Galois correspondence of subgroups of $D_4$ with subfields of $\mathbb Q (\sqrt[4]{2},i)$

Question

The Galois group of $\mathbb Q (\sqrt[4]{2},i)$ over $\mathbb Q$ is the Dihedral group $D_4$ = {$id, \sigma, \sigma^2, \sigma^3, \tau, \sigma\tau, \sigma^2\tau, \sigma^3\tau $}

Denoting $\sqrt[4]{2}$ as $\theta$, the action of the elements are $\sigma$(i) = i, $\sigma(\theta)$ = i $\theta$, $\tau$(i) = $-$i, $\tau(\theta)$ = $\theta$

Then the following are the subgroups with corresponding fixed fields that I have been able to conclude are correctly associated:

1. $H_0$ = {id} with $\mathbb Q (\theta,i)$
2. $H_8$ = $D_4$ with $\mathbb Q$
3. $H_1$ = {id, $\tau$} with $\mathbb Q(\theta)$
4. $H_5$ = {id,$\tau, \sigma^2, \sigma^2\tau$ } with $\mathbb Q(\theta^2)$
5. $H_7$ = {id, $\sigma, \sigma^2, \sigma^3$} with $\mathbb Q(i)$

Further I think these two are also correctly associated:

6. $H_6$ = {id, $ \sigma\tau, \sigma^2, \sigma^3\tau $ } with $\mathbb Q(i\theta^2)$
7. $H_2$ = {id, $\sigma^2\tau $} with $\mathbb Q(i\theta)$

Are these two also correct?

Assuming the above are correct, it still leaves me to find the corresponding fixed fields of these two subgroups:

8. $H_3$ = {id, $\sigma\tau$}
9. $H_4$ = {id, $\sigma^3\tau$}

what will be the corresponding fixed fields?

I thought the two missing subfields are $\mathbb Q(\theta^3)$ and $\mathbb Q(i\theta^3)$ but they don't seem to be fixed under $H_3$ or $H_4$

Answer 1

As $\Bbb Q(\theta^3)=\Bbb Q(\theta)$, you already have that, as the fixed field of $H_1$.

To find elements fixed by $H_3$ let's look at elements of the form $a+\sigma\tau(a)$, which will automatically be in the fixed field. We compute $$\sigma\tau(\theta)=\sigma(\tau(\theta))=\sigma(\theta)=i\theta.$$ Therefore $$\beta=\theta+i\theta=(1+i)\theta$$ lies in the fixed field $K_3$ of $H_3$. We find $$\beta^2=2i\theta^2$$ and $$\beta^4=-8.$$ Then $\beta$ has degree $4$ over $\Bbb Q$ and $K_3=\Bbb Q(\beta)=\Bbb Q((1+i)\theta)$.

The $H_4$ case is very similar.