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Does every non-empty set admit a group structure (in ZF)?Let $X$ be an arbitrary nonempty set. Can you define a multiplication making it into an abelian group?

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.