Why we can represent automorphisms in $\text{Gal}(\Bbb Q(\sqrt[4]{2},i)/\Bbb Q)$ as permutations in $S_4$ but not $S_8$? The splitting field of $x^4-2$ over $\Bbb Q$ is $G=\text{Gal}(\Bbb Q(\sqrt[4]{2},i)/\Bbb Q)$. By primitive element theorem, $K=\Bbb Q(\alpha)$ for some $\alpha$ and $[K:\Bbb Q]=8$. So I know that the minimal polynomial of $\alpha$ over $\Bbb Q$ has degree $8$. And at the same time there are eight automorphisms in $G$.
However, one thing I don't understand is that some authors labeled the roots $\sqrt[4]{2},~\sqrt[4]{2}i,~-\sqrt[4]{2},~-\sqrt[4]{2}i$ of $x^4-2$ as $1,~2,~3,~4$, and say that every automorphisms in $G$ can be represented as a permutation. For example, $\sigma_1=(1~~2~~3~~4)$. How can we do this? Why we can only analyze how the roots $\sqrt[4]{2},~\sqrt[4]{2}i,~-\sqrt[4]{2},~-\sqrt[4]{2}i$ is affected by the automorphism, and then we can decide it? Maybe there are different $\sigma,~\sigma'\in G$ such that $\forall r\in\{\sqrt[4]{2},~\sqrt[4]{2}i,~-\sqrt[4]{2},~-\sqrt[4]{2}i\},~\sigma(r)=\sigma'(r)$.
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Remember that if $F$ is the ground field, and $p(x)$ is an irreducible polynomial over $F$ of degree $n\geq 1$ without repeated roots, let $K$ be the splitting field of $p(x)$. $K/F$ is Galois. You have that the Galois group of $K/F$, say $G$, it is embedded on $S_n$, this is because $G$ permutes the roots.
In your problem, $F=\mathbb{Q}$ and $K$ is the splitting field of $p(x)=x^4-2$, since its roots are $\sqrt[4]{2}$, $-\sqrt[4]{2}$, $i\sqrt[4]{2}$, $-i\sqrt[4]{2}$.
For a better understanding. The Galois group acts transitively on the roots of that polynomial (it is important the irreducibility). If you are not clear about the Galois group yet, you can take intermediate fields, study the irredubile polynomial on the tower of fields and you will know that the Galois Group in some extension of the tower of fields acts transitively on the root and fix the elements on the base field. For example, in this case, if you want to use that method, consider the tower of fields $\mathbb{Q}\left(\sqrt[4]{2},i\right)\supset\mathbb{Q}\left(\sqrt[4]{2}\right)\supset \mathbb{Q}$, since $\mathbb{Q}\left(\sqrt[4]{2},i\right)/\mathbb{Q}\left(\sqrt[4]{2}\right)$ is Galois (splitting field of $x^2+1$), you know now that theres is a $\sigma$ in $\mbox{Gal}\left(\mathbb{Q}\left(\sqrt[4]{2},i\right)/\mathbb{Q}\left(\sqrt[4]{2}\right)\right)$ such that send $i\mapsto -i$ (acts on the roots of $x^2+1$) and leave $\mathbb{Q}\left(\sqrt[4]{2}\right)$ fixed, but the Galois group of that extension if a subgroup of the Galois group of $K/\mathbb{Q}$, so $\sigma$ lie to $\mbox{Gal}\left(K/\mathbb{Q}\right)$
