I'm stuck on the next problem:
Let $\mathfrak{g}$ a Lie Algebra. Prove that $\mathfrak{g}$ is solvable if and only if there is a chain of ideals $$\mathfrak{g}=\mathfrak{g}_{0}\supset \mathfrak{g}_{1}\supset \cdots \supset \mathfrak{g}_{k}=\{0\}$$ such that $\mathfrak{g}_{i}/\mathfrak{g}_{i+1}$ is abelian for every $i=0,1,\ldots, k$.
I proved the direction $\fbox{$\Rightarrow$}$, but I do not see how to proceed in the other direction. I will appreciate any hint.
Thanks
By hypothesis $\mathfrak g_0/\mathfrak g_1$ is abelian, i.e $\mathfrak g^{(1)} = [\mathfrak g_0, \mathfrak g_0] \subset \mathfrak g_1$. Now, since $\mathfrak g_1/\mathfrak g_2$ is abelian, we have $\mathfrak g^{(2)} \subset [\mathfrak g_1, \mathfrak g_1] \subset \mathfrak g_2$. Continuing this pattern we obtain that $\mathfrak g^{(k-1)} \subset [\mathfrak g_{k-2}, \mathfrak g_{k-2}] \subset \mathfrak g_{k-1}$ and by hypothesis $\mathfrak g_{k-1}$ is abelian so indeed $\mathfrak g^{(k)} \subset [\mathfrak g_{k-1},\mathfrak g_{k-1}] = 0$ i.e $\mathfrak g$ is solvable.