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Topic: Free group on m generators elementary extension of the free group on n generators (n < m)?
Replies: 11   Last Post: Jul 15, 2013 1:13 PM

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Butch Malahide

Posts: 894
Registered: 6/29/05
Re: Free group on m generators elementary extension of the free group
on n generators (n < m)?

Posted: Jul 15, 2013 1:35 AM
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On Sunday, July 14, 2013 11:29:23 PM UTC-5, 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. 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.

Oops, I got that backwards. I meant to say, Phi(2) holds in A, not in B.




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