Schur lemma counterexample (not simple module) I've got a question concerning this version of Schur's Lemma.
Let $V$ be an irreducible $\mathbb{C}[G]$-module.
If $\varphi : V \to V$ is a $\mathbb{C}[G]$-homomorphism, then $\varphi$ is scalar.
The proof isn't the problem, but I'm trying to find a good counterexample if $V$ is not irreducible. So far, I thought since irreducible -> indecomposable, I'd like to find a decomposable $\mathbb{C}[G]$-module, but the field has to be algebraic closed ($\mathbb{C}$). 
Suggest, I'd take the matrix $A=( \{ 0 1 \} , \{ 1 0 \} )$ as a representation matrix which is surely no scalar, but an automorphism in $V$. Do you have any idea how to construct $G$ etc. out of this?

Even with $|G|=1$, you can use the non-irreducible $\mathbb C[G]$ module $\mathbb C\times \mathbb C$ as a counterexample.
The ring of $\mathbb C$ homomorphisms of $\mathbb C\times \mathbb C$ is isomorphic to $M_2(\mathbb C) $ which contains many homomorphisms that don't act as scalars...
I think this is in line with your idea if we use $T=\begin{bmatrix}0&1\\ 1&0\end{bmatrix}$ to act on $\mathbb C\times \mathbb C$ by matrix multiplication. The transformation swaps the coordinates, and obviously you can't achieve a swap of coordinates with just scalar multiplication (just think of $(1,0)$, you can't get that to be $(0,1)$ with a scalar.)
So it is not hard to show there exist modules with nonscalar endomorphisms. A more interesting line of query is to exhibit a module which isn't indecomposable and yet all endomorphisms are scalar. That has already been done here.
