For a linear algebraic group over an algebraically closed field of characteristic zero $G$, with unipotent radical $U$, we have that $G/U$ is reductive.


When $G$ is solvable, then Lie's theorem says that irreducible representations are 1-dimensional, so the unipotent radical acts trivially, and so the irreducible representations of $G$ are in bijection with the irreducible representations of $G/U$ under the pullback functor.


I figure there must be a generalization of this to the case of general $G$. Chriss/Ginzburg proves something like this in their book using geometric methods but I feel like there should also be a more group-theoretic proof which specializes to the argument using Lie's theorem when $G$ is solvable.


It would suffice to show that on any irreducible representation of $G$, the unipotent radical acts trivially. I can't find a proof of this statement, however. Is there a quick and easy way to bootstrap Lie's theorem, or can someone point me to a reference?


Apologies if this is too easy for MO.



Irreducible representations of the reductive quotient
Let $V$ be an irreducible representation of $G$. Consider it as a representation of $U$. Since $U$ is unipotent, $V^U\neq 0$. On the other hand, since $U$ is normal in $G$, $V^U$ is a subrepresentation. But the irreducibility of $V$ implies then that $V^U=V$, which is what you wanted.


