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

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