the converse of Schur lemma
Solution 1:
Let $A=\left( \begin{array}{cc} 1 & 1 \\ 0 & 1 \\ \end{array} \right)$ and $B=\left( \begin{array}{cc} 2 & 0 \\ 0 & 1 \\ \end{array} \right)$, let $F$ be the free group on generators $a$ and $b$, and consider the representation of $F$ on $\mathbb C^2$ which maps $a$ and $b$ to $A$ and $B$. This has trivial endomorphism ring, so in particular it is indecomposable, yet it is not simple: the vector $(1,0)$ spans a proper submodule.