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Determine the number submodules of $\mathbb{Z}_6 \times \mathbb{Z}_6$ as a $\mathbb{Z}_{12}$-module.

Today our professor introduced the notion of modules and gave this problem as an exercise. I have thought over it and I have come up with the following solution. Is this is correct?

My attempt:

The abelian group $\mathbb{Z}_6 \times \mathbb{Z}_6 \simeq \mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_3 \times \mathbb{Z}_3$ has 16 subgroups. Now I know that "An abelian group $A$ is a $\mathbb{Z}_m$-module iff $mA=0$. From this I conclude that each of the subgroups of $\mathbb{Z}_6 \times \mathbb{Z}_6$ should also be a submodule. Hence there are 16 submodules.

Miz
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  • I think you missed some subgroups, but other than that the argument is correct. – Tobias Kildetoft Apr 09 '18 at 09:53
  • @TobiasKildetoft Is the number of subgroups equal to 20? – Miz Apr 09 '18 at 11:00
  • I get $30$. There are $5$ subgroups of $\mathbb{Z}_2\times\mathbb{Z}_2$ and $6$ subgroups of $\mathbb{Z}_3\times\mathbb{Z}_3$ (and these groups have coprimes orders, so all subgroups correspond to a pair of subgroups). – Tobias Kildetoft Apr 09 '18 at 11:04

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