Proof of Jordan-Hölder for Modules carries over for Groups? The Book [Auslander, Reiten - Representation theory of Artin algebras] begins with the Jordan-Hölder theorem for modules of finite length over arbitrary rings. The proof is probably quite standard - here is the idea:

Define the length of a module $M$ and the multiplicities of its composition factors as minimal length and minimal multiplicities over all (generalized) composition series. Then show that these functions are additive with respect to short exact sequences. The Jordan-Hölder theorem now follows easily by induction on the length of $M$:
For $l(M) \leq 1$ the statement holds clearly. If $l(M) \geq 2$ there is a submodule $0 \lneq U \lneq M$. Any (generalized) composition series of $M$ splits into a (generalized) composition series of $U$ and of $M/U$. By induction hypothesis, those sequences satisfy the claim, i.e. they have length $l(U)$ and $l(M/U)$, respectively, and certain factor multiplicities defined by $U$ and $M/U$. By additivity of the length function and the multiplicity functions shown before, the claim also holds for the chosen composition series of $M$.

I wonder whether this proof can be adopted verbatim to prove the Jordan-Hölder theorem for groups. At first sight, I see no reason why this cannot be done. However, I haven't seen this proof in any source concerning groups (usually, the Zassenhaus lemma is used instead).
 A: The answer to your question is that, yes, you are right. Here are some more details.
Let $G$ be a group. Define a composition series of $G$ to be a sequence of subgroups
$$\{ e \} = G_0 \triangleleft G_1 \triangleleft G_2 \triangleleft \cdots \triangleleft G_n = G$$
such that each quotient $G_{j+1}/G_j$ is simple; and define a generalized composition series similarly where we allow the factors to be either simple or trivial.  Obviously, a generalized composition series can be turned into a composition series by collapsing the repetitions.
A group $G$ is said to have finite length if it has a composition series, in which case $\ell(G)$ is the minimum length of a composition series. Given a simple group $\Gamma$ and a generalized composition series $G_{\bullet}$ for a group $G$, we define $m(\Gamma, G_{\bullet})$ to be the number of indices $j$ such that $\Gamma \cong G_{j+1}/G_j$. The Jordan-Holder theorem is that $m(\Gamma, G_{\bullet})$ depends only on the group $G$ and not on the generalized composition series $G_{\bullet}$.
Let $1 \to A \overset{\alpha}{\longrightarrow} B \overset{\beta}{\longrightarrow} C \to 1$ be a short exact sequence of groups.
The key to the proof you discuss (pages two and three here) is to have recipes to translate between (1) pairs of generalized composition series $(A_{\bullet}, C_{\bullet})$ in $A$ and $C$ and (2) generalized composition series $B_{\bullet}$ in $B$. Namely:
Proposition 1: Let $A_0 \triangleleft A_1 \triangleleft \cdots \triangleleft A_a$ and $C_0 \triangleleft C_1 \triangleleft \cdots \triangleleft C_c$ be generalized composition series for $A$ and $C$. Then
$$\{ e \} = \alpha(A_0) \triangleleft \alpha(A_1) \triangleleft \cdots \triangleleft \alpha(A_a) = \beta^{-1}(C_0) \triangleleft \beta^{-1}(C_1) \triangleleft \cdots \triangleleft \beta^{-1}(C_c) = B$$
is a generalized composition series for $B$. Moreover, $\alpha(A_{j+1})/\alpha(A_j) \cong A_{j+1}/A_j$ and $\beta^{-1}(C_{j+1})/\beta^{-1}(C_j) \cong C_{j+1}/C_j$.
Proposition 2 Let $B_0 \triangleleft B_1 \triangleleft \cdots \triangleleft B_b$ be a generalized composition series for $B$. Set $A_j = \alpha^{-1}(B_j)$ and $C_j = \beta(B_j)$. Then $A_{\bullet}$ and $C_{\bullet}$ are generalized composition series. If $B_{j+1} = B_j$ then $A_{j+1} = A_j$ and $C_{j+1}=C_j$; if $B_{j+1}/B_j$ is simple, then exactly one of $A_{j+1}/A_j$ and $C_{j+1}/C_j$ is trivial and the other is isomorphic to $B_{j+1}/B_j$.
The proofs of these results are slightly harder for groups than for modules, because you need to check at each step that the subgroups which are supposed to be normal actually are. But I think they are still at the level of reasonable exercises.
Once you have this, the rest of the proof is exactly the same in both cases. Using Prop 1, if $A$ and $C$ are finite length, then so is $B$, and $\ell(B) \leq \ell(A)+\ell(C)$. So we may inductively assume Jordan-Holder is known for $A$ and $C$ and deduce it for $B$. For any simple group $\Gamma$ and any generalized composition series $B_{\bullet}$ of $B$, Prop. 2 constructs generalized composition series $A_{\bullet}$ and $C_{\bullet}$ with $m(\Gamma, B_{\bullet}) = m(\Gamma, A_{\bullet})+m(\Gamma, C_{\bullet})$. Since, inductively, $m(\Gamma, A_{\bullet})$ and $m(\Gamma, C_{\bullet})$ depend only on $A$ and $C$, we deduce that $m(\Gamma, B_{\bullet})$ depends only on $B$. $\square$.
I like this proof! I might use it next time I teach Jordan-Holder!
