Q:

Matrix Group Isomorphism

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)?

A:

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 ...

