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Let $G$ be an infinite group with an infinite cyclic normal subgroup $H$ such that $|G/H|=2$ and is cyclic. Show that $G$ is isomorphic to one of $\mathbb{Z},\mathbb{Z}\times\mathbb{Z_2},D_\infty$.

$D_\infty$ is the infinite dihedral group, which is generated by $\langle r,s\rangle$ such that $s^2=1$ and $srs=r^{-1}$.

What I've tried so far: $H$ and $G/H$ are finitely generated, so $G$ is as well. We next look for an element $g\in G$ of order $2$. If such $g$ doesn't exist, we have to show it's isomorphic to $\mathbb{Z}$, otherwise we look for whether $G$ is abelian. If it's not, we have to show $G\simeq D_\infty$, otherwise $G\simeq\mathbb{Z}\times\mathbb{Z}_2$.

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  • $\begingroup$ You are on the right track. Hint: Consider an element $x$ such that $x^2\in H$ but $x\not\in H$. What happens if $x$ does not have order $2$? $\endgroup$ – Tobias Kildetoft Nov 25 '14 at 8:02
  • $\begingroup$ Does that mean that $xH$ is the only coset of $H$ in $G$? I'm not really sure $\endgroup$ – ant11 Nov 25 '14 at 18:57
  • $\begingroup$ Word 'normal' is extra since index is 2. $\endgroup$ – Jihad Nov 25 '14 at 20:10

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