Prove that the Galois group of $x^{4}-2x^{2}-2$ over $\mathbb{Q}$ is $D_{8}$.
Note that the order of $D_{8}$ is $8$.
How would I find the eight automorphisms of this group? Or can I just use the order of $G(x^{4}-2x^{2}-2/\mathbb{Q})$ to show that its order is $8$?
If I show the order of the Galois group is $8$, is it sufficent enough to show that, since it's nonabelian and order $8$, it's isomorphic to $D_{8}$?
First prove that $x^4-2x^2-2$ is irreducible over $\Bbb{Q}$. Then its Galois group is isomorphic to a transitive subgroup of $S_4$. That means it is isomorphic to either $\Bbb{Z}/4\Bbb{Z}$, $(\Bbb{Z}/2\Bbb{Z})^2$, $D_8$, $A_4$ or $S_4$.
If you are then able to show that the order of the Galois group is $8$, then it must be isomorphic to $D_8$.
Slightly more in line with your approach, you could note that up to isomorphism there are precisely two non-abelian groups of order $8$. They are $D_8$ and $Q_8$, the quaternion group of order $8$. But the latter is not isomorphic to a subgroup of $S_4$, so it cannot be the Galois group of $x^4-2x^2-2$.