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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.

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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.

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  • $\begingroup$ I decide to write the question in MO mathoverflow.net/questions/256763, I tried your approach and I get some form like $\beta=(−1)^{(f(q1))_{Z2 \times Z2} (n2)_{Z2}}$. I numerically check that it should be $(-1)^{\#}$, and the inflation trivialize the cocycle, but analytically it is hard to obtain the exponent # based on your previous answer -maybe I misunderstood. $\endgroup$ – miss-tery Dec 8 '16 at 20:58

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