Subgroups of a group generated by a free semigroup Let $F$ be a free semigroup (say, $2$-generated) which is embedded in a group $G$, and suppose that $G$ (as a group) is generated by $F$. The most simple such situation would be when $G$ is a free group, but there are lot of groups, besides free ones, which could occur in this situation (for example, $G$ could be solvable). Is it possible that $G$ contains a subgroup isomorphic to $\mathbb{Z} \times \mathbb{Z}$ (the direct product of two copies of the infinite cyclic group $\mathbb{Z}$)?
Update: thanks to all the people for a very interesting discussion. Sorry that I cannot award a few "accepted answers", so the only one goes to Greg who supplied the most lucid and explicit example.
 A: I don't think Richard's idea is quite right.  If G is the kernel of the map to Z^2 then ab and ba lie in the same coset of G, so G can't have property (P).
But there is a simple solvable example.  Consider the standard example of a lattice in Sol, namely the semidirect product of Z^2 (generated by a,b ) by Z (generated by t) in which tat^{-1}=2a+b and tbt^{-1}=a+b. This group, S, is generated by a and t and contains a rank-two abelian subgroup.
You can think of S as an HNN extension.  Britton's Lemma implies that any word w in t and a is reducible if and only if it 'obviously' is, ie if and only if you see something of the form tat^-1 or t^-1at.  In particular, every positive word in a and t is reduced and so a and t generate a free semigroup.
EDIT:
Greg points out that this doesn't work.  But it can be fixed with a nasty hack.  By the Milnor--Wolf Theorem, the group S does contain a non-abelian free semigroup, which we may as well take to be two-generator.  Let T be the subgroup generated by these two generators.  Now it's not hard to convince yourself that the only possibility is that T is a finite-index subgroup of S.  (Otherwise, T would be nilpotent, and so have polynomial growth.)   So T is generated by two elements that generate a free semigroup, and contains a copy of Z^2.
FURTHER EDIT:
Here are some more details, to exacerbate Tom's delight/revulsion.  The Milnor--Wolf Theorem asserts that every solvable group is either virtually nilpotent (ie has a nilpotent finite-index subgroup) or contains a free subsemigroup.  (Googling "Milnor--Wolf Theorm" gives abundant references.)  S is a solvable group that is not virtually nilpotent, so it contains a free sub-semigroup, which may as well be of rank two.  Let T be the group generated by this sub-semigroup.
Now S decomposes as a group extension -
1->Z^2->S->Z->1
and T inherits a decomposition as an extension
1->(Z^2\cap T)->T->T'->1.
Because T isn't abelian, T' is non-trivial.  If Z^2\cap T is cyclic or trivial then T is nilpotent, which contradicts the fact that T contains a free semigroup.  Therefore Z^2\cap T is finite-index in Z^2 and so T contains a copy of Z^2.
A: It would be schizophrenic to try to slice this into comments on the various answers, so:
Brandon's condition (P) does not imply that no two elements of the semigroup lie in the same coset. Let F=F2; from Richard's and Henry's answers we have the example of G = [F,F], which has (P), while nevertheless we have ab=ba. The same issue showed up in Henry's first solution.
However, continuing the construction nevertheless we obtain the universal 2-step solvable group Γ=F/[[F,F],[F,F]]. Since we know from Greg's and Henry's answers that free semigroups exist in 2-generated 2-step solvable groups, it follows from universality that the semigroup in Γ generated by a and b is free. Since Γ contains [F,F]/[[F,F],[F,F]], an infinite-rank free abelian group, we see that Brandon's construction does work, at least in this case.
(Please direct your upvotes to the people who came up with these examples, not me.)
A: It seems like you might be able to get whatever subgroup you desire. Here is an outline which hopefully you or someone else can complete. Let H be the subgroup you desire, and let F2 denote the non-abelian free group with two generators a and b. It is well known that there are group elements g1, g2, ... in F2 which generate a subgroup isomorphic to the free group with countably many generators, I don't remember how to find the gi though. You want to pick the gi carefully so that every group element in the subgroup generated by the gi has the property (P) that either an a^{-1} or b^{-1} appears in the middle of its reduced word representation. This will insure that no two elements in the semigroup generated by a and b will lie in the same coset of <{g1, g2, ...}>. Since <{g1, g2, ...}> is isomorphic to the free group with countably many generators, we get a surjective homomorphism h:<{g1, g2, ...}> -> H. Let K = kernel(h) and let K' be the normal subgroup generated by K. If you chose g1, g2, ... well, then K' will still have property (P) and the intersection of K' with <{g1, g2, ....}> will be K. Then F2/K' will have a subgroup isomorphic to H and will be generated by two elements which generate a free semigroup.
I'm not sure if you can always pick the gi to get all of the required properties. Hopefully someone can complete this proof.
A: I think that Henry's solution doesn't work either.  In that example, I get at2a = ta3t = a6b3t2.  But I think that this construction can be fixed.
Consider the group of affine linear maps f(x) = αx+β over the reals ℝ.  Let a act by multiplication by α, where α is transcendental, let b act by adding 1, let F be the semigroup that they generate, and let G be the group that they generate.  F is free because two distinct words in a and b, if they have the same degree in a, have in them two distinct polynomials in α with non-negative integer coefficients.  (It is interesting that α must be transcendental for this to work.)  Then G contains a ℤ x ℤ, generated by adding 1 and adding α.
A: Would Brandon's argument work if you take an element like w=ababAbaba (where A=a^{-1}) and then looked at the subgroup generated by the conjugates of w by powers of b?  Note that any element of this subgroup has the property that any element has an A in its reduced word.
