Let be $\zeta=\cos\left(\frac{2\pi}{55}\right)+i\sin\left(\frac{2\pi}{55}\right)$ and $K=\mathbb{Q}(\zeta)$. Obviously $G:=\text{Gal}(K/\mathbb{Q})\cong\left(\mathbb Z_2,+\right)\times\left(\mathbb Z_4,+\right)\times\left(\mathbb Z_5,+\right)=H.$ Since there are 3 subgroups of order 2 of $H$, there are 3 subfield $L_1,L_2,L_3$ of $K$ with $[K:L_i]=2,$ but (again) how can I express these subfields "explicitely" as fixed field of a subgroup of $\text{Aut}(K)$? And how can I find such subgroups? I think I'm missing something from the Fundamental theorem of Galois theory...
There are some more tasks related to this, but maybe later.
Anyway, thx for help.
Use the usual convention$\newcommand{\si}{\sigma}\newcommand{\ze}{\zeta}$ that $\si_a$ denotes the automorphism with $\si_a(\ze)=\ze^a$. The solutions of $a^2\equiv1\pmod{55}$ are $a\equiv\pm1$, $\pm21$. The elements of order $2$ are $\si_{-1}$, $\si_{21}$ and $\si_{-21}$. The $L_i$ are the fixed fields of these, viz., $\Bbb Q(\ze+\ze^{-1})$, $\Bbb Q(\ze+\ze^{21})$ and $\Bbb Q(\ze+\ze^{-21})$.