Question: $V$ is a representation of $G$ over $K$. $V$ is not necessarily semisimple. If $End_G(V) = K$, then is $V$ irreducible?
($K$ is infinite, if it matters. So one can assume that $K$ is algebraically closed, since under my hypothesis that $End_G(V)= K$, irreducibility can be checked after base changing to algebraic closure.)
If $V$ is semi-simple, then there is a projector in $End_G(V)$, which is not in $K$ because it is not zero and it is not invertible. So the interesting case is when $V$ is not semi simple.
I've looked at some examples of representations of groups which are not completely reducible and non-irreducible, and in them additional elements pop up in the commutator of $G$, $End_G(V)$. (I was looking at them hoping to compute that $End_G(V) = K$.)
(details of the examples: there is the defining representation of the unipotent upper triangular matrices in $SL_2$. This group is Abelian, so there appear other elements in $End_G(V)$. I think the Heisenberg group, say over $\mathbb{C}$, provides another example - the defining representation is not irreducible, not semi-simple, and elements in the center of the Heisenberg group act as intertwiners. One can generalize this to the defining representation of upper triangular matrices with ones along the diagonal. The subspace generated by the first $n-2$ basis elements is invariant, and the quotient is $2 \times 2$ case, which is not semi-simple. Moreover, the representation is not irreducible, which is pretty clear since for example $e_1$ is invariant. Finally, the center of group is non-trivial, since it contains those matrices with 1's along the diagonal, and the only other nonzero entry at very upper right.)
So, to summarize, I'm looking for either:
A proof that $End_G(V) = K$ implies that $V$ is irreducible, even if $V$ is not completely reducible. (Better, an example of an interesting intertwiner that always exists if the representation is not irreducible, but not necessarily completely reducible.) OR, an example of a group $G$ acting on $V$, so that $End_G(V) = K$, but $V$ is not irreducible.
(Background motivation: I'm trying to show that if $V$ and $W$ are irreducible representations of $G$ and $H$, respectively, and if $End_G(V) = K$ and $End_H(W) = K$, then $V \otimes W$ is irreducible as a $G \times H$ representation. Again, in the setting where I do not have a guarantee that the representations of completely reducible. I think I see an alternative direction of proof of this statement, if the answer to my question is false, but it seems more tedious than the argument that is motivating this question... alternatively, one could first try to show that $V \otimes W$ is completely reducible as $G \times H$ representation, but I don't see any way to do that.)
Try
$$G=\left\{\begin{pmatrix}1&x\\0&y\end{pmatrix} : x,y\in K, y\neq0\right\}.$$
The examples you looked at were all subgroups of the group of upper triangular $n\times n$ matrices with ones on the diagonal. In this case, the last basis element spans a copy of the trivial module, and the quotient by the last $n-1$ basis elements is a copy of the trivial module, so there is a rank one endomorphism.
My example is a non-trivial extension of two non-isomorphic simple modules, each with endomorphism algebra $K$, and any such extension will have endomorphism algebra $K$.
[For the question in the motivation, for general $K$-algebras, the easiest proof I know uses the Jacobson density theorem, although it's possible there's a more elementary proof for group algebras.]