{2},(ab^{n})^{2}\rangle$ in which $K=\langle
b^{n}\rangle$ is an infinite cyclic normal subgroup such that
$H/K=\mathbb{Z}_{2}\ast\mathbb{Z}_{n},n\geq 2$. Thus in all cases, $G$ is not
hyperbolic.

######  Corollary 29.

Let $G$ be a group of F-type. $G$ is hyperbolic unless $U$ is a proper power
or a product of two elements of order 2 and $V$ also is a proper power or a
product of two elements of order 2.

We just want to remark here that by the result of Dahmani and Guirardel, see
[10], the isomorphism problem for a hyperbolic group of F-type is solvable in
the class of hyperbolic groups.

It remains the question if the isomorphism problem is solvable for a group of
F-type in some class of groups. This is an open problem. There exists just a
result in the special case of cyclically pinched one-relator groups, that is,
in the case of groups of F-type with $e_{1}=e_{2}=\dots=e_{n}=0$. The
isomorphism problem for a cyclically pinched one-relator group $G$ is solvable
in the class of one-relator groups, see [22] and .

Certainly, $G$ has no faithful representation in
$\operatorname{PSL}(2,\mathbb{R})$ if $U=XY$ with $X^{2}=Y^{2}=1$ and $V=WZ$
with $W^{2}=Z^{2}=1$ or $U=U_{1}^{k}$, $V=V_{1}^{q}$ with $k,q\geq 2$.

Concerning the question if there exists a faithful representation $\rho\colon
G\to\operatorname{PSL}(2,\mathbb{R})$ we are left, up to symmetry, with the
case $U=XY$ with $X^{2}=Y^{2}=1$ and $V=V_{1}^{q}$ with $q\geq 2$. But in this
case there does not exist such a $\rho$ because

$(\rho(X))^{2}=(\rho(UY))^{2}=(\rho(VY))^{2}=1$

implies $\rho(V_{1}Y)^{2}=1$, but $(V_{1}Y)^{2}\neq 1$ in $G$ for any
essential representation $\rho\colon G\to\operatorname{PSL}(2,\mathbb{R})$.

Our next aim is to characterize the hyperbolicity of a group of F-type by
means of faithful representations in $\operatorname{PSL}(2,\mathbb{R})$. In
[13] we showed the following.

######  Theorem 30.

Let $G$ be a hyperbolic group of F-type. Then $G$ has a faithful
representation in $\operatorname{PSL}(2,\mathbb{R})$.

Hence, we get the following.

######  Corollary 31.

Let $G$ be a group of F-type. Then $G$ is hyperbolic if and only if $G$ has a
faithful representation in $\operatorname{PSL}(2,\mathbb{R})$.

This has now some additional algebraic consequences. We know that
$\operatorname{PSL}(2,\mathbb{R})$ is commutative transitive. Hence, any
hyperbolic group of F-type is commutative transitive. Recall that a group $H$
is commutative transitive if $[x,y]=1=[y,z]$ for $x,y,z\in H$, $y\neq 1$, then
$[x,z]=1$.

Closely tied to commutative transitivity is the concept of being CSA. A group
$H$ is CSA or conjugately separated Abelian if maximal Abelian subgroups are
malnormal. These concepts have played a prominent role in the studies of fully
residually free groups, limits groups and discriminating groups see [12]. They
also play a role in the solution of the Tarski problems. CSA always implies
commutative transitivity.

In general the class of CSA groups is a proven subclass of the class of
commutative transitive groups, however they are equivalent in the presence of
residual freeness. For hyperbolic groups of F-type we have the following.

######  Theorem 32.

Let $G$ be a hyperbolic group of F-type. Assume that $G$ is torsion-free or
has only odd torsion, that is, $e_{i}$ is odd if $e_{i}\geq 2$. Then $G$ is
CSA.

######  Proof 4.5.

We may assume that $G$ is already a subgroup of
$\operatorname{PSL}(2,\mathbb{R})$. Let $A$ be a maximal Abelian subgroup of
$G$. Since $G$ is a subgroup of $\operatorname{PSL}(2,\mathbb{R})$ any two
elements $a,b\in A$ have, considered as linear fractional transformations, the
same fixed points.

Assume that $G$ is not CSA. Then there are non-trivial elements $a,b\in A$ and
$x\in G\setminus A$ with $xax^{-1}=b$. This means that $x$ permutes non-
trivially the fixed points of $a$, which are also the fixed points $b$. Hence,
$x$ must have order two. This gives a contradiction, and therefore $G$ is CSA.

