I asked this in [this MSE question](https://math.stackexchange.com/questions/3789119/deciding-if-mathbbz-ltimes-a-mathbbz5-and-mathbbz-ltimes-b-mathbb) but I didn't get answers. I think maybe here someone can help me.


I have the two following groups


$G\_A=\mathbb{Z}\ltimes\_A \mathbb{Z}^5$, where $A=\begin{pmatrix} 1&0&0&1&0\\0&-1&0&0&0\\0&0&-1&0&0\\0&0&0&0&-1\\0&0&0&1&0\end{pmatrix}$ and


$G\_B=\mathbb{Z}\ltimes\_B \mathbb{Z}^5$, where $B=\begin{pmatrix} 1&0&0&1&0\\0&-1&0&1&0\\0&0&-1&0&0\\0&0&0&0&-1\\0&0&0&1&0\end{pmatrix}$.


The product is given (for example in $G\_A$) by $(k,m)\cdot(\ell,n)=(k+\ell, m+A^k n)$.



> 
> **Problem**: Decide if $G\_A$ is isomorphic to $G\_B$ or not.
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**My thoughts**: I think strongly that they are not isomorphic but I couldn't prove it. The matrices $A$ and $B$ are both of order 4, they're not conjugate in $\mathsf{GL}(n,\mathbb{Z})$ (neither $B$ and $A^{-1}$) but they are conjugate in $\mathsf{GL}(n,\mathbb{Q})$. In some other cases, I've seen that they're not isomorphic by computing the abelianization, but in this case both have the same abelianization, namely $\mathbb{Z}\oplus\mathbb{Z}\oplus\mathbb{Z}\_2\oplus\mathbb{Z}\_2$. Even worse, both have 1 as an eigenvalue.


In [my previous MO question](https://mathoverflow.net/questions/350102/isomorphism-of-mathbbz-ltimes-a-mathbbzm-and-mathbbz-ltimes-b-math) there is a counterexample for the implication "$G\_A\cong G\_B\Rightarrow A\sim B^{\pm 1}$" so I cannot use that.


Thanks!



Deciding if $\mathbb{Z}\ltimes_A \mathbb{Z}^5$ and $\mathbb{Z}\ltimes_B \mathbb{Z}^5$ are isomorphic or not

> 
> **Claim.** The groups $G\_A$ and $G\_B$ are **not** isomorphic.
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We will use the following lemmas.



> 
> **Lemma 1.** Let $A \in \text{GL}\_n(\mathbb{Z})$ and let $G\_A \Doteq \mathbb{Z} \ltimes\_A \mathbb{Z}^n$. Then the following hold:
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> 
> * The center $Z(G\_A)$ of $G\_A$ is generated by $\{0\} \times \ker(A - 1\_n)$ and $(\omega, (0, \dots, 0))$ where $1\_n$ is the $n \times n$ identity matrix and $\omega$ is the order of $A$ in $\text{GL}\_n(\mathbb{Z})$ if $A$ has finite order, zero otherwise.
> * The derived subgroup $[G\_A, G\_A]$ of $G\_A$ is $\{0\} \times (A - 1\_n)\mathbb{Z}^n$. More generally, setting $\gamma\_{i + 1}(G\_A) \Doteq [\gamma\_i(G\_A), G\_A]$ with $\gamma\_1(G\_A) \Doteq G\_A$, we have $\gamma\_{i + 1}(G\_A) = \{0\} \times (A - 1\_n)^i \mathbb{Z}^n$.
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> 
> *Proof.* Straightforward.
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For $A$ and $B$ as in OP's question, we have thus $$Z(G\_A) = 4\mathbb{Z} \times \ker(A - 1\_5), \, Z(G\_B) = 4\mathbb{Z} \times \ker(B - 1\_5)$$
with $\ker(A - 1\_5) = \ker(B - 1\_5) = \mathbb{Z} \times \{ (0, 0, 0, 0) \} \subset \mathbb{Z}^5$.



> 
> **Lemma 2.** Let $A$ and $B$ as in OP's question and set $\Gamma\_A \Doteq G\_A / Z(G\_A)$ and $\Gamma\_B \Doteq G\_B / Z(G\_B)$. Then we have $\Gamma\_A/ [\Gamma\_A, \Gamma\_A] \simeq (\mathbb{Z}/ 2 \mathbb{Z})^3 \times \mathbb{Z}/ 4 \mathbb{Z}$ and $\Gamma\_B/ [\Gamma\_B, \Gamma\_B] \simeq \mathbb{Z}/ 2 \mathbb{Z} \times (\mathbb{Z}/ 4 \mathbb{Z})^2$.
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> 
> *Proof.* Write $\Gamma\_A = \mathbb{Z} / 4 \mathbb{Z} \ltimes\_{A'} \mathbb{Z}^4$ and
> $\Gamma\_B = \mathbb{Z} / 4 \mathbb{Z} \ltimes\_{B'} \mathbb{Z}^4$ where $A', B' \in \text{GL}\_4(\mathbb{Z})$ are obtained from $A$ and $B$ by removing the first row and the first column. Use then the description of the derived subgroup of Lemma 1 which still applies to $\Gamma\_A$ and $\Gamma\_B$ if we replace $A$ by $A'$ and $B$ by $B'$.
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> 
> *Proof of the claim.* If $G\_A$ and $G\_B$ are isomorphic, then so are $\Gamma\_A$ and $\Gamma\_B$. This is impossible since the two latter groups have non-isomorphic abelianizations by Lemma 2.
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---


**Addendum.** 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 $a \mapsto (1, (0, \dots,0)) \in G\_{A^{-1}}$.


For the instances of this MO question, straightforward computations show that
$$\left\langle e\_0, e\_2, e\_3, e\_5 \, \vert \, (a - 1)e\_0 = (a + 1)e\_2 = (a + 1)e\_3 = (a^3 -a^2 + a - 1)e\_5 = 0\right\rangle$$ is a presentation of both $K\_A$ and $K\_{A^{-1}}$ and
$$\left\langle e\_0, e\_1, e\_2, e\_3, e\_5 \, \vert \, 
(a - 1)e\_0 = (a -1)e\_1 = (a + 1)e\_2 = (a + 1)e\_3 = (a^2 + 1)e\_5 + e\_1 + e\_2 = 0\right\rangle$$ is a presentation of $K\_B$.


From the above presentations, we easily infer the following isomorphisms of Abelian groups: $K\_A/(a + 1)K\_A \simeq \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/4 \mathbb{Z} \times \mathbb{Z}^2$ and $K\_B/(a + 1)K\_B \simeq (\mathbb{Z}/2\mathbb{Z})^2 \times \mathbb{Z}^2$.


As result, the groups $G\_A$ and $G\_B$ are not isomorphic.


