# First-order axiomatization of free groups

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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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).
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.