## The automorphism group of the quaterion group is isomorphic to Sym(4)

Prove that $\mathsf{Aut}(Q_8) \cong S_4$.

We already know that $|\mathsf{Aut}(Q_8)| = 24$ by counting the possible images of, say, $i$ and $j$. By a previous theorem, it suffices to show that no automorphism of $Q_8$ has order 6.

Let $\varphi$ be an automorphism of $Q_8$. Now $Q_8$ has 6 elements of order 4, one element of order 2, and one identity; $\varphi$ must fix the identity and the element of order 2, so that we can consider $\varphi$ as an element of $S_A$, where $A = \{i,j,k,-i,-j,-k\}$. Suppose now that some automorphism $\varphi$ has order 6. Now $\varphi(i) = a$, where $a \notin \{i,-i\}$, and $\varphi(a) = b$, where $b \notin \{i,-i,a,-a\}$. Clearly $\varphi(-i) = -a$ and $\varphi(-a) = -b$. If $\varphi(b) = i$, then $\varphi = (i\ a\ b)(-i\ -a\ -b)$ and $\varphi$ has order 3; thus $\varphi(b) = -i$, and we have $\varphi = (i\ a\ b\ -i\ -a\ -b)$. Now $\varphi^3 = (i\ -i)(a\ -a)(b\ -b)$. Note that $ia \in \{b,-b\}$. If $ia = b$, then $ia = (-i)(-a)$ $= \varphi^3(i)\varphi^3(a)$ $= \varphi^3(ia)$ $= \varphi^3(b) = -b$. Similarly, if $ia = -b$, then $ia = b$. Both cases yield a contradiction, so that no automorphism of order 6 exists.

In addition, we record this handy lemma.

Lemma: Let $G$ be a group and let $K \leq G$ be characteristic. Then $\mathsf{Aut}(G)$ acts on $G/K$ by $\varphi \cdot xK = \varphi(x)K$. Proof: First we need to show well-definedness: if $xK = yK$, we have $xy^{-1} \in K$. Then $\varphi(xy^{-1}) = \varphi(x)\varphi(y)^{-1} \in K$ since $K$ is characteristic in $G$, so that $\varphi(x)K = \varphi(y)K$. Moreover, $1 \cdot xK = 1(x)K = xK$ and $(\varphi \circ \psi) \cdot xK = (\varphi \circ \psi)(x)K = \varphi(\psi(x))K$ $= \varphi \cdot (\psi(x)K)$ $= \varphi \cdot (\psi \cdot xK)$. $\square$