Why is the set of units of integer quaternions isomorphic to the quaternion group of order 8?

Question

Let's say that I've got a ring $V$ of integer quaternions of the form $\mathbb{Z} + \mathbb{Z}i + \mathbb{Z}j + \mathbb{Z}k$. Now assume that there exists an element $a = a_1 + a_2i + a_3j + a_4k$ such that $ab=ba=1$ for some other element $b \in V$ and likewise that ${a_1}^2 + {a_2}^2 + {a_3}^2 + {a_4}^2 = 1$. Why would this imply that the set of all such $a \in V$, called $V^\text{x}$, is isomorphic to the quaternion group of order $8$?

Answer 1

I’m guessing from your comment that you don’t know what $Q_8$ is. It has eight elements, $\{\pm1,\pm i,\pm j, \pm k\}$ with the property that $\{\pm1\}$ commute with everything, and such that $i$, $j$, and $k$ satisfy the combination rules that you know: $ij=k=-ji$, $jk=i=-kj$, and $ki=j=-ik$. Now you see it?