Skip to main content

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].

Filter by
Sorted by
Tagged with
0 votes
0 answers
34 views

Prove that the group of unipotent matrices is nilpotent. [duplicate]

I'm trying to prove that the group of unipotent matrices (upper triangular with $1$'s on the diagonal) is nilpotent, so that the lower central series terminates in the trivial group after finite ...
Vincent Batens's user avatar
6 votes
1 answer
185 views

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 $|...
K_D's user avatar
  • 163
0 votes
1 answer
45 views

Is the group of 4 by 4 nilpotent triangular matrices a semidirect product?

Let $N$ be the Lie group of 4 by 4 nilpotent triangular matrices with 1 on the diagonal. Let us denote by $E_{ij}$ the square matrix with entry 1 where the $i$th row and $j$th column meet, all the ...
alejandro's user avatar
  • 123
0 votes
0 answers
19 views

Prove the length of LCS and UCS are equal

In my lecture notes on growth of groups, we are given two definitions for a $c$-step nilpotent group $G$. It goes as follows: A group $G$ is called $c$-step nilpotent if its lower central series, $\...
soggycornflakes's user avatar
2 votes
1 answer
85 views

Proof of a particular piece of Milnor-Wolf theorem

The Milnor-Wolf theorem says that a finitely generated solvable group that doesn't have exponential growth is virtually nilpotent. The proof I've seen is divided into two pieces: Prove that such a ...
Hempelicious's user avatar
0 votes
2 answers
133 views

Examples of nilpotent subgroups of $S_n$ which have less than say 10^9 elements?

To make some computational experiments with finite nilpotent group - it would be helpful to know the following: Question: What are the examples of nilpotent (but not commutative) subgroups in ...
Alexander Chervov's user avatar
-2 votes
1 answer
79 views

Why is the direct product of finitely many nilpotent groups nilpotent? [duplicate]

I want to ask a question, and I found it here: Why is the direct product of a finite number of nilpotent groups nilpotent? But I am struggling to understand how can we take the product of two normal ...
NotaChoice's user avatar
0 votes
1 answer
65 views

Is there an infinite nilpotent group with one Sylow subgroup that is not normal? [closed]

It is known that a finite nilpotent group has every Sylow subgroup normal in it. Does this result generalize to infinite nilpotent groups or not ? If not, why, what is a counter example?
NotaChoice's user avatar
7 votes
1 answer
223 views

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 ...
noparadise's user avatar
5 votes
1 answer
80 views

Non-nilpotent group where the union of the upper central groups equals the whole group

I am studying residually nilpotent groups. I am using the following defintions. Lower central series: Let $G$ be a group. Then we inductively define the lower central series of $G$ as $$\gamma_1(G) = ...
noparadise's user avatar
8 votes
1 answer
689 views

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: ...
Eric Vaz's user avatar
  • 397
3 votes
1 answer
75 views

The existence of a matrix $B$ with $AB=BA^c$ implies nilpotence of $A^r-I$.

My friend and me were studying some group theory, and we thought of the following problem. Let $n$ be a positive integer, and let $A\in \text{GL}_n(\mathbb{Z})$ be a matrix with integer entries that ...
Barry Allen's user avatar
1 vote
1 answer
89 views

If $G/Z(G)$ is nilpotent, then $G$ is nilpotent

If $G/Z(G)$ is nilpotent, then $G$ is nilpotent Theorem 8 (Dummit, Foote) A group $G$ is nilpotent if and only if $G^n = \{1\}$ for some $n \geq 0$. There is Quizlet solution to this problem which I ...
Grigor Hakobyan's user avatar
4 votes
0 answers
79 views

A different approach to proving a property of nilpotent injectors in solvable groups

Let $G$ be a finite solvable group. Call $J\subseteq G$ a nilpotent injector if it is a nilpotent subgroup that contains $\mathbf{F}(G)$, and that is maximal with this property (not properly contained ...
semisimpleton's user avatar
1 vote
2 answers
72 views

Is upper central series of a group unique?

Let G be a group. Its upper central series is defined inductively $$Z_0=\{1\}$$ $$Z_{i+1}/Z_i=Z(G/Z_i)$$ Existence of $Z_{i+1}$ is guaranteed by correspondence theorem. But can we say anything about ...
Anirban Sarkar's user avatar
2 votes
1 answer
180 views

If $G$ is a finite group in which every subgroup has a normal complement, then $G$ is nilpotent.

