TITLE: Can every set be a group?
QUESTION [18 upvotes]: Lets consider a set A. Can we prove that  there exists at least one operation * such that under it the set forms a group. May I know how to approach this question?

REPLY [5 votes]: The answers are already given, but since you asked how you should approach this, I'll try to address that issue. 
The first thing to understand is that if you have a bijection $f:A\to B$ between any two sets, then any group structure on $A$ can be transferred along $f$ to a group structure on $B$. More precisely, there is precisely one group structure on $B$ for which $f$ is a group homomorphism (necessarily isomorphism). If this is not clear to you, spend some time and prove it. 
Now, the question becomes equivalent to, given any cardinality $\kappa$ (other than $\kappa = 0$), is there some set of that cardinality which supports a group structure. (Under the axiom of choice,) every set has a cardinality and so if for every cardinality there is some  group of that cardinality then, using the result above, every set supports a group structure (upon choosing a bijection, so the axiom of choice is used again). 
So, it boils down to constructing groups of all possible cardinalities. Finite ones are dealt with easily by the finite cyclic groups. Infinite ones are constructed by taking products of sufficiently many finite groups, using basic rules of cardinal arithmetic (assuming the axiom of choice again).