I asked this in this MSE question 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.

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 there is a counterexample for the implication "$G_A\cong G_B\Rightarrow A\sim B^{\pm 1}$" so I cannot use that.


  • 4
    $\begingroup$ Let me just remark that the fact that these matrices are conjugate over $\mathbb{Q}$ gives rise to an embedding of $G_A$ into $G_B$ as a finite index subgroup and vice versa. In general this is still not enough to conclude that $G_A$ and $G_B$ are isomorphic. $\endgroup$ Aug 15, 2020 at 8:52
  • 8
    $\begingroup$ I did a quick Magma calculation, and the exponent 2 class 2 central factors (these are calculated by the well-known $p$-quotient algorithm) are different in the two groups, which means that they are not isomorphic. They were order $2^9$ in the first group and $2^8$ in the second. $\endgroup$
    – Derek Holt
    Aug 15, 2020 at 10:07
  • 2
    $\begingroup$ I'm also puzzled by the terminology. My guess would be: define $G^{[1]}=G$, $G^{[2]}=[G,G]G^2$, $G^3=[G,G^{[2]}](G^{[2]})^2$, then consider the quotient $G/G^{[3]}$. It's the largest quotient of $G$ that is central extension where both the kernel and quotient have exponent dividing 2. $\endgroup$
    – YCor
    Aug 15, 2020 at 11:19
  • 6
    $\begingroup$ Yes, YCor's definition is correct. It is called the lower exponent $p$ central series, and you will get plenty of hits if you search for that. It is very well known in computational circles, because there is an efficient algorithm (due to Mike Newman and George Havas) to compute it, and it has been used to compute orders of restricted Burnside groups, including $R(2,7)$ of order $7^{20416}$. There are also nice theoretical properties arising from the fact that the $p$-covering group of a group (as opposed to ordinary Schur cover) is unique. $\endgroup$
    – Derek Holt
    Aug 15, 2020 at 11:39
  • 1
    $\begingroup$ I will show you my Magma calcualtion shortly, but it you are more familiar with GAP then you could easily do the same calculation there. All of the time is taken up by typing in the presentation! $\endgroup$
    – Derek Holt
    Aug 15, 2020 at 11:40

3 Answers 3


Claim. The groups $G_A$ and $G_B$ are not isomorphic.

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:

  • 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 $\gamma_{i + 1}(G_A) = \{0\} \times (A - 1_n)^i \mathbb{Z}^n$.

Proof. Straightforward.

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 $\Gamma_B/ [\Gamma_B, \Gamma_B] \simeq \mathbb{Z}/ 2 \mathbb{Z} \times (\mathbb{Z}/ 4 \mathbb{Z})^2$.

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 $B'$.

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.

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. 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.

  • 1
    $\begingroup$ Woah, excellent! I tried to compute the invariant $G/Z(G)$ but I didn't think of the abelianization of that space. Very very helpful your answer indeed, sir. Thanks $\endgroup$ Aug 15, 2020 at 16:32
  • $\begingroup$ Excuse me, I've been doing some more computations in other similar examples. I wanted to ask, what was your idea to compute the abelianization of $G_A/Z(G_A)$?. Because I 've a similar example and the groups $\Gamma_A$ and $\Gamma_B$ are isomorphic, and I want to look to some kind of invariant like you've proposed. $\endgroup$ Aug 19, 2020 at 21:10
  • 1
    $\begingroup$ Dear Ale, what I had initially in mind was to inspect the nilpotent quotients $\Gamma_i(G) = G/ \gamma_i(G)$ of $G$ for $G \in \{G_A, \Gamma_A\}$. Here $\gamma_i$ is defined as in my answer and by means of "inspection" I would start looking at the center or the derived subgroup. It turned out that $\Gamma_1(G_A)$ was just enough. $\endgroup$
    – Luc Guyot
    Aug 20, 2020 at 6:41
  • 1
    $\begingroup$ I had also in mind to study $M_A = [G_A, G_A]$ as a module over $\mathbb{Z}[A] \subset M_n(\mathbb{Z})$, in particular the computation of the Fitting invariants of $M_A$ (this resembles Alexander's polynomial of a knot group). Note that if $A$ and $B$ are $GL_n(\mathbb{Q})$-conjugate, then $\mathbb{Z}[A] \simeq \mathbb{Z}[B]$ and an isomorphism between $G_A$ and $G_B$ induces an isomorphism between $M_A$ and $M_B$. Eventually, looking for non-isomorphic finite quotients of $G_A$ and $G_B$ by characteristic subgroups is a very common strategy; Derek Holt gave you excellent hints. $\endgroup$
    – Luc Guyot
    Aug 20, 2020 at 6:44
  • 1
    $\begingroup$ @AleTolcachier The Fitting invariants of a module $M$ over $R$ are ideals of $R$ generated by the minors of fixed size of a matrix defining a presentation of $M$ ($M$ must be isomorphic to the cokernel of this matrix). Given a finite presentation, they are easy to compute (of course the number of generators and relators do matter). However, deciding whether two ideals are equal in a ring can be a difficult problem. In the case of $M_A$, they are easy to compute. $\endgroup$
    – Luc Guyot
    Aug 24, 2020 at 16:52

