Let $R$ be a ring, $M$ is $R$-module, and $N$ is submodule of $M$. Then $M$ is called a multiplication module if for each submodule $N$ of $M$ there exists an ideal $I$ of $R$ such that $N = IM$.
Let $$I=\{r\in R\vert rM\subseteq N\}.$$ (I have proved $I$ is ideal of $R$)
In this case it is easy to prove that $$IM = N.$$
To prove that, I will proof $IM\subseteq N$ and $N\subseteq IM$.
Let $a\in IM$, then $a=rm$, for $r\in R$ such that $rM\subseteq N$ and $m\in M$.
So, we have $$a=rm\in rM\subseteq N.$$ Hence, $a\in N$. So, $IM\subseteq N$.
(1) Is this proof true?
Now I confused to proof $N\subseteq IM$. Let $n\in N$, I don't know how to get $n\in IM$.
(2) Is there any hint to proof this part?