I am looking a Lie's theorem in Lie algebra liturature but I do not fully understand one part of the proof. The following proof is given in these notes on page 12.
Thm. Let $\mathfrak{g}\subset \mathfrak{gl}(V)$ be solvable, where $V$ is a nonzero vector space. Then $V$ contains a common eigenvector for $\mathfrak{g}$.
Proof: The first step in the proof of this theorem is to find an ideal of $\mathfrak{g}$ of codimension one. Lets assume $dim(\mathfrak{g})>1$. $\mathfrak{g}$ is solvable so [$\mathfrak{g}, \mathfrak{g}$]$\subset \mathfrak{g}$. Then $\mathfrak{g/[g,g]}$ is a nonzero abelian algebra. Hence any subspace is an abelian ideal. (Following is the part I don't understand...)
Any codimension one subspace of $\mathfrak{g/[g,g]}$ then lifts to a codimension one ideal in $\mathfrak{[g,g]\subset \mathfrak{h} \subset \mathfrak{g}}$
Can someone explain plainly why this must be so?
Also, why is it possible to choose a codimension on subspace of $\mathfrak{g/[g,g]}$?