Let $G$ be an algebraic group over an algebrically closed field $\mathbb{K}$ of caracteristic $0$. We say that $G$ est semi-simple if every connected, solvable, normal, closed subgroup of $G$ is equal to $\{1\}$.
My purpose is to show that if $G$ is semi-simple then $G$ is perfect (i.e. $G$ is equal to its commutator group $(G,G)$. I already found a proof in P. Tauvel AND R. W. T. Yu "Lie Algebras and Algebraic Groups" book, but I wanted to make one that does not use Lie Algebras.
My first idea was to consider the group quotient $G/(G,G)$ and show that this group is isomorphic to $\{1\}$. But all I know about this group is that it is connected and commutative(so solvable).
Any hint or help would be greatly appreciated.
Thanks in advance.
K. Y.