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

We prove existence by induction on the composition length from [F1]. If $M=0$ or $M$ is indecomposable, there is nothing to do. Otherwise $M=M_1\oplus M_2$ with both summands nonzero and of strictly smaller length than $M$. Applying the induction hypothesis to $M_1$ and $M_2$ yields a finite decomposition of $M$ into indecomposable summands.

Let $X$ be an indecomposable finite-length module and $f\in \mathrm{End}(X)$. Since $X$ has finite length, the ascending chain of kernels and descending chain of images of the powers of $f$ stabilize. For large $n$ one has $X=\ker (f^n)\oplus \mathrm{im}(f^n)$. Because $X$ is indecomposable, either $\ker (f^n)=0$ and $f$ is invertible, or $\mathrm{im}(f^n)=0$ and $f$ is nilpotent. In the second case $1_X-f$ is invertible by the finite geometric series. So the endomorphism ring of an indecomposable finite-length module is local.

Suppose $M\cong X_1\oplus \cdots \oplus X_r\cong Y_1\oplus \cdots \oplus Y_s$ with all $X_i$ and $Y_j$ indecomposable. Restrict the identity of $M$ to $X_1$ and write it as the sum of the composites $X_1\to Y_j\to X_1$. Because $\mathrm{End}(X_1)$ is local by step 2.1, one of these composites is invertible; therefore the corresponding map $X_1\to Y_j$ is an isomorphism. Cancel that isomorphic summand from both decompositions and apply induction on the composition length of the complement. This proves $r=s$ and uniqueness up to permutation and isomorphism.