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Is there a way to axiomatize [non-abelian] free groups in first-order logic using the language of groups (which contains the binary operation symbol $\cdot$, and the constant symbol $e$)?

Is there one particular axiom, or even a schema, from which we can prove that $G$ is a free group? (Regardless to the cardinality of a generating set.)

I should clarify that I'm not interested in augmented languages where we allow additional constant symbols for the generating set (in which case we can just write a schema stating when the various strings are equal).

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up vote 23 down vote accepted

The free groups cannot be axiomized by first order axioms. If the free groups were axiomatizable by first order axioms, then the ultraproduct of free groups would be a free group. However, the group $\mathbb{Z}$ is free, but for every non-principal ultrafilter $\mathcal{U}$ on $\mathbb{N}$, the ultrapower $\mathbb{Z}^{\mathcal{U}}$ is not free since it is an abelian group of cardinality continuum. More generally, any ultrapower $G^{\mathcal{U}}$ by a non $\sigma$-complete ultrafilter $\mathcal{U}$ of any free group $G$ is not free since the ultrapower $G^{\mathcal{U}}$ contains an isomorphic copy of the non-free subgroup $\mathbb{Z}^{\mathcal{U}}$ (recall that the subgroup of a free group is always free).

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Of course! Why didn't I see that? Thank you very much! –  Asaf Karagila May 27 '13 at 10:23
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The surface group of genus $\ge 2$ has the same elementary theory as any free non-Abelian group. That follows from results of Kharlampovich-Myasnikov and Sela on the Tarski problem. In fact one can completely describe all finitely generated groups that are elementary equivalent to free non-Abelian groups. That class does not consist of free groups (since surface groups are not free), but is not too far from the class of free groups.

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Thank you, Mark. –  Asaf Karagila May 27 '13 at 11:57
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This throws some new light on why this MO question was hard to answer: mathoverflow.net/questions/93330/… –  John Stillwell May 27 '13 at 12:38
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