Finite-dimensional kG-modules decompose as finite direct sums of indecomposables uniquely up to order and isomorphism

Prove existence by induction on $d={\dim }_{k}M$. If $M=0$ or $M$ is indecomposable there is nothing to split. Otherwise $M=M_1\oplus M_2$ with both summands nonzero. By dimension additivity, $0<{\dim }_k{M}_i<d$ for each $i$, so the induction hypothesis decomposes each $M_i$ into finitely many indecomposables. Their combined list decomposes $M$.

Let $X$ be a nonzero indecomposable finite-dimensional module and $f\in \mathrm{End}(X)$. The dimensions of $\ker f^n$ increase and are bounded by ${\dim }_{k}X$, so for some $n\ge 1$, $\ker f^n=\ker f^{2n}$. If $y=f^n(x)$ lies in $\ker f^n$, then $x\in \ker f^{2n}=\ker f^n$ and $y=0$. Rank-nullity therefore gives $X=\ker f^n\oplus \mathrm{im}{f}^n$. Both are submodules. Indecomposability makes one summand zero: in the first case $f$ is injective and hence invertible in finite dimension; in the second $f^n=0$. Thus every endomorphism of $X$ is invertible or nilpotent.

A nilpotent $f$ has inverse $1+f+\cdots +f^{n-1}$ for $1-f$. If nonunits $a,b$ of $\mathrm{End}(X)$ had unit sum $u=a+b$, then $u^{-1}a$ and $u^{-1}b$ would still be nonunits and would sum to $1$; the first is nilpotent by step 1.2, so the second $1-u^{-1}a$ would be a unit, a contradiction. Thus the nonunits are closed under finite sums. This is the only endomorphism-ring property needed below; no commutativity is assumed.

For uniqueness induct on ${\dim }_{k}M$. Suppose $M=X\oplus A=Y_1\oplus \cdots \oplus Y_s$, with $X$ and all $Y_j$ nonzero indecomposable. The identity of $X$ is the sum of composites $e_j:X\hookrightarrow M\twoheadrightarrow Y_j\hookrightarrow M\twoheadrightarrow X$. By step 2.1 at least one $e_j$ is a unit. Write its two factors as $a:X\to Y_j$ and $b:Y_j\to X$, so $ba$ is invertible. Then $a(ba{)}^{-1}$ splits the surjection $b$, giving $Y_j=a(X)\oplus \ker b$. Indecomposability of $Y_j$ forces $\ker b=0$, hence $X\cong Y_j$ and $b$ is an isomorphism.

Put $B={⨁}_{i\mathrm{\ne }j}{Y}_i$. Since the projection $M\to X$ restricts to the isomorphism $b:Y_j\to X$, the projection $M\to B$ restricts to an isomorphism $A\to B$: its kernel is $A\cap Y_j=0$, and for every $z\in B$ subtract from $z$ the unique $y\in Y_j$ with the same $X$-projection to obtain a preimage in $A$. Thus $A\cong B$. Both have dimension smaller than $M$; induction matches their indecomposable decompositions. Adding $X\cong Y_j$ proves uniqueness up to permutation and isomorphism.