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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?

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  • $\begingroup$ it is possible; but I don't know. $\endgroup$ – p Groups Jan 3 '16 at 12:50
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First note that $-1$ is the unique element of order $2$ in $Q_8$ so $\sigma(-1) = -1$ for any $\sigma \in \text{Aut}(Q_8)$. Mapping $i,j,k$ to the standard basis vectors $e_1, e_2, e_3$ of $\mathbb{R}^3$ induces an injective homomorphism $\varphi: \text{Aut}(Q_8) \to \text{GL}_3(\mathbb{R})$: given $\sigma \in \text{Aut}(Q_8)$, the images of $\sigma(i),\sigma(j),\sigma(k)$ form a basis for $\mathbb{R}^3$, hence $\sigma$ induces a unique (invertible) linear transformation. Moreover, since $\pm e_1, \pm e_2, \pm e_3$ form an orthonormal basis, then the image of $\varphi$ lies inside the orthogonal group $O_3$. Since $\sigma$ fixes $-1$, then the induced linear transformation $\varphi(\sigma)$ preserves orientation, hence the image of $\varphi$ lies inside the special orthogonal group $SO_3$. By construction, $\varphi(\sigma)$ maps the cube centered at the origin with side length $2$ to itself, hence produces an orientation-preserving symmetry of the cube. We still need to show that $\varphi$ is surjective, i.e., given a symmetry $T$ of the cube, the map on $Q_8$ induced by the action of $T$ on the basis $e_1, e_2, e_3$ is actually a homomorphism, but I haven't thought through the proof yet.

(Just as an aside, I think you could also try working over $\mathbb{F}_3 = \{-1,0,1\}$ and show that $\text{Aut}(Q_8) \cong \text{SO}_3(\mathbb{F}_3)$.)

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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.

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