A Levi decomposition of the Euclidean-motion algebra of R^3

Put $V={\mathbb{R}}^3$. In the semidirect product, $[(v,A),(w,B)]=(Aw-Bv,[A,B])$. Hence $V\oplus 0$ is an abelian ideal, and the quotient by it is $\mathfrak{s}\mathfrak{o}(3)$.

Under the vector-space identification $u\mapsto A_u$, $A_u(v)=u\times v$, one has $[A_u,A_v]=A_{u\times v}$. If an ideal of $\mathfrak{s}\mathfrak{o}(3)$ contains a nonzero $A_u$, then the vectors $u\times v$ as $v$ varies span $u^{\perp }$, and a further bracket supplies the $u$-direction. Thus the ideal is all of $\mathfrak{s}\mathfrak{o}(3)$. The algebra is nonabelian and $[\mathfrak{s}\mathfrak{o}(3),\mathfrak{s}\mathfrak{o}(3)]=\mathfrak{s}\mathfrak{o}(3)$, so it is simple and not solvable.

Let $I$ be a solvable ideal of $\mathfrak{e}(3)$. Its image in the quotient is a solvable ideal of $\mathfrak{s}\mathfrak{o}(3)$, hence is zero by step 1.2. Thus $I\subseteq V$. Conversely $V$ is itself a solvable ideal by step 1.1, so it is the radical.

The subalgebra $0\oplus \mathfrak{s}\mathfrak{o}(3)$ is semisimple, meets $V$ in zero, and together with $V$ spans the whole algebra. It is therefore a Levi factor by [L1]. The zero intersections and full quotient are explicit, and no choice is used.