# Profinite genus of fundamental groups of Sol and Nil 3-manifolds

Genildo de Jesus Nery
The author held scholarship of CAPES during the elaboration of this paper.

###### Abstract

We prove that the profinite genus of the fundamental group
$\pi_{1}(M)\cong(\mathbb{Z}\times\mathbb{Z})\rtimes_{A}\mathbb{Z}$ of a Sol
3-manifold $M$ is equal to the order of the class group of the field
$\mathbb{Q}(\lambda)$, where $\lambda$ is an eigenvalue of the matrix $A$ in
$GL_{2}(\mathbb{Z})$.

Keywords: Profinite genus; Polycyclic group; Sol 3-manifold.

##  1  Introduction

Let $Sol$ be a solvable Lie group given by
$Sol=\mathbb{R}^{2}\rtimes\mathbb{R}$ with $\mathbb{R}$ acting on
$\mathbb{R}^{2}$ via the diagonal matrix with entries $e^{t}$ and $e^{-t}$ and
$\Gamma$ a co-compact discrete subgroup of $Sol$. The compact quotient
$Sol/\Gamma$ is called Sol 3-manifold. Now, let $Nil$ be the 3-dimensional Lie
group which consists of all 3x3 real upper triangular matrices i.e. the
Heisenberg group and $\Gamma$ a co-compact discrete subgroup of $Nil$. The
compact quotient $Nil/\Gamma$ is called Nil 3-manifold (see [Sco83]).

Note that, $M$ is a Sol 3-manifold if and only if $M$ is homeomorphic to a
torus bundle $M_{A}$ over the circle $S^{1}$ , with hyperbolic monodromy given
by the matrix of Anosov $A\in GL_{2}(\mathbb{Z})$ or $M$ is homeomorphic to a
torus semi-bundle $M_{B}$, with gluing map given by $B\in GL_{2}(\mathbb{Z})$,
whose entries are all non-zero [[Zem16], Theorem 6].

The homogeneous spaces of the form $M=G/\Gamma$, where $G$ is a connected
solvable Lie group and $\Gamma$ is a co-compact subgroup of $G$ can be
represented by Mostow fibration [[JKM08], Theorem 2.4]. In particular, the
corresponding Mostow fibration for

  1. (i)

Sol 3-manifold $M$ has the form $\mathbb{T}^{2}\hookrightarrow M\rightarrow
S^{1}$.

  2. (ii)

Nil 3-manifold and some Euclidean manifolds $M$ has the form
$S^{1}\hookrightarrow M\rightarrow\mathbb{T}^{2}$.

where $S^{1}$ is the unit circle and $\mathbb{T}^{2}$ is 2-dimensional torus,
which leads to the corresponding exact sequences of the fundamental groups,
respectively (see [JKM08] , p. 926)

$\displaystyle
0\rightarrow\mathbb{Z}\times\mathbb{Z}\rightarrow\pi_{1}(M)\rightarrow\mathbb{Z}\rightarrow
0\ \ \text{or}\ \
0\rightarrow\mathbb{Z}\rightarrow\pi_{1}(M)\rightarrow\mathbb{Z}\times\mathbb{Z}\rightarrow
0.$

So, the fundamental group of the Sol 3-manifolds are
$(\mathbb{Z}\times\mathbb{Z})\rtimes_{A}\mathbb{Z}$, where $A\in
GL_{2}(\mathbb{Z})$ [[JKM08], Lemma 2.6]. It is well known that a compact Sol
3-manifold is determined by its fundamental group uniquely up to
diffeomorphism [[Mos54], Theorem A].

One can describe the geometry of the torus bundles through the associated
matrix. We denote the trace of a matrix $A$ by $tr(A)$ and its determinant by
$det(A)$.

  * •

If $|tr(A)|>2$, then $A$ is hyperbolic (i.e, neither of its eigenvalues has
absolute value 1) and the torus bundle $M_{A}$ admits Sol geometry [[Sco83],
Theorem 5.5 (i)].

  * •

If $|tr(A)|\leq 2$ then the torus bundle $M_{A}$ admits Nil geometry (Sol
geometry) if $A$ has infinite order and isn’t hyperbolic ($A$ is a hyperbolic
matrix) [[Sco83], Theorem 5.5 (iii)].

