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Questions tagged [infinite-groups]

For questions about groups where the underlying set has infinite cardinality.

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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 ...
ghc1997's user avatar
  • 1,641
2 votes
1 answer
52 views

Suppose $H$ is a normal subgroup of $G$ with prime index $p$, does this implies that $H$ contains $G^p$?

The power subgroup is defined as $G^p =\left\{g^p \mid g \in G\right\}$. Question: If $H$ is a normal subgroup of $G$ with prime index $p$, does this imply that $H$ contains $G^p$? This statement is ...
ghc1997's user avatar
  • 1,641
1 vote
0 answers
88 views

Compostitum is Galois and isomorphism of Galois groups

Question: Let $K, L, F$ be subfields of a field $\Omega$, and suppose that $K \subset L$ is Galois and that $K \subset F$. Prove that $F \subset LF$ is Galois and that $\text{Gal}(LF/F) \cong \text{...
ByteBlitzer's user avatar
0 votes
0 answers
70 views

How to find the Galois group of an infinite Galois extension? For example $\mathbb{Q}(\mathbb{\sqrt{\mathbb{Q}}})$. [duplicate]

I know that this is a very broad question, but I am especially interest in the case of the Galois extension $\mathbb{Q}(\sqrt{\mathbb{Q}}) = \mathbb{Q}( \{ \sqrt{-1} \} \cup \sqrt{p} \mid p \text{ ...
Cosima's user avatar
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1 vote
1 answer
70 views

Understanding the completed group algebra of a profinite ring.

I am learning about the completed group algebra $ k[[G]] = \underset{\underset{N\unlhd_{o}G}{\leftarrow}}{\lim} $ $k[\frac{G}{N}]$ of a profinite group $G$ and a finite field $k$. As far as I ...
Facundo Canale's user avatar
1 vote
0 answers
41 views

Can we classify the structure of an infinite soluble $p$-group of bounded exponent?

By Baer-Pruffer Theorem, we know that, if $A$ is an abelian $p$-group of bounded exponent, then $A$ is a direct sum of cyclic $p$-groups. Now, assume that $A$ is an infinite soluble $p$-group of ...
Reza Fallah Moghaddam's user avatar
8 votes
3 answers
336 views

Proving that a group is infinite and nonabelian

As an exercise I am trying to prove that the group $$G = \langle a,b,c \mid ac = ba, ab=ca, bc=ab\rangle$$ is infinite and non-abelian. Moreover, the author claims that its center has finite index. I ...
user1008978's user avatar
5 votes
1 answer
52 views

Can we conclude that an infinite bounded exponent periodic locally nilpotent group $G$ is the direct product of its Sylow $p$-subgroups?

When $G$ is a finite nilpotent group, we know that $G$ is the direct product of its Sylow subgroups. Now, assume that $G$ is an infinite locally nilpotent group. Also, consider that there exists a ...
Reza Fallah Moghaddam's user avatar
1 vote
1 answer
62 views

Can there be a subgroup $K$ of a group $G$ such that for some $a \in G$, $aK \subseteq Ka$ but $Ka \nsubseteq aK$? [duplicate]

I have a question, in a sense, about how asymmetric left and right cosets can be when dealing with an infinite, non-normal subgroup $K$ of a (non-abelian) group $G$. Specifically, my question is ...
MathNeophyte's user avatar
0 votes
1 answer
69 views

Symmetric difference as a group law

Let $D$ be any set and define $\mathcal{P}(D)=\{A:A\subseteq D\}$ to be the power set of $D$. For any $A,B\in\mathcal{P}(D)$, define the symmetric difference operation $$A*B=(A\setminus B)\cup (B\...
clathratus's user avatar
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3 votes
1 answer
54 views

Prove a surjective endomorphism $\phi$ of a 1-relator group $ ⟨a, b ∣ a^{-1} b^2 a b^{-3}⟩ $ is not injective

Consider the infinite group $H$ with presentation $$ ⟨a, b ∣ a^{-1} b^2 a b^{-3}⟩ $$ so that the relation is $a^{-1} b^2 a=b^3$. The map $$ a ↦ a\\b ↦ b^2 $$ induces a surjective homomorphism $ϕ:H\to ...
hbghlyj's user avatar
  • 3,047
1 vote
1 answer
61 views

Automorphism group of d-regular trees

A (infinite) tree is $d$-regular if every vertex has $d$-many neighbors. Needless to say, the tree will be infinite in size. My question is about the automorphism group of the said tree, $\mathrm{Aut(...
Sajid Bin Mahamud's user avatar
1 vote
1 answer
64 views

Does there exist a group $G$ such that $(G\setminus S)^2=G$ for each $S\subseteq G$ with $|S|<\infty$?

