Can every nonempty set carry abelian group structure? [duplicate]

Possible Duplicate:
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.

marked as duplicate by Steven Gubkin, Theo Johnson-Freyd, Andreas Blass, Emil Jeřábek, Ryan BudneyMar 1 '12 at 5:08

This question was marked as an exact duplicate of an existing question.

• See related question mathoverflow.net/questions/12973/…, concerning the use of the axiom of choice in imposing a group structure. – Joel David Hamkins Feb 29 '12 at 16:16
• The free abelian group on an infinite set X has the same cardinality as X – Steven Gubkin Feb 29 '12 at 16:19
• You can read several answers on math.SE: math.stackexchange.com/q/105433/622 – Asaf Karagila Feb 29 '12 at 16:29
• You can take the direct sum of $|X|$ copies of $\mathbb{Z}$ (assuming the axiom of choice, at any rate). This is the set of all functions $f\colon X\to\mathbb{Z}$ of finite support, with pointwise addition. The cardinality is $|X|$. – Arturo Magidin Feb 29 '12 at 16:43
• @Arturo: Indeed if you take $\bigoplus_X\mathbb Z$ then $X$ indeed embeds into it as you said. However without the axiom of choice it is possible that the group generated by this embedding has a strictly larger cardinality. – Asaf Karagila Feb 29 '12 at 17:02