Set $f(X) = X^4 − 6X^2 − 2$ and denote by $K$ the splitting field of $f$ over $\mathbb{Q}$.
(a) Find complex numbers $α$ and $β$ such that the roots of f are $±α$ and $±β$, and show that $K = \mathbb{Q}(α, αβ)$.
(b) Show that the Galois group $Γ(K : \mathbb{Q})$ is a non-abelian group of order 8.
(c) List the subgroups of $Γ(K : \mathbb{Q})$
I'm quite stuck on this problem. I've done part (a) already and so I have that four roots where $±α$ is equal to $\pm\sqrt{3+\sqrt11}$ and $±β$ is equal to $\pm i\sqrt{-3+\sqrt11}$. The splitting field I have is then $K=\mathbb{Q}(\alpha,i\sqrt2)$.
I'm now not sure how to start (b) as while i've seen the question layout for $X^4 -2$ I'm confused whether the addition of the $-6X^2$ changes anything. For (b) I was thinking of stating that:
Any Galois automorphism of K : Q sends $i$ to $±i$ and $α$ to $±α$ or $±\beta$. This yields eight combinations. Thus $Γ(K : \mathbb{Q})$ has at most eight elements. Since $\mathbb{Q}(α, i\sqrt2) : \mathbb{Q}(α)$ is of degree 2 and $\mathbb{Q}(α) : \mathbb{Q}$ is of degree 4, we get that $[K: \mathbb{Q}] = 8$. Then we can see that because the order is 8 it is a Sylow-2 Subgroup and thus is isomorphic to the dihedral group $D_{8}$ which is known to be non-abelian. Is that enough for (b) or am I going about this worng?
I'm not really sure what else to do and I'm completely lost for (c).
