Is a group which is equal to its derived subgroup necessarily semisimple?

Question

let $G$ be either a (connected) Lie group or an (connected) algebraic group over a field (which is algebraically closed of characteristic zero). It is well known that if $G$ is semisimple then $G=G'$ when $G'=[G,G]$. Is the converse true?

Answer 1

Let $ G=SE_n:= \mathbb{R}^n \rtimes SO_n $ be the (special) Euclidean group, which is a connected Lie group. Then $ RvR^{-1}v^{-1} $, equals the pure vector element $ R(v)-v $, where $ R $ is pure rotation element and $ v $ is a pure vector element. Consider a rotation $ R $ that takes $ v $ to $ -v $. Then $ R(v)-v=-2v $. Thus we can produce all vectors this way. So $ [G,G] $ contains all the pure vector elements. On the other hand $ [SO_n,SO_n]=SO_n $ for $ n\geq 3 $ so $ [G,G] $ contains all pure rotation elements. But $ G $ is generated by the pure vector elements and the pure rotations elements. Thus $ [G,G] $ must be all of $ G $ for $ n \geq 3 $. $ G $ is clearly not semisimple since $ \mathbb{R}^n $ is a normal solvable subgroup.

Since the representation of $ SO_n $ on $ \mathbb{R}^n $ is irreducible this is really an example of the more general phenomenon linked to in the comment by Arrow above which states that if $ \pi, V $ is a (nontrivial) irreducible representation of a semisimple group $ G $ then $ V \rtimes_\pi G $ is perfect.

This is all related to Perfect semi direct products

The assumption of nontriviality is just to rule out that $ V $ is a one dimensional vector space, as stated in the comment by Patrick Da Silva here https://mathoverflow.net/questions/60498/lie-algebra-semisimple#