TITLE: First-order axiomatization of free groups
QUESTION [9 upvotes]: 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).

REPLY [4 votes]: 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.