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!

  • 7
    $\begingroup$ 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
    Commented Jan 9, 2020 at 20:09
  • 2
    $\begingroup$ 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. $\endgroup$ Commented Jan 9, 2020 at 20:52
  • 4
    $\begingroup$ @DavidHandelman I've checked this example: the abelianizations are then not isomorphic! $\endgroup$
    – YCor
    Commented Jan 9, 2020 at 21:09
  • 3
    $\begingroup$ It is known to be a necessary and sufficient condition when $n = 2$ (elementary). $\endgroup$
    – Luc Guyot
    Commented Jan 9, 2020 at 21:27
  • 4
    $\begingroup$ 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
    Commented Jan 9, 2020 at 23:00

4 Answers 4


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.

  • $\begingroup$ Do you remember how you defined or found the matrix $A$? I would have had to asked this before, sorry. I'd like to know at least if $A$ has det 1. $\endgroup$ Commented Aug 21, 2020 at 21:39
  • 3
    $\begingroup$ I am afraid I cannot find any record of the matrix $A$ that I found, but since it has order $37$ it must have determinant $1$. $\endgroup$
    – Derek Holt
    Commented Aug 23, 2020 at 10:07
  • 1
    $\begingroup$ What do you call "degree" of a matrix? $\endgroup$
    – YCor
    Commented Aug 24, 2020 at 19:32
  • 2
    $\begingroup$ By a matrix of degree $36$, I mean a $36 \times 36$ matrix. (I guess I am taking this terminology from the degree of a corresponding group representation.) $\endgroup$
    – Derek Holt
    Commented Aug 24, 2020 at 19:58
  • $\begingroup$ It is strange.. I put the matrix in Magma and the command IsGLZConjugate(E,E^2) returns "true" $\endgroup$ Commented Jun 12, 2021 at 23:30

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

  • 3
    $\begingroup$ 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
    Commented Jan 10, 2020 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 $h$.

  • 3
    $\begingroup$ 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
    Commented Jan 10, 2020 at 19:11
  • $\begingroup$ Why $\sigma(h)=h^{a^{21}}$? I couldn't understand the following either. Can you give me a reference to read more about these number theory (I suppose) things? $\endgroup$ Commented Aug 24, 2020 at 22:39

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.

  • $\begingroup$ Dear Luc, I tried to understand every line in your re-proof. I have two questions or point where I need a little bit of explanation if it is possible: (1) Why $K_A$ is isomorphic to $K_B$ or to $K_{B^{-1}}$ as $\mathbb{Z}[C]$-module? I can't see that $\phi:K_A\to K_B$ is a $\mathbb{Z}[C]$-module homomorphism (in the case of $\phi(a)=(1,z)$. (2) How follows that $1\oplus A$ is conjugate to $1\oplus B$ or $1\oplus B^{-1}$ from the isomorphism of the corresponding $\mathbb{Z}[C]$-modules? Thanks $\endgroup$ Commented Aug 25, 2020 at 21:56
  • $\begingroup$ @AleTolcachier Dear Ale, I added some details in an Addendum. $\endgroup$
    – Luc Guyot
    Commented Aug 26, 2020 at 0:49
  • 1
    $\begingroup$ Great! Now that part is very clear. Yesterday I noticed that I didn't know why $\phi(a)$ can be only $(\pm 1, z)$. It isn't true that $\phi(\mathbb{Z}^n)=\mathbb{Z}^n$, is it? $\endgroup$ Commented Aug 27, 2020 at 20:12
  • $\begingroup$ @AleTolcachier Dear Ale, this is good that you check so carefully. I have made a mistake: it is not true that $\phi(a) = (\pm 1, z)$. The claim needs to be downgraded to something much less general. $\endgroup$
    – Luc Guyot
    Commented Aug 27, 2020 at 21:27
  • 1
    $\begingroup$ @AleTolcachier If $\omega \in \{1, 2, 3, 4, 6\}$, then $f \in \mathbb{Z}$ is coprime with $\omega$ if and only if $f \equiv \pm 1 \text{ mod } \omega \mathbb{Z}$. $\endgroup$
    – Luc Guyot
    Commented Aug 28, 2020 at 6:39

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