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

**Answer**

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

The statement

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

is equivalent to the Axiom of Choice.

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

For uncountable sets, we can use the Axiom of Choice: let |X|=κ. Then the direct sum of κ copies of Z has cardinality κ, so there is a bijection

f:X→⨁i∈κZ.

Use transport of structure again to make X into a group.

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

**Attribution***Source : Link , Question Author : spohreis , Answer Author : Arturo Magidin*