I'm currently considering the following question:
1) Let $_{R}V$ be an indecomposable module with a simple submodule $W$, such that $V/W$ is also simple. Is it true that W is the only nontrivial proper submodule of V?
I've come across numerous problems of this type, which all seem highly related, and while I am familiar with the basic theory required to attack such problems, I can never seem to prove them in an efficient way. Some examples of similar statements to which I am referring are:
2) If $_{R}V$ is a module of finite length with all composition factors having multiplicity one, then any two isomorphic submodules of $V$ are equal.
3) If $_{R} V$ is a module of length two with two non-isomorphic composition factors, then $V$ is cyclic.
4) If $_{R}V_{1}$ and $_{R}V_{2}$ are simple and not isomorphic, then $_{R}(V_{1} \oplus V_{2})$ is cyclic.
I am hoping someone can suggest a quick, simple approach, or hints to such approaches, for any or all of these. For context, and to give you an idea of the sort of way I've been approaching these, here's what I've considered for 1):
First, $_{R}V$ has finite length two, as:
$0 \subseteq W \subseteq V $ is a composition series. By Jordan-Holder, any composition series for $V$ must be equivalent to this one. So suppose $0 \subseteq W' \subseteq V$ is a nontrivial, proper submodule of $V$. This is in fact already a composition series, by Jordan-Holder, since it is length two and any submodule can be included as a term in a composition series. Thus $W' \cong W$ or $W' \cong V/W$, again by Jordan Holder. I claim the second case cannot hold, since if $W' \cong_{f} V/W$, then the short exact sequence $0 \rightarrow W \rightarrow V \rightarrow V/W \rightarrow 0$ would be split on the right by $f^{-1}$, which is not possible since $V$ is indecomposable by hypothesis. Supposing we've established this, we have $W\cong W'$, so I could now prove 2) to finish the claim. When I've tried to prove 2) in turn, I've needed to use 4), which I feel comfortable with in the case of commutative rings (by a slight generalization of the proof that the direct sum of cyclic groups of distinct prime orders is cyclic), but am stuck on in the general case.
Edit: I now realize 4) can be quickly proven using Jordan-Holder.
The answer to all your questions is yes, and the proofs are all pretty similar.
Let's look at 1). Assume that $V$ is indecompsable, and that both $W$ and $V/W$ are simple. Assume that $W'$ is a non-zero submodule of $V$. If $W' \neq W$, then $(W+W')/W$ is a non-trivial submodule of $V/W$, and since the latter is simple, the former has to be equal to it. Hence $W+W'=V$. Now, either $W'$ contains $W$, in which case $W'=V$, or $W'$ does not contain $W$, in which case their intersection is trivial (since $W$ is simple) and we get $V=W\oplus W'$, contradicting the indecomposability of $V$. The claim is proved.
You can prove all the other claims in similar ways. Some indications: