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The question is what might be the elementary divisors of a subgroup $H\leq \mathbb{Z^2}$ of index 36? I need to list all the possible options.

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

Alright, so I started to look for subgroups of $\mathbb{Z^2}$ of rank $1$. For such a subgroup $H$ there is a basis $\{m_1\alpha_1\}$ when $\{\alpha_1,\alpha_2\}$ is a basis of $\mathbb{Z^2}$. I think it is pretty clear that for any two integers $p \not= q$ the left cosets $p\alpha_2+H$ and $q\alpha_2+H$ 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 $\{m_1\alpha_1,m_2\alpha_2\}$ when $\{\alpha_1,\alpha_2\}$ is a basis of $\mathbb{Z^2}$ and $m_1|m_2$. Now I proved that for two elements $z_1=x_1\alpha_1+y_1\alpha_2$ and $z_2=x_2\alpha_1+y_2\alpha_2$ in $\mathbb{Z^2}$ we have that $z_1+H=z_2+H$ iff $x_1\equiv x_2\pmod{m_1}$ and $y_1\equiv y_2\pmod{m_2}$. 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.

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  • $\begingroup$ Yes, this is the correct conclusion. $\endgroup$ – Lord Shark the Unknown Aug 26 '18 at 11:09
  • $\begingroup$ @Lord Shark the Unknown, thanks a lot. $\endgroup$ – Mark Aug 26 '18 at 11:18

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