On the converse of Schur's Lemma 
Let $G$ be a finite group and $F$ a field with $\mathrm{char}(F)=0$ or coprime to $|G|$. Let $V$ be a $FG$-module in a way that every $ FG$-homomorphism $ f : V \to V $ is given by $f(x)= \lambda x $. Then $V$ is irreducible.

I already managed to proof this by contradiction, using Maschke's Theorem. We can  write $V$ as $U\oplus W$ and get a $FG$-homomorphism $\pi:U\oplus W \to U\oplus W$, $\pi (u+w)=u$, then $\pi(u+w)=\lambda (u+w)=u $ then either $U$ or $W$ are trivial and thus $V$ is irreducible.
My question is if this statements still holds if the characterictic of $F$ divides the order of $G$, since I only used this fact in my proof to use Maschke's theorem.

There are counterexamples in general.
Suppose $V$ is a non-split extension of two non-isomorphic simple modules. I.e., $V$ has a unique simple submodule $W$ with simple quotient $U=V/W$ where $U\not\cong W$. Then any non-zero endomorphism of $V$ is an isomorphism, since the only possible kernel is $W$, but the image can't be $V/W=U$ since $V$ has no submodule isomorphic to $U$. If, further, $U$ and $W$ have $F$ as endomorphism ring, this isomorphism must be a scalar multiple of the identity.
For an explicit example, let $G$ be the group of invertible upper triangular $2\times 2$ matrices over a finite field $F$ with $\vert F\vert>2$, and let $V$ be the natural $2$-dimensional $FG$-module.
