Questions tagged [infinite-groups]

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

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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
  • 389
1 vote
1 answer
59 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
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6 votes
1 answer
163 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
  • 1,044
-1 votes
2 answers
77 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
77 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
21 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
58 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
25 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
91 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
522 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
84 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
49 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
29 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
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0 votes
0 answers
66 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
56 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
  • 2,795
3 votes
3 answers
449 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. ...
3f183201's user avatar
  • 475
-1 votes
1 answer
63 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
108 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
93 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
84 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
105 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
103 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
39 views

Finding a tangent space of an infinite-dimensional manifold given by level set of a submersion

Problem: Let's consider a linear space of smooth functions $C^k[a,b]$ and a submersion $g:C^k[a,b]\to\mathbb R$. $M = \{x\in C^k[a,b] | g(x) = 0\}$ is a differentiable manifold. What is the tangent ...
brovovar's user avatar
  • 126
0 votes
0 answers
44 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
64 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
1 vote
1 answer
65 views

The factor group $\mathbb{R} / \mathbb{Z}$ under addition has an infinite number of elements of order 4 [closed]

The factor group $\mathbb{R} / \mathbb{Z}$ under addition has an infinite number of elements of order 4. The answer is false, but I cannot still understand it. I thought that it is true since there ...
user avatar
0 votes
1 answer
61 views

If $\forall n\in\Bbb N$, if $g\in G$ with $|g|=n$ implies $\forall m\in\Bbb N,\exists h_m\in G,|h_m|=n^m,$ then can non-torsion-free $G$ be f.p.?

Motivation and Details: Thinking back a couple of years ago, when I did combinatorial group theory a lot, I decided to explore the idea that whenever an element of a group has finite order $n$, then ...
Shaun's user avatar
  • 43k
9 votes
2 answers
297 views

Can adding a single element to a Lie group make it infinite-dimensional?

Suppose that I have a finite-dimensional Lie group, $F\subset H=\text{Diff}(\mathcal{M})$, of diffeomorphisms on some smooth finite-dimensional manifold $\mathcal{M}$. Suppose that there is some ...
Daniel Grimmer's user avatar
4 votes
0 answers
52 views

Consider the group $\mathbb{Z}_2 \wr \mathbb{Z}$, what is $\mathbb{Z}_2 \wr 2 \mathbb{Z}$?

Consider the (Lamplighter) group $(\bigoplus_{n=-\infty}^{n=\infty}\mathbb{Z}_2) \rtimes_\phi\mathbb{Z}$, where $\phi(1)$ "shifts" every element in $\bigoplus_{-\infty}^{\infty}\mathbb{Z}_2$ ...
ghc1997's user avatar
  • 1,163
1 vote
0 answers
60 views

Is the extension of a torsion free group by a torsion free group necessarily torsion free?

My instinct tells me the above isn't true, although I am having trouble finding a counterexample. Hints on where to look for a counterexample would be much appreciated. Hopefully there is some ...
user1044791's user avatar
1 vote
1 answer
71 views

2-ended groups are virtually cyclic: reference

It is often said (e.g., in this post) that "It is standard that two-ended groups are virtually infinite cyclic". However, I cannot find a simple proof for this fact anywhere. Can anyone give ...
Zhiltsoff Igor's user avatar
1 vote
1 answer
132 views

If $G$ is an infinite group, how many elements of order $n$ are there?

Suppose we're given an infinite group $G$ with no additional structure. How many elements of that group have order $n$? There is a related question here, but I was surprised to see this not mentioned. ...
user avatar
3 votes
2 answers
159 views

Can every group be realized as an isometry group?

The question is actually from Löh's Geometric Group Theory: an introduction on page 40: Exercise 2.E.10: Is there for every group $G$, an $n\in \mathbb{N}$ and a subset $X\subset \mathbb{R}^n$ such ...
hinamizawa's user avatar
2 votes
2 answers
66 views

Prove that there are $g_1, g_2, \cdots, g_n \in G$ with $g_i x_1 = x_i$ such that $g_1 H, g_2 H, \cdots, g_n H$ are all the left cosets of $H$ in $G$.

Let $X = \{x_1, x_2, \cdots, x_n\}$ be a $G$-transitive set and $H = G_{x_1}$. It is required to prove that there are $g_1, g_2, \cdots, g_n \in G$ with $g_i x_1 = x_i$ such that $g_1 H, g_2 H, \cdots,...
Blue Tomato's user avatar
1 vote
0 answers
55 views

Extending a group to a ring defining a new operator using group automorphisms.

Let $\mathcal{G} = \langle A, + \rangle$ be an infinite group. I was wondering whether it is possible to extend $\mathcal{G}$ to a commutative ring $\mathcal{R}= \langle A, + , * \rangle$ by ...
Vincenzo 's user avatar
2 votes
1 answer
55 views

Show that every $x \in G\setminus \{1\}$ and every $H<G$ have infinitely many conjugates for $G$ infinite simple group.