Recently, the following question mentioned in [WZ18] attracts attention of
many specialists in geometric group theory:

Question A: Let $M$ be a compact, orientable 3-manifold. To what extent is
$\pi_{1}(M)$ determined by its profinite completion?

H. Wilton and P. Zalesskii in [WZ17] showed that the geometry of the manifold
is determined by the profinite completion of its fundamental group. In [WZ18],
they also proved that the profinite completion of the fundamental group of a
3-manifold $M$ determines the Jaco-Shalen-Johannson decomposition of $M$. In
the same work, these authors pointed out the next step in addressing the above
question based on their results and the work of Wilkes [Wil18] who presented a
complete answer to the question in relation to graph manifolds, not including
the Sol 3-manifolds. They highlight that in light of [[WZ18], Theorem B], the
next step in approaching the issue is to consider the parts of the Jaco-
Shalen-Johannson decomposition and point out that a definitive treatment of
the case of Sol 3-manifolds would be a valuable addition to the literature.
This paper is dedicated to it.

Given a group $G$ we denote by $\widehat{G}$ its profinite completion. Let
$\mathfrak{C}$ be a family of groups and $G\in\mathfrak{C}$. Following [GZ11]
we define the $\mathfrak{C}$-genus of $G$ as

$\displaystyle\mathfrak{g}(\mathfrak{C},G):=\text{IsoClasses}(\\{H\in\mathfrak{C}\
|\ \widehat{H}\cong\widehat{G}\\}).$

i.e. the $\mathfrak{C}$-genus is the set of isomorphism classes of groups
belonging to $\mathfrak{C}$ which has the same profinite completion as the
fixed group $G$. Since the fundamental groups of the Sol 3-manifolds are
polycyclic, in this paper we will restrict to the family $\mathcal{FP}$ of
polycyclic groups. So we calculate the $\mathfrak{g}(\mathcal{FP},G_{A})$ when
the fixed group $G_{A}=(\mathbb{Z}\times\mathbb{Z})\rtimes_{A}\mathbb{Z}$
where $A\in GL_{2}(\mathbb{Z})$.

In [AFW15], page 138, the following conjecture was proposed:

Conjecture: The 3-manifold genus of a fundamental group of a compact
3-manifold is finite.

F. J. Grunewald, P. F. Pickel and D. Segal in [GPS80] proved that the
$\mathcal{FP}$-genus of a polycyclic-by-finite group is finite. Since the
fundamental group of Sol 3-manifold is a polycyclic group, it follows that the
conjecture is true for Sol 3-manifolds. Louis Funar in [Fun12] proved that for
torus bundle $M_{A}$ the $\mathcal{FP}$-genus
$\mathfrak{g}(\mathcal{FP},\pi_{1}(M_{A}))$ is bounded by the order of the
class group of the field $\mathbb{Q}(\lambda)$, where $\lambda$ is an
eigenvalue of the matrix $A$. We show that this number is exactly the
$\mathcal{FP}$-genus $\mathfrak{g}(\mathcal{FP},\pi_{1}(M_{A}))$.

Our main result is the following:

######  Theorem 1.1.

Let $A$ be a matrix in $GL_{2}(\mathbb{Z})$ and consider the group
$G_{A}=(\mathbb{Z}\times\mathbb{Z})\rtimes_{A}\mathbb{Z}$. Then
$\mathfrak{g}(\mathcal{FP},G_{A})$ is equal to the order of the class group of
the field $\mathbb{Q}(\lambda)$, where $\lambda$ is an eigenvalue of the
matrix $A$.

######  Corollary 1.2.

Let $A$ be a matrix in $GL_{2}(\mathbb{Z})$ and consider the group
$G_{A}=(\mathbb{Z}\times\mathbb{Z})\rtimes_{A}\mathbb{Z}$. If $tr(A)=0$ or $A$
has equal eigenvalues then $\mathfrak{g}(\mathcal{FP},G_{A})=1$.

