If $G$ is a finite group with $H \trianglelefteq G$, and $P$ is a Sylow $p$-subgroup of $H$, then we can show that $G = N_G(P)H$. While I'm now aware of the (admittedly much simpler/nicer) Frattini argument for showing this, I was curious if the following method also works:
Proof: Firstly, as $H \trianglelefteq G$, $N_G(P)H \leq G$, so it is sufficient to show that $|G| = |N_G(P)H|$. Then by the product formula, it is sufficient to show that $$|G| = \frac{|N_G(P)||H|}{|N_G(P) \cap H|} = \frac{|N_G(P)||H|}{|N_H(P)|} \text{ as $N_G(P) \cap H = N_H(P)$}. $$ Then by Lagrange's theorem, this is equivalent to showing $|G: N_G(P)| = |H : N_H(P)| = n_p$, where $n_p$ is the number of Sylow $p$-subgroups of $H$. But then as $H \trianglelefteq G$, any conjugate $gPg^{-1}$ is also a Sylow $p$-subgroup of $H$, so by mirroring the argument from the Sylow theorems that $|H : N_H(P)| = n_p$ (specifically the group action argument by considering the conjugation action on $X = \{ hPh^{-1} \, | \, h \in H \}$ ), we're done.
While what I've written is a bit more of a sketch than the full argument, I can't see any flaws in it (the reason I doubt it's right isn't a mathematical one). Would somebody please be able to say whether it is right or not, and if it isn't, where it goes wrong?
The idea seems correct, although I am sure what you are talking about in the last part. Here is how you can see the equality:
The set of Sylow $p$-subgroups of $H$ is equal to the set of $G$-conjugates of $P$ (since $H$ is a normal subgroup). The number of such is $[G : N_G(P)]$. On the other hand the number of Sylow $p$-subgroups of $H$ is equal to $[H:N_H(P)]$ (number of $H$-conjugates of $P$). Hence the equality $[G : N_G(P)] = [H : N_H(P)]$.