Question on Lie's theorem

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 $\mathfrak{g/[g,g]}$?

• The link to the notes no longer works. – Dietrich Burde Dec 6 '17 at 13:40

If $U$ and $V$ are vector spaces, $V\subseteq U$, the subspaces of $U/V$ are in 1-1 correspondence with those subspaces of $U$ that contain $V$. A more verbose explanation follows.
By basic linear algebra (all the spaces are finite dimensional) it is possible to choose a subspace $\mathfrak{h}$ in such a way that $$[\mathfrak{g},\mathfrak{g}]\subseteq \mathfrak{h}\subseteq\mathfrak{g},$$ and $\dim \mathfrak{h}=\dim\mathfrak{g}-1$. For example, pick a basis of $[\mathfrak{g},\mathfrak{g}]$, extend that to a basis of $\mathfrak{g}$, and drop one of the basis elements that you added. What remains is a basis of such a space $\mathfrak{h}$.
Then $$[\mathfrak{h},\mathfrak{g}]\subseteq [\mathfrak{g},\mathfrak{g}]\subseteq\mathfrak{h},$$ so $\mathfrak{h}$ is an ideal.
If so desired, you can select another codimension (up to $\dim \mathfrak{g}-\dim [\mathfrak{g},\mathfrak{g}]$) by dropping an appropriate number of basis elements.