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I am doing some self-study on $\mathbb{Z}$-modules and the notes I am following propose the following problem:

Let $f : \mathbb{Z}^2 \rightarrow \mathbb{Z}^2$ be defined as $f((x; y)) = (28x + 38y; 12x + 16y)$. Find the index of $\text{Im} (f)$ in $\mathbb{Z}^2$ and describe $\mathbb{Z}^2/\text{Im} (f)$.

I am not sure where to begin - can anyone help me?

Alessio K
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mat95
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    The Smith Normal Form of a rectangular matrix over a PID, the invariant factors of a finitely generated submodule of a free module over a PID, the theory of finitely generated modules over PID's are the key areas you should seek to become familiar with in order to solve the above problem. The index of the image of $f$ will be equal to the absolute value of the determinant of $f$. – ΑΘΩ Sep 29 '20 at 13:04

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For $\begin{bmatrix}u\\ v\end{bmatrix}\in\operatorname{Im}(f)\subset\Bbb{Z}^2$, by definition there exists $\begin{bmatrix}x\\ y\end{bmatrix}\in\Bbb{Z}^2$ such that $$\begin{bmatrix}28&38\\ 12&16\end{bmatrix}\begin{bmatrix}x\\ y\end{bmatrix}=\begin{bmatrix}u\\ v\end{bmatrix}.$$ From this it follows immediately that $v\equiv0\pmod{4}$ and $u\equiv0\pmod{2}$.

Can you show conversely that every pair $\begin{bmatrix}u\\ v\end{bmatrix}\in\Bbb{Z}^2$ satisfying these congruences is contained in $\operatorname{Im}(f)$? What does this tell you about the index of $\operatorname{Im}(f)$ in $\Bbb{Z}^2$ and the structure of the quotient?

Servaes
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  • The quotient should be $\mathbb{Z}^2 / \mathbb{Z}_4 \times \mathbb{Z}_2$ and the index of Im(f) seems to be the index of $\mathbb{Z}_4 $ in $\mathbb{Z}$ multiplied by the index of $\mathbb{Z}_2$ in $\mathbb{Z}$. – mat95 Sep 29 '20 at 14:23
  • No wait, 4 and 2 are not coprime so I cannot use the chinese remainder thm to "split" the quotient – mat95 Sep 29 '20 at 14:26
  • What do you mean by $\Bbb{Z}_4$ and $\Bbb{Z}_2$? – Servaes Sep 29 '20 at 14:26
  • I mean $\mathbb{Z}/4\mathbb{Z}$ and $\mathbb{Z}/2\mathbb{Z}$ – mat95 Sep 29 '20 at 14:43
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    These aren't even subgroups of $\Bbb{Z}$... Consider what the cosets of $\operatorname{Im}(f)$ in $\Bbb{Z}^2$ look like: They can be represented as $[u,v]+\operatorname{Im}(f)$, so given the congruences above, how many cosets are there (at most)? – Servaes Sep 29 '20 at 14:57
  • we have 8 cosets – mat95 Sep 29 '20 at 15:06
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    Great! Have you convinced yourself that these 8 cosets are all distinct? Then you have the index of $\operatorname{Im}(f)$ in $\Bbb{Z}^2$. The quotient is then an abelian group of that order. There aren't many of those, and I think you can guess which one it should be. – Servaes Sep 29 '20 at 15:22
  • Oh! I understand. Thank you so much! – mat95 Sep 29 '20 at 15:24