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Questions about the branch of algebra that deals with groups.

23 votes
4 answers
2k views

Are infinite groups in which most elements have order $\leq 2$ commutative?

The starting point of this question is the following: If $G$ is a group such that all elements have order at most $2$, then $G$ is commutative. If $G$ is any group, let $G_{>2}$ denote the set o …
Dominic van der Zypen's user avatar
20 votes
3 answers
987 views

Does the hypergraph of subgroups determine a group?

A hypergraph is a pair $H=(V,E)$ where $V\neq \emptyset$ is a set and $E\subseteq{\cal P}(V)$ is a collection of subsets of $V$. We say two hypergraphs $H_i=(V_i, E_i)$ for $i=1,2$ are isomorphic if t …
Dominic van der Zypen's user avatar
10 votes
2 answers
579 views

Maximal Abelian subgroups of $S_\omega$

Let $S_\omega$ be the group of permutations (bijections) $\varphi:\omega\to\omega$, together with composition as binary operation. Zorn's Lemma implies that every commutative subgroup of $S_\omega$ is …
Dominic van der Zypen's user avatar
10 votes
2 answers
743 views

Universal group such that every finite group is a quotient

We say that a permutation $\varphi:\mathbb{N}\to\mathbb{N}$ is finitary if there is $k\in\mathbb{N}$ such that $\varphi(i) = i$ for all $i\in\mathbb{N}$ with $i\geq k$. Let $I_\mathbb{N}$ denote the g …
Dominic van der Zypen's user avatar
9 votes
0 answers
323 views

Uncountable group with no proper subgroup of maximal cardinal

The Prüfer group $\mathbb{Z}(p^\infty)$, for $p$ prime, has the interesting property that it is infinite, and every proper subgroup is finite, but can be arbitrarily large, so there is no proper subgr …
Dominic van der Zypen's user avatar
7 votes
1 answer
450 views

Surjective group homomorphism from $\text{Sym}(X)$ onto $\mathbb{Z}$

For any non-empty set $X$ let $\text{Sym}(X)$ denote the group of bijections $f:X\to X$ with composition. Is there an infinite set $X$ and a surjective group homomorphism $\pi: \text{Sym}(X)\to \math …
Dominic van der Zypen's user avatar
7 votes
1 answer
160 views

Minimal generating set for $S_\omega$

If $G$ is a group and $S\subseteq G$, let $\langle S \rangle$ be the intersection of all subgroups of $G$ containing $S$. Let $S_\omega$ denote the group of all bijections $f:\omega\to\omega$ with co …
Dominic van der Zypen's user avatar
6 votes
1 answer
1k views

Countable group with uncountable number of subgroups $< 2^{\aleph_0}$ [duplicate]

Is it consistent that there is a countable group $G$ such that the cardinality of the set of subgroups of $G$ is uncountable, but strictly less than $2^{\aleph_0}$?
Dominic van der Zypen's user avatar
6 votes
3 answers
313 views

Group such that factors in any product-decomposition are reducible

Motivation. Let us call a group $G = (G,\cdot)$ (product-)reducible if there are groups $H_1, H_2$, each having more than $1$ element, with $G \cong H_1\times H_2$. Otherwise, $G$ is said to be irredu …
Dominic van der Zypen's user avatar
6 votes
1 answer
359 views

Large subgroups of Knuth's non-associative "group" on ${\cal P}(\mathbb{N})$

Donald Knuth introduced a fast, bit-wise approximation to integer addition by $$(a,b) \mapsto a \, ^{\land} \, b \, ^{\land} \, ((a \text{ & } b) \ll 1)$$ where $a,b$ are given in binary and $\,^{\lan …
Dominic van der Zypen's user avatar
5 votes

Solving algebraic problems with topology

Using Priestley duality for distributive lattices and compact, totally disconnected ordered topological spaces, many purely algebraic questions have been solved using quite simple topological tools. F …
5 votes
0 answers
203 views

Universal group on $\kappa$ elements

It is well known that for every positive cardinal $\kappa$, every group of cardinality $\kappa$ can be embedded into $\text{Sym}(\kappa)$, the group of bijections on $\kappa$ with composition as group …
Dominic van der Zypen's user avatar
5 votes
1 answer
524 views

Maximal subgroups not containing a specific element

Given a non-trivial group $G$ and $g\in G\setminus \{e_G\}$ where $e_G$ is the neutral element, it is easy to show using Zorn's Lemma, that there is a subgroup not containing $g$ that is maximal among …
Dominic van der Zypen's user avatar
4 votes
1 answer
439 views

Non-associative commutative "group"

When dealing with some hash functions that I was trying to speed up, I toyed with a binary operation with the goal to "approximate" the addition on $\{0,1\}^*$ when seen as binary representation of th …
Dominic van der Zypen's user avatar
4 votes

coloring infinite vertex transitive graph without large cliques

No - there are triangle-free, vertex transitive graphs of infinite chromatic number, see for instance this article.
Dominic van der Zypen's user avatar

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