Why does existence of abelian tower not imply solvability? While I am reading Serge Lang's Algebra, I am confused by the definition of solvable group (page 18).
In the book, $G$ is solvable if $G$ has an abelian tower with last element being trivial subgroup. Is he trying to say a group with an abelian tower may be non-solvable? If so, what is an example of this?
It seems to me that if $G$ has an abelian tower then $G$ is solvable: say 
$$
G_{1}\subset G_2 \subset \cdots \subset G_n=G
$$
is an abelian tower of $G$. Why can't I just add $\{e\}=G_{0}$ (which is clearly normal in $G_{1}$) to the sequence? To me it looks like possessing an abelian tower implies solvability.
 A: Expanding @user98602's comment.

Suppose that you add $\{e\}$ to the end of the given abelian tower of $G$ to get
$$
\{e\} \subset G_1 \subset G_2 \subset \cdots \subset G_n = G.
$$
When will this tower be solvable? Only when every quotient is abelian. Since we already have that each $G_{i+1}/G_i$ is abelian for $i=1,\dots,n$, this tower is solvable only when $G_1/\{e\} = G_1$ is abelian. But, it need not be the case that $G_1$ is abelian, because in an abelian tower it is only the factor groups $G_{i+1}/G_i$ that are abelian, and not necessarily the individual subgroups $G_i$.
For a concrete example, take the abelian tower of $S_n$ given by
$$
A_n \subset S_n,
$$
where $n \geq 4$. Since $S_n / A_n \cong \Bbb{Z} / 2\Bbb{Z}$, this is an abelian tower, but
$$
\{e\} \subset A_n \subset S_n
$$
is not an abelian tower since $A_n$ is not abelian for $n \geq 4$. Note however that there does exist an abelian tower of $S_4$ ending at $\{e\}$, but that no such tower exists for $n \geq 5$. Said differently, $S_4$ is solvable, whereas $S_n$ is not solvable for $n \geq 5$.
