Abelianization of $\mathbb{Z}\ltimes_\varphi \mathbb{Z}^n$

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.

Thanks!

• Have you tried writing a presentation for your group, i.e. generators and relators? – 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 $$\mathbb{Z}\times\operatorname{coim}(A-I).$$
• 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? – Guido A. May 22 at 3:45
• 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 $A-I$. – user10354138 May 22 at 3:51
• 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? – Guido A. May 22 at 4:16
• 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 $A-I$. – user10354138 May 22 at 4:17