# 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
• @DonAntonio On Lang's book, it's actually not a requirement that the last item of the tower is the trivial subgroup. But I see on Wikipedia entry "subgroup series" it is actually a part of the definition. Should I just admit the Wikipedia definition and forget about Lang's? Because otherwise your argument doesn't hold since the last (or in your argument the first) element is not necessarily the trivial subgroup. – Sayako Hoshimiya Jul 24 at 20:05
• @SayakoHoshimiya I can only guess to what you mean when you say "it's actually not a requirement". Not a requirment..." for what ? For a tower? For course we don't need "the last" (or first, "smallest") group in the trivial one. Did you mean for solvable groups? Then yes: we must have that the smallest group in the abelian tower is the trivial one. – DonAntonio Jul 25 at 16:16

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.