# 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.

• The $G_i$ needn't be abelian. What's stopping you is that $G_1$, in your example, need be abelian; and if this is the case sure, you're done. – user98602 Sep 3 '14 at 22:57
• Ah, I see. That was my mistake. – user45765 Sep 3 '14 at 22:58

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 $$n \geq 5$$.