L. Funar also proved that torus bundles are not distinguished by the profinite
completions of its fundamental groups (see [Fun12]). It follows from Theorem
1.1 that the torus bundles $M_{A}$ that are determined by the profinite
completions of its fundamental groups are exactly for which
$\mathfrak{g}(\mathcal{FP},\pi_{1}(M_{A}))=1$. Note that, in 1801 Gauss
conjectured that there are infinitely many real quadratic fields with class
number one and this conjecture is still open (see [Gau66]).

We denote by $h(K)$ the order of the class group of the field $K$ and set
$K_{A}=\mathbb{Q}(\lambda)$, where $\lambda$ is an eigenvalue of the matrix
$A$.

######  Corollary 1.3.

The Torus bundle $M_{A}$ is determined by the profinite completion of its
fundamental group in the following cases:

  1. (i)

If $M_{A}$ is a Nil 3-manifold.

  2. (ii)

When $M_{A}$ is a Sol 3-manifold, if $h(K_{A})=1$.

Let us now analyze the 3-dimensional manifolds $M$ of the type
$S^{1}\hookrightarrow M\rightarrow\mathbb{T}^{2}$ . Consider the homomorphism
of groups $\sigma:\mathbb{Z}\times\mathbb{Z}\longrightarrow
Aut(\mathbb{Z})\cong\mathbb{Z}/2\mathbb{Z}$ . If the action of $\sigma$ is
trivial, then $M$ is a Heisenberg Nil 3-manifold [[JKM08], Theorem 2.11], i.e.
$M=H/\Gamma$ where

$H=\left\\{\left(\begin{array}[]{c c c}1&x&z\\\ 0&1&y\\\ 0&0&1\\\
\end{array}\right)\ ;\ x,y,z\in\mathbb{R}\right\\}$

and

$\Gamma=\left\\{\left(\begin{array}[]{c c c}1&x&z\\\ 0&1&y\\\ 0&0&1\\\
\end{array}\right)\ ;\ x,y,z\in\mathbb{Z}\right\\}$

and its fundamental group is $\Gamma$ (see [Apa00], p. 29). Denoting each
matrix of $\Gamma$ by $(x,y,z)$ we can define a map
$\varphi:\Gamma\longrightarrow(\mathbb{Z}\times\mathbb{Z})\rtimes_{A(x)}\mathbb{Z}$
by $\varphi(x,y,z)=((y,z),x)$ with the action

$A(x)=\left(\begin{array}[]{c c}1&0\\\ x&1\\\ \end{array}\right),\ \
x\in\mathbb{Z}.$

It is easy to see that $\varphi$ is a group isomorphism. Therefore, since the
eigenvalues of the matrix $A(x)$ are all equal, it follows from Corollary 1.2
that the Heisenberg Nil 3-manifolds are determined by the profinite completion
of their fundamental groups.

However, if the action of $\sigma$ is not trivial, then the fundamental group
of $M$ is one of the following groups [[JKM08], Theorem 2.11],

$\pi_{0}=\langle a,b,c\ |\ ac=ca;bc=c^{-1}b;[a,b]=1\rangle\cong\langle
c\rangle\rtimes(\langle a\rangle\times\langle b\rangle)$

or

$\pi_{1}=\langle a,b,c\ |\ ac=ca;bc=c^{-1}b;[a,b]=c\rangle.$

Note that $\mathbb{Z}^{3}\cong\langle c\rangle\times\langle
a\rangle\times\langle b^{2}\rangle$ is a subgroup of index 2 in $\pi_{0}$ and
that $\mathbb{Z}^{3}\cong\langle a,c,b^{2}\rangle$ is a subgroup of index 2 in
$\pi_{1}$. Therefore, $\pi_{0}$ and $\pi_{1}$ are fundamental groups of
Euclidean manifolds (see [AFW15], p. 19). We shall postpone the detailed study
of Euclidean manifolds for the next paper.

##  2 Preliminary Results

The results of this section are based on information extracted from [GZ11].

Let $A$ and $B$ be matrices in $GL_{2}(\mathbb{Z})$. The following result
provides a necessary and sufficient condition for the groups
$(\mathbb{Z}\times\mathbb{Z})\rtimes_{A}\mathbb{Z}$ and
$(\mathbb{Z}\times\mathbb{Z})\rtimes_{B}\mathbb{Z}$ to be isomorphic.

