When do $P$ $\subseteq$ $N$ $\subseteq$ $M$ and $P \cong M$ imply $N \cong M$? If $M$ is a module over a commutative ring $R$ and $P$, $N$ are submodules such that $P$ $\subseteq$ $N$ $\subseteq$ $M$ and $P$ $\cong M$ then it does not imply that $N \cong M$. A counterexample is given here.
So, I want to know under what conditions on $R$ or $M$ the result may be true.
I have noticed a more or less trivial observation that if $R$ is a PID and $M$ is free then it is true. Because we know in this case submodules are also free. So  if $P \cong M$ then $P$ and $M$ has same rank and also $rank\ P \leq rank\ N \leq rank\ M$. So we have $N \cong M$. But I can't find any other good conditions. So please help.
Thank you. 
 A: Let $(\dagger)$ denote the following property of a ring $R$:
If $P\leq N\leq M$ is any chain of $R$-modules, then $P\cong M$ implies $N\cong M$.
Thm. A ring satisfies $(\dagger)$ iff it is semisimple.
Proof.
[$\Rightarrow$]
Assume that $R$ satisfies $(\dagger)$ and that $A$ is an $R$-module. Let $I$ be an injective $R$-module containing $A$ as a submodule. Take the direct product of each factor of the chain $\{0\}\leq A\leq I$ with $\prod_{i<\omega} I$ to get  $$\{0\}\times \prod_{i<\omega} I\;\leq\; 
A\times \prod_{i<\omega} I\;\leq\; I\times \prod_{i<\omega} I.$$ 
In this chain, the first and last are isomorphic, 
so by $(\dagger)$ the middle factor $A\times \prod_{i<\omega} I$ must be isomorphic to them. The first and last are injective, so the middle factor must be injective. Since $A$ is a direct factor of this injective module, it too is injective. We have established that an arbitrary $R$-module $A$ is injective, so $R$ must be semisimple.
[$\Leftarrow$]
Now assume that $R$ is semisimple. Every $R$-module is a direct sum of simple modules, and the multiplicity of each type of simple summand is an invariant of the module. If we have $P\leq N\leq M$, and $S$ is a simple $R$-module, then the multiplicities of $S$ in each of these modules satisfies $\mu_{S}(P)\leq \mu_{S}(N)\leq \mu_{S}(M)$. Now if $P\cong M$, then they have the same multiplicities of simple summands, so $N$ must also have these multiplicities. Thus $N\cong M$, and $(\dagger)$ holds. $\Box$
