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I needed to check if any subgroup of cyclic abelian group is cyclic and successfully proved that "yes" using the fact that $\mathbb{Z}$ is Euclidian domain. It's easy to give an example that it doesn't hold for modules over arbitrary ring, but I am pretty sure that it does for PID. How can I prove it?

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    $\begingroup$ $M=R/I$ is a cyclic module, a submodule has the form $J/I$ with $J\supseteq I$ ideal of $R$, $J$ is principal, so... $\endgroup$
    – user26857
    Apr 30, 2017 at 15:55
  • $\begingroup$ @user26857 is absolutely correct. This is the best way to prove the statement. $\endgroup$ Apr 30, 2017 at 15:56
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    $\begingroup$ @user26857, thanks, I got it. $\endgroup$
    – Vladislav
    Apr 30, 2017 at 16:07

1 Answer 1

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This was answered in the comments, but I will write it out here to mark this as "answered" :)

Every cyclic $R$-module is of the form $R/I$ for some ideal $I$ of $R$. Submodules correspond to ideals and vice versa. Every ideal of $R/I$ is of the form $J/I$ for some ideal $J$ in $R$, and since $J$ is principal (generated by $x$, say), $J/I$ is also principal (generated by $x + I$), i.e., it is cyclic.

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