######  Lemma 2.1 ([GZ11], p. 135).

Let $A$ and $B$ in $GL_{2}(\mathbb{Z})$ and consider the corresponding semi-
direct products $G_{A}=(\mathbb{Z}\times\mathbb{Z})\rtimes_{A}\mathbb{Z}$ and
$G_{B}=(\mathbb{Z}\times\mathbb{Z})\rtimes_{B}\mathbb{Z}$. Then $G_{A}$ and
$G_{B}$ are isomorphic if and only if $A$ and $B$ are conjugate matrices in
$GL_{2}(\mathbb{Z})$.

An analogous version of Lemma 2.1 holds for profinite groups with the same
proof.

######  Lemma 2.2.

Let $A$ and $B$ in $GL_{2}(\widehat{\mathbb{Z}})$ and consider the
corresponding semi-direct products
$\widehat{G}_{A}=\widehat{(\mathbb{Z}\times\mathbb{Z})}\rtimes_{A}\widehat{\mathbb{Z}}$
and
$\widehat{G}_{B}=\widehat{(\mathbb{Z}\times\mathbb{Z})}\rtimes_{B}\widehat{\mathbb{Z}}$.
Then $\widehat{G}_{A}$ and $\widehat{G}_{B}$ are isomorphic if and only if $A$
and $B$ are conjugate matrices in $GL_{2}(\widehat{\mathbb{Z}})$.

######  Lemma 2.3 ([DK18], p. 497).

If a matrix $A$ in $GL_{n}(\mathbb{Z})$ has all eigenvalues equal to 1 then
the group $G_{A}=\mathbb{Z}^{n}\rtimes_{A}\mathbb{Z}$ is nilpotent.

We write $\mathcal{I}_{2}$ for representing the class of finitely generated
torsion-free nilpotent groups of nilpotency class 2 and $\mathfrak{F}(G)$ for
denoting the set of finite quotients of a group $G$. In relation to the
isomorphism problem F. Grunewald and R. Scharlau proved the following result.

######  Proposition 2.4 ([GS79], p. 163).

Let $G_{1},G_{2}$ in $\mathcal{I}_{2}$, such that the Hirsch number of $G_{1}$
is smaller than or equal to 5 and $\mathfrak{F}(G_{1})=\mathfrak{F}(G_{2})$.
Then $G_{1}\cong G_{2}$.

######  Proposition 2.5 ([FPDR82], p. 227).

Let $G$ and $H$ be finitely generated groups. Then $\widehat{G}$ and
$\widehat{H}$ are isomorphic as topological groups if and only if
$\mathfrak{F}(G)=\mathfrak{F}(H)$.

Proposition 2.4 then can be restated as follows.

######  Proposition 2.6.

Let $G_{1},G_{2}$ in $\mathcal{I}_{2}$, such that the Hirsch number of $G_{1}$
is smaller than or equal to 5 and $\widehat{G_{1}}\cong\widehat{G_{2}}$. Then
$G_{1}\cong G_{2}$.

######  Proposition 2.7 (Latimer-MacDffee’s Theorem, see [LM33], p. 315).

Let $f$ be a monic, irreducible polynomial of degree $n$. Then there is a one-
to-one correspondence between $\mathbb{Z}$-similarity classes of $n\times n$
matrices with characteristic polynomial $f$ and the ideal classes in the order

$\frac{\mathbb{Z}[x]}{(f(x))}.$

##  3 Profinite Genus

Let $K$ be an algebraic number field of degree $n$. We denote by
$\mathcal{O}_{K}$ the ring of integers of $K$, $I(K)$ the set of all nonzero
integral and fractional ideals of $\mathcal{O}_{K}$ and by $P(K)$ the set of
all the principal ideals of $\mathcal{O}_{K}$. Recall that $I(K)$ is an
abelian group with the operation of multiplication of ideals, $P(K)$ is a
subgroup of $I(K)$ and the quotient group $I(K)/P(K)$ has finite order. The
group $I(K)/P(K)$ is called the ideal class group. We denote this group by
$H(K)$ and its order by $h(K)$.

