Equivalence of associative Lie bracket

Question

Let $L$ be a Lie-algebra. I have to show that the Lie-bracket $[\,,]$ is associative if and only if $[x,y]\in C(L)$ for all $x,y\in L$, where $C(L)$ denotes the center of $L$.

I know that $[x,y]\in C(L)\Leftrightarrow [[x,y],z]=0\,\,\,\forall z\in L$. But from here I do not know how to continue.

Can anyone give me hint?

Answer 1

Suppose that the Lie bracket is associative: $[x,[y,z]]=[[x,y],z]$

We know that $[x,[y,z]]+[z,[x,y]]+[y,[z,x]]=0$

we have $[z,[x,y]]=-[[x,y],z]=-[x,[y,z]]=0$ replacing this in the Jacobi equality, we obtain $[y,[z,x]]=0$, which is equivalent to saying that for every $x,z, [z,x]$ is in the center of $L$.

Answer 2

Being associative means that you always have$$\bigl[x,[y,z]\bigr]=\bigl[[x,y],z\bigr].\tag1$$But, since we are dealing with a Lie algebra,we have$$\bigl[x,[y,z]\bigr]=\bigl[[x,y],z]+\bigl[y,[x,z]\bigr].\tag2$$Since we want to have both $(1)$ and $(2)$…

Answer 3

A Lie algebra $L$ is associative if and only if it is two-step nilpotent (including abelian), i.e., if and only if $[L,L]\subseteq Z(L)$, see this duplicate:

Does there exist a non-trivial, associative Lie algebra?