**EDIT**: I've made a mistake with the matrices. Now it is corrected.


A couple of days ago I asked [this question](https://mathoverflow.net/questions/369214/deciding-if-mathbbz-ltimes-a-mathbbz5-and-mathbbz-ltimes-b-mathbb). There, answerers gave me excellent hints to solve that case and others too. But I've found two matrices for which I had to distinguish the corresponding groups and I couldn't solve the problem with any of those techniques (see below).


I'm almost done with my task of analyzing these matrices and groups and I think the following are the last examples which I've to distinguish.


Let $A=\begin{pmatrix} 1&0&0&0&0\\0&0&-1&0&0 \\ 0&1&-1&0&0\\ 0&0&0&0&-1\\0&0&0&1&1\end{pmatrix}=1\oplus A'$ and $B=\begin{pmatrix} 1&0&0&0&0\\ 0&0&-1&1&0\\0&1&-1&0&0\\0&0&0&0&-1\\0&0&0&1&1\end{pmatrix}=1\oplus B'$.



> 
> Question: Are isomorphic $G\_A=\mathbb{Z}\ltimes\_A \mathbb{Z}^5$ and $G\_B=\mathbb{Z}\ltimes\_B\mathbb{Z}^5$? Well again I think they're not.
> 
> 
> 


**Thoughts and advances**:


$\bullet$ $B$ is not conjugate to $A$ or $A^{-1}$ in $\mathsf{GL}\_5(\mathbb{Z})$ but they are in $\mathsf{GL}\_5(\mathbb{Q})$. They are both of order 6 and have 1 as eigenvalue.


$\bullet$ I computed the 2 and 3 exponencial central classes up to 11 (as the answerers taught me in the previous question) and result in isomorphic pQuotients. The presentations are:



```
> GA :=  Group<a,b,c,d,e,t | (a,b), (a,c), (a,d), (a,e), (b,c), (b,d), (b,e), (c,d), (c,e), (d,e),  
> a^t=a, b^t=b^-1*c^-1, c^t=b, d^t=d*e^-1, e^t=d>;
>
> GB :=  Group<a,b,c,d,e,t | (a,b), (a,c), (a,d), (a,e), (b,c), (b,d), (b,e), (c,d), (c,e), (d,e),  
> a^t=a, b^t=b^-1*c^-1, c^t=b, d^t=b*c*d*e^-1, e^t=b*c*d>;


```

$\bullet$ I've found in [this paper](https://arxiv.org/pdf/1509.05879.pdf) Corollary 8.9 (cf Prop 4.2 and Def 4.3) that if I had $\mathbb{Z}\ltimes\_{A'} \mathbb{Z}^4$ and $\mathbb{Z}\ltimes\_{B'}\mathbb{Z}^4$ then those semidirect products wouldn't be isomorphic because $B'\not\sim A',(A')^{-1}$ in $\mathsf{GL}\_5(\mathbb{Z})$ (and because neither have 1 as eigenvalue) but I don't know how to relate these semidirect products with the original ones I have.


$\bullet$ $G\_A^{ab}\cong G\_B^{ab}\cong \mathbb{Z}\oplus\mathbb{Z}\oplus\mathbb{Z}\_3$. Also I tried to compute the quotients $G/\gamma\_i(G)$ (for $i\geq 2$) where $\gamma\_i=[\gamma\_{i-1}(G),G]$ and $\gamma\_1=[G,G]$ and all of them are isomorphic.


