TITLE: On what sets can we define a group operation?
QUESTION [9 upvotes]: A question came up into mind.
Prove that for every nonempty set $X$ an operation can be suggested such that $X$ would be a group with that operation. For example, it is obvious for finite and countable sets: $(\mathbb{Z}_n,+), (\mathbb{Q}, +)$. Also, it can be done for all sets of form $X=2^L$, as $(X, \Delta)$, which is symmetric difference on subsets of $L$.
So is seems that the question is reduced to (1) sets which are high in hierarchy of cardinals (not of form $2^L$) and (2) sets which do not exist assuming continuum hypothesis (between $\mathbb{R}$ and $2^\mathbb{R}$ for instance).
Axiom of choice is given.
Or maybe intuition is wrong and for some sets it cannot be done, then a proof of existance of such or a single example would be nice. Thank you in advance.

REPLY [9 votes]: The free group (or the free Abelian group, etc.) generated by an infinite set $X$ will have the same cardinality as $X.$ Therefore, every nonempty set is the underlying set of a group.