If two submodules are isomorphic, so is their quotient… conditions on the ring! In this question Isomorphic quotients by isomorphic normal subgroups it is shown that if we have isomorphic normal subgroups, their quotients need not be isomorphic.
Now, what if we take finite dimensional vector spaces, instead of groups, we do have that if two subspaces are isomorphic, then their quotients are isomorphic (simply these quotients have the same dimension, so their are isomorphic).
Now considering an arbitrary module $M$ over some ring $R$, what conditions do we have to impose on $R$ and $M$ for the above statement to be true? Do modules with this property have a name?
Thanks
 A: The statement holds for finitely generated modules over semisimple ring. More generally, the statement holds fo semisimlple finitely generated modules over arbitrary ring.
Consider a semisimple ring $R$. Then any right $R$-module $M$ is semisimple, which means that it is a direct sum of its simple submodules. If you further take $M$ to be finitely generated, then $M$ is finite direct sum of its simple submodules. (In the case of a general ring,  just consider a finitely generated semisimple module $M$. The rest works in both cases.)
Suppose you have $K_1, K_2$ two isomorphic submodules of $M$. Then $M=K_1\oplus C_1=K_2 \oplus C_2$ for suitable submodules $C_1,C_2$ of $M$ (by alternative definition of semisimple module). So you need to show that $C_1 \simeq C_2$. However, consider the decompositions of $K_1,K_2,C_1,C_2$ into direct sums of simple submodules, i. e. say 
$$K_1=\bigoplus_{i=1}^{k}S_i, \; K_2=\bigoplus_{i=1}^{k}S'_i, \; C_1=\bigoplus_{j=1}^{m}T_j,\; C_2=\bigoplus_{j=1}^{n}T'_j, $$
where $S_i,S'_i, T_i, T'_i$ are simple submodules of $M$.
Then
$$\bigoplus_{i=1}^{k}S_i\oplus\bigoplus_{j=1}^{m}T_j = M = \bigoplus_{i=1}^{k}S'_i\oplus\bigoplus_{j=1}^{n}T'_j, $$
are two decompositions of $M$ into sum of simple submodules, from which follows (after using the fact, that $K_1 \simeq K_2$ and, basically, Jordan-Holder theorem) that $m=n$ and the summands $T_j, T'_j$ are (in some suitable bijection of indices) isomorphic.
