Let $A,B\in \operatorname{GL}(k,\mathbb{Z})$ be matrices of finite order such that $AB=BA$ and $A\neq \operatorname{I}$, $B\neq \operatorname{I}$.

Define the group $\Sigma_{A,B}:=\mathbb{Z^2}\ltimes_{A,B} \mathbb{Z}^k$, that is, the multiplication is given by $$\left(r_1,r_2, \left(\begin{array}{c}{t_1\\ \vdots \\ t_k}\end{array}\right)\right) \cdot \left(s_1,s_2, \left(\begin{array}{c}{t'_1\\ \vdots \\ t'_k}\end{array}\right)\right)=\left(r_1+s_1,r_2+s_2, \left(\begin{array}{c}{t_1\\ \vdots \\ t_k}\end{array}\right)+A^{r_1} B^{r_2}\left(\begin{array}{c}{t'_1\\ \vdots \\ t'_k}\end{array}\right) \right).$$ Note that this multiplication is well defined since $AB=BA$.

Question: Assume that $C,D\in\operatorname{GL}(k,\mathbb{Z})$ are commuting matrices of finite order such that $\langle A,B\rangle=\langle C,D\rangle$, i.e. the pairs of matrices $\{A,B\}$ and $\{C,D\}$ generate the same subgroup. Also assume that the group $\langle A,B\rangle$ is not cyclic.

Are $\Sigma_{A,B}$ and $\Sigma_{C,D}$ isomorphic?

We have the following lemma:


(1) $\Sigma_{A,B}$ is isomorphic to $\Sigma_{B,A}$

(2) $\Sigma_{A,B}$ is isomorphic to $\Sigma_{A^{-1},B}$

(3) $\Sigma_{A,B}$ is isomorphic to $\Sigma_{A,AB}$

Proof: The isomorphisms are:

For (1), $\varphi\left(r_1,r_2, \left(\begin{array}{c}{t_1\\ \vdots \\ t_k}\end{array}\right)\right)=\left(r_2,r_1, \left(\begin{array}{c}{t_1\\ \vdots \\ t_k}\end{array}\right)\right) $

For (2), $\varphi\left(r_1,r_2, \left(\begin{array}{c}{t_1\\ \vdots \\ t_k}\end{array}\right)\right)=\left(-r_1,r_2, \left(\begin{array}{c}{t_1\\ \vdots \\ t_k}\end{array}\right)\right)$

For (3), $\varphi\left(r_1,r_2, \left(\begin{array}{c}{t_1\\ \vdots \\ t_k}\end{array}\right)\right)=\left(r_1-r_2,r_2, \left(\begin{array}{c}{t_1\\ \vdots \\ t_k}\end{array}\right)\right)$

I was thinking that if we could go from one generator set $\{A,B\}$ to another one $\{A^p B^q, A^r B^s\}$ applying the 3 operations above then the groups would be isomorphic, but I don't know if this is true, and I also don't know which properties must satisfy $p,q,r,s$ in order for $\{A^p B^q, A^r B^s\}$ to generate the subgroup $\langle A,B\rangle$.

Help would be greatly appreciated. Thanks!

  • $\begingroup$ Why do you need to assume that $\langle A,B \rangle$ is not cyclic? $\endgroup$
    – Derek Holt
    Aug 30, 2023 at 7:49
  • $\begingroup$ For cyclic groups it does not hold: Indeed, for the matrix $A\in\operatorname{GL}(36,\mathbb{Z})$ you gave me in mathoverflow.net/questions/350102/… the groups $\Sigma_A$ and $\Sigma_{A^2}$ are not isomorphic but clearly $\langle A\rangle=\langle A^2\rangle$. I should add that my principal interest is $k=5$ (so these strange counterexamples do not exist) but I wanted to say something in general. $\endgroup$ Aug 30, 2023 at 8:02
  • 1
    $\begingroup$ Oh yes I remember, the point is that there can be an isomorphism that does not fix the normal subgroup in the semidirect product. I would expect it would be possible to extend this to a counterexample in which $\langle A, B \rangle$ is not cyclic. Start with the known counterexample ${\mathbb Z}\rtimes_A {\mathbb Z}^k$ using $A$ and then take a direct product with any example${\mathbb Z}\rtimes_B {\mathbb Z}^j$ using $B$. For example $B$ could act on ${\mathbb Z}$ by inversion. $\endgroup$
    – Derek Holt
    Aug 30, 2023 at 8:27


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