A "subtle" isomorphism testing problem: $\mathbb{Z}\ltimes_{A} \mathbb{Z}^5\cong \mathbb{Z}\ltimes_{B}\mathbb{Z}^5$ or not?

Question

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

A couple of days ago I asked this question. 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 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.

Answer 1

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.

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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. 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 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.

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[1] G. Levitt and V. Metaftsis, "Rank of mapping tori and companion matrices", 2010.
[2] L. Guyot, "Generators of split extensions of Abelians groups by cyclic groups", 2018.

Answer 2

The following calculations seem to distinguish between them.

>  #LowIndexSubgroups(GA,4);
30
>  #LowIndexSubgroups(GB,4);
26

They have different numbers of homomorphisms onto $A_4$:

> #Homomorphisms(GA,Alt(4),Sym(4));
5
> #Homomorphisms(GB,Alt(4),Sym(4));
1

(The options third argument $\mathsf{Sym(4)}$ means count (surjective homomorphisms) up to conjugacy in $\mathsf{Sym(4)}$.)

Here is yet another approach:

> P,phi:=pQuotient(GA,3,1); 
> AQInvariants(Kernel(phi));
[ 2, 2, 0, 0, 0, 0 ]
> P,phi:=pQuotient(GB,3,1);
> AQInvariants(Kernel(phi));
[ 0, 0, 0, 0 ]

In fact these three methods are all detecting the same difference in finite quotients of the groups, but I included them all to give you an indication of possible techniques for proving non-isomorphism.

Ultimately, all of these techniques rely on looking at various types of computable quotients of the groups. Unfortunately there are examples of pairs of non-isomorphic finitely presented groups which cannot be distinguished in this fashion by their computable quotients (in fact the unsolvability of the general isomorphism problem implies that such examples must exist.)