Let $A,B(a,b,c,d)\in\mathsf{GL}(4,\mathbb{Z})$ be given by $$A=\begin{pmatrix} I\_2 & \begin{pmatrix} 0&0\\0&1 \end{pmatrix} \\0& -I\_2\end{pmatrix},\quad B(a,b,c,d)=\begin{pmatrix} -2a-b & 2c& 0& c\\ -\frac{1+(2a+b)^2}{2c}&2a+b&d&a\\0&0&-b& \frac{1+b^2}{2d}\\ 0&0&-2d&b\end{pmatrix}$$ (which implies $a,b,c,d\in \mathbb{Z}$, $b$ odd, $2c\mid 1+(2a+b)^2$ and $2d\mid 1+b^2$). **$A$ has order 2 and $B^2(a,b,c,d)=-Id$ so has order 4**.


In general, if we have $A,B$ we can define the group $G\_{A,B}$ by $G\_{A,B}=\mathbb{Z}^2\ltimes \mathbb{Z}^4$ where the product is given by $(k,\ell,v)\cdot (k',\ell',v')=(k+k',\ell+\ell',v+A^k B^{\ell} v')$.


**Problem**: Decide if $G\_{A,B(a,b,c,d)}$ is isomorphic to $G\_{A,B(e,f,g,h)}$ or not.


**Thoughts**: Actually, I think that the groups are isomorphic for every possible choice of parameters (I've computed some computational invariants **(Abelianization, #LowIndexSubgroups, #AutomorphismGroup(PQuotient))** of the groups given by some specific choices of the parameters and they coincide) but I can't find an isomorphism. I have proved the following (for any $A,B\in\mathsf{GL}(n,\mathbb{Z}$)):


1. $G\_{A,B}$ is isomorphic to $G\_{AB,B}$ and $G\_{A,AB}$
2. $G\_{A,B}$ is isomorphic to $G\_{A^{-1}, B}$ and $G\_{A,B^{-1}}$.
3. $G\_{A,B}$ is isomorphic to $G\_{C^{-1}AC, C^{-1}BC}$ for any invertible integer matrix $C$.


I know that all the matrices $B(a,b,c,d)$ are conjugate to $C\_4\oplus C\_4=\begin{pmatrix} 0&-1\\1&0\end{pmatrix}\oplus \begin{pmatrix}0&-1\\1&0\end{pmatrix}$ but when I conjugate, I move the matrix $A$ and there still are parameters around, and also is it too difficult (I think impossible in general) to find which matrix $P$ makes $B(a,b,c,d)$ similar to $C\_4\oplus C\_4$ for general $a,b,c,d$.


**The presentation of the group $G\_{A,B(a,b,c,d)}$ is (computed in Magma calculator)**:



> 
> G1 := Group<x1,x2,x3,x4,e1,e2 | (x1,x2), (x1,x3), (x1,x4), (x2,x3), (x2,x4),  
> 
> (x3,x4), (e1,e2), x1^(e1)=x1, x2^(e1)=x2^-1, x3^(e1)=x3, x4^(e1)=x3\*x4^-1
> , x1^(e2)=x1^(-2a-b)\*x2^(-(1+(2a+b)^2)/(2c)), x2^(e2)=x1^(2c)\*x2^(2a+b), x3^(e2)=x2^(d)\*x3^(-b)\*x4^(-2d), x4^(e2)=x1^(c)\*x2^(a)\*x3^((1+b^2)/(2d))\*x4^b>;
> 
> 
> 


Any help will be very appreciated!



Finding isomorphism between $\mathbb{Z}^2\ltimes_{A,B} \mathbb{Z}^4$ and $\mathbb{Z}^2\ltimes_{A,C} \mathbb{Z}^4$
Finally I've come with an answer to my question. I wasn't sure about answer because it changes a little bit the approach of the question but I'll do anyway.


The problem originally was about finding a classification (up to isomorphism) about the groups $G\_{A,B}$ where $A=P^{-1} \tilde{A} P$ and $B=P^{-1} \tilde{B} P$ are integer matrices and $P\in \mathsf{GL}\_4(\mathbb{R})$ and $$\tilde{A}=\begin{pmatrix} 1&0&0&0\\ 0&1&0&0 \\ 0&0&-1&0\\ 0&0&0&-1 \end{pmatrix}, \quad \tilde{B}=\begin{pmatrix} 0&-1&0&0\\1&0&0&0\\0&0&0&-1\\0&0&1&0\end{pmatrix}.$$
I was dealing with the case $A$ as in the question. Then, as $\tilde{B}$ commutes with $\tilde{A}$, $B$ must commute with $A$ so we can write (also using that $B$ will be of the form $B=\begin{pmatrix} S&\*\\0&T\end{pmatrix}$ with $S^4=T^4=I$) $B=\begin{pmatrix} a& -\frac{1+a^2}{b}&0&-\frac{1+a^2}{2b}\\ b&-a & -\frac{d}{2}& \frac{c-a}{2} \\ 0&0&c&-\frac{1+c^2}{d}\\0&0&d&-c\end{pmatrix}$ (I cleared the equations somewhat different as in the question).


Now all that I've to do is find $C\in\mathsf{GL}\_4(\mathbb{Z})$ such that $AC=CA$ and $B(a,b,c,d)C=C B(1,1,1,2)$.


First, as $AC=CA$, $C$ must be of the form $C=\begin{pmatrix} p\_1&p\_3&0&\frac{p\_3}{2}\\ p\_2&p\_4& -\frac{p\_8}{2}& \frac{p\_4-p\_{12}}{2}\\ 0&0&p\_7&p\_{11}\\ 0&0&p\_8&p\_{12}\end{pmatrix}$. As $\begin{pmatrix} p\_1&p\_3\\ p\_2&p\_4 \end{pmatrix}$ must conjugate $\begin{pmatrix} a& -\frac{1+a^2}{b}\\b&-a\end{pmatrix}$ to $\begin{pmatrix} 1&-2\\1&-1\end{pmatrix}$ we can put $p\_3, p\_4$ in terms of $p\_1$ and $p\_2$ (Note that $p\_1, p\_2, p\_3, p\_4$ exists because the two matrices are indeed conjugate) and the same for the matrix below. The condition of the determinant being equal to 1 ensures that $p\_3, p\_8$ and $p\_4-p\_{12}$ are even and it is straightforward to check that $C$ works so we're done.