$\bullet$ Thinking of $\Gamma\_A=(G\_A/Z(G\_A))$ and $\Gamma\_B=(G\_B/Z(G\_B))$ I get $\Gamma\_A\cong \mathbb{Z}\_6\ltimes\_{A'}\mathbb{Z}^4$ and $\Gamma\_B\cong \mathbb{Z}\_6\ltimes\_{B'}\mathbb{Z}^4$ and I computed the abelianization ($\mathbb{Z}\_6\oplus\mathbb{Z}\_3$) and pQuotients here too but I couldn't distinguish them either.



```
> Gamma_A :=  Group<a,b,c,d,t | (a,b), (a,c), (a,d), (b,c), (b,d),  
>      (c,d), t^6, a^t=a^-1*b^-1, b^t=a, c^t=c*d^-1, d^t=c>;
> 

> Gamma_B :=  Group<a,b,c,d,t | (a,b), (a,c), (a,d), (b,c), (b,d),  
>      (c,d),  t^6, a^t=a^-1*b^-1, b^t=a, c^t=a*b*c*d^-1, d^t=a*b*c>;

```

I hope someone can help me again with this.



A "subtle" isomorphism testing problem: $\mathbb{Z}\ltimes_{A} \mathbb{Z}^5\cong \mathbb{Z}\ltimes_{B}\mathbb{Z}^5$ or not?
**Claim.** The groups $G\_A$ and $G\_B$ are not isomorphic.


We will use the following lemma.


**Lemma.** Let $\Gamma\_A = G\_A/Z(G\_A) = C\_6 \ltimes\_{A'} \mathbb{Z}^4$ and $\Gamma\_B = G\_B/Z(G\_B) = C\_6 \ltimes\_{B'} \mathbb{Z}^4$, where $C\_6 = \langle \alpha \rangle$ is the cyclic group of order $6$ and $A'$ and $B'$ are obtained from $A$ and $B$ respectively by removing the first row and the first column.
Let $M\_A \Doteq [\Gamma\_A, \Gamma\_A]$ and $M\_B \Doteq [\Gamma\_B, \Gamma\_B]$ be the corresponding derived subgroups considered as $\mathbb{Z}[C\_6]$-modules where $\alpha$ acts as $A'$ on $M\_A$ and as $B'$ on $M\_B$. Then we have the following $\mathbb{Z}[C\_6]$-module presentations:
$$M\_A = \langle x, y \vert \, (\alpha^2 + \alpha + 1)x = (\alpha^2 - \alpha + 1)y = 0\rangle
$$ and
$$
M\_B = \langle x \,\vert \, (\alpha^4 + \alpha^2 + 1)x = 0\rangle
$$



> 
> *Proof.* Use the description of $M\_A$ as $(A' - 1\_4) \mathbb{Z}^4$ and observe how $A'$ transforms the column vectors of $A' - 1\_4$. Do the same for $M\_B$.
> 
> 
> 


We are now in position to prove the claim.



> 
> *Proof of the claim.* It suffices to show that $\Gamma\_A$ and $\Gamma\_B$ are not isomorphic. An isomorphism $\phi: \Gamma\_A \rightarrow \Gamma\_B$ would induce an isomorphism $M\_A \rightarrow M\_B$ of Abelian groups. As we necessarily have $\phi((\alpha, (0, 0, 0, 0))) = (\alpha^{\pm 1}, z)$ for some $z \in \mathbb{Z}^4$ and since the presentations of the above lemma remain unchanged if we replace $\alpha$ by $\alpha^{-1}$, the isomorphism $\phi$ would induce an isomorphism of $\mathbb{Z}[C\_6]$-modules. This is impossible because $M\_A$ cannot be generated by less than two elements whereas $M\_B$ is cyclic over $\mathbb{Z}[C\_6]$. Observe indeed that $M\_A$ surjects onto $\mathbb{F}\_4 \times \mathbb{F}\_4$ where $\mathbb{F}\_4 \simeq \mathbb{Z}[C\_6]/(2, \alpha^2 + \alpha + 1)$ is the field with four elements.
> 
> 
> 




---


**Addendum 1.**
Let $G$ be finitely generated group $G$.
We denote by $d(G)$ the *rank of $G$*, i.e., the minimum number of generators of $G$.
For these two special instances, we actually have $d(G\_A) = 4$ and $d(G\_B) = 3$.


In general, it can be difficult to compute the rank of a group, but some knowledge is available for $G\_A$ and some other split extensions by cyclic groups, see [1, Corollary 2.4] and [2, Theorem A and Section 3.1].


