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

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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:

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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:

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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 :)