In addition to the Jordan-Holder theorem for groups, there are various Jordan-Holder Theorems for other categories:
Finite dimensional representations have filtrations whose associated graded consists of irreducible representations. Any other such associated graded is the same up to permutation of its elements.
Artinian modules have filtrations whose associated graded consists of simple modules. Any other such associated graded is the same up to permutation of its elements.
Finitely generated $A$-modules have filtrations whose associated graded consists of modules of the form $A / p$ for a prime $p$. Any other such associated graded is the same up to permutation of its elements.
There is a commonality to the proofs of these as well. I am wondering if someone has come up with a categorical version of the Jordan-Holder theorem, which in a sense encompasses these ones.
For instance, one might try to look at the category of subquotients of an object $X$ in a category $C$. Objects are compositions $X \twoheadrightarrow Y \hookrightarrow Z$, and morphisms are the expected commutative diagrams.
For a moment, assume that $C$ is abelian. Then, to construct the direct sum of $X \twoheadrightarrow Y \hookrightarrow Z$ and $X \twoheadrightarrow Y' \hookrightarrow Z'$, start by taking the pushout $X \twoheadrightarrow Y''$ of $X \twoheadrightarrow Y$ and $X \twoheadrightarrow Y''$. Then take the pullback of $Z \rightarrow Y \rightarrow Y''$ and $Z' \rightarrow Y' \rightarrow Y''$ to get $Z'' \rightarrow Y''$. $X \rightarrow Y'' \rightarrow Z''$ is the desired direct sum.
This construction can be used to refine filtrations. The Jordan-Holder Theorem for modules then asserts that, for two filtrations with simple subquotients $F$ and $G$, we can take a mutual refinement $F \vee G$. The mutual refinement, after discarding its redundant elements, $F \vee G$ is the same as both $F$ and $G$.
So, this doesn't quite apply to groups because my construction used that the category is abelian. But that doesn't seem too far off. I am particularly interested to see if anyone can make something like this work for the third example above.