Determine tha intermediate fields of $\mathbb{Q}(\zeta_{20})$ I tried to find all intermediate fields of $\mathbb{Q}(\zeta_{20})$, where $\zeta=\zeta_{20}$ is the primitive  20th root of unity. I found the Galois group of $\mathbb{Q}(\zeta)$ over $\mathbb{Q}$ is isomorphic to $\mathbb{Z}/2\mathbb{Z}\times \mathbb{Z}/4\mathbb{Z}$ and it has 6 subgroups. If the generater of $\mathbb{Z}/2\mathbb{Z}$ is $\sigma$ and that of $\mathbb{Z}/4\mathbb{Z}$ is $\tau$, correspondence between subgroups and intermediate fields is following.
$$ <\sigma^2>\leftrightarrow \mathbb{Q}(\zeta+\zeta^9) $$
$$ <\tau\sigma^2>\leftrightarrow \mathbb{Q}(\zeta+\zeta^{19}) $$
$$ <\sigma>\leftrightarrow \mathbb{Q}(\zeta+\zeta^3+\zeta^7+\zeta^9) $$
$$ <\tau\sigma>\leftrightarrow \mathbb{Q}(\zeta+\zeta^{9}+\zeta^{13}+\zeta^{17}) $$
The question I have is what a subgroup $<\sigma^2,\tau>,<\tau>$corresponds.
Please give me some advice.
 A: There are apparently some confusions about the identification of the elements.
In an attempt to clarify those let's denote by $\sigma_j$ the automorphism determined by $\sigma_j(\zeta)=\zeta^j$. Here $j\in S:=\{1,3,7,9,11,13,17,19\}$,
and the Galois group consists of all the elements $\sigma_j, j\in S$.
The subgroups of order four are then


*

*$H_1=\langle\sigma_3\rangle=\{\sigma_j\mid j\in\{1,3,9,7\}\}$ (cyclic),

*$H_2=\langle\sigma_{13}\rangle=\{\sigma_j\mid j\in\{1,13,9,17\}\}$ (cyclic),

*$H_3=\langle\sigma_9,\sigma_{11}\rangle=\{\sigma_j\mid j\in\{1,9,11,19\}\}$ (isomorphic to Klein Viergruppe).


The fixed fields of these must be quadratic extensions of $\Bbb{Q}$. What could those be? Clearly $i\in L:=\Bbb{Q}(\zeta_{20})$, so $\Bbb{Q}(i)$ is one of them. It is a standard exercise to show that $\sqrt5\in\Bbb{Q}(\zeta_5)\subset\Bbb{Q}(\zeta_{20})$, so $\Bbb{Q}(\sqrt5)$ is another quadratic subfield. The third quadratic subfield is thus $\Bbb{Q}(\sqrt{-5})$.
Which subfield correspondens to which subgroup of order four? A most obvious thing is that $\sigma_{19}$ is the usual complex conjugation. Therefore its fixed points are all real numbers. This allows us to deduce that the fixed field of $H_3$ must be $\Bbb{Q}(\sqrt5)$. $i=\zeta_{20}^5$, so $\sigma_j(i)=i^j$. The exponents $1,13,9,17$ are all congruent to $1\pmod4$, so we see that $i$ is a fixed point for all of the automorphisms in $H_2$. Hence $\Bbb{Q}(i)$ is the fixed field of $H_2$. By a process of elimination we then see that the fixed field of $H_1$ must be $\Bbb{Q}(\sqrt{-5})$.
The group $G$ has three elements of order two, so there are three subgroups of order two, yielding three quartic subfields. The Galois correspondence goes as follows:


*

*The subgroup $K_1=\langle\sigma_9\rangle=H_1\cap H_2\cap H_3$, so its fixed field must be $\Bbb{Q}(i,\sqrt5)$.

*The subgroup $K_2=\langle\sigma_{19}\rangle$ has complex conjugation as its only non-trivial automorphism, so its fixed field is the intersection $L\cap\Bbb{R}$. That field is generated by the number $z=2\cos(\pi/10)=\zeta+\zeta^{-1}$. The minimal polynomial of $z$ is $x^4-5x^2+5$ (irreducible by Eisenstein).

*The automorphism $\sigma_{11}$ maps the fifth root of unity $\zeta_5=\zeta^4$ to itself, so the fixed field of $K_3=\langle\sigma_{11}\rangle$ is
the quartic field $\Bbb{Q}(\zeta_5)$.

