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I am actually having some trouble in demonstrating the following result in the book Lie algebras and Algebraic groups of P. Tauvel and R. W. T. Yu:

Let $G$ be a connected reductive algebraic group. Then the derived group $\mathcal{D}(G)$ is semi-simple.

Actually, it is a direct consequence of the corollary 20.5.5 in the same book. This corollary states thant a Lie algebra is reductive if, and only if its derived algbra is semisimple.

However, I am thinking of a demonstration that does not use Lie algebras. So this is how I started: Let $H$ is the greatest solvable, connected, normal, closed subgroup of $\mathcal{D}(G)$.

Since $\mathcal{D}(G)$ is connected, normal, closed in $G$, so is $H$ in $G$. So there is only left to use the fact that H is solvable in order to show that H is trivial. I though of showing that the radical of $\mathcal{D}(G)$ is equal to the unipotent radical of $G$ in order to use the fact that $G$ is reductive, but I did not managed to.

I am a beginner in the field of Algbraic groups so any help would be great.

Thanks a lot.

K. Y.

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$[H,H]$ is unipotent since $G$ is reductive, $[H,H]=1$, $H=1$ since a commutative connected group is unipotent.

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  • $\begingroup$ First of all, I wanted to thank you for your answer. I'm sorry for this late comment. It took me some time to analyse you answer since I'm a beginner. Tell me if I'm right: Since H is solvable so D(H) is unipotent, I found this in Corrolary 5.2.7 of relaunch.hcm.uni-bonn.de/fileadmin/perrin/ag-chap5.pdf Since G is reductive so D(H) = {e}. So H is a commutative group. Since H is commutative connected, it is unipotent in a reductive group and then H = {e}. However, I didn't find a reference for the fact that a connected commutative group is unipotent. Could you help me on this one? $\endgroup$ – Kal_Aki Mar 25 '18 at 12:02

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