TITLE: Making a proof precise of "Aut$(Q_8)\cong S_4$"
QUESTION [11 upvotes]: I know, with some machinery, how to prove that Aut$(Q_8)\cong S_4$. My question here is about not how to prove, but is about an incomplete proof (I feel) given by a student to me (it should be possibly somewhere on-line, since it is not so easy to get the idea of his proof, I think.)

Consider a cube, and label $i,-i$ on a pair of opposite faces; similarly put $j,-j$ and $k,-k$ on other faces. Since the group of rotations of cube is $S_4$, Aut$(Q_8)$ is $S_4$.

Can anyone make this argument precise?

REPLY [5 votes]: As SpamIAm says in his solution, $-1$ and $1$ must be fixed by any automorphism $\phi$. Moreover, since $i$ and $j$ generate the group, an automorphism is determined by the images of $i$ and $j$. There are six possibilities left for $\phi(i)$ and at most five for $\phi(j)$. In fact, there are at most four, since $\phi(-i) = \phi(i^3) = \phi(i)^3$ is already decided. Therefore $Q_8$ has at most $24$ automorphisms.
All that is left now is to see that every rotation of the cube is actually an automorphism of the group. But it is well known that every rotation in $\mathbb{R}^3$ corresponds to some mapping $x \mapsto qxq^{-1}$ restricted to the pure quaternions, and this mapping is an inner automorphism of the ring of quaternions.