Note: I have answered this question myself in typing it up. I thought I'd share it because it took me too long. Maybe someone will benefit from it. The Question: Does there exist a group $G$ such ...
Shaun's user avatar
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1 vote
0 answers
32 views

Riemann Integration and Supremum

Suppose we have $f(t, \mathbf{s}): \mathbb{R} \times \mathbb{R}^{g}$ where $\mathbf{s} \in [0, 1]^{g}$ with $g \in \mathbb{Z}_{+} \cup \infty$, $t \in \mathbf{t}$ where $\mathbf{t}$ is a set of tagged ...
SiegAndy's user avatar
1 vote
1 answer
87 views

Non-abelian groups with trivial socle.

Recently, my teacher told me about Socle of a group $G$. So, I know that the socle $Soc(G)$ of a group $G$ is the subgroup generated by all minimal normal subgroups of $G$. I was thinking then that ...
Priya Sarkar's user avatar
1 vote
1 answer
45 views

How to Prove the Divergence of an Improper Integral Involving Absolute Value

I'm working on understanding the convergence properties of certain improper integrals and encountered the following integral: $$\int_{0}^{\infty} \left| \frac{\cos(x)}{\sqrt{x}} \right| \, dx$$ I ...
Matan Bitton's user avatar
1 vote
0 answers
40 views

Subgroups and Quotient Groups of Infinite bounded Abelian p-groups

I am looking to a reference to a solution of Exercise 13 (page 91) of L. Fuchs, Infinite Abelian Groups, Vol. 1, Academic Press, 1970. There is no solution to this exercise in a collection Exercises ...
George's user avatar
  • 11
1 vote
2 answers
56 views

Let $K\le H\le G$ with $[G:H]<\infty$. Is $[G:K]<\infty$ necessarily?

While reading this proof for the Tower Law for Subgroups, the data for the theorem only stipulates that H is a subgroup of G with finite index, and that K is a subgroup of H. Later in the proof ...
giorgio's user avatar
  • 583
2 votes
1 answer
74 views

Doubling radii of balls in groups of polynomial growth

Let $G$ be a group of polynomial growth, let $S$ be a finite generating set for $G$ and let $B_n$ be the set of elements of $G$ given by words of length $\leq n$ in the generating set $S$. Is there an ...
Saúl RM's user avatar
  • 3,545
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
2 votes
0 answers
57 views

If every proper subgroup of an infinite group $G$ is cyclic, then $G$ is abelian [duplicate]

If every proper subgroup of an infinite group $G$ is cyclic, then $G$ is abelian ? I know there is an infinite abelian with this property(every proper subgroup is cyclic). But I was unable to find any ...
EUQAH's user avatar
  • 29
1 vote
0 answers
52 views

Comparing cardinality of set of real numbers $[2,3]$ and set of real numbers $[4,5]$

The function $f(x) = x+2$ on the given sets $([2,3]$ and $[4,5])$ is bijective which should imply that the cardinality is equal. Now, consider the function $f(x) = x^2$ on the sets of real numbers $[2,...
Vishal Sharma's user avatar
0 votes
2 answers
126 views

Infinite non cyclic group with proper subgroups being cyclic

I was thinking about the question regarding infinite groups with all of its proper subgroups being cyclic but the group itself is not cyclic. The finite case is a well known fact $(K_4)$. I have found ...
nkh99's user avatar
  • 483
4 votes
0 answers
113 views

Proving certain triangle groups are infinite

[Cross-posted to MathOverflow] Consider the Von Dyck group $$ G = \langle x,y\mid x^a=y^b=(xy)^c=1\rangle $$ where $a,b,c\ge3$. Because $G$ is infinite and residually finite, it has an infinite family ...
Steve D's user avatar
  • 4,606
5 votes
1 answer
394 views

What is the cardinality of the set of all groups (up to isomorphism) of countable order?

