Is it true that the semidirect product groups $\mathbb{Z}\ltimes_A \mathbb{Z}^n$ and $\mathbb{Z}\ltimes_B\mathbb{Z}^n$ are isomorphic if and only if $A$ is conjugate in $\operatorname{GL}(n,\mathbb{Z})$ to $B$ or $B^{-1}$? Here the matrices $A$ and $B$ are matrices with integer coefficients and the action is $\varphi(n)\cdot(m_1,\,\ldots\,,m_n)=A^n(m_1,\ldots,m_n)$. The matrices $A$ and $B$ have all complex eigenvalues in the unit circle.

It is easy to construct an isomorphism if $A$ is conjugate to $B$ or $B^{-1}$ but I don't have any ideas about how to proof the reverse implication. I've read 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.

This is a fact which would be very useful for my diploma thesis, so I would thank you if you give some references.

Thank you! and sorry for my writing, it is my first time asking something here

  • $\begingroup$ This seems like a great question, but I can't entirely understand it. Can you clarify the multiplication a little bit, please? For $Z \ltimes Z^3$, a typical element is something like $(u; v) = (u, v_1, v_2, v_3)$ with $u \in Z$ and $v \in Z^3$; how does one compute $(u, v) * (u', v')$, for instance (assuming one has the $3 \times 3$ matrix $A$)? $\endgroup$ – John Hughes Nov 5 '18 at 23:47
  • $\begingroup$ $(m,n)*(m',n')=(m+m',n+A^m*n')$ $\endgroup$ – Ale Tolcachier Nov 5 '18 at 23:50
  • $\begingroup$ Ah...thanks. That makes perfect sense. Sadly, I have no useful suggestions about your original question (yet). $\endgroup$ – John Hughes Nov 5 '18 at 23:56
  • $\begingroup$ I wonder if there's a conceptual proof rather than the one I offered, e.g. exhibiting the similitude class of $A$ as an invariant of $\mathbb{Z}\ltimes_A\mathbb{Z}^n$ $\endgroup$ – Max Nov 6 '18 at 19:45
  • 1
    $\begingroup$ If it helps, here math.stackexchange.com/questions/2376842/… my question is answered but in that case $A$ and $B$ are hyperbolic matrices. Here it isn't the case $\endgroup$ – Ale Tolcachier Nov 6 '18 at 22:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.