Problem from an old prelim exam: If $G$ is a group and $H$ is a subgroup, then a normal complement to $H$ in $G$ is a normal subgroup $N\trianglelefteq G$ such that $N\cap H = \{1\}$ and $N H = G$. ...
w.w's user avatar
  • 101
0 votes
0 answers
54 views

References on Nilpotent groups of class 2

I wanted to read about the nilpotent groups of class 2, nilpotent groups of class 2 with exponent p and nilpotent groups of class q. Can anyone suggest some references (E.g., books, thesis, or ...
Jins's user avatar
  • 564
0 votes
1 answer
71 views

Property preserved by tensor product/ epimorphic image/ extension [closed]

Let $G$ be a group. Consider $\gamma_1(G):=G$ and $\gamma_{i+1}(G):=[\gamma_i,G]$, we denote $G_{\text{ab}}$ the abelianization of $G$. Then $\gamma_{i+1}/\gamma_{i+2}(G)$ is an epimorphic image of $...
Mario's user avatar
  • 739
0 votes
1 answer
38 views

Some calculation in $F_5[D_{30}]$.

I am currently reading a research paper and have encountered a point that I am struggling to understand. In the paper, it is proven that $J(F_5[D_{30}])^5 = (0)$, indicating that the Jacobson radical ...
neelkanth's user avatar
  • 6,100
4 votes
1 answer
180 views

Commutator of two elements in group algebra $\mathbb F_{5}D_{30}.$

I want to understand how to find the commutator of two elements in the group algebra $\mathbb{F}_{5}D_{30}$ using GAP. Additionally, I would like to determine the nilpotency class of the nilpotent ...
neelkanth's user avatar
  • 6,100
2 votes
3 answers
121 views

Non-Abelian group of exponent $3$ and Nilpotent class $2$ .

Let $G$ be a non-Abelian group such that $G^3 = 1$ and $G$ is a nilpotent class $2$ group, with order $3^{32}$. Our task is to determine the structure of the group $G$ or identify any information ...
neelkanth's user avatar
  • 6,100
0 votes
0 answers
66 views

If $G$ is nilpotent of class $n$, then $G/Z(G)$ is nilpotent of class $n−1$.

I saw this here: If $G$ is nilpotent of class $c$, then $G/Z(G)$ is nilpotent of class $c-1$. However, the definition of Nilpotent I have to work with is equivalent but different. A group is Nilpotent ...
NoCap's user avatar
  • 55
1 vote
1 answer
130 views

Classification of torsion-free nilpotent groups of class 2

Some background Let $G$ be a torsion-free nilpotent group of class $2$ and rank $2$ (i.e., generated by two elements). Then, $G$ has to be isomorphic to the Heisenberg group. This is relatively easy ...
Gauss's user avatar
  • 2,663
3 votes
1 answer
141 views

What is the index of $H$ in $G=\langle H,g\rangle$, if $G$ is nilpotent and $g$ has finite order?

Suppose we have a nilpotent group $G$ that is generated by a subgroup $H$ and a single element $g$ of finite order. Is $H$ necessarily of finite index in $G$? What can be said of this index in any ...
Tim Seifert's user avatar
  • 2,243
2 votes
0 answers
100 views

Class of nilpotency of semidirect product

Let $G=N\rtimes H$ for some groups $N$ and $H$. Is it true that the nilpotency class of $G$ is at most maximum of nilpotency classes of $N$ and $H$, if it is known that both groups are nilpotent? And ...
Marja's user avatar
  • 67
1 vote
1 answer
181 views

Example of a quasi nilpotent element which is not a nilpotent element

Let $R$ be a ring with unity. An element $a\in R$ is said to be a quasi nilpotent element of $R$ if $1-ax$ is unit for all $x\in comm(a)$ where $comm(a)=\lbrace x \in R | ax=xa\rbrace $. It is obvious ...
Chaudhary's user avatar
  • 533
5 votes
1 answer
108 views

Nilpotent quotients of a residually nilpotent group

I am working on a problem that requires taking quotients by normal subgroups that do not intersect a finite set of members of a group $G$. Is there a known collection of groups which always allows ...
Ashani's user avatar
  • 524
1 vote
0 answers
68 views

Fitting length of Dihedral groups

Let $G$ be a dihedral group of order $2n$ where $n\geq 1$, denoted by $D_n$. We know that $G$ is nilpotent if and only if $n=2^i$ for all $i\geq 1$, a proof of this you can check in the below link 1. ...
MathMeet's user avatar
  • 313
1 vote
2 answers
100 views

The $i^{th}$ term in the upper central series of the dihedral group of order $2^n$ is equal to the $(n-1-i)^{th}$ term in its lower central series.

