Why are these algebras non-isomorphic?

Question

I am trying to classificate 3-dimensional complex Lie algebras, and this is the first place where i got stuck. Consider a 3-dimensional vector space with basis {$x$, $y$, $z$}. Now i have managed to find these 3 algebras: $$1).[y,z]=y \ , [x,z]=-x \ , [x,y]=0$$ $$2).[y,z]=x \ , [x,z]=y \ , [x,y]=0$$ $$3).[y,z]=x \ , [x,z]=-y \ , [x,y]=0$$ I am trying to understand now which of them are isomorphic (or not). They are all solvable, not nilpotent, and their center is trivial. If they are not isomorphic, there must be some deeper properties that can distinguish them, but i don't know which. Please help.

Answer 1

The three of them are isomorphic.

Define $\varphi_1(x)=x-y$, $\varphi_1(y)=x+y$, $\varphi_1(z)=z$. This gives you an isomorphism between the first and the second.

Now, define $\varphi_2(x)=x-iy$, $\varphi_2(y)=x+iy$, $\varphi_2(z)=-iz$. This gives you an isomorphism between the first and the third.

The key point is that $\{x,y\}$ generates an abelian subalgebra and that $ad_z$ restricts to this subalgebra and it is diagonalizable.