A metabelian group is a group, that is an extension of an abelian group by an abelian group. To be used with the tag [group-theory]
Questions tagged [metabelian-groups]
12 questions
9
votes
1 answer
Is a finite centerless metabelian group always a semidirect product of two abelian groups?
Suppose $G$ is a finite centerless metabelian group. Is it true that it is a semidirect product of two abelian groups?
It does not seem true to me, but I failed to find any counterexamples. Actually, I know quite a few examples of finite metabelian…
Chain Markov
- 16,012
4
votes
1 answer
Is there a way to classify all metabelian finite groups $G$, such that $ \operatorname{Aut}(G) \cong G$?
Is there a way to classify all metabelian finite groups $G$, such that $ \operatorname{Aut}(G) \cong G$?
I know that the trivial group is the only abelian group that satisfies this condition. I also know two non-abelian groups that satisfy this…
Chain Markov
- 16,012
3
votes
2 answers
Are the terms of the derived series of finitely generated groups finitely normally generated?
Let $S$ be a finite generating set of a finitely generated group $G$. Then the set $S'$ of $[a,b]$ for $a,b \in S$ normally generates $G'$, i.e., any element of $G'$ is a product of conjugates of commutators of generators. Indeed, denoting the…
Levi
- 4,894
2
votes
1 answer
Problem understanding Itô's proof about product of abelian groups
I am trying to understand Itô's Theorem that states the following: Let the group $G=AB$ be the product of two abelian subgroups $A$ and $B$. Then $G$ is metabelian.
I am following the book 'Product of groups' by Amberg, Franciosi and De Giovanni…
user602039
1
vote
0 answers
Criterion for $G/Z(G)$ being metabelian
I was reading an article. There is written that if $G$ is an $FC$-group which is also a soluble $\overline T$-group, then the locally finite group $G/Z(G)$ is metabelian.
$G$ is an $FC$-group if every element has only finitely many conjugates.
$G$…
Looool 2015
- 27
1
vote
0 answers
Relators in a finitely generated metabelian group
Consider the 2-generated metabelian group $G$ with finite $\mathbf{A}^2$-presentation
$$ \langle\langle a,b \,\vert\, [a,b^{-1}][a^{-1},b][a,b]^2 \rangle\rangle, $$
where $[a,b] = a^{-1}b^{-1}ab$, i.e. all relators of the forms $[[w,x],[y,z]]$ are…
kg583
- 203
1
vote
1 answer
The commutator subgroup of finite metabelian p-group
Let $G$ be a finite metabelian $p$-group, i.e. the commutator subgroup $G′$ of $G$ is abelian. In some paper of groups theory, I find This statement :
" Since $G/G'$ $\cong \mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$ then $G'$ is cyclic"
I…
Fouad El
- 433
1
vote
1 answer
On classification of groups of order $p^5$
Can someone suggest me some source where the author has classified all non-isomorphic groups of order $p^5$ ?
Edit 1 : I need complete classification (not upto isoclinism), and also in finitely presented form . I found that with increase in value of…
HIMANSHU
- 206
1
vote
1 answer
Center of a split metabelian group of order $p^nq$ is direct summand.
Let $p$ and $q$ be odd primes such that such that $q \ | \ p-1.$ Suppose we have a group $G$ which is a split extension of the cyclic group $B$ of order $q$ by a finite abelian $p$-group $A.$ Can we say that $Z(G)$ is a direct summand of $G?$
I…
math seeker
- 311
0
votes
1 answer
Is every $p$-group of order $p^4$ metabelian?
Upon learning that all $p$-groups of order $p^5$ are metabelian I wanted to investigate the case of order $p^4$. A group $G$ is metabelian if $G'$, its commutator subgroup, is abelian; equivalently, $G$ is metabelian if and only if there is an…
Alex Petzke
- 8,971
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- 96
0
votes
1 answer
Meta cyclic p-group
While studying meta cyclic p groups, I came across an interesting class of meta cyclic groups which can be written as semi-direct product of two cyclic p-groups of order $p^m$ and $p^n$ respectively. These kind of groups are called split meta cyclic…
A. Gupta
- 73
-1
votes
1 answer
Showing that every abelian group is metabelian.
Some terms before I define my problem:
Definition: A group is abelian if every pair of elements commutes.
Proposition: All subgroups of an abelian group are normal.
Definition: A group $G$ is called metabelian if it has an abelian normal subgroup…
V. Isidore
- 129