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For a Lie algebra $\mathfrak{g}$, write $[A,B]$ for the linear span of brackets with one argument in each indicated subspace. The three definitions areThese are respectively an Abelian Lie algebra, a Solvable Lie algebra and a Nilpotent Lie algebra. The second and third sequences are the derived series of a Lie algebra and the Lower central series of a Lie algebra.

An Abelian Lie algebra has ${\gamma }_2=0$, and a Nilpotent Lie algebra is a Solvable Lie algebra: induction gives ${\mathfrak{g}}^{(j)}\subseteq {\gamma }_{j+1}$. Thus all the implications are generated byNone of the reverse implications holds. The Heisenberg Lie algebra with basis $x,y,z$ and $[x,y]=z$, all other basic brackets zero, is nonabelian but has ${\gamma }_2=\mathbb{C}z$, ${\gamma }_3=0$. The two-dimensional affine Lie algebra of the line with $[h,e]=e$ has ${\mathfrak{g}}^{(1)}=\mathbb{C}e$ and ${\mathfrak{g}}^{(2)}=0$, but ${\gamma }_j=\mathbb{C}e$ for every $j\ge 2$. It is solvable and neither nilpotent nor abelian. These two examples answer all six ordered-pair comparisons.

The Lie theorem says that a finite-dimensional Lie algebra representation of a finite-dimensional Solvable Lie algebra over an algebraically closed field of characteristic zero has a common eigenvector whenever its representation space is nonzero. Equivalently it admits an invariant complete flag, or simultaneous upper triangularization. Here the field may be taken to be $\mathbb{C}$; these hypotheses are essential.

To prove the common-eigenvector assertion, induct on $\dim \mathfrak{g}$. The zero algebra is immediate. Since $\mathfrak{g}$ is nonzero and solvable, its derived algebra is proper. Choose a codimension-one Lie algebra ideal $\mathfrak{h}$ containing it, and write $\mathfrak{g}=\mathfrak{h}\oplus \mathbb{C}x$. The Lie algebra $\mathfrak{h}$ is solvable, so induction supplies $v\ne 0$ and a linear functional $\lambda :\mathfrak{h}\to \mathbb{C}$ with $hv=\lambda (h)v$.

Consider the finite-dimensional cyclic subspace $W=\mathrm{span}\{v,xv,x^{2}v,\dots \}$. Until the first linear dependence, these powers form a basis. The identityand $[h,x]\in \mathfrak{h}$ show by induction, simultaneously for all $h\in \mathfrak{h}$, that $W$ is $\mathfrak{h}$-invariant and that $h$ is upper triangular on it with every diagonal entry $\lambda (h)$. It is also $x$-invariant by construction. HenceThe trace of a commutator is zero, and characteristic zero gives $\lambda ([x,h])=0$.

The simultaneous eigenspace $E_{\lambda }=\{w:hw=\lambda (h)w\text{ for every }h\in \mathfrak{h}\}$ is nonzero and $x$-invariant, sinceOver an algebraically closed field, $x{|}_{E_{\lambda }}$ has an eigenvector, which is therefore a common eigenvector for all of $\mathfrak{g}$. This finishes induction. Apply the same assertion to the quotient representation by its invariant line, and then to successive quotients. A basis adapted to the resulting complete flag gives the stated simultaneous triangularization of a Lie algebra representation, completing the proof of the Lie theorem.

A Nilpotent Lie algebra is a Solvable Lie algebra, so the inclusion $\mathfrak{g}\subseteq \mathfrak{gl}(V)$ satisfies the Lie theorem and is upper triangular in a suitable basis, for finite-dimensional complex $V$.

This is insufficient to prove the Engel theorem. Its matrix version starts with a Lie subalgebra of nilpotent endomorphisms and concludes that they are simultaneously strictly upper triangular; its abstract version concludes nilpotence from nilpotence of every Adjoint representation endomorphism. Merely upper triangular matrices can have nonzero diagonal entries: the one-dimensional algebra $\mathbb{C}{I}_V$ is an Abelian Lie algebra and a Nilpotent Lie algebra, but acts by a nonnilpotent identity matrix. This is nilpotent Lie algebras need not act nilpotently. Moreover, in the abstract Engel theorem nilpotence of the algebra is a conclusion, so assuming it first to invoke the Lie theorem would be circular. Abstract nilpotence and nilpotence of each representing matrix are different conditions.