Let us set $G\_A = \mathbb{Z} \ltimes\_A N\_A$ with $N\_A \Doteq \mathbb{Z}^n$. Then $N\_A$ can be endowed with the structure of a $\mathbb{Z}[C]$ module where $C \subset G\_A$ is the infinite cyclic group generated by $a \Doteq (1, (0, \dots, 0)) \in G\_A$ which acts on $N\_A$ via conjugation, or equally, via multiplication by $A$.


Let $R$ be a unital ring and let $M$ be a finitely generated $R$-module.
We denote by $d\_R(M)$ the minimum number of generators of $M$ over $R$. Then the aforementioned results implies that $$d(G\_A) = d\_{\mathbb{Z}[C]}(N\_A) + 1.$$


Let us denote by $(e\_1, \dots, e\_n)$ the canonical basis of $\mathbb{Z}^n$.
For $A$ and $B$ as in OP's question, it is easy to derive the following $\mathbb{Z}[C]$-module presentations: $$N\_A = \langle e\_1, e\_2, e\_4 \, \vert (a - 1)e\_1 = (a^2 + a + 1)e\_2 = (a^2 - a + 1)e\_4 = 0 \rangle$$ and
$$N\_B = \langle e\_1, e\_5 \, \vert (a - 1)e\_1 = (a^4 + a^2 + 1)e\_5 = 0 \rangle.$$


From these presentations and the above rank formula, we can easily infer the claimed identities, that is, $d(G\_A) = 4$ and $d(G\_B) = 3$.


**Addendum 2.** Let $C\_A$ be the cyclic subgroup of $G\_A$ generated by $a \Doteq (1, (0, \dots, 0))$ and $K\_A$ the $\mathbb{Z}[C\_A]$-module defined as in Johannes Hahn's answer (and subsequently mine) to this [MO question](https://mathoverflow.net/questions/350102/isomorphism-of-mathbbz-ltimes-a-mathbbzm-and-mathbbz-ltimes-b-math?noredirect=1&lq=1).
Let $\omega(A)$ be the order of $A$ in $\text{GL}\_n(\mathbb{Z})$, that we assume to be finite, and set $e\_0 \Doteq (\omega(A), (0, \dots, 0)) \in G\_A$. Let us denote by $(e\_1, \dots, e\_n)$ the canonical basis of $\mathbb{Z}^n \triangleleft G\_A$.


It has been established that the pair $\{K\_A, K\_{A^{-1}}\}$ of $\mathbb{Z}[C]$-modules is an isomorphism invariant of $G\_A$, where $C = C\_A \simeq C\_{A^{-1}}$ with the identification $a \mapsto (1, (0, \dots,0)) \in G\_{A^{-1}}$. It can be used to address the [previous example](https://mathoverflow.net/questions/369214/deciding-if-mathbbz-ltimes-a-mathbbz5-and-mathbbz-ltimes-b-mathbb?noredirect=1&lq=1) and this one as well.


For the instances of this MO question, straightforward computations show that
$$K\_A = K\_{A^{-1}}= \langle e\_0, e\_1, e\_2, e\_4 \, \vert \, (a - 1)e\_0 = (a - 1)e\_1 = (a^2 + a + 1)e\_2 = (a^2 - a + 1)e\_4 = 0\rangle$$ and
$$K\_B = \langle e\_0, e\_1, e\_5 \, \vert \, 
(a - 1)e\_0 = (a -1)e\_1 = (a^4 + a^2 + 1)e\_5 = 0\rangle.$$
Since $d\_{\mathbb{Z}[C\_A]}(K\_A) = 4$ and $d\_{\mathbb{Z}[C\_B]}(K\_B) = 3$ the groups $G\_A$ and $G\_B$ are not isomorphic.




---


[1] G. Levitt and V. Metaftsis, ["Rank of mapping tori and companion matrices"](https://arxiv.org/pdf/1004.2649.pdf), 2010.  

[2] L. Guyot, "Generators of split extensions of Abelians groups by cyclic groups", 2018.


