Let $A$ and $B$ be matrices of finite order with integer coefficients.

Let $n\in\mathbb{N}$ and let $G_A=\mathbb{Z}\ltimes_A \mathbb{Z}^n$ be the semidirect product, where the action is $\varphi(n)\cdot (m_1,\ldots,m_n)=A^n (m_1,\ldots,m_n)$, and similarly with $B$.

It is easy to construct an isomorphism between $G_A$ and $G_B$ if $A$ is conjugate in $GL(n,\mathbb{Z})$ to $B$ or $B^{-1}$.

But, this is also a necessary condition? I mean, does $G_A\cong G_B$ implies $A\cong B$ or $A\cong B^{-1}$ in $GL(n,\mathbb{Z})$ or is there a counterexample?

I've seen in A necessary condition for two semi-direct products to be isomorphic. that it is true if A and B are hyperbolic, i.e none of their eigenvalues have module 1, but it isn't the case.

Thank you!

  • 1
    $\begingroup$ Note that your assumption implies that $A,B$ have finite order. $\endgroup$
    – YCor
    Jan 8, 2020 at 17:18
  • $\begingroup$ Yes, you're right, they have finite order. Is this a hint? $\endgroup$ Jan 8, 2020 at 17:22
  • $\begingroup$ No, it's just natural to say it. $\endgroup$
    – YCor
    Jan 8, 2020 at 17:24
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    $\begingroup$ I think it would fit on MO (I'd recommend mention explicitly that $A,B$ have finite order). Also a remark, denoting by $G_A$ this group, is that $G_A$ is virtually abelian iff $A$ has finite order. In particular, if $A$ has finite order and $G_A\simeq G_B$ then $B$ also has finite order. $\endgroup$
    – YCor
    Jan 9, 2020 at 18:51
  • $\begingroup$ @YCor thank you for your recommendations, I will do that. By the way, what's the difference between MO and MSE? $\endgroup$ Jan 9, 2020 at 19:28


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