I know there exist two real perfect and simple Lie algebras of dimension 3 up to isomorphism, let's say $L_1$ and $L_2$. A proof is performed here.
So, let $L$ be a real Lie algebra whose dimension is equal to $3$, that is simple and perfect, i.e. $L= [L, L]$.
I'm trying to perform the classification based on the following criteria $(C)$: It exists $x, y \in L$ linearly independent such that $[x, y]$ belongs to $\operatorname{span}\{x, y\}$.
I suppose that $(C)$ is satisfied, and I'm trying to prove that $L$ is isomorphic to the Lie algebra $L_1$ defined by the structure equations: $$[x, y]= x, [x, z]= -2 y, [y, z]= z.$$
As $(C)$ is satisfied and $L = [L, L]$, we can pick up $x, y \in L$ such that $[x, y] =x$. We can select $w \in L$ for $\{x, y, w\}$ being a basis of $L$. As $L =[L, L]$, for any $t \in L$, $\operatorname{tr} \operatorname{ad} t = 0$. Therefore $[x, w] = \alpha x + \beta y$ for some $\alpha, \beta \in \mathbb R$ and $[y, w] = \alpha^\prime x + \beta^\prime y + w$ for some $\alpha^\prime, \beta^\prime \in \mathbb R$.
But here, I don't know how to follow on...
Therefore my question: is criteria $(C)$ sufficient to separate between the two Lie algebras $L_1, L_2$? If yes how to prove it?