There is a certain relation between CSA groups and RG groups. A group $H$ is a
restricted Gromov group or an RG group, if $H$ satisfies the following
property: If $g$ and $h$ are elements in $H$ then either the subgroup $\langle
g,h\rangle$ is cyclic or there exists a positive integer $t$ with $g^{t}\neq
h^{t}$ and $\langle g^{t},h^{t}\rangle=\langle g^{t}\rangle\ast\langle
h^{t}\rangle$.

An RG group $H$ has the property that every Abelian subgroup is locally
cyclic. Hence, it is often convenient to assume that every element of $H$ is
contained in a maximal cyclic subgroup of $H$. This last property is of course
satisfied in hyperbolic groups. Certainly, an RG group is commutative
transitive. Using Theorem 3.9.18 from we easily get the following.

######  Theorem 33.

Let $G$ be a hyperbolic group of F-type. Assume that $G$ is torsion-free or
has only odd torsion. Then $G$ is an RG group.

We present the following new result and sketch its proof.

######  Theorem 34.

Let $G$ be a group of F-Type which is not a free product of cyclic groups.
Assume that each subgroup of $G$ of infinite index is a free product of cyclic
groups. Then $G$ is a co-compact planar discontinuous group.

For the proof we just consider $G$ as a quotient of a cyclically pinched one-
relator group and extend the arguments in [7]. Then we use the solution of
Nielsen’s realization problem, see [26].

We end this section with the following conjecture.

######  Conjecture 35.

Let $H$ be a finitely generated, non-elementary subgroup of
$\operatorname{PSL}(2,\mathbb{R})$. Then $H$ is finitely presented.

##  5\. One-Relator Amalgamated Products and some Algebraic Consequences

If $A,B\in\operatorname{PSL}(2,\mathbb{C})$ we say that the pair $\\{A,B\\}$
is irreducible if $A,B$, regarded as linear fractional transformations, have
no common fixed point, that is, $\operatorname{tr}[A,B]\neq 2$.

######  Theorem 36 (Freiheitssatz).

Suppose $G=G_{1}\Asterisk_{A}G_{2}$ with $A=\langle a\rangle$ cyclic. Let
$R\in G\setminus A$ be given in a reduced form $R=c_{1}d_{1}\cdots c_{k}d_{k}$
with $k\geq 1$ and $c_{i}\in G_{i}\setminus A$, $d_{i}\in G_{2}\setminus A$
for $i=1,\dots,k$. Assume that there exists a representation $\phi\colon
G\to\operatorname{PSL}(2,\mathbb{C})$ such that $\phi|_{G_{1}}$ and
$\phi|_{G_{2}}$ are faithful and the pairs $\\{\phi(c_{i}),\phi(a)\\}$ and
$\\{\phi(d_{i}),\phi(a)\\}$are irreducible for $i=1,\dots,k$.

Then the group $H=G/N(R^{m})$, $m\geq 2$, admits a representation $\rho\colon
H\to\operatorname{PSL}(2,\mathbb{C})$ such that $G_{1}\to
H\stackrel{{\scriptstyle\rho}}{{\to}}\operatorname{PSL}(2,\mathbb{C})$ and
$G_{2}\to
H\stackrel{{\scriptstyle\rho}}{{\to}}\operatorname{PSL}(2,\mathbb{C})$ are
faithful and $\rho(R)$ has order $m$. In particular, $G_{1}\to H$ and
$G_{2}\to H$ are injective.

######  Proof 5.1.

Let $\phi\colon G\to\operatorname{PSL}(2,\mathbb{C})$ be the given
representation of $G$ such that $\phi|_{G_{1}}$ and $\phi|_{G_{2}}$ are
faithful and the pairs $\\{\phi(c_{i}),\phi(a)\\}$ and
$\\{\phi(d_{i}),\phi(a)\\}$ are irreducible for $i=1,\dots,k$.

We may assume that $\phi(a)$ has the form $\phi(a)=\begin{pmatrix}s&0\\\
0&s^{-1}\end{pmatrix}$ or $\phi(a)=\begin{pmatrix}1&1\\\ 0&1\end{pmatrix}$.

Suppose first that $\phi(a)=\begin{pmatrix}s&0\\\ 0&s^{-1}\end{pmatrix}$ and
let $T=\begin{pmatrix}t&0\\\ 0&t^{-1}\end{pmatrix}$ with $t$ a variable whose
value in $\mathbb{C}$ is to be determined.

