I'm trying to justify that $Gal(\mathbb{Q}(e^{2\pi i/6}):\mathbb{Q})$ is isomorphic to $C_2$, the cyclic group of order $2$. So far, I've easily shown that the minimal polynomial of $r=e^{2\pi i/6}\ $ is $\ p(x)=x^4+x^2+1$, which has distinct roots $r$ and $r^5$.
From this, one has that $Gal(\mathbb{Q}(r):\mathbb{Q})=\{\sigma_1,\sigma_2\}$, where $\sigma_1$ is the unique automorphism that maps $r$ to $r$ (in this case, the identity mapping), and $\sigma_2$ is the unique automorphism that maps $r$ to $r^5$. I'm very rusty in my group theory, so I'm not sure on how to proceed from this point onward. Any help or guidance would be great!
Best Answer
Once you have that the Galois group is of order $2$, you are basically done showing that it is isomorphic to the cyclic group of order $2$, because any group of order $2$ is isomorphic to $C_2$. (For one thing, any group of prime order is cyclic.)