Number of fields between $\mathbb{Q}$ and $\mathbb{Q}(\zeta_n)$ If $\zeta_n$ is the $n$-th primitive root of unity then $Gal(\mathbb{Q}(\zeta_n)/\mathbb{Q}) \simeq Z_n^*$ due to the following map
$$\tau(\zeta_n)=\zeta^n$$

I was wondering if we could use this and the Galois Correspondence to find the number of fields between $\mathbb{Q}$ and $\mathbb{Q}(\zeta_n)$. How could we determine this?

I know that if $n$ is prime then $\mathbb{Q}(\zeta_n)$ contains subfields of degree $d$, where $d$ are the divisors of $p-1$. 
For example, $Gal(\mathbb{Q}(\zeta_3)/\mathbb{Q}) \simeq Z_3^*$ and since $3-1=2$ has divisors $1$ and $2$, the number of fields between $\mathbb{Q}$ and $\mathbb{Q}(\zeta_3)$ is two... is that right? Do these 'sub'fields actually corresponds to the fields themselves in this case?

What can we say if $n$ is not prime?
 A: Recall by the fundamental theorem of Galois theory that the number of fields between $\mathbb Q$ and $\mathbb Q(\zeta_n)$ is the number of subgroups of $\text{Gal}(\mathbb Q(\zeta_n))\cong (\mathbb Z/n\mathbb Z)^\times$. If $n$ is a power of a prime, i.e. $n=p^k$ for some $p$ prime and $k\in\mathbb N$, then $(\mathbb Z/n\mathbb Z)^\times$ is cyclic of order $\varphi(p^k)=p^{k}-p^{k-1}$. By the fundamental theorem of cyclic groups, for all $d|\varphi(p^k)$ there is exactly one subgroup of order $d$. Hence, the number of intermediate fields between $\mathbb Q$ and $\mathbb Q(\zeta_n)$ is the number of divisors of $p^k-p^{k-1}$. (For example, if $n$ is prime, then this is the number of divisors of $n-1$ as you stated.)
If $n$ is not a power of a prime, this question becomes much more difficult (see https://mathoverflow.net/questions/46115/subgroups-of-a-finite-abelian-group). By the way, for any finite abelian group $G$, there exists some cyclotomic field $\mathbb Q(\zeta_n)$ and some intermediate field $\mathbb Q\subset F\subset \mathbb Q(\zeta_n)$ such that $\text{Gal}(F/\mathbb Q)\cong G$. This provides a trivial partial answer to the inverse Galois problem.
