I am trying to understand the proof of the proposition below. There are other discussion of this here and here but I find them fragmented and unclear, so I am writing it all out here in perhaps aggravating detail in order to try to fully understand it. Questions and points of confusion are in bold. I appreciate verifications of what I've written and responses to the text in bold.
Excerpt:
Let's first prove that if $G$ is finite and abelian, then $G$ admits a cyclic tower ending in $\{ e \}$. That this implies the results we want can be dealt with after.
We use induction on the order of $G$. If $G = \{ e \}$, then the cyclic tower is $G \supset \{ e \}$, and this is cyclic because $G / \{ e \} \cong G = \{ e \}$ using the first isomorphism theorem (is that right?).
Now we use strong induction and assume that finite and abelian $G$ with $1 \leq i \leq n – 1$ elements admits a cyclic tower ending in $\{ e \}$, and consider finite and abelian $G$ with $n$ elements. We consider element $x \neq e \in G$ because $G$ has at least two elements, and if we took $x = e$ then our quotient group construction would have $n$ elements and so we couldn't take advantage of our inductive assumption.
Let $X$ be the cyclic group generated by $x$. Since $G$ is abelian, $X$ is normal in $G$ and $G / X$ is abelian, and it also has at most $n – 1$ elements (how would I prove that?) which means that $G / X$ admits a cyclic tower ending in $\{ e \}$. This tower would look something like
\begin{align*}
G/X := G_1/X \supset G_2/X \supset \dots \supset G_{m – 1}/X \supset G_m/X := \{ e \}.
\end{align*}
(How is the trivial group a subgroup of quotient groups? The quotient groups are comprised of cosets, not individual elements.)
If we consider the quotient homomorphism $f \colon G \to G/X$, then the preimages of the subgroups $G_j/X$ under $f$ are subgroups of $G$, and moreover normality and cyclicity are preserved, which means that if $f^{-1} (G_j/X) := G_j$, we get a cyclic tower
\begin{align*}
G := G_1 \supset G_2 \supset \dots \supset G_{m – 1} \supset G_m := X.
\end{align*}
(I don't understand how the preimage of $\{ e \}$ under $f$ is $X$.)
We can refine the tower immediately above by adding $\{ e \}$ at the end, which works because $\{ e \}$ is normal in $X$, $X / \{ e \} \cong X$ so the additional quotient group formed by the addition of $\{ e \}$ to the cyclic tower is cyclic, which means that the cyclic tower with $\{ e \}$ added is still cyclic, as desired.
Now we need to address why what we've proven implies the statement of the proposition. Assume we have a finite group $G$ with an abelian tower. We want to show that this tower can be refined into a cyclic tower. The abelian tower of $G$ looks like
\begin{align*}
G := G_1 \supset G_2 \supset \dots \supset G_{n – 1} \supset G_n.
\end{align*}
Consider the ultimate quotient $G_{n – 1}/G_n$. This is a finite abelian group and by what we've shown above, it admits a cyclic tower ending in $\{ e \}$:
\begin{align*}
G_{n – 1}/G_n := G_{n – 1, 1}/G_n \supset G_{n – 1, 2}/G_n \supset \dots \supset G_{n – 1, n_n}/G_n := \{ e \}.
\end{align*}
We use the quotient homomorphism $g \colon G_{n – 1} \to G_{n – 1}/G_n$ to deduce that there is a “preimage cyclic tower'' beginning with $G_{n – 1}$ and ending in $G_n$:
\begin{align*}
G_{n – 1} \supset \dots \supset G_n.
\end{align*}
We can use this to refine the abelian tower for $G$:
\begin{align*}
G := G_1 \supset G_2 \supset \dots \supset G_{n – 1} \supset \dots \supset G_n
\end{align*}
and the final segment of this abelian tower is cyclic. Repeat this process for the penultimate quotient $G_{n – 2}/G_{n – 1}$ and you will get a cyclic tower beginning with $G_{n – 2}$ and ending in $G_{n – 1}$, which can be inserted as a refinement into the already partially refined tower for $G$. Do this for all the other quotients and you've got a cyclic refinement of the abelian tower of $G$.
Lastly, if $G$ is finite and solvable, that means its abelian tower ends in $\{ e \}$. The above process works just the same way, giving us a cyclic refinement of this tower that begins with $G$ and ends with $\{ e \}$.
Best Answer
You correctly understand the first point.
$\vert G/X \vert = \frac{\vert G \vert}{\vert X \vert} \lt \vert G \vert$ (because $\vert X \vert \gt 1$) answers your second question.
The trivial group being a quotient group is a slight abuse of notation. You're correct that the elements of the quotient group are cosets, but if the only coset is the one containing the identity, we simply say that the quotient group is the (unique) one-element group, $\{ e \}$.
The preimage of $e$ is $X$ because the preimage of any element is the coset containing that element. The coset containing $e$ is $X$ itself.