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This must be known inside out by model theorists by I have no cluse whether the following is true or not:

Denote by $F_n$ the free group on $n$ generators. Suppose that $n\neq m$. Are the groups $F_n$ and $F_m$ elementarily equivalent? What if we allow the set of generators to be countably infinite?

I'd appreciate any hints for the proof of the answer to this question. Looking at ultrapowers seem to me to be intractable, but who knows?

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Note that we want to ask when $m,n \ge 2$, since $m=1$ can be distinguished by abelianness. –  hardmath Jul 16 at 19:16

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up vote 8 down vote accepted

Yes they are. This is know as the Tarski Problem, and was recently solved by Sela and Kharlampovich-Myasnikov. The solution however is very difficult and spans more than 100 pages. Sela's approach at least uses sophisticated ideas from geometric group theory.

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Thank you! Does it include the free group on infinitely many generators? –  TMK Jul 16 at 19:54
    
It includes free groups on infinitely many generators. Though for them it is much easier. For infinite $X \subseteq Y$ it is easy to show that $F_X \preceq F_Y$ by Tarski-Vaught test. –  Levon Haykazyan Jul 17 at 9:24
    
@LevonHaykazyan Do you know whether 1) $F_X \equiv F_Y$ when $2 < |X| < \aleph_0$ and $Y$ is infinite? 2) $F_X\preceq F_Y$ when $X\subseteq Y$ and $X$ and $Y$ are finite? Of course, a positive answer to (2) implies a positive answer to (1)... –  Alex Kruckman Jul 19 at 1:37
    
@AlexKruckman Yes! (2) is true and hence so is (1). –  Levon Haykazyan Jul 20 at 19:25
    
Oh good, that's comforting. Is there a good reason for this (i.e. the theory of the free group is model complete), or is it just a hard consequence of Sela's work? –  Alex Kruckman Jul 20 at 19:29

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