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Questions about the branch of algebra that deals with groups.
3
votes
1
answer
309
views
Fundamental group of the grid on $\mathbb{R}^\mathbb{N}$
The grid on $\mathbb{R}^2$ is defined by the set of points such that at most one coordinate is not an integer. With this in mind, e endow $\mathbb{R}^\mathbb{N}$ with the product topology, where $\mat …
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 …
3
votes
2
answers
613
views
Non-simple groups $G$ with only non-trivial quotient isomorphic to $G$
If $G$ is a group such that every non-trivial subgroup is isomorphic to $G$ itself, then $G= \mathbb{Z}$ is the only infinite group with that property (up to isomorphism). Amongst the finite groups we …
-1
votes
1
answer
235
views
Is $\text{Sym}(\omega)/\text{(fin)}$ embeddable in $\text{Sym}(\omega)$? [duplicate]
Let $\omega$ denote the set of natural numbers, let $\text{Sym}(\omega)$ be the collection of bijections $\psi:\omega\to\omega$, and let $\text{(fin)}$ be the set of members of $\text{Sym}(\omega)$ ha …
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 …
3
votes
2
answers
341
views
Are there $2^{\aleph_0}$ pairwise non-isomorphic countable groups containing every finite gr...
Let us call a group $(G,\cdot)$ finitarily complete if $G$ is countable, and every finite group is isomorphic to a subgroup of $(G,\cdot)$.
Is there a collection of $2^{\aleph_0}$ pairwise non-isomorp …
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 …
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 …
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 …
3
votes
1
answer
94
views
Graph structure on $S_\omega$ induced by fixed points on compositions
Let $S_\omega$ denote the collection of bijections $f:\omega\to\omega$. We say that $f \in S_\omega$ has a fixed point if there is $x\in \omega$ with $f(x) = x$.
It is a short exercise to show that i …
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 …
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 …
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 …
-1
votes
1
answer
207
views
Union of an ascending chain of subgroups in group $G$ isomorphic to subgroup $S_0\subseteq G$ [closed]
Let $G$ be an infinite group, let $S_0\subseteq G$ be a subgroup and suppose that ${\frak C}$ is a collection of subgroups of $G$ such that
$C \cong S_0$ for all $C\in {\frak C}$, and
for all $C, C' …
2
votes
1
answer
133
views
Is $S_\omega/F_\omega$ embeddable to $S_\omega$?
Let $S_\omega$ be the group of bijections $f:\omega\to\omega$, and let $F_\omega = \{\pi\in S_\omega: \exists N\in \omega(\pi(k) = k \text{ for all } k\geq N)\}$. It is easy to see that $F_\omega$ is …