Vtyjkyuk

The question is what might be the elementary divisors of a subgroup $Hleq mathbbZ^2$ of index 36? I need to list all the possible options.

My solution: The theorem states that if $A$ is an abelian free group, $Bleq A$ a subgroup then we can find a basis $alpha_1,...,alpha_n$ of $A$, a number $0leq kleq n$ and natural numbers $m_1,...,m_k$ such that $m_1|m_2|...|m_k$ and $m_1alpha_1,...,m_kalpha_k$ is a basis of $B$. The numbers $m_1,...,m_k$ are called elementary divisors.

Alright, so I started to look for subgroups of $mathbbZ^2$ of rank $1$. For such a subgroup $H$ there is a basis $m_1alpha_1$ when $alpha_1,alpha_2$ is a basis of $mathbbZ^2$. I think it is pretty clear that for any two integers $p not= q$ the left cosets $palpha_2+H$ and $qalpha_2+H$ are different, so the index of $H$ is infinite. (and not 36)

So now I look for subgroups of rank $2$. Let $H$ be such a subgroup. Then it has a basis $m_1alpha_1,m_2alpha_2$ when $alpha_1,alpha_2$ is a basis of $mathbbZ^2$ and $m_1|m_2$. Now I proved that for two elements $z_1=x_1alpha_1+y_1alpha_2$ and $z_2=x_2alpha_1+y_2alpha_2$ in $mathbbZ^2$ we have that $z_1+H=z_2+H$ iff $x_1equiv x_2pmodm_1$ and $y_1equiv y_2pmodm_2$. So that way we get that the index of $H$ (the number of different left cosets) equals to $m_1m_2$.

So the elementary divisors of a subgroup of index 36 must be $m_1,m_2$ such that $m_1|m_2$ and $m_1m_2=36$. That way we get that all the possible options for $(m_1,m_2)$ are $(1,36),(2,18),(3,12),(6,6)$.

Is my solution correct? I'm asking because this is the first time I see such a question. It is from a group theory exam and there are no solutions there.