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

Solution 1:

It's not true.

Start with the Jacobi identity: $[a,[b,c]] + [b,[c,a]] + [c,[a,b]] = 0.$

If $[,]$ is associative then $[a,[b,c]] = [[a,b],c] = -[c,[a,b]]$, so $[b,[c,a]] = 0$ for all $a,b,c$. In other words, all brackets must lie in the center of the Lie algebra.

Conversely, if we define a bilinear skew-symmetric $[,]$ for which $[b,[c,a]]=0$ holds identically, then we have a Lie algebra because the Jacobi identity is trivially satisfied.

So take a three dimensional vector space with basis $x,y,z$ and define $[x,y]=z$ and $[x,z]=0$ and $[y,z]=0$. This gives a counterexample.

It is true (and we have proved above), that if the Lie bracket is associative then we have $[[a,b],c] = [a,[b,c]] = 0$ for all $a,b,c$.