Recently I've been looking through Lang's Algebra, and I encountered a problem in the proof of Proposition 3.1 in Chapter I Groups.
Proposition 3.1. Let $G$ be a finite group. An abelian tower of $G$ admits a cyclic refinement. Let $G$ be a finite solvable group. Then $G$ admits a cyclic tower, whose last element is $\{e\}$.
Proof. The second assertion is an immediate consequence of the first, and it clearly suffices to prove that if $G$ is finite, abelian, then $G$ admits a cyclic tower. We use induction on the order of $G$. Let $x$ be an element of $G$. We may assume that $x \neq e$. Let $X$ be the cyclic group generated by $x$. Let $G' = G/X$. By induction, we can find a cyclic tower in $G'$, and its inverse image is a cyclic tower in $G$ whose last element is $X$. If we refine this tower by inserting $\{e\}$ at the end, we obtain the desired cyclic tower.
I don't understand why it suffices to prove that if $G$ is finite, abelian, then $G$ admits a cyclic tower. In the statement of the proposition $G$ is not assumed to be an abelian group.
Moroever, even assuming that we do prove that if $G$ is finite, abelian, then $G$ admits a cyclic tower, I don't see how can we use this in proving Proposition 3.1.
Maybe this question is very easy, but currently I can't understand it. Any help would be appreciated.
Best Answer
You are assuming we have an abelian tower for the finite group $\;G\;$ :
$$(**)\;\;\;1=G_m\lhd G_{m-1}\lhd\ldots\lhd G_1\lhd G_0:=G\;,\;\;s.t.\;\;G_i/G_{i+1}\;\;\text{abelian}\;\;\forall\,1=0,1,...,m-1 $$
The above means in particular that $\;G_{m-1}\cong G_{m-1}/G_m\;$ is abelian, so by the part marked in red in the proof, there's a cyclic refinement of it:
$$1= A_0\lhd A_1\lhd\ldots\lhd A_{m_1}:=G_{m-1}\;,\;\;A_k/A_{k+1}\;\;\text{cyclic}$$
But also $\;G_{m-2}/G_{m-1}\;$ is abelian, so again by the red part we've a cyclic refinement
$$G_{m-1}=:B_0\lhd B_1\lhd\ldots\lhd B_{m_2}:=G_{m-2}\;,\;\;B_i/B_{i+1}\;\;\text{cyclic}$$
Observe now that the subrefinement ("sub" because it is a refinement of part of the original tower)
$$1=G_m:=A_0\lhd A_1\lhd\ldots\lhd A_{m_1}=G_{m_1}=B_0\lhd B_1\lhd\ldots\lhd B_{m_2}=G_{m_2}$$
is cyclic! Well, go on like this inductively up through the whole first, original tower (**) ...