I want to show that for submodules $M,M'$, the submodule generated by $M\cup M'$ is equal to the submodule $M+M'$. The inclusion $M+M'\subseteq$ is easy from the closedness of the submodules under addition. However, I don't have a clue how to do the reverse inclusion.
How can I show that the set generated by $M\cup M'\subseteq M+M'$? Or is there an easier way to get this result?
The generated submodule by a set $S$ is the smallest submodule (by inclusion) which contains $S$. Now, $M+M'$ is a submodule which contains the set $M\cup M'$ (for example, if $x\in M$ then $x=x+0\in M+M'$. Similarly for $x\in M'$), so it must contain the submodule which is generated by $M\cup M'$.