here it's a question that I've posted in MSE but unfortunately got no answers:

Let $A$ and $B$ be matrices of finite order with integer coefficients.

Let $n\in\mathbb{N}$ and let $G_A=\mathbb{Z}\ltimes_A \mathbb{Z}^n$ be the semidirect product, where the action is $\varphi(n)\cdot (m_1,\ldots,m_n)=A^n (m_1,\ldots,m_n)$, and similarly with $B$.

It is easy to construct an isomorphism between $G_A$ and $G_B$ if $A$ is conjugate in $\mathrm{GL}(n,\mathbb{Z})$ to $B$ or $B^{-1}$.

But, this is also a necessary condition? I mean, does $G_A\cong G_B$ implies $A\cong B$ or $A\cong B^{-1}$ in $\mathrm{GL}(n,\mathbb{Z})$ or is there a counterexample?

I've seen at this MSE question that it is true if $A$ and $B$ are hyperbolic, i.e none of their eigenvalues have module 1, but it isn't the case.

Thank you very much!

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    $\begingroup$ I think it is a necessary condition, but I haven't had time to write down a proof. If the isomorphism $\phi:G_A \to G_B$ maps the normal subgroup $Z^n$ in the semidirect product to the $Z^n$ in $G_B$ then $\phi$ on $Z^n$ is essentially the required conjugating element of ${\rm GL}(n,Z)$. Otherwise $\phi(Z^n)$ must be a normal subgroup that intersects the $Z^n$ in $G_B$ in a subgroup $Z^{n-1}$. But then that subgroup would be centralized by $B$ and since $B$ has finite order, that would force $B=I_n$. Similarly $A=I_n$, so they are conjugate as required. $\endgroup$ – Derek Holt Jan 9 at 20:09
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    $\begingroup$ Did you try the simplest possible example: $n = 2$, $A = {\rm diag} (1,-1)$ and $B$ the other conjugacy class for a matrix of order two and size two (the two by two permutation matrix)? The fact that one of them is diagonalizable and the other isn't, probably allows the semidirect products to be distinguished. $\endgroup$ – David Handelman Jan 9 at 20:52
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    $\begingroup$ @DavidHandelman I've checked this example: the abelianizations are then not isomorphic! $\endgroup$ – YCor Jan 9 at 21:09
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    $\begingroup$ It is known to be a necessary and sufficient condition when $n = 2$ (elementary). $\endgroup$ – Luc Guyot Jan 9 at 21:27
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    $\begingroup$ Another easy case is when $1$ is not an eigenvalue of $A$ (which means that the abelianization of $G_A$ has rank 1) — not assuming $A$ has finite order by the way. In this case the normal copy of $\mathbf{Z}^m$ in $G_A$ is characterized as the inverse image of the torsion from the abelianization. $\endgroup$ – YCor Jan 9 at 23:00

I believe now that David Speyer's example can be adapted to provide a counterexample to the original question. (So I retract my earlier comment on the question and will delete it soon.)

In David's example, $A$ is a degree $\phi(m)$ matrix of order $m$ defining the action by multiplication of $\zeta_m$ on the ideal $I$ of the number field ${\mathbb Q}[\zeta_m]$, and $B$ is the action on the ideal $\sigma(I)$, and $A$ and $B$ are not conjugate to each other or to their inverses in ${\rm GL}_{\phi(m)}({\mathbb Z})$. A specific example is $m=37$, $\phi(m)=36$.

We define degree $n:=\phi(m)+1$ matrices $A'$ and $B'$ as the diagonal joins of $A$ and $B$ with the identity matrix $I_1$. So the corresponding ${\mathbb Z}$-modules can be thought of as $I \oplus \langle y \rangle$ and $\sigma(I) \oplus \langle z \rangle$, with trivial action on the second factors. These modules cannot be isomorphic, because an isomorphism would have to map the fixed points submodule $\langle y \rangle$ onto $\langle z \rangle$ and then their quotients $I$ and $\sigma(I)$ would be isomorphic, which they are not. So $A'$ and $B'$ are not conjugate in ${\rm GL}_{n}({\mathbb Z})$.

I claim (at least in some cases) that we can choose $A$ and $B$ such that the corresponding semidirect products $\langle \alpha \rangle \ltimes_{A'} {\mathbb Z}^n$ and $\langle \beta \rangle \ltimes_{B'} {\mathbb Z}^n$ are isomorphic, where $\alpha$ and $\beta$ generate infinite cyclic groups. We can (in some cases?) choose $A = B^a$ with $a$ coprime to $m$ and $2 \le a < \phi(m)-1$ such that $B$ is not conjugate in ${\rm GL}_{\phi(m)}({\mathbb Z})$ to $A$ or to $A^{-1}$, and choose integers $r,s$ with $ra-sm=1$.

Then we can define a isomorphism from $\langle \alpha \rangle \ltimes_{A'} {\mathbb Z}^n$ to $\langle \beta \rangle \ltimes_{B'} {\mathbb Z}^n$ by mapping $I$ to $\sigma(I)$ as in David's example, $y$ to $\beta^m z^r$ and $\alpha$ to $\beta^a z^s$. Note that this induces an isomorphism from the free abelian group $\langle \alpha, y \rangle$ to $\langle \beta, z \rangle$, such that the image of $y$ centralizes $\sigma(I)$.

