All Questions
17 questions
113
votes
2
answers
16k
views
Does every non-empty set admit a group structure (in ZF)?
It is easy to see that in ZFC, any non-empty set $S$ admits a group structure: for finite $S$ identify $S$ with a cyclic group, and for infinite $S$, the set of finite subsets of $S$ with the binary ...
- 3,511
39
votes
5
answers
4k
views
A “mother of all groups”? What kind of structures have "mother of all"s?
For ordered fields, we have a “mother of all ordered fields”, the surreal numbers $\mathbf{No}$, a proper-class “field” which includes (an isomorphic copy of) every other ordered field as a subfield. ...
- 1,687
35
votes
7
answers
4k
views
Paradoxical Mathematical Objects Pending for Construction [duplicate]
The possible properties and applications of some mathematical objects have been described far before their rigorous mathematical definition. Some of them even had a seemingly paradoxical description ...
35
votes
2
answers
3k
views
Is Lagrange's Theorem equivalent to AC?
Lagrange's Theorem is most often stated for finite groups, but it has a natural formation for infinite groups too: if $G$ is a group and $H$ a subgroup of $G$, then $|G| = |G:H| \times |H|$.
If we ...
- 643
22
votes
2
answers
2k
views
What is the largest Laver table which has been computed?
Richard Laver proved that there is a unique binary operation $*$ on $\{1,\ldots,2^n\}$ which satisfies $$a*1 \equiv a+1 \mod 2^n$$
$$a* (b* c) = (a* b) * (a * c).$$
This is the $n$th Laver table $(A_n,...
- 3,547
15
votes
3
answers
2k
views
Construction of a proper uncountable subgroup of $\mathbb{R}$ without Choice.
It is straightforward to construct proper uncountable subgroups of $\mathbb{R}$. One can construst a basis for $\mathbb{R}$ over $\mathbb{Q}$, and then there are many possibilities (just consider the ...
- 2,441
14
votes
2
answers
1k
views
If every definable class admits a group structure, must global choice hold?
It is a remarkable fact, due to Hajnal and Kertész and explained very well in this MathOverflow answer by user Ashutosh, that the axiom of choice is equivalent to the assertion that every nonempty set ...
- 236k
11
votes
0
answers
564
views
Isomorphic free groups have bijective generating sets
Let $F(X)$ be the free group on a set $X$. Classically, we can prove the statement:
$F(X) \cong F(Y)$ if and only if $|X|=|Y|$.
The proofs (that I have seen) consist of turning the group ...
- 1,185
10
votes
2
answers
581
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 ...
- 48.9k
10
votes
1
answer
1k
views
The Tall Tale of Terminating Transfinite Towers
The transfinite tower of iterative automorphisms of a group $G$ is simply definied to be the following chain of the groups where $G_{\alpha+1}=Aut(G_{\alpha})$ for each ordinal $\alpha$ and the direct ...
10
votes
0
answers
438
views
On the Number of Parallel Automorphism Lines
Given a group $G$, one can define the transfinite line of iterative automorphisms of $G$ to be the following chain of the groups where $G_{\alpha+1}=Aut(G_{\alpha})$ for each ordinal $\alpha$ and the ...
7
votes
1
answer
601
views
Action of infinite symmetric groups on iterated power sets
Let $X$ be an infinite set, and $k \ge 1$ be a natural number. We work without the axiom of choice.
Let $G_0$ be the full symmetric group on $X$, and let $G_1$ be the
full symmetric group on ${\cal ...
- 1,615
6
votes
2
answers
461
views
Automorphism of the transfinite rooted binary tree
I was studying combinatorical group theory recently, and I came across the infinite regular rooted binary tree and its automorphism group $Aut(T^{(2)})$with the Grigorchuk subgroup.
Let me now ...
- 3,738
6
votes
0
answers
294
views
Independence results on pure algebra
I think that the most celebrated result in this direction is Shelah's famous work on Whitehead's Problem:
Is every abelian group $A$ such that $Ext^1(A, \mathbb{Z})=0$ free?
This is known to be ...
- 3,318
5
votes
0
answers
170
views
How much choice is required for a countably-infinite index subgroup of the real additive group?
The existence of such subgroups implies the existence of a non-measurable set; simply intersect each of the cosets with $[0,1]$. The results will all have equal outer measure, but their union will be ...
- 1,252
3
votes
2
answers
347
views
Are there $2^{\aleph_0}$ pairwise non-isomorphic countable groups containing every finite group?
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-...
- 48.9k
3
votes
1
answer
383
views
Universe-sized groups with only set-sized normal subgroups, their cardinality in a certain range
Let $\kappa$ be an inaccessible cardinal, and let $G$ be a group with $|G| \geq \kappa$. For any cardinal $\lambda \le \kappa$ (regular, say, but not necessary), say $G$ is $\lambda$-simple if for all ...