The Problem: Prove that $Z_i(D_{2^n})=D_{2^n}^{n-1-i}$, where $D_{2^n}$ is the dihedral group of order $2^n$. $Z_i(G)$ is the $i^{th}$ term in the upper central series of $G$, which is inductively ...
Dick Grayson's user avatar
  • 1,467
0 votes
0 answers
139 views

Proof that if $G/Z(G)$ is nilpotent then $G$ is nilpotent.

As a reference my definition of nilpotence is as follows. Definition: A group $G$ is said to be nilpotent if it has a subnormal series $$ G = G_1 \trianglerighteq G_2 \trianglerighteq \dots \...
Irving Rabin's user avatar
  • 2,673
4 votes
1 answer
279 views

If $G$ is nilpotent then so is $G/Z(G)$.

The Problem: If $G$ is nilpotent then so is $G/Z(G)$. My Background: Chapter 1-5 of Abstract Algebra $\mathit{3^{rd}}$ edition by Dummit and Foote. For any group $G$ define the following subgroups ...
Dick Grayson's user avatar
  • 1,467
3 votes
1 answer
205 views

All maximal subgroups are Sylow subgroups

Let $G$ be a group in which all maximal non-trivial subgroups are Sylow subgroups. Then $G$ isn't a simple non-abelian group. I know how to prove this by relying on the theorem that if all maximal ...
Mr. Nobody's user avatar
6 votes
1 answer
145 views

Alternative proof that maximal abelian normal subgroups of a nilpotent group are self-centralizing

I was writing lecture notes on nilpotent groups, and wanted to show the following well-known basic result. Theorem: Let $G$ be a nilpotent group and let $A$ be an abelian normal subgroup of $G$, ...
David A. Craven's user avatar
2 votes
1 answer
112 views

Nilradical in an Artinian ring

I am reading a book that suggests to find what $N(R)$ is - the nilradical of $R$ where $R$ is Artinian you just need to find a nilpotent ideal $I $ of $R$ and then show that $R/I $ has no nonzero ...
Anonmath101's user avatar
  • 1,828
1 vote
1 answer
42 views

Nilpotent ideals in a ring proof

I am trying to understand a proof I am reading but it doesn’t make much sense to me. If $R$ is a ring with $I$ a nilpotent left ideal then $I$ is contained in a nilpotent two-sided ideal. The proof ...
Anonmath101's user avatar
  • 1,828
2 votes
0 answers
87 views

Nilpotent of subgroup implies nilpotent of group

Suppose that I have two normal subgroups $L, M$ in $H$ such that $L<M$ and $M / L$ is nilpotent of class $2$. Suppose also that $[H,[H,H]]$ is contained in $[M,M]$. Can I then conclude that $H / L$ ...
J.L.'s user avatar
  • 309
2 votes
0 answers
61 views

Finite group where product of elements of coprime order has product order is nilpotent

Let $G$ be a finite group where for all $a, b \in G$ with $\operatorname{gcd}(o(a), o(b)) = 1$, $o(ab) = o(a)o(b)$. Is $G$ nilpotent? My try: Let's work by induction on $\lvert G \rvert$. Since every ...
Ris's user avatar
  • 1,292
1 vote
0 answers
122 views

Lie algebras of upper central series of connected Lie group

Let \begin{equation*} 1=Z_0(G)\lhd Z_1(G)\lhd ...\lhd Z_{n-1}(G)\lhd Z_n(G)=G \end{equation*} be the upper central series of a connected nilpotent Lie group $G$, i.e. $Z_i(G)=\{g\in G:[g,G]\subset Z_{...
kringelton4000's user avatar
3 votes
1 answer
48 views

An alternative notion of nilpotency class for $p$-groups

Let $G$ be a finite $p$-group for some prime $p$, and let $\rho: G \to \text{Aut}(A)$ a faithful representation of $G$ for some finite abelian $p$-group $A$ (which exists because, for example, we may ...
user113019's user avatar
1 vote
1 answer
49 views

Can we say anything about $\gamma_n(G) \cap \zeta_{n-1}(G)$, where $\gamma_n(G)$ is lower central series and $\zeta_{n-1}(G)$ is upper central series? [closed]

Can we say anything about $\gamma_n(G) \cap \zeta_{n-1}(G)$, where $\gamma_n(G)$ is lower central series and $\zeta_{n-1}(G)$ is upper central series? For example one can say $Z(G)\cap [G,G]$ is a ...
Cloud JR K's user avatar
  • 2,506
1 vote
0 answers
58 views

Reference request-dimension of projective representations of nilpotent groups.

