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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
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@user1729: Yes you're right. I missed there's not a 1, however; it is infinite. :-) –  B. S. Sep 6 '13 at 9:30
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@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
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(@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

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up vote -1 down vote accepted

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 $(x y)^x = (x y)^y = y x = (x y)^{-1}$.

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This is a non-answer - how can you say "no further information (can be derived)"? I mean, there are many, many things which we can deduce. Simply saying "we know what the group is" is pointless and meaningless! (Although, of course, the question is currently un-answerable, but then why answer it?!) –  user1729 Sep 8 '13 at 13:19

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