Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
@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
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 $(x y)^x = (x y)^y = y x = (x y)^{-1}$.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.