Prove that $[G : H] \equiv [N_G(H) : H] \bmod p$ [closed]

Hint: show that $|N_G(H):H|$ equals the number of right cosets of $H$ in $G$ fixed under the right multiplication action by $H$. And use that the number of right cosets is of course $|G:H|$.

Let $\Omega$ be the set of right cosets of $H$. $H$ acts on $\Omega$ by right multiplication. Denoting the set of fixed point by $\Omega_0$, we have

$$\Omega_0=\{Ha: Hah=Ha \text{ for all } h \in H \}$$ $$=\{Ha: aha^{-1} \in H \text{ for all } h \in H \}$$ $$=\{Ha: aHa^{-1} \subseteq H\}$$ $$=\{Ha: H^a=H\}$$ $$=\{Ha: a \in N_G(H)\}.$$ So, $\#\Omega_0=|N_G(H):H|$. Can you finish? Use the Orbit-Stabilizer Theorem!

Let $S$ be the Sylow $p$-subgroup of $G$ containing $H$. Then $S>H$. Every proper subgroup of a nilpotent group is properly contained in its normalizer, so $N_G(H)\ge N_S(H)>H$, hence $|N_G(H):H|\equiv 0\pmod{p}$. Since $N_G(H)\le G$, we have $|G:H|$ is divisible by $p$ too. So $|G:H|\equiv |N_G(H):H|\equiv 0\pmod{p}$.