Does $[\mathfrak{h},[\mathfrak{h},\mathfrak{h}]]=0$ imply $[\mathfrak{h},\mathfrak{h}]=0$?

Question

Let $\mathfrak{g}$ be a Lie algebra. Suppose there exists a subspace $\mathfrak{h}\subset \mathfrak{g}$ such that $$ [X,[Y,Z]] =0 $$ for all $X,Y,Z \in \mathfrak{h}$.

Is it true that $\mathfrak{h}$ is an abelian subalgebra of $\mathfrak{g}$, i.e., $[X,Y]=0$ for all $X,Y \in \mathfrak{h}$?

Answer 1

No, it is not true in general. Let $\mathfrak{g}$ be the $3$-dimensional Heisenberg Lie algebra with basis $(x,y,z)$ and Lie brackets determined by $[x,y]=z$. Then $\mathfrak{h}=\mathfrak{g}$ satisfies $[X,[Y,Z]]=0$ for all $X,Y,Z$, but the Heisenberg Lie algebra is not abelian.