Let $d$ be an arbitrary real number and $M_d=\left\{ \begin{pmatrix} a& db\\ b& a \end{pmatrix}\in\mathcal{M}_2(\mathbb{R}), \text{where } a^2-db^2\neq 0\right\}. $
The problem is: show that $(M_d, \cdot)$ is a group("$\cdot$" is here the matrix multiplication), and find the values of $d$ such that $(M_d, \cdot)\cong (\mathbb{C}^*, \cdot)$.
I've managed to show that $(M_d, \cdot)$ is a group, it's just about checking the group axioms. But I can't find the all values of $d$.
I know that it's well-known fact that $(M_{-1}, \cdot)\cong (\mathbb{C}^*, \cdot)$, where the isomorphism function is just $ \begin{pmatrix} a& db\\ b& a \end{pmatrix}\mapsto a+bi $, but I have no idea how I can prove that $d=-1$ is only one solution, or find another solutions.
For any non-zero $d,$ consider the map $\varphi_d$ on $M_d$ defined as $\varphi_d:\begin{pmatrix} a& db\\ b& a \end{pmatrix}\rightarrow a-dbi.$
Then, $\varphi_d$ is clearly surjective. Also, for
$$A=\begin{pmatrix} a& db\\ b& a \end{pmatrix},\quad B=\begin{pmatrix} x& dy\\ y& x \end{pmatrix}, $$ we have that $$\varphi_d(AB)=\varphi\left(\begin{pmatrix} ax+dby& day+dbx\\ bx+ay& bdy+ax \end{pmatrix} \right)=ax+bdy-d(ay+bx)i=(a-dbi)(x-dyi)=\varphi_d(A)\varphi_d(B).$$
So, $\varphi_d$ is an endomorphism between groups, and by the first isomorphism theorem, $M_d/\ker(\varphi_d)\cong C^*.$
Now, $\ker(\varphi_d)$ is those set of matrices $A=\begin{pmatrix} a& db\\ b& a \end{pmatrix}$ such that $a-dbi=1.$ Comparing imaginary and real parts, we must have that the kernel has those matrices such that $a=1,b=0,$ that is, the identiy matrix.
Thus, $\ker(\varphi_d)=\{I\}$ and so, $M_d\cong C^*.$
Exercise: What about when $d=0?$ Is there an isomorphism?