# Infinite Dihedral Group.

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 $G$.

• What is $\;G\;$ in question 1? – DonAntonio Oct 5 '13 at 17:15

• 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; $Z(G)$? – R2D2 Oct 5 '13 at 12:40
• @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. – user1729 Oct 5 '13 at 12:57
• @R2D2 No, another presentation is $G=\langle a, b| a^{-1}ba=b^{-1}, a^2=1\rangle$, where $a=x,b=xy$. Indeed $Z(G)=1$. – Boris Novikov Oct 5 '13 at 13:11
• @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 $a^2b=ba^2$. – R2D2 Oct 5 '13 at 13:32
• @R2D2 $a^2$ is central since it is equal to $1$. – Boris Novikov Oct 5 '13 at 13:37