Explicit computation of a Galois group

Solution 1:

By definition of the splitting field, it is generated by the roots of $x^6 - 2$. These roots form the vertices of a regular hexagon on the complex plane. Any element of the Galois group must permute the roots, and is completely determined by what it does to the roots, since the splitting group is generated by them.

Let $\sigma$ be an element of the Galois group. Now consider what happens to two adjacent vertices, $\zeta_6^i\sqrt[6]{2}$ and $\zeta_6^{i+1}\sqrt[6]{2}$. By considering $$\frac{\zeta_6^{i+1}\sqrt[6]{2}}{\zeta_6^{i}\sqrt[6]{2}} = \zeta_6,$$ the action of $\sigma$ on the two adjacent vertices determines the action on $\zeta_6$. Since $\zeta_6$ satisfies $\Phi_6(x) = x^2-x+1$, the image of $\zeta_6$ must be another root of $x^2-x+1$, which is either $\zeta_6$ or $\zeta_6^5$, as you note. And if you know what happens to $\zeta_6^i\sqrt[6]{2}$ and to $\zeta_6$, then you know what happens to $\sqrt[6]{2}$ as well. And of course, if you know what happens to both of these, then you know what happens to the adjacent vertices $\sqrt[6]{2}$ and $\zeta_6\sqrt[6]{2}$. So knowing the action on at least two adjacent vertices is equivalent to knowing the action on $\sqrt[6]{2}$ and on $\zeta_6$, and vice versa, and this in turn tells you what is happening on all the vertices. This shows that what you are saying is correct.

Now, to finish off, try showing that two adjacent vertices are always mapped to two adjacent vertices, so that any permutation of the hexagon induced by $\sigma\in\mathrm{Gal}(E/\mathbb{Q})$ is actually a rigid motion of the hexagon. This will show the Galois group is contained in the dihedral group, and the computation of the order that you have already done will finish off the problem.