Isomorphism of $\mathbb{Z}\ltimes_A \mathbb{Z}^m$ and $\mathbb{Z}\ltimes_B \mathbb{Z}^m$

Question

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!

Answer 1

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 $2 \le a \le 36$.

For completeness, here is the matrix $A$ in machine readable format. I used the Magma function $\mathsf{AreGLConjugate}$ to test $A$ for conjugacy with $A^i$. This uses a fairly new algorithm published in Bettina Eick, Tommy Hofmann, and E.A. O'Brien. The conjugacy problem in ${\rm GL}(n,{\mathbb Z})$. J. London Math. Soc., 2019.

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]

Answer 2

$\newcommand{\IZ}{\mathbb{Z}}$ 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.

Answer 3

$\def\QQ{\mathbb{Q}}\def\ZZ{\mathbb{Z}}$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 $h$.

Answer 4

This is a complement to Johannes Hahn's answer.

Corrigendum. In the previous version of this answer, I have made an erroneous claim, allowing $\omega$, the order of $A$ and $B$, to be any positive number. The claim below is valid only if $$\omega \in \{1, 2, 3, 4, 6 \},$$ which is sufficient to address OP's subsequent examples.

Following Johannes Hahn's approach, we can prove the following:

Claim. Assume that $G_A$ and $G_B$ are isomorphic. Then $\begin{pmatrix} 1 & 0 \\ 0 & A \end{pmatrix}$ is a conjugate of $\begin{pmatrix} 1 & 0 \\ 0 & B \end{pmatrix}$ or $\begin{pmatrix} 1 & 0 \\ 0 & B^{-1} \end{pmatrix}$ in $\text{GL}_{n + 1}(\mathbb{Z})$. In particular $A$ is a conjugate of $B$ or $B^{-1}$ in $\text{GL}_{n}(\mathbb{Q})$.

Proof. Let $K_A$ be the centraliser of the derived subgroup $G_A' = [G_A, G_A]$ of $G_A$. It is clearly a characteristic subgroup of $G_A$. Let $C_A$ be the infinite cyclic subgroup of $G_A$ generated by $a \Doteq (1, (0, \dots, 0))$. The conjugation by $a$, or equivalently, the multiplication by $\begin{pmatrix} 1 & 0 \\ 0 & A \end{pmatrix}$ induces a structure of $\mathbb{Z}[C_A]$-module on $K_A$. This structure is almost invariant under isomorphism in the following sense: if $\phi:G_A \rightarrow G_B$ is a group isomorphism, and if we identify $C_A$ with $C_B$ via $a \mapsto b = (1, (0, \dots, 0)) \in G_B$ then $K_A$ is isomorphic to $K_B$ or to $K_{B^{-1}}$ as a $\mathbb{Z}[C]$-module with $C = C_A \simeq C_B$, depending on whether $\phi(a) = bk$ or $b^{-1}k$ for some $k \in K_B$. This is so because conjugation by $bz$ induces a group action on $K_B$ which is independent of $k$. Now the claimed result immediately follows.

Thus the pair of modules $\{K_A, K_{A^{-1}}\}$ is a group isomorphism invariant of $G_A$ It turns out to be useful for this example and this one.

---

Addendum. Here are some details on the module $K_A$.

An element of $\mathbb{Z}[C_A]$ is a Laurent polynomial with coefficients in $\mathbb{Z}$ of the form $P(a) = \sum_{i = 0}^d c_i a^{e_i}$ where $e_i \in \mathbb{Z}$ for every $i$. The structure of $\mathbb{Z}[C_A]$-module of $K_A$ is defined in the following way: $$P(a) \cdot k = (a^{e_0}k^{c_0}a^{-e_0}) \cdots (a^{e_d}k^{c_d}a^{-e_d})$$ for $k \in K_A$. Assume now that there is an isomorphism $\phi: G_A \rightarrow G_B$. As $\phi$ is surjective and $\phi(K_A) = K_B$, there is $f \in \mathbb{Z}$ coprime with $\omega$ and $z \in \mathbb{Z}^n \triangleleft G_B$, such that $\phi(a) = b^f z$. Since $\omega \in \{1, 2, 3, 4, 6\}$, we infer that $\phi(a) = b^{\epsilon}k'$ for some $\epsilon \in \{\pm 1\}$ and some $k' \in K_B$. Thus $\phi(a^e) = b^{\epsilon e}k''$ where $k'' \in K_B $ depends on $e$, $k$ and $\epsilon$. The image of $P(a) \cdot k$ by $\phi$, after substituting $\phi(a^{e_i})$ with $b^{\epsilon e_i}k_i''$, and after simplification ($K_B$ is Abelian), results in $$(b^{\epsilon e_0}\phi(k)^{c_0}b^{- \epsilon e_0}) \cdots (b^{\epsilon e_d}\phi(k)^{c_d}b^{- \epsilon e_d}) = P(b^{\epsilon}) \cdot \phi(k).$$ Therefore $\phi$ induces an isomorphism of $\mathbb{Z}[C]$-module if $\epsilon = 1$, where $C = C_A \simeq C_B$. Let $e_0 \Doteq (\omega, (0, \dots, 0))$. Let $(e_1, \dots, e_n)$ denote the canonical basis of $\mathbb{Z}^n$ and let $C \in \text{GL}_{n + 1}(\mathbb{Z})$ be the matrix of $\phi$ with respect to $(e_0, e_1, \dots, e_n)$. If $\epsilon = 1$, then the following identity $\phi(a \cdot k) = b \cdot \phi(k)$ holds true and translates into $$C \begin{pmatrix} 1 & 0 \\ 0 & A \end{pmatrix} k = \begin{pmatrix} 1 & 0 \\ 0 & B \end{pmatrix}C k$$ simply because of the way we defined the action of $a$ and $b$ on $K_A$ and $K_B$ respectively. The claimed result on the matrix conjugation follows.