 Solution

For the Euclidean motion Lie algebra in two dimensions, use the ordered basis $(J,E_1,E_2)$ and keep the printed rotation signs. The Adjoint representation matrices areOnly the square of ${\mathrm{ad}}_J$ has a nonzero trace. Hence the Killing form of the planar Euclidean motion Lie algebra has matrixIts radical of the Killing form is $\mathrm{span}\{E_1,E_2\}$, precisely the translation subspace. This is a nonzero abelian ideal of a Lie algebra, so the Lie algebra is neither simple nor semisimple. In fact its first derived series of a Lie algebra term is this translation ideal and its next is zero. Thus the degeneracy and the failure of semisimplicity agree with the structural proofs above; the adjoint action nevertheless remains nontrivial.