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PropositionStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-14
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  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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Subalgebras, quotients, and extensions of solvable Lie algebras

Statement

Subalgebras and quotients of solvable Lie algebras are solvable. If i is a solvable ideal of g and g/i is solvable, then g is solvable; more precisely,

dl⁡(g)≤dl⁡(i)+dl⁡(g/i).

Facts & Assumptions

Given: A Lie algebra g, a subalgebra h, and, for the quotient and extension assertions, an ideal i.

[L1]

Solvability is termination of the derived series (Derived series and solvable Lie algebras).

[L2]

An ideal defines a quotient Lie algebra and a surjective canonical projection π:g→g/i (Quotient Lie algebras).

[L3]

The kernel of a Lie homomorphism is an ideal and its quotient by the kernel identifies with its image (Kernels, images, and the first isomorphism theorem for Lie algebras).

Proof

technique · direct
1.1givenL1algebra

Induction on r gives h(r)⊆g(r): it is clear at r=0, and bracketing a subspace with itself preserves containment. Therefore a vanishing derived term of g forces the corresponding term of h to vanish.

1.2L1L2L3algebra

Bracket preservation and surjectivity of the canonical map in [L2] give (g/i)(r)=π(g(r)) for every r. Thus every solvable quotient of g terminates no later than g; [L3] records the same calculation for any homomorphic image.

2.1L1L2L3step 1.2∎

Suppose m=dl⁡(g/i) and n=dl⁡(i). Step 1.2 gives π(g(m))=0, so g(m)⊆ker⁡π=i. Iterating the derived operation n further times gives g(m+n)=(g(m))(n)⊆i(n)=0. This proves solvability and the bound, including m=0 or n=0.

Depends on

Dependency tree · 2 levels

Sources