Galois group of $x^8+2$ over the rationals? As to check whether I got the theory right I tried figuring out $ G=\Gamma(L:\mathbb{Q}) $ where $L$ is the splitting field of $f=x^8+2$ over the rationals. I then considered the intermediate field $\mathbb{Q}(\omega)$ where $\omega$ is a primitive 8th root of unity because if $\alpha$ is a root of $f$ then the other roots are $\omega^k \alpha$.
I found out $ H=\Gamma (\mathbb{Q}(\omega):\mathbb{Q}) = C_2\times C_2$ and that $N=\Gamma(L:\mathbb{Q}(\omega)) = C_4. $
This should imply that there are three possibilities for $G$ which are $ C_2\times C_2\times C_4;\;C_2\times C_8;\;C_2\times D_4 $.
Is it correct and which is the fastest method to decide between those?
 A: Let $\zeta$ be a primitive 8th root of unity.  Then the splitting field $L$ of $X^8+2$ is
$$L = \mathbb{Q}(\zeta,\sqrt[8]{-2}).$$
Let $\sigma \in G$ be an automorphism of $L$ fixing $\mathbb{Q}$.  How must $\sigma$ act on the generators $\zeta$ and $\sqrt[8]{-2}$ ?
We must have $\sigma(\zeta) = \zeta^a$ with $a \in \{1,3,5,7\}$ and $\sigma(\sqrt[8]{-2}) = \zeta^b \sqrt[8]{-2}$ with $b \in \{0,1,2,3,4,5,6,7\}$.
However, we must also satisfy a consistency condition that restricts our choice of $a$ and $b$, since $(\sqrt[8]{-2})^4 = \sqrt{-2} = \zeta + \zeta^3$.  Applying $\sigma$ to this relation gives the condition
$$\zeta^{4b} \sqrt{-2} = \zeta^a + \zeta^{3a}.$$
We find that if $a \in \{1,3\}$ then $b \in \{0,2,4,6\}$, and if $a \in \{5,7\}$ then $b \in \{1,3,5,7\}$.
Now let $\sigma_1$ be the automorphism with $\sigma_1(\zeta)=\zeta^3$ and $\sigma_1(\sqrt[8]{-2}) = \sqrt[8]{-2}.$  Also, let $\sigma_2$ be the automorphism with $\sigma_2(\zeta) = \zeta$ and $\sigma_2(\sqrt[8]{-2}) = \zeta^2 \sqrt[8]{-2}$.
Then
$$(\sigma_2 \circ \sigma_1)(\sqrt[8]{-2}) = \sigma_2(\sqrt[8]{-2}) = \zeta^2 \sqrt[8]{-2}$$
but
$$(\sigma_1 \circ \sigma_2)(\sqrt[8]{-2}) = \sigma_1(\zeta^2 \sqrt[8]{-2}) = \zeta^6 \sqrt[8]{-2} = -\zeta^2 \sqrt[8]{-2}.$$
Thus $\sigma_2 \circ \sigma_1 \neq \sigma_1 \circ \sigma_2$.  Thus, $G$ is nonabelian.
There are 9 different nonabelian groups of order 16 (see http://en.wikipedia.org/wiki/List_of_small_groups#List_of_small_non-abelian_groups).
To narrow down further, we can use Dedekind's theorem.  Since $X^8 + 2$ is irreducible modulo $5$, $G$ contains a permutation of the roots of the form $(12345678)$, thus $G$ contains a cyclic subgroup of order 8.
Now we can play with automorphisms to find an element of $G$ of order 8.  In fact, let $\alpha \in G$ be such that $\alpha(\zeta) = \zeta^5$ and $\alpha(\sqrt[8]{-2}) = \zeta \sqrt[8]{-2}$.  Then $\alpha$ has order 8 in $G$.
Also, let $\beta \in G$ be the automorphism with $\beta(\zeta)=\zeta$ and $\beta(\sqrt[8]{-2}) = \zeta^4 \sqrt[8]{-2}$.  Since $\beta$ is not in the subgroup generated by $\alpha$, we have that $\alpha$ and $\beta$ generate $G$.  Furthermore, we have the relations $\alpha^8 = 1$, $\beta^2 = 1$ and $(\alpha\beta)^2 = 1$.  So $G$ must be a quotient of $D_8$.  But since the order of $G$ is 16, we have that $G = D_8$, the dihedral group of order 16.
