Finding the Galois group over $\Bbb{Q}$. If K is the splitting field of $X^8-2$ over $\Bbb{Q}$, I want to find the galois group. 
We know that $K=\Bbb{Q}(2^{1/8}, \zeta_8)$.
So first I want to look at $Gal(\Bbb{Q}(\zeta_8)/\Bbb{Q})$ and then look at $Gal(\Bbb{Q}(2^{1/8})/\Bbb{Q})$, since there is a homomorphism $\rho: Gal(\Bbb{Q}(2^{1/8},\zeta_8)/\Bbb{Q}) \rightarrow Gal(\Bbb{Q}(\zeta_8)/\Bbb{Q})$. 
Since $Gal(\Bbb{Q}(\zeta_8)/\Bbb{Q}) \cong Aut(<\zeta_8>) \cong Z^{\times}_8 = \{1, 3, 5, 7\}$, we know that we have four subgroups in $Gal(Q(\zeta_8)/\Bbb{Q})$ (Let $\zeta_8 = \zeta$):
$\sigma_1(\zeta) = \zeta$
$\sigma_3(\zeta) = \zeta^3$
$\sigma_5(\zeta) = \zeta^5$
$\sigma_7(\zeta) = \zeta^7$
And now I have to relate this to the homomorphism $\rho$ in order to find the rest of the permutations, right? But I'm a bit confused. I've spent hours trying to do it and I'm seriously stuck...could anybody help me with this? 
Thanks in advance 
 A: $\zeta:=\zeta_8$ 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.
