Isomorphism between $SU(2)$ and $U(1, \mathbb H)$ Question:

Prove $SU(2)$ is isomorphic to the group of quaternions of norm $1$, that is, $U(1,\mathbb H) \simeq SU(2)$.

Attempt:
How could I start finding the isomorphism? Intuitively, a quaternion can be expressed as two complex numbers $a+bi+cj+dk=a+bi+(c+di)j$, and as an element of $SU(2)$ is $\left[
  \begin{array}{ c c }
     \alpha & \beta \\
     -\bar{\beta} & \bar{\alpha}
  \end{array} \right]$ such that $\mid{\alpha}\mid^2+\mid\beta\mid^2=1$ with $\alpha,\beta \in\mathbb{C} $ then we can map $\alpha=a +bi$ and $\beta=c +di$.
$a^2 + b^2 +c^2 +d^2 = 1$ holds for quaternions of norm $1$, hence $\mid{\alpha}\mid^2+\mid\beta\mid^2$ also holds.
 A: Just to formalize your idea for future reference. (I came looking for an answer but it wasn't complete). 
Consider the map $$\begin{align}\varphi: U(1, \mathbb H) &\to SU(2)\\x = z + wj &\mapsto \begin{pmatrix}z & w \\ -\overline w &\overline z\end{pmatrix}\end{align}$$
where $x = x^0 \textbf 1 + x^1 \textbf i + x^2 \textbf j + x^3 \textbf k  $, $z = x^0 + x^1 \textbf i \in \mathbb C$ and $w = x^2 + x^3 \textbf i \in \mathbb C$. 
It's clear that $$|x|^2 = |z|^2 + |w|^2 = \det \begin{pmatrix} z & w \\-\overline w & \overline z \end{pmatrix}$$
So $x \in U(1, \mathbb H)$ if, and only if, $\begin{pmatrix} z & w \\-\overline w & \overline z \end{pmatrix}$ has determinant $1$. Plus, such matrices account for all elements of $SU(2)$. 
Now we must check $\varphi$ is in fact  a group isomorphism.


*

*$\varphi$ is a group homomorphism. 


A simple calculation shows that $jz = \overline z j$ and $j w = \overline w j$. Thus if $x,y \in U(1,\mathbb H)$ are given we have $$\begin{align} xy &= (u + vj)(z + w j)\\&= uz + uw j + v(jz) + v(jw)j\\&= (uz - v\overline w) + (uw + v\overline z)j  \end{align}$$
Applying $$\begin{align}\varphi(xy) &= \begin{pmatrix}uz - v\overline w & uw + v\overline z \\-\overline{uw + v \overline z} & \overline{uz - v\overline w}\end{pmatrix} \\&=\begin{pmatrix}uz - v\overline w & uw + v\overline z \\- \overline u \,\overline w -  \overline v z &  \overline u \,  \overline z -  \overline v \,w\end{pmatrix} \\&=\begin{pmatrix}u & u \\-\overline v & \overline u\end{pmatrix} \begin{pmatrix}z & w \\-\overline w & \overline z\end{pmatrix}\\&= \varphi(x)\,\, \varphi(y)\end{align}$$


*

*$\varphi$ is clearly a bijection.


Thus we conclude that $U(1,\mathbb H) \simeq SU(2)$ as we wanted. 
Remark: As $S^3 = U(1,\mathbb H) \simeq SU(2)$ we get that $SU(2)$ is simply connected. 
