# An abelian tower of a finite group admits a cyclic refinement - Proposition I.3.1, Lang's *Algebra*

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.

• Because each quotient in the tower is abelian, so you just refine each one. – Tobias Kildetoft Oct 29 '13 at 11:22
• Thanks for your help.I already understand it. – Abel Oct 30 '13 at 11:42
• – Brahadeesh Mar 6 at 14:03

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 (**) ...

• Firstly,thanks for your answer.Your answer help me to understand previous question.But I think there are some points in your answer which are wrong. $G_{m-1}\cong G_{m-1}/G_m$~is correct if only if~$G_m=\{e\}$ So your following proof has some problems. – Abel Oct 30 '13 at 12:31
• No, it hasn't: the given isomorphism is true only for the specific case $\;G_m=1\;$ , as written there. After that we don't use more this. – DonAntonio Oct 30 '13 at 15:40

This is my attempt to answer the question using the help I received from Tobias Kildetoft and Don Antonio.

Suppose that we have proved the assertion that if $$G$$ is finite, abelian, then $$G$$ admits a cyclic tower. And we already have an abelian tower for the finite group $$G$$ $$1 = G_m \triangleleft G_{m-1} \triangleleft \dots \triangleleft G_1 \triangleleft G_0 :=G,$$ such that $$G_i/G_{i+1}$$ is abelian for all $$i = 0,1,\dots,m-1$$.

For every abelian group $$G_i/G_{i+1}$$, there exists a canonical homomorphism $$G_i \to G_i/G_{i+1}$$. One of the isomorphism theorems says that this map establishes a bijection between subgroups of $$G/X$$ and subgroups of $$G$$ that contain $$X$$. Moreover, this bijection preserves inclusions, normality and quotients. So $$G_i/G_{i+1}$$ admits a cyclic tower. Then $$G_i$$ admits a cyclic tower whose last element is $$G_{i+1}$$. So, for every group $$G_i$$ ($$i=0,\dots,m-1$$) there is a cyclic tower whose last element is $$G_{i+1}$$. Hence, we can refine an abelian tower to a cyclic tower.