# Prove (or disprove) that two pairs of matrices generating the same subgroup give rise to isomorphic groups

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:

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!

• Why do you need to assume that $\langle A,B \rangle$ is not cyclic? Aug 30, 2023 at 7:49
• 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. Aug 30, 2023 at 8:02
• 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. Aug 30, 2023 at 8:27