Apologies if this is a known result, I have looked around and could not find it. There is a pretty vast literature on the number of groups up to isomorphism of order $n$, for any natural number $n$. ...
Jonathan Beer's user avatar
6 votes
3 answers
495 views

Has every infinite simple group a faithful irreducible representation?

Has every infinite simple group a faithful irreducible representation? This question solves the finite case. However, the proof requires a non-trivial linear representation of a finite group. I want ...
wer's user avatar
  • 399
1 vote
1 answer
96 views

Using the Orbit-Stabiliser Theorem to prove a set of matrices is a subgroup of $\operatorname{GL}_2(\mathbb{R})$

I'm working on an unassessed course problem, Use the Orbit-Stabiliser Theorem to show that the set of matrices $$ M = \begin{pmatrix} a&b \\ c&d \end{pmatrix} \in \operatorname{GL}_2(\...
mjc's user avatar
  • 2,281
6 votes
1 answer
187 views

If an infinite group $G$ is generated by two elements $a,b$ such that $a^n=b^n=e$, must $x^n=e$ have infinitely many solutions?

Suppose that $G$ is an infinite group, $a,b\in G$ satisfy $G=\langle a,b\rangle$ and $a^n=b^n=e$ for some $n\in\mathbb{N}^*$, where $e$ is the identity of $G$. Must $G$ contain infinitely many $x$ ...
Jianing Song's user avatar
  • 1,923
-1 votes
2 answers
111 views

Notation $\pi(G)$ for a group $G$ [closed]

For a finite group $G$, I've encountered the notation $\pi(G)$ that is the set of all primes that divide the order of $G$. If the group $G$ is infinite, what does $\pi(G)$ mean?
Brendon's user avatar
  • 21
2 votes
1 answer
158 views

Show that $S$ has Lebesgue measure zero.

Consider the set $S \subseteq \mathbb R$ defined by $S : = \left \{x \in \mathbb R\ \bigg |\ \lim\limits_{n \to \infty} \sin (n! \pi x) = 0 \right \}.$ Show that Lebesgue measure of $S$ is zero. I ...
Akiro Kurosawa's user avatar
1 vote
1 answer
27 views

Does $B(BE\dot{\cup} E)\subset BE\dot{\cup} E$ imply that $B$ is a sub-semigroup?

Let $B, E$ be nonempty subsets of a group or semigroup $X$ such that $E\cap BE=\emptyset$, and put $A=BE\cup E$. Then (a) if $B$ is a sub-semigroup of $X$, then $BA\subset A$ (i.e., $BA$ is a proper ...
M.H.Hooshmand's user avatar
3 votes
1 answer
91 views

Decomposition of infinite abelian groups

A colleague recently mentioned a "Prüfer Decomposition Theorem", claiming that every abelian group $A$ could be expressed as the direct sum $A = T(A) \oplus F(A)$, where $T(A)$ is the ...
alexchandel's user avatar
0 votes
1 answer
27 views

Upper Bound On the Average Of $(\phi_n-n)$ For $\phi\in S_\infty$

Let $\phi:\mathbb{N}\rightarrow\mathbb{N}$ be a bijection. Consider $D_N=\biggr\lvert\sum_{n=0}^N(\phi_n-n)\biggr\lvert.$ Are there any well established upper bounds on this value in the limit? It ...
Miles Gould's user avatar
2 votes
1 answer
132 views

Continuous action by profinite completion of a group [closed]

I'm working on a problem involving diagram dependence of a specific group action. It turns out that taking the profinite completion of the group will allow me to speak from a point of universality in ...
Alex Byard's user avatar
2 votes
1 answer
555 views

Subgroups of the $p$-adics that miss $\mathbb{Z}$

I am trying to understand the complementation of $\mathbb{Z}$ in the $p$-adics. The motivation for this question is that it will help me understand the situation in $\text{Tak}(\mathbb{Z}_p)$, where $\...
Alex Byard's user avatar
2 votes
1 answer
190 views

Is every closed subgroup of a profinite group contained in a maximal subgroup?

I recently asked this question about whether or not every subgroup of a profinite group is contained in a maximal subgroup. The answer is no, and a counterexample is given by $\mathbb{Z}<\mathbb{Z}...
Alex Byard's user avatar
3 votes
1 answer
59 views

Is every subgroup of a profinite group contained in a maximal subgroup?

