Galois group of $x^6-2$ over $\Bbb Q$ I want to find the Galois group of $x^6-2$ over $\Bbb Q$.

I have posted my attempt in an answer below. Is there a better way? Alternative proofs are greatly appreciated(especially shorter ones ;)$\quad$ ).
 A: In this answer I show that $x^6-2$ splits over $[\Bbb Q(\alpha,\zeta):\Bbb Q]=12$ where $\alpha$ is the real $6^{th}$ root of $2$ and $\zeta$ is a primitive $6^{th}$ root of unity.
Since $\Bbb Q(\alpha,\zeta)$ is the normal closure of $\Bbb Q(\alpha)$(as shown in that answer), it follows that $\Bbb Q(\alpha,\zeta)$ is the splitting field of $f(x)=x^6-2$, which is a separable polynomial and hence $\Bbb Q(\alpha,\zeta)/\Bbb Q$ is a Galois extension. Since it is Galois, $|G|=[\Bbb Q(\alpha,\zeta):\Bbb Q]=12$
Denote $L=\Bbb Q(\alpha,\zeta)$
We can fully determine elements of $\text{Gal}(L/\Bbb Q)$ by finding where $\zeta$ and $\alpha$ are mapped. Since $\zeta$ has minimal polynomial $x^2-x+1$, it can map $\zeta \mapsto \zeta$ or $\overline{\zeta}$
$\alpha$ has minimal polynomial $x^6-2$ and $\alpha\mapsto \zeta^j \alpha,j=0,\cdots,5$
So we have:
$$\begin{matrix}\text{Aut}&\text{Image of }\alpha& \text{Image of } \zeta\\ 1&\alpha&\zeta\\2&\alpha\zeta&\zeta\\3&\alpha\zeta^2&\zeta\\4&\alpha\zeta^3&\zeta\\5&\alpha\zeta^4&\zeta\\6&\alpha\zeta^5&\zeta\\7&\alpha&\overline\zeta\\8&\alpha\zeta&\overline\zeta\\9&\alpha\zeta^2&\overline\zeta\\10&\alpha\zeta^3&\overline\zeta\\11&\alpha\zeta^4&\overline\zeta\\12&\alpha\zeta^5&\overline\zeta\\\end{matrix}$$
Let Aut $1$, mean map $\sigma_1$.
So these are my $12$ maps. Now we check how these interact: $\sigma_1=\text{Id}$, $\sigma_2^2=\sigma_3$, $\sigma_2^3=\sigma_4$, $\sigma_2^4=\sigma_5$ and $\sigma_2^5=\sigma_6$
Here I realise this looks like it might be $D_6$.
$\sigma_2^6=\text{Id}$ and $\sigma_7^2=\text{Id}$ and $\sigma_2\sigma_7\sigma_2\sigma_7=\text{Id}$
It satisfies the three properties of the group presentation Not sure this means that it is infact $D_6$ yet
A: I was doing the same thing for a Galois Theory course today and I came up with the following solution:
Considering the notation you've already used we can check that $\sigma_2 \circ \sigma_7 \neq \sigma_7 \circ \sigma_2$ since the first sends $\alpha$ to $\alpha \zeta$ and the second sends $\alpha$ to $\alpha \overline{\zeta}$, therefore our group is not abelian. 
Now, you've already checked that it has a cyclic subgroup of order six, namely $\langle \sigma_2 \rangle$. Knowing that, we have only two other options (look at the classification of order 12 groups here https://en.wikipedia.org/wiki/List_of_small_groups). The options are
\begin{gather*}
Z_3 \ltimes Z_4 && D_6
\end{gather*}
the 2-sylow subgroups of the first are all $Z_4$ and the second has a subgroup of order 4 isomorphic to $Z_2 \times Z_2$, so the way to go here is find a subgroup of order 4 isomorphic to $Z_2 \times Z_2$. To do this it suffices to show that some subgroup of order 4 has 3 elements of order 2 ($Z_4$ has only 2 elements of order 2). I tried $\langle \sigma_4, \sigma_7 \rangle$ and it worked! It has $\sigma_4,\sigma_7$ and $\sigma_4 \circ \sigma_7$ as elements of order 2.
