Subalgebras of nipotent algebras

[jostpuur]

I know that if $\mathfrak{g}$ is a finite dimensional nilpotent Lie algebra, then a subalgebra $\mathfrak{h}$ with precisely one dimension less exists.

If $\mathfrak{g}_1=[\mathfrak{g},\mathfrak{g}]$ has the codimension of one, then choose it. If instead
$$\textrm{dim}(\mathfrak{g})-\textrm{dim}(\mathfrak{g}_1)=n>1$$,
then add some arbitrary vectors to it like this
$$\mathfrak{h}=\mathfrak{g}_1\oplus X_1\oplus\cdots\oplus X_{n-1}$$
and this is a subalgebra since
$$[\mathfrak{h},\mathfrak{h}]\subset [\mathfrak{g},\mathfrak{g}]=\mathfrak{g}_1\subset \mathfrak{h}.$$

My question is, that is it possible to easily prove that such subalgebra with codimension of one exists also, if we don't know the nilpotency of the algebra, but instead only know that each element of the algebra is nilpotent, so that $X^n=0$ for all $X\in\mathfrak{g}$ for some n.

This should be true, because it can be shown that if an element is nilpotent, it is also ad-nilpotent, and the Engel's theorem says, that if each element of the Lie algebra is ad-nilpotent, then the algebra is nilpotent too. However, I'm now struggling with the proof of the Engel's theorem, so I would like some other way.

 

[fresh_42]

Yes, you already have proven it. Your construction works regardless of any properties of the Lie algebra. Since $\mathfrak{g}/[\mathfrak{g},\mathfrak{g}]$ is Abelian, you only needed that $[\mathfrak{g},\mathfrak{g}]\subsetneq \mathfrak{g}$ is a proper subalgebra. Such a subalgebra is even an ideal.

Only if $[\mathfrak{g},\mathfrak{g}]=\mathfrak{g}$ as e.g. for semisimple Lie algebras we get empty statements which are trivially true, but the construction always yields the entire Lie algebra.