######  Lemma 3.1.

Let $A$ be a matrix in $GL_{2}(\mathbb{Z})$. If $tr(A)\neq 0$ and $A$ has
distinct eigenvalues $\lambda_{1}$ and $\lambda_{2}$ then
$\mathbb{Q}(\lambda_{1},\lambda_{2})$ is a quadratic field.

###### Proof.

Let $A$ in $GL_{2}(\mathbb{Z})$. It is easy to see that the characteristic
polynomial of $A$ has the form $p(\lambda)=\lambda^{2}-tr(A)\lambda+det(A)$
and that $\lambda_{\pm}=(tr(A)\pm\sqrt{tr(A)^{2}-4det(A)})/2$ are the
eigenvalues. It follows directly from the hypotheses of the Lemma that the
possible values for the trace and the determinant of the matrix $A$ are:

  1. (i)

$tr(A)=\pm 1$ and $det(A)=\pm 1$ or

  2. (ii)

$tr(A)=\pm 2$ and $det(A)=-1$ or

  3. (iii)

$|tr(A)|>2$ and $det(A)=\pm 1$.

The cases (i) and (ii) are trivial. If $A$ satisfies (iii), then the
eigenvalues of $A$ are irrational. Indeed, suppose that there is an integer
$k$ such that $tr(A)^{2}-4det(A)=k^{2}$. So, $tr(A)^{2}-k^{2}=\pm 4$, which
implies $k=tr(A)\pm 1$ and so $2(\pm tr(A))-1=\pm 4$, a contradiction.
Therefore, we conclude that $K=\mathbb{Q}(\lambda_{+},\lambda_{-})$ is a
quadratic field, i.e., there is a square-free integer $d$ such that
$K=\mathbb{Q}(\sqrt{d})$. ∎

######  Lemma 3.2.

Let $\mathcal{O}_{K}$ be the ring of integers of a field
$K=\mathbb{Q}(\lambda_{1},\lambda_{2})$, where $\lambda_{1}$ and $\lambda_{2}$
are as in Lemma 3.1. Then $\mathcal{O}_{K}$ has a unit $u\neq\pm 1$.

###### Proof.

We have $\mathcal{O}_{K}=\mathbb{Z}+\mathbb{Z}\sqrt{d}$ if $d\not\equiv 1(mod\
4)$ or $\mathcal{O}_{K}=\mathbb{Z}+\mathbb{Z}(\frac{1+\sqrt{d}}{2})$ if
$d\equiv 1(mod\ 4)$ ([AW04], Theorem 5.4.2). Hence, $\lambda_{1},\
\lambda_{2}\in\mathcal{O}_{K}$. Note that $\lambda_{2}$ is the conjugate of
$\lambda_{1}$ and $|\lambda_{1}\lambda_{2}|=1$ because $\lambda_{1}$ and
$\lambda_{2}$ are eigenvalues of a matrix $A$ in $GL_{2}(\mathbb{Z})$.
Therefore, $\lambda_{1}$ is a unit of $\mathcal{O}_{K}$ and
$\lambda_{1}\neq\pm 1$. ∎

The next three Lemmas are extracted from the proof of Proposition 1 in
[Seg83], page 256.

######  Lemma 3.3.

Let $K$ be a quadratic field. If $\mathcal{O}_{K}$ has a unit $u\neq\pm 1$
then the subring $S=\mathbb{Z}[u,u^{-1}]$ has finite index in
$\mathcal{O}_{K}$.

###### Proof.

We know that $u=\alpha+\beta\sqrt{d}$ where $\alpha,\beta,d\in\mathbb{Z}$ with
$\beta\neq 0$ and $d$ square-free. Note that an element of $S$ is of the form
$r+s\beta\sqrt{d}$, where $r,s\in\mathbb{Z}$. Suppose $x=a+b\sqrt{d}$ and
$y=e+f\sqrt{d}$ are elements of $\mathcal{O}_{K}$, $a,b,e,f\in\mathbb{Z}$,
congruent modulo $S$, i.e. $x-y\in S$. Then, $x-y$ can be written in the form
$r+s\beta\sqrt{d}$, which gives $a-e=r$ and $b-f=s\beta$. Since each integer
is in $S$, it is sufficient to analyze the condition $b-f=s\beta$, which is
equivalent to: $b$ is congruent to $f$ modulo $\beta$. This proves that the
index of $S$ in $\mathcal{O}_{K}$ is finite. ∎

######  Lemma 3.4.

Let $K$ be a quadratic field and $S=\mathbb{Z}[u,u^{-1}]$, where $u\neq\pm 1$
is a unit of $\mathcal{O}_{K}$. Then any two ideals in distinct classes of the
group $H(K)$ are not isomorphic as $S$-modules.

