Is the following property true？ Let $L$ be a finite-dimensional Lie algebra over $\mathbb{C}$. Then $L$ is semisimple if and only if for every $x \in L$, there exists $y, z\in L$, such that $x=[y, z]$. Remark. For Lie algebra of type $A$, I know that this result is true. However, I do not know other cases.
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3$\begingroup$ The Lie algebra $\mathfrak{sl}_2(\mathbf{C})\ltimes\mathbf{C}^2$ is not semisimple but every element is a commutator (it has only 5 orbits under the action by automorphisms of $\mathrm{GL}_2(\mathbf{C})\ltimes\mathbf{C}^2$, so this is quite easy). $\endgroup$– YCorCommented Feb 27, 2020 at 5:49
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5$\begingroup$ It seems anyway that you're primarily interested in the direct implication. Yes, every element is a Lie commutator in a semisimple Lie algebra. This is already asked and answered on MathSE: math.stackexchange.com/questions/769881. $\endgroup$– YCorCommented Feb 27, 2020 at 5:50
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$\begingroup$ @YCor Thank you. $\endgroup$– user11090426Commented Feb 27, 2020 at 9:02
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1$\begingroup$ PS in my first comment I should have said "only 5 orbits modulo nonzero scalar multiplication" $\endgroup$– YCorCommented Feb 27, 2020 at 9:05
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