Let $G$ be a group and $H$ a normal subgroup. Then $G$ is solvable if and only if $H$ and $G/H$ are solvable. (Lang's Algebra, p. 19) There are questions about this topic elsewhere but I haven't seen any other question addressing this particular proof.
Excerpt:



Discussion and questions:
I will type out what I think I understand of this proof, and write points of confusion in bold.
The reason that $H_{i + 1}$ is normal in $H_i$ is because $xHx^{-1} = H$ for all $x \in G$, and $yG_{i + 1}y^{-1} = G_{i + 1}$ for all $y \in G_i$, therefore $z(H \cap G_{i + 1})z^{-1} = H \cap G_{i + 1}$ for all $z \in H \cap G_i$. Is that right?
The embedding comes from the following diagram:
.
The embedding preserves commutativity, whence $H$ is solvable.
To prove that solvability of $G$ implies solvability of $G/H$, I am told that this passage essentially provides the proof:

.

I can see that, again, an injective homomorphism will preserve commutativity. So I need an injective homomorphism along the lines of $G/H \to \ ?$. But I'm not sure how to come up with this homomorphism.
Lastly, I need to prove that solvability of $H$ and $G/H$ imply solvability of $G$. Consider the abelian tower for $G/H$:
\begin{align*}
G/H := G_1/H \supset G_2/H \supset \dots \supset G_{m - 1}/H \supset G_m/H := \{ e \}.
\end{align*}
(It's my understanding that $G_m/H$ is actually $H$, not $\{ e \}$, but I digress.)
If we consider the quotient homomorphism $f \colon G \to G/H$, then the preimages of the subgroups $G_i/H$ of $G/H$, $f^{-1}(G_i/H) := \mathcal{G_i}$, are subgroups of $G$, and the preimages preserve normality. Thus we have the normal tower
\begin{align*}
G = \mathcal{G_1} \supset \mathcal{G_2} \supset \dots \supset \mathcal{G_n} = H.
\end{align*}
Using one of the isomorphism theorems, we have that $G_i/H \bigg/ G_{i + 1}/H \cong \mathcal{G_i} / \mathcal{G_{i + 1}}$, and the isomorphism preserves commutativity. The last thing we have to do is show that this tower for $G$ can be refined so it ends with $\{ e \}$. This is clear because solvability of $H$ ensures that we can tack on its abelian tower, which ends with $\{ e \}$ as desired.
I appreciate any help.
 A: I found some course notes here that seem to help me prove that solvability of $G$ implies solvability of $G/H$. I'm not sure this approach uses the hint given, but a solution is still a solution.
Given the abelian tower
\begin{align*}
G \supset G_1 \supset G_2 \supset \dots \supset G_n = \{ e \},
\end{align*}
consider the quotient homomorphism $f \colon G \to G/H$. Because homomorphisms send subgroups to subgroups, we get a tower
\begin{align*}
f(G) = G/H \supset f(G_1) \supset \dots \supset f(G_n) = H.
\end{align*}
We want to show that this tower is abelian, which will show that $G/H$ is solvable.
First we show normality of $f(G_{i + 1})$ in $f(G_i)$. Given $g_i \in f(G_i)$ and $g_{i + 1} \in f(G_{i + 1})$, we want to show that $g_i g_{i + 1} g_i^{-1} \in f(G_{i + 1})$. There is some $g_i' \in G_i$ such that $f(g_i') = g_i$, and some $g_{i + 1}' \in G_{i + 1}$ such that $f(g_{i + 1}') = g_{i + 1}$. Thus
\begin{align*}
g_i g_{i + 1} g_i^{-1} &= f(g_i') f(g_{i + 1}') f(g_i')^{-1}\\
&= f(g_i' g_{i + 1}' g_i'^{-1})
\end{align*}
and since $G_{i + 1}$ is normal in $G_i$, we see that $g_i' g_{i + 1}' g_i'^{-1} \in G_{i + 1}$ which shows that $g_i g_{i + 1} g_i^{-1} \in f(G_{i + 1})$ as desired.
Now we need to show commutativity of $f(G_i)/f(G_{i + 1})$. Consider the diagram below:
.
Because the homomorphism across the top is surjective, and $G_{i + 1}$ is the kernel, we get an isomorphism between $G_i/G_{i + 1}$ and $f(G_i)/f(G_{i + 1})$, which establishes commutativity, as desired.
