Today, I'm reading lemma 2.2.c of an article by Antonio Beltran.
Lemma 2.2. Suppose that A is a finite group acting coprimely on a finite group G, and let $C = C_G(A)$. Then, for every prime p,
c) if $N$ is an $A$-invariant normal subgroup of $G$, then $ν_p^A (N)$ and $ν_p^A (G/N)$ divide $ν_p^A (G)$.
I don't know why "if $v_p^A(G/N)=|CN/N:N_C(P)N/N|$, which certainly divides $v_p^A(G)=|C:N_C(P)|$".
Thank you very much.
Observe that $v^A_p(G/N)=|CN/N:N_C(P)N/N|=|CN:N_C(P)N|$ and that $CN=(N_C(P)N)C$. But $N_C(P) \subseteq C \cap N_C(P)N \subseteq C$, whence, see diagram, $$v^A_p(G)=|C:N_C(P)|=|C:C \cap N_C(P)N| \cdot |C \cap N_C(P)N:N_C(P)|=$$
$$v^A_p(G/N) \cdot |C \cap N_C(P)N:N_C(P)|.$$ From this you also see that $v^A_p(G)=v^A_p(G/N)$ if and only if $N \cap C \subseteq N_C(P)$.