There is folklore theorem concerning dimensions of complex projective representations of nilpotent groups that I want a reference for. I have searched Karpilovsky monographs but no success. The ...
Ofir Schnabel's user avatar
2 votes
1 answer
171 views

Let $G$ be a finite group. Suppose that there exists $n\in\mathbb N_0$ such that $G_n=G_{n+1}$ and $Z(G)=1$. Show that $C(G_n)\subseteq G_n$.

Let $G$ be a finite group. Recursively define the sequence $G_1=G$ and $G_{n+1}=[G_n,G]$. Suppose that there exists $n\in\mathbb N_0$ such that $G_n=G_{n+1}$ and $Z(G)=1$. Show that $C(G_n)\subseteq ...
Guest's user avatar
  • 1,599
3 votes
1 answer
324 views

Classification of nilpotent groups via $p$-groups

First some background: Some time ago I learned that since any finite abelian group is a direct product of (finite) cyclic groups, the finite cyclic groups ($\mathbb{Z}_n$) are the key to understanding ...
John Doe's user avatar
  • 3,309
2 votes
1 answer
149 views

Proof of Baer's theorem

Baer's Theorem: Let $x$ be a $p$-element of a finite group $G$. Suppose that $\langle x,x^g\rangle$ is a $p$-subgroup for every $g\in G$. Then $x\in O_p(G)$. Here, $O_p(G)$ denotes the largest normal ...
Guest's user avatar
  • 1,599
1 vote
1 answer
82 views

Let $N \unlhd G$. Then $G/N$ is nilpotent of class $c\in\Bbb{N}$ iff $c$ is the smallest natural number such that $\gamma_c(G) \subset N$

I think I have established the following proposition: Theorem: Let $G$ be a group and $N \unlhd G$. Then, $G/N$ is nilpotent of class $c \in \mathbb{N}$ if, and only if, $c$ is the smallest natural ...
Gauss's user avatar
  • 2,663
6 votes
1 answer
178 views

Property of Fitting subgroup: Let $G$ be a finite group and $C:= C_G(F(G))$. Then $O_p(C/C\cap F(G))=1$ for every prime $p$.

I'm trying to understand the proof of the following which is stated in Kurzweil and Stellmacher: Let $G$ be a finite group and $C:= C_G(F(G))$. Then $$O_p(C/C\cap F(G))=1$$ for every prime $p$. Here,...
Guest's user avatar
  • 1,599
3 votes
0 answers
63 views

Characterization of finite nilpotent groups

Can someone please check my proof of the following "evident" equivalence stated in Kurzweil and Stellmacher. A finite group $G$ is nilpotent if and only if $$U<N(U) \text{ whenever } U&...
Guest's user avatar
  • 1,599
3 votes
1 answer
113 views

How do subgroups of the inner automorphisms of a group look like?

I'm trying to prove the following proposition: Let $G$ be a group. Then $G$ is nilpotent iff ${\rm Inn}(G)$ is nilpotent. I've proven that if $G$ is nilpotent then ${\rm Inn}(G)$ is nilpotent as ...
M_N1's user avatar
  • 149
0 votes
2 answers
220 views

Is it true that nilpotent always has some eigenvalue?

I understand that if a nilpotent matrix has some $\lambda$ eigenvector, then it implies that $\lambda=0$ because if $$Ax = \lambda x \\ A^2x = \lambda^2x \\ A^3x=\lambda^3x \\ \vdots \\ 0=A^k=\lambda^...
Eran's user avatar
  • 31
-1 votes
2 answers
68 views

an Equivalent condition for nilpotency of a finite group like of finite solvable group

We know that these two conditions are equivalent for finite groups : i) $G$ is solvable. ii) If $\left|G\right|=mn\;$ and $(m,n)=1$ then G has a subgroup of order $m$. Now are the following two ...
Amirhossein Haddadian's user avatar

1
2 3 4 5