# Lie algebra 3 Dimensional with 2 Dimensional derived lie algebra #2

I read the book of Karin Erdmann and Mark Wildon: “An introduction to Lie algebras”. In page 22 they say that:

If $\dim (L) = 3$, $\dim (L’) = 2$ then (a) $L’$ abelian and (b) $\operatorname{ad} x : L’ \rightarrow L’$ is an isomorphism

I have problem now to prove part (b) of lemma 3. $\operatorname{ad} x : L’ \rightarrow L’$. I stuck to show $\operatorname{ad} x : L’ \rightarrow L$ is a homomorphism. How to show that $\operatorname{ad} x([y,z]) = [\operatorname{ad} x(y), \operatorname{ad}x(z)]$ where $y,z \in L’$.

Since the first part showed that $L’$ is abelian, we have $\rm{ad}_x([y,z]) = \rm{ad}_x(0) = 0$ and on the other hand, we have that both $\rm{ad}_x(y) = [x,y]$ and $\rm{ad}_x(z) = [x,z]$ are elements of $L’$, so the bracket $[\rm{ad}_x(y),\rm{ad}_x(z)] = 0$, and hence the map is a homomorphism.