Consider $x^{4}-2=(x+\sqrt[4]{2})(x-\sqrt[4]{2})(x+i\sqrt[4]{2})(x-i\sqrt[4]{2}) \in \mathbb{Q}[x]$. Let $K=\mathbb{Q}(\sqrt[4]{2},i)$ be the splitting field of $x^{4}-2$. Since $K$ is a splitting field and we are in characteristic 0, it follows that $K/\mathbb{Q}$ is Galois. Finally, $[K\colon\mathbb{Q}]=8$.
I want to compute the Galois group $K/\mathbb{Q}$. Since the extension is Galois, there are 8 elements in this group. It turns out that this group is isomorphic to the dihedral group of order 8 (I've seen examples of this but don't have a reference).
What steps would I take to reach the conclusion that this group is the dihedral group?
Specifically I know that each automorphism of the Galois group permutes the roots, but I don't see how to make the connection that these elements are the same as the dihedral group of order 8. I would appreciate a detailed analysis because I think this would allow me to apply these techniques to many other problems in Galois theory. Thanks.