# Is every submodule of cyclic module over PID cyclic?

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?

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

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.