[proofplan]
We compare the derived series of the quotient $\mathfrak g/\mathfrak a$ with the image of the derived series of $\mathfrak g$ under the quotient homomorphism. The key point is that the quotient map preserves Lie brackets, so taking commutators and passing to the quotient commute. Since the derived series of $\mathfrak g$ reaches $0$, the derived series of the quotient reaches $0$ at the same or an earlier stage.
[/proofplan]
[step:Define the derived series and the quotient homomorphism]
For any Lie algebra $\mathfrak h$ over $k$, define its derived series $(\mathfrak h^{(m)})_{m \geq 0}$ by
\begin{align*}
\mathfrak h^{(0)} &:= \mathfrak h, \\
\mathfrak h^{(m+1)} &:= [\mathfrak h^{(m)}, \mathfrak h^{(m)}],
\end{align*}
where $[\mathfrak u,\mathfrak v]$ denotes the $k$-linear span of all brackets $[u,v]$ with $u \in \mathfrak u$ and $v \in \mathfrak v$.
Since $\mathfrak a \trianglelefteq \mathfrak g$ is a Lie ideal, the quotient [vector space](/page/Vector%20Space) $\mathfrak g/\mathfrak a$ has the quotient Lie bracket
\begin{align*}
[x+\mathfrak a, y+\mathfrak a]_{\mathfrak g/\mathfrak a}
:= [x,y]_{\mathfrak g}+\mathfrak a
\end{align*}
for $x,y \in \mathfrak g$. Define the quotient map
\begin{align*}
\pi: \mathfrak g &\to \mathfrak g/\mathfrak a \\
x &\mapsto x+\mathfrak a .
\end{align*}
This map is a surjective Lie algebra homomorphism, because for all $x,y \in \mathfrak g$,
\begin{align*}
\pi([x,y]_{\mathfrak g})
= [x,y]_{\mathfrak g}+\mathfrak a
= [x+\mathfrak a, y+\mathfrak a]_{\mathfrak g/\mathfrak a}
= [\pi(x),\pi(y)]_{\mathfrak g/\mathfrak a}.
\end{align*}
[/step]
[step:Identify each derived algebra of the quotient with the image of the corresponding derived algebra]
We prove by induction on $m \geq 0$ that
\begin{align*}
(\mathfrak g/\mathfrak a)^{(m)} = \pi(\mathfrak g^{(m)}).
\end{align*}
For $m=0$,
\begin{align*}
(\mathfrak g/\mathfrak a)^{(0)}
= \mathfrak g/\mathfrak a
= \pi(\mathfrak g)
= \pi(\mathfrak g^{(0)}),
\end{align*}
using surjectivity of $\pi$.
Assume for some $m \geq 0$ that $(\mathfrak g/\mathfrak a)^{(m)} = \pi(\mathfrak g^{(m)})$. Then
\begin{align*}
(\mathfrak g/\mathfrak a)^{(m+1)}
&= [(\mathfrak g/\mathfrak a)^{(m)},(\mathfrak g/\mathfrak a)^{(m)}] \\
&= [\pi(\mathfrak g^{(m)}),\pi(\mathfrak g^{(m)})] \\
&= \pi([\mathfrak g^{(m)},\mathfrak g^{(m)}]) \\
&= \pi(\mathfrak g^{(m+1)}).
\end{align*}
The third equality holds because $\pi$ is a Lie algebra homomorphism: the bracket of two elements $\pi(x),\pi(y)$ with $x,y \in \mathfrak g^{(m)}$ is $\pi([x,y])$, and taking $k$-linear spans gives equality of the two subspaces. Thus the induction is complete.
[guided]
We want to show that the quotient derived series is controlled by the original derived series. The precise assertion is that, for every $m \geq 0$,
\begin{align*}
(\mathfrak g/\mathfrak a)^{(m)} = \pi(\mathfrak g^{(m)}).
\end{align*}
The base case is the statement that the zeroth derived algebra of the quotient is the whole quotient:
\begin{align*}
(\mathfrak g/\mathfrak a)^{(0)}
= \mathfrak g/\mathfrak a.
\end{align*}
Since $\pi: \mathfrak g \to \mathfrak g/\mathfrak a$ is surjective, this is exactly $\pi(\mathfrak g)$. Since $\mathfrak g^{(0)}=\mathfrak g$, we obtain
\begin{align*}
(\mathfrak g/\mathfrak a)^{(0)}=\pi(\mathfrak g^{(0)}).
\end{align*}
Now assume the equality holds at level $m$. The next derived algebra is obtained by taking all brackets inside the previous one and then taking their $k$-linear span:
\begin{align*}
(\mathfrak g/\mathfrak a)^{(m+1)}
= [(\mathfrak g/\mathfrak a)^{(m)},(\mathfrak g/\mathfrak a)^{(m)}].
\end{align*}
Using the induction hypothesis, this becomes
\begin{align*}
(\mathfrak g/\mathfrak a)^{(m+1)}
= [\pi(\mathfrak g^{(m)}),\pi(\mathfrak g^{(m)})].
\end{align*}
The quotient map preserves brackets, so for $x,y \in \mathfrak g^{(m)}$,
\begin{align*}
[\pi(x),\pi(y)]_{\mathfrak g/\mathfrak a}
= \pi([x,y]_{\mathfrak g}).
\end{align*}
Therefore the $k$-linear span of all brackets of elements from $\pi(\mathfrak g^{(m)})$ is exactly the image under $\pi$ of the $k$-linear span of all brackets of elements from $\mathfrak g^{(m)}$:
\begin{align*}
[\pi(\mathfrak g^{(m)}),\pi(\mathfrak g^{(m)})]
= \pi([\mathfrak g^{(m)},\mathfrak g^{(m)}]).
\end{align*}
By the definition of the derived series,
\begin{align*}
[\mathfrak g^{(m)},\mathfrak g^{(m)}] = \mathfrak g^{(m+1)}.
\end{align*}
Thus
\begin{align*}
(\mathfrak g/\mathfrak a)^{(m+1)}
= \pi(\mathfrak g^{(m+1)}).
\end{align*}
This proves the induction step, and hence the formula holds for all $m \geq 0$.
[/guided]
[/step]
[step:Use solvability of $\mathfrak g$ to terminate the quotient derived series]
Since $\mathfrak g$ is solvable, there exists an integer $N \geq 0$ such that
\begin{align*}
\mathfrak g^{(N)} = 0.
\end{align*}
Using the identity from the previous step at $m=N$, we get
\begin{align*}
(\mathfrak g/\mathfrak a)^{(N)}
= \pi(\mathfrak g^{(N)})
= \pi(0)
= 0.
\end{align*}
Thus the derived series of $\mathfrak g/\mathfrak a$ terminates at $0$, so $\mathfrak g/\mathfrak a$ is solvable.
[/step]
