Suppose $K$ and $L$ are two number fields and $L/K$ is a Galois extension with Galois group $G$. Let $H$ be a subgroup of $G$, which corresponds to intermediate number field $M$.
\begin{equation}
K \subset M \subset L
\end{equation}
Now suppose $\mathfrak{p}$ is a finite prime of $K$, which splits into
\begin{equation}
\mathfrak{p}=\prod_{i=1}^{n_M} \mathfrak{P}_i^{e_i}
\end{equation}
while the residue field of $\mathfrak{P}_i$ is a degree-$f_i$ extension of the residue field of $\mathfrak{p}$. So we have
\begin{equation}
[M:K]=\sum_{i=1}^{n_M}e_i f_i
\end{equation}
Now suppose each $\mathfrak{P}_i$ decomposes into
\begin{equation}
\mathfrak{P}_i=\prod_{j=1}^{n_{i,L}}\mathfrak{Q}_{i,j}^{e_{i,j}}
\end{equation}
Let $D:=D_{i,j} \subset G$ be the decomposition group of the prime $\mathfrak{Q}_{i,j}$ and $I:=I_{i,j}$ be its inertia group. The group $H$ acts on the the primes $\mathfrak{Q}_{i,j}$, whose orbits are in one-to-one bijection with the primes $\mathfrak{P}_i$.
On the other hand, $G$ acts transitively on the primes $\mathfrak{Q}_{i,j}$ and the total number of such primes is $\#(D\setminus G)$. By considering the right action of $H$ on $D\setminus G$, it is easy to see that the primes $\mathfrak{P}_i$ are in one-to-one bijection with the double cosets of $D \setminus G /H$.
We can also consider the action of left action of $D$ on $G/H$, each orbit corresponds to a double coset of $D \setminus G/H$, which then corresponds to a prime $\mathfrak{P}_i$. So they are in total $n_M$ orbits, which is the number of primes that appears in the decomposition of $\mathfrak{p}$. From a lecture note I am reading, the following statements are true, but I could not prove them.
Question 1: The orbit of $G/H$ corresponds to $\mathfrak{P}_i$ has $e_if_i$ points.
Question 2: The inertia group $I \triangleleft D$ acts on this orbit, and the each $I$-orbit consists of $e_i$ points and there are $f_i$ $I$-orbit in this orbit of $D$. While a choice of the Frobenius element permutes these $f_i$ $I$-orbits.