$G$ is a finite group with order $|G|=p^am$, where $p$ is prime and $p\nmid m$. $H$ is a p-subgroup of $G$, so $|H|=p^i$, where $1\le i\le a$. Define: $N=\{g\in G| gHg^{-1}=H\}$. Prove: $(G:H)\equiv (N:H) \pmod{p}$.
The textbook (also in this post) proves this by using the method of group action on a set and using Orbit Stabilizer Theorem. I am wondering if there is a straightforward method (such as a computational way) to prove this property without using Orbit Stabilizer Theorem or Sylow Theorem. Here is my attempt:
Since $N$ is a subgroup of $G$, so we have $|N|=p^j m_1$, where $j\le a$, and $m_1|m$.
Since $H$ is a subgroup of $N$, so we have $i\le j\Rightarrow i\le j\le a$
Case (1), if $i=a$
Then we have $j=i=a$, and $(G:H)=m, (N:H)=m_1$, where $m_1|m$, but how to show $m\equiv m_1 \pmod{p}$?
Case (2), if $i<a$
Then we have $(G:H)=p^{a-i}m, (N:H)=p^{j-i}m_1$, where $m_1|m$. Since $i<a$, we have $p|(G:H)$, but how to show $p|(N:H)?$ or equivalently, how to show $i<j$?