I did some calculations in Magma in the case $m=37$, and found a degree 36 integer matrix $A$ that is not conjugate to $A^a$ for any $a$ with $2 \le a \le 36$.

  • $\begingroup$ Excellent answer, I really apprecciate your efforts and all of the commentators and other answers, it was a very interesting interchange of ideas. I met the question studying lattices in almost abelian groups a year ago (in my Diploma thesis), and unluckly I don't have the full knowledge to understand every line on the answer. But I will sit and try to do it. Again, thanks! $\endgroup$ – Ale Tolcachier Jan 13 at 23:52

$\newcommand{\IZ}{\mathbb{Z}}$ One can easily verify that $G_A' = \{0\}\times \operatorname{im}(A-1_{m\times m})$. Moreover $G_A$ acts on $G_A'$ by conjugation. The elements of $\IZ^m$ act trivially and the extra $\IZ$ acts by multiplication with $A$. The normal subgroup $K_A:=\operatorname{ord}(A)\IZ \times \IZ^m$ is the kernel of this action, i.e. the subgroup of all elements that act trivially on $G_A'$.

Therefore any isomorphism $G_A \to G_B$ must map $K_A$ to $K_B$. In particular $ord(A)=|G_A/K_A| = |G_B/K_B|=\operatorname{ord}(B)$, let's call that $n$, and $G_A/K_A \cong G_B/K_B \cong \IZ/n\IZ$.

Now consider the conjugation action of $G_A$ on $K_A$ instead of $G_A'$. Since $K_A$ is abelian, this is really an action of $G_A/K_A\cong \IZ/n\IZ$ on $K_A\cong \IZ \times\IZ^m$ given by multiplication with the block matrix $A':=\begin{pmatrix}1&\\&A\end{pmatrix}$.

By considering the induced action on $K_A \otimes \mathbb{Q}$, we find that the two $\mathbb{Q}[\IZ/n]$-modules $K_A \otimes \mathbb{Q}$ and $K_B\otimes \mathbb{Q}$ must be isomorphic. That means that $A'$ and $B'$ are $\mathrm{GL}_{1+m}(\mathbb{Q})$-conjugated at the very least. I'm not sure how one would go from there.

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    $\begingroup$ Since $A'$ and $B'$ have the same canonical Jordan form, the matrices $A$ and $B$ have also the same canonical Jordan form. Therefore $A$ and $B$ are $GL_m(\mathbb{C})$-conjugated and hence $GL_m(\mathbb{Q})$-conjugated. $\endgroup$ – Luc Guyot Jan 10 at 12:29

$\def\QQ{\mathbb{Q}}\def\ZZ{\mathbb{Z}}$I misread the question as asking about $C_m \ltimes_A \ZZ^n$ and $C_m \ltimes_B \ZZ^n$, where $m$ is the order of $A$ and $B$. If we work with $\ZZ \ltimes_A \ZZ^n$ and $\ZZ \ltimes_B \ZZ^n$, I'm not sure what happens.

Working with $C_m \ltimes_A \ZZ^n$, this is not true. Let $m$ be the order of $A$ and $B$, let $\zeta_m$ be a primitive $m$-th root of unity, let $K$ be the cylotomic field $\QQ(\zeta_m)$. Let $G$ be the Galois group of $K$ over $\QQ$, so $G \cong (\ZZ/m \ZZ)^{\times}$. Let $H$ be the class group of $K$. Suppose that $H$ contains a class $h$ whose $G$-orbit is larger than $h^{\pm 1}$; say $\sigma(h) \neq h^{\pm 1}$.

Let $I$ be an ideal representing the class $h$, so $I$ is a free $\ZZ$-module of rank $\phi(m)$. Let $A$ be the matrix of multiplication by $\zeta_m$ on $I$, and let $B$ be the matrix of multiplication by $\zeta_m$ on $\sigma(I)$. Since $I^{\pm 1}$ and $\sigma(I)$ are not isomorphic as $\ZZ[\zeta_m]$ modules, $A^{\pm 1}$ and $B$ are not conjugate.

However, $C_m \ltimes_A \ZZ^{\phi(m)} \cong \langle \zeta \rangle \ltimes I$ and $C_m \ltimes_B \ZZ^{\phi(m)} \cong \langle \zeta \rangle \ltimes \sigma(I)$, and these are isomorphic by $(\zeta^j, x) \mapsto (\sigma(\zeta)^j, \sigma(x))$.

This occurs for $m=37$, where $H \cong \ZZ/37 \ZZ$. If I recall correctly, if $\sigma(\zeta) = \zeta^a$ then $\sigma(h) = h^{a^{21}}$. Since $\mathrm{GCD(21,36)} = 3$, the monomial $a^{21}$ takes $12$ different values modulo $37$ so, taking $h$ a generator of the class group, there are values of than $h^{\pm 1}$ in the $G$ orbit of $h$.

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    $\begingroup$ Of course the significant difference between $C_m$ and ${\mathbb Z}$ is that the only isomorphisms of ${\mathbb Z}$ are the identity and inversion. $\endgroup$ – Derek Holt Jan 10 at 19:11

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