Index of image subgroup of $\mathbb{Z}^2$ by a linear transform

Question

The image of $\mathbb{Z}^2$ by a linear transform given in canonical basis by $T = \begin{bmatrix}1&-3\\1&2\end{bmatrix}$ is a subgroup of $\mathbb{Z}^2$ of index:

Answer: 5.

How do I do this? For what I searched, $\rm{ind}(T)=\rm{dimKer}T - \rm{dimCok}T$, but by gaussian elimination I have that $\rm{Ker}T=\{0\}$, so this difference can't be 5.I think index of linear transform is different than index of image as a subgroup. But then I'm really lost, allI know that may help is Lagrange theorem..

Thanks.

Answer 1

In our case try and look at point not in the image, we have here that a point $(c,d)$ is the image of $(a,b)$ iff $T(a,b)=(c,d)$ this leads us the the equation $c+5b=d$ i.e. $c\sim d$ mod$_5$.

So we obtain that $(c,d)$ is in Im($T$) iff $c\sim d$ mod$_5$.

This leads us to co-sets $(0,i)$ for $i\in \{0,1,2,3,4\}$ obviously $(0,i) \nsim (0,j)$ for $i\ne j$ and any $(a,b)\in \mathbb{Z}^2$ is conjugate to one of then since $(0,b)\sim (i_a,i_b)\sim (0,i_a-i_b)$ where $i_a+5k=a$