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.

  • 3
    $\begingroup$ Because each quotient in the tower is abelian, so you just refine each one. $\endgroup$ – Tobias Kildetoft Oct 29 '13 at 11:22
  • $\begingroup$ Thanks for your help.I already understand it. $\endgroup$ – Abel Oct 30 '13 at 11:42
  • $\begingroup$ Related: math.stackexchange.com/q/39609/279515 $\endgroup$ – 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 (**) ...

  • $\begingroup$ 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. $\endgroup$ – Abel Oct 30 '13 at 12:31
  • 1
    $\begingroup$ 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. $\endgroup$ – 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.


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