This is a article which Antonio Beltran. I'm reading lemma 2.2.c). I see that:
"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,
(b) $ν_p(C)$ divides $ν_p^A (G)$ and $ν_p^A (G)$ divides $ν_p(G)$.
(c) if N is an A-invariant normal subgroup of G, then $ν_p^A (N)$ and $ν_p^A (G/N)$ divide $ν_p^A (G)$. "
In the proof c) below, I don't understand "If P is an A-invariant Sylow p-subgroup of G, then $PN/N$ is an A-invariant Sylow subgroup of G/N "
Thank you very much.
You asked in your comment why $$|CN/N:N_{CN/N}(PN/N)|=|CN/N:\color {red}{N_C(P)N/N}|.$$ Well, I think there is a typo here, it should read $$|CN/N:N_{CN/N}(PN/N)|=|CN/N:\color {red}{N_{CN}(PN)/N}|.$$ Let me explain. In general, if you have a group $G$ with $N \unlhd G$, $H \leq G$, then $N_{\overline{G}}(\overline{H})=\overline{N_G(HN)}$, for a proof see Corollary 1 here. Now replace in this formula $"G"$ by $"CN"$ and $"H"$ by $"P"$, and you get $$N_{CN/N}(PN/N)=N_{CN}(PN)N/N.$$ And $N_{CN}(PN)=CN \cap N_G(PN)$ obviously contains $N$, hence $N_{CN/N}(PN/N)=N_{CN}(PN)/N$. It follows that $n_p^A(G/N)=|CN:N_{CN}(PN)|$. Now observe that $N \subseteq N_G(PN) \cap CN = N_{CN}(PN) \subseteq CN$, hence $CN=C \cdot N_{CN}(PN)$. Also, $C \cap N_{CN}(PN)=C \cap CN \cap N_G(PN)=N_C(PN).$ This implies that $\color {blue}{n_p^A(G/N)}=|CN:N_{CN}(PN)|=\color {blue}{|C:N_C(PN)|}$ and since $N_C(P) \subseteq N_C(PN)$, we see that $n_p^A(G/N)$ divides $n_p^A(G)=|C:N_C(P)|.$
As a corollary we get $n_p^A(G)=n_p^A(G/N)$ if and only if $N_C(P)=N_C(PN)$.