Solvable Lie algebra

Remarks

Each $\mathfrak g^{(i)}$ is a Lie ideal of $\mathfrak g$, and $\mathfrak g^{(i)} / \mathfrak g^{(i+1)}$ is abelian. An equivalent condition: there is a chain of subalgebras $\mathfrak g = \mathfrak a_0 \supseteq \mathfrak a_1 \supseteq \cdots \supseteq \mathfrak a_n = 0$ in which each $\mathfrak a_{i+1}$ is an ideal of $\mathfrak a_i$ and each $\mathfrak a_i / \mathfrak a_{i+1}$ is abelian. Subalgebras and quotients of solvable Lie algebras are solvable; if an ideal $\mathfrak a$ and the quotient $\mathfrak g / \mathfrak a$ are solvable, then $\mathfrak g$ is solvable.

Example: the two-dimensional Lie algebra with basis $x, y$ and $[x, y] = y$. Its derived algebra $\mathfrak g^{(1)}$ is spanned by $y$, and $\mathfrak g^{(2)} = [ky, ky] = 0$.

Non-example: the special linear Lie algebra $\mathfrak{sl}_2$ over a field of characteristic not $2$, with basis $e, h, f$ and $[h, e] = 2e$, $[h, f] = -2f$, $[e, f] = h$. The brackets $e = \tfrac12 [h, e]$, $f = -\tfrac12 [h, f]$ and $h = [e, f]$ lie in $\mathfrak g^{(1)}$, so $\mathfrak g^{(1)} = \mathfrak{sl}_2$ and the derived series never reaches $0$.