Free abelian subgroup of index 2.

Let $G$ be a group with the following presentation $G=gp(x,y \mid x^2=y^2=1)$. I need to know, what further information about $G$ can be derived from knowing that $G$ has a free abelian subgroup of index 2.

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@BabakS. This group is not infinite dihedral. You would need $x^2=y^2=1$ for that to be so. – user1729 Sep 6 '13 at 9:07
@user1729: Yes you're right. I missed there's not a 1, however; it is infinite. :-) – Babak S. Sep 6 '13 at 9:30
@BabakS. Indeed, it is a free product with amalgamation $\mathbb{Z}\ast_{2\mathbb{Z}}\mathbb{Z}$...(actually, re-reading the question, I wonder if the infinite dihedral group was what they were after?) – user1729 Sep 6 '13 at 9:59
(@BabakS. Actually, this group has a subgroup of index two isomorphic to $\mathbb{Z}\times\mathbb{Z}$. This is the subgroup normally generated by $xy^{-1}$. So perhaps they weren't after $D_{\infty}$ after all...) – user1729 Sep 6 '13 at 10:06
@user1729 @ BabakS. Yes, we are dealing with the infinite dihedral group. Sorry, for incorrectness. Edited it. – R2D2 Sep 6 '13 at 12:05

No further information, as this group, as noted in the comments, is (isomorphic to) the infinite dihedral group, and thus has a free abelian subgroup $F$ of rank one (isomorphic to the integers, that is) and index $2$, and thus nornal.
This is $\langle x y \rangle$, as $(x y)^x = (x y)^y = y x = (x y)^{-1}$.