I know that if $f$ is a polynomial over a subfield $F$ of $\mathbb{C}$ and $f$ is solvable by radicals, then the Galois group of $f$ over $F$ is solvable. I’ve also seen many applications of this fact in demonstrating that certain quintic polynomials over $\mathbb{Q}$ are not solvable by radicals. My question is this: […]

Prove a group of order $48$ must have a normal subgroup of order $8$ or $16$. Solution: The number of Sylow $2$-subgroups is $1$ or $3$. In the first case, there is a normal subgroup of order $16$ so we are done. In the second case, let $G$ act by conjugation on the Sylow $2$-subgroups. […]

I’ve come across an exercise saying the following: If $G$ is a finite group which contains a maximal subgroup $M$ which is abelian, show that $G$ is solvable and that $G^{(3)}$ (the third term in the derived series) equals 1. I can see that it’s sufficient to show that $G^{(2)}$ is a subgroup of $M$, […]

We’ll denote $A \lhd G$ for $A$ a normal subgroup of $G$; and $A \leq G$ to mean $A$ is a subgroup of $G$. I don’t know if it’s a tricky question. But it seems strange to me. The question asks for an example of $A \le G$ solvable, $B \lhd G$ solvable, but $AB$ […]

I am looking for examples of finitely generated solvable groups that are not polycyclic. In Wikipedia Baumslag-Solitar group $BS(1,2)$ is an example. But how to prove this fact?

Can one characterize groups $G$ for which there is a unique normal subgroup of order $d$ for every divisor of the order of $G$? For example must they be solvable?

My question is straightforward to pose: given a polynomial $f$ over a subfield of $\mathbb{C}$, are there conditions which guarantee the existence of a closed formula for the roots of $f$ in terms of its coefficients? $\textit{Edit:}$ By “closed formula”, I am referring to a formula using addition, subtraction, multiplication, division, and extraction of roots, […]

If $G$ is a connected solvable Lie-group, then $[G,G]$ is nilpotent. The corresponding statement for Lie algebras follows from Lie’s theorem, and it then follows from connected Lie groups by exponentiation. Is the statement also true for finite groups? I can’t find a counterexample, but I didn’t try that hard. Motivation: I’m prepping to teach […]

I am trying to prove that a group of order $p^2q^2$ where $p$ and $q$ are primes is solvable, without using Burnside’s theorem. Here’s what I have for the moment: If $p = q$, then $G$ is a $p$-group and therefore it is solvable. If $p \neq q$, we shall look at the Sylow $p$-subgroups […]

Show that every minimal normal subgroup of a finite solvable group is an elementary abelian $p$-group for some prime $p$. I’m stuck on this one, any idea is appreciated.

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