Why $\operatorname{Gal}(\mathbb{Q}_q(\zeta_p)/\mathbb{Q}_q)$ is isomorphic to $\mathbb{Z}/f\mathbb{Z}$ I would like to ask a question on this answer. I don't understand why $\operatorname{Gal}(\mathbb{Q}_q(\zeta_p)/\mathbb{Q}_q)$ is isomorphic to $\mathbb{Z}/f\mathbb{Z}$, where $f$ is the degree of the cyclotomic polynomial. How could one calculate this degree?
And a second question: If I know the degree's of the irreducible factors of the cyclotomic polynomial, why is there a connection to the cardinality of a galois gorup?

edit: I found a similar statement here in Corollary 3.10, but without a proof. Maybe the context can help
 A: The key ingredients are:


*

*The polynomial $x^p-1$ is separable over the finite field $\mathbb{F}_q$.

*The simple Galois theory of finite fields.

*Hensel's Lemma.

*An extension $\mathbb{F}_{q^m}$ of $\mathbb{F}_q$ contains $p^{th}$ roots of unity, if and only if $p\mid q^m-1$ (the multiplicative group of a finite field is cyclic).


The last point allows you to deduce that the minimal polynomial of $\zeta_p$ over $\mathbb{F}_q$ has degree $f$ = the order of $q$ in $\mathbb{Z}/p\mathbb{Z}^*$. Therefore the cyclotomic polynomial $\Phi_p(x)=(x^p-1)/(x-1)$ has a factor $\overline{m}(x)$ of degree $f$ in
$\mathbb{F}_q[x]$. Hensel's Lemma then tells you that the cyclotomic polynomial 
also has a factor $m(x)$ of degree $f$ in $\mathbb{Z}_q[x]$. Now you know that 
$$[\mathbb{Q}_q(\zeta_p):\mathbb{Q}_q]=f.$$
So the Galois group is also of order $f$ (the extension is obviously the splitting field of $\Phi_p(x)$).
The roots of $\overline{m}(x)$ are (by the Galois theory of finite fields)
$\zeta_p^{q^i}, 0\le i<f$. Thus the same holds for $m(x)$. Thus there is a
unique automorphism $F$ in the Galois group that has the property $F(\zeta_p)=\zeta_p^q$. From our knowledge of the roots of $m(x)$ we see that the lowest power of $F$ that maps $\zeta_p$ to itself is $F^f$. Hence the Galois group must be cyclic of order $f$.
