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).
 A: More can be said than non-first-order axiomatizability. Since the free group $  \mathbb{Z}^{(\omega)}$ is an $L_{\infty, \omega}$-elementary substructure of the non-$\aleph_{2}$-free group $\mathbb{Z}^{\omega}$, there is no axiomatization of the class of free groups (or of $\aleph_{2}$-free groups) in $L_{\infty, \omega}$. 
If large cardinals exist, the outcome is very different. Mekler proved that if there is a strongly compact cardinal $\kappa$, then the class of free abelian groups is definable in $L_{\infty, \infty}$ (and in $L_{\kappa, \kappa}$ in fact). Furthermore, if the class of free abelian groups is definable in $L_{\infty, \infty}$, then there is an inner model with a measurable cardinal.
A: 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).
A: 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. 
