Definition 12.9.1. Let $\mathcal{A}$ be an abelian category. An object $A$ of $\mathcal{A}$ is said to be simple if it is nonzero and the only subobjects of $A$ are $0$ and $A$.
12.9 Jordan-Hölder
The Jordan-Hölder lemma is Lemma 12.9.7. First we state some definitions.
Definition 12.9.2. Let $\mathcal{A}$ be an abelian category.
We say an object $A$ of $\mathcal{A}$ is Artinian if and only if it satisfies the descending chain condition for subobjects.
We say $\mathcal{A}$ is Artinian if every object of $\mathcal{A}$ is Artinian.
Definition 12.9.3. Let $\mathcal{A}$ be an abelian category.
We say an object $A$ of $\mathcal{A}$ is Noetherian if and only if it satisfies the ascending chain condition for subobjects.
We say $\mathcal{A}$ is Noetherian if every object of $\mathcal{A}$ is Noetherian.
Lemma 12.9.4. Let $\mathcal{A}$ be an abelian category. Let $0 \to A_1 \to A_2 \to A_3 \to 0$ be a short exact sequence of $\mathcal{A}$. Then $A_2$ is Artinian if and only if $A_1$ and $A_3$ are Artinian.
Proof. Omitted. $\square$
Lemma 12.9.5. Let $\mathcal{A}$ be an abelian category. Let $0 \to A_1 \to A_2 \to A_3 \to 0$ be a short exact sequence of $\mathcal{A}$. Then $A_2$ is Noetherian if and only if $A_1$ and $A_3$ are Noetherian.
Proof. Omitted. $\square$
Lemma 12.9.6. Let $\mathcal{A}$ be an abelian category. Let $A$ be an object of $\mathcal{A}$. The following are equivalent
$A$ is Artinian and Noetherian, and
there exists a filtration $0 \subset A_1 \subset A_2 \subset \ldots \subset A_ n = A$ by subobjects such that $A_ i/A_{i - 1}$ is simple for $i = 1, \ldots , n$.
Proof. Assume (1). If $A$ is zero, then (2) holds. If $A$ is not zero, then there exists a smallest nonzero object $A_1 \subset A$ by the Artinian property. Of course $A_1$ is simple. If $A_1 = A$, then we are done. If not, then we can find $A_1 \subset A_2 \subset A$ minimal with $A_2 \not= A_1$. Then $A_2/A_1$ is simple. Continuing in this way, we can find a sequence $0 \subset A_1 \subset A_2 \subset \ldots $ of subobjects of $A$ such that $A_ i/A_{i - 1}$ is simple. Since $A$ is Noetherian, we conclude that the process stops. Hence (2) follows.
Assume (2). We will prove (1) by induction on $n$. If $n = 1$, then $A$ is simple and clearly Noetherian and Artinian. If the result holds for $n - 1$, then we use the short exact sequence $0 \to A_{n - 1} \to A_ n \to A_ n/A_{n - 1} \to 0$ and Lemmas 12.9.4 and 12.9.5 to conclude for $n$. $\square$
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 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\} $.
\[ 0 \subset A_1 \subset A_2 \subset \ldots \subset A_ n = A \quad \text{and}\quad 0 \subset B_1 \subset B_2 \subset \ldots \subset B_ m = A \]
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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