The derived group of a reductive connected group is semisimple

Question

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.

Answer 1

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