Krull-Schmidt theorem

To discuss the Krull-Schmidt theorem, we must first define the ascending chain condition and descending chain condition.

\begin{definition} A group $G$ is said to satisfy the ascending chain condition (ACC) on subgroups iff for every chain $G_1 \leq G_2 \leq \cdots$ of subgroups of $G$ there is an integer $n$ such that $G_i = G_n$ for all $i \geq n$. $G$ is said to satisfy the descending chain condition (DCC) on subgroups iff for every chain $G_1 \geq G_2 \geq \cdots$ of subgroups of $G$ there is an integer $n$ such that $G_i = G_n$ for all $ i \geq n$. We say $G$ satisfies ACC on normal subgroups if every increasing sequence of normal subgroups of $G$ eventually becomes constant. Similarly with DCC on normal subgroups. \end{definition} \begin{theorem} If a group $G$ satisfies either ACC or DCC on normal subgroups, then $G$ is the direct product of a finite number of indecomposable subgroups. \end{theorem}

Proof. Suppose $G$ is not a finite direct product of indecomposable subgroups. Let $S$ be the set of all normal subgroups $H$ of $G$ such that $H$ is a direct factor of $G$ and $H$ is not a finite direct product of indecomposable subgroups. Clearly $G \in S$. If $H \in S$, then $H$ is not indecomposable, whence there exist proper subgroups $K_H$ and $J_H$ of $H$ such that $H = K_H \times J_H$. Furthermore, one of these groups, say $K_H$, must lie in $S$. Let $f: S \to S$ defined by $f(H) = K_H$. There exists a function $\phi: \mathbb{N} \cup \{0\} \to S$ such that $\phi(0) = G$ and $\phi(n+1) = f(\phi(n)) = K_{\phi(n)}$. Denoting $\phi (n)$ by $G_n$, we have a sequence of normal subgroups $G_0, G_1, G_2, \cdots$ of $G$ such that $G > G_1 > G_2 > \cdots$. If $G$ satisfies the DCC on normal subgroups, this is a contradiction. A routine inductive argument shows $\forall n \in \mathbb{N}$, $$G = G_n \times J_{G_{n-1}} \times J_{G_{n-2}} \times \cdots \times J_{G_0}$$ with each $J_{G_i}$ a proper subgroup of $G$. Thus there is a properly ascending chain of normal subgroups: $$J_{G_0} < J_{G_1} \times J_{G_0} < J_{G_2} \times J_{G_1} \times J_{G_0} < \cdots$$ If $G$ satisfies the ACC on normal subgroups, this is a contradiction. ◻

While it was easy to prove the existence of such a decomposition, proving that such a decomposition is unique turns out to be much more challenging. Unlike in the existence case, proving uniqueness requires both ACC and DCC on normal subgroups to hold.

\begin{definition} An endomorphism $f$ of a group $G$ is called a normal endomorphism iff $\forall a, b \in G, af(b) a^{-1} = f(aba^{-1})$. \end{definition} \begin{lemma} Let $G$ be a group satisfying the ACC (resp. DCC) on normal subgroups and $f$ a normal endomorphism of $G$. Then $f$ is an automorphism iff $f$ is an epimorphism (resp. monomorphism). \end{lemma}

Proof. Suppose $G$ satisfies the ACC and $f$ is an epimorphism. The ascending chain of normal subgroups $$\langle e \rangle \leq \operatorname{ker} f \leq \operatorname{ker} f^2 \leq \cdots$$ must eventually become constant, say at $n$. Since $f$ is an epimorphism, so is $f^n$. If $a \in G, f(a) = e, a = f^n (b)$ for some $b \in G$ and $e = f(a) = f^{n+1}(b)$. $b \in \operatorname{ker} f^{n+1} = \operatorname{ker} f^n$ so $a = f^n (b) = e$. Thus $f$ is a monomorphism. Next suppose $G$ satisfies the DCC and $f$ is a monomorphism. $\forall k \in \mathbb{N}, \operatorname{im} f^k$ is normal since $f$ is a normal endomorphism. The descending chain of normal subgroups $$G \geq \operatorname{im} f \geq \operatorname{im} f^2 \geq \cdots$$ must become constant, say at $n$. $\forall a \in G, f^n(a) = f^{n+1}(b)$ for some $b \in G$. Since $f$ is a monomorphism, so is $f^n$ and hence $f^n (a) = f^n (f(b)), a = f(b)$. Thus $f$ is an epimorphism. ◻

\begin{lemma} If $G$ is a group that satisfies both the ACC and DCC on normal subgroups and $f$ is a normal endomorphism of $G$, then for some $n \in \mathbb{N}, G = \operatorname{ker} f^n \times \operatorname{im} f^n$. \end{lemma}

