This is something more like a small doubt than some problem that I need help with.
I'm doing and exercise that is asking me to find subextensions of a given extension of all posible orders, and find the exact number of them.
To place the question in concrete terms, I've been given the field extension $\mathbb{Q}(\sqrt[6]{6},\sqrt{3}i)\vert \mathbb{Q}$ which I know is galois. I've found its degree, which is $12$, and I know that the Galois group $G$ is the dihedral group of order 12, $D_6$, and now I need to find subextensions of all posible orders. I know that the Fundamental Theorem of Galois theory says that the intermediate fields are in bijection to the subgroups of the Galois group, but I'm not completely sure that I've understood it correctly, so the first question is
I need to find field extensions whose degree is a divisor of the order of the Galois group (order $12$), don't I?
Or in more general terms
If the Galois group of a field extension $E\vert K$ has order $n$, do all subextensions of $E \vert K$ need to be of degree $d$, being $d$ such that $d \vert n$?
Now, the next section asks me to find the number of extensions of degree $3$ and $4$, so my last question would be
Should I seek for subgroups of the Galois group of orders $4$ and $3$ to find the number of subextensions of orders $3$ and $4$ respectively?
Or in more general terms
Does the number of subextensions of degre $d$ equal to the order of the Galois group, $G$, divided by the number of subgroups of $G$ of order $d$?
Thanks in advance for you help.