Extend solvable Lie subalgebra to subalgebra This is a question I've been pondering for a course and have made solid progress on, but am lacking in the last step. Let $L$ be a Lie algebra, and $S$ is a proper, solvable Lie subalgebra of dimension $d$, how can I construct a Lie subalgebra $S'$ of dimension $d+1$ such that $S \subset S'$?
Progress so far is that given the normalizer of $S$, $S \subset N(S)$:
$$N(S)= \{x \in L | [x,s] \in S ~\forall s\in S \}$$
If $S$ is not all of $N(S)$, appending an element from $N(S)$ suffices (the new Lie algebra is also solvable, but this is not possible in general). However, in the case where $N(S)=S$ things get more difficult. Some avenues of attack so far:
Lie's Theorem gives me a common eigenvector in the adjoint representation, that is, there exists some $x\in N(S)$ such that $[s,x]=\lambda(s)x$ for all $s\in S=N(S)$. I'd like to somehow extend or bracket this special element with another in $L$ that keeps it to be a subalgebra. 
Alternatively, there could be some utility in thinking of $S$ as a subset of upper triangular matrices in the adjoint representation (also Lie's theorem). But then, I am struggling for what assumptions I may place on what the rest of $L$ looks like as an adjoint representation. 
Any advice or directions is much appreciated!
 A: I too had this problem in a course, and I too struggled with it for a while. Eventually, with much help, I come to a solution:
Consider the Lie algebra homomorphism
\begin{align*}
\Phi:S &\to \mathfrak{gl}(L/S)\\
s &\mapsto \phi_s
\end{align*}
where $\phi_s$ is the natural action of $\text{ad}_s$ on $L/S$ (where $\text{ad}_x$ is the endomorphism of $L$ defined my $y\mapsto [x,y]$); this action is well defined since for every $a,b\in S$ and $x\in L$ we have $[a,x+b]=[a,x]+[a,b]$, noting that $[a,b]\in S$.
As $S$ is solvable, its homomorphic image $\Phi(S)$ is also solvable.  
Now recall the following result:

Theorem: (Humphreys' Introduction to Lie Algebras and Representation Theorem Theorem 4.1)
Let $L$ be a solvable subalgebra of $\mathfrak{gl}(V)$, $V$ finite dimensional. If $V\neq 0$, then $V$ contains a common eigenvector for all the endomorphisms in $L$.

(Remark: Lie's theorem is a corollary of this theorem).
Applying the above theorem to $\Phi(S)$, we have that there exists a nonzero element $v\in L/S$ which is an eigenvector of the endomorphisms $\Phi(s)$, for all $s\in S$. In particular $v\not\in S$ (since $v$ is nonzero in $L/S$) and
$$ \Phi(a)(v)=[s,v]=\lambda_s v$$
for all $s\in S$, where $\lambda_s\in \mathbb{C}$ is a scalar. Setting $S'=S\oplus \mathbb{C}_v$, where $\mathbb{C}_v$ is the one dimensional Lie algebra generated by $v$, we have that $\text{dim}(S')=\text{dim}(S)+1$ and is a subalgebra of $L$.
