Lie algebras containing $ so_3(\mathbb{R}) $ Every subalgebra of a solvable Lie algebra is solvable. So a Lie algebra $ \mathfrak{g} $ containing a subalgebra isomorphic to $ \mathfrak{so}_3(\mathbb{R}) $ cannot be solvable.
Is the converse true? In other words, is it true that a Lie algebra $ \mathfrak{g} $ contains a subalgebra isomorphic to $ \mathfrak{so}_3(\mathbb{R}) $ if and only if $ \mathfrak{g} $ is not solvable?
 A: As coudy already said, the answer to your question is no, and as pointed out in the comments by Moishe Kohan, if you modify your question to also consider $\mathfrak{sl}(2,\mathbb{R})$ then the answer becomes yes. So we have
Theorem: A real finite-dimensional Lie algebra is not solvable if and only if it has a subalgebra isomorphism to either $\mathfrak{sl}(2,\mathbb{R})$ or $\mathfrak{so}(3)$.
Here's a proof sketch: You already proved one direction. For the other direction, note that by the Levi decomposition, every real finite-dimensional Lie algebra is a semidirect product of a solvable Lie algebra and a semisimple Lie algebra. But every real semisimple Lie algebra has a subalgebra isomorphic to $\mathfrak{sl}(2,\mathbb{R})$ or $\mathfrak{so}(3)$.
You can complete the proof sketch without the full power of the classification. For example, the complexification of your real semisimple Lie algebra will be a complex semisimple Lie algebra, and this complex semisimple Lie algebra will have a subalgebra isomorphic to $\mathfrak{sl}(2,\mathbb{C})$ which is preserved by the complex conjugation map. So your original real semisimple Lie algebra will contain a real form of $\mathfrak{sl}(2,\mathbb{C})$, and there are only two of these: $\mathfrak{sl}(2,\mathbb{R})$ and $\mathfrak{so}(3)$.
A: No, the Lie algebra of $SO_{2,1}$ is not solvable. The Tits alternative may help you decide if a group is virtually solvable or not.
Let G  be a finitely generated linear group over a field. Then two following possibilities occur:
either G  is virtually solvable (i.e. has a solvable subgroup of finite index)
or it contains a nonabelian free group (i.e. it has a subgroup isomorphic to the free group on two generators).
