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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.


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

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

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    $\begingroup$ See related question mathoverflow.net/questions/12973/…, concerning the use of the axiom of choice in imposing a group structure. $\endgroup$ – Joel David Hamkins Feb 29 '12 at 16:16
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    $\begingroup$ The free abelian group on an infinite set X has the same cardinality as X $\endgroup$ – Steven Gubkin Feb 29 '12 at 16:19
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    $\begingroup$ You can read several answers on math.SE: math.stackexchange.com/q/105433/622 $\endgroup$ – Asaf Karagila Feb 29 '12 at 16:29
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    $\begingroup$ 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 $|X|$. $\endgroup$ – Arturo Magidin Feb 29 '12 at 16:43
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    $\begingroup$ @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. $\endgroup$ – Asaf Karagila Feb 29 '12 at 17:02

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 to the question I mention in the comment.


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