Consider a semisimple Lie algebra S and a vector space V - considered as Abelian Lie algebra - with a non-zero irreducible representation $$\rho: S \rightarrow End(V).$$ $L$ and $V$ are finite-dimensional over the base field $\mathbb{R}$ or $\mathbb{C}$.
Then the semidirect product $$L:= V \rtimes_\rho S$$ is a perfect Lie algebra, i.e. $[L,L]=L$.
- How to prove this result?
Solution thanks to the comments of @YCor:
Assume $L=V \rtimes_{\rho}S$. Then $$L=V+S$$ with $S\subset L$ subalgebra and $V\subset L$ ideal. By definition of the commutator in $L$ holds
$$[S,S]_S=[S,S]_L.$$ Semisimpleness of $S$ implies $$S=[S,S]_S.$$ Moreover $$\rho(S)(V)=[S,V]_L.$$ Irreducucibility of $\rho \neq 0$ implies $$V=\rho(S)(V)=[S,V]_L.$$ Then $$L=S+V=[S,S]_L+[S,V]_L\subset [L,L]_L$$ The other inclusion is obvious.