solvable Lie algebra
A Lie algebra is solvable when its non-commutativity unravels in stages: take the brackets of everything, then the brackets of those, and so on, and you eventually reach zero. The name echoes solvable groups in Galois theory, and for good reason — solvability of a Lie algebra is the infinitesimal counterpart of being built from abelian layers, the property behind solving polynomials by radicals.
Define the derived series by L^{(0)} = L and L^{(k+1)} = [L^{(k)}, L^{(k)}]. Each term is an ideal of L (indeed an ideal in the previous term). The algebra L is solvable if L^{(m)} = 0 for some m. Equivalently, L has a finite chain of subalgebras L = L_0 > L_1 > ... > L_m = 0 in which each L_{i+1} is an ideal of L_i with abelian quotient.
Every nilpotent Lie algebra is solvable; the converse fails. Over an algebraically closed field of characteristic zero, Lie's theorem says solvable algebras of matrices can be simultaneously put in upper-triangular form, so the upper-triangular matrices are the universal example. There is also a largest solvable ideal, the radical, whose quotient is semisimple — this is the Levi decomposition.
The 2-dimensional non-abelian Lie algebra with basis x, y and bracket [x, y] = y satisfies L^{(1)} = [L, L] = span(y) and L^{(2)} = 0. It is solvable but not nilpotent, since ad(x) acts as the identity on y, hence is not nilpotent.
The unique 2-dimensional non-abelian Lie algebra: solvable, not nilpotent.
Cartan's solvability criterion gives a clean trace test: a Lie algebra of matrices is solvable if and only if trace(xy) = 0 for every x in [L, L] and every y in L.