Let $G$ be an infinite simple group. It is requested to prove: $(i)$ Every $x \in G\setminus \{1\}$ has infinitely many conjugates. $(ii)$ Every non-trivial proper subgroup $H < G$ has infinitely ...
Blue Tomato's user avatar
3 votes
1 answer
60 views

Index of an infinite cyclic subgroup of a free group of infinite rank is infinite

Question: Let $G$ be free of rank at least $2$, and let $H$ be an infinite cyclic subgroup of $G$. Is it true that the index $[G: H]$ of $H$ in $G$ is infinite? My thought: I know the case when the ...
Random's user avatar
  • 533
1 vote
1 answer
150 views

On infinite permutations of a countable set acting on a set: Does a transposition always witness inequality? [closed]

Let $\phi \in Sym(\mathbb{N})$, where $\mathbb{N}$ is a set of countable atoms which can be represented by natural numbers and $Sym(\mathbb{N})$ is the symmetric group of $\mathbb{N}$. Let $X$ be a ...
Lisa T's user avatar
  • 69
1 vote
0 answers
52 views

Showing that the direct sum is an infinite $p$-group which is not nilpotent.

I am trying to solve problem 5.45 from Rotman's book: For each $n\geq 1$ , let $G_n$ be a finite $p$-group of class $n$. Define $H$ to be the group of all sequences $(g_1, g_2, ... )$, with $g_n\in ...
Milan Rashed's user avatar
4 votes
0 answers
133 views

Suppose $G$ and $H$ are two countably infinite abelian groups s.t. every nontrivial element of $G\times H$ has order $7$. Then $G\cong H$.

Suppose $G$ and $H$ are two countably infinite abelian groups such that every nontrivial element of $G \times H$ has order $7$. Then $G$ is isomorphic to $H$. My idea is that each non-trivial element ...
nkh99's user avatar
  • 391
1 vote
0 answers
80 views

Question about "Infinite Group with all subgroups having finite index" [duplicate]

Since my reputation is not enough to comment on the original question, I have to post a question here, I will sincerely apologize if this behavior is not appropriate. In the comments and the answer, @...
Quzs's user avatar
  • 153
2 votes
1 answer
287 views

Infinite Group with all subgroups having finite index [duplicate]

Does there exist a infinite group besides $\Bbb Z$ that all its nontrivial subgroups have finite index? I know $\Bbb Z$ works, but is there any other examples? Or a proof that $\Bbb Z$ is the only one ...
Q Lee's user avatar
  • 78
-1 votes
1 answer
43 views

Prove that there exist equal number of irrational numbers between any 2 rational numbers, when the difference between the 2 rational numbers is same. [closed]

Prove that there exist equal number of irrational numbers between any 2 rational numbers, when the difference between the 2 rational numbers is same. If the assertion is not true then please prove ...
Ash_Tag's user avatar
1 vote
1 answer
72 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
  • 2,207
3 votes
2 answers
131 views

What happens if I add an $\omega$th digit to the $p$-adic numbers?

If you add $3$-adic numbers like $\dots 111111 + \dots121212 = \dots 010100$ the digits all carry over, so it seems intuitively like you lose a digit at the $\omega$th place, as it's missing an $\...
Zoe Allen's user avatar
  • 3,684
7 votes
1 answer
134 views

Which cyclic groups are automorphism groups?

It is easy to prove, e.g. here that any group $G$ with cyclic automoprhism group must be Abelian. Cyclic groups of order $\phi(p^n) = (p-1)p^{n-1}$ for $p \ne 2$ are obviously automorphism groups, as ...
Zoe Allen's user avatar
  • 3,684
0 votes
0 answers
33 views

The regular representation of a compact (connected) group

The regular representation of a compact group could be expressed following the Peter-Weyl theorem, as the sum of all the irreducible representations with a multiplicity equal to their respective ...
HitMan01's user avatar
11 votes
0 answers
172 views

Proof check: If $G$ is an infinite group, then it has infinite number of subgroups. [duplicate]

Prob.48 in Section.6 of "A First Course in Abstract Algebra", Edition 7, Fraleigh. "Show that a group that has only a finite number of subgroups must be a finite group." Here is my ...
MathFail's user avatar
2 votes
0 answers
87 views

Is Lagrange group theorem still valid for infinite groups?

So it is a well know result that the quotient space $G/\cal R$ corresponding to right congruent relation $\cal R$ is equipotent to the quotient space $G/\cal L$ corresponding to the left congruent ...
Antonio Maria Di Mauro's user avatar
1 vote
1 answer
86 views

Does $[G:H]=|G/H|$ hold for infinite groups?

I'm working on an unassessed course problem, We have showed that $$\text{det}:\text{O}_n\rightarrow\{\pm1\}$$ is a surjective homomorphism. Use the First Isomorphism Theorem to deduce from this that $...
mjc's user avatar
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