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This is Exercise 5.4.7 of Robinson's "A Course in the Theory of Groups (Second Edition)". According to this search, it is new to MSE.

The Details:

(This can be skipped.)

Denote the derived subgroup of a group $G$ by $G'$. It is the subgroup of $G$ generated by the commutators $[g,h]=g^{-1}h^{-1}gh$ of $G$ for all $g,h\in G$.

Since definitions vary, the definition of a normal subgroup $H\unlhd G$ of a group $G$ I use is a subgroup of $G$ such that $$aH=Ha$$ for all $a\in G$.


Robinson's definition of the following is equivalent to the one given on proof wiki:

Let $G$ be a group whose identity is $e$.

A normal series for $G$ is a sequence of normal subgroups of $G$:

$$\{e\}=G_0\lhd G_1\lhd\dots\lhd G_n=G,$$

where $G_{i-1}\lhd G_i$ denotes that $G_{i-1}$ is a proper normal subgroup of $G_i$.

On page 122 of Robinson's book,

Definition: A group $G$ is called nilpotent if it has a central series, that is, a normal series $1=G_0\le G_1\le \dots \le G_n=G$ such that $G_{i+1}/G_i$ is contained in the centre of $G/G_i$ for all $i$.


A supersolvable group $G$ is a group with a normal series

$$1=N_0\unlhd N_1\unlhd N_2\unlhd \dots\unlhd N_n=G$$

such that each factor $N_{i+1}/N_i$ is cyclic and $N_i\unlhd G$.

The Question:

Paraphrasing,

If $N\unlhd G$ is nilpotent and $G/N'$ is supersolvable, then $G$ is supersolvable.

Thoughts:

(This can be skipped.)

Usually, when stuck on a problem, I would try out an example; I couldn't think of a nilpotent subgroup $N\unlhd G$ such that $G/N'$ is supersolvable. I had a look here for examples of supersolvable groups, hoping to reverse engineer something to experiment with, but to no avail.


If it helps, I know that each supersolvable group is solvable (but the converse doesn't hold).


I can see roughly why the result holds. The quotient $G/N'$, being supersolvable, has a normal series whose quotients are cyclic, so we might be able to exploit one or more of the isomorphism theorems to get such a series for $G$; I don't see where $N$ being nilpotent is needed though.


I had a look in:

  • Rotman's, "An Introduction to the Theory of Groups (Fourth Edition)".
  • Roman's, "Fundamentals of Group Theory: An Advanced Approach".
  • Hall's, "The Theory of Groups".
  • Rose's, "A Course in Group Theory".

There is nothing obvious in them that I could find to help out here.


Here are some previous questions of mine on supersolvable groups:

Previous questions of mine about nilpotent groups are listed here:


Please help :)

Shaun
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    Why the downvote? – Shaun Jan 26 '22 at 19:24
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    I don’t know but you have an overflow of information. From me +1, but perhaps try to be more concise. – Nicky Hekster Jan 26 '22 at 20:19
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    Thank you, @NickyHekster. I like to be thorough. – Shaun Jan 26 '22 at 20:21
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    The first part of your question is not relevant: this site is intended for asking a question, and unless you expect users to review and read all your linked questions, it is irrelevant to your current question. And if you expect users to read through five previous questions from you, in order to answer, that is inappropriate to expect from users. – amWhy Jan 26 '22 at 20:32
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    I don't expect other users to read through them in detail, @amWhy; I am just supplying context. – Shaun Jan 26 '22 at 20:35
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    I was responding to your request in Constructive Feedback, but can not yet chat again. I think the questions, if you don't expect users to read them, are self-serving, or at least it may very well appear to be so. Please focus your question. I appreciate your goal of being thorough, but you cite the source, explain the problem, add your thoughts. No need to worry about context: those inclusions add plenty of context. Just a suggestion for how you can be more concise and focused, and not wear out potential answerers. – amWhy Jan 26 '22 at 20:38
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    It is appreciated, @amWhy. – Shaun Jan 26 '22 at 21:46
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    Thank you, @PenAndPaperMathematics. – Shaun Jan 26 '22 at 21:46
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    You say that you cannot see where $N$ is nilpotent is needed, but it's obviously false without that assumption. Let $G=N=A_5$, for example. – Derek Holt Jan 27 '22 at 11:00
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    To solve this, you need to read carefully the section "Tensor Products and Lower Central Factors" in Section 5.2 of Robinson's book, and apply it with $G = N$, noting that the $N$ in the question is an $\Omega$-operator group with $\Omega = G$. – Derek Holt Jan 28 '22 at 09:04

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