Let $M$ and $N$ be two left modules over some ring, both with finite length, so they admit composite series and we denote the set of Jordan-Hölder factors (I mean the graded pieces modulo isomorphism classes) of $M$ and $N$ by $JH(M)$ and $JH(N)$, respectively.
If $f:M \to N$ is a nonzero homomorphism, then we can show easily that $JH(M)\cap JH(N)\neq \emptyset$. My question is: does the converse hold?
If the underlying ring is $\mathbb{Z}$, i.e. $M, N$ are finite abelian groups, then the result holds easily by the structure theorem. Note that the analogous questions for finite (nonabelian) groups does not hold, for example $\mathbb{Z}/3\mathbb{Z}$ and $S_3$.
There are such examples with no nonzero module homomorphisms between them. Here is an example - there may be easier ones.
We let $M$ and $N$ be $KG$-modules of dimension $4$, with $K$ the field of order $2$ and $G = {\rm GL}(3,2)$, the simple group of order $168$.
Both $M$ and $N$ are uniserial and have the trivial $1$-dimensional module $T$ as their unique minimal submodule (so $M$ and $N$ have a common composition factor), and $M/T$ and $N/T$ are simple of dimension $3$ but not isomorphic to each other: the natural module for $G$ and its dual.
Alternatively we could take the duals of $M$ and $N$, which are the other way up.