Define

  * •

$\rho(h_{1})=\phi(h_{1})$ for $h_{1}\in H_{1}$ and

  * •

$\rho(h_{2})=T\phi(h_{2})T^{-1}$ for $h_{2}\in H_{2}$.

Since $T$ commutes with $\phi(a)$ for any $t$ the map $\rho\colon
H\to\operatorname{PSL}(2,\mathbb{C})$ will define a representation with the
desired properties if there exists a value $t$ such that $\rho(R)$ has order
$m$.

Recall that a complex projective matrix $B$ in
$\operatorname{PSL}(2,\mathbb{C})$ will have finite order $m\geq 2$ if
$\operatorname{tr}B=\pm 2\cos(\frac{\pi}{m})$. As in the statement of the
theorem assume $R=c_{1}d_{1}\cdots c_{k}d_{k}$ with $k\geq 1$ and $c_{i}\in
G_{1}\setminus A$, $d_{i}\in G_{2}\setminus A$ for $i=1,\dots,k$, and assume
that the pairs $\\{\phi(c_{i}),\phi(a)\\}$ and $\\{\phi(d_{i}),\phi(a)\\}$ are
irreducible for $i=1,\dots,k$. Define

$f(t)=\operatorname{tr}(\phi(c_{1})T\phi(d_{1})T^{-1}\cdots\phi(c_{k})T\phi(d_{k})T^{-1}),$

$f(t)$ is a Laurent polynomial in $t$ of degree $2k$ in both $t$ and $t^{-1}$.
The coefficients of $t^{2k}$ and $t^{-2k}$ are non-zero because the pairs
$\\{\phi(c_{i}),\phi(a)\\}$ and $\\{\phi(d_{i}),\phi(a)\\}$ are irreducible.

Therefore by the Fundamental Theorem of Algebra there exists a $t_{0}$ with
$f(t_{0})=2\cos\frac{\pi}{m}.$ With this choice of $t_{0}$ we have
$\operatorname{tr}(\rho(R))=2\cos\frac{\pi}{m}$ and this $\rho(R)$ has order
$m$. Therefore $\rho$ is a representation with the desired properties. Now
assume $\phi(a)=\begin{pmatrix}1&1\\\ 0&1\end{pmatrix}$. In this case, define
$T=\begin{pmatrix}1&t\\\ 0&1\end{pmatrix}$ with $t$ again a variable. Again
$T$ commutes with $\rho(a)$ and the proof goes through analogously as above
giving the desired representation.

######  Corollary 37.

Let $G=G_{1}\Asterisk_{A}G_{2}$ be a group of F-type. Assume further that
$n\geq 4$, $2\leq p\leq n-2$ and neither $U=U(a_{1},\dots,a_{p})$ nor
$V=V(a_{p+1},\dots,a_{n})$ is a proper power in the free product on the
generators they involve.

Assume that $UV$ involves all the generators. Let $R=c_{1}d_{1}\cdots
c_{k}d_{k}$, $k\geq 1$, with $c_{i}\in\langle
a_{1},\dots,a_{p}\rangle\setminus\langle U\rangle$, $d_{i}\in\langle
a_{p+1},\dots,a_{n}\rangle\setminus\langle V\rangle$ for $i=1,\dots,k$ and let
$m\geq 2$. Then the conclusion of Theorem 36 holds for $H=G/N(R^{m})$ with

$G_{1}=\langle a_{1},\dots,a_{p}\mid
a_{1}^{e_{1}}=\dots=a_{p}^{e_{p}}=1\rangle,$ $G_{2}=\langle
a_{p+1},\dots,a_{n}\mid a_{p+1}^{e_{p+1}}=\dots=a_{n}^{e_{n}}=1\rangle,$

and

$A=\langle U^{-1}\rangle=\langle V\rangle.$

######  Proof 5.2.

By Theorem 3, the group $G$ admits a faithful representation $\phi$ in
$\operatorname{PSL}(2,\mathbb{C})$ such that the pairs
$\\{\phi(c_{i}),\phi(U)\\}$ and $\\{\phi(d_{i}),\phi(V)\\}$ are irreducible
for $i=1,\dots,k$.

######  Corollary 38.

Let $G$ be a group of F-type as in Corollary 37 and $R$ the relator as in
Corollary 37. Suppose $m\geq 8$. Then $H=G/N(R^{m})$ is virtually torsion-
free.

