# What is the the fundamental group of $H_{\mathbb{R}}/H_{\mathbb{Z}}$

Consider $M = H_{\mathbb{R}}/H_{\mathbb{Z}}$, where $H_{\mathbb{R}} = \lbrace\begin{bmatrix} 1 & x & z \\ 0 & 1 & y \\ 0 & 0 & 1\\ \end{bmatrix}|x,y,z \in \mathbb{R}\rbrace$ and $H_{\mathbb{Z}} = \lbrace\begin{bmatrix} 1 & x & z \\ 0 & 1 & y \\ 0 & 0 & 1\\ \end{bmatrix}|x,y,z \in \mathbb{Z}\rbrace$

If we regards $M$ as a manifold (Only need to consider $M$ as topological manifold).

What is the fundamental group of $M$, $\pi_1(M)$? How to deduce that?

Thank you very much!

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I'm curious: did you lose interest in the question, are you not satisfied with the answer or what else is going on? –  t.b. Apr 27 '12 at 3:16
Sorry~ because there something bother me which makes me forget to login the forum so many days. I really appreciate your help and your answer, and I can understand your answer! :) –  Peter Hu Apr 29 '12 at 16:34

If a group $G$ acts properly discontinuously and freely on a simply connected manifold $M$ then $\pi_1(M/G) \cong G$. Apply this to the present situation with $M = H_\mathbb{R}$ (note that $M$ is diffeomorphic to $\mathbb{R}^3$) and $G = H_\mathbb{Z}$ to see that $\pi_1(H_\mathbb{R}/ H_\mathbb{Z}) \cong H_\mathbb{Z}$ — Note that a discrete subgroup of a Lie group always acts properly discontinuously on the surrounding Lie group.