In addition to the Jordan-Holder theorem for groups, there are various Jordan-Holder Theorems for other categories:


1. 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.
2. 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.
3. (I have changed this example from the original one) [Finite dimensional hopf algebras](https://arxiv.org/pdf/1407.0931.pdf) have filtrations whose associated graded consists of simple hopf algebras. 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 $\text{SubQuot}(X)$ of an object $X$ in a category $C$. Objects in this category are pairs $(Z, Y)$ with $Y \in \text{Quot}(X)$ and $Z \in \text{Sub}(Y)$. Morphisms $f : (Y, Z) \rightarrow (Y', Z')$ are pairs of maps $Y \twoheadrightarrow Y'$ in $\text{Quot}(X)$ and $Z' \rightarrow \text{im}(Z \rightarrow Y \rightarrow Y')$ in $\text{Sub}(Y')$.


Say a filtration is then a sequences of subquotients $(Y\_i, Z\_i)$ where $0 \rightarrow Z\_i \rightarrow Y\_i \rightarrow Y\_{i+1} \rightarrow 0$ is exact. Say a simple object is one without nontrivial quotient objects.


We can represent a pair $(Z, Y)$ with $(Z\_X, Y)$, where $Z\_X$ is the pullback of $Z \rightarrow Y$ along $X \rightarrow Y$. To construct the coproduct $(Y, Z) \amalg (Y', Z')$ of $(Y, Z)$ and $(Y', Z')$, we simply take the pushout $Y''$ of $Y$ and $Y'$ in $C$ (coproduct in $\text{Quot}(X)$), and the pullback $Z''\_X$ of $Z\_X$ and $Z'\_X$ in $C$ (product in $\text{Sub}(X)$). The coproduct is represented by the pair $(Y'', Z''\_X)$.


Coproducts can be used to refine filtrations. The Jordan-Holder Theorem for modules then asserts that, for two filtrations $\{ (Z\_i, Y\_i) \}\_{i = 1}^n$ and $\{ (Z\_j', Y\_j') \}\_{j = 1}^m$ with simple (no nontrivial quotient objects) subquotients $Z\_i$ and $Z\_j'$, we can take a mutual refinement $\{ (Z\_i, Y\_i) \amalg (Z\_j', Y\_j') \}\_{1 \leq i \leq n, 1 \leq j \leq m }$. The mutual refinement, after discarding its redundant elements, has the same subquotients as both $\{ (Z\_i, Y\_i) \}\_{i = 1}^n$ and $\{ (Z\_j', Y\_j') \}\_{j = 1}^m$, since a filtration by simple subquotients should have only the trivial refinements.


I think this works for groups and modules. However, I am particularly interested to see if anyone can make something like this work for the third example above, which seems harder, because I don't quite see how it fits in with the rest.



Categorical Unification of Jordan Holder Theorems
This does not really involve any category theory, but perhaps it is useful to note the following general setting for the Jordan-Hölder theorem.


For $G$ a group and $\Omega$ a set, a *group with operators* is $(G, \Omega)$ equipped with an action $\Omega \times G \rightarrow G$: $(\omega,g) \mapsto g^\omega$ such that $(gh)^{\omega} = g^{\omega} h^{\omega}$ for all $\omega \in \Omega$ and $g, h \in G$. 


See the Wikipedia article for more information about groups with operators: [link](https://en.wikipedia.org/wiki/Group_with_operators).


The point is that the Jordan-Hölder theorem holds for a group with operators, and it seems to have most Jordan-Hölder type theorems as a special case. These include for example the Jordan-Hölder theorems for groups and modules over a ring. Also, by taking $\Omega = G$ with conjugation action, you get results about chief series and chief factors of $G$.


