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Let $D_{\infty}= \langle x,y \mid x^2=y^2=1\rangle$ be the infinite dihedral group. Are the following statements true?

  1. Since $G$ is not torsionfree, $\mathbb{Q}[G]$ is not a domain.

  2. $D_{\infty}$ is an infinite subgroup of $G= \langle x,y \mid x^2=y^2\rangle$.

  3. The infinite dihedral group has a free abelian subgroup $F$ generated by $\langle x y \rangle$ of rank one and index $2$, and thus normal. $F$ is also subgroup of $G$.

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    $\begingroup$ What is $\;G\;$ in question 1? $\endgroup$ – DonAntonio Oct 5 '13 at 17:15
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  1. Yes, it is not a domain.

  2. No, it is a factor group.

  3. Yes.

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  • $\begingroup$ In this context one more question. There is another presentation of $G=\langle a,b \mid a^{-1}b a=b \rangle$. Is $a^2$ in the center of G; $Z(G)$? $\endgroup$ – R2D2 Oct 5 '13 at 12:40
  • $\begingroup$ @R2D2 None of the groups you have mentioned have the presentation $G=\langle a, b; a^{-1}ba=b\rangle$. this group is abelian, but neither of the groups in your question are abelian. $\endgroup$ – user1729 Oct 5 '13 at 12:57
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    $\begingroup$ @R2D2 No, another presentation is $G=\langle a, b| a^{-1}ba=b^{-1}, a^2=1\rangle$, where $a=x,b=xy$. Indeed $Z(G)=1$. $\endgroup$ – Boris Novikov Oct 5 '13 at 13:11
  • $\begingroup$ @BorisNovikov. Sorry I meant $G=\langle a,b \mid a^{-1}ba=b^{-1}\rangle$ for $a=x$ and $b=xy$. Would you still say, that $a^2$ is not central? I would. But I am working with something, where they claim $a^2$ to be central, or at least $a^2b=ba^2$. $\endgroup$ – R2D2 Oct 5 '13 at 13:32
  • $\begingroup$ @R2D2 $a^2$ is central since it is equal to $1$. $\endgroup$ – Boris Novikov Oct 5 '13 at 13:37

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