Proof. Consider the two chains of normal subgroups: $$G \geq \operatorname{im} f \geq \operatorname{im} f^2 \geq \cdots, \quad \langle e \rangle \leq \operatorname{ker} f \leq \operatorname{ker} f^2 \leq \cdots$$ By hypothesis there is an $n$ such that $\operatorname{im} f^k = \operatorname{im} f^n, \operatorname{ker} f^k = \operatorname{ker} f^n$ for all $k \geq n$. Suppose $a \in \operatorname{ker} f^n \cap \operatorname{im} f^n$. Then $a = f^n (b)$ for some $b \in G, f^{2n}(b) = f^n(f^n(b)) = f^n(a) = e$. Thus $b \in \operatorname{ker} f^{2n} = \operatorname{ker} f^n$ so $a = f^n(b) = e$. Thus $\operatorname{ker} f^n \cap \operatorname{im} f^n = \langle e \rangle$. $\forall c \in G, f^n (c) \in \operatorname{im} f^n = \operatorname{im} f^{2n}$ so $f^n(c) = f^{2n}(d)$ for some $d \in G$. $f^n (c f^n ( d^{-1})) = f^n (c) f^{2n} (d)^{-1} = e$ thus $c f^n (d^{-1}) \in \operatorname{ker} f^n$. Since $c = c f^n (d^{-1}) f^n (d)$, $G = \operatorname{ker} f^n \times \operatorname{im} f^n$. ◻

\begin{definition} An endomorphism $f$ of a group $G$ is said to be nilpotent iff $\exists n \in \mathbb{N}, \forall g \in G, f^n (g) = e$. \end{definition} \begin{corollary} If $G$ is an indecomposable group that satisfies both the ACC and DCC on normal subgroups and $f$ is a normal endomorphism of $G$, then either $f$ is nilpotent or $f$ is an automorphism. \end{corollary}

Proof. $\exists n \in \mathbb{N}, G = \operatorname{ker} f^n \times \operatorname{im} f^n$. Since $G$ is indecomposable, either $\operatorname{ker} f^n = \langle e \rangle$ or $\operatorname{im} f^n = \langle e \rangle$. The latter implies $f$ is nilpotent. If $\operatorname{ker} f^n = \langle e \rangle$, then $\operatorname{ker} f = \langle e \rangle$ and $f$ is a monomorphism, which implies that $f$ is an automorphism. ◻

We define some unconventional notation: if $G$ is a group and $f, g: G \to G$ are functions, let $f + g: G \to G$ be defined by $a \mapsto f(a) g(a)$. With $0_G: G \to G$ given by $a \mapsto e$, $G^G$ is a group under $+$.

\begin{corollary} Let $G \neq \langle e \rangle$ be an indecomposable group satisfying both ACC and DCC on normal subgroups. If $f_1, \cdots, f_n$ are normal nilpotent endomorphisms of $G$ such that every $f_{i_1} + \cdots + f_{i_r} (1 \leq i_1 < i_2 < \cdots < i_r \leq n)$ is an endomorphism, then $f_1 + f_2 + \cdots + f_n$ is nilpotent. \end{corollary}

Proof. Since each $f_{i_1} + \cdots + f_{i_r}$ is a normal endomorphism, the proof will follow by induction once the $n=2$ case is established. If $f_1 + f_2$ is not nilpotent, it is an automorphism. Note that the inverse $g$ of $f_1 + f_2$ is a normal automorphism. If $g_1 = f_1 \circ g, g_2 = f_2 \circ g$, then $\text{id}_G = g_1 + g_2$ and $\forall x \in G, x^{-1} = (g_1 + g_2)(x^{-1}) = g_1(x^{-1})g_2(x^{-1})$. Thus $x = g_2(x) g_1 (x) = (g_2 + g_1)(x)$ and $\text{id}_G = g_2 + g_1$. Thus $g_1 + g_2 = g_2 + g_1 = \text{id}_G$ and $g_1 \circ (g_1 + g_2) = (g_1 + g_2) \circ g_1$ implying $g_1 \circ g_2 = g_2 \circ g_1$. An inductive argument shows that $(g_1 + g_2)^m = \sum_{i=0}^m \binom{m}{i} g_1^i g_2^{m-i}$. Since each $f_i$ is nilpotent, $g_i = f_i \circ g$ has a nontrivial kernel, whence