Exercise 4.4.4 ($|\mathrm{Aut}(Q_8)| \leq 24$)

Proof. Let $\sigma$ be any automorphism of $Q_8$. We know that $Q_8=⟨i,j⟩$, where $|i|=|j|=4$. Therefore $|\sigma (i)|=4$, so (cf. Exercise 1.5.1)

Moreover, since $-1$ has order $2$, where $-1$ is the only element of order $2$ in $Q_8$,

Therefore $\sigma (-i)=\sigma ((-1)i)=\sigma (-1)\sigma (i)=-\sigma (i)$.

Note that

otherwise $j=\pm i$, which is false. Thus

where $\{\sigma (i),\sigma (-i)\}=\{\sigma (i),-\sigma (i)\}\subset \{i,-i,j,-j,k,-k\}$. Thus there are only 6 possibilities for the choice of $\sigma (j)$, as soon as $\sigma (i)$ is given.

So $(\sigma (i),\sigma (j))$ can take at most $24$ values. Since $Q_8=⟨i,j⟩$,

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