Derived subalgebra is an ideal

For any Lie algebra , the commutator subalgebra is an ideal of .
Derived subalgebra is an ideal

Let g\mathfrak g be a .

Lemma

The [g,g][\mathfrak g,\mathfrak g] is an in g\mathfrak g; equivalently,

[g,[g,g]][g,g]. [\mathfrak g,\,[\mathfrak g,\mathfrak g]] \subseteq [\mathfrak g,\mathfrak g].

Proof

Take xgx\in\mathfrak g and an element of [g,g][\mathfrak g,\mathfrak g] of the form [y,z][y,z]. Using the Jacobi identity for the ,

[x,[y,z]]=[[x,y],z]+[y,[x,z]]. [x,[y,z]] = [[x,y],z] + [y,[x,z]].

Each term on the right is a commutator of two elements of g\mathfrak g, hence lies in [g,g][\mathfrak g,\mathfrak g]. By linearity, [x,w][g,g][x,w]\in[\mathfrak g,\mathfrak g] for all w[g,g]w\in[\mathfrak g,\mathfrak g], proving the claim.

Context. This lemma ensures that iterating [,][\cdot,\cdot] produces characteristic ideals (e.g. the ), making solvability a robust isomorphism-invariant notion.