Please help clarify this Galois proof

I more or less copied the following notes from my professor:

Quote:

Find the Galois group of $\displaystyle x^4-2$ over $\displaystyle \mathbb{Q}$.

Solution: $\displaystyle x^4-2$ has roots $\displaystyle \pm\sqrt[4]{2},\pm i\sqrt[4]{2}$. **Note that since $\displaystyle G$ acts on the four roots, $\displaystyle G$ is isomorphic to a subgroup of $\displaystyle S_4$.** The splitting field is $\displaystyle \mathbb{Q}(\sqrt[4]{2},i)$, so $\displaystyle |G|=8$. By Sylow, any two subgroups of $\displaystyle S_4$ of order $\displaystyle 8$ must be isomorphic. So $\displaystyle G\cong D_8$.

Everything about this proof makes perfect sense to me except the part in bold: Why is $\displaystyle G$ a subgroup (isomorphic) of $\displaystyle S_4$? It cannot be simply because $\displaystyle G$ acts on the four roots (because for instance any group of any order acts on any set trivially). All I can tell from that is that there is a homomorphism from $\displaystyle G$ into $\displaystyle S_4$. So, how do I conclude that $\displaystyle G$ is a subgroup (isomorphic) of $\displaystyle S_4$?

Also, I'm a bit confused as to how it can be that $\displaystyle |G|=8$. For each $\displaystyle \alpha\in G$ is completely determined by $\displaystyle \alpha(i)$ and $\displaystyle \alpha(\sqrt[4]{2})$. Now, $\displaystyle 1=\alpha(1)=\alpha(-i^2)=-\alpha(i)^2$, which seems to imply $\displaystyle \alpha(i)=i$. But then $\displaystyle \alpha(\sqrt[4]{2})$ must be one of the four roots. So $\displaystyle |G|\leq 4$, a contradiction. Where did I go wrong, I wonder?

Any help would be much appreciated!