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 $X$ is a nonempty set, then there is a binary operation $\cdot$ such that $(X,\cdot)$ 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 $f\colon X\to\{0,1,\ldots,n-1\}$ 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 $\mathbb{Z}$ and use transport of structure. 

For uncountable sets, we can use the Axiom of Choice: let $|X|=\kappa$. Then the direct sum of $\kappa$ copies of $\mathbb{Z}$ has cardinality $\kappa$, so there is a bijection
$$f\colon X\to \bigoplus_{i\in\kappa}\mathbb{Z}.$$
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.
