# 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