Center of a simple Lie algebra is trivial

Proposition. Let $\mathfrak g$ be a simple Lie algebra. Then

Z(\mathfrak g)=\{0\},

where $Z(\mathfrak g)$ is the center of $\mathfrak g$ .

Proof

The center is an ideal, so simplicity implies that it is either $0$ or all of $\mathfrak g$ . The second possibility would make $\mathfrak g$ abelian, contrary to the definition of a simple Lie algebra. Equivalently, this follows from the identity $\ker(\operatorname{ad})=Z(\mathfrak g)$ .