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.

  • 6
    $\begingroup$ I do believe that this is a perfectly fine question and I don't think that there is any reason to apologize. The answer to the last question gave you one obstruction and now it is the most natural thing to look for examples, where this obstruction does not help. You provided many details showing that you first thought about it yourself. I would encourage such questions. $\endgroup$ Aug 21, 2020 at 7:00
  • 1
    $\begingroup$ I perfectly agree with Henrik. By the way, I edited the question deleting the apology. $\endgroup$ Aug 21, 2020 at 8:32
  • 1
    $\begingroup$ Great! Thank you for your words and comments, they're very motivational. $\endgroup$ Aug 21, 2020 at 14:03

2 Answers 2


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. 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.

[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.

  • 1
    $\begingroup$ It is great to know a non-computational proof. I'll keep it and I thank you for all the facts and insights you shared with me! $\endgroup$ Aug 21, 2020 at 19:42
  • $\begingroup$ Excellent! I understand a little bit more with the Addendum. Thanks for the references, I'll keep them. I wonder if there is a general method to distinguish all of them at the same time, I mean, I gave a classification of the conjugacy classes of integer matrices with determinant 1 in $\mathsf{GL}_5(\mathbb{Z})$ of finite order, and I had to decide if the corresponding groups $G_A=\mathbb{Z}\ltimes_A \mathbb{Z}^n$ are isomorphic or not. I've found that the isomorphism class depends on the conjugacy class of $A$ but only by computing invariants. $\endgroup$ Aug 23, 2020 at 0:44
  • 1
    $\begingroup$ @AleTolcachier Dear Ale, I was looking for an invariant which could address both questions on the non-existence of an isomorphism. It turns out that the module $K_A$ defined by Johannes Hahn in his answer to your initial question (mathoverflow.net/questions/350102/…) fits the bill. $\endgroup$
    – Luc Guyot
    Aug 24, 2020 at 16:48
  • $\begingroup$ Yes, and it seems useful for other examples too, it is an excellent situation. Anyway, I've learned another techniques and things about modules, groups, etc; with the other answers in this and my previous question. It has been a very rich question/problem indeed! Now, I have to sit and understand that answer of yours in my initial question. I suppose I can leave a comment if there is something that I don't understand. Thank you very much, sir $\endgroup$ Aug 24, 2020 at 16:52

The following calculations seem to distinguish between them.

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

They have different numbers of homomorphisms onto $A_4$:

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

(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.)

  • 1
    $\begingroup$ Was not the first computation enough to conclude? $\endgroup$ Aug 21, 2020 at 8:12
  • 3
    $\begingroup$ Yes, but the different methods available could be useful with other examples. $\endgroup$
    – Derek Holt
    Aug 21, 2020 at 8:15
  • 2
    $\begingroup$ @LSpice Yes, in fact the isomorphism problem is known to be decidable for polycyclic-by-finite groups (which includes these examples), but I don't think there is a practical algorithm to solve it in that case. So you tend to be limited to what you can do with finite quotients. $\endgroup$
    – Derek Holt
    Aug 21, 2020 at 13:43
  • 2
    $\begingroup$ @DerekHolt I'm very grateful, sir. Thanks for your answers but also for your teachings about how to deal with these kind of finitely presented groups, the invariants and also with Magma computations. It's a very rich way to learn. $\endgroup$ Aug 21, 2020 at 14:12
  • 5
    $\begingroup$ Perhaps it's helpful to mention an explicit pair of "non-isomorphic finitely presented groups which cannot be distinguished in this fashion by their [finite] quotients". Baumslag exhibited such a pair of metacyclic groups in "Residually finite groups with the same finite images." Compositio Math., 29:249–252, 1974. $\endgroup$
    – HJRW
    Aug 21, 2020 at 17:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.