Q:

Dimension of the derived algebra $L'$ of a 3-dimensional Lie algebra over a field $\Bbb{F}$.

From Erdmann and Wildon's Introduction to Lie algebras.

If $L$ is a non-abelian 3-dimensional Lie algebra over a field $F$, then we know only that the derived algebra $L'$ is non-zero. It might have dimension 1 or 2 or even 3.

I don't understand how $L'$ could have less than 3 dimensions. I did the following computations:
Consider a Lie algebra over the vector space $L$ over $F$ such that $B=\{x,y,z\}$ spans $L$, i.e. $\operatorname{dim}L=3$. Note $[x,y],[x,z],[y,z] \neq 0$ because $L$ is non-abelian, i.e. $[a,b]=0 \Leftrightarrow b=ca$, for some $c \in F$; that is, $a$ and $b$ are pairwise dependent of each other. However, if $[x,y]=\alpha[x,z]$, then $[x,y]-\alpha[x,z]=0 \Leftrightarrow [x,y]-[x,\alpha z]=0 \Leftrightarrow [x, y-\alpha z]=0$. Since $L$ is non-abelian, $x=y-\alpha z$. This means $B$ is not linearly independent which is a contradiction. Hence, $\forall \alpha \in F$, $[x,y] \neq \alpha[x,z]$.

For this reason, I don't understand how $L'$ could be of dimension 1.

But looking at $$[x,y]=c_1[x,z]+c_2[y,z]=[c_1x+c_2y,z]$$ makes me think that there is no reason why $L'$ cannot have 2 dimensions.
Can anyone please elaborate this passage on the book. Thanks!

A:

For one dimension, one can have generators $x$, $y$, $z$ with $[x,y]=y$, $[x,z]=[y,z]=0$. For two dimensions: $[x,y]=y$, $[x,z]=z$, $[y,z]=0$.

