Semisimple Lie algebras are perfect.

Question

Can anyone explain why a semi-simple finite dimensional Lie algebra $\mathfrak{g}$ has to be perfect ?

The natural way to prove something like that would be to look to the algebra generated by the Lie brackets, which when $\mathfrak{g}$ is not perfect would be expected to be solvable. But it doesn't seem to work.

Answer 1

One possible way (and I have not been able to think of an easier way) is the following:

Note that here I will assume this to be over $\mathbb{C}$.

That semisimple Lie algebras are perfect follows as a corollary to the theorem that such a Lie algebra is a direct sum of simple ideals. It is then clear as no non-trivial quotient of such a direct sum can be abelian.

Answer 2

here you can find some relevant discussion.

https://mathoverflow.net/questions/60498/lie-algebra-semisimple

Answer 3

If you are allowed to use the fact that a quotient of $\mathfrak{g}$ (a finite dimension semisimple Lie algebra over a field with characteristic $0$) is semisimple, then there is an easy answer. The usual way to prove this fact is to use Cartan's criterion: see here.

If $\mathfrak{g}$ is semisimple, then its quotient $\mathfrak{g}/[\mathfrak{g},\mathfrak{g}]$ is also semisimple. But this is abelian, so $\mathfrak{g}/[\mathfrak{g},\mathfrak{g}] = 0$, then $\mathfrak{g} =[\mathfrak{g},\mathfrak{g}]$.