I have the following exercise.
Consider a homomorphism given by a multiplication by an integer matrix $A$ given by $\varphi : \mathbb{Z}^k \to \mathbb{Z}^k$. I have to prove that the image of $\varphi$ is of finite index if and only if $A$ is non-singular and the index is equal to $|\det(A)|$.
My attempt:
- "$\rightarrow$" $\operatorname{Im}(\varphi)$ is a subgroup of $\mathbb{Z}^k$ so since $\mathbb{Z}^k$ is abelian the index of the group is equal to $|\mathbb{Z}^k/\operatorname{Im}(\varphi)|=n$ with $n$ a natural number. Now I can say that, using Artin´s interpretation, $A=Q^{-1}A'P$. So $A'=QAP^{-1}$ where $A'$ is a diagonal matrix and $d_i$ with $i \in \{1, \ldots, r\}$ is a diagonal entry in row $i$ of $A'$ and $di|dj$ for each k>=j>i followed by, eventually, $k-r$ zeros. The next step is to write down $\mathbb{Z}^k/\operatorname{Im}(\varphi)=\mathbb{Z}^k/A\mathbb{Z}^k$ because I know this is true from the hypothesis. I'm getting stuck here, I don't know how could I follow. Another problem is that could possibly exist $k-r$ rows full of zeros of $A'$. Am I on the right track?
Any comment would be appreciated!
