Find Galois group and all intermediate fields of the extension $L=\mathbb{Q}(\sqrt[4]{2},i)\supseteq \mathbb{Q}=K$

Find Galois group, all subgroups and the corresponding subfields of the following extension :

$$L=\mathbb{Q}(\sqrt[4]{2},i)\supseteq \mathbb{Q}=K$$

1) I found the degree of the extension
$$[L:K]=8$$ Thus $L$ is a splitting field for $$x^4-2=(x-\sqrt[4]{2})(x+\sqrt[4]{2})(x-i\sqrt[4]{2})(x+i\sqrt[4]{2})$$ and as $L\supseteq K$ is Galois then $8=[L:K]=|Gal(L:K)|$ So the group is cyclic and will have 8 automorphisms $Gal(L:K))=\{\sigma_{0},\sigma_{1},\sigma_{2},\sigma_{3},\sigma_{4},\sigma_{5},\sigma_{6},\sigma_{7}\}$

Then any $\sigma\in Gal(L:K)$ is determined by $\sigma(\sqrt[4]{2})$ and $\sigma(i)$

So I have a table of possible root permutations with orders of elements from $Gal(L:K)$

$\begin{array}{|c|c|c|c|} \hline & \sqrt[4]{2} & i& o \\ \hline \sigma_{0}& \sqrt[4]{2} & i & 1 \\ \hline \sigma_{1} & -\sqrt[4]{2} & i & 2 \\ \hline \sigma_{2} & i\sqrt[4]{2} & i & 4 \\ \hline \sigma_{3} &-i\sqrt[4]{2} & i & 4 \\ \hline \sigma_{4} &\sqrt[4]{2} &-i & 2 \\ \hline \sigma_{5} & -\sqrt[4]{2} &-i & 2 \\ \hline \sigma_{6} & i\sqrt[4]{2} & -i & 2 \\ \hline \sigma_{7}& -i\sqrt[4]{2} & -i & 2 \\ \hline \end{array}$

By the above table $Gal(L:K)$ will have subgroups of orders that are divisors of 8 i.e $1,2,4,8$

It is easy to find subgroups but I have problems with identifying intermediate fields that are associated with them.

1) trivial subgroup $I=\{\sigma_{0}\}$ will correspond to $L^{I}=L$

2) $H_{1}=\{\sigma_{0},\sigma_{1}\}$

so by def $$L^{H_{1}}=\{l\in L,\;\;\sigma(l)=l,\;\;\forall \sigma\in H_{1}\}$$

The first automorphism (identity) $\sigma_{0}$ fixes both $\sqrt[4]{2}$ and $i$ but $\sigma_{2}$ fixes only $i$ this means that $i\in L^{H_{1}}$ right?

In the solution I found something like this

''If $H=\{l,\sigma\}$ then $l+\sigma(l)$ and $l\sigma(l)$ are both in $L^{H}$'' The first interediate field is obtained by $$\sqrt[4]{2}\sigma_{1}(\sqrt[4]{2})=-\sqrt{2}\in L^{H_{1}}=L^{\sigma_{1}}$$

Why the norm is taken, and why $\sqrt[4]{2}$ and not $i$ instead? The resulting intermediate field contains both $\sqrt{2}$ and $i$ and equals $L^{H_{1}}=\mathbb{Q}(\sqrt{2},i)$

Going on to next subgroup

$H_{2}=\{\sigma_{0},\sigma_{4}\}$ and $L^{H_{2}}=L^{\sigma_{4}}=\mathbb{Q}(\sqrt[4]{2})$

This is okay as $H_{2}$ fixes only $(\sqrt[4]{2}$

Also

$H_{3}=\{\sigma_{0},\sigma_{5}\}$ and $L^{H_{3}}=L^{\sigma_{5}}=\mathbb{Q}(i\sqrt[4]{2})$

since $\sigma_{5}$ fixes the product $i\sqrt[4]{2}$ which means $i\sqrt[4]{2}\in L^{H_{3}}=\mathbb{Q}(i\sqrt[4]{2})$

$\sigma_{5}(i\sqrt[4]{2})=\sigma_{5}(i)\sigma_{5}(\sqrt[4]{2})=(-i)(-\sqrt[4]{2}))$

In case of $H_{4}=\{\sigma_{0},\sigma_{6}\}$ How to find the root that will be fixed? The solution given is as follows: $\sqrt[4]{2}+\sigma_{6}(\sqrt[4]{2})=(1+i)\sqrt[4]{2}\in L^{\sigma_{6}}$ And gain trace is taken, why

Can someone please enlighten me? When do we use trace and when do we use norm?