Date: Jul 15, 2013 4:38 AM
Author: Sandy
Subject: Re: Free group on m generators elementary extension of the free group<br> on n generators (n < m)?
Butch Malahide wrote:

> On Sunday, July 14, 2013 5:15:13 AM UTC-5, peps...@gmail.com 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.

Indeed so, Chang and Keisler (reference elsewhere in the thread) pose

the problem as follows:

Conjecture: If 2 <= p < n then A_p elementarily equivalent to A_n,

and in fact A_p elementary submodel of A_n (Tarski).

where A_m is the free group on m <= omega generators.

I see that you use Fred Galvin's e-mail address. That must be the Fred

Galvin who has four items in the C&C References. Respec'! (In my

country the youth seem not to pronounce the 't' in 'respect' :-).)

> 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.

>

> 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.

>