I am reading the notes of Victor Kac (Introduction to Lie Algebras). After Cartan's theorem there is an application, the classification (up to isomorphism) of 3-dimentional Lie algebras.
We use the Cartan's theorem to write
$$ \mathfrak{g} = \bigoplus_{\lambda\in \mathbb{F}}\mathfrak{g}^{a}_{\lambda} $$
where $a \in \mathfrak{g}$ is a regular element and $\mathfrak{h}=\mathfrak{g}_{0}^{a}$ is a subalgebra de Cartan.
I have proven when $\mbox{rank }\mathfrak{g}=\dim\mathfrak{h}=1$ one has 3 possibilities for $\mathfrak{g}=\mbox{span}\{a,b,c\}$:
\begin{eqnarray} 1 &-& [a,b]=b, [a,c] = c + b, [b,c] = 0\\ 2 &-& [a,b]=b, [a,c] = \lambda c, [b,c] = 0, \mbox{ for a non null $\lambda$} \\ 3 &-& [a,b]=b, [a,c] = -c, [b,c] = a \end{eqnarray}
So my question is: How proof that these 3 Lie algebras are non isomorphic?
I would be happy with any help!
Already considering the invariant $\dim ([\mathfrak{g},\mathfrak{g}])$ shows that algebras of type $3$ cannot be isomorphic to the other two types. In fact, we have $[\mathfrak{g},\mathfrak{g}]=\mathfrak{g}$ for type $3$, which implies that $\mathfrak{g}$ is a simple Lie algebra, because any non-trivial ideal $I$ would be of dimension $\le 2$, hence solvable with quotient of dimension $\le 2$. Then $\mathfrak{g}$ would be solvable, which is impossible with $[\mathfrak{g},\mathfrak{g}]=\mathfrak{g}$. It follows that $\mathfrak{g}\cong \mathfrak{sl}_2(\mathbb{C})$. Algebras of type $1$ and $2$ can be classified as Yves indicated in the comment. For a detailed proof see Jacobson's book on Lie algebras, chapter I, section 4, pages $11-13$.