I recently asked this question about whether or not profinite groups admit maximal subgroups: And indeed, profinite groups admit subgroups of finite index, so taking any minimum index subgroup ...
Alex Byard's user avatar
0 votes
0 answers
31 views

$\prod_p \mathbb{Z}/p\mathbb{Z}$ is not the direct sum of $\bigoplus_p \mathbb{Z}/p\mathbb{Z}$ and a torsion-free subgroup

While I was reading "Abelian Groups" by Fuchs $(2015)$, I encountered Example $1.2$ in the chapter on Mixed Groups, which stated the following: Let $p_1,p_2,\dots,p_n,\dots$ denote different ...
John123's user avatar
  • 69
0 votes
0 answers
73 views

Proving object cannot be profinite without Stone topology

A profinite group $G$ arises as an inverse limit $G=\varprojlim G_\lambda$ of finite discrete groups $G_\lambda$. There is a nice property which states that a group $G$ is profinite if and only if it ...
Alex Byard's user avatar
3 votes
1 answer
76 views

Finitely generated group with a unique maximal subgroup. Is it cyclic?

Prove or disprove: If $G$ is a finitely generated group which has a unique maximal subgroup, then $G$ is a finite cyclic group. I am well aware that the result is true when $G$ is finite (see here). ...
user2345678's user avatar
  • 2,895
3 votes
3 answers
467 views

Does this group with prime order elements exist?

Does there exist a group such that each non trivial element has prime order and for each prime $p$ exactly $p-1$ elements are there with order $p$? Such group say $G,$ if exist is obviously infinite. ...
Eloon_Mask_P's user avatar
-2 votes
1 answer
68 views

Is a sufficient condition that $H$, $K$ $\leq G$ such that $|H|=|K|$ and $[G:H]$ are finites to conclude that $[G:K]=[G:H]$? [duplicate]

Let $G$ an infinite group, $H$ and $K$ proper subgroups of $G$ such that $|H|$ and $|K|$ are finite and $|H|=|K|$, also $[G:H]$ is finite. The question is: With this hypothesis can I conclude that $[G:...
AugustoLP's user avatar
3 votes
1 answer
132 views

Do profinite groups admit maximal subgroups

I've been looking into profinite groups, their topological subgroup lattices, etc. I asked the question does every profinite group admit maximal subgroups? I can't find an example of a profinite group ...
Alex Byard's user avatar
6 votes
1 answer
103 views

Basic closed subsets of Stone topological group presented as inverse limit, inner automorphism group of profinite group

Take a profinite group $G=\varprojlim G_\alpha$. We know that the inner automorphism group $\text{Inn}(G)$ of $G$ is profinite since $\text{Inn}(G)=G/Z(G)$, and the quotient of a profinite group by a ...
Alex Byard's user avatar
3 votes
1 answer
120 views

Positive-negative partition of groups!

Let $G$ be a group and denote by $G_2$ the set of all $x$ such that $x^2=1$. By a positive-negative partition of $G$, I mean a disjoint union $G=G_+\dot{\cup} G_- \dot{\cup} G_2$ such that $G_-=(G_+)^{...
M.H.Hooshmand's user avatar
3 votes
1 answer
163 views

Inner automorphism group of a profinite group is profinite

I'm continuing to investigate my professor's statement regarding profinite groups and Stone topological groups. See this post for more information. A profinite group $G$ is an inverse limit of an ...
Alex Byard's user avatar
6 votes
1 answer
177 views

A group is profinite if and only if it is a Stone space

A profinite group is an inverse limit of an inverse system of discrete finite groups. Alternatively, a profinite group is a topological group that is also a Stone space. Under the second, axiomatic ...
Alex Byard's user avatar
0 votes
0 answers
49 views

Condition which implies that finitely generated subgroups of a finitely generated group have fewer generators

I'm still working on quandles, and have made some progress. I now want to understand finitely generated things. To this end, I want to, again, consider finitely generated groups. Let $G$ be a finitely ...
Alex Byard's user avatar
2 votes
0 answers
67 views

Complementation of infinite groups

I am continuing to work on the problem of complementation of infinite groups in the sense of Zacher. The problem was motivated by an attempt to understand subquandles, about which very little can be ...
Alex Byard's user avatar

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