TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-14

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Engel's theorem

Statement

A finite-dimensional Lie algebra $g$ over any field is nilpotent if and only if ${ad}_{x}$ is a nilpotent endomorphism of $g$ for every $x \in g$ .

Facts & Assumptions

Given: A finite-dimensional Lie algebra $g$ .

[L1]

Nilpotence is termination of the lower central series (Lower central series and nilpotent Lie algebras).

[L2]

A nil finite-dimensional representation has a flag lowered by every represented operator (Engel triangularization theorem).

[L3]

Inner derivations form the adjoint representation, with ${ad}_{x} (y) = [x, y]$ (Derivations form a Lie algebra and inner derivations an ideal).

Proof

technique · direct

1.1givenL1L3algebra

Suppose $γ_{c + 1} (g) = 0$ . For $x, y \in g$ , the vector $({ad}_{x})^{c} (y)$ is a left-nested bracket with $c$ copies of $x$ , hence lies in $γ_{c + 1} = 0$ by [L1] and [L3]. Thus every ${ad}_{x}$ is nilpotent. This also covers $g = 0$ .

2.1givenL1L2L3algebra∎

Conversely suppose every ${ad}_{x}$ is nilpotent. The adjoint representation in [L3] is nil, so [L2] gives $0 = V_{0} \subset V_{1} \subset \dots \subset V_{n} = g$ with $[g, V_{i}] \subseteq V_{i - 1}$ . Induction then gives $γ_{r + 1} (g) \subseteq V_{n - r}$ for $0 \leq r \leq n$ , and in particular $γ_{n + 1} \subseteq V_{0} = 0$ . Hence $g$ is nilpotent by [L1].

Depends on

Dependency tree · 2 levels

Sources

Milne, Lie Algebras, Corollary 2.11