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.



Subgroups of a group generated by a free semigroup
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.


