Finding the intermediate fields of $\Bbb{Q}(\zeta_7)$.

Finding the intermediate fields of $\Bbb{Q}(\zeta_7)$.

Let $F= \Bbb{Q}(\zeta_7)$ with $\zeta_7 = e^{2\pi/7}$.
a) What is the Galois group of $F$ over $\Bbb{Q}$?
b)Find all intermediate fields between $\Bbb{Q}$ and $F$. (Write each in
the form $\Bbb{Q}(\alpha)$ for some specific $\alpha \in F$.)
c) For each intermediate field $E$ above, give the Galois group of $E$
over $\Bbb{Q}$.
a) Since $\Bbb{Q}(\zeta_7)$ is the splitting field for the cyclotomic
polynomial $x^6+x^5+...+1$, $Gal(\Bbb{Q}(\zeta_7)/\Bbb{Q}) \cong U_7 \cong
\Bbb{Z}^{\times}_7 \cong \Bbb{Z}_6$.
b) $Gal(\Bbb{Q}(\zeta_7)/\Bbb{Q}) \cong \Bbb{Z}_6$
So we have the following diagram for $\Bbb{Z}_6$,

And a corresponding diagram for $\Bbb{Q}(\zeta_7)$ for some $\alpha$ and
$\beta$
Since $\Bbb{Q}(\zeta_7)$ has complex numbers, we can look at $\Bbb{Q}(i)$
to see if it has degree 2 or 3 over $\Bbb{Q}$. Since $x^2+1$ is its
minimal polynomial, we see that $[\Bbb{Q(i)} : \Bbb{Q}]=2$ and so it must
correspond to $\Bbb{Z}_3$, and $\beta=i$.
Sending every number to its conjugate is one possible automorphism. Of
course, it has order 2 and so is isomorphic to $\Bbb{Z}_2$. Since it fixes
every real number, the corresponding intermediate field must be of the
form $\Bbb{Q}(\alpha)$ for some real number $\alpha$, and must have degree
3 over $\Bbb{Q}$.
If we look at $\zeta_7 + \zeta_7^6$, we see that
$\cos(2\pi/7) + i\sin(2\pi/7) + cos(12\pi/7) + i\sin(12\pi/7)$
$= \cos(2\pi/7) + i\sin(2\pi/7) + \cos(-2\pi/7) + i\sin(-2\pi/7)$
$= 2\cos(2\pi/7)$
which is irrational. So $\Bbb{Q}(\zeta_7 + \zeta_7^6)$ is either a proper
subfield of $\Bbb{Q}(\alpha)$ or is equal to it. But from the diagram, we
know that there is only one other intermediate field, so $\Bbb{Q}(\zeta_7
+ \zeta_7^6) = \Bbb{Q}(\alpha)$.
c) The corresponding Galois group of $\Bbb{Q}(i)$ is isomorphic to
$\Bbb{Z}_3$ and that of $\Bbb{Q}(\zeta_7 + \zeta_7^6)$ is isomorphic to
$\Bbb{Z}_2$.
Do you think this is correct?
Thanks in advance