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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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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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  • $\begingroup$ Of course! Why didn't I see that? Thank you very much! $\endgroup$
    – Asaf Karagila ♦
    Commented May 27, 2013 at 10:23
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    $\begingroup$ The third sentence might be rephrased; there are of course free abelian groups of cardinality continuum. The non-principal ultrapower is not $\aleph_{2}$-free. $\endgroup$
    – Avshalom
    Commented Oct 20, 2014 at 21:30
  • $\begingroup$ So any group with an infinite Abelian subgroup has an elementary extension with an uncountable Abelian subgroup. You used an ultrapower, but I guess you can also do this with the compactness theorem or the upward Löwenheim–Skolem theorem? $\endgroup$
    – bof
    Commented Oct 20, 2014 at 22:19
  • $\begingroup$ @bof That is an attractive observation. I imagine one could expand the language with uncountable many new constants and appropriate sentences asserting commutativity; then considering the elementary diagram together with the sentences, use compactness. But ultrapowers appeal to me. $\endgroup$
    – Avshalom
    Commented Oct 20, 2014 at 22:44
  • $\begingroup$ @Avshalom That's the sort of construction I had in mind, but I'm not a logician so I wasn't sure. Anyway, I guess the general compactness theorem is just as nonconstructive as an ultrafilter? Either way, I guess you don't quite need an infinite Abelian subgroup in the base group, unbounded finite Abelian subgroups would do as well. $\endgroup$
    – bof
    Commented Oct 20, 2014 at 22:55
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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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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.

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  • $\begingroup$ Interesting. The strongly compact cardinal enters from the Magidor-Vaananen work about Lowenheim-Skolem-Tarski numbers for second-order logic. Right? $\endgroup$
    – Asaf Karagila ♦
    Commented Oct 20, 2014 at 22:01
  • $\begingroup$ If $\kappa$ is strongly compact, then $\kappa$-freeness implies freeness (like singular compactness), so there is a $L_{\kappa, \kappa}$-sentence saying a group is $\kappa$-free. For the converse, a downward Loewenheim Skolem property is used, as you say, but I am not sure of all names of the people involved. $\endgroup$
    – Avshalom
    Commented Oct 20, 2014 at 22:34

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