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Questions tagged [nilpotent-groups]

A nilpotent group is:

  1. a group, whose upper central series stabilizes after a finite length at the whole group.
  2. a group, whose lower central series stabilizes after a finite length at the trivial subgroup.
  3. a group, that possesses a central series.

All those definitions are equivalent. To be used with the tag [group-theory].

252 questions
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Is the dihedral group $D_n$ nilpotent? solvable?

Is the dihedral group $D_n$ nilpotent? solvable? I'm trying to solve this problem but I've been trying to apply a couple of theorems but have been unsuccessful so far. Can anyone help me?
nomadicmathematician
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Why is the direct product of a finite number of nilpotent groups nilpotent?

I read that a direct product of a finite number of nilpotent groups is nilpotent. Here the definition of a nilpotent group is one that has a central series. A comment in my book following this claim says If $G_{ij}$ is the $i^{th}$ term of a…
yunone
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A question about Frattini subgroup of specific form

Suppose $p$ is a prime number and $G$ is a finite group, such that $\Phi(G) = C_p \times C_p$, where $\Phi$ denotes the Frattini subgroup. Is it always true, that $p^4$ divides $|G|$? This statement can be easily proved for $p$-groups by seeing that…
Chain Markov
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Is there an analogue of the abelianization of a group for nilpotent groups?

If $G$ is an arbitrary group we denote its abelianization as $G^{ab} := \frac{G}{[G,G]}$, where $[G,G]$ is the commutator. As an abelian group it is characterized by the following universal property: Let $\pi : G \to G^{ab}$ be the quotient…
Eric Vaz
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A generalizaton of nilpotent groups

‎‎‎‎‎‎‎Let ‎$‎G‎$ ‎be a‎ ‎group ‎and ‎‎$‎‎\alpha\in Aut(G)$ ‎be a‎ ‎fixed ‎automorphism ‎of ‎‎$‎G‎$‎. An ‎$‎‎\alpha$-commutator ‎of ‎elements ‎‎$‎‎x, y\in G$ ‎is ‎‎$‎‎[x, y]_{\alpha}= x^{-1}y^{-1}xy^{\alpha}$. ‎The ‎‎$‎‎\alpha$-center ‎subgroup ‎of…
Banoo
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Let $G$ be a nilpotent group generated by a finite set of torsion (i.e. finite order) elements. Show that $G$ is finite.

Given: Let $G$ be a nilpotent group generated by a finite set of torsion (i.e. finite order) elements. Show that $G$ is finite. Also would love to know if it's possible to show that an infinite finitely generated nilpotent group has an infinite…
Ilan Aizelman WS
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2 answers

Nilpotent groups are solvable

I know this should be obvious but somehow I can't seem to figure it out and it annoys me! My definition of nilpotent groups is the following: A group $G$ is nilpotent if every subgroup of $G$ is subnormal in $G$, or equivalently if $U
Helen
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Finite index subgroup with free abelianization

Suppose $G$ is a nilpotent, finitely generated group such that it's abelianization has rank $r$. How does one go on proving that $G$ has a finite-index subgroup $H$ with free abelianization of rank $r$?
Matthew Clayton
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Iterated commutator of powers in torsion-free nilpotent group

Suppose that $G$ is a torsion-free nilpotent group. It is a classical fact that if $g,h \in G$ have powers that commute, then $g$ and $h$ must commute. So if there are integers $m,n$ such that $[g^m,h^n] = 1$, then $[g,h] = 1$. This is usually…
SFSH
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Finitely generated, nilpotent, torsion-free group that is also radicable

I am currently working with Mal'cev completions, using the following definition: Let $N$ be group that is Nilpotent Torsion-free Finitely generated Then the Mal'cev completion or radicable hull is the unique group $N^\mathbb{Q}$ (up to…
noparadise
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3 answers

An explicit representation of a free nilpotent group by unitriangular matrices

P. Hall proved that every finitely generated torsion-free nilpotent group can be faithfully represented by upper unitriangular matrices over $\mathbb Z$. The most famous example is the integral Heisenberg group, which is free nilpotent, of rank $2$…
Ashot Minasyan
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7
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Bound for the rank of a nilpotent group

Let $G$ be a nilpotent group generated by $d$ elements. Is there a function $r(d)$ such that every (necessarily finitely generated) subgroup $H$ of $G$ can be generated by at most $r(d)$ elements? Thanks in advance!
Florian Pei
  • 297
7
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3 answers

Let $G$ be a finite nilpotent group and $G'$ its commutator subgroup. Show that if $G/G'$ is cyclic then $G$ is cyclic.

So I thought the cleanest way to do this was to simply prove $G' = 1$ since if $G$ is cyclic $G' = 1$ and then $G \cong G/G'$, but I got no where with this. My next idea was since $G$ is nilpotent I know it's the direct product of normal Sylow…
Samantha Wyler
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6
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Prove Fitting's theorem for finite groups

Fitting's Lemma is: Let $M$ and $N$ be normal nilpotent subgroups of a group $G$. If $c$ and $d$ are the nilpotent classes of $M$ and $N$, the $L = MN$ is nilpotent of class at most $c+d$. There is an exercise on Derek J.S. Robinson's A Course in…
ShinyaSakai
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6
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1 answer

Classifying finite groups where order is multiplicative on elements with coprime orders

It's well known that $|g_1 g_2| = |g_1||g_2|$ whenever $g_1$ and $g_2$ are commuting elements of a group with $\gcd(|g_1|, |g_2|) = 1$. So, for example, $\gcd(|g_1|, |g_2|) = 1$ always implies that $|g_1 g_2| = |g_1||g_2|$ in an abelian group. I am…
K_D
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