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> **Possible Duplicate:**  
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> [Does every non-empty set admit a group structure (in ZF)?](https://mathoverflow.net/questions/12973/does-every-non-empty-set-admit-a-group-structure-in-zf) 
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> Let $X$ be an arbitrary nonempty set. Can you define a multiplication making it into an abelian group?
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If $X$ is finite, say $|X|=n$, we can just use $X \cong \mathbb{Z}/n\mathbb{Z}$. What if $X$ is infinite?


If I'm not mistaken, the group of permutations of $X$ with finite support has the same cardinality as $X$. So at least any nonempty set carries a group structure. But abelianizing this particular group structure changes the cardinality.


Apologies if it is obvious, my group theory knowledge is just insufficient.



Can every nonempty set carry abelian group structure?
If the axiom of choice holds, then this is an immediate consequence of the upward Lowenheim-Skolem theorem. Any first order theory in a finite language with an infinite model, such as the theory of the infinite cyclic group, admits models of every infinite cardinality. Thus, one can find infinite abelian groups of any given size satisfying exactly the same theory, in the language of group theory, as your favorite infinite abelian group.


Without the axiom of choice, there are sets admitting no group structure at all, as show in [Ashutosh's answer](https://mathoverflow.net/questions/12973/does-every-non-empty-set-admit-a-group-structure-in-zf) to the question I mention in the comment. 


