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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.

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Chris Heunen
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    See related question http://mathoverflow.net/questions/12973/does-every-non-empty-set-admit-a-group-structure-in-zf, concerning the use of the axiom of choice in imposing a group structure. – Joel David Hamkins Feb 29 '12 at 16:16
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    The free abelian group on an infinite set X has the same cardinality as X – Steven Gubkin Feb 29 '12 at 16:19
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    You can read several answers on math.SE: http://math.stackexchange.com/q/105433/622 – Asaf Karagila Feb 29 '12 at 16:29
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    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
  • P.S. I believe that you only need the axiom of choice in order to prove that the resulting group is bijectable to $X$ (and thus use transfer of structure to endow $X$ with an abelian group structure); the fact that the group exists given a set $X$, and that you can embed $X$ into the group (via mapping $x\in X$ to the characteristic function of ${x}$), does not seem to me to require AC, but if I'm wrong I'm sure the experts on this will correct me. – Arturo Magidin Feb 29 '12 at 16:56
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    @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
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    (Arturo’s and Steven’s examples are one and the same.) – Emil Jeřábek Feb 29 '12 at 18:10
  • @AsafKaragila I believe that such questions (as do essentially all of mathematics outside set theory) assume AC implicitly. – Boaz Tsaban Jul 31 '15 at 03:33

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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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Joel David Hamkins
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