###### Proof.

Let $I$ and $J$ be ideals in distinct classes of the group $H(K)$. Suppose
that $\varphi:I\longrightarrow J$ is a $S$-module isomorphism. By Lemma 3.3
the index of $S$ in $\mathcal{O}_{K}$ is finite, so for every
$b\in\mathcal{O}_{K}$ there is a natural number $t$ such that $tb\in S$. Note
that for every $a\in I$ we have to $t\varphi(ab)=t\varphi(a)b$. Since
$\mathcal{O}_{K}$ is an integral domain then $\varphi(ab)=\varphi(a)b$. Then
$\varphi:I\longrightarrow J$ is an $\mathcal{O}_{k}$-module isomorphism. Since
$I$ is a nonzero ideal of $\mathcal{O}_{K}$, there is $\alpha\in I$ with
$\alpha\neq 0$. So we have that
$I\cong\varphi(I)=\varphi(I\alpha)=\varphi(I)\alpha=J\alpha$. We have a
contradiction, because $I$ and $J$ are taken in distinct classes of $H(K)$. ∎

Fix a matrix $A$ in $GL_{2}(\mathbb{Z})$, we call the set of matrices $B$ in
$GL_{2}(\mathbb{Z})$ such that $B$ is conjugate to $A$ in
$GL_{2}(\widehat{\mathbb{Z}})$ of $\approx_{A}$-class.

######  Lemma 3.5.

Let $A$ be a matrix in $GL_{2}(\mathbb{Z})$. If $tr(A)\neq 0$ and $A$ has
distinct eigenvalues $\lambda_{1}$ and $\lambda_{2}$, then there are at least
$h(\mathbb{Q}(\lambda_{1},\lambda_{2}))$ matrices in $\approx_{A}$-class that
are not conjugate to each other.

###### Proof.

Let $I_{1}=\mathcal{O}_{K},\cdots,I_{r}$ be ideals in distinct classes of the
group $H(K)$. By Lemma 3.2 there is a unit $u$ of $\mathcal{O}_{K}$ such that
$u\neq\pm 1$. The subring $S=\mathbb{Z}[u,u^{-1}]$ of $\mathcal{O}_{K}$ has
finite index in $\mathcal{O}_{K}$ (Lemma 3.3). For $i\neq j,\
i,j\in\\{1,\cdots,r\\}$, the ideals $I_{i},\cdots,I_{j}$ are non-isomorphic as
$S$-modules (Lemma 3.4). On the other hand, for each positive integer $m$ and
each $i\in\\{1,\cdots,r\\}$ there is an
$\mathcal{O}_{K}/m\mathcal{O}_{K}$-module isomorphism

$\displaystyle\frac{I_{i}}{mI_{i}}\cong\frac{\mathcal{O}_{K}}{m\mathcal{O}_{K}}.$
(1)

This is because $I_{i}/mI_{i}$ is generated by one element and the order of
$I_{i}/mI_{i}$ and $\mathcal{O}_{K}/m\mathcal{O}_{K}$ are equal to $m^{2}$.

