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Consider the group $A = \left\{\begin{pmatrix} e^{t}&&\\& e^{\alpha t}&\\&&1\end{pmatrix}: t \in \mathbb{C}\right\}$. Why is it that when $\alpha \not\in \mathbb{Q}$, $A \cong \mathbb{C}$ and when $\alpha \in \mathbb{Q}$ we have $A \cong \mathbb{C}^{\ast}$ (and hence is not simply connected)?

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    And by $\mathbb{C}^*$ you mean $\mathbb{C}\backslash 0$? What makes you sure that this is right?2012-03-29

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When $\alpha \notin \mathbb Q, t \rightarrow A(t)$ is an isomorhism. To be in the kernel u have to satisfy $e^t = 1$ making $t = 2 \pi i j $ etc. When $\alpha \in \mathbb Q$, say $\alpha = 1$ there is a non trivial kernel equal to $\pi i j, j \in \mathbb Z$. I think you must be $\mathbb R \times $ the circle in this case, and that either that or $C^*$ are simply connected but have fundamental group ...