i would like to ask how to compute the abelianization of the semidirect product $\mathbb{Z}\ltimes_\varphi\mathbb{Z}^n$ where the action is $\varphi(k)v=A^k v$ where $A$ is a fixed invertible matrix in $\mathbb{Z}$. I have read from here a general case https://mathoverflow.net/questions/35713/abelianization-of-a-semidirect-product, but I don't understand how to particularize to this case.

Reading the general case it seems to me that the abelianization would be trivial as $H^{ab}$ is trivial since in my case $H=\mathbb{Z}^n$ is abelian.


  • 2
    $\begingroup$ Have you tried writing a presentation for your group, i.e. generators and relators? $\endgroup$ – Lee Mosher May 22 at 3:00

$H^{ab}=\mathbb{Z}^n$, not trivial. Then you need to take the coinvariants $(H^{ab})_G=\mathbb{Z}^n/\operatorname{image}(A-I)$ Finally take product with $G^{ab}=\mathbb{Z}$, so the end result is $$ \mathbb{Z}\times\operatorname{coim}(A-I). $$

  • $\begingroup$ Would you mind expanding on why the coinvariants are $\mathbb{Z}^n/\ker(A-I)$? More specifically, what happens with the kernels of $A^k-I$ for larger $k$? Shouldn't we be dividing by that as well? $\endgroup$ – Guido A. May 22 at 3:45
  • $\begingroup$ Oops, should be image not kernel. We are modding out the subgroup generated by $h^g-h=(A-I)h$, so we are modding out the image of $A-I$. $\endgroup$ – user10354138 May 22 at 3:51
  • $\begingroup$ That I can see, but isn't the action $kv = A^kv$? Hence one should divide by the subgroup generated by the union of $\ker A^n-I$ for all $n$, no? $\endgroup$ – Guido A. May 22 at 4:16
  • $\begingroup$ No, we are modding out the image, and the image of $A^n-I$ is contained in the image of $A-I$ since $A^n-I=(A-I)(A^{n-1}+\dots+I)$. Similarly $A^{-1}-I=(A-I)(-A^{-1})$ so the image o $A^{-k}-I$ is also contained in the image of $A-I$. $\endgroup$ – user10354138 May 22 at 4:17
  • $\begingroup$ I see it know. Got it. Thanks! $\endgroup$ – Guido A. May 22 at 5:34

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