Now, for every $i\in\\{1,\cdots,r\\}$, choose a $\mathbb{Z}$-basis for the
ideal $I_{i}$ and denote by $A_{i}$ the multiplication matrix by $u$ with
respect to this basis. By what has already been proven in the previous
paragraph, we have that $A_{i}$ is not conjugated to $A_{j}$ in
$GL_{2}(\mathbb{Z})$ for $i\neq j,\ i,j\in\\{1,\cdots,r\\}$, because otherwise
the ideas $I_{i}$ and $I_{j}$ would be isomorphic as $S$-modules. Moreover,
(1) implies that $A_{i}$ is conjugated to $A_{1}$ in
$GL_{2}(\widehat{\mathbb{Z}})$ for all $i\in\\{1,\cdots,r\\}$. ∎

Now we are ready to prove Theorem 1.1.
Proof of Theorem 1.1. Let $A$ and $B$ in $GL_{2}(\mathbb{Z})$ and let $\zeta$
be an eigenvalue of $A$. Consider the semi-direct products
$G_{A}=(\mathbb{Z}\times\mathbb{Z})\rtimes_{A}\mathbb{Z}$ and
$G_{B}=(\mathbb{Z}\times\mathbb{Z})\rtimes_{B}\mathbb{Z}$ and suppose that

$\displaystyle\widehat{G}_{B}\cong\widehat{G}_{A}.$ (2)

Suppose initially that $tr(A)\neq 0$ and that $A$ has distinct eigenvalues
$\lambda_{1}$ and $\lambda_{2}$. Since
$\widehat{G}_{B}\cong\widehat{(\mathbb{Z}\times\mathbb{Z})}\rtimes_{B}\widehat{\mathbb{Z}}$
and
$\widehat{G}_{A}\cong\widehat{(\mathbb{Z}\times\mathbb{Z})}\rtimes_{A}\widehat{\mathbb{Z}}$,
it follows from Lemma 2.2 that the matrices $B$ and $A$ are conjugate in
$GL_{2}(\widehat{\mathbb{Z}})$, so $B$ is in $\approx_{A}$-class. By Lemma 3.5
there are at least $h(\mathbb{Q}(\zeta))$ matrices in $\approx_{A}$-class that
are not conjugate to each other in $GL_{2}(\mathbb{Z})$ but are conjugate in
$GL_{2}(\widehat{\mathbb{Z}})$, pairwise. By Lemma 2.1 we conclude that
$\mathfrak{g}(\mathcal{FP},G_{A})\geq h(\mathbb{Q}(\zeta))$. On the other
hand, since the matrices $B$ and $A$ are conjugate in
$GL_{2}(\widehat{\mathbb{Z}})$, then for each positive integer $m$, $tr(B)$ is
congruent to $tr(A)$ modulo $m$ and $det(B)$ is congruent to $det(A)$ modulo
$m$. So we can conclude that $tr(B)=tr(A)$ and $det(B)=det(A)$ and therefore,
the characteristic polynomials
$p_{B}(\lambda)=\lambda^{2}-tr(B)\lambda+det(B)$ and
$p_{A}(\lambda)=\lambda^{2}-tr(A)\lambda+det(A)$ of $B$ and $A$, respectively,
are equal. By Proposition 2.7 there is a bijection between the conjugacy
classes of matrices in $GL_{2}(\mathbb{Z})$ that has characteristic polynomial
$p_{A}(\lambda)$ and the elements of the class group of the ring

$\displaystyle\frac{\mathbb{Z}[\lambda]}{(p_{A}(\lambda))}\cong\mathcal{O}_{K}.$

In other words, there are $h(K)$ conjugacy classes of matrices in
$GL_{2}(\mathbb{Z})$ with the same characteristic polynomial $p_{A}(\lambda)$.
Denote by $\mathcal{C}=\\{[A_{1}],\cdots,[A_{h(K)}]\\}$ these conjugacy
classes. It follows from Lemma 2.1 that

$(\mathbb{Z}\times\mathbb{Z})\rtimes_{A_{i}}\mathbb{Z}\ncong(\mathbb{Z}\times\mathbb{Z})\rtimes_{A_{j}}\mathbb{Z},$

for $i,\ j$ distinct in $\\{1,\cdots,h(K)\\}$. This shows that
$\mathfrak{g}(\mathcal{FP},G_{A})\leq h(K)$. Therefore,
$\mathfrak{g}(\mathcal{FP},G_{A})=h(K)$ in this case.

