Finding the number of automorphisms in $G_{\Bbb{Q}(\sqrt[4]{2},i),\Bbb{Q}}$? I am trying to solve the following problem:

Given $K$, the splitting field of $x^4-2$ over $\Bbb{Q}$, consider $\sigma,\tau\in G_{K / \Bbb{Q}}$ such that
$$\sigma(\sqrt[4]{2})=\sqrt[4]{2}i\hspace{2cm} \sigma(i)=i\hspace{2cm}\tau(\sqrt[4]{2})=\sqrt[4]{2} \hspace{2cm}\tau(i)=-i$$
Show that $|\sigma|=4$ and $|\tau|=2$.

I think we must count all the $\Bbb{Q}$-automorphisms $\sigma, \tau$ with the given conditions, is that correct? I tried to write the expression for the elements in $K$:
$$a+b\sqrt[4]{2}+c\sqrt[4]{2}^2+d\sqrt[4]{2}^3+ei+fi\sqrt[4]{2}+gi\sqrt[4]{2}^2+hi\sqrt[4]{2}^3$$
And check for  the automorphisms but this looks a bit unwieldy. Is there a more practical way to do it? I suspect there is because the $\sigma$ and $\tau$ looks a bit specific and I have found before that $K=\Bbb{Q}(\sqrt[4]{2},i)$ and $2,4$ are the degrees of the extensions. I have also seen other similar questions to mine but they don't seem to address exactly what I am looking for.
 A: The question gives you two specific automorphisms $\sigma, \tau\in\textrm{Gal} (K/\mathbb{Q}) $ and asks you to verify their order (as members of the Galois group).
Since you already know that $K=\mathbb {Q} (\sqrt[4]{2},i)$, observe that any automorphism in $\textrm{Gal} (K /\mathbb {Q}) $ is determined / specified by its action on just these two members $i, \sqrt [4]{2}$. Whenever the field is specified like $K=F(a_1,a_2,\dots,a_n)$ you don't need to check all members of $K$ (as you have tried in question by expressing in terms of a basis) for guessing any automorphism. It is just sufficient to check for the members $a_i$.
You are trying to interpret the question as if we need to find (and count) automorphisms which have properties like $\sigma, \tau$. No! Those properties determine $\sigma, \tau $ uniquely as explained in previous paragraph.
Finding order of $\sigma, \tau$ is rather straightforward. Just keep on applying $\sigma$ (similarly for $\tau $) to both $i, \sqrt[4]{2}$ and see when you get the identity map.
Thus $$\sigma^2(\sqrt[4]{2})=\sigma(\sigma(\sqrt[4]{2}))=\sigma(i\sqrt [4]{2})=\sigma(i)\sigma(\sqrt[4]{2})=i^2\sqrt[4]{2}$$ and in general $\sigma^n(\sqrt [4]{2})=i^n\sqrt[4]{2}$. Further $\sigma^n(i) =i$. Thus we see that $\sigma^4$ is the identity map so that $|\sigma|=4$ and in a like manner you can show that $|\tau|=2$.
A: Letting $K=\mathbf Q(i,\sqrt [4] 2)=$ your splitting field and $k=\mathbf Q (i)$, it's easier to determine the whole Galois group $G=G(K/\mathbf Q)$ without using brute force. Noting that $i$ is a primitive 4th root of unity, Kummer theory tells you that $G(K/k)$ is cyclic of order 4, generated by your $\sigma$ (even if you don't know about Kummer, it's easy here to keep track of the action of $\sigma$ and use the bound $[K:k]\le 4$ to conclude). In the same vein, introduce the subextension $k'=\mathbf Q (\sqrt [4] 2)$. Because $k'$ is totally real and $i$ is imaginary, it's immediate that $G(K/k')$ is cyclic of order 2, generated by $\tau$. By the same token, you get the bonus that $G$ is generated by $\sigma$ and $\tau$. Subsidiary question: is $G$ isomorphic to $D_8$ or $H_8$ ?
