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Let $G$ be a connected algebraic group of positive dimension. Prove that $G$ is semisimple if and only if $G$ has no closed connected commutative normal subgroup except $e$.

It is clear in one direction, given the definition of semisimple algebraic groups. In the other direction, suppose $G$ is not semisimple, and let $H=R(G) \neq \{e\}$, the radical of $G$. As $H$ is solvable, it has a lower central series:

$H=H_1 \unrhd H_2=(H_1,H_1) \unrhd \cdots \unrhd H_{n+1}=(H_n,H_n)=e$.

Hence, $H_n$ is commutative and $\neq \{e\}$. The contradiction can be found if $H_n \unlhd G$. But in fact, I have only $H \unlhd G$. Normality is in general not liftable. So, I don't know what to do. Maybe other nontrivial closed connected commutative normal subgroup can be found?

Many thanks.

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In this case normality is liftable because commutator subgroups are characteristic, and this property is liftable. – Tobias Kildetoft Aug 1 '11 at 9:37
Thank you very much for the hint. – ShinyaSakai Aug 2 '11 at 13:11

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