Elements of the subgroup for Galois group over $\mathbb{Q}$ (Cyclotomic extension) Well... this question looks like silly though, I have a curious about the below.
Let $K= \mathbb{Q}(\omega)$ for $\omega=e^{2\pi i \over n}$ 
My book said $G(K/\mathbb{Q}) = \{\sigma_i \vert \sigma_i : \omega \to \omega^{i}$ for $(i,n)=1$$ \} \simeq  \mathbb{Z}_n^*$
The notation $\mathbb{Z}_n^*$ is a multiplicative group for $\mathbb{Z}_n$
But It doesn't say the elements of the subgroup of the $G(K/Q)$, So I couldn't help only guessing about that.
Here is the my guess.
Let the subfield $E(\leq K)$(or the fixed field for $G(K/E)$) $s.t.$ $\vert G(K/E) \vert = [K;E] =d$ and $d \vert \phi(n)$
Then $G(K/E) = \{\sigma_j \vert \sigma : \omega \to \omega^j$ for $(j,n)=1\}$  for $G(K/E)\leq G(K/\mathbb{Q}) $
(Here the $j(\in Z_n^*)$ is $\vert j \vert \vert d$, $\vert j \vert$ means the order of the $j$)
Is my guess right? If my guess incorrect, how do I revise that? Any answer or comment would be appreciated. 
 A: No, this is not a valid way to describe $G(K/E)$. The reason is that when $\Bbb{Z}_n^*$ is not cyclic we will not get all the subgroups by listing the elements of orders $\mid d$.
The smallest counterexample is when $n=8$. In that case we have
$$
\Bbb{Z}_8^*=\{1,3,5,7\}.
$$
Note that all the elements of $\Bbb{Z}_8^*$ have order one or two. 
The field $K=\Bbb{Q}(e^{\pi i/4})=\Bbb{Q}(i,\sqrt2)$ has three distinct subfields that are degree two extensions of the rationals. Namely
$$
E_1=\Bbb{Q}(i),\quad E_2=\Bbb{Q}(\sqrt2),\quad E_3=\Bbb{Q}(\sqrt{-2}).
$$
The corresponding Galois groups $G(K/E_i)$ are (leaving that as an exercise for now)
$$
\begin{aligned}
G(K/E_1)&=\{\sigma_1,\sigma_5\},\\
G(K/E_2)&=\{\sigma_1,\sigma_7\},\\
G(K/E_3)&=\{\sigma_1,\sigma_3\}.
\end{aligned}
$$
Observe that $d=2$ for all these intermediate fields.

Knowing $[E:\Bbb{Q}]=d$ is insufficient information to describe $E$, and hence insufficient to describe $G(K/E)$.

The lattice of subgroups and intermediate fields becomes more complicated roughly as a function of the number of prime factors of $n$. The Chinese Remainder Theorem lets us describe $\Bbb{Z}_n^*$ as a direct product of groups
$\Bbb{Z}_{p_i^{\ell_i}}^*$ with $p_i^{\ell_i}$ ranging over the prime powers appearing in the factorization of $n$. With the exception of $p_i=2$ these are always cyclic, but enumerating all the subgroups, while possible, is not very simple.

How to revise your guess? The general facts are that any intermediate field $E$ is of the form $E=\Bbb{Q}(\rho)$ for some element $\rho$. The trivial thing is then that
$$
G(K/E)=\{\sigma\in G(K/\Bbb{Q})\mid \sigma(\rho)=\rho\}.
$$
I doubt this is very useful for you in that it amounts to regurgitating the basic idea of Galois correspondence. The intermediate fields can be described in terms of Gaussian periods. Wait for Will Jagy to show up with sources to more information :-)
