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 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.

  • 8
    $\begingroup$ I believe many Jordan-Hölder type theorems can be deduced from the version for modular lattices. $\endgroup$ Feb 25, 2019 at 21:32
  • $\begingroup$ That's interesting, Benjamin. Do you think by chance that it could be made to include example 3 above? $\endgroup$ Feb 26, 2019 at 2:19
  • $\begingroup$ I am not so familiar with example 3 to know if the modular lattice result applies $\endgroup$ Feb 26, 2019 at 3:00
  • $\begingroup$ @DeanYoung: In example 3, what is $A$? $\endgroup$ Feb 26, 2019 at 22:11
  • 2
    $\begingroup$ The statement 3 is not correct. Just take $A=M=\mathbb Z$ and the two filtrations $0\subset M$ and $0\subset 2M\subset M$. There is the primary decomposition but that's different. $\endgroup$ Feb 27, 2019 at 18:03

3 Answers 3


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.

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 $G$.


There are various generalizations of the Jordan-Hölder theorem. Beyond groups and groups with operators it holds for any equational theory which contains a Mal'cev operation. This means that from the given operations one can form a ternary operation $m(x,y,z)$ satisfying $m(x,x,z)=z$ and $m(x,z,z)=x$ for all $x,y,z$. Every equational theory containing a group operation is an example since $m(x,y,z)=xy^{-1}z$ has the Mal'cev property. Another example not of that type would be the class of Heyting algebras. A non-example is the class of lattices.

The fact that every equational theory containing a Mal'cev operation satisfies the Jordan-Hölder theorem has been proved by Lambek in Goursat's theorem and the Zassenhaus lemma. Canad. J. Math. 10 (1958) 45–56. It is my impression that the Mal'cev condition is pretty much optimal in this context.

The JH-theorem clearly also holds in abelian categories but that is not covered by Lambek's theorem above. A common generalization has been achieved by Carboni-Lambek-Pedicchio with the notion of a Mal'cev category (in Diagram chasing in Mal’cev categories, J. Pure Appl. Algebra 69 (1991) 271–284). These are exact categories (in the sense of Barr, not Quillen!) which have the curious property that every reflexive relation is already an equivalence relation. The category of models of an equationial theory with a Mal'cev operation would be an example (that's an amusing exercise). The category should also have a $0$-object in order to be able to define kernels. Then one can show that then all objects of finite length have the JH-property. A proof has been given in F. Knop: Tensor envelopes of regular categories. Advances in Mathematics 214 (2007) 571-617 even though the fact must have been at least folklore.

  • $\begingroup$ Beautiful! Two questions: Let $X$ be a finite-length object in a Barr-exact Mal'cev category $C$. (1) If $Z_1,\dots,Z_n$ are the ordered factors of a composition series for $X$, then is any permutation $Z_{\sigma 1},\dots,Z_{\sigma n}$ also the ordered factors of a composition series for $X$? (2) Suppose $Y_1,\dots, Y_k$ each appear as the factors in various normal series for $X$. If we concatenate the factors for composition series for the $Y_i$'s and get the factors for a composition series for $X$, then is $Y_1,\dots,Y_k$ the ordered factors for a normal series for $X$? $\endgroup$
    – Tim Campion
    Feb 27, 2019 at 18:41

There's also the work of Francis Borceux and Marco Grandis,

Jordan-Hölder, modularity and distributivity in non-commutative algebra, J. Pure Appl. Algebra 208 (2007), no. 2, 665-689, doi.

There the authors prove a Jordan-Hölder theorem (4.4) for all 'weakly exact' categories (1.2).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.