$\zeta:=\zeta_8$ is of degree $2$ over $\mathbb{Q}(\sqrt[8]{2})$, hence $K:=\mathbb{Q}(\sqrt[8]{2},\zeta)$ is of degree $16$ over $\mathbb{Q}$ and of degree $4$ over $\mathbb{Q}(\zeta)$. The minimal polynomial of $\sqrt[8]{2}$ over $\mathbb{Q}(\zeta)$ is $X^4-\sqrt{2} = X^4 - (\zeta + \zeta^{-1})$.
Hence $Gal(K/\mathbb{Q})$ is an extension of $(\mathbb Z_8)^\times$ by $\mathbb Z_4$.
If $\sigma \in Gal(K/\mathbb{Q})$, then $\sigma$ satisfies: $\sigma(\zeta) = \zeta^a$ and $\sigma(\sqrt[8]{2})=\zeta^b \sqrt[8]{2}$ for some $a \in \mathbb Z_8^\times$, $b \in \mathbb Z_8$ such that $\zeta^{4b} \sqrt{2}= \zeta^a + \zeta^{-a}$ which means $b = \tfrac{a-1}{2} \pmod 2$.
EDIT: Note that $\zeta_8 = \exp(2i\pi/8) = \exp(2i\pi/8)$. We know that take any value in $\{ 1,3,5,7 \} =\mathbb Z_8^\times$, and we have :
- $\exp(1 \times i \pi/4) + \exp(-1.i \pi/4) =\sqrt{2}$.
- $\exp(7 \times i \pi/4) + \exp(-7.i \pi/4) =\sqrt{2}$.
- $\exp(3 \times i \pi/4) + \exp(-3.i \pi/4) =-\sqrt{2}$.
- $\exp(5 \times i \pi/4) + \exp(-5.i \pi/4) =-\sqrt{2}$.
Depending on $a$, the value $(\zeta^b)^4$ must be $+1$ or $-1$. In any case they are exactly $4$ values of $b$ allowed.
Here it might be best to realize the Galois group as a group of permutations of the roots of the polynomial of interest (as in user64494's comment under the OP). You have already observed that the roots of $x^p-2$ are $x_i=\zeta_p^{i-1}\root p\of 2, i=1,2,\ldots,p.$ You also know that the splitting field is of degree $p(p-1)$, so that is also the order of the Galois group.
Let us consider the action of your automorphism $\tau$ defined by $\tau(\zeta_p)=\zeta_p$ and $\tau(\root p\of 2)=\zeta_p\root p\of 2$. So we see that
$\tau(x_i)=x_{i+1}$, if $i<p$, and $\tau(x_p)=x_1$. The action of $\tau$ on the chosen indexing of the roots thus corresponds to the $p$-cycle $\tau=(123\cdots p).$
On the other hand the automorphism $\sigma_a:x_1\mapsto x_1, \zeta_p\mapsto \zeta_p^a,1\le a<p,$ keeps the real root $x_1$ fixed, and permutes the others according to the rule $x_i=x_1\zeta_p^{i-1}\mapsto x_1\zeta_p^{a(i-1)}=x_{1+a(i-1)}$, where the subscript is calculated modulo $p$. You see that all these share $x_1$ as a fixed point (this was also clear from your construction of $\sigma$:s as elements of the Galois group $G_2=Gal(\mathbb{Q}(x_1,\zeta_p)/\mathbb{Q}(\zeta_p))$.
You can either look at all these automorphisms as elements of $S_p$. This works beautifully, once you have found a generator of $G_2$. This is equivalent to finding a generator of the multiplicative group $\mathbb{Z}_p^*$, i.e. a primitive root. There is no general formula for such a generator, so I won't say much about that (this may be a cause of your difficulties). We simply know that one exists! But if $p$ is fixed, say $p=5$ or another smallish prime, then I recommend this way, as you can easily calculate with permutations.
PVAL is strongly hinting at the possibility that you may get a semi-direct product of $G_1$ and $G_2$. Indeed, you will see that one of the two subgroups is stable under conjugation by elements of the other. Which way does it work? I'm a mean dude and won't tell you! But you do remember that the fixed field of a normal subgroup is itself Galois over the base field. So which of the fields $\mathbb{Q}(\zeta_p)$ or $\mathbb{Q}(\root p\of2)$ is Galois over the rationals?
