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May 26, 2015 at 4:36 history edited Duchamp Gérard H. E. CC BY-SA 3.0
put "An interesting alternative..." paragraph at the end and withdrawed "I suspect the answer will be YES for the two questions.""
May 25, 2015 at 20:44 comment added Duchamp Gérard H. E. @EmilJeřábek I think you are right. Then, I modified my text. Thanks. As I did it (too) quickly, I did not explore the fact that any free group can be embedded in a two generator free group. Vote +1 for Bjorn's answer
May 25, 2015 at 20:43 history edited Duchamp Gérard H. E. CC BY-SA 3.0
Modification of the construction oof the projective limit in order to take the infinite alphabet into account
May 24, 2015 at 21:21 comment added Christian Remling However, for the first sentence, the free group with one generator is enough: If $\forall x\exists y\: w=1$ holds in $\mathbb Z$ and $x\in G$ is given, map $1\mapsto x$ and take as $y$ the image (under this homomorphism) of the $y'\in \mathbb Z$ that exists by assumption. So the (first) sentence holds in all groups precisely if it holds in $\mathbb Z$.
May 24, 2015 at 14:41 comment added Emil Jeřábek @BjørnKjos-Hanssen: The sentences in question are positive, hence preserved by homomorphic images, and every group is a homomorphic image of a free group. One would need a free group over infinitely many generators for this to work, though.
May 24, 2015 at 9:02 comment added Bjørn Kjos-Hanssen I don't really see, how do you relate the free group to what's true of all groups?
May 24, 2015 at 6:52 history edited Duchamp Gérard H. E. CC BY-SA 3.0
positioning
May 24, 2015 at 6:43 history undeleted Duchamp Gérard H. E.
May 24, 2015 at 3:55 history deleted Duchamp Gérard H. E. via Vote
May 24, 2015 at 3:54 history edited Duchamp Gérard H. E. CC BY-SA 3.0
"Yes" made explicit in the beginning
May 24, 2015 at 3:46 history answered Duchamp Gérard H. E. CC BY-SA 3.0