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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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3 Answers 3

up vote 24 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
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. –  Avshalom Oct 20 '14 at 21:30
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? –  bof Oct 20 '14 at 22:19
@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. –  Avshalom Oct 20 '14 at 22:44
@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. –  bof Oct 20 '14 at 22:55

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
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

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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Interesting. The strongly compact cardinal enters from the Magidor-Vaananen work about Lowenheim-Skolem-Tarski numbers for second-order logic. Right? –  Asaf Karagila Oct 20 '14 at 22:01
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. –  Avshalom Oct 20 '14 at 22:34

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