Date: Jul 15, 2013 1:13 PM
Author: David C. Ullrich
Subject: Re: Free group on m generators elementary extension of the free group on n generators (n < m)?

On Sun, 14 Jul 2013 21:29:23 -0700 (PDT), Butch Malahide
<> wrote:

>On Sunday, July 14, 2013 5:15:13 AM UTC-5, wrote:
> I think I get it now. When a non-abelian group has an abelian subgroup, this is a non-elementary extension because the statement "For all x, For all y, xy = yx" is false in the larger group but true in the subgroup. I think the question with which you opened the post is basically equivalent to asking whether the theorem cited by David is correct. Am I more on target now?
>Closer but not there yet. Davild cited a result saying that all free groups (on two or more generators, I guess) are elementarily equivalent, which is apparently weaker than saying that one is an elementary extension of the other.
>Two structures A and B are "elementarily equivalent" if they satisfy the same first order *sentences* (formulas without free variables); thus, if A is an Abelian group, so is B.
>A is an "elementary extension" of B (in other words B is an "elementary substructure" of A" if, for any first order *formula* phi (which may have free variables), and any
>assignment of values in B to the free variables, phi is satisfied in B if and only if it's satisfied in A.

Ah. I missed that. (Being aware of my own ignorance, I _was_ careful
to avoid claiming that I'd actually answered the question...)

>For example, consider the group A = (Z,+) of all integers, and the subgroup B = (2Z,+) of the even integers. B is isomorphic (and a fortiori elementarily equivalent) to B, but A is not an elementary extension of B. To see this, consider the formula phi(x) := (exist y)(y + y = x), and observe that phi(2) holds in B but does not hold in A.