Number of Sylow subgroups in $p$-solvable groups - paper of G. Navarro This is a article which Gabriel Navarro wrote in 2003. I'm reading lemma 2.2. I see that 
So we may assume that there is a proper normal subgroup N of $G$ such that $G/N$
is a $p$-group $\cdots $ divides $|N:H|=|G:M|.$ 
I don't understand why $|N:H|=|G:M|.$ 
https://www.ams.org/journals/proc/2003-131-10/S0002-9939-03-06884-9/S0002-9939-03-06884-9.pdf
 A: In general, if $K,L \leq G$, with gcd$(|G:K|,|G:L|)=1$, then $G=KL$ (try to prove this yourself!). Since in the paper of Navarro, the Sylow $p$-subgroup $P \subseteq M$, we have $|G:M|$ is not divisible by $p$. And the $N \lhd G$ is chosen with $G/N$ to be a $p$-group. So, $G=MN$ and hence $|G:M|=|N:N \cap M|$. Navarro puts $H=N \cap M$.
Note The result of Navarro is very nice indeed and implies that for solvable groups $G$ the number $n_p(H)$ of Sylow $p$-subgroups of a subgroup $H$ with $p$ dividing $|H|$, divides the number of Sylow $p$-subgroups $n_p(G)$ of $G$. (A group is solvable iff it is $p$-solvable for every prime $p$). If you want to lift the $p$-solvability of $G$, then the result holds in case $H$ is subnormal. Without any condition, if $N \unlhd G$, then $n_p(G/N)$ divides $n_p(G)$. For a proof see here. 
Here are some explanations for the proof of Lemma(2.2),we need some preparatory results, which are interesting in their own right.
Proposition Let $G$ be a finite group, $P \in Syl_p(G)$, and $N \unlhd G$. 
$(a)$ If $G/N$ is a $p$-group, then $N_G(P)/N_N(P)$ is a $p$-group.
$(b)$ If $G/N$ is a $p'$-group, then $N_G(P)/N_N(P)$ s a $p'$-group.

Proof note that since $N \unlhd G$, always $N \cap N_G(P)=N_N(P) \unlhd N_G(P)$. $(a)$ We have $P \subseteq N_G(P)$, whence $|G:N_G(P)|$ is not divisible by $p$. Hence gcd$(|G:N_G(P)|,|G:N|)=1$ yielding $G=NN_G(P)$. But then $|G:N|=|N_G(P):N_N(P)|$, being a power of $p$.$(b)$ $G/N$ is a $p'$-group, so we must have $P \subseteq N$, whence $G=NN_G(P)$ by the Frattini argument. Again, $|G:N|=|N_G(P):N_N(P)|$ is a $p'$-group.

Theorem Let $G$ be a finite group, $P \in Syl_p(G)$, $N \unlhd G$ and $H \leq G$ with $P \subseteq H$. 
If $G/N$ is a $p$-group or a $p'$-group, then
$(a)$ $N_{HN}(P)=N_H(P)N_N(P)$
$(b)$ $|N_{HN}(P):N_H(P)|=|N_N(P):N_{H \cap N}(P)|$.
Proof Let us first assume that $G/N$ is a $p$-group. Since $P \subseteq H$, we have $P \subseteq N_H(P)$, whence $|N_{HN}(P):N_H(P)|$ is not divisible by $p$. By the Proposition $(a)$ $N_G(P)/N_N(P)$ is a $p$-group, so $|N_{HN}(P):N_N(P)|$ is a power of $p$. Hence gcd$(|N_{HN}(P):N_H(P)|,|N_{HN}(P):N_N(P)|)=1$ and $(a)$ follows.
$(b)$ This follows from $(a)$ and the fact that $N_H(P) \cap N_N(P)=H \cap N_G(P) \cap N \cap N_G(P)=H \cap N \cap N_G(P)=N_{H \cap N}(P)$. 
Now, let $G/N$ be a $p'$-group, then $P \subseteq H \cap N \unlhd H$, so by the Frattini argument we have $H=(H \cap N)N_H(P)$, which implies $HN=NN_H(P)$. But then $N_{HN}(P)=HN \cap N_G(P)=NN_H(P) \cap N_G(P)=N_H(P)(N \cap N_G(P))=N_H(P)N_N(P)$, which is $(a)$. And $(b)$ follows by the same argument as above.

We are now ready to prove the statements at the end of the proof of Navarro's Lemma (2.2). Recall that the situation is as follows: $P \in Syl_p(G)$, $N \unlhd G$, even $P \cap N \unlhd G$, and so $P \cap N \in Syl_p(N)$ and $\overline{P}$ (the paper states $P$) acts coprimely (via conjugation) on $N/(P \cap N)$. We are given $P \subseteq M$. Note that since $G/N$ is a $p$-group, $PN/N=G/N$, so $G=PN=MN$.
 $H=M \cap N$ and $\overline{H}=(M \cap N)/(N \cap P)=\overline{M} \cap \overline{N}$ is $\overline{P}$-invariant, since $\overline{P} \subseteq \overline{M}$ and $N$ is normal. Observe that $|G:M|=|MN:M|=|N:M \cap N|=|\overline{N}:\overline{H}|$$\text{    }(*)$. To apply Lemma(2.1) (which is by the way Exercise 3E.4 in Finite Group Theory, I.M. Isaacs) we need to identify the $\overline{P}$-fixed points in $\overline{N}$, being $C_{\overline{N}}(\overline{P})$. It is not hard to see that this is the subgroup $\overline{N_G(P) \cap N}=\overline{C}$ in the notation of the paper. Then owing to Lemma(2.1), we obtain: $|\overline{C}:\overline{C \cap H}|$ divides $|\overline{N}:\overline{H}|=|G:M|$ by formula $(*)$ above.
Finally, $$|\overline{C}:\overline{C \cap H}|=|(N_G(P) \cap N)/(P \cap N):(N_G(P) \cap N \cap M \cap N)/(P \cap N)| \\ =|N_G(P) \cap N:N_G(P) \cap M \cap N| \\ = |N_N(P):N_{M \cap N}(P)|= (\text{by Theorem (b) above}) \\ =|N_{MN}(P):N_M(P)|=|N_G(P):N_M(P)|.$$
This proves the Lemma(2.2).
