The article Number of Sylow subgroups in p-solvable groups - Navarro This is a article which Gabriel Navarro wrote. I'm reading lemma 2.1. I see that 
"$|C ∩ P : C ∩ Q|≤|P : Q|$, and therefore $|C ∩ P : C ∩ Q|$ divides $|P : Q|$. We deduce that $|C|q/|C ∩ H|q$ divides $|G|_q/|H|_q$.
Hence, $|C : C ∩ H|$ divides $|G : H|$."
I don't understand why 
$|C|q/|C ∩ H|q$ divides $|G|_q/|H|_q$ 
and
$|C : C ∩ H|$ divides $|G : H|$
Thank you very much. 
https://www.ams.org/journals/proc/2003-131-10/S0002-9939-03-06884-9/S0002-9939-03-06884-9.pdf
 A: The following appears as Problem 3E.4 in Marty Isaacs' Finite Group Theory book, and as Lemma (2.1) in the wonderful paper of Gabriel Navarro Number of Sylow subgroups in $p$-solvable groups, Proc. AMS 131, 10 (2013).

Lemma Suppose that $A$ is a finite group acting co-primely on a finite group $G$, and let $H$ be an $A$-invariant subgroup of $G$.Then the following hold.
$(a) |C_G(A):C_H(A)|$ divides $|G:H|$.
$(b) |H:C_H(A)|$ divides $|G:C_G(A)|$.

Proof All references will be to Isaacs' book mentioned above. We will only prove $(a)$, since from the proof it will be clear that $(b)$ can be proved in a similar fashion. We will show that for any prime $q$, the highest $q$-power $|C_G(A):C_H(A)|_q$ of $|C_G(A):C_H(A)|$ divides $|G:H|_q$, the highest $q$-power of $|G:H|$. Of course this implies (a). 
Now fix a prime $q$ and let $Q \in Syl_q(H)$ be $A$-invariant. Such a $Q$ exists, see for example 3.23 Theorem (a). By 3.25 Corollary, $Q$ must be contained in some $A$-invariant $P \in Syl_q(G)$. It follows that $Q=P \cap H$. Put $C=C_G(A)$. Then also (3.32 Lemma) $P \cap C \in Syl_q(C)$ and $Q \cap (H \cap C)=Q \cap C \in Syl_q(H \cap C)$. Hence we have the following cube-shaped diagram of subgroups and the corresponding indices.

X.11 Corollary implies $|P \cap C: Q\cap C| \leq |P:Q|$, and since these numbers are $q$-powers we get that $|P \cap C: Q\cap C|$ divides $|P:Q|$.
Since its vertical edges are $q'$-numbers, the right side of the cube shows that
$$|P \cap C: Q \cap C| \text{ is the highest } q \text {-power of } |C:H \cap C|$$
and looking at the left side of the cube, similarly
$$|P: Q| \text{ is the highest } q \text {-power of } |G:H|.$$
But $|P \cap C: Q \cap C|$ divides $|P:Q|$, whence $|C:H \cap C|_q$ divides $|G:H|_q$, and we are done.$\square$