The group of automorphisms associated to that field should be a normal subgroup of the big Galois group. After figuring that out, you can start studying the effect of either $\sigma_a\tau\sigma_a^{-1}$ or $\tau\sigma_a\tau^{-1}$ all according to which feels more interesting...
Best Answer
Well note that first we need to extend into $\mathbb{Q}(\sqrt{3})$ so that we write $$ x^4-3 = (x^2 - \sqrt{3})(x^2 + \sqrt{3}) $$ and then we need to extend into $\mathbb{Q}(\sqrt{3},\sqrt[4]{3}) = \mathbb{Q}(\sqrt[4]{3})$ since $$ (x^2 - \sqrt{3})(x^2 + \sqrt{3}) = (x - \sqrt[4]{3})(x + \sqrt[4]{3})(x^2 + \sqrt{3}) $$ Now if for the sake of seeing where to extend to next, let's work in $\mathbb{C}$ and use the quadratic formula to find the linear factors of $x^2 + \sqrt{3}$: $$ x_0 = \frac{\pm\sqrt{-4\sqrt{3}}}{2} = \pm i \sqrt[4]{3} $$ so clearly we need to extend into $\mathbb{Q}(\sqrt[4]{3},i)$.
Now note that this is a separable polynomial (since in it's splitting field there are no repeated roots). This tells us that, denoting $E$ the extension field and $F$ the base field we have $$ \left| \mathrm{Gal}(E/F) \right| = [E : F] $$ Since for this polynomial $E = \mathbb{Q}(\sqrt[4]{3},i), F = \mathbb{Q}$ and we have $$ [\mathbb{Q}(\sqrt[4]{3}) : \mathbb{Q}] = 4, [\mathbb{Q}(\sqrt[4]{3},i) : \mathbb{Q}(\sqrt[4]{3})] = 2 \implies [\mathbb{Q}(\sqrt[4]{3},i) : \mathbb{Q}] = 8 $$ Now an explanation for these above equalities would be needed for a valid proof, I'll leave that to you, but a hint would be to think of the minimal polynomials of the elements that are being adjoined.
So now we know that the size of $\mathrm{Gal}(\mathbb{Q}(\sqrt[4]{3},i) / \mathbb{Q})$ is $8$.
Now for the automorphisms, note that $i \to \pm i$ so you automatically have two automorphisms. Now we have that $\sqrt[4]{3}$ can be sent to any other fourth root of $3$. So since there are four choices for our roots, and two choices for $i$ we have our $8$ automorphisms. Note that defining the automorphisms for just $i$ and $\sqrt[4]{3}$ are sufficient for defining the entire automorphisms (why?). These automorphisms are as follow:
$$ \begin{pmatrix} 3^{1 / 4} \to 3^{1 / 4} \\ i \to i \end{pmatrix} \begin{pmatrix} 3^{1 / 4} \to i3^{1 / 4} \\ i \to i \end{pmatrix} \begin{pmatrix} 3^{1 / 4} \to -3^{1 / 4} \\ i \to i \end{pmatrix} \begin{pmatrix} 3^{1 / 4} \to -i3^{1 / 4} \\ i \to i \end{pmatrix} \\ \begin{pmatrix} 3^{1 / 4} \to 3^{1 / 4} \\ i \to -i \end{pmatrix} \begin{pmatrix} 3^{1 / 4} \to i3^{1 / 4} \\ i \to -i \end{pmatrix} \begin{pmatrix} 3^{1 / 4} \to -3^{1 / 4} \\ i \to -i \end{pmatrix} \begin{pmatrix} 3^{1 / 4} \to -i3^{1 / 4} \\ i \to -i \end{pmatrix} \\ $$
I hope this helps you understand how to determine the group. I usually find the size of the group to see how many elements I need to look for. Next I would think about where the adjoined elements can go. From there we can create any possible automorphism.
Also another comment: If you put $s = \begin{pmatrix} \sqrt[4]{3} \to i\sqrt[4]{3} \\ i \to -i \end{pmatrix}, r = \begin{pmatrix} \sqrt[4]{3} \to i\sqrt[4]{3} \\ i \to i \end{pmatrix}$ we see that this group is isomorphic to $D_4$!