Here is my Magma calculation - I did the $2$-quotient calculations to class 3. Please check that I have entered the group presentations correctly. Note that $(a,b)$ is Magma's notation for the commutator $a^{-1}b^{-1}ab$, and $a^t$ means $t^{-1}at$.

> G1 :=  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^t=c^-1, d^t=e*a, e^t=d^-1 >;
> G2 :=  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^t=c^-1, d^t=e*a*b, e^t=d^-1 >;
> P1 := pQuotient(G1,2,3 : Print:=1);

Lower exponent-2 central series for G1
Group: G1 to lower exponent-2 central class 1 has order 2^4
Group: G1 to lower exponent-2 central class 2 has order 2^9
Group: G1 to lower exponent-2 central class 3 has order 2^14

> P2 := pQuotient(G2,2,3 : Print:=1);

Lower exponent-2 central series for G2   
Group: G2 to lower exponent-2 central class 1 has order 2^4
Group: G2 to lower exponent-2 central class 2 has order 2^8
Group: G2 to lower exponent-2 central class 3 has order 2^13
  • 1
    $\begingroup$ Thanks for showing this. Unfortunately, I have some technical complications. I've tried to download Magma but it seems that it isn't a free Software. I didn't find how to do it with GAP either :/ $\endgroup$ Aug 15, 2020 at 12:57
  • 8
    $\begingroup$ You can do it on GAP by using the Package "anupq" $\endgroup$ Aug 15, 2020 at 13:05
  • 6
    $\begingroup$ You should be able to do it with the Magma calculator you have to copy and paste all of the input into the wondow, and then click on "submit". $\endgroup$
    – Derek Holt
    Aug 15, 2020 at 14:07

Here is Derek Holt's computation done in GAP:

gap> LoadPackage("anupq");
gap> F := FreeGroup("a","b","c","d","e","t");;
gap> AssignGeneratorVariables(F);
gap> comms := List(Combinations(GeneratorsOfGroup(F){[1..5]},2),Comm);;
gap> G1 := F/Concatenation(comms,
>                          [Comm(a,t),b^t*b,c^t*c,d^t*a^-1*e^-1,     e^t*d]);;
gap> G2 := F/Concatenation(comms,
>                          [Comm(a,t),b^t*b,c^t*c,d^t*b^-1*a^-1*e^-1,e^t*d]);;
gap> Pq(G1:Prime:=2,ClassBound:=2);
<pc group of size 512 with 9 generators>
gap> StructureDescription(last);
"(C4 x C4 x C4 x C2) : C4"
gap> Pq(G2:Prime:=2,ClassBound:=2);
<pc group of size 256 with 8 generators>
gap> StructureDescription(last);   
"C2 x ((C4 x C4 x C2) : C4)"

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.