Lemma 12.9.7 (Jordan-Hölder). Let $\mathcal{A}$ be an abelian category. Let $A$ be an object of $\mathcal{A}$ satisfying the equivalent conditions of Lemma 12.9.6. Given two filtrations

$0 \subset A_1 \subset A_2 \subset \ldots \subset A_ n = A \quad \text{and}\quad 0 \subset B_1 \subset B_2 \subset \ldots \subset B_ m = A$

with $S_ i = A_ i/A_{i - 1}$ and $T_ j = B_ j/B_{j - 1}$ simple objects we have $n = m$ and there exists a permutation $\sigma$ of $\{ 1, \ldots , n\}$ such that $S_ i \cong T_{\sigma (i)}$ for all $i \in \{ 1, \ldots , n\}$.

Proof. Let $j$ be the smallest index such that $A_1 \subset B_ j$. Then the map $S_1 = A_1 \to B_ j/B_{j - 1} = T_ j$ is an isomorphism. Moreover, the object $A/A_1 = A_ n/A_1 = B_ m/A_1$ has the two filtrations

$0 \subset A_2/A_1 \subset A_3/A_1 \subset \ldots \subset A_ n/A_1$

and

$0 \subset (B_1 + A_1)/A_1 \subset \ldots \subset (B_{j - 1} + A_1)/A_1 = B_ j/A_1 \subset B_{j + 1}/A_1 \subset \ldots \subset B_ m/A_1$

We conclude by induction. $\square$

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