Suppose $M$ is a completely reducible left $R$-module for a ring $R$ and $I$ is an ideal of $R$. Then prove that the following are equivalent:
$M=IM$
If $I$ annihilates an element $x \in M$, then $x=0$
For all $x\in M, x$ is contained in $Ix$
The implication $(3)\rightarrow(1)$ is simple. For $(2)\rightarrow(3)$, I took a projection from $Rx$ to $Ix$ ($Rx$ and $Ix$ are themselves completely reducible and hence this map is surjective), and tried showing it is 1-1, and hence a module isomorphism. For $(1)\rightarrow(2)$, I tried showing that the ideal $I$ cannot annihilate any non-zero $x\in IM$.
In both these cases, I have been stuck for hours with no progress. Any hints would be appreciated.