Cyclotomic field Let be $\zeta=\cos\left(\frac{2\pi}{253}\right)+i\sin\left(\frac{2\pi}{253}\right)$ and $K=\mathbb{Q}(\zeta)$. We know that $G:=\text{Gal}(K/\mathbb{Q})\cong\left(\mathbb{Z}^\times_{253},\cdot\right)\cong\left(\mathbb{Z}_{10},+\right)\times\left(\mathbb{Z}_{22},+\right)\cong\left(\mathbb Z_2,+\right)\times\left(\mathbb Z_5,+\right)\times\left(\mathbb Z_2,+\right)\times\left(\mathbb Z_{11},+\right)=:H.$
From this we can see there are 3 subfields $L_1,L_2,L_3$ of $K$ with $[K:L_i]=2,$ but how can I express these subfields "explicitely" (I mean $L_1=\text{Fix}(\sigma_1)$ for an automorphism $\sigma_1\in G$, etc.)?
And how do I decide, which of these extensions $L_i/\mathbb Q$ are cyclic? I know it means to find cyclic subgroups of $H$, but how can I find them?
Ty for any help.
 A: The subgroups of index $2$ in $G \cong \mathbb{Z}_2 \times \mathbb{Z}_2 \times \mathbb{Z}_{55}$ are $\{0\} \times \mathbb{Z}_2 \times \mathbb{Z}_{55}$, $\mathbb{Z}_2 \times \{0\} \times \mathbb{Z}_{55}$ and $\{(0, 0), (1, 1)\} \times \mathbb{Z}_{55}$.
The elements of order $2$ in $\mathbb{Z}_{253}^\times$ are the solutions to the equation $x^2 = 1 \pmod {253}$ which, by the Chinese remainder theorem can be found to be $45$, $208$ and $252$. Moreover, the element $3$ has order $55$ in $\mathbb{Z}_{253}^\times$.
The orders of $3 \times 45 \equiv 135$, $3 \times 208 \equiv 118$ and $3 \times 252 \equiv 250$ will all be $110$ and thus the index-$2$ subgroups of $\mathbb{Z}_{253}^\times$ are generated by these elements and are cyclic.
(Note that what we are looking for is generators for the subgroups of order $110$. But there might be simpler ones, I just found those by the method above. Since the order $253$ is pretty small, one could also brute force the generators.).
Under the isomorphism with $\mathbb{Z}_{253}^{\times}$, the subgroups given above correspond to the subgroups $\left<135\right>$, $\left<118\right>$ and $\left<250\right>$ respectively.
These generators correspond to the automorphisms $\sigma_1:\zeta \rightarrow \zeta^{135}$, $\sigma_2: \zeta \rightarrow \zeta^{118}$ and $\sigma_3: \zeta \rightarrow \zeta^{250}$.
So the fixed fields are $L_1 = \text{Fix}(\sigma_1)$, $L_2 = \text{Fix}(\sigma_2)$ and $L_3 = \text{Fix}(\sigma_3)$.
Note that the degrees $[L_i : \mathbb{Q}] = [G : G_i] = 2$ by Galois theory (where $G_i = \text{Stab}_G(L_i))$, and so each of these is a quadratic extension of $\mathbb{Q}$, so they must also be cyclic extensions.
I don't know if there is an easy way to compute squarefree $d_i \in \mathbb{Z}$ such that $L_i = \mathbb{Q}(\sqrt{d_i})$ (which must exist since the extensions are quadratic).
EDIT:
One can at least easily find which primes can divide $d_i$. Primes dividing $d_i$ ramify in $\mathbb{Q}(\sqrt{d_i})$ and thus also in $\mathbb{Q}(\zeta)$. The primes that ramify are those dividing $\text{disc}(\mathcal{O}_K)$. But the minimal polynomial of $\zeta$ is $\Phi_{253}(x)$ with discriminant $11^{198} \times 23^{210}$ (see Wolfram Alpha). We know from algebraic number theory that $\mathcal{O}_K = \mathbb{Z}[\zeta]$ and thus this is also $\text{disc}(\mathcal{O}_K)$.
So the primes that can divide $d_i$ are $11$ and $23$. So $d_i \in \{\pm 11, \pm 23, \pm 11 \times 23\}$. Probably one sign for each will work. (This might probably also be obtained from knowing that $253 = 11 \times 23$ in a simpler way.)
A: Because $253=23\cdot11$ the field $\Bbb{Q}(\zeta_{253})$ is the compositum of the fields $K_1=\Bbb{Q}(\zeta_{23})$ and $K_2=\Bbb{Q}(\zeta_{11})$. 
Because $K_i/\Bbb{Q}$ is cyclic for $i=1,2$, they each have a unique subfield $M_i, i=1,2,$ such that $[K_i:M_i]=2$. We easily see that the real subfields
have this property so the uniqueness allows us to conclude that
$M_1=K_1\cap\Bbb{R}=\Bbb{Q}(\cos(2\pi/23))$ and $M_2=K_2\cap\Bbb{R}=\Bbb{Q}(\cos(2\pi/11))$.
You already figured  out that there are three subfields of the desired type. From the above consideration we see that they must be


*

*$L_1=M_1K_2=\Bbb{Q}(\cos(2\pi/23),\zeta_{11}),$

*$L_2=M_2K_1=\Bbb{Q}(\cos(2\pi/11),\zeta_{23}),$ and

*$L_3=K\cap \Bbb{R}=\Bbb{Q}(\cos(2\pi/253))$.


The Galois group is $G\simeq (C_2\times C_2)\times C_{55}$. If $H\le G$ is of order two, we see that $H$ is a subgroup of that unique Sylow $2$-subgroup of $G$., so for $j=1,2,3$ we have 
$$Gal(L_i/\Bbb{Q})\simeq G/H\simeq C_2\times C_{55}\simeq C_{110}.$$
In other words, all these three fields are cyclic.
