12
votes
1answer
400 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 ...
10
votes
3answers
743 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 ...
11
votes
2answers
2k 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 ...
57
votes
3answers
5k 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 ...