######  Proof 5.3.

In $G$ any element of finite order is conjugate to a power of a generator
$a_{i}$. Since $m\geq 8$ from the torsion theorem for small cancellation
products, see [21] or [8], any element of finite order in $G$ is conjugate to
a power of a generator or a power of $R$.

Since $\rho(R)$ has exact order $m$ it follows that (the constructed) $\rho$
is essentially faithful and therefore $G$ is virtually torsion-free.

######  Remark 39.

Let $G$ be a group of F-type with the presentation

$G=G_{1}\Asterisk_{A}G_{2}$

with $G_{1}\neq A\neq G_{2}$,

$G_{1}=\langle a_{1},\dots,a_{p}\mid
a_{1}^{e_{1}}=\dots=a_{p}^{e_{p}}=1\rangle,$ $G_{2}=\langle
a_{p+1},\dots,a_{n}\mid a_{p+1}^{e_{p+1}}=\dots=a_{n}^{e_{n}}=1\rangle,$

$U=U(a_{1},\dots,a_{p})$, $V=(a_{p+1},\dots,a_{n})$, and $A=\langle
U^{-1}\rangle=\langle V\rangle$. We call $G$ a special group of F-type if $UV$
involves all the generators, $n\geq 4$, $2\leq p\leq n-2$, and neither $U$ nor
$V$ is a proper power in the free product on the generators they involve.

Now let $G$ be a special group of F-type and $R\in G\setminus A$ be given in a
reduced form $R=c_{1}d_{1}\cdots c_{k}d_{k}$ with $k\geq 1$ and $c_{i}\in
G_{1}\setminus A$, $d_{i}\in G_{2}\setminus A$ for $i=1,\dots,k$. Then we may
apply the theory of one-relator quotients $H=G/N(R^{m})$ with $m\geq 2$ as
described in [17] and the deficiancy arguments in Section 2 analogously in
this more general context. If we mirror the respective proofs there, then we
easily get the following.

######  Theorem 40.

Let $G$ be a special group of F-type, $R$ the relator as in Corollary 37, and
$H=G/N(R^{m})$, $m\geq 2$. Then the following hold.

  1. 1.

For $i=1,\dots,n$ let $\alpha_{i}=0$ if $e_{i}=0$ and
$\alpha_{i}=\frac{1}{e_{i}}$, if $e_{i}\geq 2$. Then

    1. (i)

if $\sum_{i=1}^{n}\alpha_{i}+\frac{1}{m}\geq 2$, then $H$ has a subgroup of
finite index mapping homomorphically on $\mathbb{Z}$. In particular, $H$ is
infinite.

    2. (ii)

if $\sum_{i=1}^{n}\alpha_{i}+\frac{1}{m}<n-2$ then $H$ has a subgroup of
finite index mapping homomorphically onto a free group of rank 2. In
particular, $H$ is SQ-universal.

  2. 2.

$H$ is a non-trivial free product with amalgamation.

  3. 3.

If $n\geq 5$ or $n=4$ and at least one of the $e_{i}$ is not equal to 2, then
$H$ has a free subgroup of rank 2.

The details can be found in [11].

We have considered special groups $G$ of F-type. We can make similar
calculations for the hyperbolic groups of F-type which also have a faithful
representation on $\operatorname{PSL}(2,\mathbb{C})$.

If, for instance, $V=V_{1}^{q}$, $q\geq 2$, $V_{1}$ not a proper power in
$\langle a_{p+q},\dots,a_{n}\rangle$, one has to be careful with the relator
$R=c_{1}d_{1}\cdots c_{k}d_{k}$, $k\geq 1$ with $c_{i}\in\langle
a_{1},\dots,a_{p}\rangle\setminus\langle U\rangle$, $d_{i}\in\langle
a_{p+1},\dots,a_{n}\rangle\setminus\langle V\rangle$, for $i=1,\dots,k$. $G$
admits a faithful representation $\phi$ in $\operatorname{PSL}(2,\mathbb{C})$
but then a pair $\\{\phi(d_{i}),\phi(V)\\}$ is not irreducible if
$d_{i}=V_{1}^{t}$ with $q\nmid t$. But everything goes through analogously if
we consider only relators $R$ of the form as above with $d_{i}\in\langle
a_{p+1},\dots,a_{n}\rangle\setminus\langle V_{1}\rangle$.

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