Questions tagged [solvable-groups]

For questions on solvable groups, their properties, and structure.

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Example of a finitely generated metabelian group whose Fitting subgroup is not nilpotent

It is known that the Fitting subgroup of a finitely generated polycyclic-by-finite group is nilpotent, but this statement is not true for the solvable group. It is clear that both Lamplighter groups ...
• 1,631
1 vote
32 views

Let $G$ be a group of order $p^nq$ where $p$ and $q$ are distinct primes and suppose $q \nmid p^i-1$ for $1 \leq i \leq n-1$. Prove $G$ is solvable

This is an extension of this post. Let $G$ be a group of order $p^nq$ where $p$ and $q$ are distinct primes and suppose $q \nmid p^i-1$ for $1 \leq i \leq n-1$. Prove that $G$ is solvable. This can be ...
• 2,756
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Real forms of a solvable Lie algebra

Consider the Lie algebra $\mathfrak{r}_2=\mathfrak{aff}(\mathbb{C})$ of the group of affine maps of $\mathbb{C}$, and let $\mathfrak{g}=\mathfrak{r}_2 \oplus \mathfrak{r}_2$. I am interested in ...
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Finite solvable Frattini-free group having a unique minimal normal subgroup N implies that N is the Fitting subgroup

This is exercise 6.1.6 of Kurzweil and Stellmacher. A restatement is: Let $G$ be a finite solvable group with $\Phi(G)=1$, and assume that $G$ has a unique minimal normal subgroup $N$. Then $N=F(G)$....
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Abelian groups that can be extended with non abelian solvable groups [closed]

Let $H$ be a non-trivial finite abelian group. Is it true (in some cases) that there exist a non-abelian solvable group $G$ and a normal abelian subgroup $K$ of $G$ such that $G/K$ isomorphic to $H$? ...
1 vote
46 views

$G$ is solvable and $H \trianglelefteq G$. Does $H$'s chain contain a tail end of $G$'s chain?

Let $G$ be a solvable group and let $H \lhd G$ be a normal subgroup of $G$. As $G$ is solvable, then $H$ is solvable. One equivalence of solvability (for $H$) is: there exists a chain of normal ...
• 2,756
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Given a finitely generated metablian group $G$ with exponential growth, does the ratio of sizes of consecutive balls have a limit?

Suppose $G$ is a metabelian (hence solvable and amenable) group with exponential growth that is finitely generated by a symmetric set S. Given $k \geq 0$, let $B_k$ be the ball centred at the identity ...
• 1,631
86 views

Solvability by radicals, but you don't get to choose the roots

It is well-known that a polynomial equation $P(X)=0$ over a field $K$ is solvable by radicals if and only its Galois group is solvable. Here, "solvabile by radicals" is taken to mean that ...
61 views

Example of 3-solvable group of large 3-length

I would like to construct examples of $3$-solvable groups with large $3$-length. That means that the Sylow $3$-subgroups need to be large in a sense. Is there any general construction of such examples?...
• 220
1 vote
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Using Fredholm Alternative theorem to solve PDEs [closed]

I am trying to solve a coupled PDE system using the perturbation theorem, where parameter $\epsilon$ is small and all the variables are expanded in even powers of $\epsilon$. The set of equations and ...
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What does "proper" mean in this context?

Part a) of Problem 2C.1 of Isaacs' Finite Group Theory reads Show that every proper homomorphic image of an $N$-group is solvable. What does "proper" mean here? Please note that I'm not ...
• 1,970
1 vote
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• 497
1 vote
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If $G_f\cong D_5$ for $f(x)=x^5+ax+b\in \mathbb{Q}[x]$, show that these criteria are true.

As a followup to this post, I was wondering if the converse of this statement was true. That is, if $G_f$ (the galois group of $f(x)=x^5+ax+b$ for $a,b\in\mathbb{Q}$) is isomorphic to $D_5$ (the ...
• 1,516
132 views

Affine type primitive group

By the definition it looks like 'Every solvable primitive groups are of affine type primitive groups only.' Is the converse true ? i.e. Is it true that every affine primitive permutation groups are ...
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1 vote
A group is solvable if it has only one $p$-Sylow subgroup for each $p$
The problem is to prove that a group is solvable if it has just one $p$-Sylow subgroup for each prime $p$ dividing its order. My solution: If$|G|=p_1^{\alpha_1}p_2^{\alpha_2}\dots p_n^{\alpha_n}$ ...