I will begin by stating my question, and then write down some related thoughts.

Let $\mathfrak{g}$ be a finite dimensional nilpotent Lie algebra over $\mathbb{C}$. Choose an ideal $\mathfrak{h}$ in $\mathfrak{g}$ of codimension 2. The quotient $\mathfrak{g}/\mathfrak{h}$ is then abelian. If $L$ is any 1-dimensional subspace of $\mathfrak{g}/\mathfrak{h}$, we can form its preimage in $\mathfrak{g}$, which is a Lie subalgebra.

True/False? The set of isomorphism classes of Lie algebras obtained in this way is finite. (Here I am disregarding the embedding into $\mathfrak{g}$ and asking about isomorphisms as abstract Lie algebras.)

EDIT. To clarify the statement: both $\mathfrak{g}$ and $\mathfrak{h}$ are fixed. Only the 1-dimensional subspace $L$ of the quotient $\mathfrak{g}/\mathfrak{h}$ varies.

Remark 1. A counterexample, if there is one, could only exist in dimension $\geq 8$. This is why it's pretty difficult to "get my hands on" this problem. I don't see an easy way to prove the statement, nor do I see any obvious counterexamples.

Remark 1'. I tried to assume that the answer is positive and get some kind of contradiction with the statement that the set of isomorphism classes of nilpotent Lie algebras of dimension $n$ (where $n\geq 7$) is infinite, but I couldn't find one.

Remark 2. This is not very helpful, but at least the answer to my question is positive when $\mathfrak{h}$ is abelian. (This does severely limit the possibilities for $\mathfrak{g}$, but at least it does not limit the nilpotence class of $\mathfrak{g}$.)

Any information about this would be much appreciated!

`$\mathfrak g$`

, the number of isomorphism classes of subalgebras that are the preimages of one-dimensional subalgebras of a two-dimensional quotient algebra. But it's not clear to me what is fixed and what is varying. – Theo Johnson-Freyd Apr 14 '10 at 5:50