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This question already has an answer here:

Let $\mathfrak{g}$ and $\mathfrak{h}$ be two-dimensional non abelian Lie-algebras. Show that they are isomorphic.

Suppose that $\{E_1,E_2\}$ and $\{F_1,F_2\}$ be basis for $\mathfrak{g}$ and $\mathfrak{h}$ respectively. Since both $\mathfrak{g}$ and $\mathfrak{h}$ are nonabelian, $[E_1,E_2] \ne 0$ and $[F_1,F_2] \ne 0$. Suppose that $[E_1,E_2]=aE_1+bE_2$. If $a=0$, then dividing by nonzero $b$ we have $[E_1,\frac{1}{b}E_2]=E_2$. If $a \ne 0$, then dividing by $a$ throughout we have $[E_1,\frac{1}{a}E_2]=E_1+\frac{b}{a}E_2$. Then $[E_1+\frac{b}{a}E_2-\frac{b}{a}E_2,\frac{1}{a}E_2]=[E_1+\frac{b}{a}E_2,\frac{1}{a}E_2]=E_1+\frac{b}{a}E_2$. So to start with, we can choose $\{E_1,E_2\}$ and $\{F_1,F_2\}$ such that $[E_1,E_2]=E_1$ and $[F_1,F_2]=F_1$.

Now define $\phi: \mathfrak{g} \to \mathfrak{h}$ by $\phi(aE_1+bE_2)=aF_1+bF_2$. Then $\phi$ is linear, one-to-one and onto. Moreover, we have $\phi([E_1,E_2])=\phi(E_1)=F_1=[\phi(E_1),\phi(E_2)]$. Thus, $\phi$ is an isomorphism.

I was wondering if I could do this more directy, without any assumption on the Lie brackets of both sides. If I do that and define $\phi$ as above, then I just have to show the nice behavior of $\phi$ with the Lie brackets. Now $\phi([E_1,E_2])=aF_1+bF_2$ but I don't know how to show that $[F_1,F_2]=aF_1+bF_2$.

Thanks for the help!!

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marked as duplicate by Dietrich Burde, dantopa, Lord Shark the Unknown, user370967, user91500 Jun 21 '17 at 9:31

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This just doesn't work. Indeed, given any pair of scalars $c$ and $d$ (not both zero), $[F_1,F_2]=cF_1+dF_2$ defines a two-dimensional non-abelian Lie algebra. So $[F_1,F_2]$ could be any nonzero linear combination of $F_1$ and $F_2$ at all, and certainly doesn't have to be the same linear combination as $[E_1,E_2]$ is of $E_1$ and $E_2$.

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