# Does every set have a group structure?

I know that there is no vector space having precisely $6$ elements. Does every set have a group structure?

The trivial answer is “no”: the empty set does not admit a group structure.

The statement

If $$XX$$ is a nonempty set, then there is a binary operation $$⋅\cdot$$ such that $$(X,⋅)(X,\cdot)$$ is a group.

is equivalent to the Axiom of Choice.

It is not needed for finite or countable sets: if $$XX$$ is finite, with $$nn$$ elements, then let $$f:X→{0,1,…,n−1}f\colon X\to\{0,1,\ldots,n-1\}$$ be a bijection, and use transport of structure to give $$XX$$ the structure of a cyclic group of order $$nn$$. If $$XX$$ is denumerably infinite, biject with $$Z\mathbb{Z}$$ and use transport of structure.

For uncountable sets, we can use the Axiom of Choice: let $$|X|=κ|X|=\kappa$$. Then the direct sum of $$κ\kappa$$ copies of $$Z\mathbb{Z}$$ has cardinality $$κ\kappa$$, so there is a bijection
$$f:X→⨁i∈κZ.f\colon X\to \bigoplus_{i\in\kappa}\mathbb{Z}.$$
Use transport of structure again to make $$XX$$ into a group.

That the converse holds (the statement implies the Axiom of Choice), is proven in this Math Overflow post.