In Dummit and Foote, section 14.5, p.597, he considers the generators of $\mathbb{Q}(\zeta_{13})$ which corresponds to the subgroups of $(\mathbb{Z}/13\mathbb{Z})^{\times}\cong \mathbb{Z}/12\mathbb{Z}$. A generator for this cyclic group is the automorphism $\sigma = \sigma_{2}$ which maps $\zeta_{13}$ to $\zeta_{13}^{2}$. The nontrivial subgroups correspond to the nontrivial divisor of 12, i.e. (2,3,4,6) with generators $\sigma^{6}, \sigma^{4}, \sigma^{3},\sigma^{2}$. Could somebody explain me why? I am missing something. Moreover, for the case $\mathbb{Q}(\zeta_{7})$, what are the nontrivial subgroups?
Thanks.
Nontrivial subgroups of $G(\mathbb{Q}[\zeta_n]/\mathbb{Q})$ are easy to find when $n$ is prime. This is because $G(\mathbb{Q}[\zeta_n]/\mathbb{Q}) \cong \mathbb{Z}_n^\times$, so when $n$ is prime, $\mathbb{Z}_p^\times \cong \mathbb{Z}_{p-1}$. (This fact can be proven using the Structure theorem for finitely generated abelian groups).
So $\mathbb{Q}[\zeta_7]$ has a Galois group isomorphic to $\mathbb{Z}_6$. We know that if a group is cyclic, then there will be exactly one subgroup of each order that divides the order of the group. Therefore, this particular Galois group will have subgroups of order $2$ and $3$.
To find the cyclic subgroups, I first try to find the generator of the group. The way I go about this is by noting that, for any $k$, there will be a legitimate $\mathbb{Q}$-automorphism that will send $\zeta$ to $\zeta^k$ (since $Gal(f)$ acts transitively on the roots of $f$ when $f$ is irreducible). Therefore, I simply experiment as follows:
Consider $\phi$ such that $\phi(\zeta) = \zeta^2$. How many times must we compose $\phi$ with itself until we wrap back around to $\zeta$? Now let's try $\phi(\zeta) = \zeta^3$. I continue in this manner, testing automorphisms until I have found the generator. After that, it is very easy to find the subgroups.