Regarding the unique subfields of a cyclotomic field extension What are the unique subfields $L$ of $\mathbb{Q}(\zeta_{13})$ such that $[L:\mathbb{Q}]=2, 3, 4$, and $6$? I get that the Euler's totient function of $13$ is $12$ and that must have some relation, as its factors are listed above. Let's take the first case, $[L:\mathbb{Q}]=2$ and I'll show you why I'm confused.
I would assume for $[L:\mathbb{Q}]=2$ we have $\mathbb{Q}$-basis $\{e,\sigma\}$ where $\sigma: \zeta^x\mapsto\zeta^{x+6}$, because this cycles round our $\zeta^1$ to $\zeta^{12}$ in $2$ steps, however, why can we just miss out $\zeta^{13}=1$? Surely when it cycles from, say, $\zeta^6$ it goes $\zeta^6\mapsto \zeta^{12}\mapsto \zeta^{18}=\zeta^5\mapsto \zeta^{11}$ etc and we generate the whole group? Then $[L:\mathbb{Q}]\neq2$?
Sorry I'm not phrasing myself very well, but I honestly don't know how to. Thanks to all in advance!

The subfields of $\mathbb{Q}(\zeta_p)$ for $p$ prime correspond to the subgroups of its Galois group $U(p)\cong C_{p-1}$. For $p=13$ we have to find the subgroups of the cyclic group $C_{12}$. We know that for each divisor $d\mid 12$ there exists exactly one subgroup of order $d$, because $C_{12}$ is cyclic - see here.
