Find all the middle fields of the splitting field of $x^4-2$ over $\mathbb{Q}$ The question:

Find all the middle fields of the splitting field of $x^4-2$ over $\mathbb{Q}$.

My attampt:
Let $E$ be the splitting field of $f(x):=x^4-2$ over $F:=\mathbb{Q}$.
The roots of $f$ are $\pm\alpha,\pm\alpha i$ where $\alpha =\sqrt[4]{2}$. I know that
$$
G:=\text{Aut}(E/F)=\{\sigma^i\tau^j:0\leq i\leq 1,0\leq j\leq3\}
$$
where $$  \sigma: \alpha \mapsto i\alpha\\ \tau: i\mapsto -i$$
And because $\sigma\tau\sigma=\tau$, we get $G\cong D_8$.
The subgroups of $G$ are
$$\begin{align}
&H_1=\{1,\sigma^2\},&&H_2=\{1,\tau\}\\&H_3=\{1,\sigma^2\tau\},&&H_4=\{1,\sigma\tau\}\\&H_5=\{1,\sigma^3\tau\},&& H_6=\{1,\tau,\sigma^2,\sigma^2\tau\}\\&H_7=\{1,\sigma\tau,\sigma^2,\sigma^3\tau\},&&H_8=\{1,\sigma,\sigma^2,\sigma^3\}
\end{align}$$
The required middle fields are $K_i=E^{H_i}$.
Observe that $K_1=\mathbb{Q}(i)$ and $K_2=\mathbb{Q}(\alpha)$, but I don't think I calculate $K_i$ correctly. For example:
I want to calculate $K_3$: Lex $K_3\ni x=a+\alpha b+ic$. We know $\sigma^2\tau(x)=x$ then $$a=b\alpha i\\ \Rightarrow x=b(\alpha(1+i))+ci$$ I don't know how to calculate $K_3$ from here.
Another example: Let $K_4\ni x=a+\alpha b+ci$. Then $$x=\sigma\tau(x)\\ \Rightarrow x=a+b\alpha-b({i-\alpha\over 1+i})\\ =a+{1\over 2}b(5\alpha-\alpha i-i-1)$$.
I'm not sure how to proceed and I think that's not the way to solve the question. Thanks in advence.
 A: It seems like you have listed all the subgroups. Now all you need to do is find the corresponding fixed fields using the fundamental theorem of galois theory. Let's start with the subgroups of order 4 (index 2), which will give rise to intermediate field extensions of degree 2. You could calculate them by hand, but in this case it's hardly necessary, as you have 3 obvious subfields of degree 2, namely $\mathbb{Q}(\sqrt{2}), \mathbb{Q}(\sqrt{2}i), \mathbb{Q}(i)$. This takes care of all subgroups of order 4.
You now have to take care of all subgroups of order 2, which correspond to subfields of degree 4.
Again, we have 3 obvious ones $\mathbb{Q}(\alpha), \mathbb{Q}(\alpha i), \mathbb{Q}(i,\sqrt{2})$.
So there's 2 subfields of order 4, which are not that obvious. 
Looking at the subgroup generated by $\tau\sigma$, this is a subgroup of order 2, and therefore the corresponding subfield will be of degree 4. To be precise, $\alpha - \alpha i$ is fixed by the action of the generator of this subgroup, as $\tau\sigma(\alpha-\alpha i$)=$-\alpha i+\alpha$. Therefore $\mathbb{Q(\alpha-\alpha i)}$ is the required fixed field.
Finally, looking at the subgroups which sends $\alpha$ to $\alpha i$, (I believe that is the subgroup generated by $\tau \sigma^3$), then $\alpha+\alpha i$ is fixed by the action of this group, and thus $\mathbb{Q(\alpha+\alpha i)}$ is the required subfield of degree 4.