Suppose now that all eigenvalues of the matrix $A$ are equal to 1. Then
$G_{A}$ is a finitely generated torsion-free nilpotent group of nilpotency
class 2 with Hirsch number 3 and so by Proposition 2.6
$\mathfrak{g}(\mathcal{FP},G_{A})=1$. Note that in this case, $K=\mathbb{Q}$
and $\mathcal{O}_{K}=\mathbb{Z}$. Since $\mathbb{Z}$ is a principal ideal
domain, then $h(K)=1$ ([AW04], Theorem 12.1.1). Therefore,
$\mathfrak{g}(\mathcal{FP},G_{A})=h(K)$.

Now suppose that the matrix $A$ has all its eigenvalues equal to $-1$. Note
that $tr(A)=-2$ and $det(A)=1$. Assume that (2) holds. It is easy to see that
the groups $G_{B^{2}}=(\mathbb{Z}\times\mathbb{Z})\rtimes_{B^{2}}\mathbb{Z}$
and $G_{A^{2}}=(\mathbb{Z}\times\mathbb{Z})\rtimes_{A^{2}}\mathbb{Z}$ are
subgroup in $G_{B}$ and $G_{A}$, respectively. Since the matrices $B$ and $A$
are conjugate in $GL_{2}(\widehat{\mathbb{Z}})$, so are $B^{2}$ and $A^{2}$.
Then $\widehat{G}_{B^{2}}\cong\widehat{G}_{A^{2}}$ by Lemma 2.2. Note that
$A^{2}$ has all its eigenvalues equal to 1, hence $G_{B^{2}}\cong G_{A^{2}}$,
by what already been proved in the preceding paragraph, so $B^{2}$ and $A^{2}$
are conjugate in $GL_{2}(\mathbb{Z})$ (Lemma 2.1). By $(\ref{G_N iso G_M})$ we
have that $B$ and $A$ are conjugate in $GL_{2}(\widehat{\mathbb{Z}})$. This
implies that for each positive integer $m$, the trace of $B$ is congruent to
$-2$ modulo $m$ and the determinant of $B$ is congruent to 1 modulo $m$. So we
can conclude that $tr(B)=tr(A)$ and $det(B)=det(A)$. Then $B$ and $A$ are
conjugate in $GL_{2}(\mathbb{Z})$ [[CT91], Lemma 3.2]. By Lemma 2.1,
$G_{B}\cong G_{A}$, therefore $\mathfrak{g}(\mathcal{FP},G_{A})=1$. Therefore,
as before, we have that $\mathfrak{g}(\mathcal{FP},G_{A})=h(K)$.

Finally, suppose that $tr(A)=0$. Then by (2) $det(B)=det(A)$ and
$tr(B)=tr(A)$. So by [[Hen97], Theorem 3], we have

  * •

if $det(A)=1$ then $A$ is conjugate over $GL_{2}(\mathbb{Z})$ to

$\left(\begin{array}[]{c c}0&-1\\\ 1&0\\\ \end{array}\right).$

  * •

if $det(A)=-1$ the $A$ is conjugate over $GL_{2}(\mathbb{Z})$ to

$\left(\begin{array}[]{c c}0&1\\\ 1&0\\\ \end{array}\right).$

Therefore $\mathfrak{g}(\mathcal{FP},G_{A})=1$. It is easy to see that the
characteristic polynomial of $A$ is $p_{A}(\lambda)=\lambda^{2}-1$, if
$det(A)=-1$, or $p_{A}(\lambda)=\lambda^{2}+1$, if $det(A)=1$. If
$p_{A}(\lambda)=\lambda^{2}-1$, we have that the eigenvalues of $A$ are 1 and
$-1$, so $h(K)=1$. If $p_{A}(\lambda)=\lambda^{2}+1$, the eigenvalues of $A$
are $i$ and $-i$. So, $K=\mathbb{Q}(\sqrt{-1})$ and then $h(K)=1$ (see [AW04],
p. 325). Therefore $\mathfrak{g}(\mathcal{FP},G_{A})=h(K)$.

$\square$

The proof of Corollary 1.2 follows from the third, fourth and fifth paragraphs
of the proof of Theorem 1.1.

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Genildo de Jesus Nery, Department of Mathematics,

University of Brasilia, Brasilia, Brazil

E-mail address<EMAIL_ADDRESS>