On page 125 of Humphreys' "Linear Algebraic Groups," it states the following: "An arbitrary algebraic group G possesses a unique largest normal solvable subgroup." However, I am currently questioning this assertion.
Finite groups have unique largest solvable normal subgroup? I have read it, which is about finite groups.
Let $G$ be an arbitrary algebraic group. If we set $H=\bigcup_{N\in\Gamma}N$ where $\Gamma$ is the set of normal solvable subgroup of $G$, then $H$ has to be the unique largest normal solvable subgroup. It is easy to see that $H$ is a group and normal as in the above link. But how can we show that $H$ is solvable? I tried to address this problem as follows:
Set $H_1=(H,H)$ and $H_{i+1}=(H_i,H_i)$ for $i$, and similarly for $N$. Then, $$H_i=\bigcup_{N\in\Gamma}N_i.$$
$N_{r}=\{e\}$ for some $r$. But can we uniformize this $r$? Namely, I need a positive integer $M$ such that $r<M$ for all $N\in\Gamma$. That seems to be necessary in order to prove that H is solvable. But I doubt whether such an $r$ can be chosen.
If the condition of connectedness were added, then I think we could choose that $r=\dim(G)$. But I doubt whether we can choose $r$ without the condition of connectedness.