Prove that $\operatorname{Gal}(\mathbb{Q}(\sqrt[8]{2}, i)/\mathbb{Q}(\sqrt{-2})) \cong Q_8$ I seem to have reached a contradiction.  I am trying to prove that $\operatorname{Gal}(\mathbb{Q}(\sqrt[8]{2}, i)/\mathbb{Q}(\sqrt{-2})) \cong Q_8$.
I could not think of a clever way to do this, so I decided to just list all the automorphisms of $\mathbb{Q}(\sqrt[8]{2}, i)$ that fix $\mathbb{Q}$ and hand-pick the ones that fix $i\sqrt{2}$.  By the Fundamental Theorem of Galois Theory, those automorphisms should be a subgroup of the ones that fix $\mathbb{Q}$.  I proved earlier that those automorphisms are given by $\sigma: \sqrt[8]{2} \mapsto \zeta^n\sqrt[8]{2}, i \mapsto \pm i$, where $n \in [0, 7]$ and $\zeta = e^\frac{2\pi i}{8}$.
However, I am getting too many automorphisms.  One automorphism that fixes $i\sqrt{2}$ is $\sigma: \sqrt[8]{2} \mapsto \zeta\sqrt[8]{2}, i \mapsto -i$.  However, this means all powers of $\sigma$ fix $i\sqrt{2}$, and I know $Q_8$ does not contain a cyclic subgroup of order $8$.  What am I doing wrong?
(Please do not give me the answer.  I have classmates for that.)

Would it be easier to notice that extension $\mathbb{Q}(\sqrt[8]{2},i)$ is equal to $\mathbb{Q}(\sqrt[8]{2},\zeta)$ which is a cyclotomic extension followed by Kummer extension? You can then work out which elements of its Galois group fix $\sqrt{-2}$.
