Solvable Lie group

Remarks

A connected Lie group is solvable if and only if its Lie algebra $\mathfrak{g}$ is a solvable Lie algebra, that is, its derived series $\mathfrak{g} \supseteq [\mathfrak{g}, \mathfrak{g}] \supseteq \cdots$ reaches $0$.

Example: the group $G$ of affine maps $x \mapsto a x + b$ of $\R$ with $a > 0$ and $b \in \R$, written $(a, b)$, with $(a, b)(a', b') = (a a', a b' + b)$. It is connected, being homeomorphic to $(0, \infty) \times \R$. The commutator of $(a, 0)$ and $(1, t)$ is $(1, (a - 1)t)$, and every commutator has first entry $1$, so $G^{(1)}$ is the abelian group of translations $(1, b)$. Then $G^{(2)} = \set{e}$, and $G$ is solvable.

Non-example: $G = SL(2, \R)$. Its Lie algebra $\mathfrak{sl}(2, \R)$ satisfies $[\mathfrak{sl}(2, \R), \mathfrak{sl}(2, \R)] = \mathfrak{sl}(2, \R)$, since the basis elements $H, E, F$ with $[H, E] = 2E$, $[H, F] = -2F$, $[E, F] = H$ are all brackets. So the derived series of the Lie algebra never reaches $0$, and $G$ is not solvable.