All Questions
Tagged with axiom-of-choice gr.group-theory
13
questions
7
votes
2
answers
603
views
Involutions in the absolute Galois group (and the Axiom of Choice)
It is known that the only elementary abelian $2$-groups (finite and nonfinite) in $\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$ are in fact finite and cyclic – that is to say, they are of order $2$....
- 4,503
6
votes
0
answers
449
views
Groups $G$ such that $\mathrm{Aut}(G) \simeq \mathbb{Z}/2\mathbb{Z}$
$\DeclareMathOperator\Aut{Aut}\DeclareMathOperator\Inn{Inn}$First I hope my question belongs here, please let me know if it doesn't.
It isn't too hard to show there is no groups $G$ such that $\Aut(G) ...
- 69
8
votes
0
answers
307
views
A $\mathsf{ZF}$ example of a nonreflexive group which is isomorphic to its double dual?
Given a group $G$ denote by $G^\ast=\mathrm{Hom}(G,\Bbb Z)$ its dual and by $j\colon G\to G^{\ast\ast}$ the canonical homomorphism $g\mapsto (f\mapsto f(g))$. A group is reflexive iff $j$ is an ...
- 3,011
5
votes
0
answers
169
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
8
votes
0
answers
368
views
Can the set of endomorphisms of $(\mathbb R,+)$ have cardinality strictly between $\frak c$ and $\frak{c^c}$?
Let $\frak c$ be the cardinality of the reals. I know that in ZF the set of endomorphisms of $(\mathbb R,+)$ can have at least two different cardinalites:
If we allow the axiom of choice, you can ...
user35370
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 ...
20
votes
2
answers
1k
views
Can There be a 1 dimensional Banach-Tarski paradox in the absence of choice
Let $\mathbb{R}$ act on itself by translation. Then there is no finite decomposition of a unit interval into pieces which, when translated, yields two distinct unit intervals.
More formally does ...
- 545
4
votes
2
answers
448
views
Number of torsion-free abelian groups
Let $\mathfrak{c}$ be the cardinality of the continuum. How much Choice, if any, is needed to prove that there are $2^{\mathfrak{c}}$ distinct (mutually nonisomorphic) torsion-free abelian groups of ...
- 2,019
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
19
votes
2
answers
2k
views
Without choice, can every homomorphism from a profinite group to a finite group be continuous?
In ZFC, some homomorphisms from profinite groups to finite groups are discontinuous. For instance, see the examples in this question. However, all three constructions given use consequences of the ...
- 141k
15
votes
3
answers
1k
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 ...
- 2,421
13
votes
2
answers
8k
views
AC in group isomorphism between R and R^2
Using the axiom of choice, one can show that $\mathbb{R}$ and $\mathbb{R}^2$ are isomorphic as additive groups. In particular, they are both vector spaces over $\mathbb{Q}$ and AC gives bases of ...
- 8,433
111
votes
2
answers
15k
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 ...
- 3,481