The plane Euclidean-motion Lie algebra is solvable

The standard rotation action gives $[R,P]=Q$, $[R,Q]=-P$, and $[P,Q]=0$ via [L1]. Therefore every bracket lies in the translation ideal $T=\mathbb{R}P\oplus \mathbb{R}Q$, while the first two displayed brackets span $T$. Hence $\mathfrak{e}(2{)}^{(1)}=T$. Since $T$ is abelian, $\mathfrak{e}(2{)}^{(2)}=0$. The first derived term is nonzero, so [L2] gives derived length exactly two.

The same calculation gives ${\gamma }_2(\mathfrak{e}(2))=T$. Moreover $[\mathfrak{e}(2),T]=T$, because bracketing $R$ with $P,Q$ again yields $Q,-P$. Induction gives ${\gamma }_r(\mathfrak{e}(2))=T\mathrm{\ne }0$ for every $r\ge 2$, so the algebra is not nilpotent by [L3].

Thus the abelian translation ideal is responsible for solvability, while the nontrivial rotation action prevents lower-central termination. Both endpoint computations are exact, and the finite explicit example uses no choice.